3.6.78 \(\int \frac {1}{(a+b \sec (c+d x))^{7/2}} \, dx\) [578]

3.6.78.1 Optimal result

Integrand size = 14, antiderivative size = 535 \[ \int \frac {1}{(a+b \sec (c+d x))^{7/2}} \, dx=\frac {2 \left (58 a^4-41 a^2 b^2+15 b^4\right ) \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{15 a^3 (a-b)^2 (a+b)^{5/2} d}-\frac {2 \left (45 a^4-13 a^3 b-36 a^2 b^2+5 a b^3+15 b^4\right ) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{15 a^3 (a-b)^2 (a+b)^{5/2} d}-\frac {2 \sqrt {a+b} \cot (c+d x) \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{a^4 d}+\frac {2 b^2 \tan (c+d x)}{5 a \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{5/2}}+\frac {2 b^2 \left (13 a^2-5 b^2\right ) \tan (c+d x)}{15 a^2 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))^{3/2}}+\frac {2 b^2 \left (58 a^4-41 a^2 b^2+15 b^4\right ) \tan (c+d x)}{15 a^3 \left (a^2-b^2\right )^3 d \sqrt {a+b \sec (c+d x)}} \]

output

2/15*(58*a^4-41*a^2*b^2+15*b^4)*cot(d*x+c)*EllipticE((a+b*sec(d*x+c))^(1/2 
)/(a+b)^(1/2),((a+b)/(a-b))^(1/2))*(b*(1-sec(d*x+c))/(a+b))^(1/2)*(-b*(1+s 
ec(d*x+c))/(a-b))^(1/2)/a^3/(a-b)^2/(a+b)^(5/2)/d-2/15*(45*a^4-13*a^3*b-36 
*a^2*b^2+5*a*b^3+15*b^4)*cot(d*x+c)*EllipticF((a+b*sec(d*x+c))^(1/2)/(a+b) 
^(1/2),((a+b)/(a-b))^(1/2))*(b*(1-sec(d*x+c))/(a+b))^(1/2)*(-b*(1+sec(d*x+ 
c))/(a-b))^(1/2)/a^3/(a-b)^2/(a+b)^(5/2)/d-2*cot(d*x+c)*EllipticPi((a+b*se 
c(d*x+c))^(1/2)/(a+b)^(1/2),(a+b)/a,((a+b)/(a-b))^(1/2))*(a+b)^(1/2)*(b*(1 
-sec(d*x+c))/(a+b))^(1/2)*(-b*(1+sec(d*x+c))/(a-b))^(1/2)/a^4/d+2/5*b^2*ta 
n(d*x+c)/a/(a^2-b^2)/d/(a+b*sec(d*x+c))^(5/2)+2/15*b^2*(13*a^2-5*b^2)*tan( 
d*x+c)/a^2/(a^2-b^2)^2/d/(a+b*sec(d*x+c))^(3/2)+2/15*b^2*(58*a^4-41*a^2*b^ 
2+15*b^4)*tan(d*x+c)/a^3/(a^2-b^2)^3/d/(a+b*sec(d*x+c))^(1/2)
 

3.6.78.2 Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(1790\) vs. \(2(535)=1070\).

Time = 13.38 (sec) , antiderivative size = 1790, normalized size of antiderivative = 3.35 \[ \int \frac {1}{(a+b \sec (c+d x))^{7/2}} \, dx =\text {Too large to display} \]

input

Integrate[(a + b*Sec[c + d*x])^(-7/2),x]
 

