Integrand size = 42, antiderivative size = 315 \[ \int (a+b \cos (c+d x))^{5/2} \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=-\frac {\left (9 a b B+4 a^2 C-8 b^2 C\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{4 d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}+\frac {\left (11 a^2 b B+8 b^3 B+4 a^3 C+16 a b^2 C\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{4 d \sqrt {a+b \cos (c+d x)}}+\frac {a \left (4 a^2 B+15 b^2 B+20 a b C\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticPi}\left (2,\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{4 d \sqrt {a+b \cos (c+d x)}}+\frac {a (7 b B+4 a C) \sqrt {a+b \cos (c+d x)} \tan (c+d x)}{4 d}+\frac {a B (a+b \cos (c+d x))^{3/2} \sec (c+d x) \tan (c+d x)}{2 d} \] Output:
-1/4*(9*B*a*b+4*C*a^2-8*C*b^2)*(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))/d/((a+b*cos(d*x+c))/(a+b))^(1/2)+1/4*(11 *B*a^2*b+8*B*b^3+4*C*a^3+16*C*a*b^2)*((a+b*cos(d*x+c))/(a+b))^(1/2)*Invers eJacobiAM(1/2*d*x+1/2*c,2^(1/2)*(b/(a+b))^(1/2))/d/(a+b*cos(d*x+c))^(1/2)+ 1/4*a*(4*B*a^2+15*B*b^2+20*C*a*b)*((a+b*cos(d*x+c))/(a+b))^(1/2)*EllipticP i(sin(1/2*d*x+1/2*c),2,2^(1/2)*(b/(a+b))^(1/2))/d/(a+b*cos(d*x+c))^(1/2)+1 /4*a*(7*B*b+4*C*a)*(a+b*cos(d*x+c))^(1/2)*tan(d*x+c)/d+1/2*a*B*(a+b*cos(d* x+c))^(3/2)*sec(d*x+c)*tan(d*x+c)/d
Result contains complex when optimal does not.
Time = 6.27 (sec) , antiderivative size = 451, normalized size of antiderivative = 1.43 \[ \int (a+b \cos (c+d x))^{5/2} \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\frac {\frac {8 b \left (a^2 B+4 b^2 B+12 a b C\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{\sqrt {a+b \cos (c+d x)}}+\frac {2 \left (8 a^3 B+21 a b^2 B+36 a^2 b C+8 b^3 C\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticPi}\left (2,\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{\sqrt {a+b \cos (c+d x)}}+\frac {2 i \left (-9 a b B-4 a^2 C+8 b^2 C\right ) \sqrt {-\frac {b (-1+\cos (c+d x))}{a+b}} \sqrt {\frac {b (1+\cos (c+d x))}{-a+b}} \csc (c+d x) \left (-2 a (a-b) E\left (i \text {arcsinh}\left (\sqrt {-\frac {1}{a+b}} \sqrt {a+b \cos (c+d x)}\right )|\frac {a+b}{a-b}\right )+b \left (-2 a \operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt {-\frac {1}{a+b}} \sqrt {a+b \cos (c+d x)}\right ),\frac {a+b}{a-b}\right )+b \operatorname {EllipticPi}\left (\frac {a+b}{a},i \text {arcsinh}\left (\sqrt {-\frac {1}{a+b}} \sqrt {a+b \cos (c+d x)}\right ),\frac {a+b}{a-b}\right )\right )\right )}{a b \sqrt {-\frac {1}{a+b}}}+4 a \sqrt {a+b \cos (c+d x)} (2 a B+(9 b B+4 a C) \cos (c+d x)) \sec (c+d x) \tan (c+d x)}{16 d} \] Input:
Integrate[(a + b*Cos[c + d*x])^(5/2)*(B*Cos[c + d*x] + C*Cos[c + d*x]^2)*S ec[c + d*x]^4,x]
Output:
((8*b*(a^2*B + 4*b^2*B + 12*a*b*C)*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*Elli pticF[(c + d*x)/2, (2*b)/(a + b)])/Sqrt[a + b*Cos[c + d*x]] + (2*(8*a^3*B + 21*a*b^2*B + 36*a^2*b*C + 8*b^3*C)*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*El lipticPi[2, (c + d*x)/2, (2*b)/(a + b)])/Sqrt[a + b*Cos[c + d*x]] + ((2*I) *(-9*a*b*B - 4*a^2*C + 8*b^2*C)*Sqrt[-((b*(-1 + Cos[c + d*x]))/(a + b))]*S qrt[(b*(1 + Cos[c + d*x]))/(-a + b)]*Csc[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)])))/(a*b*Sqrt[-(a + b)^(-1)]) + 4*a*Sqrt[a + b*Cos[c + d*x]]*(2*a*B + (9*b*B + 4*a*C)*Cos[c + d*x])*Sec [c + d*x]*Tan[c + d*x])/(16*d)
Time = 2.82 (sec) , antiderivative size = 323, normalized size of antiderivative = 1.03, number of steps used = 23, number of rules used = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.548, Rules used = {3042, 3508, 3042, 3468, 27, 3042, 3526, 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.
