\(\int \frac {1}{(a+b \cos (c+d x))^3 \sqrt {\sec (c+d x)}} \, dx\) [727]

Optimal result
Mathematica [A] (warning: unable to verify)
Rubi [A] (verified)
Maple [B] (verified)
Fricas [F(-1)]
Sympy [F]
Maxima [F]
Giac [F]
Mupad [F(-1)]
Reduce [F]

Optimal result

Integrand size = 23, antiderivative size = 317 \[ \int \frac {1}{(a+b \cos (c+d x))^3 \sqrt {\sec (c+d x)}} \, dx=\frac {\left (5 a^2+b^2\right ) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{4 a \left (a^2-b^2\right )^2 d}+\frac {3 \left (a^2+b^2\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{4 b \left (a^2-b^2\right )^2 d}-\frac {\left (3 a^4+10 a^2 b^2-b^4\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{4 a (a-b)^2 b (a+b)^3 d}+\frac {b^2 \sqrt {\sec (c+d x)} \sin (c+d x)}{2 a \left (a^2-b^2\right ) d (b+a \sec (c+d x))^2}-\frac {b \left (7 a^2-b^2\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{4 a \left (a^2-b^2\right )^2 d (b+a \sec (c+d x))} \] Output:

1/4*(5*a^2+b^2)*cos(d*x+c)^(1/2)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))*sec 
(d*x+c)^(1/2)/a/(a^2-b^2)^2/d+3/4*(a^2+b^2)*cos(d*x+c)^(1/2)*InverseJacobi 
AM(1/2*d*x+1/2*c,2^(1/2))*sec(d*x+c)^(1/2)/b/(a^2-b^2)^2/d-1/4*(3*a^4+10*a 
^2*b^2-b^4)*cos(d*x+c)^(1/2)*EllipticPi(sin(1/2*d*x+1/2*c),2*b/(a+b),2^(1/ 
2))*sec(d*x+c)^(1/2)/a/(a-b)^2/b/(a+b)^3/d+1/2*b^2*sec(d*x+c)^(1/2)*sin(d* 
x+c)/a/(a^2-b^2)/d/(b+a*sec(d*x+c))^2-1/4*b*(7*a^2-b^2)*sec(d*x+c)^(1/2)*s 
in(d*x+c)/a/(a^2-b^2)^2/d/(b+a*sec(d*x+c))
 

Mathematica [A] (warning: unable to verify)

Time = 3.94 (sec) , antiderivative size = 395, normalized size of antiderivative = 1.25 \[ \int \frac {1}{(a+b \cos (c+d x))^3 \sqrt {\sec (c+d x)}} \, dx=\frac {-\frac {4 b \left (7 a^3-a b^2+b \left (5 a^2+b^2\right ) \cos (c+d x)\right ) \sin (c+d x)}{a \left (a^2-b^2\right )^2 (a+b \cos (c+d x))^2 \sqrt {\sec (c+d x)}}+\frac {2 \cot (c+d x) \left (5 a^3 b \sec ^{\frac {3}{2}}(c+d x)+a b^3 \sec ^{\frac {3}{2}}(c+d x)-5 a^3 b \cos (2 (c+d x)) \sec ^{\frac {3}{2}}(c+d x)-a b^3 \cos (2 (c+d x)) \sec ^{\frac {3}{2}}(c+d x)-2 a b \left (5 a^2+b^2\right ) E\left (\left .\arcsin \left (\sqrt {\sec (c+d x)}\right )\right |-1\right ) \sqrt {-\tan ^2(c+d x)}+2 b \left (5 a^3-7 a^2 b+a b^2+b^3\right ) \operatorname {EllipticF}\left (\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right ) \sqrt {-\tan ^2(c+d x)}+6 a^4 \operatorname {EllipticPi}\left (-\frac {a}{b},\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right ) \sqrt {-\tan ^2(c+d x)}+20 a^2 b^2 \operatorname {EllipticPi}\left (-\frac {a}{b},\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right ) \sqrt {-\tan ^2(c+d x)}-2 b^4 \operatorname {EllipticPi}\left (-\frac {a}{b},\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right ) \sqrt {-\tan ^2(c+d x)}\right )}{a^2 (a-b)^2 b (a+b)^2}}{16 d} \] Input:

Integrate[1/((a + b*Cos[c + d*x])^3*Sqrt[Sec[c + d*x]]),x]
 

