Integrand size = 45, antiderivative size = 535 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {3}{2}}(c+d x)}{(a+b \cos (c+d x))^{5/2}} \, dx=\frac {2 \left (8 A b^4+6 a^3 b B-2 a b^3 B+3 a^4 (A-C)-a^2 b^2 (15 A+C)\right ) \sqrt {\cos (c+d x)} \csc (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{3 a^4 \sqrt {a+b} \left (a^2-b^2\right ) d \sqrt {\sec (c+d x)}}+\frac {2 \left (8 A b^3+2 a b^2 (3 A-B)-3 a^3 (A-B-C)-a^2 b (9 A+3 B+C)\right ) \sqrt {\cos (c+d x)} \csc (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{3 a^3 \sqrt {a+b} \left (a^2-b^2\right ) d \sqrt {\sec (c+d x)}}+\frac {2 \left (A b^2-a (b B-a C)\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{3 a \left (a^2-b^2\right ) d (a+b \cos (c+d x))^{3/2}}-\frac {2 \left (4 A b^4+5 a^3 b B-a b^3 B-2 a^4 C-2 a^2 b^2 (4 A+C)\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{3 a^2 \left (a^2-b^2\right )^2 d \sqrt {a+b \cos (c+d x)}} \] Output:
2/3*(8*A*b^4+6*B*a^3*b-2*B*a*b^3+3*a^4*(A-C)-a^2*b^2*(15*A+C))*cos(d*x+c)^ (1/2)*csc(d*x+c)*EllipticE((a+b*cos(d*x+c))^(1/2)/(a+b)^(1/2)/cos(d*x+c)^( 1/2),(-(a+b)/(a-b))^(1/2))*(a*(1-sec(d*x+c))/(a+b))^(1/2)*(a*(1+sec(d*x+c) )/(a-b))^(1/2)/a^4/(a+b)^(1/2)/(a^2-b^2)/d/sec(d*x+c)^(1/2)+2/3*(8*A*b^3+2 *a*b^2*(3*A-B)-3*a^3*(A-B-C)-a^2*b*(9*A+3*B+C))*cos(d*x+c)^(1/2)*csc(d*x+c )*EllipticF((a+b*cos(d*x+c))^(1/2)/(a+b)^(1/2)/cos(d*x+c)^(1/2),(-(a+b)/(a -b))^(1/2))*(a*(1-sec(d*x+c))/(a+b))^(1/2)*(a*(1+sec(d*x+c))/(a-b))^(1/2)/ a^3/(a+b)^(1/2)/(a^2-b^2)/d/sec(d*x+c)^(1/2)+2/3*(A*b^2-a*(B*b-C*a))*sec(d *x+c)^(1/2)*sin(d*x+c)/a/(a^2-b^2)/d/(a+b*cos(d*x+c))^(3/2)-2/3*(4*A*b^4+5 *B*a^3*b-B*a*b^3-2*a^4*C-2*a^2*b^2*(4*A+C))*sec(d*x+c)^(1/2)*sin(d*x+c)/a^ 2/(a^2-b^2)^2/d/(a+b*cos(d*x+c))^(1/2)
Leaf count is larger than twice the leaf count of optimal. \(4821\) vs. \(2(535)=1070\).
