\(\int (e \csc (c+d x))^{5/2} (a+a \sec (c+d x))^2 \, dx\) [287]

Optimal result

Integrand size = 25, antiderivative size = 270 \[ \int (e \csc (c+d x))^{5/2} (a+a \sec (c+d x))^2 \, dx=-\frac {2 a^2 e^2 \cot (c+d x) \sqrt {e \csc (c+d x)}}{3 d}-\frac {4 a^2 e^2 \csc (c+d x) \sqrt {e \csc (c+d x)}}{3 d}-\frac {2 a^2 e^2 \csc (c+d x) \sqrt {e \csc (c+d x)} \sec (c+d x)}{3 d}+\frac {2 a^2 e^2 \arctan \left (\sqrt {\sin (c+d x)}\right ) \sqrt {e \csc (c+d x)} \sqrt {\sin (c+d x)}}{d}+\frac {2 a^2 e^2 \text {arctanh}\left (\sqrt {\sin (c+d x)}\right ) \sqrt {e \csc (c+d x)} \sqrt {\sin (c+d x)}}{d}+\frac {7 a^2 e^2 \sqrt {e \csc (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right ),2\right ) \sqrt {\sin (c+d x)}}{3 d}+\frac {5 a^2 e^2 \sqrt {e \csc (c+d x)} \tan (c+d x)}{3 d} \] Output:

-2/3*a^2*e^2*cot(d*x+c)*(e*csc(d*x+c))^(1/2)/d-4/3*a^2*e^2*csc(d*x+c)*(e*c 
sc(d*x+c))^(1/2)/d-2/3*a^2*e^2*csc(d*x+c)*(e*csc(d*x+c))^(1/2)*sec(d*x+c)/ 
d+2*a^2*e^2*arctan(sin(d*x+c)^(1/2))*(e*csc(d*x+c))^(1/2)*sin(d*x+c)^(1/2) 
/d+2*a^2*e^2*arctanh(sin(d*x+c)^(1/2))*(e*csc(d*x+c))^(1/2)*sin(d*x+c)^(1/ 
2)/d+7/3*a^2*e^2*(e*csc(d*x+c))^(1/2)*InverseJacobiAM(1/2*c-1/4*Pi+1/2*d*x 
,2^(1/2))*sin(d*x+c)^(1/2)/d+5/3*a^2*e^2*(e*csc(d*x+c))^(1/2)*tan(d*x+c)/d
 

Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 13.04 (sec) , antiderivative size = 195, normalized size of antiderivative = 0.72 \[ \int (e \csc (c+d x))^{5/2} (a+a \sec (c+d x))^2 \, dx=-\frac {a^2 e^2 \cos ^4\left (\frac {1}{2} (c+d x)\right ) \sqrt {e \csc (c+d x)} \left (-7+6 \arctan \left (\sqrt {\csc (c+d x)}\right ) \sqrt {\cos ^2(c+d x)} \sqrt {\csc (c+d x)}-6 \text {arctanh}\left (\sqrt {\csc (c+d x)}\right ) \sqrt {\cos ^2(c+d x)} \sqrt {\csc (c+d x)}+4 \csc ^2(c+d x)+4 \sqrt {\cos ^2(c+d x)} \csc ^2(c+d x)+7 \sqrt {-\cot ^2(c+d x)} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},\csc ^2(c+d x)\right )\right ) \sec ^4\left (\frac {1}{2} \csc ^{-1}(\csc (c+d x))\right ) \tan (c+d x)}{3 d} \] Input:

Integrate[(e*Csc[c + d*x])^(5/2)*(a + a*Sec[c + d*x])^2,x]
 

Output:

-1/3*(a^2*e^2*Cos[(c + d*x)/2]^4*Sqrt[e*Csc[c + d*x]]*(-7 + 6*ArcTan[Sqrt[ 
Csc[c + d*x]]]*Sqrt[Cos[c + d*x]^2]*Sqrt[Csc[c + d*x]] - 6*ArcTanh[Sqrt[Cs 
c[c + d*x]]]*Sqrt[Cos[c + d*x]^2]*Sqrt[Csc[c + d*x]] + 4*Csc[c + d*x]^2 + 
4*Sqrt[Cos[c + d*x]^2]*Csc[c + d*x]^2 + 7*Sqrt[-Cot[c + d*x]^2]*Hypergeome 
tric2F1[1/4, 1/2, 5/4, Csc[c + d*x]^2])*Sec[ArcCsc[Csc[c + d*x]]/2]^4*Tan[ 
c + d*x])/d
 

Rubi [A] (verified)

Time = 0.68 (sec) , antiderivative size = 189, normalized size of antiderivative = 0.70, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {3042, 4366, 3042, 4360, 3042, 3352, 2009}

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.

