\(\int \frac {\sec (c+d x) (B \sec (c+d x)+C \sec ^2(c+d x))}{(a+b \sec (c+d x))^{5/2}} \, dx\) [851]

Optimal result

Integrand size = 40, antiderivative size = 387 \[ \int \frac {\sec (c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{5/2}} \, dx=\frac {2 \left (a^2 b B+3 b^3 B+2 a^3 C-6 a b^2 C\right ) \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{3 (a-b) b^3 (a+b)^{3/2} d}+\frac {2 \left (2 a^2 C-3 b^2 (B+C)+a b (B+3 C)\right ) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{3 b^2 \sqrt {a+b} \left (a^2-b^2\right ) d}+\frac {2 a (b B-a C) \tan (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}+\frac {2 \left (a^2 b B+3 b^3 B+2 a^3 C-6 a b^2 C\right ) \tan (c+d x)}{3 b \left (a^2-b^2\right )^2 d \sqrt {a+b \sec (c+d x)}} \] Output:

2/3*(B*a^2*b+3*B*b^3+2*C*a^3-6*C*a*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+sec(d*x+c))/(a-b))^(1/2)/(a-b)/b^3/(a+b)^(3/2)/d+2/3*(2*C*a^2-3*b^2 
*(B+C)+a*b*(B+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*b-C*a)*tan(d*x+c)/b/(a^2- 
b^2)/d/(a+b*sec(d*x+c))^(3/2)+2/3*(B*a^2*b+3*B*b^3+2*C*a^3-6*C*a*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 = 16.83 (sec) , antiderivative size = 587, normalized size of antiderivative = 1.52 \[ \int \frac {\sec (c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{5/2}} \, dx=\frac {(b+a \cos (c+d x))^3 \sec ^3(c+d x) \left (-\frac {2 \left (a^2 b B+3 b^3 B+2 a^3 C-6 a b^2 C\right ) \sin (c+d x)}{3 b^2 \left (-a^2+b^2\right )^2}+\frac {2 (b B \sin (c+d x)-a C \sin (c+d x))}{3 \left (-a^2+b^2\right ) (b+a \cos (c+d x))^2}+\frac {2 \left (2 a^2 b B \sin (c+d x)+2 b^3 B \sin (c+d x)+a^3 C \sin (c+d x)-5 a b^2 C \sin (c+d x)\right )}{3 b \left (-a^2+b^2\right )^2 (b+a \cos (c+d x))}\right )}{d (a+b \sec (c+d x))^{5/2}}+\frac {2 (b+a \cos (c+d x))^2 \sec ^{\frac {5}{2}}(c+d x) \sqrt {\cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x)} \left (2 (a+b) \left (a^2 b B+3 b^3 B+2 a^3 C-6 a b^2 C\right ) \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sqrt {\frac {b+a \cos (c+d x)}{(a+b) (1+\cos (c+d x))}} E\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {a-b}{a+b}\right )-2 b (a+b) \left (a b (B-3 C)+3 b^2 (B-C)+2 a^2 C\right ) \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sqrt {\frac {b+a \cos (c+d x)}{(a+b) (1+\cos (c+d x))}} \operatorname {EllipticF}\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {a-b}{a+b}\right )+\left (a^2 b B+3 b^3 B+2 a^3 C-6 a b^2 C\right ) \cos (c+d x) (b+a \cos (c+d x)) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \tan \left (\frac {1}{2} (c+d x)\right )\right )}{3 \left (-a^2 b+b^3\right )^2 d \sqrt {\sec ^2\left (\frac {1}{2} (c+d x)\right )} (a+b \sec (c+d x))^{5/2}} \] Input:

Integrate[(Sec[c + d*x]*(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*(a^2*b*B + 3*b^3*B + 2*a^3*C - 
 6*a*b^2*C)*Sin[c + d*x])/(3*b^2*(-a^2 + b^2)^2) + (2*(b*B*Sin[c + d*x] - 
a*C*Sin[c + d*x]))/(3*(-a^2 + b^2)*(b + a*Cos[c + d*x])^2) + (2*(2*a^2*b*B 
*Sin[c + d*x] + 2*b^3*B*Sin[c + d*x] + a^3*C*Sin[c + d*x] - 5*a*b^2*C*Sin[ 
c + d*x]))/(3*b*(-a^2 + b^2)^2*(b + a*Cos[c + d*x]))))/(d*(a + b*Sec[c + d 
*x])^(5/2)) + (2*(b + a*Cos[c + d*x])^2*Sec[c + d*x]^(5/2)*Sqrt[Cos[(c + d 
*x)/2]^2*Sec[c + d*x]]*(2*(a + b)*(a^2*b*B + 3*b^3*B + 2*a^3*C - 6*a*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)] 
 - 2*b*(a + b)*(a*b*(B - 3*C) + 3*b^2*(B - C) + 2*a^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]) 
)]*EllipticF[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)] + (a^2*b*B + 3*b^3 
*B + 2*a^3*C - 6*a*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*Sqrt[Sec[(c + d*x)/2]^2]*( 
a + b*Sec[c + d*x])^(5/2))
 

Rubi [A] (verified)

Time = 1.58 (sec) , antiderivative size = 402, normalized size of antiderivative = 1.04, number of steps used = 13, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.325, Rules used = {3042, 4560, 3042, 4497, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4560

