\(\int \frac {\sec ^3(c+d x) (B \sec (c+d x)+C \sec ^2(c+d x))}{(a+b \sec (c+d x))^{5/2}} \, dx\) [849]

Optimal result

Integrand size = 42, antiderivative size = 509 \[ \int \frac {\sec ^3(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{5/2}} \, dx=-\frac {2 \left (8 a^4 b B-15 a^2 b^3 B+3 b^5 B-16 a^5 C+28 a^3 b^2 C-8 a b^4 C\right ) \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{3 (a-b) b^5 (a+b)^{3/2} d}-\frac {2 \left (a^3 b (8 B-12 C)-9 a b^3 (B-C)-b^4 (3 B-C)-16 a^4 C+2 a^2 b^2 (3 B+8 C)\right ) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{3 b^4 \sqrt {a+b} \left (a^2-b^2\right ) d}+\frac {2 a (b B-a C) \sec ^2(c+d x) \tan (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}-\frac {2 a^2 \left (3 a^2 b B-7 b^3 B-6 a^3 C+10 a b^2 C\right ) \tan (c+d x)}{3 b^3 \left (a^2-b^2\right )^2 d \sqrt {a+b \sec (c+d x)}}-\frac {2 \left (a b B-2 a^2 C+b^2 C\right ) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{3 b^3 \left (a^2-b^2\right ) d} \] Output:

-2/3*(8*B*a^4*b-15*B*a^2*b^3+3*B*b^5-16*C*a^5+28*C*a^3*b^2-8*C*a*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+sec(d*x+c))/(a-b))^(1/2)/(a-b)/b^5/(a 
+b)^(3/2)/d-2/3*(a^3*b*(8*B-12*C)-9*a*b^3*(B-C)-b^4*(3*B-C)-16*a^4*C+2*a^2 
*b^2*(3*B+8*C))*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)/b^4/(a+b)^(1/2)/(a^2-b^2)/d+2/3*a*(B*b-C*a)*sec(d*x+c)^2*tan(d*x+c 
)/b/(a^2-b^2)/d/(a+b*sec(d*x+c))^(3/2)-2/3*a^2*(3*B*a^2*b-7*B*b^3-6*C*a^3+ 
10*C*a*b^2)*tan(d*x+c)/b^3/(a^2-b^2)^2/d/(a+b*sec(d*x+c))^(1/2)-2/3*(B*a*b 
-2*C*a^2+C*b^2)*(a+b*sec(d*x+c))^(1/2)*tan(d*x+c)/b^3/(a^2-b^2)/d
 

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(4342\) vs. \(2(509)=1018\).

Time = 24.35 (sec) , antiderivative size = 4342, normalized size of antiderivative = 8.53 \[ \int \frac {\sec ^3(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{5/2}} \, dx=\text {Result too large to show} \] Input:

Integrate[(Sec[c + d*x]^3*(B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(a + b*Sec[ 
c + d*x])^(5/2),x]
 

Output:

((b + a*Cos[c + d*x])^3*Sec[c + d*x]^3*((2*(8*a^4*b*B - 15*a^2*b^3*B + 3*b 
^5*B - 16*a^5*C + 28*a^3*b^2*C - 8*a*b^4*C)*Sin[c + d*x])/(3*b^4*(-a^2 + b 
^2)^2) + (2*(a^2*b*B*Sin[c + d*x] - a^3*C*Sin[c + d*x]))/(3*b^2*(-a^2 + b^ 
2)*(b + a*Cos[c + d*x])^2) + (2*(-4*a^4*b*B*Sin[c + d*x] + 8*a^2*b^3*B*Sin 
[c + d*x] + 7*a^5*C*Sin[c + d*x] - 11*a^3*b^2*C*Sin[c + d*x]))/(3*b^3*(-a^ 
2 + b^2)^2*(b + a*Cos[c + d*x])) + (2*C*Tan[c + d*x])/(3*b^3)))/(d*(a + b* 
Sec[c + d*x])^(5/2)) + (2*(b + a*Cos[c + d*x])^2*((5*a^2*B)/((-a^2 + b^2)^ 
2*Sqrt[b + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) - (8*a^4*B)/(3*b^2*(-a^2 + 
b^2)^2*Sqrt[b + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) - (b^2*B)/((-a^2 + b^2 
)^2*Sqrt[b + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) + (16*a^5*C)/(3*b^3*(-a^2 
 + b^2)^2*Sqrt[b + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) - (28*a^3*C)/(3*b*( 
-a^2 + b^2)^2*Sqrt[b + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) + (8*a*b*C)/(3* 
(-a^2 + b^2)^2*Sqrt[b + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) - (8*a^5*B*Sqr 
t[Sec[c + d*x]])/(3*b^3*(-a^2 + b^2)^2*Sqrt[b + a*Cos[c + d*x]]) + (17*a^3 
*B*Sqrt[Sec[c + d*x]])/(3*b*(-a^2 + b^2)^2*Sqrt[b + a*Cos[c + d*x]]) - (3* 
a*b*B*Sqrt[Sec[c + d*x]])/((-a^2 + b^2)^2*Sqrt[b + a*Cos[c + d*x]]) + (5*a 
^2*C*Sqrt[Sec[c + d*x]])/((-a^2 + b^2)^2*Sqrt[b + a*Cos[c + d*x]]) + (16*a 
^6*C*Sqrt[Sec[c + d*x]])/(3*b^4*(-a^2 + b^2)^2*Sqrt[b + a*Cos[c + d*x]]) - 
 (32*a^4*C*Sqrt[Sec[c + d*x]])/(3*b^2*(-a^2 + b^2)^2*Sqrt[b + a*Cos[c + d* 
x]]) + (b^2*C*Sqrt[Sec[c + d*x]])/(3*(-a^2 + b^2)^2*Sqrt[b + a*Cos[c + ...
 

Rubi [A] (verified)

Time = 2.47 (sec) , antiderivative size = 525, normalized size of antiderivative = 1.03, number of steps used = 16, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.381, Rules used = {3042, 4560, 3042, 4517, 27, 3042, 4578, 27, 3042, 4570, 27, 3042, 4493, 3042, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4560

\(\displaystyle \int \frac {\sec ^4(c+d x) (B+C \sec (c+d x))}{(a+b \sec (c+d x))^{5/2}}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^4 \left (B+C \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{5/2}}dx\)

\(\Big \downarrow \) 4517

\(\displaystyle \frac {2 \int \frac {\sec ^2(c+d x) \left (-3 \left (-2 C a^2+b B a+b^2 C\right ) \sec ^2(c+d x)-3 b (b B-a C) \sec (c+d x)+4 a (b B-a C)\right )}{2 (a+b \sec (c+d x))^{3/2}}dx}{3 b \left (a^2-b^2\right )}+\frac {2 a (b B-a C) \tan (c+d x) \sec ^2(c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {\sec ^2(c+d x) \left (-3 \left (-2 C a^2+b B a+b^2 C\right ) \sec ^2(c+d x)-3 b (b B-a C) \sec (c+d x)+4 a (b B-a C)\right )}{(a+b \sec (c+d x))^{3/2}}dx}{3 b \left (a^2-b^2\right )}+\frac {2 a (b B-a C) \tan (c+d x) \sec ^2(c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^2 \left (-3 \left (-2 C a^2+b B a+b^2 C\right ) \csc \left (c+d x+\frac {\pi }{2}\right )^2-3 b (b B-a C) \csc \left (c+d x+\frac {\pi }{2}\right )+4 a (b B-a C)\right )}{\left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2}}dx}{3 b \left (a^2-b^2\right )}+\frac {2 a (b B-a C) \tan (c+d x) \sec ^2(c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}\)