\(\displaystyle \int (a \sec (c+d x)+a)^2 (e \csc (c+d x))^{5/2} \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4366

\(\displaystyle e^2 \sqrt {\sin (c+d x)} \sqrt {e \csc (c+d x)} \int \frac {(\sec (c+d x) a+a)^2}{\sin ^{\frac {5}{2}}(c+d x)}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle e^2 \sqrt {\sin (c+d x)} \sqrt {e \csc (c+d x)} \int \frac {\left (a-a \csc \left (c+d x-\frac {\pi }{2}\right )\right )^2}{\cos \left (c+d x-\frac {\pi }{2}\right )^{5/2}}dx\)

\(\Big \downarrow \) 4360

\(\displaystyle e^2 \sqrt {\sin (c+d x)} \sqrt {e \csc (c+d x)} \int \frac {(-\cos (c+d x) a-a)^2 \sec ^2(c+d x)}{\sin ^{\frac {5}{2}}(c+d x)}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle e^2 \sqrt {\sin (c+d x)} \sqrt {e \csc (c+d x)} \int \frac {\left (a \sin \left (c+d x-\frac {\pi }{2}\right )-a\right )^2}{\cos \left (c+d x-\frac {\pi }{2}\right )^{5/2} \sin \left (c+d x-\frac {\pi }{2}\right )^2}dx\)

\(\Big \downarrow \) 3352

\(\displaystyle e^2 \sqrt {\sin (c+d x)} \sqrt {e \csc (c+d x)} \int \left (\frac {\sec ^2(c+d x) a^2}{\sin ^{\frac {5}{2}}(c+d x)}+\frac {2 \sec (c+d x) a^2}{\sin ^{\frac {5}{2}}(c+d x)}+\frac {a^2}{\sin ^{\frac {5}{2}}(c+d x)}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle e^2 \sqrt {\sin (c+d x)} \sqrt {e \csc (c+d x)} \left (\frac {2 a^2 \arctan \left (\sqrt {\sin (c+d x)}\right )}{d}+\frac {2 a^2 \text {arctanh}\left (\sqrt {\sin (c+d x)}\right )}{d}-\frac {4 a^2}{3 d \sin ^{\frac {3}{2}}(c+d x)}+\frac {7 a^2 \operatorname {EllipticF}\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),2\right )}{3 d}-\frac {2 a^2 \cos (c+d x)}{3 d \sin ^{\frac {3}{2}}(c+d x)}-\frac {2 a^2 \sec (c+d x)}{3 d \sin ^{\frac {3}{2}}(c+d x)}+\frac {5 a^2 \sqrt {\sin (c+d x)} \sec (c+d x)}{3 d}\right )\)

Input:

Int[(e*Csc[c + d*x])^(5/2)*(a + a*Sec[c + d*x])^2,x]
 

Output:

e^2*Sqrt[e*Csc[c + d*x]]*((2*a^2*ArcTan[Sqrt[Sin[c + d*x]]])/d + (2*a^2*Ar 
cTanh[Sqrt[Sin[c + d*x]]])/d + (7*a^2*EllipticF[(c - Pi/2 + d*x)/2, 2])/(3 
*d) - (4*a^2)/(3*d*Sin[c + d*x]^(3/2)) - (2*a^2*Cos[c + d*x])/(3*d*Sin[c + 
 d*x]^(3/2)) - (2*a^2*Sec[c + d*x])/(3*d*Sin[c + d*x]^(3/2)) + (5*a^2*Sec[ 
c + d*x]*Sqrt[Sin[c + d*x]])/(3*d))*Sqrt[Sin[c + d*x]]
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3352

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n 
_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Int[ExpandTrig 
[(g*cos[e + f*x])^p, (d*sin[e + f*x])^n*(a + b*sin[e + f*x])^m, x], x] /; F 
reeQ[{a, b, d, e, f, g, n, p}, x] && EqQ[a^2 - b^2, 0] && IGtQ[m, 0]
 

rule 4360

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + 
(a_))^(m_.), x_Symbol] :> Int[(g*Cos[e + f*x])^p*((b + a*Sin[e + f*x])^m/Si 
n[e + f*x]^m), x] /; FreeQ[{a, b, e, f, g, p}, x] && IntegerQ[m]
 

rule 4366

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.)*((g_.)*sec[(e_.) + (f_.)*( 
x_)])^(p_), x_Symbol] :> Simp[g^IntPart[p]*(g*Sec[e + f*x])^FracPart[p]*Cos 
[e + f*x]^FracPart[p]   Int[(a + b*Csc[e + f*x])^m/Cos[e + f*x]^p, x], x] / 
; FreeQ[{a, b, e, f, g, m, p}, x] &&  !IntegerQ[p]
 

