\(\int \frac {(e \csc (c+d x))^{5/2}}{(a+a \sec (c+d x))^2} \, dx\) [300]

Optimal result

Integrand size = 25, antiderivative size = 268 \[ \int \frac {(e \csc (c+d x))^{5/2}}{(a+a \sec (c+d x))^2} \, dx=-\frac {4 e^2 \cot (c+d x) \sqrt {e \csc (c+d x)}}{231 a^2 d}+\frac {16 e^2 \cot (c+d x) \csc ^2(c+d x) \sqrt {e \csc (c+d x)}}{77 a^2 d}-\frac {2 e^2 \cot ^3(c+d x) \csc ^2(c+d x) \sqrt {e \csc (c+d x)}}{11 a^2 d}-\frac {4 e^2 \csc ^3(c+d x) \sqrt {e \csc (c+d x)}}{7 a^2 d}-\frac {2 e^2 \cot (c+d x) \csc ^4(c+d x) \sqrt {e \csc (c+d x)}}{11 a^2 d}+\frac {4 e^2 \csc ^5(c+d x) \sqrt {e \csc (c+d x)}}{11 a^2 d}+\frac {4 e^2 \sqrt {e \csc (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right ),2\right ) \sqrt {\sin (c+d x)}}{231 a^2 d} \] Output:

-4/231*e^2*cot(d*x+c)*(e*csc(d*x+c))^(1/2)/a^2/d+16/77*e^2*cot(d*x+c)*csc( 
d*x+c)^2*(e*csc(d*x+c))^(1/2)/a^2/d-2/11*e^2*cot(d*x+c)^3*csc(d*x+c)^2*(e* 
csc(d*x+c))^(1/2)/a^2/d-4/7*e^2*csc(d*x+c)^3*(e*csc(d*x+c))^(1/2)/a^2/d-2/ 
11*e^2*cot(d*x+c)*csc(d*x+c)^4*(e*csc(d*x+c))^(1/2)/a^2/d+4/11*e^2*csc(d*x 
+c)^5*(e*csc(d*x+c))^(1/2)/a^2/d+4/231*e^2*(e*csc(d*x+c))^(1/2)*InverseJac 
obiAM(1/2*c-1/4*Pi+1/2*d*x,2^(1/2))*sin(d*x+c)^(1/2)/a^2/d
 

Mathematica [A] (verified)

Time = 1.98 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.43 \[ \int \frac {(e \csc (c+d x))^{5/2}}{(a+a \sec (c+d x))^2} \, dx=-\frac {e^3 \csc ^2\left (\frac {1}{2} (c+d x)\right ) \sec ^6\left (\frac {1}{2} (c+d x)\right ) \left (52+97 \cos (c+d x)+4 \cos (2 (c+d x))+\cos (3 (c+d x))+\csc ^4\left (\frac {1}{2} (c+d x)\right ) \operatorname {EllipticF}\left (\frac {1}{4} (-2 c+\pi -2 d x),2\right ) \sin ^{\frac {11}{2}}(c+d x)\right )}{3696 a^2 d \sqrt {e \csc (c+d x)}} \] Input:

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

Output:

-1/3696*(e^3*Csc[(c + d*x)/2]^2*Sec[(c + d*x)/2]^6*(52 + 97*Cos[c + d*x] + 
 4*Cos[2*(c + d*x)] + Cos[3*(c + d*x)] + Csc[(c + d*x)/2]^4*EllipticF[(-2* 
c + Pi - 2*d*x)/4, 2]*Sin[c + d*x]^(11/2)))/(a^2*d*Sqrt[e*Csc[c + d*x]])
 

Rubi [A] (verified)

Time = 0.90 (sec) , antiderivative size = 202, normalized size of antiderivative = 0.75, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.360, Rules used = {3042, 4366, 3042, 4360, 3042, 3354, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4366

\(\displaystyle e^2 \sqrt {\sin (c+d x)} \sqrt {e \csc (c+d x)} \int \frac {1}{(\sec (c+d x) a+a)^2 \sin ^{\frac {5}{2}}(c+d x)}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle e^2 \sqrt {\sin (c+d x)} \sqrt {e \csc (c+d x)} \int \frac {1}{\cos \left (c+d x-\frac {\pi }{2}\right )^{5/2} \left (a-a \csc \left (c+d x-\frac {\pi }{2}\right )\right )^2}dx\)