| \(\displaystyle \int \sec ^4(c+d x) (a+b \cos (c+d x))^{5/2} \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{5/2} \left (B \sin \left (c+d x+\frac {\pi }{2}\right )+C \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^4}dx\) |
| \(\Big \downarrow \) 3508 |
| \(\displaystyle \int \sec ^3(c+d x) (a+b \cos (c+d x))^{5/2} (B+C \cos (c+d x))dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{5/2} \left (B+C \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^3}dx\) |
| \(\Big \downarrow \) 3468 |
| \(\displaystyle \frac {1}{2} \int \frac {1}{2} \sqrt {a+b \cos (c+d x)} \left (-b (a B-4 b C) \cos ^2(c+d x)+2 \left (B a^2+4 b C a+2 b^2 B\right ) \cos (c+d x)+a (7 b B+4 a C)\right ) \sec ^2(c+d x)dx+\frac {a B \tan (c+d x) \sec (c+d x) (a+b \cos (c+d x))^{3/2}}{2 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{4} \int \sqrt {a+b \cos (c+d x)} \left (-b (a B-4 b C) \cos ^2(c+d x)+2 \left (B a^2+4 b C a+2 b^2 B\right ) \cos (c+d x)+a (7 b B+4 a C)\right ) \sec ^2(c+d x)dx+\frac {a B \tan (c+d x) \sec (c+d x) (a+b \cos (c+d x))^{3/2}}{2 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{4} \int \frac {\sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )} \left (-b (a B-4 b C) \sin \left (c+d x+\frac {\pi }{2}\right )^2+2 \left (B a^2+4 b C a+2 b^2 B\right ) \sin \left (c+d x+\frac {\pi }{2}\right )+a (7 b B+4 a C)\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^2}dx+\frac {a B \tan (c+d x) \sec (c+d x) (a+b \cos (c+d x))^{3/2}}{2 d}\) |
| \(\Big \downarrow \) 3526 |
| \(\displaystyle \frac {1}{4} \left (\int \frac {\left (-b \left (4 C a^2+9 b B a-8 b^2 C\right ) \cos ^2(c+d x)+2 b \left (B a^2+12 b C a+4 b^2 B\right ) \cos (c+d x)+a \left (4 B a^2+20 b C a+15 b^2 B\right )\right ) \sec (c+d x)}{2 \sqrt {a+b \cos (c+d x)}}dx+\frac {a (4 a C+7 b B) \tan (c+d x) \sqrt {a+b \cos (c+d x)}}{d}\right )+\frac {a B \tan (c+d x) \sec (c+d x) (a+b \cos (c+d x))^{3/2}}{2 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{4} \left (\frac {1}{2} \int \frac {\left (-b \left (4 C a^2+9 b B a-8 b^2 C\right ) \cos ^2(c+d x)+2 b \left (B a^2+12 b C a+4 b^2 B\right ) \cos (c+d x)+a \left (4 B a^2+20 b C a+15 b^2 B\right )\right ) \sec (c+d x)}{\sqrt {a+b \cos (c+d x)}}dx+\frac {a (4 a C+7 b B) \tan (c+d x) \sqrt {a+b \cos (c+d x)}}{d}\right )+\frac {a B \tan (c+d x) \sec (c+d x) (a+b \cos (c+d x))^{3/2}}{2 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{4} \left (\frac {1}{2} \int \frac {-b \left (4 C a^2+9 b B a-8 b^2 C\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2+2 b \left (B a^2+12 b C a+4 b^2 B\right ) \sin \left (c+d x+\frac {\pi }{2}\right )+a \left (4 B a^2+20 b C a+15 b^2 B\right )}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {a (4 a C+7 b B) \tan (c+d x) \sqrt {a+b \cos (c+d x)}}{d}\right )+\frac {a B \tan (c+d x) \sec (c+d x) (a+b \cos (c+d x))^{3/2}}{2 d}\) |
| \(\Big \downarrow \) 3538 |
