3.7.26 \(\int \frac {1}{\sqrt {\sec (c+d x)} (a+b \sec (c+d x))^3} \, dx\) [626]

3.7.26.1 Optimal result
3.7.26.2 Mathematica [B] (warning: unable to verify)
3.7.26.3 Rubi [A] (verified)
3.7.26.4 Maple [B] (verified)
3.7.26.5 Fricas [F(-1)]
3.7.26.6 Sympy [F]
3.7.26.7 Maxima [F]
3.7.26.8 Giac [F]
3.7.26.9 Mupad [F(-1)]

3.7.26.1 Optimal result

Integrand size = 23, antiderivative size = 342 \[ \int \frac {1}{\sqrt {\sec (c+d x)} (a+b \sec (c+d x))^3} \, dx=\frac {\left (8 a^4-29 a^2 b^2+15 b^4\right ) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{4 a^3 \left (a^2-b^2\right )^2 d}-\frac {3 b \left (8 a^4-11 a^2 b^2+5 b^4\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{4 a^4 \left (a^2-b^2\right )^2 d}+\frac {b^2 \left (35 a^4-38 a^2 b^2+15 b^4\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticPi}\left (\frac {2 a}{a+b},\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{4 a^4 (a-b)^2 (a+b)^3 d}+\frac {b^2 \sqrt {\sec (c+d x)} \sin (c+d x)}{2 a \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}+\frac {b^2 \left (11 a^2-5 b^2\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{4 a^2 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))} \]

output
1/2*b^2*sin(d*x+c)*sec(d*x+c)^(1/2)/a/(a^2-b^2)/d/(a+b*sec(d*x+c))^2+1/4*b 
^2*(11*a^2-5*b^2)*sin(d*x+c)*sec(d*x+c)^(1/2)/a^2/(a^2-b^2)^2/d/(a+b*sec(d 
*x+c))+1/4*(8*a^4-29*a^2*b^2+15*b^4)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2* 
d*x+1/2*c)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))*cos(d*x+c)^(1/2)*sec(d*x+ 
c)^(1/2)/a^3/(a^2-b^2)^2/d-3/4*b*(8*a^4-11*a^2*b^2+5*b^4)*(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)/a^4/(a^2-b^2)^2/d+1/4*b^2*(35*a^4-38*a^2*b^2 
+15*b^4)*(cos(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*a/(a+b),2^(1/2))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/a^4/(a-b 
)^2/(a+b)^3/d
 
3.7.26.2 Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(707\) vs. \(2(342)=684\).