\(\Big \downarrow \) 4578

\(\displaystyle \frac {\frac {2 \int \frac {\sec (c+d x) \left (-3 b \left (a^2-b^2\right ) \left (-2 C a^2+b B a+b^2 C\right ) \sec ^2(c+d x)+\left (-12 C a^5+6 b B a^4+22 b^2 C a^3-13 b^3 B a^2-6 b^4 C a+3 b^5 B\right ) \sec (c+d x)+a b \left (-6 C a^3+3 b B a^2+10 b^2 C a-7 b^3 B\right )\right )}{2 \sqrt {a+b \sec (c+d x)}}dx}{b^2 \left (a^2-b^2\right )}-\frac {2 a^2 \left (-6 a^3 C+3 a^2 b B+10 a b^2 C-7 b^3 B\right ) \tan (c+d x)}{b^2 d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{3 b \left (a^2-b^2\right )}+\frac {2 a (b B-a C) \tan (c+d x) \sec ^2(c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\int \frac {\sec (c+d x) \left (-3 b \left (a^2-b^2\right ) \left (-2 C a^2+b B a+b^2 C\right ) \sec ^2(c+d x)+\left (-12 C a^5+6 b B a^4+22 b^2 C a^3-13 b^3 B a^2-6 b^4 C a+3 b^5 B\right ) \sec (c+d x)+a b \left (-6 C a^3+3 b B a^2+10 b^2 C a-7 b^3 B\right )\right )}{\sqrt {a+b \sec (c+d x)}}dx}{b^2 \left (a^2-b^2\right )}-\frac {2 a^2 \left (-6 a^3 C+3 a^2 b B+10 a b^2 C-7 b^3 B\right ) \tan (c+d x)}{b^2 d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{3 b \left (a^2-b^2\right )}+\frac {2 a (b B-a C) \tan (c+d x) \sec ^2(c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (-3 b \left (a^2-b^2\right ) \left (-2 C a^2+b B a+b^2 C\right ) \csc \left (c+d x+\frac {\pi }{2}\right )^2+\left (-12 C a^5+6 b B a^4+22 b^2 C a^3-13 b^3 B a^2-6 b^4 C a+3 b^5 B\right ) \csc \left (c+d x+\frac {\pi }{2}\right )+a b \left (-6 C a^3+3 b B a^2+10 b^2 C a-7 b^3 B\right )\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{b^2 \left (a^2-b^2\right )}-\frac {2 a^2 \left (-6 a^3 C+3 a^2 b B+10 a b^2 C-7 b^3 B\right ) \tan (c+d x)}{b^2 d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{3 b \left (a^2-b^2\right )}+\frac {2 a (b B-a C) \tan (c+d x) \sec ^2(c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}\)