| \(\displaystyle \frac {1}{4} \left (\frac {1}{2} \left (-\left (\left (4 a^2 C+9 a b B-8 b^2 C\right ) \int \sqrt {a+b \cos (c+d x)}dx\right )-\frac {\int -\frac {\left (a b \left (4 B a^2+20 b C a+15 b^2 B\right )+b \left (4 C a^3+11 b B a^2+16 b^2 C a+8 b^3 B\right ) \cos (c+d x)\right ) \sec (c+d x)}{\sqrt {a+b \cos (c+d x)}}dx}{b}\right )+\frac {a (4 a C+7 b B) \tan (c+d x) \sqrt {a+b \cos (c+d x)}}{d}\right )+\frac {a B \tan (c+d x) \sec (c+d x) (a+b \cos (c+d x))^{3/2}}{2 d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {1}{4} \left (\frac {1}{2} \left (\frac {\int \frac {\left (a b \left (4 B a^2+20 b C a+15 b^2 B\right )+b \left (4 C a^3+11 b B a^2+16 b^2 C a+8 b^3 B\right ) \cos (c+d x)\right ) \sec (c+d x)}{\sqrt {a+b \cos (c+d x)}}dx}{b}-\left (4 a^2 C+9 a b B-8 b^2 C\right ) \int \sqrt {a+b \cos (c+d x)}dx\right )+\frac {a (4 a C+7 b B) \tan (c+d x) \sqrt {a+b \cos (c+d x)}}{d}\right )+\frac {a B \tan (c+d x) \sec (c+d x) (a+b \cos (c+d x))^{3/2}}{2 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{4} \left (\frac {1}{2} \left (\frac {\int \frac {a b \left (4 B a^2+20 b C a+15 b^2 B\right )+b \left (4 C a^3+11 b B a^2+16 b^2 C a+8 b^3 B\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}-\left (4 a^2 C+9 a b B-8 b^2 C\right ) \int \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx\right )+\frac {a (4 a C+7 b B) \tan (c+d x) \sqrt {a+b \cos (c+d x)}}{d}\right )+\frac {a B \tan (c+d x) \sec (c+d x) (a+b \cos (c+d x))^{3/2}}{2 d}\) |
| \(\Big \downarrow \) 3134 |
| \(\displaystyle \frac {1}{4} \left (\frac {1}{2} \left (\frac {\int \frac {a b \left (4 B a^2+20 b C a+15 b^2 B\right )+b \left (4 C a^3+11 b B a^2+16 b^2 C a+8 b^3 B\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}-\frac {\left (4 a^2 C+9 a b B-8 b^2 C\right ) \sqrt {a+b \cos (c+d x)} \int \sqrt {\frac {a}{a+b}+\frac {b \cos (c+d x)}{a+b}}dx}{\sqrt {\frac {a+b \cos (c+d x)}{a+b}}}\right )+\frac {a (4 a C+7 b B) \tan (c+d x) \sqrt {a+b \cos (c+d x)}}{d}\right )+\frac {a B \tan (c+d x) \sec (c+d x) (a+b \cos (c+d x))^{3/2}}{2 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{4} \left (\frac {1}{2} \left (\frac {\int \frac {a b \left (4 B a^2+20 b C a+15 b^2 B\right )+b \left (4 C a^3+11 b B a^2+16 b^2 C a+8 b^3 B\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}-\frac {\left (4 a^2 C+9 a b B-8 b^2 C\right ) \sqrt {a+b \cos (c+d x)} \int \sqrt {\frac {a}{a+b}+\frac {b \sin \left (c+d x+\frac {\pi }{2}\right )}{a+b}}dx}{\sqrt {\frac {a+b \cos (c+d x)}{a+b}}}\right )+\frac {a (4 a C+7 b B) \tan (c+d x) \sqrt {a+b \cos (c+d x)}}{d}\right )+\frac {a B \tan (c+d x) \sec (c+d x) (a+b \cos (c+d x))^{3/2}}{2 d}\) |
| \(\Big \downarrow \) 3132 |
| \(\displaystyle \frac {1}{4} \left (\frac {1}{2} \left (\frac {\int \frac {a b \left (4 B a^2+20 b C a+15 b^2 B\right )+b \left (4 C a^3+11 b B a^2+16 b^2 C a+8 b^3 B\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}-\frac {2 \left (4 a^2 C+9 a b B-8 b^2 C\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}\right )+\frac {a (4 a C+7 b B) \tan (c+d x) \sqrt {a+b \cos (c+d x)}}{d}\right )+\frac {a B \tan (c+d x) \sec (c+d x) (a+b \cos (c+d x))^{3/2}}{2 d}\) |
