Integrand size = 42, antiderivative size = 437 \[ \int \frac {\left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{(a+b \cos (c+d x))^{5/2}} \, dx=-\frac {\left (3 a^4 B-26 a^2 b^2 B+15 b^4 B+14 a^3 b C-6 a b^3 C\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{3 a^3 \left (a^2-b^2\right )^2 d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}+\frac {\left (3 a^2 B-5 b^2 B+2 a b C\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{3 a^2 \left (a^2-b^2\right ) d \sqrt {a+b \cos (c+d x)}}-\frac {(5 b B-2 a C) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticPi}\left (2,\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{a^3 d \sqrt {a+b \cos (c+d x)}}+\frac {b \left (3 a^2 B-5 b^2 B+2 a b C\right ) \sin (c+d x)}{3 a^2 \left (a^2-b^2\right ) d (a+b \cos (c+d x))^{3/2}}+\frac {b \left (3 a^4 B-26 a^2 b^2 B+15 b^4 B+14 a^3 b C-6 a b^3 C\right ) \sin (c+d x)}{3 a^3 \left (a^2-b^2\right )^2 d \sqrt {a+b \cos (c+d x)}}+\frac {B \tan (c+d x)}{a d (a+b \cos (c+d x))^{3/2}} \] Output:
-1/3*(3*B*a^4-26*B*a^2*b^2+15*B*b^4+14*C*a^3*b-6*C*a*b^3)*(a+b*cos(d*x+c)) ^(1/2)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2)*(b/(a+b))^(1/2))/a^3/(a^2-b^2) ^2/d/((a+b*cos(d*x+c))/(a+b))^(1/2)+1/3*(3*B*a^2-5*B*b^2+2*C*a*b)*((a+b*co s(d*x+c))/(a+b))^(1/2)*InverseJacobiAM(1/2*d*x+1/2*c,2^(1/2)*(b/(a+b))^(1/ 2))/a^2/(a^2-b^2)/d/(a+b*cos(d*x+c))^(1/2)-(5*B*b-2*C*a)*((a+b*cos(d*x+c)) /(a+b))^(1/2)*EllipticPi(sin(1/2*d*x+1/2*c),2,2^(1/2)*(b/(a+b))^(1/2))/a^3 /d/(a+b*cos(d*x+c))^(1/2)+1/3*b*(3*B*a^2-5*B*b^2+2*C*a*b)*sin(d*x+c)/a^2/( a^2-b^2)/d/(a+b*cos(d*x+c))^(3/2)+1/3*b*(3*B*a^4-26*B*a^2*b^2+15*B*b^4+14* C*a^3*b-6*C*a*b^3)*sin(d*x+c)/a^3/(a^2-b^2)^2/d/(a+b*cos(d*x+c))^(1/2)+B*t an(d*x+c)/a/d/(a+b*cos(d*x+c))^(3/2)
Result contains complex when optimal does not.
Time = 7.30 (sec) , antiderivative size = 750, normalized size of antiderivative = 1.72 \[ \int \frac {\left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{(a+b \cos (c+d x))^{5/2}} \, dx=\frac {\frac {2 \left (36 a^3 b^2 B-20 a b^4 B-24 a^4 b C+8 a^2 b^3 C\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{\sqrt {a+b \cos (c+d x)}}+\frac {2 \left (-33 a^4 b B+86 a^2 b^3 B-45 b^5 B+12 a^5 C-38 a^3 b^2 C+18 a b^4 C\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticPi}\left (2,\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{\sqrt {a+b \cos (c+d x)}}-\frac {2 i \left (-3 a^4 b B+26 a^2 b^3 B-15 b^5 B-14 a^3 b^2 C+6 a b^4 C\right ) \sqrt {\frac {b-b \cos (c+d x)}{a+b}} \sqrt {-\frac {b+b \cos (c+d x)}{a-b}} \cos (2 (c+d x)) \left (2 a (a-b) E\left (i \text {arcsinh}\left (\sqrt {-\frac {1}{a+b}} \sqrt {a+b \cos (c+d x)}\right )|\frac {a+b}{a-b}\right )+b \left (2 a \operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt {-\frac {1}{a+b}} \sqrt {a+b \cos (c+d x)}\right ),\frac {a+b}{a-b}\right )-b \operatorname {EllipticPi}\left (\frac {a+b}{a},i \text {arcsinh}\left (\sqrt {-\frac {1}{a+b}} \sqrt {a+b \cos (c+d x)}\right ),\frac {a+b}{a-b}\right )\right )\right ) \sin (c+d x)}{a \sqrt {-\frac {1}{a+b}} \sqrt {1-\cos ^2(c+d x)} \sqrt {-\frac {a^2-b^2-2 a (a+b \cos (c+d x))+(a+b \cos (c+d x))^2}{b^2}} \left (2 a^2-b^2-4 a (a+b \cos (c+d x))+2 (a+b \cos (c+d x))^2\right )}}{12 a^3 (-a+b)^2 (a+b)^2 d}+\frac {\sqrt {a+b \cos (c+d x)} \left (\frac {2 \left (-b^3 B \sin (c+d x)+a b^2 C \sin (c+d x)\right )}{3 a^2 \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}+\frac {2 \left (-10 a^2 b^3 B \sin (c+d x)+6 b^5 B \sin (c+d x)+7 a^3 b^2 C \sin (c+d x)-3 a b^4 C \sin (c+d x)\right )}{3 a^3 \left (a^2-b^2\right )^2 (a+b \cos (c+d x))}+\frac {B \tan (c+d x)}{a^3}\right )}{d} \] Input:
Integrate[((B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^3)/(a + b*Cos[ c + d*x])^(5/2),x]
Output:
((2*(36*a^3*b^2*B - 20*a*b^4*B - 24*a^4*b*C + 8*a^2*b^3*C)*Sqrt[(a + b*Cos [c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*b)/(a + b)])/Sqrt[a + b*Cos[ c + d*x]] + (2*(-33*a^4*b*B + 86*a^2*b^3*B - 45*b^5*B + 12*a^5*C - 38*a^3* b^2*C + 18*a*b^4*C)*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*EllipticPi[2, (c + d*x)/2, (2*b)/(a + b)])/Sqrt[a + b*Cos[c + d*x]] - ((2*I)*(-3*a^4*b*B + 26 *a^2*b^3*B - 15*b^5*B - 14*a^3*b^2*C + 6*a*b^4*C)*Sqrt[(b - b*Cos[c + d*x] )/(a + b)]*Sqrt[-((b + b*Cos[c + d*x])/(a - b))]*Cos[2*(c + d*x)]*(2*a*(a - b)*EllipticE[I*ArcSinh[Sqrt[-(a + b)^(-1)]*Sqrt[a + b*Cos[c + d*x]]], (a + b)/(a - b)] + b*(2*a*EllipticF[I*ArcSinh[Sqrt[-(a + b)^(-1)]*Sqrt[a + b *Cos[c + d*x]]], (a + b)/(a - b)] - b*EllipticPi[(a + b)/a, I*ArcSinh[Sqrt [-(a + b)^(-1)]*Sqrt[a + b*Cos[c + d*x]]], (a + b)/(a - b)]))*Sin[c + d*x] )/(a*Sqrt[-(a + b)^(-1)]*Sqrt[1 - Cos[c + d*x]^2]*Sqrt[-((a^2 - b^2 - 2*a* (a + b*Cos[c + d*x]) + (a + b*Cos[c + d*x])^2)/b^2)]*(2*a^2 - b^2 - 4*a*(a + b*Cos[c + d*x]) + 2*(a + b*Cos[c + d*x])^2)))/(12*a^3*(-a + b)^2*(a + b )^2*d) + (Sqrt[a + b*Cos[c + d*x]]*((2*(-(b^3*B*Sin[c + d*x]) + a*b^2*C*Si n[c + d*x]))/(3*a^2*(a^2 - b^2)*(a + b*Cos[c + d*x])^2) + (2*(-10*a^2*b^3* B*Sin[c + d*x] + 6*b^5*B*Sin[c + d*x] + 7*a^3*b^2*C*Sin[c + d*x] - 3*a*b^4 *C*Sin[c + d*x]))/(3*a^3*(a^2 - b^2)^2*(a + b*Cos[c + d*x])) + (B*Tan[c + d*x])/a^3))/d
Time = 4.01 (sec) , antiderivative size = 471, normalized size of antiderivative = 1.08, number of steps used = 26, number of rules used = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.619, Rules used = {3042, 3508, 3042, 3479, 27, 3042, 3535, 27, 3042, 3534, 27, 3042, 3538, 25, 3042, 3134, 3042, 3132, 3481, 3042, 3142, 3042, 3140, 3286, 3042, 3284}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \frac {\sec ^3(c+d x) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^{5/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {B \sin \left (c+d x+\frac {\pi }{2}\right )+C \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\sin \left (c+d x+\frac {\pi }{2}\right )^3 \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{5/2}}dx\) |
| \(\Big \downarrow \) 3508 |
| \(\displaystyle \int \frac {\sec ^2(c+d x) (B+C \cos (c+d x))}{(a+b \cos (c+d x))^{5/2}}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {B+C \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^2 \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{5/2}}dx\) |
| \(\Big \downarrow \) 3479 |
| \(\displaystyle \frac {\int -\frac {\left (-3 b B \cos ^2(c+d x)+5 b B-2 a C\right ) \sec (c+d x)}{2 (a+b \cos (c+d x))^{5/2}}dx}{a}+\frac {B \tan (c+d x)}{a d (a+b \cos (c+d x))^{3/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {B \tan (c+d x)}{a d (a+b \cos (c+d x))^{3/2}}-\frac {\int \frac {\left (-3 b B \cos ^2(c+d x)+5 b B-2 a C\right ) \sec (c+d x)}{(a+b \cos (c+d x))^{5/2}}dx}{2 a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {B \tan (c+d x)}{a d (a+b \cos (c+d x))^{3/2}}-\frac {\int \frac {-3 b B \sin \left (c+d x+\frac {\pi }{2}\right )^2+5 b B-2 a C}{\sin \left (c+d x+\frac {\pi }{2}\right ) \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{5/2}}dx}{2 a}\) |
| \(\Big \downarrow \) 3535 |
| \(\displaystyle \frac {B \tan (c+d x)}{a d (a+b \cos (c+d x))^{3/2}}-\frac {\frac {2 \int \frac {\left (-b \left (3 B a^2+2 b C a-5 b^2 B\right ) \cos ^2(c+d x)-6 a b (b B-a C) \cos (c+d x)+3 \left (a^2-b^2\right ) (5 b B-2 a C)\right ) \sec (c+d x)}{2 (a+b \cos (c+d x))^{3/2}}dx}{3 a \left (a^2-b^2\right )}-\frac {2 b \left (3 a^2 B+2 a b C-5 b^2 B\right ) \sin (c+d x)}{3 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{3/2}}}{2 a}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {B \tan (c+d x)}{a d (a+b \cos (c+d x))^{3/2}}-\frac {\frac {\int \frac {\left (-b \left (3 B a^2+2 b C a-5 b^2 B\right ) \cos ^2(c+d x)-6 a b (b B-a C) \cos (c+d x)+3 \left (a^2-b^2\right ) (5 b B-2 a C)\right ) \sec (c+d x)}{(a+b \cos (c+d x))^{3/2}}dx}{3 a \left (a^2-b^2\right )}-\frac {2 b \left (3 a^2 B+2 a b C-5 b^2 B\right ) \sin (c+d x)}{3 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{3/2}}}{2 a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {B \tan (c+d x)}{a d (a+b \cos (c+d x))^{3/2}}-\frac {\frac {\int \frac {-b \left (3 B a^2+2 b C a-5 b^2 B\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2-6 a b (b B-a C) \sin \left (c+d x+\frac {\pi }{2}\right )+3 \left (a^2-b^2\right ) (5 b B-2 a C)}{\sin \left (c+d x+\frac {\pi }{2}\right ) \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2}}dx}{3 a \left (a^2-b^2\right )}-\frac {2 b \left (3 a^2 B+2 a b C-5 b^2 B\right ) \sin (c+d x)}{3 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{3/2}}}{2 a}\) |