\(\Big \downarrow \) 4570

\(\displaystyle \frac {\frac {\frac {2 \int \frac {3 \sec (c+d x) \left (\left (-4 C a^4+2 b B a^3+7 b^2 C a^2-6 b^3 B a+b^4 C\right ) b^2+\left (-16 C a^5+8 b B a^4+28 b^2 C a^3-15 b^3 B a^2-8 b^4 C a+3 b^5 B\right ) \sec (c+d x) b\right )}{2 \sqrt {a+b \sec (c+d x)}}dx}{3 b}-\frac {2 \left (a^2-b^2\right ) \left (-2 a^2 C+a b B+b^2 C\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{d}}{b^2 \left (a^2-b^2\right )}-\frac {2 a^2 \left (-6 a^3 C+3 a^2 b B+10 a b^2 C-7 b^3 B\right ) \tan (c+d x)}{b^2 d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{3 b \left (a^2-b^2\right )}+\frac {2 a (b B-a C) \tan (c+d x) \sec ^2(c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\frac {\int \frac {\sec (c+d x) \left (\left (-4 C a^4+2 b B a^3+7 b^2 C a^2-6 b^3 B a+b^4 C\right ) b^2+\left (-16 C a^5+8 b B a^4+28 b^2 C a^3-15 b^3 B a^2-8 b^4 C a+3 b^5 B\right ) \sec (c+d x) b\right )}{\sqrt {a+b \sec (c+d x)}}dx}{b}-\frac {2 \left (a^2-b^2\right ) \left (-2 a^2 C+a b B+b^2 C\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{d}}{b^2 \left (a^2-b^2\right )}-\frac {2 a^2 \left (-6 a^3 C+3 a^2 b B+10 a b^2 C-7 b^3 B\right ) \tan (c+d x)}{b^2 d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{3 b \left (a^2-b^2\right )}+\frac {2 a (b B-a C) \tan (c+d x) \sec ^2(c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\left (-4 C a^4+2 b B a^3+7 b^2 C a^2-6 b^3 B a+b^4 C\right ) b^2+\left (-16 C a^5+8 b B a^4+28 b^2 C a^3-15 b^3 B a^2-8 b^4 C a+3 b^5 B\right ) \csc \left (c+d x+\frac {\pi }{2}\right ) b\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}-\frac {2 \left (a^2-b^2\right ) \left (-2 a^2 C+a b B+b^2 C\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{d}}{b^2 \left (a^2-b^2\right )}-\frac {2 a^2 \left (-6 a^3 C+3 a^2 b B+10 a b^2 C-7 b^3 B\right ) \tan (c+d x)}{b^2 d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{3 b \left (a^2-b^2\right )}+\frac {2 a (b B-a C) \tan (c+d x) \sec ^2(c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}\)

\(\Big \downarrow \) 4493

\(\displaystyle \frac {\frac {\frac {b \left (-16 a^5 C+8 a^4 b B+28 a^3 b^2 C-15 a^2 b^3 B-8 a b^4 C+3 b^5 B\right ) \int \frac {\sec (c+d x) (\sec (c+d x)+1)}{\sqrt {a+b \sec (c+d x)}}dx-b (a-b) \left (-16 a^4 C+a^3 b (8 B-12 C)+2 a^2 b^2 (3 B+8 C)-9 a b^3 (B-C)-b^4 (3 B-C)\right ) \int \frac {\sec (c+d x)}{\sqrt {a+b \sec (c+d x)}}dx}{b}-\frac {2 \left (a^2-b^2\right ) \left (-2 a^2 C+a b B+b^2 C\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{d}}{b^2 \left (a^2-b^2\right )}-\frac {2 a^2 \left (-6 a^3 C+3 a^2 b B+10 a b^2 C-7 b^3 B\right ) \tan (c+d x)}{b^2 d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{3 b \left (a^2-b^2\right )}+\frac {2 a (b B-a C) \tan (c+d x) \sec ^2(c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {b \left (-16 a^5 C+8 a^4 b B+28 a^3 b^2 C-15 a^2 b^3 B-8 a b^4 C+3 b^5 B\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-b (a-b) \left (-16 a^4 C+a^3 b (8 B-12 C)+2 a^2 b^2 (3 B+8 C)-9 a b^3 (B-C)-b^4 (3 B-C)\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}-\frac {2 \left (a^2-b^2\right ) \left (-2 a^2 C+a b B+b^2 C\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{d}}{b^2 \left (a^2-b^2\right )}-\frac {2 a^2 \left (-6 a^3 C+3 a^2 b B+10 a b^2 C-7 b^3 B\right ) \tan (c+d x)}{b^2 d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{3 b \left (a^2-b^2\right )}+\frac {2 a (b B-a C) \tan (c+d x) \sec ^2(c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}\)

\(\Big \downarrow \) 4319

\(\displaystyle \frac {\frac {\frac {b \left (-16 a^5 C+8 a^4 b B+28 a^3 b^2 C-15 a^2 b^3 B-8 a b^4 C+3 b^5 B\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 (a-b) \sqrt {a+b} \left (-16 a^4 C+a^3 b (8 B-12 C)+2 a^2 b^2 (3 B+8 C)-9 a b^3 (B-C)-b^4 (3 B-C)\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{d}}{b}-\frac {2 \left (a^2-b^2\right ) \left (-2 a^2 C+a b B+b^2 C\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{d}}{b^2 \left (a^2-b^2\right )}-\frac {2 a^2 \left (-6 a^3 C+3 a^2 b B+10 a b^2 C-7 b^3 B\right ) \tan (c+d x)}{b^2 d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{3 b \left (a^2-b^2\right )}+\frac {2 a (b B-a C) \tan (c+d x) \sec ^2(c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}\)

