\(\int \frac {\sec (c+d x) (A+C \sec ^2(c+d x))}{(a+b \sec (c+d x))^{5/2}} \, dx\) [750]

Optimal result

Integrand size = 33, antiderivative size = 378 \[ \int \frac {\sec (c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{5/2}} \, dx=-\frac {4 a \left (2 A b^2-a^2 C+3 b^2 C\right ) \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{3 (a-b) b^3 (a+b)^{3/2} d}+\frac {2 \left (2 a^2 C+3 a b (A+C)-b^2 (A+3 C)\right ) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{3 b^2 \sqrt {a+b} \left (a^2-b^2\right ) d}-\frac {2 \left (A b^2+a^2 C\right ) \tan (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}-\frac {4 a \left (2 A b^2-a^2 C+3 b^2 C\right ) \tan (c+d x)}{3 b \left (a^2-b^2\right )^2 d \sqrt {a+b \sec (c+d x)}} \] Output:

-4/3*a*(2*A*b^2-C*a^2+3*C*b^2)*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+se 
c(d*x+c))/(a-b))^(1/2)/(a-b)/b^3/(a+b)^(3/2)/d+2/3*(2*C*a^2+3*a*b*(A+C)-b^ 
2*(A+3*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^2/(a+b)^(1/2)/(a^2-b^2)/d-2/3*(A*b^2+C*a^2)*tan(d*x+c)/b/(a^2-b^2)/d/ 
(a+b*sec(d*x+c))^(3/2)-4/3*a*(2*A*b^2-C*a^2+3*C*b^2)*tan(d*x+c)/b/(a^2-b^2 
)^2/d/(a+b*sec(d*x+c))^(1/2)
 

Mathematica [A] (warning: unable to verify)

Time = 16.15 (sec) , antiderivative size = 633, normalized size of antiderivative = 1.67 \[ \int \frac {\sec (c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{5/2}} \, dx=\frac {(b+a \cos (c+d x))^3 \sec (c+d x) \left (A+C \sec ^2(c+d x)\right ) \left (-\frac {8 a \left (-2 A b^2+a^2 C-3 b^2 C\right ) \sin (c+d x)}{3 b^2 \left (a^2-b^2\right )^2}+\frac {4 \left (A b^2 \sin (c+d x)+a^2 C \sin (c+d x)\right )}{3 a \left (a^2-b^2\right ) (b+a \cos (c+d x))^2}+\frac {4 \left (-5 a^2 A b^2 \sin (c+d x)+A b^4 \sin (c+d x)+a^4 C \sin (c+d x)-5 a^2 b^2 C \sin (c+d x)\right )}{3 a b \left (-a^2+b^2\right )^2 (b+a \cos (c+d x))}\right )}{d (A+2 C+A \cos (2 c+2 d x)) (a+b \sec (c+d x))^{5/2}}+\frac {8 (b+a \cos (c+d x))^2 \sqrt {\sec (c+d x)} \sqrt {\cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x)} \left (A+C \sec ^2(c+d x)\right ) \left (2 a (a+b) \left (-2 A b^2+\left (a^2-3 b^2\right ) C\right ) \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sqrt {\frac {b+a \cos (c+d x)}{(a+b) (1+\cos (c+d x))}} E\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {a-b}{a+b}\right )+b (a+b) \left (-2 a^2 C+3 a b (A+C)+b^2 (A+3 C)\right ) \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sqrt {\frac {b+a \cos (c+d x)}{(a+b) (1+\cos (c+d x))}} \operatorname {EllipticF}\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {a-b}{a+b}\right )+a \left (-2 A b^2+\left (a^2-3 b^2\right ) C\right ) \cos (c+d x) (b+a \cos (c+d x)) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \tan \left (\frac {1}{2} (c+d x)\right )\right )}{3 \left (-a^2 b+b^3\right )^2 d (A+2 C+A \cos (2 c+2 d x)) \sqrt {\sec ^2\left (\frac {1}{2} (c+d x)\right )} (a+b \sec (c+d x))^{5/2}} \] Input:

Integrate[(Sec[c + d*x]*(A + 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]*(A + C*Sec[c + d*x]^2)*((-8*a*(-2*A*b 
^2 + a^2*C - 3*b^2*C)*Sin[c + d*x])/(3*b^2*(a^2 - b^2)^2) + (4*(A*b^2*Sin[ 
c + d*x] + a^2*C*Sin[c + d*x]))/(3*a*(a^2 - b^2)*(b + a*Cos[c + d*x])^2) + 
 (4*(-5*a^2*A*b^2*Sin[c + d*x] + A*b^4*Sin[c + d*x] + a^4*C*Sin[c + d*x] - 
 5*a^2*b^2*C*Sin[c + d*x]))/(3*a*b*(-a^2 + b^2)^2*(b + a*Cos[c + d*x]))))/ 
(d*(A + 2*C + A*Cos[2*c + 2*d*x])*(a + b*Sec[c + d*x])^(5/2)) + (8*(b + a* 
Cos[c + d*x])^2*Sqrt[Sec[c + d*x]]*Sqrt[Cos[(c + d*x)/2]^2*Sec[c + d*x]]*( 
A + C*Sec[c + d*x]^2)*(2*a*(a + b)*(-2*A*b^2 + (a^2 - 3*b^2)*C)*Sqrt[Cos[c 
 + d*x]/(1 + Cos[c + d*x])]*Sqrt[(b + a*Cos[c + d*x])/((a + b)*(1 + Cos[c 
+ d*x]))]*EllipticE[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)] + b*(a + b) 
*(-2*a^2*C + 3*a*b*(A + C) + b^2*(A + 3*C))*Sqrt[Cos[c + d*x]/(1 + Cos[c + 
 d*x])]*Sqrt[(b + a*Cos[c + d*x])/((a + b)*(1 + Cos[c + d*x]))]*EllipticF[ 
ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)] + a*(-2*A*b^2 + (a^2 - 3*b^2)*C 
)*Cos[c + d*x]*(b + a*Cos[c + d*x])*Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2]))/ 
(3*(-(a^2*b) + b^3)^2*d*(A + 2*C + A*Cos[2*c + 2*d*x])*Sqrt[Sec[(c + d*x)/ 
2]^2]*(a + b*Sec[c + d*x])^(5/2))
 

Rubi [A] (verified)

Time = 1.44 (sec) , antiderivative size = 393, normalized size of antiderivative = 1.04, number of steps used = 11, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 4569, 27, 3042, 4491, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4569

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4491

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4493

\(\displaystyle \frac {\frac {(a-b) \left (2 a^2 C+3 a b (A+C)-b^2 (A+3 C)\right ) \int \frac {\sec (c+d x)}{\sqrt {a+b \sec (c+d x)}}dx+2 a \left (a^2 (-C)+2 A b^2+3 b^2 C\right ) \int \frac {\sec (c+d x) (\sec (c+d x)+1)}{\sqrt {a+b \sec (c+d x)}}dx}{a^2-b^2}-\frac {4 a \left (2 A b^2-C \left (a^2-3 b^2\right )\right ) \tan (c+d x)}{d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{3 b \left (a^2-b^2\right )}-\frac {2 \left (a^2 C+A b^2\right ) \tan (c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4319

\(\displaystyle \frac {\frac {2 a \left (a^2 (-C)+2 A b^2+3 b^2 C\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 (a-b) \sqrt {a+b} \left (2 a^2 C+3 a b (A+C)-b^2 (A+3 C)\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{b d}}{a^2-b^2}-\frac {4 a \left (2 A b^2-C \left (a^2-3 b^2\right )\right ) \tan (c+d x)}{d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{3 b \left (a^2-b^2\right )}-\frac {2 \left (a^2 C+A b^2\right ) \tan (c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}\)

\(\Big \downarrow \) 4492

\(\displaystyle \frac {\frac {\frac {2 (a-b) \sqrt {a+b} \left (2 a^2 C+3 a b (A+C)-b^2 (A+3 C)\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{b d}-\frac {4 a (a-b) \sqrt {a+b} \left (a^2 (-C)+2 A b^2+3 b^2 C\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{b^2 d}}{a^2-b^2}-\frac {4 a \left (2 A b^2-C \left (a^2-3 b^2\right )\right ) \tan (c+d x)}{d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{3 b \left (a^2-b^2\right )}-\frac {2 \left (a^2 C+A b^2\right ) \tan (c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}\)

Input:

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

Output:

(-2*(A*b^2 + a^2*C)*Tan[c + d*x])/(3*b*(a^2 - b^2)*d*(a + b*Sec[c + d*x])^ 
(3/2)) + (((-4*a*(a - b)*Sqrt[a + b]*(2*A*b^2 - a^2*C + 3*b^2*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^2*d) + (2*(a - b)*Sqrt[a + b]*(2*a^2*C + 3*a*b*(A + C) - b^2*(A 
+ 3*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 + Se 
c[c + d*x]))/(a - b))])/(b*d))/(a^2 - b^2) - (4*a*(2*A*b^2 - (a^2 - 3*b^2) 
*C)*Tan[c + d*x])/((a^2 - b^2)*d*Sqrt[a + b*Sec[c + d*x]]))/(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 4491

