Integrand size = 33, antiderivative size = 284 \[ \int \frac {A+B \cos (c+d x)}{(a+b \cos (c+d x))^2 \sec ^{\frac {3}{2}}(c+d x)} \, dx=-\frac {\left (a A b-3 a^2 B+2 b^2 B\right ) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{b^2 \left (a^2-b^2\right ) d}+\frac {\left (a^2 A b-2 A b^3-3 a^3 B+4 a b^2 B\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{b^3 \left (a^2-b^2\right ) d}-\frac {a \left (a^2 A b-3 A b^3-3 a^3 B+5 a b^2 B\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{(a-b) b^3 (a+b)^2 d}+\frac {a (A b-a B) \sqrt {\sec (c+d x)} \sin (c+d x)}{b \left (a^2-b^2\right ) d (b+a \sec (c+d x))} \]
a*(A*b-B*a)*sin(d*x+c)*sec(d*x+c)^(1/2)/b/(a^2-b^2)/d/(b+a*sec(d*x+c))-(A* a*b-3*B*a^2+2*B*b^2)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*Ellip ticE(sin(1/2*d*x+1/2*c),2^(1/2))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/b^2/(a^ 2-b^2)/d+(A*a^2*b-2*A*b^3-3*B*a^3+4*B*a*b^2)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/ cos(1/2*d*x+1/2*c)*EllipticF(sin(1/2*d*x+1/2*c),2^(1/2))*cos(d*x+c)^(1/2)* sec(d*x+c)^(1/2)/b^3/(a^2-b^2)/d-a*(A*a^2*b-3*A*b^3-3*B*a^3+5*B*a*b^2)*(co s(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticPi(sin(1/2*d*x+1/2*c) ,2*b/(a+b),2^(1/2))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/(a-b)/b^3/(a+b)^2/d
Time = 6.52 (sec) , antiderivative size = 461, normalized size of antiderivative = 1.62 \[ \int \frac {A+B \cos (c+d x)}{(a+b \cos (c+d x))^2 \sec ^{\frac {3}{2}}(c+d x)} \, dx=\frac {\frac {4 a b^2 (-A b+a B) \sin (c+d x)}{-a^2+b^2}+\frac {\cos (c+d x) \cot (c+d x) (b+a \sec (c+d x)) \left (-2 b^2 \left (-a A b-a^2 B+2 b^2 B\right ) \left (\operatorname {EllipticF}\left (\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right )-\operatorname {EllipticPi}\left (-\frac {a}{b},\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right )\right ) \sqrt {\sec (c+d x)} \sqrt {-\tan ^2(c+d x)}+8 a b^2 (-A b+a B) \operatorname {EllipticPi}\left (-\frac {a}{b},\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right ) \sqrt {\sec (c+d x)} \sqrt {-\tan ^2(c+d x)}+\left (a A b-3 a^2 B+2 b^2 B\right ) \left (4 a b-4 a b \sec ^2(c+d x)+4 a b E\left (\left .\arcsin \left (\sqrt {\sec (c+d x)}\right )\right |-1\right ) \sqrt {\sec (c+d x)} \sqrt {-\tan ^2(c+d x)}-2 (2 a-b) b \operatorname {EllipticF}\left (\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right ) \sqrt {\sec (c+d x)} \sqrt {-\tan ^2(c+d x)}+4 a^2 \operatorname {EllipticPi}\left (-\frac {a}{b},\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right ) \sqrt {\sec (c+d x)} \sqrt {-\tan ^2(c+d x)}-2 b^2 \operatorname {EllipticPi}\left (-\frac {a}{b},\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right ) \sqrt {\sec (c+d x)} \sqrt {-\tan ^2(c+d x)}\right )\right )}{a (a-b) (a+b)}}{4 b^3 d (a+b \cos (c+d x)) \sqrt {\sec (c+d x)}} \]
((4*a*b^2*(-(A*b) + a*B)*Sin[c + d*x])/(-a^2 + b^2) + (Cos[c + d*x]*Cot[c + d*x]*(b + a*Sec[c + d*x])*(-2*b^2*(-(a*A*b) - a^2*B + 2*b^2*B)*(Elliptic F[ArcSin[Sqrt[Sec[c + d*x]]], -1] - EllipticPi[-(a/b), ArcSin[Sqrt[Sec[c + d*x]]], -1])*Sqrt[Sec[c + d*x]]*Sqrt[-Tan[c + d*x]^2] + 8*a*b^2*(-(A*b) + a*B)*EllipticPi[-(a/b), ArcSin[Sqrt[Sec[c + d*x]]], -1]*Sqrt[Sec[c + d*x] ]*Sqrt[-Tan[c + d*x]^2] + (a*A*b - 3*a^2*B + 2*b^2*B)*(4*a*b - 4*a*b*Sec[c + d*x]^2 + 4*a*b*EllipticE[ArcSin[Sqrt[Sec[c + d*x]]], -1]*Sqrt[Sec[c + d *x]]*Sqrt[-Tan[c + d*x]^2] - 2*(2*a - b)*b*EllipticF[ArcSin[Sqrt[Sec[c + d *x]]], -1]*Sqrt[Sec[c + d*x]]*Sqrt[-Tan[c + d*x]^2] + 4*a^2*EllipticPi[-(a /b), ArcSin[Sqrt[Sec[c + d*x]]], -1]*Sqrt[Sec[c + d*x]]*Sqrt[-Tan[c + d*x] ^2] - 2*b^2*EllipticPi[-(a/b), ArcSin[Sqrt[Sec[c + d*x]]], -1]*Sqrt[Sec[c + d*x]]*Sqrt[-Tan[c + d*x]^2])))/(a*(a - b)*(a + b)))/(4*b^3*d*(a + b*Cos[ c + d*x])*Sqrt[Sec[c + d*x]])
