\(\int \frac {(e \csc (c+d x))^{5/2}}{a+a \sec (c+d x)} \, dx\) [293]

Optimal result

Integrand size = 25, antiderivative size = 155 \[ \int \frac {(e \csc (c+d x))^{5/2}}{a+a \sec (c+d x)} \, dx=-\frac {4 e^2 \cot (c+d x) \sqrt {e \csc (c+d x)}}{21 a d}+\frac {2 e^2 \cot (c+d x) \csc ^2(c+d x) \sqrt {e \csc (c+d x)}}{7 a d}-\frac {2 e^2 \csc ^3(c+d x) \sqrt {e \csc (c+d x)}}{7 a d}+\frac {4 e^2 \sqrt {e \csc (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right ),2\right ) \sqrt {\sin (c+d x)}}{21 a d} \] Output:

-4/21*e^2*cot(d*x+c)*(e*csc(d*x+c))^(1/2)/a/d+2/7*e^2*cot(d*x+c)*csc(d*x+c 
)^2*(e*csc(d*x+c))^(1/2)/a/d-2/7*e^2*csc(d*x+c)^3*(e*csc(d*x+c))^(1/2)/a/d 
+4/21*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)/a/d
 

Mathematica [A] (verified)

Time = 1.49 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.85 \[ \int \frac {(e \csc (c+d x))^{5/2}}{a+a \sec (c+d x)} \, dx=-\frac {\csc ^2\left (\frac {1}{2} (c+d x)\right ) (e \csc (c+d x))^{5/2} \sec ^4\left (\frac {1}{2} (c+d x)\right ) \left ((2+\cos (c+d x)-2 \cos (2 (c+d x))-\cos (3 (c+d x))) \operatorname {EllipticF}\left (\frac {1}{4} (-2 c+\pi -2 d x),2\right )+2 (4+2 \cos (c+d x)+\cos (2 (c+d x))) \sqrt {\sin (c+d x)}\right ) \sin ^{\frac {5}{2}}(c+d x)}{168 a d} \] Input:

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

Output:

-1/168*(Csc[(c + d*x)/2]^2*(e*Csc[c + d*x])^(5/2)*Sec[(c + d*x)/2]^4*((2 + 
 Cos[c + d*x] - 2*Cos[2*(c + d*x)] - Cos[3*(c + d*x)])*EllipticF[(-2*c + P 
i - 2*d*x)/4, 2] + 2*(4 + 2*Cos[c + d*x] + Cos[2*(c + d*x)])*Sqrt[Sin[c + 
d*x]])*Sin[c + d*x]^(5/2))/(a*d)
 

Rubi [A] (verified)

Time = 0.82 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.82, number of steps used = 19, number of rules used = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.720, Rules used = {3042, 4366, 3042, 4360, 25, 25, 3042, 25, 3318, 25, 3042, 3044, 15, 3047, 3042, 3116, 3042, 3120}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4366

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4360

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3318

\(\displaystyle -e^2 \sqrt {\sin (c+d x)} \sqrt {e \csc (c+d x)} \left (\frac {\int \frac {\cos ^2(c+d x)}{\sin ^{\frac {9}{2}}(c+d x)}dx}{a}+\frac {\int -\frac {\cos (c+d x)}{\sin ^{\frac {9}{2}}(c+d x)}dx}{a}\right )\)

\(\Big \downarrow \) 25

\(\displaystyle -e^2 \sqrt {\sin (c+d x)} \sqrt {e \csc (c+d x)} \left (\frac {\int \frac {\cos ^2(c+d x)}{\sin ^{\frac {9}{2}}(c+d x)}dx}{a}-\frac {\int \frac {\cos (c+d x)}{\sin ^{\frac {9}{2}}(c+d x)}dx}{a}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle -e^2 \sqrt {\sin (c+d x)} \sqrt {e \csc (c+d x)} \left (\frac {\int \frac {\cos (c+d x)^2}{\sin (c+d x)^{9/2}}dx}{a}-\frac {\int \frac {\cos (c+d x)}{\sin (c+d x)^{9/2}}dx}{a}\right )\)