Maple [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 2.68 (sec) , antiderivative size = 462, normalized size of antiderivative = 1.71

method	result	size
parts	\(\frac {a^{2} \sqrt {2}\, e^{2} \sqrt {e \csc \left (d x +c \right )}\, \left (i \left (1+\cos \left (d x +c \right )\right ) \sqrt {1-i \cot \left (d x +c \right )+i \csc \left (d x +c \right )}\, \sqrt {-i \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right )}\, \operatorname {EllipticF}\left (\sqrt {1+i \cot \left (d x +c \right )-i \csc \left (d x +c \right )}, \frac {\sqrt {2}}{2}\right ) \sqrt {1+i \cot \left (d x +c \right )-i \csc \left (d x +c \right )}-\sqrt {2}\, \cot \left (d x +c \right )\right )}{3 d}+\frac {a^{2} \sqrt {2}\, e^{2} \sqrt {e \csc \left (d x +c \right )}\, \left (i \left (5 \cos \left (d x +c \right )+5\right ) \sqrt {1-i \cot \left (d x +c \right )+i \csc \left (d x +c \right )}\, \sqrt {-i \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right )}\, \operatorname {EllipticF}\left (\sqrt {1+i \cot \left (d x +c \right )-i \csc \left (d x +c \right )}, \frac {\sqrt {2}}{2}\right ) \sqrt {1+i \cot \left (d x +c \right )-i \csc \left (d x +c \right )}+\sqrt {2}\, \left (-5 \cot \left (d x +c \right )+3 \sec \left (d x +c \right ) \csc \left (d x +c \right )\right )\right )}{6 d}+\frac {2 a^{2} \left (\left (3 \cos \left (d x +c \right )-3\right ) \operatorname {arctanh}\left (\frac {\sin \left (d x +c \right ) \sqrt {\frac {\sin \left (d x +c \right )}{\left (1+\cos \left (d x +c \right )\right )^{2}}}}{-1+\cos \left (d x +c \right )}\right )+\left (-3 \cos \left (d x +c \right )+3\right ) \arctan \left (\frac {\sin \left (d x +c \right ) \sqrt {\frac {\sin \left (d x +c \right )}{\left (1+\cos \left (d x +c \right )\right )^{2}}}}{-1+\cos \left (d x +c \right )}\right )-2 \sqrt {\frac {\sin \left (d x +c \right )}{\left (1+\cos \left (d x +c \right )\right )^{2}}}\right ) \sqrt {e \csc \left (d x +c \right )}\, e^{2} \csc \left (d x +c \right )}{3 d \sqrt {\frac {\sin \left (d x +c \right )}{\left (1+\cos \left (d x +c \right )\right )^{2}}}}\)	\(462\)
default	\(\frac {a^{2} \sqrt {2}\, \left (1+\cos \left (d x +c \right )\right ) \left (6 i \cos \left (d x +c \right ) \sin \left (d x +c \right ) \sqrt {1-i \cot \left (d x +c \right )+i \csc \left (d x +c \right )}\, \sqrt {i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}\, \sqrt {1+i \cot \left (d x +c \right )-i \csc \left (d x +c \right )}\, \operatorname {EllipticPi}\left (\sqrt {1+i \cot \left (d x +c \right )-i \csc \left (d x +c \right )}, \frac {1}{2}-\frac {i}{2}, \frac {\sqrt {2}}{2}\right )-5 i \cos \left (d x +c \right ) \sin \left (d x +c \right ) \sqrt {1-i \cot \left (d x +c \right )+i \csc \left (d x +c \right )}\, \sqrt {i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}\, \sqrt {1+i \cot \left (d x +c \right )-i \csc \left (d x +c \right )}\, \operatorname {EllipticF}\left (\sqrt {1+i \cot \left (d x +c \right )-i \csc \left (d x +c \right )}, \frac {\sqrt {2}}{2}\right )+6 i \cos \left (d x +c \right ) \sin \left (d x +c \right ) \sqrt {1-i \cot \left (d x +c \right )+i \csc \left (d x +c \right )}\, \sqrt {i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}\, \sqrt {1+i \cot \left (d x +c \right )-i \csc \left (d x +c \right )}\, \operatorname {EllipticPi}\left (\sqrt {1+i \cot \left (d x +c \right )-i \csc \left (d x +c \right )}, \frac {1}{2}+\frac {i}{2}, \frac {\sqrt {2}}{2}\right )-6 \cos \left (d x +c \right ) \sin \left (d x +c \right ) \sqrt {1-i \cot \left (d x +c \right )+i \csc \left (d x +c \right )}\, \sqrt {i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}\, \sqrt {1+i \cot \left (d x +c \right )-i \csc \left (d x +c \right )}\, \operatorname {EllipticPi}\left (\sqrt {1+i \cot \left (d x +c \right )-i \csc \left (d x +c \right )}, \frac {1}{2}-\frac {i}{2}, \frac {\sqrt {2}}{2}\right )+6 \cos \left (d x +c \right ) \sin \left (d x +c \right ) \sqrt {1-i \cot \left (d x +c \right )+i \csc \left (d x +c \right )}\, \sqrt {i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}\, \sqrt {1+i \cot \left (d x +c \right )-i \csc \left (d x +c \right )}\, \operatorname {EllipticPi}\left (\sqrt {1+i \cot \left (d x +c \right )-i \csc \left (d x +c \right )}, \frac {1}{2}+\frac {i}{2}, \frac {\sqrt {2}}{2}\right )-7 \sqrt {2}\, \cos \left (d x +c \right )+3 \sqrt {2}\right ) \sqrt {e \csc \left (d x +c \right )}\, e^{2} \sec \left (d x +c \right ) \csc \left (d x +c \right )}{6 d}\)	\(607\)