\(\Big \downarrow \) 4492

\(\displaystyle \frac {2 a (b B-a C) \tan (c+d x) \sec ^2(c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}+\frac {\frac {\frac {-\frac {2 (a-b) \sqrt {a+b} \left (-16 a^4 C+a^3 b (8 B-12 C)+2 a^2 b^2 (3 B+8 C)-9 a b^3 (B-C)-b^4 (3 B-C)\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{d}-\frac {2 (a-b) \sqrt {a+b} \left (-16 a^5 C+8 a^4 b B+28 a^3 b^2 C-15 a^2 b^3 B-8 a b^4 C+3 b^5 B\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{b d}}{b}-\frac {2 \left (a^2-b^2\right ) \left (-2 a^2 C+a b B+b^2 C\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{d}}{b^2 \left (a^2-b^2\right )}-\frac {2 a^2 \left (-6 a^3 C+3 a^2 b B+10 a b^2 C-7 b^3 B\right ) \tan (c+d x)}{b^2 d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{3 b \left (a^2-b^2\right )}\)

Input:

Int[(Sec[c + d*x]^3*(B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(a + b*Sec[c + d* 
x])^(5/2),x]
 

Output:

(2*a*(b*B - a*C)*Sec[c + d*x]^2*Tan[c + d*x])/(3*b*(a^2 - b^2)*d*(a + b*Se 
c[c + d*x])^(3/2)) + ((-2*a^2*(3*a^2*b*B - 7*b^3*B - 6*a^3*C + 10*a*b^2*C) 
*Tan[c + d*x])/(b^2*(a^2 - b^2)*d*Sqrt[a + b*Sec[c + d*x]]) + (((-2*(a - b 
)*Sqrt[a + b]*(8*a^4*b*B - 15*a^2*b^3*B + 3*b^5*B - 16*a^5*C + 28*a^3*b^2* 
C - 8*a*b^4*C)*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))])/(b*d) - (2*(a - b)*Sqrt[a + b]*(a^3*b*(8*B 
- 12*C) - 9*a*b^3*(B - C) - b^4*(3*B - C) - 16*a^4*C + 2*a^2*b^2*(3*B + 8* 
C))*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)/b - (2*(a^2 - b^2)*(a*b*B - 2*a^2*C + b^2*C)*Sqrt[a 
 + b*Sec[c + d*x]]*Tan[c + d*x])/d)/(b^2*(a^2 - b^2)))/(3*b*(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 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 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 4493

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

rule 4517

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

rule 4560

Int[((a_.) + csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*((A_.) + csc[(e_.) + (f_. 
)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*((c_.) + csc[(e_.) + (f_.) 
*(x_)]*(d_.))^(n_.), x_Symbol] :> Simp[1/b^2   Int[(a + b*Csc[e + f*x])^(m 
+ 1)*(c + d*Csc[e + f*x])^n*(b*B - a*C + b*C*Csc[e + f*x]), x], x] /; FreeQ 
[{a, b, c, d, e, f, A, B, C, m, n}, x] && EqQ[A*b^2 - a*b*B + a^2*C, 0]
 

rule 4570

Int[csc[(e_.) + (f_.)*(x_)]*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e 
_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_S 
ymbol] :> Simp[(-C)*Cot[e + f*x]*((a + b*Csc[e + f*x])^(m + 1)/(b*f*(m + 2) 
)), x] + Simp[1/(b*(m + 2))   Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^m*Simp[ 
b*A*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Csc[e + f*x], x], x], x] /; 
 FreeQ[{a, b, e, f, A, B, C, m}, x] &&  !LtQ[m, -1]
 