Output:

((-4*b*(7*a^3 - a*b^2 + b*(5*a^2 + b^2)*Cos[c + d*x])*Sin[c + d*x])/(a*(a^ 
2 - b^2)^2*(a + b*Cos[c + d*x])^2*Sqrt[Sec[c + d*x]]) + (2*Cot[c + d*x]*(5 
*a^3*b*Sec[c + d*x]^(3/2) + a*b^3*Sec[c + d*x]^(3/2) - 5*a^3*b*Cos[2*(c + 
d*x)]*Sec[c + d*x]^(3/2) - a*b^3*Cos[2*(c + d*x)]*Sec[c + d*x]^(3/2) - 2*a 
*b*(5*a^2 + b^2)*EllipticE[ArcSin[Sqrt[Sec[c + d*x]]], -1]*Sqrt[-Tan[c + d 
*x]^2] + 2*b*(5*a^3 - 7*a^2*b + a*b^2 + b^3)*EllipticF[ArcSin[Sqrt[Sec[c + 
 d*x]]], -1]*Sqrt[-Tan[c + d*x]^2] + 6*a^4*EllipticPi[-(a/b), ArcSin[Sqrt[ 
Sec[c + d*x]]], -1]*Sqrt[-Tan[c + d*x]^2] + 20*a^2*b^2*EllipticPi[-(a/b), 
ArcSin[Sqrt[Sec[c + d*x]]], -1]*Sqrt[-Tan[c + d*x]^2] - 2*b^4*EllipticPi[- 
(a/b), ArcSin[Sqrt[Sec[c + d*x]]], -1]*Sqrt[-Tan[c + d*x]^2]))/(a^2*(a - b 
)^2*b*(a + b)^2))/(16*d)
 

Rubi [A] (verified)

Time = 2.21 (sec) , antiderivative size = 314, normalized size of antiderivative = 0.99, number of steps used = 20, number of rules used = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.870, Rules used = {3042, 3717, 3042, 4332, 27, 3042, 4588, 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.

\(\displaystyle \int \frac {1}{\sqrt {\sec (c+d x)} (a+b \cos (c+d x))^3} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^3}dx\)

\(\Big \downarrow \) 3717

\(\displaystyle \int \frac {\sec ^{\frac {5}{2}}(c+d x)}{(a \sec (c+d x)+b)^3}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{5/2}}{\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+b\right )^3}dx\)

\(\Big \downarrow \) 4332

\(\displaystyle \frac {\int -\frac {b^2+4 a \sec (c+d x) b-\left (4 a^2-b^2\right ) \sec ^2(c+d x)}{2 \sqrt {\sec (c+d x)} (b+a \sec (c+d x))^2}dx}{2 a \left (a^2-b^2\right )}+\frac {b^2 \sin (c+d x) \sqrt {\sec (c+d x)}}{2 a d \left (a^2-b^2\right ) (a \sec (c+d x)+b)^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {b^2 \sin (c+d x) \sqrt {\sec (c+d x)}}{2 a d \left (a^2-b^2\right ) (a \sec (c+d x)+b)^2}-\frac {\int \frac {b^2+4 a \sec (c+d x) b-\left (4 a^2-b^2\right ) \sec ^2(c+d x)}{\sqrt {\sec (c+d x)} (b+a \sec (c+d x))^2}dx}{4 a \left (a^2-b^2\right )}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {b^2 \sin (c+d x) \sqrt {\sec (c+d x)}}{2 a d \left (a^2-b^2\right ) (a \sec (c+d x)+b)^2}-\frac {\int \frac {b^2+4 a \csc \left (c+d x+\frac {\pi }{2}\right ) b+\left (b^2-4 a^2\right ) \csc \left (c+d x+\frac {\pi }{2}\right )^2}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \left (b+a \csc \left (c+d x+\frac {\pi }{2}\right )\right )^2}dx}{4 a \left (a^2-b^2\right )}\)