| \(\Big \downarrow \) 3481 |
| \(\displaystyle \frac {1}{4} \left (\frac {1}{2} \left (\frac {a b \left (4 a^2 B+20 a b C+15 b^2 B\right ) \int \frac {\sec (c+d x)}{\sqrt {a+b \cos (c+d x)}}dx+b \left (4 a^3 C+11 a^2 b B+16 a b^2 C+8 b^3 B\right ) \int \frac {1}{\sqrt {a+b \cos (c+d x)}}dx}{b}-\frac {2 \left (4 a^2 C+9 a b B-8 b^2 C\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}\right )+\frac {a (4 a C+7 b B) \tan (c+d x) \sqrt {a+b \cos (c+d x)}}{d}\right )+\frac {a B \tan (c+d x) \sec (c+d x) (a+b \cos (c+d x))^{3/2}}{2 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{4} \left (\frac {1}{2} \left (\frac {a b \left (4 a^2 B+20 a b C+15 b^2 B\right ) \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx+b \left (4 a^3 C+11 a^2 b B+16 a b^2 C+8 b^3 B\right ) \int \frac {1}{\sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}-\frac {2 \left (4 a^2 C+9 a b B-8 b^2 C\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}\right )+\frac {a (4 a C+7 b B) \tan (c+d x) \sqrt {a+b \cos (c+d x)}}{d}\right )+\frac {a B \tan (c+d x) \sec (c+d x) (a+b \cos (c+d x))^{3/2}}{2 d}\) |
| \(\Big \downarrow \) 3142 |
| \(\displaystyle \frac {1}{4} \left (\frac {1}{2} \left (\frac {a b \left (4 a^2 B+20 a b C+15 b^2 B\right ) \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {b \left (4 a^3 C+11 a^2 b B+16 a b^2 C+8 b^3 B\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \int \frac {1}{\sqrt {\frac {a}{a+b}+\frac {b \cos (c+d x)}{a+b}}}dx}{\sqrt {a+b \cos (c+d x)}}}{b}-\frac {2 \left (4 a^2 C+9 a b B-8 b^2 C\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}\right )+\frac {a (4 a C+7 b B) \tan (c+d x) \sqrt {a+b \cos (c+d x)}}{d}\right )+\frac {a B \tan (c+d x) \sec (c+d x) (a+b \cos (c+d x))^{3/2}}{2 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{4} \left (\frac {1}{2} \left (\frac {a b \left (4 a^2 B+20 a b C+15 b^2 B\right ) \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {b \left (4 a^3 C+11 a^2 b B+16 a b^2 C+8 b^3 B\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \int \frac {1}{\sqrt {\frac {a}{a+b}+\frac {b \sin \left (c+d x+\frac {\pi }{2}\right )}{a+b}}}dx}{\sqrt {a+b \cos (c+d x)}}}{b}-\frac {2 \left (4 a^2 C+9 a b B-8 b^2 C\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}\right )+\frac {a (4 a C+7 b B) \tan (c+d x) \sqrt {a+b \cos (c+d x)}}{d}\right )+\frac {a B \tan (c+d x) \sec (c+d x) (a+b \cos (c+d x))^{3/2}}{2 d}\) |
| \(\Big \downarrow \) 3140 |
| \(\displaystyle \frac {1}{4} \left (\frac {1}{2} \left (\frac {a b \left (4 a^2 B+20 a b C+15 b^2 B\right ) \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 b \left (4 a^3 C+11 a^2 b B+16 a b^2 C+8 b^3 B\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{d \sqrt {a+b \cos (c+d x)}}}{b}-\frac {2 \left (4 a^2 C+9 a b B-8 b^2 C\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}\right )+\frac {a (4 a C+7 b B) \tan (c+d x) \sqrt {a+b \cos (c+d x)}}{d}\right )+\frac {a B \tan (c+d x) \sec (c+d x) (a+b \cos (c+d x))^{3/2}}{2 d}\) |