Time = 6.67 (sec) , antiderivative size = 707, normalized size of antiderivative = 2.07 \[ \int \frac {1}{\sqrt {\sec (c+d x)} (a+b \sec (c+d x))^3} \, dx=\frac {\frac {2 \left (8 a^4-7 a^2 b^2+5 b^4\right ) \cos ^2(c+d x) \left (\operatorname {EllipticF}\left (\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right )-\operatorname {EllipticPi}\left (-\frac {b}{a},\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right )\right ) (a+b \sec (c+d x)) \sqrt {1-\sec ^2(c+d x)} \sin (c+d x)}{b (b+a \cos (c+d x)) \left (1-\cos ^2(c+d x)\right )}+\frac {2 \left (-32 a^3 b+8 a b^3\right ) \cos ^2(c+d x) \operatorname {EllipticPi}\left (-\frac {b}{a},\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right ) (a+b \sec (c+d x)) \sqrt {1-\sec ^2(c+d x)} \sin (c+d x)}{a (b+a \cos (c+d x)) \left (1-\cos ^2(c+d x)\right )}+\frac {\left (8 a^4-29 a^2 b^2+15 b^4\right ) \cos (2 (c+d x)) (a+b \sec (c+d x)) \left (-4 a b+4 a b \sec ^2(c+d x)-4 a b E\left (\left .\arcsin \left (\sqrt {\sec (c+d x)}\right )\right |-1\right ) \sqrt {\sec (c+d x)} \sqrt {1-\sec ^2(c+d x)}-2 a (a-2 b) \operatorname {EllipticF}\left (\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right ) \sqrt {\sec (c+d x)} \sqrt {1-\sec ^2(c+d x)}+2 a^2 \operatorname {EllipticPi}\left (-\frac {b}{a},\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right ) \sqrt {\sec (c+d x)} \sqrt {1-\sec ^2(c+d x)}-4 b^2 \operatorname {EllipticPi}\left (-\frac {b}{a},\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right ) \sqrt {\sec (c+d x)} \sqrt {1-\sec ^2(c+d x)}\right ) \sin (c+d x)}{a^2 b (b+a \cos (c+d x)) \left (1-\cos ^2(c+d x)\right ) \sqrt {\sec (c+d x)} \left (2-\sec ^2(c+d x)\right )}}{16 a^2 (a-b)^2 (a+b)^2 d}+\frac {\sqrt {\sec (c+d x)} \left (-\frac {b^2 \left (-13 a^2+7 b^2\right ) \sin (c+d x)}{4 a^3 \left (-a^2+b^2\right )^2}+\frac {b^4 \sin (c+d x)}{2 a^3 \left (a^2-b^2\right ) (b+a \cos (c+d x))^2}+\frac {3 \left (-5 a^2 b^3 \sin (c+d x)+3 b^5 \sin (c+d x)\right )}{4 a^3 \left (a^2-b^2\right )^2 (b+a \cos (c+d x))}\right )}{d} \]

input
Integrate[1/(Sqrt[Sec[c + d*x]]*(a + b*Sec[c + d*x])^3),x]
 
output
((2*(8*a^4 - 7*a^2*b^2 + 5*b^4)*Cos[c + d*x]^2*(EllipticF[ArcSin[Sqrt[Sec[ 
c + d*x]]], -1] - EllipticPi[-(b/a), ArcSin[Sqrt[Sec[c + d*x]]], -1])*(a + 
 b*Sec[c + d*x])*Sqrt[1 - Sec[c + d*x]^2]*Sin[c + d*x])/(b*(b + a*Cos[c + 
d*x])*(1 - Cos[c + d*x]^2)) + (2*(-32*a^3*b + 8*a*b^3)*Cos[c + d*x]^2*Elli 
pticPi[-(b/a), ArcSin[Sqrt[Sec[c + d*x]]], -1]*(a + b*Sec[c + d*x])*Sqrt[1 
 - Sec[c + d*x]^2]*Sin[c + d*x])/(a*(b + a*Cos[c + d*x])*(1 - Cos[c + d*x] 
^2)) + ((8*a^4 - 29*a^2*b^2 + 15*b^4)*Cos[2*(c + d*x)]*(a + b*Sec[c + d*x] 
)*(-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[1 - Sec[c + d*x]^2] - 2*a*(a - 2*b)*Ellip 
ticF[ArcSin[Sqrt[Sec[c + d*x]]], -1]*Sqrt[Sec[c + d*x]]*Sqrt[1 - Sec[c + d 
*x]^2] + 2*a^2*EllipticPi[-(b/a), ArcSin[Sqrt[Sec[c + d*x]]], -1]*Sqrt[Sec 
[c + d*x]]*Sqrt[1 - Sec[c + d*x]^2] - 4*b^2*EllipticPi[-(b/a), ArcSin[Sqrt 
[Sec[c + d*x]]], -1]*Sqrt[Sec[c + d*x]]*Sqrt[1 - Sec[c + d*x]^2])*Sin[c + 
d*x])/(a^2*b*(b + a*Cos[c + d*x])*(1 - Cos[c + d*x]^2)*Sqrt[Sec[c + d*x]]* 
(2 - Sec[c + d*x]^2)))/(16*a^2*(a - b)^2*(a + b)^2*d) + (Sqrt[Sec[c + d*x] 
]*(-1/4*(b^2*(-13*a^2 + 7*b^2)*Sin[c + d*x])/(a^3*(-a^2 + b^2)^2) + (b^4*S 
in[c + d*x])/(2*a^3*(a^2 - b^2)*(b + a*Cos[c + d*x])^2) + (3*(-5*a^2*b^3*S 
in[c + d*x] + 3*b^5*Sin[c + d*x]))/(4*a^3*(a^2 - b^2)^2*(b + a*Cos[c + d*x 
]))))/d
 