| \(\Big \downarrow \) 3534 |
| \(\displaystyle \frac {B \tan (c+d x)}{a d (a+b \cos (c+d x))^{3/2}}-\frac {\frac {\frac {2 \int \frac {\left (3 (5 b B-2 a C) \left (a^2-b^2\right )^2+b \left (3 B a^4+14 b C a^3-26 b^2 B a^2-6 b^3 C a+15 b^4 B\right ) \cos ^2(c+d x)-2 a b \left (-6 C a^3+9 b B a^2+2 b^2 C a-5 b^3 B\right ) \cos (c+d x)\right ) \sec (c+d x)}{2 \sqrt {a+b \cos (c+d x)}}dx}{a \left (a^2-b^2\right )}-\frac {2 b \left (3 a^4 B+14 a^3 b C-26 a^2 b^2 B-6 a b^3 C+15 b^4 B\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}}{3 a \left (a^2-b^2\right )}-\frac {2 b \left (3 a^2 B+2 a b C-5 b^2 B\right ) \sin (c+d x)}{3 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{3/2}}}{2 a}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {B \tan (c+d x)}{a d (a+b \cos (c+d x))^{3/2}}-\frac {\frac {\frac {\int \frac {\left (3 (5 b B-2 a C) \left (a^2-b^2\right )^2+b \left (3 B a^4+14 b C a^3-26 b^2 B a^2-6 b^3 C a+15 b^4 B\right ) \cos ^2(c+d x)-2 a b \left (-6 C a^3+9 b B a^2+2 b^2 C a-5 b^3 B\right ) \cos (c+d x)\right ) \sec (c+d x)}{\sqrt {a+b \cos (c+d x)}}dx}{a \left (a^2-b^2\right )}-\frac {2 b \left (3 a^4 B+14 a^3 b C-26 a^2 b^2 B-6 a b^3 C+15 b^4 B\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}}{3 a \left (a^2-b^2\right )}-\frac {2 b \left (3 a^2 B+2 a b C-5 b^2 B\right ) \sin (c+d x)}{3 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{3/2}}}{2 a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {B \tan (c+d x)}{a d (a+b \cos (c+d x))^{3/2}}-\frac {\frac {\frac {\int \frac {3 (5 b B-2 a C) \left (a^2-b^2\right )^2+b \left (3 B a^4+14 b C a^3-26 b^2 B a^2-6 b^3 C a+15 b^4 B\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2-2 a b \left (-6 C a^3+9 b B a^2+2 b^2 C a-5 b^3 B\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{a \left (a^2-b^2\right )}-\frac {2 b \left (3 a^4 B+14 a^3 b C-26 a^2 b^2 B-6 a b^3 C+15 b^4 B\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}}{3 a \left (a^2-b^2\right )}-\frac {2 b \left (3 a^2 B+2 a b C-5 b^2 B\right ) \sin (c+d x)}{3 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{3/2}}}{2 a}\) |
| \(\Big \downarrow \) 3538 |
| \(\displaystyle \frac {B \tan (c+d x)}{a d (a+b \cos (c+d x))^{3/2}}-\frac {\frac {\frac {\left (3 a^4 B+14 a^3 b C-26 a^2 b^2 B-6 a b^3 C+15 b^4 B\right ) \int \sqrt {a+b \cos (c+d x)}dx-\frac {\int -\frac {\left (3 b \left (a^2-b^2\right )^2 (5 b B-2 a C)-a b \left (a^2-b^2\right ) \left (3 B a^2+2 b C a-5 b^2 B\right ) \cos (c+d x)\right ) \sec (c+d x)}{\sqrt {a+b \cos (c+d x)}}dx}{b}}{a \left (a^2-b^2\right )}-\frac {2 b \left (3 a^4 B+14 a^3 b C-26 a^2 b^2 B-6 a b^3 C+15 b^4 B\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}}{3 a \left (a^2-b^2\right )}-\frac {2 b \left (3 a^2 B+2 a b C-5 b^2 B\right ) \sin (c+d x)}{3 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{3/2}}}{2 a}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {B \tan (c+d x)}{a d (a+b \cos (c+d x))^{3/2}}-\frac {\frac {\frac {\frac {\int \frac {\left (3 b \left (a^2-b^2\right )^2 (5 b B-2 a C)-a b \left (a^2-b^2\right ) \left (3 B a^2+2 b C a-5 b^2 B\right ) \cos (c+d x)\right ) \sec (c+d x)}{\sqrt {a+b \cos (c+d x)}}dx}{b}+\left (3 a^4 B+14 a^3 b C-26 a^2 b^2 B-6 a b^3 C+15 b^4 B\right ) \int \sqrt {a+b \cos (c+d x)}dx}{a \left (a^2-b^2\right )}-\frac {2 b \left (3 a^4 B+14 a^3 b C-26 a^2 b^2 B-6 a b^3 C+15 b^4 B\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}}{3 a \left (a^2-b^2\right )}-\frac {2 b \left (3 a^2 B+2 a b C-5 b^2 B\right ) \sin (c+d x)}{3 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{3/2}}}{2 a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {B \tan (c+d x)}{a d (a+b \cos (c+d x))^{3/2}}-\frac {\frac {\frac {\frac {\int \frac {3 b \left (a^2-b^2\right )^2 (5 b B-2 a C)-a b \left (a^2-b^2\right ) \left (3 B a^2+2 b C a-5 b^2 B\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}+\left (3 a^4 B+14 a^3 b C-26 a^2 b^2 B-6 a b^3 C+15 b^4 B\right ) \int \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{a \left (a^2-b^2\right )}-\frac {2 b \left (3 a^4 B+14 a^3 b C-26 a^2 b^2 B-6 a b^3 C+15 b^4 B\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}}{3 a \left (a^2-b^2\right )}-\frac {2 b \left (3 a^2 B+2 a b C-5 b^2 B\right ) \sin (c+d x)}{3 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{3/2}}}{2 a}\) |
| \(\Big \downarrow \) 3134 |
| \(\displaystyle \frac {B \tan (c+d x)}{a d (a+b \cos (c+d x))^{3/2}}-\frac {\frac {\frac {\frac {\int \frac {3 b \left (a^2-b^2\right )^2 (5 b B-2 a C)-a b \left (a^2-b^2\right ) \left (3 B a^2+2 b C a-5 b^2 B\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}+\frac {\left (3 a^4 B+14 a^3 b C-26 a^2 b^2 B-6 a b^3 C+15 b^4 B\right ) \sqrt {a+b \cos (c+d x)} \int \sqrt {\frac {a}{a+b}+\frac {b \cos (c+d x)}{a+b}}dx}{\sqrt {\frac {a+b \cos (c+d x)}{a+b}}}}{a \left (a^2-b^2\right )}-\frac {2 b \left (3 a^4 B+14 a^3 b C-26 a^2 b^2 B-6 a b^3 C+15 b^4 B\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}}{3 a \left (a^2-b^2\right )}-\frac {2 b \left (3 a^2 B+2 a b C-5 b^2 B\right ) \sin (c+d x)}{3 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{3/2}}}{2 a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {B \tan (c+d x)}{a d (a+b \cos (c+d x))^{3/2}}-\frac {\frac {\frac {\frac {\int \frac {3 b \left (a^2-b^2\right )^2 (5 b B-2 a C)-a b \left (a^2-b^2\right ) \left (3 B a^2+2 b C a-5 b^2 B\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}+\frac {\left (3 a^4 B+14 a^3 b C-26 a^2 b^2 B-6 a b^3 C+15 b^4 B\right ) \sqrt {a+b \cos (c+d x)} \int \sqrt {\frac {a}{a+b}+\frac {b \sin \left (c+d x+\frac {\pi }{2}\right )}{a+b}}dx}{\sqrt {\frac {a+b \cos (c+d x)}{a+b}}}}{a \left (a^2-b^2\right )}-\frac {2 b \left (3 a^4 B+14 a^3 b C-26 a^2 b^2 B-6 a b^3 C+15 b^4 B\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}}{3 a \left (a^2-b^2\right )}-\frac {2 b \left (3 a^2 B+2 a b C-5 b^2 B\right ) \sin (c+d x)}{3 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{3/2}}}{2 a}\) |
| \(\Big \downarrow \) 3132 |
| \(\displaystyle \frac {B \tan (c+d x)}{a d (a+b \cos (c+d x))^{3/2}}-\frac {\frac {\frac {\frac {\int \frac {3 b \left (a^2-b^2\right )^2 (5 b B-2 a C)-a b \left (a^2-b^2\right ) \left (3 B a^2+2 b C a-5 b^2 B\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}+\frac {2 \left (3 a^4 B+14 a^3 b C-26 a^2 b^2 B-6 a b^3 C+15 b^4 B\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}}{a \left (a^2-b^2\right )}-\frac {2 b \left (3 a^4 B+14 a^3 b C-26 a^2 b^2 B-6 a b^3 C+15 b^4 B\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}}{3 a \left (a^2-b^2\right )}-\frac {2 b \left (3 a^2 B+2 a b C-5 b^2 B\right ) \sin (c+d x)}{3 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{3/2}}}{2 a}\) |
| \(\Big \downarrow \) 3481 |
| \(\displaystyle \frac {B \tan (c+d x)}{a d (a+b \cos (c+d x))^{3/2}}-\frac {\frac {\frac {\frac {3 b \left (a^2-b^2\right )^2 (5 b B-2 a C) \int \frac {\sec (c+d x)}{\sqrt {a+b \cos (c+d x)}}dx-a b \left (a^2-b^2\right ) \left (3 a^2 B+2 a b C-5 b^2 B\right ) \int \frac {1}{\sqrt {a+b \cos (c+d x)}}dx}{b}+\frac {2 \left (3 a^4 B+14 a^3 b C-26 a^2 b^2 B-6 a b^3 C+15 b^4 B\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}}{a \left (a^2-b^2\right )}-\frac {2 b \left (3 a^4 B+14 a^3 b C-26 a^2 b^2 B-6 a b^3 C+15 b^4 B\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}}{3 a \left (a^2-b^2\right )}-\frac {2 b \left (3 a^2 B+2 a b C-5 b^2 B\right ) \sin (c+d x)}{3 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{3/2}}}{2 a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {B \tan (c+d x)}{a d (a+b \cos (c+d x))^{3/2}}-\frac {\frac {\frac {\frac {3 b \left (a^2-b^2\right )^2 (5 b B-2 a C) \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx-a b \left (a^2-b^2\right ) \left (3 a^2 B+2 a b C-5 b^2 B\right ) \int \frac {1}{\sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}+\frac {2 \left (3 a^4 B+14 a^3 b C-26 a^2 b^2 B-6 a b^3 C+15 b^4 B\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}}{a \left (a^2-b^2\right )}-\frac {2 b \left (3 a^4 B+14 a^3 b C-26 a^2 b^2 B-6 a b^3 C+15 b^4 B\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}}{3 a \left (a^2-b^2\right )}-\frac {2 b \left (3 a^2 B+2 a b C-5 b^2 B\right ) \sin (c+d x)}{3 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{3/2}}}{2 a}\) |
| \(\Big \downarrow \) 3142 |
| \(\displaystyle \frac {B \tan (c+d x)}{a d (a+b \cos (c+d x))^{3/2}}-\frac {\frac {\frac {\frac {3 b \left (a^2-b^2\right )^2 (5 b B-2 a C) \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {a b \left (a^2-b^2\right ) \left (3 a^2 B+2 a b C-5 b^2 B\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \int \frac {1}{\sqrt {\frac {a}{a+b}+\frac {b \cos (c+d x)}{a+b}}}dx}{\sqrt {a+b \cos (c+d x)}}}{b}+\frac {2 \left (3 a^4 B+14 a^3 b C-26 a^2 b^2 B-6 a b^3 C+15 b^4 B\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}}{a \left (a^2-b^2\right )}-\frac {2 b \left (3 a^4 B+14 a^3 b C-26 a^2 b^2 B-6 a b^3 C+15 b^4 B\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}}{3 a \left (a^2-b^2\right )}-\frac {2 b \left (3 a^2 B+2 a b C-5 b^2 B\right ) \sin (c+d x)}{3 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{3/2}}}{2 a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {B \tan (c+d x)}{a d (a+b \cos (c+d x))^{3/2}}-\frac {\frac {\frac {\frac {3 b \left (a^2-b^2\right )^2 (5 b B-2 a C) \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {a b \left (a^2-b^2\right ) \left (3 a^2 B+2 a b C-5 b^2 B\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \int \frac {1}{\sqrt {\frac {a}{a+b}+\frac {b \sin \left (c+d x+\frac {\pi }{2}\right )}{a+b}}}dx}{\sqrt {a+b \cos (c+d x)}}}{b}+\frac {2 \left (3 a^4 B+14 a^3 b C-26 a^2 b^2 B-6 a b^3 C+15 b^4 B\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}}{a \left (a^2-b^2\right )}-\frac {2 b \left (3 a^4 B+14 a^3 b C-26 a^2 b^2 B-6 a b^3 C+15 b^4 B\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}}{3 a \left (a^2-b^2\right )}-\frac {2 b \left (3 a^2 B+2 a b C-5 b^2 B\right ) \sin (c+d x)}{3 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{3/2}}}{2 a}\) |
| \(\Big \downarrow \) 3140 |
| \(\displaystyle \frac {B \tan (c+d x)}{a d (a+b \cos (c+d x))^{3/2}}-\frac {\frac {\frac {\frac {3 b \left (a^2-b^2\right )^2 (5 b B-2 a C) \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 a b \left (a^2-b^2\right ) \left (3 a^2 B+2 a b C-5 b^2 B\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{d \sqrt {a+b \cos (c+d x)}}}{b}+\frac {2 \left (3 a^4 B+14 a^3 b C-26 a^2 b^2 B-6 a b^3 C+15 b^4 B\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}}{a \left (a^2-b^2\right )}-\frac {2 b \left (3 a^4 B+14 a^3 b C-26 a^2 b^2 B-6 a b^3 C+15 b^4 B\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}}{3 a \left (a^2-b^2\right )}-\frac {2 b \left (3 a^2 B+2 a b C-5 b^2 B\right ) \sin (c+d x)}{3 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{3/2}}}{2 a}\) |
| \(\Big \downarrow \) 3286 |