Time = 26.63 (sec) , antiderivative size = 4821, normalized size of antiderivative = 9.01 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {3}{2}}(c+d x)}{(a+b \cos (c+d x))^{5/2}} \, dx=\text {Result too large to show} \] Input:
Integrate[((A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^(3/2))/(a + b*Cos[c + d*x])^(5/2),x]
Output:
(Sqrt[a + b*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*((2*(3*a^4*A - 15*a^2*A*b^2 + 8*A*b^4 + 6*a^3*b*B - 2*a*b^3*B - 3*a^4*C - a^2*b^2*C)*Sin[c + d*x])/(3*a ^3*(a^2 - b^2)^2) + (2*(A*b^2*Sin[c + d*x] - a*b*B*Sin[c + d*x] + a^2*C*Si n[c + d*x]))/(3*a*(a^2 - b^2)*(a + b*Cos[c + d*x])^2) + (2*(8*a^2*A*b^2*Si n[c + d*x] - 4*A*b^4*Sin[c + d*x] - 5*a^3*b*B*Sin[c + d*x] + a*b^3*B*Sin[c + d*x] + 2*a^4*C*Sin[c + d*x] + 2*a^2*b^2*C*Sin[c + d*x]))/(3*a^2*(a^2 - b^2)^2*(a + b*Cos[c + d*x]))))/d + (2*(-((a^2*A)/((a^2 - b^2)^2*Sqrt[a + b *Cos[c + d*x]]*Sqrt[Sec[c + d*x]])) + (5*A*b^2)/((a^2 - b^2)^2*Sqrt[a + b* Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) - (8*A*b^4)/(3*a^2*(a^2 - b^2)^2*Sqrt[a + b*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) - (2*a*b*B)/((a^2 - b^2)^2*Sqrt[a + b*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) + (2*b^3*B)/(3*a*(a^2 - b^2)^2*Sqrt[a + b*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) + (a^2*C)/((a^2 - b^2)^2*Sqrt[a + b* Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) + (b^2*C)/(3*(a^2 - b^2)^2*Sqrt[a + b*Co s[c + d*x]]*Sqrt[Sec[c + d*x]]) - (3*a*A*b*Sqrt[Sec[c + d*x]])/((a^2 - b^2 )^2*Sqrt[a + b*Cos[c + d*x]]) + (17*A*b^3*Sqrt[Sec[c + d*x]])/(3*a*(a^2 - b^2)^2*Sqrt[a + b*Cos[c + d*x]]) - (8*A*b^5*Sqrt[Sec[c + d*x]])/(3*a^3*(a^ 2 - b^2)^2*Sqrt[a + b*Cos[c + d*x]]) + (a^2*B*Sqrt[Sec[c + d*x]])/((a^2 - b^2)^2*Sqrt[a + b*Cos[c + d*x]]) - (5*b^2*B*Sqrt[Sec[c + d*x]])/(3*(a^2 - b^2)^2*Sqrt[a + b*Cos[c + d*x]]) + (2*b^4*B*Sqrt[Sec[c + d*x]])/(3*a^2*(a^ 2 - b^2)^2*Sqrt[a + b*Cos[c + d*x]]) - (a*b*C*Sqrt[Sec[c + d*x]])/(3*(a...
Time = 2.37 (sec) , antiderivative size = 534, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.289, Rules used = {3042, 4709, 3042, 3534, 27, 3042, 3534, 27, 3042, 3477, 3042, 3295, 3473}
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.
| \(\displaystyle \int \frac {\sec ^{\frac {3}{2}}(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^{5/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sec (c+d x)^{3/2} \left (A+B \cos (c+d x)+C \cos (c+d x)^2\right )}{(a+b \cos (c+d x))^{5/2}}dx\) |
| \(\Big \downarrow \) 4709 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {C \cos ^2(c+d x)+B \cos (c+d x)+A}{\cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))^{5/2}}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {C \sin \left (c+d x+\frac {\pi }{2}\right )^2+B \sin \left (c+d x+\frac {\pi }{2}\right )+A}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{5/2}}dx\) |
| \(\Big \downarrow \) 3534 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 \int -\frac {-\left ((3 A-C) a^2\right )-b B a+3 (A b+C b-a B) \cos (c+d x) a+4 A b^2-2 \left (A b^2-a (b B-a C)\right ) \cos ^2(c+d x)}{2 \cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))^{3/2}}dx}{3 a \left (a^2-b^2\right )}+\frac {2 \sin (c+d x) \left (A b^2-a (b B-a C)\right )}{3 a d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))^{3/2}}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 \sin (c+d x) \left (A b^2-a (b B-a C)\right )}{3 a