rule 4578

Int[csc[(e_.) + (f_.)*(x_)]^2*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[ 
(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x 
_Symbol] :> Simp[a*(A*b^2 - a*b*B + a^2*C)*Cot[e + f*x]*((a + b*Csc[e + f*x 
])^(m + 1)/(b^2*f*(m + 1)*(a^2 - b^2))), x] - Simp[1/(b^2*(m + 1)*(a^2 - b^ 
2))   Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^(m + 1)*Simp[b*(m + 1)*((-a)*(b 
*B - a*C) + A*b^2) + (b*B*(a^2 + b^2*(m + 1)) - a*(A*b^2*(m + 2) + C*(a^2 + 
 b^2*(m + 1))))*Csc[e + f*x] - b*C*(m + 1)*(a^2 - b^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 
]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(3490\) vs. \(2(475)=950\).

Time = 68.24 (sec) , antiderivative size = 3491, normalized size of antiderivative = 6.86

method	result	size
default	\(\text {Expression too large to display}\)	\(3491\)
parts	\(\text {Expression too large to display}\)	\(3515\)

Input:

int(sec(d*x+c)^3*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^(5/2),x,me 
thod=_RETURNVERBOSE)
 

Output:

2/3/d/(a-b)^2/(a+b)^2/b^4*(a+b*sec(d*x+c))^(1/2)/(cos(d*x+c)+1)/(a^2*cos(d 
*x+c)^2+2*a*b*cos(d*x+c)+b^2)*(sin(d*x+c)*(-15*cos(d*x+c)^2-5*cos(d*x+c)+3 
)*B*a^4*b^3+3*(cos(d*x+c)^3+3*cos(d*x+c)^2+3*cos(d*x+c)+1)*B*(1/(a+b)*(b+a 
*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*a*b^6 
*EllipticE(-csc(d*x+c)+cot(d*x+c),((a-b)/(a+b))^(1/2))+(-15*cos(d*x+c)^3-2 
2*cos(d*x+c)^2+cos(d*x+c)+8)*B*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^( 
1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*a^4*b^3*EllipticE(-csc(d*x+c)+cot(d 
*x+c),((a-b)/(a+b))^(1/2))+(15*cos(d*x+c)^3+28*cos(d*x+c)^2+11*cos(d*x+c)- 
2)*B*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+ 
c)+1))^(1/2)*a^3*b^4*EllipticF(-csc(d*x+c)+cot(d*x+c),((a-b)/(a+b))^(1/2)) 
+(-cos(d*x+c)^2-2*cos(d*x+c)-1)*C*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1) 
)^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*b^7*EllipticF(-csc(d*x+c)+cot(d* 
x+c),((a-b)/(a+b))^(1/2))+8*sin(d*x+c)*cos(d*x+c)*(cos(d*x+c)-3)*C*b*a^6+1 
6*C*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c 
)+1))^(1/2)*a^7*EllipticE(-csc(d*x+c)+cot(d*x+c),((a-b)/(a+b))^(1/2))*(-co 
s(d*x+c)^3-2*cos(d*x+c)^2-cos(d*x+c))+8*B*a^6*b*cos(d*x+c)^2*sin(d*x+c)+6* 
B*a*b^6*cos(d*x+c)*sin(d*x+c)+3*sin(d*x+c)*(cos(d*x+c)^2+3*cos(d*x+c)-2)*a 
^2*b^5*B+(-13*cos(d*x+c)^3+43*cos(d*x+c)^2+cos(d*x+c)+1)*C*a^4*b^3*tan(d*x 
+c)+(cos(d*x+c)^3-15*cos(d*x+c)^2-2*cos(d*x+c)-2)*C*a^2*b^5*tan(d*x+c)+2*s 
in(d*x+c)*(-3+14*cos(d*x+c)^2+5*cos(d*x+c))*a^5*b^2*C+4*sin(d*x+c)*(3-2...
 

