\(\int \frac {1}{(e \csc (c+d x))^{5/2} (a+a \sec (c+d x))} \, dx\) [298]

Optimal result

Integrand size = 25, antiderivative size = 120 \[ \int \frac {1}{(e \csc (c+d x))^{5/2} (a+a \sec (c+d x))} \, dx=-\frac {4 E\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right )}{5 a d e^2 \sqrt {e \csc (c+d x)} \sqrt {\sin (c+d x)}}+\frac {2 \sin (c+d x)}{3 a d e^2 \sqrt {e \csc (c+d x)}}-\frac {2 \cos (c+d x) \sin (c+d x)}{5 a d e^2 \sqrt {e \csc (c+d x)}} \] Output:

4/5*EllipticE(cos(1/2*c+1/4*Pi+1/2*d*x),2^(1/2))/a/d/e^2/(e*csc(d*x+c))^(1 
/2)/sin(d*x+c)^(1/2)+2/3*sin(d*x+c)/a/d/e^2/(e*csc(d*x+c))^(1/2)-2/5*cos(d 
*x+c)*sin(d*x+c)/a/d/e^2/(e*csc(d*x+c))^(1/2)
 

Mathematica [C] (verified)

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

Time = 1.34 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.83 \[ \int \frac {1}{(e \csc (c+d x))^{5/2} (a+a \sec (c+d x))} \, dx=\frac {8 \sqrt {1-e^{2 i (c+d x)}} (i+\cot (c+d x)) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},e^{2 i (c+d x)}\right )+20 \sin (c+d x)-6 (4 i+\sin (2 (c+d x)))}{30 a d e^2 \sqrt {e \csc (c+d x)}} \] Input:

Integrate[1/((e*Csc[c + d*x])^(5/2)*(a + a*Sec[c + d*x])),x]
 

Output:

(8*Sqrt[1 - E^((2*I)*(c + d*x))]*(I + Cot[c + d*x])*Hypergeometric2F1[1/2, 
 3/4, 7/4, E^((2*I)*(c + d*x))] + 20*Sin[c + d*x] - 6*(4*I + Sin[2*(c + d* 
x)]))/(30*a*d*e^2*Sqrt[e*Csc[c + d*x]])
 

Rubi [A] (verified)

Time = 0.71 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.82, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.560, Rules used = {3042, 4366, 3042, 4360, 25, 25, 3042, 3318, 3042, 3044, 15, 3049, 3042, 3119}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4366

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4360

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3318

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3044

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

\(\Big \downarrow \) 15

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

\(\Big \downarrow \) 3049

\(\displaystyle \frac {\frac {2 \sin ^{\frac {3}{2}}(c+d x)}{3 a d}-\frac {\frac {2}{5} \int \sqrt {\sin (c+d x)}dx+\frac {2 \sin ^{\frac {3}{2}}(c+d x) \cos (c+d x)}{5 d}}{a}}{e^2 \sqrt {\sin (c+d x)} \sqrt {e \csc (c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {2 \sin ^{\frac {3}{2}}(c+d x)}{3 a d}-\frac {\frac {2}{5} \int \sqrt {\sin (c+d x)}dx+\frac {2 \sin ^{\frac {3}{2}}(c+d x) \cos (c+d x)}{5 d}}{a}}{e^2 \sqrt {\sin (c+d x)} \sqrt {e \csc (c+d x)}}\)

\(\Big \downarrow \) 3119

\(\displaystyle \frac {\frac {2 \sin ^{\frac {3}{2}}(c+d x)}{3 a d}-\frac {\frac {4 E\left (\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right )}{5 d}+\frac {2 \sin ^{\frac {3}{2}}(c+d x) \cos (c+d x)}{5 d}}{a}}{e^2 \sqrt {\sin (c+d x)} \sqrt {e \csc (c+d x)}}\)

Input:

Int[1/((e*Csc[c + d*x])^(5/2)*(a + a*Sec[c + d*x])),x]
 

Output:

((2*Sin[c + d*x]^(3/2))/(3*a*d) - ((4*EllipticE[(c - Pi/2 + d*x)/2, 2])/(5 
*d) + (2*Cos[c + d*x]*Sin[c + d*x]^(3/2))/(5*d))/a)/(e^2*Sqrt[e*Csc[c + d* 
x]]*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 3049