Time = 1.91 (sec) , antiderivative size = 275, normalized size of antiderivative = 0.97, number of steps used = 17, number of rules used = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.515, Rules used = {3042, 3439, 3042, 4518, 27, 3042, 4594, 3042, 4274, 3042, 4258, 3042, 3119, 3120, 4336, 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.
\(\displaystyle \int \frac {A+B \cos (c+d x)}{\sec ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {A+B \sin \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^2}dx\) |
\(\Big \downarrow \) 3439 |
\(\displaystyle \int \frac {A \sec (c+d x)+B}{\sqrt {\sec (c+d x)} (a \sec (c+d x)+b)^2}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {A \csc \left (c+d x+\frac {\pi }{2}\right )+B}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \left (a \csc \left (c+d x+\frac {\pi }{2}\right )+b\right )^2}dx\) |
\(\Big \downarrow \) 4518 |
\(\displaystyle \frac {\int -\frac {-3 B a^2-(A b-a B) \sec ^2(c+d x) a+A b a+2 b^2 B+2 b (A b-a B) \sec (c+d x)}{2 \sqrt {\sec (c+d x)} (b+a \sec (c+d x))}dx}{b \left (a^2-b^2\right )}+\frac {a (A b-a B) \sin (c+d x) \sqrt {\sec (c+d x)}}{b d \left (a^2-b^2\right ) (a \sec (c+d x)+b)}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {a (A b-a B) \sin (c+d x) \sqrt {\sec (c+d x)}}{b d \left (a^2-b^2\right ) (a \sec (c+d x)+b)}-\frac {\int \frac {-3 B a^2-(A b-a B) \sec ^2(c+d x) a+A b a+2 b^2 B+2 b (A b-a B) \sec (c+d x)}{\sqrt {\sec (c+d x)} (b+a \sec (c+d x))}dx}{2 b \left (a^2-b^2\right )}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {a (A b-a B) \sin (c+d x) \sqrt {\sec (c+d x)}}{b d \left (a^2-b^2\right ) (a \sec (c+d x)+b)}-\frac {\int \frac {-3 B a^2-(A b-a B) \csc \left (c+d x+\frac {\pi }{2}\right )^2 a+A b a+2 b^2 B+2 b (A b-a B) \csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \left (b+a \csc \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{2 b \left (a^2-b^2\right )}\) |
\(\Big \downarrow \) 4594 |
\(\displaystyle \frac {a (A b-a B) \sin (c+d x) \sqrt {\sec (c+d x)}}{b d \left (a^2-b^2\right ) (a \sec (c+d x)+b)}-\frac {\frac {a \left (-3 a^3 B+a^2 A b+5 a b^2 B-3 A b^3\right ) \int \frac {\sec ^{\frac {3}{2}}(c+d x)}{b+a \sec (c+d x)}dx}{b^2}+\frac {\int \frac {b \left (-3 B a^2+A b a+2 b^2 B\right )-\left (-3 B a^3+A b a^2+4 b^2 B a-2 A b^3\right ) \sec (c+d x)}{\sqrt {\sec (c+d x)}}dx}{b^2}}{2 b \left (a^2-b^2\right )}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {a (A b-a B) \sin (c+d x) \sqrt {\sec (c+d x)}}{b d \left (a^2-b^2\right ) (a \sec (c+d x)+b)}-\frac {\frac {a \left (-3 a^3 B+a^2 A b+5 a b^2 B-3 A b^3\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}{b+a \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{b^2}+\frac {\int \frac {b \left (-3 B a^2+A b a+2 b^2 B\right )+\left (3 B a^3-A b a^2-4 b^2 B a+2 A b^3\right ) \csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{b^2}}{2 b \left (a^2-b^2\right )}\) |