| \(\Big \downarrow \) 3286 |
| \(\displaystyle \frac {1}{4} \left (\frac {1}{2} \left (\frac {\frac {a b \left (4 a^2 B+20 a b C+15 b^2 B\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \int \frac {\sec (c+d x)}{\sqrt {\frac {a}{a+b}+\frac {b \cos (c+d x)}{a+b}}}dx}{\sqrt {a+b \cos (c+d x)}}+\frac {2 b \left (4 a^3 C+11 a^2 b B+16 a b^2 C+8 b^3 B\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{d \sqrt {a+b \cos (c+d x)}}}{b}-\frac {2 \left (4 a^2 C+9 a b B-8 b^2 C\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}\right )+\frac {a (4 a C+7 b B) \tan (c+d x) \sqrt {a+b \cos (c+d x)}}{d}\right )+\frac {a B \tan (c+d x) \sec (c+d x) (a+b \cos (c+d x))^{3/2}}{2 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{4} \left (\frac {1}{2} \left (\frac {\frac {a b \left (4 a^2 B+20 a b C+15 b^2 B\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {\frac {a}{a+b}+\frac {b \sin \left (c+d x+\frac {\pi }{2}\right )}{a+b}}}dx}{\sqrt {a+b \cos (c+d x)}}+\frac {2 b \left (4 a^3 C+11 a^2 b B+16 a b^2 C+8 b^3 B\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{d \sqrt {a+b \cos (c+d x)}}}{b}-\frac {2 \left (4 a^2 C+9 a b B-8 b^2 C\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}\right )+\frac {a (4 a C+7 b B) \tan (c+d x) \sqrt {a+b \cos (c+d x)}}{d}\right )+\frac {a B \tan (c+d x) \sec (c+d x) (a+b \cos (c+d x))^{3/2}}{2 d}\) |
| \(\Big \downarrow \) 3284 |
| \(\displaystyle \frac {1}{4} \left (\frac {1}{2} \left (\frac {\frac {2 a b \left (4 a^2 B+20 a b C+15 b^2 B\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticPi}\left (2,\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{d \sqrt {a+b \cos (c+d x)}}+\frac {2 b \left (4 a^3 C+11 a^2 b B+16 a b^2 C+8 b^3 B\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{d \sqrt {a+b \cos (c+d x)}}}{b}-\frac {2 \left (4 a^2 C+9 a b B-8 b^2 C\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}\right )+\frac {a (4 a C+7 b B) \tan (c+d x) \sqrt {a+b \cos (c+d x)}}{d}\right )+\frac {a B \tan (c+d x) \sec (c+d x) (a+b \cos (c+d x))^{3/2}}{2 d}\) |
Input:
Int[(a + b*Cos[c + d*x])^(5/2)*(B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^4,x]
Output:
(a*B*(a + b*Cos[c + d*x])^(3/2)*Sec[c + d*x]*Tan[c + d*x])/(2*d) + (((-2*( 9*a*b*B + 4*a^2*C - 8*b^2*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*b*(11*a^2* b*B + 8*b^3*B + 4*a^3*C + 16*a*b^2*C)*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*E llipticF[(c + d*x)/2, (2*b)/(a + b)])/(d*Sqrt[a + b*Cos[c + d*x]]) + (2*a* b*(4*a^2*B + 15*b^2*B + 20*a*b*C)*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*Ellip ticPi[2, (c + d*x)/2, (2*b)/(a + b)])/(d*Sqrt[a + b*Cos[c + d*x]]))/b)/2 + (a*(7*b*B + 4*a*C)*Sqrt[a + b*Cos[c + d*x]]*Tan[c + d*x])/d)/4