\(\Big \downarrow \) 4588

\(\displaystyle \frac {b^2 \sin (c+d x) \sqrt {\sec (c+d x)}}{2 a d \left (a^2-b^2\right ) (a \sec (c+d x)+b)^2}-\frac {\frac {\int -\frac {-\left (\left (7 a^2-b^2\right ) \sec ^2(c+d x) b^2\right )+\left (5 a^2+b^2\right ) b^2+4 a \left (2 a^2+b^2\right ) \sec (c+d x) b}{2 \sqrt {\sec (c+d x)} (b+a \sec (c+d x))}dx}{b \left (a^2-b^2\right )}+\frac {b \left (7 a^2-b^2\right ) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \left (a^2-b^2\right ) (a \sec (c+d x)+b)}}{4 a \left (a^2-b^2\right )}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {b^2 \sin (c+d x) \sqrt {\sec (c+d x)}}{2 a d \left (a^2-b^2\right ) (a \sec (c+d x)+b)^2}-\frac {\frac {b \left (7 a^2-b^2\right ) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \left (a^2-b^2\right ) (a \sec (c+d x)+b)}-\frac {\int \frac {-\left (\left (7 a^2-b^2\right ) \sec ^2(c+d x) b^2\right )+\left (5 a^2+b^2\right ) b^2+4 a \left (2 a^2+b^2\right ) \sec (c+d x) b}{\sqrt {\sec (c+d x)} (b+a \sec (c+d x))}dx}{2 b \left (a^2-b^2\right )}}{4 a \left (a^2-b^2\right )}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {b^2 \sin (c+d x) \sqrt {\sec (c+d x)}}{2 a d \left (a^2-b^2\right ) (a \sec (c+d x)+b)^2}-\frac {\frac {b \left (7 a^2-b^2\right ) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \left (a^2-b^2\right ) (a \sec (c+d x)+b)}-\frac {\int \frac {-\left (\left (7 a^2-b^2\right ) \csc \left (c+d x+\frac {\pi }{2}\right )^2 b^2\right )+\left (5 a^2+b^2\right ) b^2+4 a \left (2 a^2+b^2\right ) \csc \left (c+d x+\frac {\pi }{2}\right ) b}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \left (b+a \csc \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{2 b \left (a^2-b^2\right )}}{4 a \left (a^2-b^2\right )}\)

\(\Big \downarrow \) 4594

\(\displaystyle \frac {b^2 \sin (c+d x) \sqrt {\sec (c+d x)}}{2 a d \left (a^2-b^2\right ) (a \sec (c+d x)+b)^2}-\frac {\frac {b \left (7 a^2-b^2\right ) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \left (a^2-b^2\right ) (a \sec (c+d x)+b)}-\frac {\frac {\int \frac {\left (5 a^2+b^2\right ) b^3+3 a \left (a^2+b^2\right ) \sec (c+d x) b^2}{\sqrt {\sec (c+d x)}}dx}{b^2}-\left (3 a^4+10 a^2 b^2-b^4\right ) \int \frac {\sec ^{\frac {3}{2}}(c+d x)}{b+a \sec (c+d x)}dx}{2 b \left (a^2-b^2\right )}}{4 a \left (a^2-b^2\right )}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {b^2 \sin (c+d x) \sqrt {\sec (c+d x)}}{2 a d \left (a^2-b^2\right ) (a \sec (c+d x)+b)^2}-\frac {\frac {b \left (7 a^2-b^2\right ) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \left (a^2-b^2\right ) (a \sec (c+d x)+b)}-\frac {\frac {\int \frac {\left (5 a^2+b^2\right ) b^3+3 a \left (a^2+b^2\right ) \csc \left (c+d x+\frac {\pi }{2}\right ) b^2}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{b^2}-\left (3 a^4+10 a^2 b^2-b^4\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}{b+a \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{2 b \left (a^2-b^2\right )}}{4 a \left (a^2-b^2\right )}\)

\(\Big \downarrow \) 4274

\(\displaystyle \frac {b^2 \sin (c+d x) \sqrt {\sec (c+d x)}}{2 a d \left (a^2-b^2\right ) (a \sec (c+d x)+b)^2}-\frac {\frac {b \left (7 a^2-b^2\right ) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \left (a^2-b^2\right ) (a \sec (c+d x)+b)}-\frac {\frac {3 a b^2 \left (a^2+b^2\right ) \int \sqrt {\sec (c+d x)}dx+b^3 \left (5 a^2+b^2\right ) \int \frac {1}{\sqrt {\sec (c+d x)}}dx}{b^2}-\left (3 a^4+10 a^2 b^2-b^4\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}{b+a \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{2 b \left (a^2-b^2\right )}}{4 a \left (a^2-b^2\right )}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {b^2 \sin (c+d x) \sqrt {\sec (c+d x)}}{2 a d \left (a^2-b^2\right ) (a \sec (c+d x)+b)^2}-\frac {\frac {b \left (7 a^2-b^2\right ) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \left (a^2-b^2\right ) (a \sec (c+d x)+b)}-\frac {\frac {3 a b^2 \left (a^2+b^2\right ) \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}dx+b^3 \left (5 a^2+b^2\right ) \int \frac {1}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{b^2}-\left (3 a^4+10 a^2 b^2-b^4\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}{b+a \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{2 b \left (a^2-b^2\right )}}{4 a \left (a^2-b^2\right )}\)