\(\Big \downarrow \) 3044

\(\displaystyle -e^2 \sqrt {\sin (c+d x)} \sqrt {e \csc (c+d x)} \left (\frac {\int \frac {\cos (c+d x)^2}{\sin (c+d x)^{9/2}}dx}{a}-\frac {\int \frac {1}{\sin ^{\frac {9}{2}}(c+d x)}d\sin (c+d x)}{a d}\right )\)

\(\Big \downarrow \) 15

\(\displaystyle -e^2 \sqrt {\sin (c+d x)} \sqrt {e \csc (c+d x)} \left (\frac {\int \frac {\cos (c+d x)^2}{\sin (c+d x)^{9/2}}dx}{a}+\frac {2}{7 a d \sin ^{\frac {7}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 3047

\(\displaystyle -e^2 \sqrt {\sin (c+d x)} \sqrt {e \csc (c+d x)} \left (\frac {-\frac {2}{7} \int \frac {1}{\sin ^{\frac {5}{2}}(c+d x)}dx-\frac {2 \cos (c+d x)}{7 d \sin ^{\frac {7}{2}}(c+d x)}}{a}+\frac {2}{7 a d \sin ^{\frac {7}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle -e^2 \sqrt {\sin (c+d x)} \sqrt {e \csc (c+d x)} \left (\frac {-\frac {2}{7} \int \frac {1}{\sin (c+d x)^{5/2}}dx-\frac {2 \cos (c+d x)}{7 d \sin ^{\frac {7}{2}}(c+d x)}}{a}+\frac {2}{7 a d \sin ^{\frac {7}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 3116

\(\displaystyle -e^2 \sqrt {\sin (c+d x)} \sqrt {e \csc (c+d x)} \left (\frac {-\frac {2}{7} \left (\frac {1}{3} \int \frac {1}{\sqrt {\sin (c+d x)}}dx-\frac {2 \cos (c+d x)}{3 d \sin ^{\frac {3}{2}}(c+d x)}\right )-\frac {2 \cos (c+d x)}{7 d \sin ^{\frac {7}{2}}(c+d x)}}{a}+\frac {2}{7 a d \sin ^{\frac {7}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle -e^2 \sqrt {\sin (c+d x)} \sqrt {e \csc (c+d x)} \left (\frac {-\frac {2}{7} \left (\frac {1}{3} \int \frac {1}{\sqrt {\sin (c+d x)}}dx-\frac {2 \cos (c+d x)}{3 d \sin ^{\frac {3}{2}}(c+d x)}\right )-\frac {2 \cos (c+d x)}{7 d \sin ^{\frac {7}{2}}(c+d x)}}{a}+\frac {2}{7 a d \sin ^{\frac {7}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 3120

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 15

Int[(a_.)*(x_)^(m_.), x_Symbol] :> Simp[a*(x^(m + 1)/(m + 1)), x] /; FreeQ[ 
{a, m}, x] && NeQ[m, -1]
 

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 3042

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

rule 3044

Int[cos[(e_.) + (f_.)*(x_)]^(n_.)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_ 
Symbol] :> Simp[1/(a*f)   Subst[Int[x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a 
*Sin[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2] &&  !(I 
ntegerQ[(m - 1)/2] && LtQ[0, m, n])
 

rule 3047

Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_)*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n 
_), x_Symbol] :> Simp[a*(a*Cos[e + f*x])^(m - 1)*((b*Sin[e + f*x])^(n + 1)/ 
(b*f*(n + 1))), x] + Simp[a^2*((m - 1)/(b^2*(n + 1)))   Int[(a*Cos[e + f*x] 
)^(m - 2)*(b*Sin[e + f*x])^(n + 2), x], x] /; FreeQ[{a, b, e, f}, x] && GtQ 
[m, 1] && LtQ[n, -1] && (IntegersQ[2*m, 2*n] || EqQ[m + n, 0])
 