output

((b + a*Cos[c + d*x])^4*Sec[c + d*x]^4*((2*b*(58*a^4 - 41*a^2*b^2 + 15*b^4 
)*Sin[c + d*x])/(15*a^3*(-a^2 + b^2)^3) + (2*b^4*Sin[c + d*x])/(5*a^3*(a^2 
 - b^2)*(b + a*Cos[c + d*x])^3) + (2*(-19*a^2*b^3*Sin[c + d*x] + 11*b^5*Si 
n[c + d*x]))/(15*a^3*(a^2 - b^2)^2*(b + a*Cos[c + d*x])^2) + (2*(74*a^4*b^ 
2*Sin[c + d*x] - 65*a^2*b^4*Sin[c + d*x] + 23*b^6*Sin[c + d*x]))/(15*a^3*( 
a^2 - b^2)^3*(b + a*Cos[c + d*x]))))/(d*(a + b*Sec[c + d*x])^(7/2)) + (2*( 
b + a*Cos[c + d*x])^(7/2)*Sec[c + d*x]^(7/2)*Sqrt[(a + b - a*Tan[(c + d*x) 
/2]^2 + b*Tan[(c + d*x)/2]^2)/(1 + Tan[(c + d*x)/2]^2)]*(58*a^5*b*Tan[(c + 
 d*x)/2] + 58*a^4*b^2*Tan[(c + d*x)/2] - 41*a^3*b^3*Tan[(c + d*x)/2] - 41* 
a^2*b^4*Tan[(c + d*x)/2] + 15*a*b^5*Tan[(c + d*x)/2] + 15*b^6*Tan[(c + d*x 
)/2] - 116*a^5*b*Tan[(c + d*x)/2]^3 + 82*a^3*b^3*Tan[(c + d*x)/2]^3 - 30*a 
*b^5*Tan[(c + d*x)/2]^3 + 58*a^5*b*Tan[(c + d*x)/2]^5 - 58*a^4*b^2*Tan[(c 
+ d*x)/2]^5 - 41*a^3*b^3*Tan[(c + d*x)/2]^5 + 41*a^2*b^4*Tan[(c + d*x)/2]^ 
5 + 15*a*b^5*Tan[(c + d*x)/2]^5 - 15*b^6*Tan[(c + d*x)/2]^5 + 30*a^6*Ellip 
ticPi[-1, ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)]*Sqrt[1 - Tan[(c + d*x 
)/2]^2]*Sqrt[(a + b - a*Tan[(c + d*x)/2]^2 + b*Tan[(c + d*x)/2]^2)/(a + b) 
] - 90*a^4*b^2*EllipticPi[-1, ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)]*S 
qrt[1 - Tan[(c + d*x)/2]^2]*Sqrt[(a + b - a*Tan[(c + d*x)/2]^2 + b*Tan[(c 
+ d*x)/2]^2)/(a + b)] + 90*a^2*b^4*EllipticPi[-1, ArcSin[Tan[(c + d*x)/2]] 
, (a - b)/(a + b)]*Sqrt[1 - Tan[(c + d*x)/2]^2]*Sqrt[(a + b - a*Tan[(c ...
 

3.6.78.3 Rubi [A] (verified)

Time = 2.44 (sec) , antiderivative size = 584, normalized size of antiderivative = 1.09, number of steps used = 17, number of rules used = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.214, Rules used = {3042, 4272, 27, 3042, 4548, 27, 3042, 4548, 27, 3042, 4546, 3042, 4409, 3042, 4271, 4319, 4492}

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.

\(\displaystyle \int \frac {1}{(a+b \sec (c+d x))^{7/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{7/2}}dx\)

\(\Big \downarrow \) 4272

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4548

\(\displaystyle \frac {\frac {2 b^2 \left (13 a^2-5 b^2\right ) \tan (c+d x)}{3 a d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}-\frac {2 \int -\frac {15 \left (a^2-b^2\right )^2+b^2 \left (13 a^2-5 b^2\right ) \sec ^2(c+d x)-6 a b \left (5 a^2-b^2\right ) \sec (c+d x)}{2 (a+b \sec (c+d x))^{3/2}}dx}{3 a \left (a^2-b^2\right )}}{5 a \left (a^2-b^2\right )}+\frac {2 b^2 \tan (c+d x)}{5 a d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{5/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\int \frac {15 \left (a^2-b^2\right )^2+b^2 \left (13 a^2-5 b^2\right ) \sec ^2(c+d x)-6 a b \left (5 a^2-b^2\right ) \sec (c+d x)}{(a+b \sec (c+d x))^{3/2}}dx}{3 a \left (a^2-b^2\right )}+\frac {2 b^2 \left (13 a^2-5 b^2\right ) \tan (c+d x)}{3 a d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}}{5 a \left (a^2-b^2\right )}+\frac {2 b^2 \tan (c+d x)}{5 a d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\int \frac {15 \left (a^2-b^2\right )^2+b^2 \left (13 a^2-5 b^2\right ) \csc \left (c+d x+\frac {\pi }{2}\right )^2-6 a b \left (5 a^2-b^2\right ) \csc \left (c+d x+\frac {\pi }{2}\right )}{\left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2}}dx}{3 a \left (a^2-b^2\right )}+\frac {2 b^2 \left (13 a^2-5 b^2\right ) \tan (c+d x)}{3 a d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}}{5 a \left (a^2-b^2\right )}+\frac {2 b^2 \tan (c+d x)}{5 a d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{5/2}}\)