\(\Big \downarrow \) 4274 |
\(\displaystyle \frac {a (A b-a B) \sin (c+d x) \sqrt {\sec (c+d x)}}{b d \left (a^2-b^2\right ) (a \sec (c+d x)+b)}-\frac {\frac {a \left (-3 a^3 B+a^2 A b+5 a b^2 B-3 A b^3\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}{b+a \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{b^2}+\frac {b \left (-3 a^2 B+a A b+2 b^2 B\right ) \int \frac {1}{\sqrt {\sec (c+d x)}}dx-\left (-3 a^3 B+a^2 A b+4 a b^2 B-2 A b^3\right ) \int \sqrt {\sec (c+d x)}dx}{b^2}}{2 b \left (a^2-b^2\right )}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {a (A b-a B) \sin (c+d x) \sqrt {\sec (c+d x)}}{b d \left (a^2-b^2\right ) (a \sec (c+d x)+b)}-\frac {\frac {b \left (-3 a^2 B+a A b+2 b^2 B\right ) \int \frac {1}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\left (-3 a^3 B+a^2 A b+4 a b^2 B-2 A b^3\right ) \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}dx}{b^2}+\frac {a \left (-3 a^3 B+a^2 A b+5 a b^2 B-3 A b^3\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}{b+a \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{b^2}}{2 b \left (a^2-b^2\right )}\) |
\(\Big \downarrow \) 4258 |
\(\displaystyle \frac {a (A b-a B) \sin (c+d x) \sqrt {\sec (c+d x)}}{b d \left (a^2-b^2\right ) (a \sec (c+d x)+b)}-\frac {\frac {a \left (-3 a^3 B+a^2 A b+5 a b^2 B-3 A b^3\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}{b+a \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{b^2}+\frac {b \left (-3 a^2 B+a A b+2 b^2 B\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\cos (c+d x)}dx-\left (-3 a^3 B+a^2 A b+4 a b^2 B-2 A b^3\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)}}dx}{b^2}}{2 b \left (a^2-b^2\right )}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {a (A b-a B) \sin (c+d x) \sqrt {\sec (c+d x)}}{b d \left (a^2-b^2\right ) (a \sec (c+d x)+b)}-\frac {\frac {a \left (-3 a^3 B+a^2 A b+5 a b^2 B-3 A b^3\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}{b+a \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{b^2}+\frac {b \left (-3 a^2 B+a A b+2 b^2 B\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx-\left (-3 a^3 B+a^2 A b+4 a b^2 B-2 A b^3\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b^2}}{2 b \left (a^2-b^2\right )}\) |
\(\Big \downarrow \) 3119 |
\(\displaystyle \frac {a (A b-a B) \sin (c+d x) \sqrt {\sec (c+d x)}}{b d \left (a^2-b^2\right ) (a \sec (c+d x)+b)}-\frac {\frac {a \left (-3 a^3 B+a^2 A b+5 a b^2 B-3 A b^3\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}{b+a \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{b^2}+\frac {\frac {2 b \left (-3 a^2 B+a A b+2 b^2 B\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}-\left (-3 a^3 B+a^2 A b+4 a b^2 B-2 A b^3\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b^2}}{2 b \left (a^2-b^2\right )}\) |
\(\Big \downarrow \) 3120 |
\(\displaystyle \frac {a (A b-a B) \sin (c+d x) \sqrt {\sec (c+d x)}}{b d \left (a^2-b^2\right ) (a \sec (c+d x)+b)}-\frac {\frac {a \left (-3 a^3 B+a^2 A b+5 a b^2 B-3 A b^3\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}{b+a \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{b^2}+\frac {\frac {2 b \left (-3 a^2 B+a A b+2 b^2 B\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}-\frac {2 \left (-3 a^3 B+a^2 A b+4 a b^2 B-2 A b^3\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}}{b^2}}{2 b \left (a^2-b^2\right )}\) |