| \(\displaystyle \frac {B \tan (c+d x)}{a d (a+b \cos (c+d x))^{3/2}}-\frac {\frac {\frac {\frac {\frac {3 b \left (a^2-b^2\right )^2 (5 b B-2 a C) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \int \frac {\sec (c+d x)}{\sqrt {\frac {a}{a+b}+\frac {b \cos (c+d x)}{a+b}}}dx}{\sqrt {a+b \cos (c+d x)}}-\frac {2 a b \left (a^2-b^2\right ) \left (3 a^2 B+2 a b C-5 b^2 B\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{d \sqrt {a+b \cos (c+d x)}}}{b}+\frac {2 \left (3 a^4 B+14 a^3 b C-26 a^2 b^2 B-6 a b^3 C+15 b^4 B\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}}{a \left (a^2-b^2\right )}-\frac {2 b \left (3 a^4 B+14 a^3 b C-26 a^2 b^2 B-6 a b^3 C+15 b^4 B\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}}{3 a \left (a^2-b^2\right )}-\frac {2 b \left (3 a^2 B+2 a b C-5 b^2 B\right ) \sin (c+d x)}{3 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{3/2}}}{2 a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {B \tan (c+d x)}{a d (a+b \cos (c+d x))^{3/2}}-\frac {\frac {\frac {\frac {\frac {3 b \left (a^2-b^2\right )^2 (5 b B-2 a C) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {\frac {a}{a+b}+\frac {b \sin \left (c+d x+\frac {\pi }{2}\right )}{a+b}}}dx}{\sqrt {a+b \cos (c+d x)}}-\frac {2 a b \left (a^2-b^2\right ) \left (3 a^2 B+2 a b C-5 b^2 B\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{d \sqrt {a+b \cos (c+d x)}}}{b}+\frac {2 \left (3 a^4 B+14 a^3 b C-26 a^2 b^2 B-6 a b^3 C+15 b^4 B\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}}{a \left (a^2-b^2\right )}-\frac {2 b \left (3 a^4 B+14 a^3 b C-26 a^2 b^2 B-6 a b^3 C+15 b^4 B\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}}{3 a \left (a^2-b^2\right )}-\frac {2 b \left (3 a^2 B+2 a b C-5 b^2 B\right ) \sin (c+d x)}{3 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{3/2}}}{2 a}\) |
| \(\Big \downarrow \) 3284 |
| \(\displaystyle \frac {B \tan (c+d x)}{a d (a+b \cos (c+d x))^{3/2}}-\frac {\frac {\frac {\frac {\frac {6 b \left (a^2-b^2\right )^2 (5 b B-2 a C) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticPi}\left (2,\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{d \sqrt {a+b \cos (c+d x)}}-\frac {2 a b \left (a^2-b^2\right ) \left (3 a^2 B+2 a b C-5 b^2 B\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{d \sqrt {a+b \cos (c+d x)}}}{b}+\frac {2 \left (3 a^4 B+14 a^3 b C-26 a^2 b^2 B-6 a b^3 C+15 b^4 B\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}}{a \left (a^2-b^2\right )}-\frac {2 b \left (3 a^4 B+14 a^3 b C-26 a^2 b^2 B-6 a b^3 C+15 b^4 B\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}}{3 a \left (a^2-b^2\right )}-\frac {2 b \left (3 a^2 B+2 a b C-5 b^2 B\right ) \sin (c+d x)}{3 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{3/2}}}{2 a}\) |
Input:
Int[((B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^3)/(a + b*Cos[c + d* x])^(5/2),x]
Output:
-1/2*((-2*b*(3*a^2*B - 5*b^2*B + 2*a*b*C)*Sin[c + d*x])/(3*a*(a^2 - b^2)*d *(a + b*Cos[c + d*x])^(3/2)) + (((2*(3*a^4*B - 26*a^2*b^2*B + 15*b^4*B + 1 4*a^3*b*C - 6*a*b^3*C)*Sqrt[a + b*Cos[c + d*x]]*EllipticE[(c + d*x)/2, (2* b)/(a + b)])/(d*Sqrt[(a + b*Cos[c + d*x])/(a + b)]) + ((-2*a*b*(a^2 - b^2) *(3*a^2*B - 5*b^2*B + 2*a*b*C)*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*Elliptic F[(c + d*x)/2, (2*b)/(a + b)])/(d*Sqrt[a + b*Cos[c + d*x]]) + (6*b*(a^2 - b^2)^2*(5*b*B - 2*a*C)*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*EllipticPi[2, (c + d*x)/2, (2*b)/(a + b)])/(d*Sqrt[a + b*Cos[c + d*x]]))/b)/(a*(a^2 - b^2) ) - (2*b*(3*a^4*B - 26*a^2*b^2*B + 15*b^4*B + 14*a^3*b*C - 6*a*b^3*C)*Sin[ c + d*x])/(a*(a^2 - b^2)*d*Sqrt[a + b*Cos[c + d*x]]))/(3*a*(a^2 - b^2)))/a + (B*Tan[c + d*x])/(a*d*(a + b*Cos[c + d*x])^(3/2))
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[2*(Sqrt[a + b]/d)*EllipticE[(1/2)*(c - Pi/2 + d*x), 2*(b/(a + b))], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]
Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[Sqrt[a + b*Sin[c + d*x]]/Sqrt[(a + b*Sin[c + d*x])/(a + b)] Int[Sqrt[a/(a + b) + ( b/(a + b))*Sin[c + d*x]], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2 , 0] && !GtQ[a + b, 0]
Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/(d*S qrt[a + b]))*EllipticF[(1/2)*(c - Pi/2 + d*x), 2*(b/(a + b))], x] /; FreeQ[ {a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]
Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[Sqrt[(a + b*Sin[c + d*x])/(a + b)]/Sqrt[a + b*Sin[c + d*x]] Int[1/Sqrt[a/(a + b) + (b/(a + b))*Sin[c + d*x]], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && !GtQ[a + b, 0]
Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp[(2/(f*(a + b)*Sqrt[c + d]))*EllipticPi[ 2*(b/(a + b)), (1/2)*(e - Pi/2 + f*x), 2*(d/(c + d))], x] /; FreeQ[{a, b, c , d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[c + d, 0]
Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp[Sqrt[(c + d*Sin[e + f*x])/(c + d)]/Sqrt [c + d*Sin[e + f*x]] Int[1/((a + b*Sin[e + f*x])*Sqrt[c/(c + d) + (d/(c + d))*Sin[e + f*x]]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a* d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && !GtQ[c + d, 0]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si mp[(-(A*b^2 - a*b*B))*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1)*((c + d*Sin [e + f*x])^(1 + n)/(f*(m + 1)*(b*c - a*d)*(a^2 - b^2))), x] + Simp[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)) Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[(a*A - b*B)*(b*c - a*d)*(m + 1) + b*d*(A*b - a*B)*(m + n + 2) + (A*b - a*B)*(a*d*(m + 1) - b*c*(m + 2))*Sin[e + f*x] - b*d*(A*b - a*B) *(m + n + 3)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n }, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && Rat ionalQ[m] && m < -1 && ((EqQ[a, 0] && IntegerQ[m] && !IntegerQ[n]) || !(I ntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n] && !IntegerQ[m]) || EqQ[a, 0]) ))
Int[(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)]))/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[ B/d Int[(a + b*Sin[e + f*x])^m, x], x] - Simp[(B*c - A*d)/d Int[(a + b* Sin[e + f*x])^m/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_ .) + (f_.)*(x_)]^2), x_Symbol] :> Simp[1/b^2 Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*(b*B - a*C + b*C*Sin[e + f*x]), x], x] /; FreeQ [{a, b, c, d, e, f, A, B, C, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[A*b^2 - a*b*B + a^2*C, 0]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-(A*b^2 - a*b*B + a^2*C))*Cos[e + f*x ]*(a + b*Sin[e + f*x])^(m + 1)*((c + d*Sin[e + f*x])^(n + 1)/(f*(m + 1)*(b* c - a*d)*(a^2 - b^2))), x] + Simp[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)) Int [(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[(m + 1)*(b*c - a* d)*(a*A - b*B + a*C) + d*(A*b^2 - a*b*B + a^2*C)*(m + n + 2) - (c*(A*b^2 - a*b*B + a^2*C) + (m + 1)*(b*c - a*d)*(A*b - a*B + b*C))*Sin[e + f*x] - d*(A *b^2 - a*b*B + a^2*C)*(m + n + 3)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b , c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -1] && ((EqQ[a, 0] && IntegerQ[m] && !IntegerQ [n]) || !(IntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n] && !IntegerQ[m]) | | EqQ[a, 0])))
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-(A*b^2 + a^2*C))*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1)*((c + d*S in[e + f*x])^(n + 1)/(f*(m + 1)*(b*c - a*d)*(a^2 - b^2))), x] + Simp[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)) Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin [e + f*x])^n*Simp[a*(m + 1)*(b*c - a*d)*(A + C) + d*(A*b^2 + a^2*C)*(m + n + 2) - (c*(A*b^2 + a^2*C) + b*(m + 1)*(b*c - a*d)*(A + C))*Sin[e + f*x] - d *(A*b^2 + a^2*C)*(m + n + 3)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -1] && ((EqQ[a, 0] && IntegerQ[m] && !IntegerQ[n]) || !(IntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n] && !IntegerQ[m]) || EqQ[a, 0])))
Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^ 2)/(Sqrt[(a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])), x_Symbol] :> Simp[C/(b*d) Int[Sqrt[a + b*Sin[e + f*x]], x] , x] - Simp[1/(b*d) Int[Simp[a*c*C - A*b*d + (b*c*C - b*B*d + a*C*d)*Sin[ e + f*x], x]/(Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])), x], x] /; Fre eQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0 ] && NeQ[c^2 - d^2, 0]
Leaf count of result is larger than twice the leaf count of optimal. \(1345\) vs. \(2(425)=850\).