\(\Big \downarrow \) 4548

\(\displaystyle \frac {\frac {\frac {2 b^2 \left (58 a^4-41 a^2 b^2+15 b^4\right ) \tan (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}-\frac {2 \int -\frac {15 \left (a^2-b^2\right )^3-b^2 \left (58 a^4-41 b^2 a^2+15 b^4\right ) \sec ^2(c+d x)-a b \left (45 a^4-23 b^2 a^2+10 b^4\right ) \sec (c+d x)}{2 \sqrt {a+b \sec (c+d x)}}dx}{a \left (a^2-b^2\right )}}{3 a \left (a^2-b^2\right )}+\frac {2 b^2 \left (13 a^2-5 b^2\right ) \tan (c+d x)}{3 a d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}}{5 a \left (a^2-b^2\right )}+\frac {2 b^2 \tan (c+d x)}{5 a d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{5/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\frac {\int \frac {15 \left (a^2-b^2\right )^3-b^2 \left (58 a^4-41 b^2 a^2+15 b^4\right ) \sec ^2(c+d x)-a b \left (45 a^4-23 b^2 a^2+10 b^4\right ) \sec (c+d x)}{\sqrt {a+b \sec (c+d x)}}dx}{a \left (a^2-b^2\right )}+\frac {2 b^2 \left (58 a^4-41 a^2 b^2+15 b^4\right ) \tan (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{3 a \left (a^2-b^2\right )}+\frac {2 b^2 \left (13 a^2-5 b^2\right ) \tan (c+d x)}{3 a d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}}{5 a \left (a^2-b^2\right )}+\frac {2 b^2 \tan (c+d x)}{5 a d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {\int \frac {15 \left (a^2-b^2\right )^3-b^2 \left (58 a^4-41 b^2 a^2+15 b^4\right ) \csc \left (c+d x+\frac {\pi }{2}\right )^2-a b \left (45 a^4-23 b^2 a^2+10 b^4\right ) \csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{a \left (a^2-b^2\right )}+\frac {2 b^2 \left (58 a^4-41 a^2 b^2+15 b^4\right ) \tan (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{3 a \left (a^2-b^2\right )}+\frac {2 b^2 \left (13 a^2-5 b^2\right ) \tan (c+d x)}{3 a d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}}{5 a \left (a^2-b^2\right )}+\frac {2 b^2 \tan (c+d x)}{5 a d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{5/2}}\)

\(\Big \downarrow \) 4546

\(\displaystyle \frac {\frac {\frac {\int \frac {15 \left (a^2-b^2\right )^3+\left (b^2 \left (58 a^4-41 b^2 a^2+15 b^4\right )-a b \left (45 a^4-23 b^2 a^2+10 b^4\right )\right ) \sec (c+d x)}{\sqrt {a+b \sec (c+d x)}}dx-b^2 \left (58 a^4-41 a^2 b^2+15 b^4\right ) \int \frac {\sec (c+d x) (\sec (c+d x)+1)}{\sqrt {a+b \sec (c+d x)}}dx}{a \left (a^2-b^2\right )}+\frac {2 b^2 \left (58 a^4-41 a^2 b^2+15 b^4\right ) \tan (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{3 a \left (a^2-b^2\right )}+\frac {2 b^2 \left (13 a^2-5 b^2\right ) \tan (c+d x)}{3 a d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}}{5 a \left (a^2-b^2\right )}+\frac {2 b^2 \tan (c+d x)}{5 a d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {\int \frac {15 \left (a^2-b^2\right )^3+\left (b^2 \left (58 a^4-41 b^2 a^2+15 b^4\right )-a b \left (45 a^4-23 b^2 a^2+10 b^4\right )\right ) \csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-b^2 \left (58 a^4-41 a^2 b^2+15 b^4\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{a \left (a^2-b^2\right )}+\frac {2 b^2 \left (58 a^4-41 a^2 b^2+15 b^4\right ) \tan (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{3 a \left (a^2-b^2\right )}+\frac {2 b^2 \left (13 a^2-5 b^2\right ) \tan (c+d x)}{3 a d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}}{5 a \left (a^2-b^2\right )}+\frac {2 b^2 \tan (c+d x)}{5 a d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{5/2}}\)