\(\Big \downarrow \) 4258

\(\displaystyle \frac {b^2 \sin (c+d x) \sqrt {\sec (c+d x)}}{2 a d \left (a^2-b^2\right ) (a \sec (c+d x)+b)^2}-\frac {\frac {b \left (7 a^2-b^2\right ) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \left (a^2-b^2\right ) (a \sec (c+d x)+b)}-\frac {\frac {3 a b^2 \left (a^2+b^2\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)}}dx+b^3 \left (5 a^2+b^2\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\cos (c+d x)}dx}{b^2}-\left (3 a^4+10 a^2 b^2-b^4\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}{b+a \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{2 b \left (a^2-b^2\right )}}{4 a \left (a^2-b^2\right )}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {b^2 \sin (c+d x) \sqrt {\sec (c+d x)}}{2 a d \left (a^2-b^2\right ) (a \sec (c+d x)+b)^2}-\frac {\frac {b \left (7 a^2-b^2\right ) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \left (a^2-b^2\right ) (a \sec (c+d x)+b)}-\frac {\frac {3 a b^2 \left (a^2+b^2\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+b^3 \left (5 a^2+b^2\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx}{b^2}-\left (3 a^4+10 a^2 b^2-b^4\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}{b+a \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{2 b \left (a^2-b^2\right )}}{4 a \left (a^2-b^2\right )}\)

\(\Big \downarrow \) 3119

\(\displaystyle \frac {b^2 \sin (c+d x) \sqrt {\sec (c+d x)}}{2 a d \left (a^2-b^2\right ) (a \sec (c+d x)+b)^2}-\frac {\frac {b \left (7 a^2-b^2\right ) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \left (a^2-b^2\right ) (a \sec (c+d x)+b)}-\frac {\frac {3 a b^2 \left (a^2+b^2\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 b^3 \left (5 a^2+b^2\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}}{b^2}-\left (3 a^4+10 a^2 b^2-b^4\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}{b+a \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{2 b \left (a^2-b^2\right )}}{4 a \left (a^2-b^2\right )}\)

\(\Big \downarrow \) 3120

\(\displaystyle \frac {b^2 \sin (c+d x) \sqrt {\sec (c+d x)}}{2 a d \left (a^2-b^2\right ) (a \sec (c+d x)+b)^2}-\frac {\frac {b \left (7 a^2-b^2\right ) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \left (a^2-b^2\right ) (a \sec (c+d x)+b)}-\frac {\frac {\frac {6 a b^2 \left (a^2+b^2\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}+\frac {2 b^3 \left (5 a^2+b^2\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}}{b^2}-\left (3 a^4+10 a^2 b^2-b^4\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}{b+a \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{2 b \left (a^2-b^2\right )}}{4 a \left (a^2-b^2\right )}\)

\(\Big \downarrow \) 4336

\(\displaystyle \frac {b^2 \sin (c+d x) \sqrt {\sec (c+d x)}}{2 a d \left (a^2-b^2\right ) (a \sec (c+d x)+b)^2}-\frac {\frac {b \left (7 a^2-b^2\right ) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \left (a^2-b^2\right ) (a \sec (c+d x)+b)}-\frac {\frac {\frac {6 a b^2 \left (a^2+b^2\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}+\frac {2 b^3 \left (5 a^2+b^2\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}}{b^2}-\left (3 a^4+10 a^2 b^2-b^4\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{2 b \left (a^2-b^2\right )}}{4 a \left (a^2-b^2\right )}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {b^2 \sin (c+d x) \sqrt {\sec (c+d x)}}{2 a d \left (a^2-b^2\right ) (a \sec (c+d x)+b)^2}-\frac {\frac {b \left (7 a^2-b^2\right ) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \left (a^2-b^2\right ) (a \sec (c+d x)+b)}-\frac {\frac {\frac {6 a b^2 \left (a^2+b^2\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}+\frac {2 b^3 \left (5 a^2+b^2\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}}{b^2}-\left (3 a^4+10 a^2 b^2-b^4\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{2 b \left (a^2-b^2\right )}}{4 a \left (a^2-b^2\right )}\)