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[(-(b*c - a*d))*(B*c - A*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(n + 1)*(c^2 - d^2))), x] + Simp[1/(d*(n + 1)*(c^2 - d^2)) Int[(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x])^(n + 1)*Simp[b*(b*c - a*d)*(B*c - A*d)*(m - 1) + a*d*(a*A*c + b*B*c - (A*b + a *B)*d)*(n + 1) + (b*(b*d*(B*c - A*d) + a*(A*c*d + B*(c^2 - 2*d^2)))*(n + 1) - a*(b*c - a*d)*(B*c - A*d)*(n + 2))*Sin[e + f*x] + b*(d*(A*b*c + a*B*c - a*A*d)*(m + n + 1) - b*B*(c^2*m + d^2*(n + 1)))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2 , 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 1] && LtQ[n, -1]
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[(-(c^2*C - B*c*d + A*d^2))*Cos[e + f*x ]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(n + 1)*(c^2 - d^2))), x] + Simp[1/(d*(n + 1)*(c^2 - d^2)) Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^(n + 1)*Simp[A*d*(b*d*m + a*c*(n + 1)) + (c*C - B* d)*(b*c*m + a*d*(n + 1)) - (d*(A*(a*d*(n + 2) - b*c*(n + 1)) + B*(b*d*(n + 1) - a*c*(n + 2))) - C*(b*c*d*(n + 1) - a*(c^2 + d^2*(n + 1))))*Sin[e + f*x ] + b*(d*(B*c - A*d)*(m + n + 2) - C*(c^2*(m + 1) + d^2*(n + 1)))*Sin[e + f *x]^2, x], x], x] /; FreeQ[{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] && GtQ[m, 0] && LtQ[n, -1]
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. \(1741\) vs. \(2(303)=606\).
Time = 92.67 (sec) , antiderivative size = 1742, normalized size of antiderivative = 5.53
| method | result | size |
| default | \(\text {Expression too large to display}\) | \(1742\) |
| parts | \(\text {Expression too large to display}\) | \(2096\) |
Input:
int((a+b*cos(d*x+c))^(5/2)*(B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^4,x,me thod=_RETURNVERBOSE)
Output:
-(-(-2*cos(1/2*d*x+1/2*c)^2*b-a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*B*b^3*(s in(1/2*d*x+1/2*c)^2)^(1/2)*((2*cos(1/2*d*x+1/2*c)^2*b+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)*EllipticF(cos(1 /2*d*x+1/2*c),(-2*b/(a-b))^(1/2))+2*B*a^3*(-1/2*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)/(-1+2*cos(1/2*d*x+ 1/2*c)^2)^2+3/4*b/a^2*cos(1/2*d*x+1/2*c)*(-2*b*sin(1/2*d*x+1/2*c)^4+(a+b)* sin(1/2*d*x+1/2*c)^2)^(1/2)/(-1+2*cos(1/2*d*x+1/2*c)^2)-1/8*b/a*(sin(1/2*d *x+1/2*c)^2)^(1/2)*((2*cos(1/2*d*x+1/2*c)^2*b+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)*EllipticF(cos(1/2*d*x+1 /2*c),(-2*b/(a-b))^(1/2))+3/8/a*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*cos(1/2*d *x+1/2*c)^2*b+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))-3/8* b^2/a^2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*cos(1/2*d*x+1/2*c)^2*b+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)*Ellip ticE(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*cos(1/2*d*x+1/2*c)^2*b+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))-3/8/a^2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*cos(1/2*d*x+1/2*c)^2 *b+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))*b^2)+2*a^2*...