\(\Big \downarrow \) 4409

\(\displaystyle \frac {\frac {\frac {15 \left (a^2-b^2\right )^3 \int \frac {1}{\sqrt {a+b \sec (c+d x)}}dx-b^2 \left (58 a^4-41 a^2 b^2+15 b^4\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-b (a-b) \left (45 a^4-13 a^3 b-36 a^2 b^2+5 a b^3+15 b^4\right ) \int \frac {\sec (c+d x)}{\sqrt {a+b \sec (c+d x)}}dx}{a \left (a^2-b^2\right )}+\frac {2 b^2 \left (58 a^4-41 a^2 b^2+15 b^4\right ) \tan (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{3 a \left (a^2-b^2\right )}+\frac {2 b^2 \left (13 a^2-5 b^2\right ) \tan (c+d x)}{3 a d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}}{5 a \left (a^2-b^2\right )}+\frac {2 b^2 \tan (c+d x)}{5 a d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {15 \left (a^2-b^2\right )^3 \int \frac {1}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-b^2 \left (58 a^4-41 a^2 b^2+15 b^4\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-b (a-b) \left (45 a^4-13 a^3 b-36 a^2 b^2+5 a b^3+15 b^4\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{a \left (a^2-b^2\right )}+\frac {2 b^2 \left (58 a^4-41 a^2 b^2+15 b^4\right ) \tan (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{3 a \left (a^2-b^2\right )}+\frac {2 b^2 \left (13 a^2-5 b^2\right ) \tan (c+d x)}{3 a d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}}{5 a \left (a^2-b^2\right )}+\frac {2 b^2 \tan (c+d x)}{5 a d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{5/2}}\)

\(\Big \downarrow \) 4271

\(\displaystyle \frac {\frac {\frac {-b^2 \left (58 a^4-41 a^2 b^2+15 b^4\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-b (a-b) \left (45 a^4-13 a^3 b-36 a^2 b^2+5 a b^3+15 b^4\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {30 \sqrt {a+b} \left (a^2-b^2\right )^3 \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{a d}}{a \left (a^2-b^2\right )}+\frac {2 b^2 \left (58 a^4-41 a^2 b^2+15 b^4\right ) \tan (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{3 a \left (a^2-b^2\right )}+\frac {2 b^2 \left (13 a^2-5 b^2\right ) \tan (c+d x)}{3 a d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}}{5 a \left (a^2-b^2\right )}+\frac {2 b^2 \tan (c+d x)}{5 a d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{5/2}}\)

\(\Big \downarrow \) 4319

\(\displaystyle \frac {\frac {\frac {-b^2 \left (58 a^4-41 a^2 b^2+15 b^4\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {30 \sqrt {a+b} \left (a^2-b^2\right )^3 \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{a d}-\frac {2 (a-b) \sqrt {a+b} \left (45 a^4-13 a^3 b-36 a^2 b^2+5 a b^3+15 b^4\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{d}}{a \left (a^2-b^2\right )}+\frac {2 b^2 \left (58 a^4-41 a^2 b^2+15 b^4\right ) \tan (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{3 a \left (a^2-b^2\right )}+\frac {2 b^2 \left (13 a^2-5 b^2\right ) \tan (c+d x)}{3 a d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}}{5 a \left (a^2-b^2\right )}+\frac {2 b^2 \tan (c+d x)}{5 a d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{5/2}}\)