\(\Big \downarrow \) 4360

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3354

\(\displaystyle \frac {e^2 \sqrt {\sin (c+d x)} \sqrt {e \csc (c+d x)} \int \frac {\cos ^2(c+d x) (a-a \cos (c+d x))^2}{\sin ^{\frac {13}{2}}(c+d x)}dx}{a^4}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {e^2 \sqrt {\sin (c+d x)} \sqrt {e \csc (c+d x)} \int \frac {\sin \left (c+d x-\frac {\pi }{2}\right )^2 \left (\sin \left (c+d x-\frac {\pi }{2}\right ) a+a\right )^2}{\cos \left (c+d x-\frac {\pi }{2}\right )^{13/2}}dx}{a^4}\)

\(\Big \downarrow \) 3352

\(\displaystyle \frac {e^2 \sqrt {\sin (c+d x)} \sqrt {e \csc (c+d x)} \int \left (\frac {a^2 \cos ^4(c+d x)}{\sin ^{\frac {13}{2}}(c+d x)}-\frac {2 a^2 \cos ^3(c+d x)}{\sin ^{\frac {13}{2}}(c+d x)}+\frac {a^2 \cos ^2(c+d x)}{\sin ^{\frac {13}{2}}(c+d x)}\right )dx}{a^4}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {e^2 \sqrt {\sin (c+d x)} \sqrt {e \csc (c+d x)} \left (-\frac {4 a^2}{7 d \sin ^{\frac {7}{2}}(c+d x)}+\frac {4 a^2}{11 d \sin ^{\frac {11}{2}}(c+d x)}+\frac {4 a^2 \operatorname {EllipticF}\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),2\right )}{231 d}-\frac {2 a^2 \cos ^3(c+d x)}{11 d \sin ^{\frac {11}{2}}(c+d x)}-\frac {4 a^2 \cos (c+d x)}{231 d \sin ^{\frac {3}{2}}(c+d x)}+\frac {16 a^2 \cos (c+d x)}{77 d \sin ^{\frac {7}{2}}(c+d x)}-\frac {2 a^2 \cos (c+d x)}{11 d \sin ^{\frac {11}{2}}(c+d x)}\right )}{a^4}\)

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]]*((4*a^2*EllipticF[(c - Pi/2 + d*x)/2, 2])/(231*d 
) + (4*a^2)/(11*d*Sin[c + d*x]^(11/2)) - (2*a^2*Cos[c + d*x])/(11*d*Sin[c 
+ d*x]^(11/2)) - (2*a^2*Cos[c + d*x]^3)/(11*d*Sin[c + d*x]^(11/2)) - (4*a^ 
2)/(7*d*Sin[c + d*x]^(7/2)) + (16*a^2*Cos[c + d*x])/(77*d*Sin[c + d*x]^(7/ 
2)) - (4*a^2*Cos[c + d*x])/(231*d*Sin[c + d*x]^(3/2)))*Sqrt[Sin[c + d*x]]) 
/a^4
 

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 3354

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

Result contains complex when optimal does not.

Time = 1.22 (sec) , antiderivative size = 211, normalized size of antiderivative = 0.79

method	result	size
default	\(\frac {\sqrt {2}\, e^{2} \sqrt {e \csc \left (d x +c \right )}\, \left (i \left (2 \cos \left (d x +c \right )^{3}+6 \cos \left (d x +c \right )^{2}+6 \cos \left (d x +c \right )+2\right ) \sqrt {1+i \cot \left (d x +c \right )-i \csc \left (d x +c \right )}\, \sqrt {1-i \cot \left (d x +c \right )+i \csc \left (d x +c \right )}\, \sqrt {-i \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right )}\, \operatorname {EllipticF}\left (\sqrt {1+i \cot \left (d x +c \right )-i \csc \left (d x +c \right )}, \frac {\sqrt {2}}{2}\right )+\left (-2 \cos \left (d x +c \right )^{3}-4 \cos \left (d x +c \right )^{2}-47 \cos \left (d x +c \right )-24\right ) \sqrt {2}\, \csc \left (d x +c \right )\right )}{231 a^{2} d \left (\cos \left (d x +c \right )^{2}+2 \cos \left (d x +c \right )+1\right )}\)	\(211\)