d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))^{3/2}}-\frac {\int \frac {-\left ((3 A-C) a^2\right )-b B a+3 (A b+C b-a B) \cos (c+d x) a+4 A b^2-2 \left (A b^2-a (b B-a C)\right ) \cos ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))^{3/2}}dx}{3 a \left (a^2-b^2\right )}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 \sin (c+d x) \left (A b^2-a (b B-a C)\right )}{3 a d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))^{3/2}}-\frac {\int \frac {-\left ((3 A-C) a^2\right )-b B a+3 (A b+C b-a B) \sin \left (c+d x+\frac {\pi }{2}\right ) a+4 A b^2-2 \left (A b^2-a (b B-a C)\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2}}dx}{3 a \left (a^2-b^2\right )}\right )\) |
| \(\Big \downarrow \) 3534 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 \sin (c+d x) \left (A b^2-a (b B-a C)\right )}{3 a d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))^{3/2}}-\frac {\frac {2 \int -\frac {3 (A-C) a^4+6 b B a^3-b^2 (15 A+C) a^2-2 b^3 B a+\left (3 B a^3-2 b (3 A+2 C) a^2+b^2 B a+2 A b^3\right ) \cos (c+d x) a+8 A b^4}{2 \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}dx}{a \left (a^2-b^2\right )}+\frac {2 \sin (c+d x) \left (-2 a^4 C+5 a^3 b B-2 a^2 b^2 (4 A+C)-a b^3 B+4 A b^4\right )}{a d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}}{3 a \left (a^2-b^2\right )}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 \sin (c+d x) \left (A b^2-a (b B-a C)\right )}{3 a d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))^{3/2}}-\frac {\frac {2 \sin (c+d x) \left (-2 a^4 C+5 a^3 b B-2 a^2 b^2 (4 A+C)-a b^3 B+4 A b^4\right )}{a d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}-\frac {\int \frac {3 (A-C) a^4+6 b B a^3-b^2 (15 A+C) a^2-2 b^3 B a+\left (3 B a^3-2 b (3 A+2 C) a^2+b^2 B a+2 A b^3\right ) \cos (c+d x) a+8 A b^4}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}dx}{a \left (a^2-b^2\right )}}{3 a \left (a^2-b^2\right )}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 \sin (c+d x) \left (A b^2-a (b B-a C)\right )}{3 a d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))^{3/2}}-\frac {\frac {2 \sin (c+d x) \left (-2 a^4 C+5 a^3 b B-2 a^2 b^2 (4 A+C)-a b^3 B+4 A b^4\right )}{a d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}-\frac {\int \frac {3 (A-C) a^4+6 b B a^3-b^2 (15 A+C) a^2-2 b^3 B a+\left (3 B a^3-2 b (3 A+2 C) a^2+b^2 B a+2 A b^3\right ) \sin \left (c+d x+\frac {\pi }{2}\right ) a+8 A b^4}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{a \left (a^2-b^2\right )}}{3 a \left (a^2-b^2\right )}\right )\) |
| \(\Big \downarrow \) 3477 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 \sin (c+d x) \left (A b^2-a (b B-a C)\right )}{3 a d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))^{3/2}}-\frac {\frac {2 \sin (c+d x) \left (-2 a^4 C+5 a^3 b B-2 a^2 b^2 (4 A+C)-a b^3 B+4 A b^4\right )}{a d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}-\frac {(a-b) \left (-3 a^3 (A-B-C)-a^2 b (9 A+3 B+C)+2 a b^2 (3 A-B)+8 A b^3\right ) \int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}dx+\left (3 a^4 (A-C)+6 a^3 b B-a^2 b^2 (15 A+C)-2 a b^3 B+8 A b^4\right ) \int \frac {\cos (c+d x)+1}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}dx}{a \left (a^2-b^2\right )}}{3 a \left (a^2-b^2\right )}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 \sin (c+d x) \left (A b^2-a (b B-a C)\right )}{3 a d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))^{3/2}}-\frac {\frac {2 \sin (c+d x) \left (-2 a^4 C+5 a^3 b B-2 a^2 b^2 (4 A+C)-a b^3 B+4 A b^4\right )}{a d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}-\frac {(a-b) \left (-3 a^3 (A-B-C)-a^2 b (9 A+3 B+C)+2 a b^2 (3 A-B)+8 A b^3\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\left (3 a^4 (A-C)+6 a^3 b B-a^2 b^2 (15 A+C)-2 a b^3 B+8 A b^4\right ) \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )+1}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{a \left (a^2-b^2\right )}}{3 a \left (a^2-b^2\right )}\right )\) |