\(\Big \downarrow \) 4336 |
\(\displaystyle \frac {a (A b-a B) \sin (c+d x) \sqrt {\sec (c+d x)}}{b d \left (a^2-b^2\right ) (a \sec (c+d x)+b)}-\frac {\frac {a \left (-3 a^3 B+a^2 A b+5 a b^2 B-3 A b^3\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{b^2}+\frac {\frac {2 b \left (-3 a^2 B+a A b+2 b^2 B\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}-\frac {2 \left (-3 a^3 B+a^2 A b+4 a b^2 B-2 A b^3\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}}{b^2}}{2 b \left (a^2-b^2\right )}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {a (A b-a B) \sin (c+d x) \sqrt {\sec (c+d x)}}{b d \left (a^2-b^2\right ) (a \sec (c+d x)+b)}-\frac {\frac {a \left (-3 a^3 B+a^2 A b+5 a b^2 B-3 A b^3\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{b^2}+\frac {\frac {2 b \left (-3 a^2 B+a A b+2 b^2 B\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}-\frac {2 \left (-3 a^3 B+a^2 A b+4 a b^2 B-2 A b^3\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}}{b^2}}{2 b \left (a^2-b^2\right )}\) |
\(\Big \downarrow \) 3284 |
\(\displaystyle \frac {a (A b-a B) \sin (c+d x) \sqrt {\sec (c+d x)}}{b d \left (a^2-b^2\right ) (a \sec (c+d x)+b)}-\frac {\frac {2 a \left (-3 a^3 B+a^2 A b+5 a b^2 B-3 A b^3\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{b^2 d (a+b)}+\frac {\frac {2 b \left (-3 a^2 B+a A b+2 b^2 B\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}-\frac {2 \left (-3 a^3 B+a^2 A b+4 a b^2 B-2 A b^3\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}}{b^2}}{2 b \left (a^2-b^2\right )}\) |
-1/2*(((2*b*(a*A*b - 3*a^2*B + 2*b^2*B)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/d - (2*(a^2*A*b - 2*A*b^3 - 3*a^3*B + 4*a*b ^2*B)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/d)/ b^2 + (2*a*(a^2*A*b - 3*A*b^3 - 3*a^3*B + 5*a*b^2*B)*Sqrt[Cos[c + d*x]]*El lipticPi[(2*b)/(a + b), (c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(b^2*(a + b)*d ))/(b*(a^2 - b^2)) + (a*(A*b - a*B)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(b*(a ^2 - b^2)*d*(b + a*Sec[c + d*x]))
3.6.76.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)* (c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2 )*(c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
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[(csc[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*((a_.) + (b_.)*sin[(e_.) + (f_.)* (x_)])^(m_.)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Sim p[g^(m + n) Int[(g*Csc[e + f*x])^(p - m - n)*(b + a*Csc[e + f*x])^m*(d + c*Csc[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && NeQ[b*c - a*d, 0] && !IntegerQ[p] && IntegerQ[m] && IntegerQ[n]
Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(b*Csc[c + d*x] )^n*Sin[c + d*x]^n Int[1/Sin[c + d*x]^n, x], x] /; FreeQ[{b, c, d}, x] && EqQ[n^2, 1/4]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[a Int[(d*Csc[e + f*x])^n, x], x] + Simp[b/d In t[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(3/2)/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[d*Sqrt[d*Sin[e + f*x]]*Sqrt[d*Csc[e + f*x]] Int[ 1/(Sqrt[d*Sin[e + f*x]]*(b + a*Sin[e + f*x])), x], x] /; FreeQ[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[b*(A*b - a*B)*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m + 1)*((d*Csc[e + f*x])^n/(a*f*( m + 1)*(a^2 - b^2))), x] + Simp[1/(a*(m + 1)*(a^2 - b^2)) Int[(a + b*Csc[ e + f*x])^(m + 1)*(d*Csc[e + f*x])^n*Simp[A*(a^2*(m + 1) - b^2*(m + n + 1)) + a*b*B*n - a*(A*b - a*B)*(m + 1)*Csc[e + f*x] + b*(A*b - a*B)*(m + n + 2) *Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f, A, B, n}, x] && NeQ[A* b - a*B, 0] && NeQ[a^2 - b^2, 0] && LtQ[m, -1] && !(ILtQ[m + 1/2, 0] && IL tQ[n, 0])
Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_. ))/(Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a _))), x_Symbol] :> Simp[(A*b^2 - a*b*B + a^2*C)/(a^2*d^2) Int[(d*Csc[e + f*x])^(3/2)/(a + b*Csc[e + f*x]), x], x] + Simp[1/a^2 Int[(a*A - (A*b - a *B)*Csc[e + f*x])/Sqrt[d*Csc[e + f*x]], x], x] /; FreeQ[{a, b, d, e, f, A, B, C}, x] && NeQ[a^2 - b^2, 0]
Leaf count of result is larger than twice the leaf count of optimal. \(848\) vs. \(2(348)=696\).
Time = 9.53 (sec) , antiderivative size = 849, normalized size of antiderivative = 2.99
-(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(2/b^3/(-2*sin( 1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1 /2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(A*b*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2) )-2*B*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*a-B*b*EllipticE(cos(1/2*d*x+1/ 2*c),2^(1/2)))+2*a^2*(A*b-B*a)/b^3*(-1/a*b^2/(a^2-b^2)*cos(1/2*d*x+1/2*c)* (-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/(2*b*cos(1/2*d*x+1/2* c)^2+a-b)-1/2/a/(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)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF( cos(1/2*d*x+1/2*c),2^(1/2))-1/2/(a^2-b^2)*b/a*(sin(1/2*d*x+1/2*c)^2)^(1/2) *(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/ 2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+1/2/(a^2-b^2)*b/a*(sin (1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x +1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2) )-3*a/(a^2-b^2)/(-2*a*b+2*b^2)*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)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2) *EllipticPi(cos(1/2*d*x+1/2*c),-2*b/(a-b),2^(1/2))+1/a/(a^2-b^2)/(-2*a*b+2 *b^2)*b^3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/( -2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticPi(cos(1/2*d*x +1/2*c),-2*b/(a-b),2^(1/2)))+4*a/b^2*(2*A*b-3*B*a)/(-2*a*b+2*b^2)*(sin(1/2 *d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+...
Timed out. \[ \int \frac {A+B \cos (c+d x)}{(a+b \cos (c+d x))^2 \sec ^{\frac {3}{2}}(c+d x)} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {A+B \cos (c+d x)}{(a+b \cos (c+d x))^2 \sec ^{\frac {3}{2}}(c+d x)} \, dx=\text {Timed out} \]
\[ \int \frac {A+B \cos (c+d x)}{(a+b \cos (c+d x))^2 \sec ^{\frac {3}{2}}(c+d x)} \, dx=\int { \frac {B \cos \left (d x + c\right ) + A}{{\left (b \cos \left (d x + c\right ) + a\right )}^{2} \sec \left (d x + c\right )^{\frac {3}{2}}} \,d x } \]
\[ \int \frac {A+B \cos (c+d x)}{(a+b \cos (c+d x))^2 \sec ^{\frac {3}{2}}(c+d x)} \, dx=\int { \frac {B \cos \left (d x + c\right ) + A}{{\left (b \cos \left (d x + c\right ) + a\right )}^{2} \sec \left (d x + c\right )^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {A+B \cos (c+d x)}{(a+b \cos (c+d x))^2 \sec ^{\frac {3}{2}}(c+d x)} \, dx=\int \frac {A+B\,\cos \left (c+d\,x\right )}{{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{3/2}\,{\left (a+b\,\cos \left (c+d\,x\right )\right )}^2} \,d x \]