3.7.26.3 Rubi [A] (verified)

Time = 2.28 (sec) , antiderivative size = 342, normalized size of antiderivative = 1.00, number of steps used = 18, number of rules used = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.783, Rules used = {3042, 4334, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4334

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4588

\(\displaystyle \frac {\frac {b^2 \left (11 a^2-5 b^2\right ) \sin (c+d x) \sqrt {\sec (c+d x)}}{a d \left (a^2-b^2\right ) (a+b \sec (c+d x))}-\frac {\int -\frac {8 a^4-29 b^2 a^2-4 b \left (4 a^2-b^2\right ) \sec (c+d x) a+15 b^4+b^2 \left (11 a^2-5 b^2\right ) \sec ^2(c+d x)}{2 \sqrt {\sec (c+d x)} (a+b \sec (c+d x))}dx}{a \left (a^2-b^2\right )}}{4 a \left (a^2-b^2\right )}+\frac {b^2 \sin (c+d x) \sqrt {\sec (c+d x)}}{2 a d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\int \frac {8 a^4-29 b^2 a^2-4 b \left (4 a^2-b^2\right ) \sec (c+d x) a+15 b^4+b^2 \left (11 a^2-5 b^2\right ) \sec ^2(c+d x)}{\sqrt {\sec (c+d x)} (a+b \sec (c+d x))}dx}{2 a \left (a^2-b^2\right )}+\frac {b^2 \left (11 a^2-5 b^2\right ) \sin (c+d x) \sqrt {\sec (c+d x)}}{a d \left (a^2-b^2\right ) (a+b \sec (c+d x))}}{4 a \left (a^2-b^2\right )}+\frac {b^2 \sin (c+d x) \sqrt {\sec (c+d x)}}{2 a d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\int \frac {8 a^4-29 b^2 a^2-4 b \left (4 a^2-b^2\right ) \csc \left (c+d x+\frac {\pi }{2}\right ) a+15 b^4+b^2 \left (11 a^2-5 b^2\right ) \csc \left (c+d x+\frac {\pi }{2}\right )^2}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{2 a \left (a^2-b^2\right )}+\frac {b^2 \left (11 a^2-5 b^2\right ) \sin (c+d x) \sqrt {\sec (c+d x)}}{a d \left (a^2-b^2\right ) (a+b \sec (c+d x))}}{4 a \left (a^2-b^2\right )}+\frac {b^2 \sin (c+d x) \sqrt {\sec (c+d x)}}{2 a d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}\)

\(\Big \downarrow \) 4594

\(\displaystyle \frac {\frac {\frac {b^2 \left (35 a^4-38 a^2 b^2+15 b^4\right ) \int \frac {\sec ^{\frac {3}{2}}(c+d x)}{a+b \sec (c+d x)}dx}{a^2}+\frac {\int \frac {a \left (8 a^4-29 b^2 a^2+15 b^4\right )-3 b \left (8 a^4-11 b^2 a^2+5 b^4\right ) \sec (c+d x)}{\sqrt {\sec (c+d x)}}dx}{a^2}}{2 a \left (a^2-b^2\right )}+\frac {b^2 \left (11 a^2-5 b^2\right ) \sin (c+d x) \sqrt {\sec (c+d x)}}{a d \left (a^2-b^2\right ) (a+b \sec (c+d x))}}{4 a \left (a^2-b^2\right )}+\frac {b^2 \sin (c+d x) \sqrt {\sec (c+d x)}}{2 a d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {b^2 \left (35 a^4-38 a^2 b^2+15 b^4\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{a^2}+\frac {\int \frac {a \left (8 a^4-29 b^2 a^2+15 b^4\right )-3 b \left (8 a^4-11 b^2 a^2+5 b^4\right ) \csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{a^2}}{2 a \left (a^2-b^2\right )}+\frac {b^2 \left (11 a^2-5 b^2\right ) \sin (c+d x) \sqrt {\sec (c+d x)}}{a d \left (a^2-b^2\right ) (a+b \sec (c+d x))}}{4 a \left (a^2-b^2\right )}+\frac {b^2 \sin (c+d x) \sqrt {\sec (c+d x)}}{2 a d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}\)