\(\Big \downarrow \) 4492

\(\displaystyle \frac {2 b^2 \tan (c+d x)}{5 a d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{5/2}}+\frac {\frac {2 b^2 \left (13 a^2-5 b^2\right ) \tan (c+d x)}{3 a d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}+\frac {\frac {2 b^2 \left (58 a^4-41 a^2 b^2+15 b^4\right ) \tan (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}+\frac {-\frac {30 \sqrt {a+b} \left (a^2-b^2\right )^3 \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{a d}+\frac {2 (a-b) \sqrt {a+b} \left (58 a^4-41 a^2 b^2+15 b^4\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{d}-\frac {2 (a-b) \sqrt {a+b} \left (45 a^4-13 a^3 b-36 a^2 b^2+5 a b^3+15 b^4\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{d}}{a \left (a^2-b^2\right )}}{3 a \left (a^2-b^2\right )}}{5 a \left (a^2-b^2\right )}\)

input

Int[(a + b*Sec[c + d*x])^(-7/2),x]
 

output

(2*b^2*Tan[c + d*x])/(5*a*(a^2 - b^2)*d*(a + b*Sec[c + d*x])^(5/2)) + ((2* 
b^2*(13*a^2 - 5*b^2)*Tan[c + d*x])/(3*a*(a^2 - b^2)*d*(a + b*Sec[c + d*x]) 
^(3/2)) + (((2*(a - b)*Sqrt[a + b]*(58*a^4 - 41*a^2*b^2 + 15*b^4)*Cot[c + 
d*x]*EllipticE[ArcSin[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a + b]], (a + b)/(a - 
b)]*Sqrt[(b*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[-((b*(1 + Sec[c + d*x]))/(a 
- b))])/d - (2*(a - b)*Sqrt[a + b]*(45*a^4 - 13*a^3*b - 36*a^2*b^2 + 5*a*b 
^3 + 15*b^4)*Cot[c + d*x]*EllipticF[ArcSin[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a 
 + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[-((b*(1 
 + Sec[c + d*x]))/(a - b))])/d - (30*Sqrt[a + b]*(a^2 - b^2)^3*Cot[c + d*x 
]*EllipticPi[(a + b)/a, ArcSin[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a + b]], (a + 
 b)/(a - b)]*Sqrt[(b*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[-((b*(1 + Sec[c + d 
*x]))/(a - b))])/(a*d))/(a*(a^2 - b^2)) + (2*b^2*(58*a^4 - 41*a^2*b^2 + 15 
*b^4)*Tan[c + d*x])/(a*(a^2 - b^2)*d*Sqrt[a + b*Sec[c + d*x]]))/(3*a*(a^2 
- b^2)))/(5*a*(a^2 - b^2))
 

3.6.78.3.1 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 4271

Int[1/Sqrt[csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[2*(Rt[a 
 + b, 2]/(a*d*Cot[c + d*x]))*Sqrt[b*((1 - Csc[c + d*x])/(a + b))]*Sqrt[(-b) 
*((1 + Csc[c + d*x])/(a - b))]*EllipticPi[(a + b)/a, ArcSin[Sqrt[a + b*Csc[ 
c + d*x]]/Rt[a + b, 2]], (a + b)/(a - b)], x] /; FreeQ[{a, b, c, d}, x] && 
NeQ[a^2 - b^2, 0]
 

rule 4272

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> Simp[b^2*Cot[ 
c + d*x]*((a + b*Csc[c + d*x])^(n + 1)/(a*d*(n + 1)*(a^2 - b^2))), x] + Sim 
p[1/(a*(n + 1)*(a^2 - b^2))   Int[(a + b*Csc[c + d*x])^(n + 1)*Simp[(a^2 - 
b^2)*(n + 1) - a*b*(n + 1)*Csc[c + d*x] + b^2*(n + 2)*Csc[c + d*x]^2, x], x 
], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && LtQ[n, -1] && Integ 
erQ[2*n]
 

rule 4319

Int[csc[(e_.) + (f_.)*(x_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_S 
ymbol] :> Simp[-2*(Rt[a + b, 2]/(b*f*Cot[e + f*x]))*Sqrt[(b*(1 - Csc[e + f* 
x]))/(a + b)]*Sqrt[(-b)*((1 + Csc[e + f*x])/(a - b))]*EllipticF[ArcSin[Sqrt 
[a + b*Csc[e + f*x]]/Rt[a + b, 2]], (a + b)/(a - b)], x] /; FreeQ[{a, b, e, 
 f}, x] && NeQ[a^2 - b^2, 0]
 

rule 4409

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_ 
.) + (a_)], x_Symbol] :> Simp[c   Int[1/Sqrt[a + b*Csc[e + f*x]], x], x] + 
Simp[d   Int[Csc[e + f*x]/Sqrt[a + b*Csc[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]
 