| \(\Big \downarrow \) 3295 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 \sin (c+d x) \left (A b^2-a (b B-a C)\right )}{3 a d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))^{3/2}}-\frac {\frac {2 \sin (c+d x) \left (-2 a^4 C+5 a^3 b B-2 a^2 b^2 (4 A+C)-a b^3 B+4 A b^4\right )}{a d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}-\frac {\left (3 a^4 (A-C)+6 a^3 b B-a^2 b^2 (15 A+C)-2 a b^3 B+8 A b^4\right ) \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )+1}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 (a-b) \sqrt {a+b} \cot (c+d x) \left (-3 a^3 (A-B-C)-a^2 b (9 A+3 B+C)+2 a b^2 (3 A-B)+8 A b^3\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{a d}}{a \left (a^2-b^2\right )}}{3 a \left (a^2-b^2\right )}\right )\) |
| \(\Big \downarrow \) 3473 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 \sin (c+d x) \left (A b^2-a (b B-a C)\right )}{3 a d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))^{3/2}}-\frac {\frac {2 \sin (c+d x) \left (-2 a^4 C+5 a^3 b B-2 a^2 b^2 (4 A+C)-a b^3 B+4 A b^4\right )}{a d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}-\frac {\frac {2 (a-b) \sqrt {a+b} \cot (c+d x) \left (-3 a^3 (A-B-C)-a^2 b (9 A+3 B+C)+2 a b^2 (3 A-B)+8 A b^3\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{a d}+\frac {2 (a-b) \sqrt {a+b} \cot (c+d x) \left (3 a^4 (A-C)+6 a^3 b B-a^2 b^2 (15 A+C)-2 a b^3 B+8 A b^4\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right )}{a^2 d}}{a \left (a^2-b^2\right )}}{3 a \left (a^2-b^2\right )}\right )\) |
Input:
Int[((A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^(3/2))/(a + b*Co s[c + d*x])^(5/2),x]
Output:
Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*((2*(A*b^2 - a*(b*B - a*C))*Sin[c + d*x])/(3*a*(a^2 - b^2)*d*Sqrt[Cos[c + d*x]]*(a + b*Cos[c + d*x])^(3/2)) - (-(((2*(a - b)*Sqrt[a + b]*(8*A*b^4 + 6*a^3*b*B - 2*a*b^3*B + 3*a^4*(A - C ) - a^2*b^2*(15*A + C))*Cot[c + d*x]*EllipticE[ArcSin[Sqrt[a + b*Cos[c + d *x]]/(Sqrt[a + b]*Sqrt[Cos[c + d*x]])], -((a + b)/(a - b))]*Sqrt[(a*(1 - S ec[c + d*x]))/(a + b)]*Sqrt[(a*(1 + Sec[c + d*x]))/(a - b)])/(a^2*d) + (2* (a - b)*Sqrt[a + b]*(8*A*b^3 + 2*a*b^2*(3*A - B) - 3*a^3*(A - B - C) - a^2 *b*(9*A + 3*B + C))*Cot[c + d*x]*EllipticF[ArcSin[Sqrt[a + b*Cos[c + d*x]] /(Sqrt[a + b]*Sqrt[Cos[c + d*x]])], -((a + b)/(a - b))]*Sqrt[(a*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[(a*(1 + Sec[c + d*x]))/(a - b)])/(a*d))/(a*(a^2 - b^2))) + (2*(4*A*b^4 + 5*a^3*b*B - a*b^3*B - 2*a^4*C - 2*a^2*b^2*(4*A + C) )*Sin[c + d*x])/(a*(a^2 - b^2)*d*Sqrt[Cos[c + d*x]]*Sqrt[a + b*Cos[c + d*x ]]))/(3*a*(a^2 - b^2)))
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[1/(Sqrt[(d_.)*sin[(e_.) + (f_.)*(x_)]]*Sqrt[(a_) + (b_.)*sin[(e_.) + (f _.)*(x_)]]), x_Symbol] :> Simp[-2*(Tan[e + f*x]/(a*f))*Rt[(a + b)/d, 2]*Sqr t[a*((1 - Csc[e + f*x])/(a + b))]*Sqrt[a*((1 + Csc[e + f*x])/(a - b))]*Elli pticF[ArcSin[Sqrt[a + b*Sin[e + f*x]]/Sqrt[d*Sin[e + f*x]]/Rt[(a + b)/d, 2] ], -(a + b)/(a - b)], x] /; FreeQ[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0] && PosQ[(a + b)/d]