Fricas [F]

\[ \int \frac {\sec ^3(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{5/2}} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right )\right )} \sec \left (d x + c\right )^{3}}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}}} \,d x } \] Input:

integrate(sec(d*x+c)^3*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^(5/2 
),x, algorithm="fricas")
 

Output:

integral((C*sec(d*x + c)^5 + B*sec(d*x + c)^4)*sqrt(b*sec(d*x + c) + a)/(b 
^3*sec(d*x + c)^3 + 3*a*b^2*sec(d*x + c)^2 + 3*a^2*b*sec(d*x + c) + a^3), 
x)
 

Sympy [F]

\[ \int \frac {\sec ^3(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{5/2}} \, dx=\int \frac {\left (B + C \sec {\left (c + d x \right )}\right ) \sec ^{4}{\left (c + d x \right )}}{\left (a + b \sec {\left (c + d x \right )}\right )^{\frac {5}{2}}}\, dx \] Input:

integrate(sec(d*x+c)**3*(B*sec(d*x+c)+C*sec(d*x+c)**2)/(a+b*sec(d*x+c))**( 
5/2),x)
 

Output:

Integral((B + C*sec(c + d*x))*sec(c + d*x)**4/(a + b*sec(c + d*x))**(5/2), 
 x)
 

Maxima [F(-1)]

Timed out. \[ \int \frac {\sec ^3(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{5/2}} \, dx=\text {Timed out} \] Input:

integrate(sec(d*x+c)^3*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^(5/2 
),x, algorithm="maxima")
 

Output:

Timed out
 

Giac [F]

\[ \int \frac {\sec ^3(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{5/2}} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right )\right )} \sec \left (d x + c\right )^{3}}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}}} \,d x } \] Input:

integrate(sec(d*x+c)^3*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^(5/2 
),x, algorithm="giac")
 

Output:

integrate((C*sec(d*x + c)^2 + B*sec(d*x + c))*sec(d*x + c)^3/(b*sec(d*x + 
c) + a)^(5/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\sec ^3(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{5/2}} \, dx=\int \frac {\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}}{{\cos \left (c+d\,x\right )}^3\,{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^{5/2}} \,d x \] Input:

int((B/cos(c + d*x) + C/cos(c + d*x)^2)/(cos(c + d*x)^3*(a + b/cos(c + d*x 
))^(5/2)),x)
 

Output:

int((B/cos(c + d*x) + C/cos(c + d*x)^2)/(cos(c + d*x)^3*(a + b/cos(c + d*x 
))^(5/2)), x)
 

Reduce [F]

\[ \int \frac {\sec ^3(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{5/2}} \, dx=\left (\int \frac {\sqrt {\sec \left (d x +c \right ) b +a}\, \sec \left (d x +c \right )^{5}}{\sec \left (d x +c \right )^{3} b^{3}+3 \sec \left (d x +c \right )^{2} a \,b^{2}+3 \sec \left (d x +c \right ) a^{2} b +a^{3}}d x \right ) c +\left (\int \frac {\sqrt {\sec \left (d x +c \right ) b +a}\, \sec \left (d x +c \right )^{4}}{\sec \left (d x +c \right )^{3} b^{3}+3 \sec \left (d x +c \right )^{2} a \,b^{2}+3 \sec \left (d x +c \right ) a^{2} b +a^{3}}d x \right ) b \] Input:

int(sec(d*x+c)^3*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^(5/2),x)
 

Output:

int((sqrt(sec(c + d*x)*b + a)*sec(c + d*x)**5)/(sec(c + d*x)**3*b**3 + 3*s 
ec(c + d*x)**2*a*b**2 + 3*sec(c + d*x)*a**2*b + a**3),x)*c + int((sqrt(sec 
(c + d*x)*b + a)*sec(c + d*x)**4)/(sec(c + d*x)**3*b**3 + 3*sec(c + d*x)** 
2*a*b**2 + 3*sec(c + d*x)*a**2*b + a**3),x)*b