\(\Big \downarrow \) 3284

\(\displaystyle \frac {b^2 \sin (c+d x) \sqrt {\sec (c+d x)}}{2 a d \left (a^2-b^2\right ) (a \sec (c+d x)+b)^2}-\frac {\frac {b \left (7 a^2-b^2\right ) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \left (a^2-b^2\right ) (a \sec (c+d x)+b)}-\frac {\frac {\frac {6 a b^2 \left (a^2+b^2\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}+\frac {2 b^3 \left (5 a^2+b^2\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}}{b^2}-\frac {2 \left (3 a^4+10 a^2 b^2-b^4\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{d (a+b)}}{2 b \left (a^2-b^2\right )}}{4 a \left (a^2-b^2\right )}\)

Input:

Int[1/((a + b*Cos[c + d*x])^3*Sqrt[Sec[c + d*x]]),x]
 

Output:

(b^2*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(2*a*(a^2 - b^2)*d*(b + a*Sec[c + d* 
x])^2) - (-1/2*(((2*b^3*(5*a^2 + b^2)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d* 
x)/2, 2]*Sqrt[Sec[c + d*x]])/d + (6*a*b^2*(a^2 + b^2)*Sqrt[Cos[c + d*x]]*E 
llipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/d)/b^2 - (2*(3*a^4 + 10*a^2*b 
^2 - b^4)*Sqrt[Cos[c + d*x]]*EllipticPi[(2*b)/(a + b), (c + d*x)/2, 2]*Sqr 
t[Sec[c + d*x]])/((a + b)*d))/(b*(a^2 - b^2)) + (b*(7*a^2 - b^2)*Sqrt[Sec[ 
c + d*x]]*Sin[c + d*x])/((a^2 - b^2)*d*(b + a*Sec[c + d*x])))/(4*a*(a^2 - 
b^2))
 

Defintions of rubi rules used

rule 27
Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 3042
Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3119
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]
 

rule 3120
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]
 

rule 3284
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]
 

rule 3717
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(m_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x 
_)]^(n_.))^(p_.), x_Symbol] :> Simp[d^(n*p)   Int[(d*Csc[e + f*x])^(m - n*p 
)*(b + a*Csc[e + f*x]^n)^p, x], x] /; FreeQ[{a, b, d, e, f, m, n, p}, x] && 
  !IntegerQ[m] && IntegersQ[n, p]
 

rule 4258
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]
 

rule 4274
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]
 

rule 4332
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( 
a_))^(m_), x_Symbol] :> Simp[(-a^2)*d^3*Cot[e + f*x]*(a + b*Csc[e + f*x])^( 
m + 1)*((d*Csc[e + f*x])^(n - 3)/(b*f*(m + 1)*(a^2 - b^2))), x] + Simp[d^3/ 
(b*(m + 1)*(a^2 - b^2))   Int[(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x]) 
^(n - 3)*Simp[a^2*(n - 3) + a*b*(m + 1)*Csc[e + f*x] - (a^2*(n - 2) + b^2*( 
m + 1))*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f}, x] && NeQ[a^2 
- b^2, 0] && LtQ[m, -1] && (IGtQ[n, 3] || (IntegersQ[n + 1/2, 2*m] && GtQ[n 
, 2]))
 

rule 4336
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]
 

rule 4588
Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_. 
))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a 
_))^(m_), x_Symbol] :> Simp[(A*b^2 - a*b*B + a^2*C)*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] + Sim 
p[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*A - b*B + a*C)*(m + 1) - (A*b^2 - a*b*B + a^2*C)*(m + n + 
1) - a*(A*b - a*B + b*C)*(m + 1)*Csc[e + f*x] + (A*b^2 - a*b*B + a^2*C)*(m 
+ n + 2)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f, A, B, C, n}, x 
] && NeQ[a^2 - b^2, 0] && LtQ[m, -1] &&  !(ILtQ[m + 1/2, 0] && ILtQ[n, 0])
 

rule 4594
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]
 
Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1735\) vs. \(2(296)=592\).