rule 3116

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[Cos[c + d*x]*(( 
b*Sin[c + d*x])^(n + 1)/(b*d*(n + 1))), x] + Simp[(n + 2)/(b^2*(n + 1))   I 
nt[(b*Sin[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1] && 
 IntegerQ[2*n]
 

rule 3120

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2 
)*(c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
 

rule 3318

Int[((cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^( 
n_.))/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[g^2/a   Int 
[(g*Cos[e + f*x])^(p - 2)*(d*Sin[e + f*x])^n, x], x] - Simp[g^2/(b*d)   Int 
[(g*Cos[e + f*x])^(p - 2)*(d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, 
d, e, f, g, n, p}, x] && EqQ[a^2 - b^2, 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 = 1.00 (sec) , antiderivative size = 181, normalized size of antiderivative = 1.17

method	result	size
default	\(\frac {\sqrt {2}\, e^{2} \sqrt {e \csc \left (d x +c \right )}\, \left (i \left (2 \cos \left (d x +c \right )^{2}+4 \cos \left (d x +c \right )+2\right ) \sqrt {1-i \cot \left (d x +c \right )+i \csc \left (d x +c \right )}\, \sqrt {i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}\, \operatorname {EllipticF}\left (\sqrt {1+i \cot \left (d x +c \right )-i \csc \left (d x +c \right )}, \frac {\sqrt {2}}{2}\right ) \sqrt {1+i \cot \left (d x +c \right )-i \csc \left (d x +c \right )}+\left (-2 \cos \left (d x +c \right )^{2}-2 \cos \left (d x +c \right )-3\right ) \sqrt {2}\, \csc \left (d x +c \right )\right )}{21 a d \left (1+\cos \left (d x +c \right )\right )}\)	\(181\)

Input:

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

Output:

1/21/a/d*2^(1/2)*e^2*(e*csc(d*x+c))^(1/2)/(1+cos(d*x+c))*(I*(2*cos(d*x+c)^ 
2+4*cos(d*x+c)+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))*( 
1+I*cot(d*x+c)-I*csc(d*x+c))^(1/2)+(-2*cos(d*x+c)^2-2*cos(d*x+c)-3)*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 = 163, normalized size of antiderivative = 1.05 \[ \int \frac {(e \csc (c+d x))^{5/2}}{a+a \sec (c+d x)} \, dx=-\frac {2 \, {\left ({\left (i \, e^{2} \cos \left (d x + c\right ) + i \, e^{2}\right )} \sqrt {2 i \, e} \sin \left (d x + c\right ) {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + {\left (-i \, e^{2} \cos \left (d x + c\right ) - i \, e^{2}\right )} \sqrt {-2 i \, e} \sin \left (d x + c\right ) {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + {\left (2 \, e^{2} \cos \left (d x + c\right )^{2} + 2 \, e^{2} \cos \left (d x + c\right ) + 3 \, e^{2}\right )} \sqrt {\frac {e}{\sin \left (d x + c\right )}}\right )}}{21 \, {\left (a d \cos \left (d x + c\right ) + a d\right )} \sin \left (d x + c\right )} \] Input:

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

Output:

-2/21*((I*e^2*cos(d*x + c) + I*e^2)*sqrt(2*I*e)*sin(d*x + c)*weierstrassPI 
nverse(4, 0, cos(d*x + c) + I*sin(d*x + c)) + (-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)^2 + 2*e^2*cos(d*x + c) + 3*e^2)*sqrt(e/si 
n(d*x + c)))/((a*d*cos(d*x + c) + a*d)*sin(d*x + c))
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F(-1)]

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

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

Output:

Timed out
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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