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

rule 3119

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(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 = 1.14 (sec) , antiderivative size = 272, normalized size of antiderivative = 2.27

method	result	size
default	\(\frac {\sqrt {2}\, \left (\left (12 \cos \left (d x +c \right )+12\right ) \sqrt {i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}\, \operatorname {EllipticE}\left (\sqrt {1+i \cot \left (d x +c \right )-i \csc \left (d x +c \right )}, \frac {\sqrt {2}}{2}\right ) \sqrt {1-i \cot \left (d x +c \right )+i \csc \left (d x +c \right )}\, \sqrt {1+i \cot \left (d x +c \right )-i \csc \left (d x +c \right )}+\left (-6 \cos \left (d x +c \right )-6\right ) \sqrt {i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}\, \sqrt {1-i \cot \left (d x +c \right )+i \csc \left (d x +c \right )}\, \operatorname {EllipticF}\left (\sqrt {1+i \cot \left (d x +c \right )-i \csc \left (d x +c \right )}, \frac {\sqrt {2}}{2}\right ) \sqrt {1+i \cot \left (d x +c \right )-i \csc \left (d x +c \right )}+\left (3 \cos \left (d x +c \right )^{3}-5 \cos \left (d x +c \right )^{2}+3 \cos \left (d x +c \right )-1\right ) \sqrt {2}\right ) \csc \left (d x +c \right )}{15 a d \sqrt {e \csc \left (d x +c \right )}\, e^{2}}\)	\(272\)

Input:

int(1/(e*csc(d*x+c))^(5/2)/(a+a*sec(d*x+c)),x,method=_RETURNVERBOSE)
 

Output:

1/15/a/d*2^(1/2)*((12*cos(d*x+c)+12)*(I*(csc(d*x+c)-cot(d*x+c)))^(1/2)*Ell 
ipticE((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)*(1+I*cot(d*x+c)-I*csc(d*x+c))^(1/2)+(-6*cos(d*x+c)-6)*(I 
*(csc(d*x+c)-cot(d*x+c)))^(1/2)*(1-I*cot(d*x+c)+I*csc(d*x+c))^(1/2)*Ellipt 
icF((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)+(3*cos(d*x+c)^3-5*cos(d*x+c)^2+3*cos(d*x+c)-1)*2^(1/2))/(e* 
csc(d*x+c))^(1/2)/e^2*csc(d*x+c)
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.09 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.92 \[ \int \frac {1}{(e \csc (c+d x))^{5/2} (a+a \sec (c+d x))} \, dx=\frac {2 \, {\left ({\left (3 \, \cos \left (d x + c\right )^{3} - 5 \, \cos \left (d x + c\right )^{2} - 3 \, \cos \left (d x + c\right ) + 5\right )} \sqrt {\frac {e}{\sin \left (d x + c\right )}} - 3 \, \sqrt {2 i \, e} {\rm weierstrassZeta}\left (4, 0, {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 3 \, \sqrt {-2 i \, e} {\rm weierstrassZeta}\left (4, 0, {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right )\right )}}{15 \, a d e^{3}} \] Input:

integrate(1/(e*csc(d*x+c))^(5/2)/(a+a*sec(d*x+c)),x, algorithm="fricas")
 

Output:

2/15*((3*cos(d*x + c)^3 - 5*cos(d*x + c)^2 - 3*cos(d*x + c) + 5)*sqrt(e/si 
n(d*x + c)) - 3*sqrt(2*I*e)*weierstrassZeta(4, 0, weierstrassPInverse(4, 0 
, cos(d*x + c) + I*sin(d*x + c))) - 3*sqrt(-2*I*e)*weierstrassZeta(4, 0, w 
eierstrassPInverse(4, 0, cos(d*x + c) - I*sin(d*x + c))))/(a*d*e^3)
 

Sympy [F(-1)]

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

integrate(1/(e*csc(d*x+c))**(5/2)/(a+a*sec(d*x+c)),x)
 

Output:

Timed out
                                                                                    
                                                                                    
 

Maxima [F]

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

integrate(1/(e*csc(d*x+c))^(5/2)/(a+a*sec(d*x+c)),x, algorithm="maxima")
 

Output:

integrate(1/((e*csc(d*x + c))^(5/2)*(a*sec(d*x + c) + a)), x)
 

Giac [F]

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

integrate(1/(e*csc(d*x+c))^(5/2)/(a+a*sec(d*x+c)),x, algorithm="giac")
 

Output:

integrate(1/((e*csc(d*x + c))^(5/2)*(a*sec(d*x + c) + a)), x)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {1}{(e \csc (c+d x))^{5/2} (a+a \sec (c+d x))} \, dx=\frac {\sqrt {e}\, \left (\int \frac {\sqrt {\csc \left (d x +c \right )}}{\csc \left (d x +c \right )^{3} \sec \left (d x +c \right )+\csc \left (d x +c \right )^{3}}d x \right )}{a \,e^{3}} \] Input:

int(1/(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)**3*sec(c + d*x) + csc(c + d* 
x)**3),x))/(a*e**3)