Timed out. \[ \int (a+b \cos (c+d x))^{5/2} \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\text {Timed out} \] Input:
integrate((a+b*cos(d*x+c))^(5/2)*(B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^ 4,x, algorithm="fricas")
Output:
Timed out
Timed out. \[ \int (a+b \cos (c+d x))^{5/2} \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\text {Timed out} \] Input:
integrate((a+b*cos(d*x+c))**(5/2)*(B*cos(d*x+c)+C*cos(d*x+c)**2)*sec(d*x+c )**4,x)
Output:
Timed out
\[ \int (a+b \cos (c+d x))^{5/2} \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\int { {\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right )\right )} {\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \sec \left (d x + c\right )^{4} \,d x } \] Input:
integrate((a+b*cos(d*x+c))^(5/2)*(B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^ 4,x, algorithm="maxima")
Output:
integrate((C*cos(d*x + c)^2 + B*cos(d*x + c))*(b*cos(d*x + c) + a)^(5/2)*s ec(d*x + c)^4, x)
\[ \int (a+b \cos (c+d x))^{5/2} \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\int { {\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right )\right )} {\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \sec \left (d x + c\right )^{4} \,d x } \] Input:
integrate((a+b*cos(d*x+c))^(5/2)*(B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^ 4,x, algorithm="giac")
Output:
integrate((C*cos(d*x + c)^2 + B*cos(d*x + c))*(b*cos(d*x + c) + a)^(5/2)*s ec(d*x + c)^4, x)
Timed out. \[ \int (a+b \cos (c+d x))^{5/2} \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\int \frac {\left (C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )\right )\,{\left (a+b\,\cos \left (c+d\,x\right )\right )}^{5/2}}{{\cos \left (c+d\,x\right )}^4} \,d x \] Input:
int(((B*cos(c + d*x) + C*cos(c + d*x)^2)*(a + b*cos(c + d*x))^(5/2))/cos(c + d*x)^4,x)
Output:
int(((B*cos(c + d*x) + C*cos(c + d*x)^2)*(a + b*cos(c + d*x))^(5/2))/cos(c + d*x)^4, x)
\[ \int (a+b \cos (c+d x))^{5/2} \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\left (\int \sqrt {\cos \left (d x +c \right ) b +a}\, \cos \left (d x +c \right ) \sec \left (d x +c \right )^{4}d x \right ) a^{2} b +\left (\int \sqrt {\cos \left (d x +c \right ) b +a}\, \cos \left (d x +c \right )^{4} \sec \left (d x +c \right )^{4}d x \right ) b^{2} c +2 \left (\int \sqrt {\cos \left (d x +c \right ) b +a}\, \cos \left (d x +c \right )^{3} \sec \left (d x +c \right )^{4}d x \right ) a b c +\left (\int \sqrt {\cos \left (d x +c \right ) b +a}\, \cos \left (d x +c \right )^{3} \sec \left (d x +c \right )^{4}d x \right ) b^{3}+\left (\int \sqrt {\cos \left (d x +c \right ) b +a}\, \cos \left (d x +c \right )^{2} \sec \left (d x +c \right )^{4}d x \right ) a^{2} c +2 \left (\int \sqrt {\cos \left (d x +c \right ) b +a}\, \cos \left (d x +c \right )^{2} \sec \left (d x +c \right )^{4}d x \right ) a \,b^{2} \] Input:
int((a+b*cos(d*x+c))^(5/2)*(B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^4,x)
Output:
int(sqrt(cos(c + d*x)*b + a)*cos(c + d*x)*sec(c + d*x)**4,x)*a**2*b + int( sqrt(cos(c + d*x)*b + a)*cos(c + d*x)**4*sec(c + d*x)**4,x)*b**2*c + 2*int (sqrt(cos(c + d*x)*b + a)*cos(c + d*x)**3*sec(c + d*x)**4,x)*a*b*c + int(s qrt(cos(c + d*x)*b + a)*cos(c + d*x)**3*sec(c + d*x)**4,x)*b**3 + int(sqrt (cos(c + d*x)*b + a)*cos(c + d*x)**2*sec(c + d*x)**4,x)*a**2*c + 2*int(sqr t(cos(c + d*x)*b + a)*cos(c + d*x)**2*sec(c + d*x)**4,x)*a*b**2