rule 4492

Int[(csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)))/Sqrt[c 
sc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[-2*(A*b - a*B)*Rt[a 
 + b*(B/A), 2]*Sqrt[b*((1 - Csc[e + f*x])/(a + b))]*(Sqrt[(-b)*((1 + Csc[e 
+ f*x])/(a - b))]/(b^2*f*Cot[e + f*x]))*EllipticE[ArcSin[Sqrt[a + b*Csc[e + 
 f*x]]/Rt[a + b*(B/A), 2]], (a*A + b*B)/(a*A - b*B)], x] /; FreeQ[{a, b, e, 
 f, A, B}, x] && NeQ[a^2 - b^2, 0] && EqQ[A^2 - B^2, 0]
 

rule 4546

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_. 
))/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Int[(A + (B - C 
)*Csc[e + f*x])/Sqrt[a + b*Csc[e + f*x]], x] + Simp[C   Int[Csc[e + f*x]*(( 
1 + Csc[e + f*x])/Sqrt[a + b*Csc[e + f*x]]), x], x] /; FreeQ[{a, b, e, f, A 
, B, C}, x] && NeQ[a^2 - b^2, 0]
 

rule 4548

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_. 
))*(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)/(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)*Simp[A*(a^2 - b^2)*(m + 1) - a*(A*b - a*B + b*C)*(m + 1)*Csc[e + f*x 
] + (A*b^2 - a*b*B + a^2*C)*(m + 2)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, 
 b, e, f, A, B, C}, x] && NeQ[a^2 - b^2, 0] && LtQ[m, -1]
 

3.6.78.4 Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(9414\) vs. \(2(492)=984\).

Time = 8.58 (sec) , antiderivative size = 9415, normalized size of antiderivative = 17.60

method	result	size
default	\(\text {Expression too large to display}\)	\(9415\)

input

int(1/(a+b*sec(d*x+c))^(7/2),x,method=_RETURNVERBOSE)
 

output

result too large to display
 

3.6.78.5 Fricas [F]

\[ \int \frac {1}{(a+b \sec (c+d x))^{7/2}} \, dx=\int { \frac {1}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {7}{2}}} \,d x } \]

input

integrate(1/(a+b*sec(d*x+c))^(7/2),x, algorithm="fricas")
 

output

integral(sqrt(b*sec(d*x + c) + a)/(b^4*sec(d*x + c)^4 + 4*a*b^3*sec(d*x + 
c)^3 + 6*a^2*b^2*sec(d*x + c)^2 + 4*a^3*b*sec(d*x + c) + a^4), x)
 

3.6.78.6 Sympy [F]

\[ \int \frac {1}{(a+b \sec (c+d x))^{7/2}} \, dx=\int \frac {1}{\left (a + b \sec {\left (c + d x \right )}\right )^{\frac {7}{2}}}\, dx \]

input

integrate(1/(a+b*sec(d*x+c))**(7/2),x)
 

output

Integral((a + b*sec(c + d*x))**(-7/2), x)
 

3.6.78.7 Maxima [F]

\[ \int \frac {1}{(a+b \sec (c+d x))^{7/2}} \, dx=\int { \frac {1}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {7}{2}}} \,d x } \]

input

integrate(1/(a+b*sec(d*x+c))^(7/2),x, algorithm="maxima")
 

output

integrate((b*sec(d*x + c) + a)^(-7/2), x)
 

3.6.78.8 Giac [F]

\[ \int \frac {1}{(a+b \sec (c+d x))^{7/2}} \, dx=\int { \frac {1}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {7}{2}}} \,d x } \]

input

integrate(1/(a+b*sec(d*x+c))^(7/2),x, algorithm="giac")
 

output

integrate((b*sec(d*x + c) + a)^(-7/2), x)
 

3.6.78.9 Mupad [F(-1)]

Timed out. \[ \int \frac {1}{(a+b \sec (c+d x))^{7/2}} \, dx=\int \frac {1}{{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^{7/2}} \,d x \]

input

int(1/(a + b/cos(c + d*x))^(7/2),x)
 

output

int(1/(a + b/cos(c + d*x))^(7/2), x)