Input:

int((e*csc(d*x+c))^(5/2)*(a+a*sec(d*x+c))^2,x,method=_RETURNVERBOSE)
 

Output:

1/3*a^2/d*2^(1/2)*e^2*(e*csc(d*x+c))^(1/2)*(I*(1+cos(d*x+c))*(1-I*cot(d*x+ 
c)+I*csc(d*x+c))^(1/2)*(-I*(-csc(d*x+c)+cot(d*x+c)))^(1/2)*EllipticF((1+I* 
cot(d*x+c)-I*csc(d*x+c))^(1/2),1/2*2^(1/2))*(1+I*cot(d*x+c)-I*csc(d*x+c))^ 
(1/2)-2^(1/2)*cot(d*x+c))+1/6*a^2/d*2^(1/2)*e^2*(e*csc(d*x+c))^(1/2)*(I*(5 
*cos(d*x+c)+5)*(1-I*cot(d*x+c)+I*csc(d*x+c))^(1/2)*(-I*(-csc(d*x+c)+cot(d* 
x+c)))^(1/2)*EllipticF((1+I*cot(d*x+c)-I*csc(d*x+c))^(1/2),1/2*2^(1/2))*(1 
+I*cot(d*x+c)-I*csc(d*x+c))^(1/2)+2^(1/2)*(-5*cot(d*x+c)+3*sec(d*x+c)*csc( 
d*x+c)))+2/3*a^2/d*((3*cos(d*x+c)-3)*arctanh(sin(d*x+c)*(sin(d*x+c)/(1+cos 
(d*x+c))^2)^(1/2)/(-1+cos(d*x+c)))+(-3*cos(d*x+c)+3)*arctan(sin(d*x+c)*(si 
n(d*x+c)/(1+cos(d*x+c))^2)^(1/2)/(-1+cos(d*x+c)))-2*(sin(d*x+c)/(1+cos(d*x 
+c))^2)^(1/2))*(e*csc(d*x+c))^(1/2)*e^2/(sin(d*x+c)/(1+cos(d*x+c))^2)^(1/2 
)*csc(d*x+c)
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.23 (sec) , antiderivative size = 806, normalized size of antiderivative = 2.99 \[ \int (e \csc (c+d x))^{5/2} (a+a \sec (c+d x))^2 \, dx=\text {Too large to display} \] Input:

integrate((e*csc(d*x+c))^(5/2)*(a+a*sec(d*x+c))^2,x, algorithm="fricas")
 

Output:

[-1/12*(6*a^2*sqrt(-e)*e^2*arctan(-1/4*(cos(d*x + c)^2 - 6*sin(d*x + c) - 
2)*sqrt(-e)*sqrt(e/sin(d*x + c))/(e*sin(d*x + c) + e))*cos(d*x + c)*sin(d* 
x + c) - 3*a^2*sqrt(-e)*e^2*cos(d*x + c)*log((e*cos(d*x + c)^4 - 72*e*cos( 
d*x + c)^2 + 8*(cos(d*x + c)^4 - 9*cos(d*x + c)^2 + (7*cos(d*x + c)^2 - 8) 
*sin(d*x + c) + 8)*sqrt(-e)*sqrt(e/sin(d*x + c)) + 28*(e*cos(d*x + c)^2 - 
2*e)*sin(d*x + c) + 72*e)/(cos(d*x + c)^4 - 8*cos(d*x + c)^2 - 4*(cos(d*x 
+ c)^2 - 2)*sin(d*x + c) + 8))*sin(d*x + c) + 14*I*a^2*sqrt(2*I*e)*e^2*cos 
(d*x + c)*sin(d*x + c)*weierstrassPInverse(4, 0, cos(d*x + c) + I*sin(d*x 
+ c)) - 14*I*a^2*sqrt(-2*I*e)*e^2*cos(d*x + c)*sin(d*x + c)*weierstrassPIn 
verse(4, 0, cos(d*x + c) - I*sin(d*x + c)) + 4*(7*a^2*e^2*cos(d*x + c)^2 + 
 4*a^2*e^2*cos(d*x + c) - 3*a^2*e^2)*sqrt(e/sin(d*x + c)))/(d*cos(d*x + c) 
*sin(d*x + c)), -1/12*(6*a^2*e^(5/2)*arctan(1/4*(cos(d*x + c)^2 + 6*sin(d* 
x + c) - 2)*sqrt(e)*sqrt(e/sin(d*x + c))/(e*sin(d*x + c) - e))*cos(d*x + c 
)*sin(d*x + c) - 3*a^2*e^(5/2)*cos(d*x + c)*log((e*cos(d*x + c)^4 - 72*e*c 
os(d*x + c)^2 + 8*(cos(d*x + c)^4 - 9*cos(d*x + c)^2 - (7*cos(d*x + c)^2 - 
 8)*sin(d*x + c) + 8)*sqrt(e)*sqrt(e/sin(d*x + c)) - 28*(e*cos(d*x + c)^2 
- 2*e)*sin(d*x + c) + 72*e)/(cos(d*x + c)^4 - 8*cos(d*x + c)^2 + 4*(cos(d* 
x + c)^2 - 2)*sin(d*x + c) + 8))*sin(d*x + c) + 14*I*a^2*sqrt(2*I*e)*e^2*c 
os(d*x + c)*sin(d*x + c)*weierstrassPInverse(4, 0, cos(d*x + c) + I*sin(d* 
x + c)) - 14*I*a^2*sqrt(-2*I*e)*e^2*cos(d*x + c)*sin(d*x + c)*weierstra...
 

Sympy [F(-1)]

Timed out. \[ \int (e \csc (c+d x))^{5/2} (a+a \sec (c+d x))^2 \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Maxima [F(-1)]

Timed out. \[ \int (e \csc (c+d x))^{5/2} (a+a \sec (c+d x))^2 \, dx=\text {Timed out} \] Input:

integrate((e*csc(d*x+c))^(5/2)*(a+a*sec(d*x+c))^2,x, algorithm="maxima")
 

Output:

Timed out
 

Giac [F]

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

integrate((e*csc(d*x+c))^(5/2)*(a+a*sec(d*x+c))^2,x, algorithm="giac")
 

Output:

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

Mupad [F(-1)]

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

int((a + a/cos(c + d*x))^2*(e/sin(c + d*x))^(5/2),x)
 

Output:

int((a + a/cos(c + d*x))^2*(e/sin(c + d*x))^(5/2), x)
 

Reduce [F]

\[ \int (e \csc (c+d x))^{5/2} (a+a \sec (c+d x))^2 \, dx=\sqrt {e}\, a^{2} e^{2} \left (\int \sqrt {\csc \left (d x +c \right )}\, \csc \left (d x +c \right )^{2} \sec \left (d x +c \right )^{2}d x +2 \left (\int \sqrt {\csc \left (d x +c \right )}\, \csc \left (d x +c \right )^{2} \sec \left (d x +c \right )d x \right )+\int \sqrt {\csc \left (d x +c \right )}\, \csc \left (d x +c \right )^{2}d x \right ) \] Input:

int((e*csc(d*x+c))^(5/2)*(a+a*sec(d*x+c))^2,x)
 

Output:

sqrt(e)*a**2*e**2*(int(sqrt(csc(c + d*x))*csc(c + d*x)**2*sec(c + d*x)**2, 
x) + 2*int(sqrt(csc(c + d*x))*csc(c + d*x)**2*sec(c + d*x),x) + int(sqrt(c 
sc(c + d*x))*csc(c + d*x)**2,x))