Input:

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

Output:

1/231/a^2/d*2^(1/2)*e^2*(e*csc(d*x+c))^(1/2)/(cos(d*x+c)^2+2*cos(d*x+c)+1) 
*(I*(2*cos(d*x+c)^3+6*cos(d*x+c)^2+6*cos(d*x+c)+2)*(1+I*cot(d*x+c)-I*csc(d 
*x+c))^(1/2)*(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))+(-2* 
cos(d*x+c)^3-4*cos(d*x+c)^2-47*cos(d*x+c)-24)*2^(1/2)*csc(d*x+c))
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.09 (sec) , antiderivative size = 220, normalized size of antiderivative = 0.82 \[ \int \frac {(e \csc (c+d x))^{5/2}}{(a+a \sec (c+d x))^2} \, dx=-\frac {2 \, {\left ({\left (i \, e^{2} \cos \left (d x + c\right )^{2} + 2 i \, e^{2} \cos \left (d x + c\right ) + i \, e^{2}\right )} \sqrt {2 i \, e} \sin \left (d x + c\right ) {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + {\left (-i \, e^{2} \cos \left (d x + c\right )^{2} - 2 i \, e^{2} \cos \left (d x + c\right ) - i \, e^{2}\right )} \sqrt {-2 i \, e} \sin \left (d x + c\right ) {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + {\left (2 \, e^{2} \cos \left (d x + c\right )^{3} + 4 \, e^{2} \cos \left (d x + c\right )^{2} + 47 \, e^{2} \cos \left (d x + c\right ) + 24 \, e^{2}\right )} \sqrt {\frac {e}{\sin \left (d x + c\right )}}\right )}}{231 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )} \sin \left (d x + c\right )} \] Input:

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

Output:

-2/231*((I*e^2*cos(d*x + c)^2 + 2*I*e^2*cos(d*x + c) + I*e^2)*sqrt(2*I*e)* 
sin(d*x + c)*weierstrassPInverse(4, 0, cos(d*x + c) + I*sin(d*x + c)) + (- 
I*e^2*cos(d*x + c)^2 - 2*I*e^2*cos(d*x + c) - I*e^2)*sqrt(-2*I*e)*sin(d*x 
+ c)*weierstrassPInverse(4, 0, cos(d*x + c) - I*sin(d*x + c)) + (2*e^2*cos 
(d*x + c)^3 + 4*e^2*cos(d*x + c)^2 + 47*e^2*cos(d*x + c) + 24*e^2)*sqrt(e/ 
sin(d*x + c)))/((a^2*d*cos(d*x + c)^2 + 2*a^2*d*cos(d*x + c) + a^2*d)*sin( 
d*x + c))
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Giac [F]

\[ \int \frac {(e \csc (c+d x))^{5/2}}{(a+a \sec (c+d x))^2} \, dx=\int { \frac {\left (e \csc \left (d x + c\right )\right )^{\frac {5}{2}}}{{\left (a \sec \left (d x + c\right ) + a\right )}^{2}} \,d x } \] 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. \[ \int \frac {(e \csc (c+d x))^{5/2}}{(a+a \sec (c+d x))^2} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^2\,{\left (\frac {e}{\sin \left (c+d\,x\right )}\right )}^{5/2}}{a^2\,{\left (\cos \left (c+d\,x\right )+1\right )}^2} \,d x \] Input:

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

Output:

int((cos(c + d*x)^2*(e/sin(c + d*x))^(5/2))/(a^2*(cos(c + d*x) + 1)^2), x)
                                                                                    
                                                                                    
 

Reduce [F]

\[ \int \frac {(e \csc (c+d x))^{5/2}}{(a+a \sec (c+d x))^2} \, dx=\frac {\sqrt {e}\, \left (\int \frac {\sqrt {\csc \left (d x +c \right )}\, \csc \left (d x +c \right )^{2}}{\sec \left (d x +c \right )^{2}+2 \sec \left (d x +c \right )+1}d x \right ) e^{2}}{a^{2}} \] Input:

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

Output:

(sqrt(e)*int((sqrt(csc(c + d*x))*csc(c + d*x)**2)/(sec(c + d*x)**2 + 2*sec 
(c + d*x) + 1),x)*e**2)/a**2