Int[((A_) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(((b_.)*sin[(e_.) + (f_.)*(x_)]) ^(3/2)*Sqrt[(c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp[-2*A* (c - d)*(Tan[e + f*x]/(f*b*c^2))*Rt[(c + d)/b, 2]*Sqrt[c*((1 + Csc[e + f*x] )/(c - d))]*Sqrt[c*((1 - Csc[e + f*x])/(c + d))]*EllipticE[ArcSin[Sqrt[c + d*Sin[e + f*x]]/Sqrt[b*Sin[e + f*x]]/Rt[(c + d)/b, 2]], -(c + d)/(c - d)], x] /; FreeQ[{b, c, d, e, f, A, B}, x] && NeQ[c^2 - d^2, 0] && EqQ[A, B] && PosQ[(c + d)/b]
Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(((a_.) + (b_.)*sin[(e_.) + (f_ .)*(x_)])^(3/2)*Sqrt[(c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> S imp[(A - B)/(a - b) Int[1/(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f* x]]), x], x] - Simp[(A*b - a*B)/(a - b) Int[(1 + Sin[e + f*x])/((a + b*Si n[e + f*x])^(3/2)*Sqrt[c + d*Sin[e + f*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] && NeQ[A, B]
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[(u_)*((c_.)*sec[(a_.) + (b_.)*(x_)])^(m_.), x_Symbol] :> Simp[(c*Sec[a + b*x])^m*(c*Cos[a + b*x])^m Int[ActivateTrig[u]/(c*Cos[a + b*x])^m, x], x] /; FreeQ[{a, b, c, m}, x] && !IntegerQ[m] && KnownSineIntegrandQ[u, x]
Leaf count of result is larger than twice the leaf count of optimal. \(4080\) vs. \(2(489)=978\).
Time = 41.87 (sec) , antiderivative size = 4081, normalized size of antiderivative = 7.63
| method | result | size |
| default | \(\text {Expression too large to display}\) | \(4081\) |
| parts | \(\text {Expression too large to display}\) | \(4220\) |
Input:
int((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^(3/2)/(a+b*cos(d*x+c))^(5/2 ),x,method=_RETURNVERBOSE)
Output:
2/3/d/(a-b)^2/(a+b)^2/a^3*((-6*cos(d*x+c)^3-13*cos(d*x+c)^2-8*cos(d*x+c)-1 )*B*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(1/(a+b)*(a+b*cos(d*x+c))/(1+cos(d*x +c)))^(1/2)*a^4*b^2*EllipticF(-csc(d*x+c)+cot(d*x+c),(-(a-b)/(a+b))^(1/2)) +sin(d*x+c)*cos(d*x+c)*(2*cos(d*x+c)-4)*C*a^5*b+sin(d*x+c)*cos(d*x+c)*(6*c os(d*x+c)+2)*B*a^3*b^3+sin(d*x+c)*cos(d*x+c)*(-3+cos(d*x+c))*B*a^2*b^4+A*( cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(1/(a+b)*(a+b*cos(d*x+c))/(1+cos(d*x+c))) ^(1/2)*b^6*EllipticE(-csc(d*x+c)+cot(d*x+c),(-(a-b)/(a+b))^(1/2))*(8*cos(d *x+c)^3+16*cos(d*x+c)^2+8*cos(d*x+c))+sin(d*x+c)*cos(d*x+c)*(-3*cos(d*x+c) +1)*a^4*b^2*C+(-3*cos(d*x+c)^2-6*cos(d*x+c)-3)*A*(cos(d*x+c)/(1+cos(d*x+c) ))^(1/2)*(1/(a+b)*(a+b*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*a^6*EllipticF(-cs c(d*x+c)+cot(d*x+c),(-(a-b)/(a+b))^(1/2))+(-3*cos(d*x+c)^2-6*cos(d*x+c)-3) *B*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(1/(a+b)*(a+b*cos(d*x+c))/(1+cos(d*x+ c)))^(1/2)*a^6*EllipticF(-csc(d*x+c)+cot(d*x+c),(-(a-b)/(a+b))^(1/2))+(3*c os(d*x+c)^2+6*cos(d*x+c)+3)*C*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(1/(a+b)*( a+b*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*a^6*EllipticF(-csc(d*x+c)+cot(d*x+c) ,(-(a-b)/(a+b))^(1/2))+(3*cos(d*x+c)^2+6*cos(d*x+c)+3)*A*(cos(d*x+c)/(1+co s(d*x+c)))^(1/2)*(1/(a+b)*(a+b*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*a^6*Ellip ticE(-csc(d*x+c)+cot(d*x+c),(-(a-b)/(a+b))^(1/2))+(-3*cos(d*x+c)^2-6*cos(d *x+c)-3)*C*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(1/(a+b)*(a+b*cos(d*x+c))/(1+ cos(d*x+c)))^(1/2)*a^6*EllipticE(-csc(d*x+c)+cot(d*x+c),(-(a-b)/(a+b))^...