Time = 17.58 (sec) , antiderivative size = 1346, normalized size of antiderivative = 3.08
| method | result | size |
| default | \(\text {Expression too large to display}\) | \(1346\) |
| parts | \(\text {Expression too large to display}\) | \(2175\) |
Input:
int((B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^3/(a+b*cos(d*x+c))^(5/2),x,me thod=_RETURNVERBOSE)
Output:
-(-(-2*b*cos(1/2*d*x+1/2*c)^2-a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*B/a^2*(- cos(1/2*d*x+1/2*c)/a*(-2*b*sin(1/2*d*x+1/2*c)^4+(a+b)*sin(1/2*d*x+1/2*c)^2 )^(1/2)/(2*cos(1/2*d*x+1/2*c)^2-1)+1/2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*b* cos(1/2*d*x+1/2*c)^2+a-b)/(a-b))^(1/2)/(-2*b*sin(1/2*d*x+1/2*c)^4+(a+b)*si n(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2)) -1/2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*b*cos(1/2*d*x+1/2*c)^2+a-b)/(a-b))^( 1/2)/(-2*b*sin(1/2*d*x+1/2*c)^4+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*Elliptic E(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))+1/2/a*(sin(1/2*d*x+1/2*c)^2)^(1/2 )*((2*b*cos(1/2*d*x+1/2*c)^2+a-b)/(a-b))^(1/2)/(-2*b*sin(1/2*d*x+1/2*c)^4+ (a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*b*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a- b))^(1/2))+1/2*b/a*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*b*cos(1/2*d*x+1/2*c)^2 +a-b)/(a-b))^(1/2)/(-2*b*sin(1/2*d*x+1/2*c)^4+(a+b)*sin(1/2*d*x+1/2*c)^2)^ (1/2)*EllipticPi(cos(1/2*d*x+1/2*c),2,(-2*b/(a-b))^(1/2)))+2*(2*B*b-C*a)/a ^3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*b*cos(1/2*d*x+1/2*c)^2+a-b)/(a-b))^(1/ 2)/(-2*b*sin(1/2*d*x+1/2*c)^4+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticPi (cos(1/2*d*x+1/2*c),2,(-2*b/(a-b))^(1/2))-2*b*(2*B*b-C*a)/a^3/sin(1/2*d*x+ 1/2*c)^2/(2*b*sin(1/2*d*x+1/2*c)^2-a-b)/(a^2-b^2)*(-2*b*sin(1/2*d*x+1/2*c) ^4+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2 *c)^2*b+(-2*b/(a-b)*sin(1/2*d*x+1/2*c)^2+(a+b)/(a-b))^(1/2)*(sin(1/2*d*x+1 /2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a-(-2*b...
Timed out. \[ \int \frac {\left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{(a+b \cos (c+d x))^{5/2}} \, dx=\text {Timed out} \] Input:
integrate((B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^3/(a+b*cos(d*x+c))^(5/2 ),x, algorithm="fricas")
Output:
Timed out
Timed out. \[ \int \frac {\left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{(a+b \cos (c+d x))^{5/2}} \, dx=\text {Timed out} \] Input:
integrate((B*cos(d*x+c)+C*cos(d*x+c)**2)*sec(d*x+c)**3/(a+b*cos(d*x+c))**( 5/2),x)
Output:
Timed out
\[ \int \frac {\left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{(a+b \cos (c+d x))^{5/2}} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right )\right )} \sec \left (d x + c\right )^{3}}{{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {5}{2}}} \,d x } \] Input:
integrate((B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^3/(a+b*cos(d*x+c))^(5/2 ),x, algorithm="maxima")
Output:
integrate((C*cos(d*x + c)^2 + B*cos(d*x + c))*sec(d*x + c)^3/(b*cos(d*x + c) + a)^(5/2), x)
\[ \int \frac {\left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{(a+b \cos (c+d x))^{5/2}} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right )\right )} \sec \left (d x + c\right )^{3}}{{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {5}{2}}} \,d x } \] Input:
integrate((B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^3/(a+b*cos(d*x+c))^(5/2 ),x, algorithm="giac")
Output:
integrate((C*cos(d*x + c)^2 + B*cos(d*x + c))*sec(d*x + c)^3/(b*cos(d*x + c) + a)^(5/2), x)
Timed out. \[ \int \frac {\left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{(a+b \cos (c+d x))^{5/2}} \, dx=\int \frac {C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )}{{\cos \left (c+d\,x\right )}^3\,{\left (a+b\,\cos \left (c+d\,x\right )\right )}^{5/2}} \,d x \] Input:
int((B*cos(c + d*x) + C*cos(c + d*x)^2)/(cos(c + d*x)^3*(a + b*cos(c + d*x ))^(5/2)),x)
Output:
int((B*cos(c + d*x) + C*cos(c + d*x)^2)/(cos(c + d*x)^3*(a + b*cos(c + d*x ))^(5/2)), x)
\[ \int \frac {\left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{(a+b \cos (c+d x))^{5/2}} \, dx=\left (\int \frac {\sqrt {\cos \left (d x +c \right ) b +a}\, \cos \left (d x +c \right ) \sec \left (d x +c \right )^{3}}{\cos \left (d x +c \right )^{3} b^{3}+3 \cos \left (d x +c \right )^{2} a \,b^{2}+3 \cos \left (d x +c \right ) a^{2} b +a^{3}}d x \right ) b +\left (\int \frac {\sqrt {\cos \left (d x +c \right ) b +a}\, \cos \left (d x +c \right )^{2} \sec \left (d x +c \right )^{3}}{\cos \left (d x +c \right )^{3} b^{3}+3 \cos \left (d x +c \right )^{2} a \,b^{2}+3 \cos \left (d x +c \right ) a^{2} b +a^{3}}d x \right ) c \] Input:
int((B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^3/(a+b*cos(d*x+c))^(5/2),x)
Output:
int((sqrt(cos(c + d*x)*b + a)*cos(c + d*x)*sec(c + d*x)**3)/(cos(c + d*x)* *3*b**3 + 3*cos(c + d*x)**2*a*b**2 + 3*cos(c + d*x)*a**2*b + a**3),x)*b + int((sqrt(cos(c + d*x)*b + a)*cos(c + d*x)**2*sec(c + d*x)**3)/(cos(c + d* x)**3*b**3 + 3*cos(c + d*x)**2*a*b**2 + 3*cos(c + d*x)*a**2*b + a**3),x)*c