Time = 7.41 (sec) , antiderivative size = 1736, normalized size of antiderivative = 5.48

method result size
default \(\text {Expression too large to display}\) \(1736\)

Input:

int(1/(a+cos(d*x+c)*b)^3/sec(d*x+c)^(1/2),x,method=_RETURNVERBOSE)
 

Output:

-(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(2/b*(-b^2/a/(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*b/(a^2-b^2)/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*b/(a^2-b^2)/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)))-2/b*a*(-1/2*b^2/a/ 
(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)^2-3/4*b^2*(3*a^2-b^2)/a^2/(a^2-b^2) 
^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)-7/8/(a+b)/(a^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+1/2*c)^4+sin(1...
 

Fricas [F(-1)]

Timed out. \[ \int \frac {1}{(a+b \cos (c+d x))^3 \sqrt {\sec (c+d x)}} \, dx=\text {Timed out} \] Input:

integrate(1/(a+b*cos(d*x+c))^3/sec(d*x+c)^(1/2),x, algorithm="fricas")
 

Output:

Timed out
 

Sympy [F]

\[ \int \frac {1}{(a+b \cos (c+d x))^3 \sqrt {\sec (c+d x)}} \, dx=\int \frac {1}{\left (a + b \cos {\left (c + d x \right )}\right )^{3} \sqrt {\sec {\left (c + d x \right )}}}\, dx \] Input:

integrate(1/(a+b*cos(d*x+c))**3/sec(d*x+c)**(1/2),x)
 

Output:

Integral(1/((a + b*cos(c + d*x))**3*sqrt(sec(c + d*x))), x)
 

Maxima [F]

\[ \int \frac {1}{(a+b \cos (c+d x))^3 \sqrt {\sec (c+d x)}} \, dx=\int { \frac {1}{{\left (b \cos \left (d x + c\right ) + a\right )}^{3} \sqrt {\sec \left (d x + c\right )}} \,d x } \] Input:

integrate(1/(a+b*cos(d*x+c))^3/sec(d*x+c)^(1/2),x, algorithm="maxima")
 

Output:

integrate(1/((b*cos(d*x + c) + a)^3*sqrt(sec(d*x + c))), x)
 

Giac [F]

\[ \int \frac {1}{(a+b \cos (c+d x))^3 \sqrt {\sec (c+d x)}} \, dx=\int { \frac {1}{{\left (b \cos \left (d x + c\right ) + a\right )}^{3} \sqrt {\sec \left (d x + c\right )}} \,d x } \] Input:

integrate(1/(a+b*cos(d*x+c))^3/sec(d*x+c)^(1/2),x, algorithm="giac")
 

Output:

integrate(1/((b*cos(d*x + c) + a)^3*sqrt(sec(d*x + c))), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{(a+b \cos (c+d x))^3 \sqrt {\sec (c+d x)}} \, dx=\int \frac {1}{\sqrt {\frac {1}{\cos \left (c+d\,x\right )}}\,{\left (a+b\,\cos \left (c+d\,x\right )\right )}^3} \,d x \] Input:

int(1/((1/cos(c + d*x))^(1/2)*(a + b*cos(c + d*x))^3),x)
 

Output:

int(1/((1/cos(c + d*x))^(1/2)*(a + b*cos(c + d*x))^3), x)
 

Reduce [F]

\[ \int \frac {1}{(a+b \cos (c+d x))^3 \sqrt {\sec (c+d x)}} \, dx=\int \frac {\sqrt {\sec \left (d x +c \right )}}{\cos \left (d x +c \right )^{3} \sec \left (d x +c \right ) b^{3}+3 \cos \left (d x +c \right )^{2} \sec \left (d x +c \right ) a \,b^{2}+3 \cos \left (d x +c \right ) \sec \left (d x +c \right ) a^{2} b +\sec \left (d x +c \right ) a^{3}}d x \] Input:

int(1/(a+b*cos(d*x+c))^3/sec(d*x+c)^(1/2),x)
 

Output:

int(sqrt(sec(c + d*x))/(cos(c + d*x)**3*sec(c + d*x)*b**3 + 3*cos(c + d*x) 
**2*sec(c + d*x)*a*b**2 + 3*cos(c + d*x)*sec(c + d*x)*a**2*b + sec(c + d*x 
)*a**3),x)