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

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 4569

Int[csc[(e_.) + (f_.)*(x_)]*((A_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[ 
(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(-(A*b^2 + a^2*C) 
)*Cot[e + f*x]*((a + b*Csc[e + f*x])^(m + 1)/(b*f*(m + 1)*(a^2 - b^2))), x] 
 + Simp[1/(b*(m + 1)*(a^2 - b^2))   Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^( 
m + 1)*Simp[a*b*(A + C)*(m + 1) - (A*b^2 + a^2*C + b*(A*b + b*C)*(m + 1))*C 
sc[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, C}, x] && LtQ[m, -1] && Ne 
Q[a^2 - b^2, 0]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(2172\) vs. \(2(348)=696\).

Time = 37.54 (sec) , antiderivative size = 2173, normalized size of antiderivative = 5.75

method	result	size
default	\(\text {Expression too large to display}\)	\(2173\)
parts	\(\text {Expression too large to display}\)	\(2217\)

Input:

int(sec(d*x+c)*(A+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^(5/2),x,method=_RETURNV 
ERBOSE)
 

Output:

2/3/d/(a-b)^2/(a+b)^2/b^2*(sin(d*x+c)*cos(d*x+c)*(-5*cos(d*x+c)+7)*C*a^2*b 
^3+sin(d*x+c)*cos(d*x+c)*(cos(d*x+c)-3)*a^4*b*C+sin(d*x+c)*cos(d*x+c)*(-5* 
cos(d*x+c)+3)*a^2*A*b^3+(-3*cos(d*x+c)^2-6*cos(d*x+c)-3)*C*(1/(a+b)*(b+a*c 
os(d*x+c))/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*b^5*Ell 
ipticF(-csc(d*x+c)+cot(d*x+c),((a-b)/(a+b))^(1/2))+sin(d*x+c)*cos(d*x+c)*( 
6*cos(d*x+c)+2)*C*a^3*b^2+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^5*EllipticE(-csc(d*x+c)+cot(d*x+c),( 
(a-b)/(a+b))^(1/2))*(-2*cos(d*x+c)^3-4*cos(d*x+c)^2-2*cos(d*x+c))+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^4*b*EllipticF(-csc(d*x+c)+cot(d*x+c),((a-b)/(a+b))^(1/2))*(2*cos(d*x+c 
)^3+4*cos(d*x+c)^2+2*cos(d*x+c))-6*C*a*b^4*cos(d*x+c)*sin(d*x+c)+4*A*a^3*b 
^2*cos(d*x+c)^2*sin(d*x+c)+sin(d*x+c)*cos(d*x+c)*(cos(d*x+c)+1)*A*b^5+(-co 
s(d*x+c)^2-2*cos(d*x+c)-1)*A*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b 
+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*b^5*EllipticF(-csc(d*x+c)+cot(d*x+c), 
((a-b)/(a+b))^(1/2))-4*A*a*b^4*cos(d*x+c)*sin(d*x+c)+(-cos(d*x+c)^3+3*cos( 
d*x+c)+2)*C*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(c 
os(d*x+c)+1))^(1/2)*a^3*b^2*EllipticF(-csc(d*x+c)+cot(d*x+c),((a-b)/(a+b)) 
^(1/2))+(-6*cos(d*x+c)^3-13*cos(d*x+c)^2-8*cos(d*x+c)-1)*C*(1/(a+b)*(b+a*c 
os(d*x+c))/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*a^2*b^3 
*EllipticF(-csc(d*x+c)+cot(d*x+c),((a-b)/(a+b))^(1/2))+(4*cos(d*x+c)^3+...
 

Fricas [F]

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

integrate(sec(d*x+c)*(A+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^(5/2),x, algorith 
m="fricas")
 

Output:

integral((C*sec(d*x + c)^3 + A*sec(d*x + c))*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]

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

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

Output:

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

Maxima [F(-1)]

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

integrate(sec(d*x+c)*(A+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^(5/2),x, algorith 
m="maxima")
 

Output:

Timed out
 

Giac [F]

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

integrate(sec(d*x+c)*(A+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^(5/2),x, algorith 
m="giac")
 

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

int(sec(d*x+c)*(A+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)**3)/(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))/(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)*a