\(\Big \downarrow \) 4274

\(\displaystyle \frac {\frac {\frac {b^2 \left (35 a^4-38 a^2 b^2+15 b^4\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{a^2}+\frac {a \left (8 a^4-29 a^2 b^2+15 b^4\right ) \int \frac {1}{\sqrt {\sec (c+d x)}}dx-3 b \left (8 a^4-11 a^2 b^2+5 b^4\right ) \int \sqrt {\sec (c+d x)}dx}{a^2}}{2 a \left (a^2-b^2\right )}+\frac {b^2 \left (11 a^2-5 b^2\right ) \sin (c+d x) \sqrt {\sec (c+d x)}}{a d \left (a^2-b^2\right ) (a+b \sec (c+d x))}}{4 a \left (a^2-b^2\right )}+\frac {b^2 \sin (c+d x) \sqrt {\sec (c+d x)}}{2 a d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {b^2 \left (35 a^4-38 a^2 b^2+15 b^4\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{a^2}+\frac {a \left (8 a^4-29 a^2 b^2+15 b^4\right ) \int \frac {1}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx-3 b \left (8 a^4-11 a^2 b^2+5 b^4\right ) \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}dx}{a^2}}{2 a \left (a^2-b^2\right )}+\frac {b^2 \left (11 a^2-5 b^2\right ) \sin (c+d x) \sqrt {\sec (c+d x)}}{a d \left (a^2-b^2\right ) (a+b \sec (c+d x))}}{4 a \left (a^2-b^2\right )}+\frac {b^2 \sin (c+d x) \sqrt {\sec (c+d x)}}{2 a d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}\)

\(\Big \downarrow \) 4258

\(\displaystyle \frac {\frac {\frac {b^2 \left (35 a^4-38 a^2 b^2+15 b^4\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{a^2}+\frac {a \left (8 a^4-29 a^2 b^2+15 b^4\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\cos (c+d x)}dx-3 b \left (8 a^4-11 a^2 b^2+5 b^4\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)}}dx}{a^2}}{2 a \left (a^2-b^2\right )}+\frac {b^2 \left (11 a^2-5 b^2\right ) \sin (c+d x) \sqrt {\sec (c+d x)}}{a d \left (a^2-b^2\right ) (a+b \sec (c+d x))}}{4 a \left (a^2-b^2\right )}+\frac {b^2 \sin (c+d x) \sqrt {\sec (c+d x)}}{2 a d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {b^2 \left (35 a^4-38 a^2 b^2+15 b^4\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{a^2}+\frac {a \left (8 a^4-29 a^2 b^2+15 b^4\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx-3 b \left (8 a^4-11 a^2 b^2+5 b^4\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{a^2}}{2 a \left (a^2-b^2\right )}+\frac {b^2 \left (11 a^2-5 b^2\right ) \sin (c+d x) \sqrt {\sec (c+d x)}}{a d \left (a^2-b^2\right ) (a+b \sec (c+d x))}}{4 a \left (a^2-b^2\right )}+\frac {b^2 \sin (c+d x) \sqrt {\sec (c+d x)}}{2 a d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}\)