\[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {3}{2}}(c+d x)}{(a+b \cos (c+d x))^{5/2}} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )^{\frac {3}{2}}}{{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {5}{2}}} \,d x } \] Input:
integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^(3/2)/(a+b*cos(d*x+c) )^(5/2),x, algorithm="fricas")
Output:
integral((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*sqrt(b*cos(d*x + c) + a)* sec(d*x + c)^(3/2)/(b^3*cos(d*x + c)^3 + 3*a*b^2*cos(d*x + c)^2 + 3*a^2*b* cos(d*x + c) + a^3), x)
Timed out. \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {3}{2}}(c+d x)}{(a+b \cos (c+d x))^{5/2}} \, dx=\text {Timed out} \] Input:
integrate((A+B*cos(d*x+c)+C*cos(d*x+c)**2)*sec(d*x+c)**(3/2)/(a+b*cos(d*x+ c))**(5/2),x)
Output:
Timed out
\[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {3}{2}}(c+d x)}{(a+b \cos (c+d x))^{5/2}} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )^{\frac {3}{2}}}{{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {5}{2}}} \,d x } \] Input:
integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^(3/2)/(a+b*cos(d*x+c) )^(5/2),x, algorithm="maxima")
Output:
integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*sec(d*x + c)^(3/2)/(b*co s(d*x + c) + a)^(5/2), x)
\[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {3}{2}}(c+d x)}{(a+b \cos (c+d x))^{5/2}} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )^{\frac {3}{2}}}{{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {5}{2}}} \,d x } \] Input:
integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^(3/2)/(a+b*cos(d*x+c) )^(5/2),x, algorithm="giac")
Output:
integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*sec(d*x + c)^(3/2)/(b*co s(d*x + c) + a)^(5/2), x)
Timed out. \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {3}{2}}(c+d x)}{(a+b \cos (c+d x))^{5/2}} \, dx=\int \frac {{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{3/2}\,\left (C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )+A\right )}{{\left (a+b\,\cos \left (c+d\,x\right )\right )}^{5/2}} \,d x \] Input:
int(((1/cos(c + d*x))^(3/2)*(A + B*cos(c + d*x) + C*cos(c + d*x)^2))/(a + b*cos(c + d*x))^(5/2),x)
Output:
int(((1/cos(c + d*x))^(3/2)*(A + B*cos(c + d*x) + C*cos(c + d*x)^2))/(a + b*cos(c + d*x))^(5/2), x)
\[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {3}{2}}(c+d x)}{(a+b \cos (c+d x))^{5/2}} \, dx=\left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \sqrt {\cos \left (d x +c \right ) b +a}\, \cos \left (d x +c \right ) \sec \left (d x +c \right )}{\cos \left (d x +c \right )^{3} b^{3}+3 \cos \left (d x +c \right )^{2} a \,b^{2}+3 \cos \left (d x +c \right ) a^{2} b +a^{3}}d x \right ) b +\left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \sqrt {\cos \left (d x +c \right ) b +a}\, \cos \left (d x +c \right )^{2} \sec \left (d x +c \right )}{\cos \left (d x +c \right )^{3} b^{3}+3 \cos \left (d x +c \right )^{2} a \,b^{2}+3 \cos \left (d x +c \right ) a^{2} b +a^{3}}d x \right ) c +\left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \sqrt {\cos \left (d x +c \right ) b +a}\, \sec \left (d x +c \right )}{\cos \left (d x +c \right )^{3} b^{3}+3 \cos \left (d x +c \right )^{2} a \,b^{2}+3 \cos \left (d x +c \right ) a^{2} b +a^{3}}d x \right ) a \] Input:
int((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^(3/2)/(a+b*cos(d*x+c))^(5/2 ),x)
Output:
int((sqrt(sec(c + d*x))*sqrt(cos(c + d*x)*b + a)*cos(c + d*x)*sec(c + d*x) )/(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(sec(c + d*x))*sqrt(cos(c + d*x)*b + a)*cos(c + d *x)**2*sec(c + d*x))/(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 + int((sqrt(sec(c + d*x))*sqrt(cos(c + d* x)*b + a)*sec(c + d*x))/(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)*a