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4497

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4491

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4493

\(\displaystyle \frac {2 a (b B-a C) \tan (c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}-\frac {\frac {\left (2 a^3 C+a^2 b B-6 a b^2 C+3 b^3 B\right ) \int \frac {\sec (c+d x) (\sec (c+d x)+1)}{\sqrt {a+b \sec (c+d x)}}dx-(a-b) \left (2 a^2 C+a b (B+3 C)-3 b^2 (B+C)\right ) \int \frac {\sec (c+d x)}{\sqrt {a+b \sec (c+d x)}}dx}{a^2-b^2}-\frac {2 \left (2 a^3 C+a^2 b B-6 a b^2 C+3 b^3 B\right ) \tan (c+d x)}{d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{3 b \left (a^2-b^2\right )}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 a (b B-a C) \tan (c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}-\frac {\frac {\left (2 a^3 C+a^2 b B-6 a b^2 C+3 b^3 B\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-(a-b) \left (2 a^2 C+a b (B+3 C)-3 b^2 (B+C)\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{a^2-b^2}-\frac {2 \left (2 a^3 C+a^2 b B-6 a b^2 C+3 b^3 B\right ) \tan (c+d x)}{d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{3 b \left (a^2-b^2\right )}\)

\(\Big \downarrow \) 4319

\(\displaystyle \frac {2 a (b B-a C) \tan (c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}-\frac {\frac {\left (2 a^3 C+a^2 b B-6 a b^2 C+3 b^3 B\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 (a-b) \sqrt {a+b} \left (2 a^2 C+a b (B+3 C)-3 b^2 (B+C)\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{b d}}{a^2-b^2}-\frac {2 \left (2 a^3 C+a^2 b B-6 a b^2 C+3 b^3 B\right ) \tan (c+d x)}{d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{3 b \left (a^2-b^2\right )}\)

\(\Big \downarrow \) 4492

\(\displaystyle \frac {2 a (b B-a C) \tan (c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}-\frac {\frac {-\frac {2 (a-b) \sqrt {a+b} \left (2 a^2 C+a b (B+3 C)-3 b^2 (B+C)\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{b d}-\frac {2 (a-b) \sqrt {a+b} \left (2 a^3 C+a^2 b B-6 a b^2 C+3 b^3 B\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{b^2 d}}{a^2-b^2}-\frac {2 \left (2 a^3 C+a^2 b B-6 a b^2 C+3 b^3 B\right ) \tan (c+d x)}{d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{3 b \left (a^2-b^2\right )}\)

Input:

Int[(Sec[c + d*x]*(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)*Tan[c + d*x])/(3*b*(a^2 - b^2)*d*(a + b*Sec[c + d*x])^(3/ 
2)) - (((-2*(a - b)*Sqrt[a + b]*(a^2*b*B + 3*b^3*B + 2*a^3*C - 6*a*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*b^2*(B + C) 
+ a*b*(B + 3*C))*Cot[c + d*x]*EllipticF[ArcSin[Sqrt[a + b*Sec[c + d*x]]/Sq 
rt[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))/(a^2 - b^2) - (2*(a^2*b*B + 3*b^3* 
B + 2*a^3*C - 6*a*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 4497

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(2414\) vs. \(2(357)=714\).

Time = 35.29 (sec) , antiderivative size = 2415, normalized size of antiderivative = 6.24

method	result	size
default	\(\text {Expression too large to display}\)	\(2415\)
parts	\(\text {Expression too large to display}\)	\(2465\)

Input:

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

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+(3*cos(d*x+c)^2+6*cos(d*x+c)+3)*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)*b^5*EllipticF(-csc(d*x+c)+cot 
(d*x+c),((a-b)/(a+b))^(1/2))+sin(d*x+c)*cos(d*x+c)*(-3*cos(d*x+c)+1)*B*a^2 
*b^3+sin(d*x+c)*cos(d*x+c)*(2*cos(d*x+c)-4)*a*b^4*B+(-3*cos(d*x+c)^2-6*cos 
(d*x+c)-3)*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)*b^5*EllipticE(-csc(d*x+c)+cot(d*x+c),((a-b)/(a+b))^(1 
/2))+sin(d*x+c)*cos(d*x+c)*(cos(d*x+c)-3)*a^4*b*C+(-3*cos(d*x+c)^2-6*cos(d 
*x+c)-3)*C*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(co 
s(d*x+c)+1))^(1/2)*b^5*EllipticF(-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*Elliptic 
E(-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))-B*a^4*b*cos(d*x 
+c)^2*sin(d*x+c)+2*B*a^3*b^2*cos(d*x+c)^2*sin(d*x+c)-6*C*a*b^4*cos(d*x+c)* 
sin(d*x+c)+(4*cos(d*x+c)^3+9*cos(d*x+c)^2+6*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^2*b^ 
3*EllipticF(-csc(d*x+c)+cot(d*x+c),((a-b)/(a+b))^(1/2))+(-cos(d*x+c)^3+...
 

Fricas [F]

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

integrate(sec(d*x+c)*(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)^3 + B*sec(d*x + c)^2)*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]

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

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

Maxima [F(-1)]

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

integrate(sec(d*x+c)*(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]

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

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

Mupad [F(-1)]

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

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

Reduce [F]

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

int(sec(d*x+c)*(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)**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)**2)/(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