\(\Big \downarrow \) 3119

\(\displaystyle \frac {\frac {\frac {b^2 \left (35 a^4-38 a^2 b^2+15 b^4\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{a^2}+\frac {\frac {2 a \left (8 a^4-29 a^2 b^2+15 b^4\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}-3 b \left (8 a^4-11 a^2 b^2+5 b^4\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{a^2}}{2 a \left (a^2-b^2\right )}+\frac {b^2 \left (11 a^2-5 b^2\right ) \sin (c+d x) \sqrt {\sec (c+d x)}}{a d \left (a^2-b^2\right ) (a+b \sec (c+d x))}}{4 a \left (a^2-b^2\right )}+\frac {b^2 \sin (c+d x) \sqrt {\sec (c+d x)}}{2 a d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}\)

\(\Big \downarrow \) 3120

\(\displaystyle \frac {\frac {\frac {b^2 \left (35 a^4-38 a^2 b^2+15 b^4\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{a^2}+\frac {\frac {2 a \left (8 a^4-29 a^2 b^2+15 b^4\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}-\frac {6 b \left (8 a^4-11 a^2 b^2+5 b^4\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}}{a^2}}{2 a \left (a^2-b^2\right )}+\frac {b^2 \left (11 a^2-5 b^2\right ) \sin (c+d x) \sqrt {\sec (c+d x)}}{a d \left (a^2-b^2\right ) (a+b \sec (c+d x))}}{4 a \left (a^2-b^2\right )}+\frac {b^2 \sin (c+d x) \sqrt {\sec (c+d x)}}{2 a d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}\)

\(\Big \downarrow \) 4336

\(\displaystyle \frac {\frac {\frac {b^2 \left (35 a^4-38 a^2 b^2+15 b^4\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)} (b+a \cos (c+d x))}dx}{a^2}+\frac {\frac {2 a \left (8 a^4-29 a^2 b^2+15 b^4\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}-\frac {6 b \left (8 a^4-11 a^2 b^2+5 b^4\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}}{a^2}}{2 a \left (a^2-b^2\right )}+\frac {b^2 \left (11 a^2-5 b^2\right ) \sin (c+d x) \sqrt {\sec (c+d x)}}{a d \left (a^2-b^2\right ) (a+b \sec (c+d x))}}{4 a \left (a^2-b^2\right )}+\frac {b^2 \sin (c+d x) \sqrt {\sec (c+d x)}}{2 a d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {b^2 \left (35 a^4-38 a^2 b^2+15 b^4\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (b+a \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{a^2}+\frac {\frac {2 a \left (8 a^4-29 a^2 b^2+15 b^4\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}-\frac {6 b \left (8 a^4-11 a^2 b^2+5 b^4\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}}{a^2}}{2 a \left (a^2-b^2\right )}+\frac {b^2 \left (11 a^2-5 b^2\right ) \sin (c+d x) \sqrt {\sec (c+d x)}}{a d \left (a^2-b^2\right ) (a+b \sec (c+d x))}}{4 a \left (a^2-b^2\right )}+\frac {b^2 \sin (c+d x) \sqrt {\sec (c+d x)}}{2 a d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}\)

\(\Big \downarrow \) 3284

\(\displaystyle \frac {b^2 \sin (c+d x) \sqrt {\sec (c+d x)}}{2 a d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}+\frac {\frac {b^2 \left (11 a^2-5 b^2\right ) \sin (c+d x) \sqrt {\sec (c+d x)}}{a d \left (a^2-b^2\right ) (a+b \sec (c+d x))}+\frac {\frac {2 b^2 \left (35 a^4-38 a^2 b^2+15 b^4\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticPi}\left (\frac {2 a}{a+b},\frac {1}{2} (c+d x),2\right )}{a^2 d (a+b)}+\frac {\frac {2 a \left (8 a^4-29 a^2 b^2+15 b^4\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}-\frac {6 b \left (8 a^4-11 a^2 b^2+5 b^4\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}}{a^2}}{2 a \left (a^2-b^2\right )}}{4 a \left (a^2-b^2\right )}\)

input
Int[1/(Sqrt[Sec[c + d*x]]*(a + b*Sec[c + d*x])^3),x]
 
output
(b^2*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(2*a*(a^2 - b^2)*d*(a + b*Sec[c + d* 
x])^2) + ((((2*a*(8*a^4 - 29*a^2*b^2 + 15*b^4)*Sqrt[Cos[c + d*x]]*Elliptic 
E[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/d - (6*b*(8*a^4 - 11*a^2*b^2 + 5*b^4 
)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/d)/a^2 
+ (2*b^2*(35*a^4 - 38*a^2*b^2 + 15*b^4)*Sqrt[Cos[c + d*x]]*EllipticPi[(2*a 
)/(a + b), (c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(a^2*(a + b)*d))/(2*a*(a^2 
- b^2)) + (b^2*(11*a^2 - 5*b^2)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(a*(a^2 - 
 b^2)*d*(a + b*Sec[c + d*x])))/(4*a*(a^2 - b^2))
 

3.7.26.3.1 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 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 4334
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( 
a_))^(m_), x_Symbol] :> Simp[b^2*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*(a^2*(m + 1) 
 - b^2*(m + n + 1) - a*b*(m + 1)*Csc[e + f*x] + b^2*(m + n + 2)*Csc[e + f*x 
]^2), x], x] /; FreeQ[{a, b, d, e, f, n}, x] && NeQ[a^2 - b^2, 0] && LtQ[m, 
 -1] && IntegersQ[2*m, 2*n]
 

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]
 
3.7.26.4 Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1956\) vs. \(2(394)=788\).

Time = 29.91 (sec) , antiderivative size = 1957, normalized size of antiderivative = 5.72

method result size
default \(\text {Expression too large to display}\) \(1957\)

input
int(1/(a+b*sec(d*x+c))^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/a^4/(-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)*(3*b*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2 
))+a*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))+2/a^4*b^4*(1/2*a^2/b/(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*a*cos(1/2*d*x+1/2*c)^2-a+b)^2+3/4*a^2*(a^2-3*b^2)/b^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*a*cos 
(1/2*d*x+1/2*c)^2-a+b)-3/8/(a+b)/(a^2-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+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))*a^2-1/4/(a+b)/(a^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*si 
n(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))*a+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(1/2*d*x+1/2*c)^2)^(1/2) 
*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+3/8*a^3/b^2/(a^2-b^2)^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(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-9/8*a 
/(a^2-b^2)^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(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d 
*x+1/2*c),2^(1/2))-3/8*a^3/b^2/(a^2-b^2)^2*(sin(1/2*d*x+1/2*c)^2)^(1/2)...
 
3.7.26.5 Fricas [F(-1)]

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

input
integrate(1/(a+b*sec(d*x+c))^3/sec(d*x+c)^(1/2),x, algorithm="fricas")
 
output
Timed out
 
3.7.26.6 Sympy [F]

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

input
integrate(1/(a+b*sec(d*x+c))**3/sec(d*x+c)**(1/2),x)
 
output
Integral(1/((a + b*sec(c + d*x))**3*sqrt(sec(c + d*x))), x)
 
3.7.26.7 Maxima [F]

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

input
integrate(1/(a+b*sec(d*x+c))^3/sec(d*x+c)^(1/2),x, algorithm="maxima")
 
output
integrate(1/((b*sec(d*x + c) + a)^3*sqrt(sec(d*x + c))), x)
 
3.7.26.8 Giac [F]

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

input
integrate(1/(a+b*sec(d*x+c))^3/sec(d*x+c)^(1/2),x, algorithm="giac")
 
output
integrate(1/((b*sec(d*x + c) + a)^3*sqrt(sec(d*x + c))), x)
 
3.7.26.9 Mupad [F(-1)]

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

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