\(\int \frac {1}{x \sin ^{\frac {5}{2}}(a+b \log (c x^n))} \, dx\) [68]

Optimal result

Integrand size = 19, antiderivative size = 68 \[ \int \frac {1}{x \sin ^{\frac {5}{2}}\left (a+b \log \left (c x^n\right )\right )} \, dx=\frac {2 \operatorname {EllipticF}\left (\frac {1}{2} \left (a-\frac {\pi }{2}+b \log \left (c x^n\right )\right ),2\right )}{3 b n}-\frac {2 \cos \left (a+b \log \left (c x^n\right )\right )}{3 b n \sin ^{\frac {3}{2}}\left (a+b \log \left (c x^n\right )\right )} \]

[Out]

-2/3*(sin(1/2*a+1/4*Pi+1/2*b*ln(c*x^n))^2)^(1/2)/sin(1/2*a+1/4*Pi+1/2*b*ln(c*x^n))*EllipticF(cos(1/2*a+1/4*Pi+
1/2*b*ln(c*x^n)),2^(1/2))/b/n-2/3*cos(a+b*ln(c*x^n))/b/n/sin(a+b*ln(c*x^n))^(3/2)

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {2716, 2720} \[ \int \frac {1}{x \sin ^{\frac {5}{2}}\left (a+b \log \left (c x^n\right )\right )} \, dx=\frac {2 \operatorname {EllipticF}\left (\frac {1}{2} \left (a+b \log \left (c x^n\right )-\frac {\pi }{2}\right ),2\right )}{3 b n}-\frac {2 \cos \left (a+b \log \left (c x^n\right )\right )}{3 b n \sin ^{\frac {3}{2}}\left (a+b \log \left (c x^n\right )\right )} \]

[In]

Int[1/(x*Sin[a + b*Log[c*x^n]]^(5/2)),x]

[Out]

(2*EllipticF[(a - Pi/2 + b*Log[c*x^n])/2, 2])/(3*b*n) - (2*Cos[a + b*Log[c*x^n]])/(3*b*n*Sin[a + b*Log[c*x^n]]
^(3/2))

Rule 2716

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] + Dist[(n + 2)/(b^2*(n + 1)), Int[(b*Sin[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1
] && IntegerQ[2*n]

Rule 2720

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]

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {1}{\sin ^{\frac {5}{2}}(a+b x)} \, dx,x,\log \left (c x^n\right )\right )}{n} \\ & = -\frac {2 \cos \left (a+b \log \left (c x^n\right )\right )}{3 b n \sin ^{\frac {3}{2}}\left (a+b \log \left (c x^n\right )\right )}+\frac {\text {Subst}\left (\int \frac {1}{\sqrt {\sin (a+b x)}} \, dx,x,\log \left (c x^n\right )\right )}{3 n} \\ & = \frac {2 \operatorname {EllipticF}\left (\frac {1}{2} \left (a-\frac {\pi }{2}+b \log \left (c x^n\right )\right ),2\right )}{3 b n}-\frac {2 \cos \left (a+b \log \left (c x^n\right )\right )}{3 b n \sin ^{\frac {3}{2}}\left (a+b \log \left (c x^n\right )\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.90 \[ \int \frac {1}{x \sin ^{\frac {5}{2}}\left (a+b \log \left (c x^n\right )\right )} \, dx=\frac {2 \left (\operatorname {EllipticF}\left (\frac {1}{4} \left (2 a-\pi +2 b \log \left (c x^n\right )\right ),2\right )-\frac {\cos \left (a+b \log \left (c x^n\right )\right )}{\sin ^{\frac {3}{2}}\left (a+b \log \left (c x^n\right )\right )}\right )}{3 b n} \]

[In]

Integrate[1/(x*Sin[a + b*Log[c*x^n]]^(5/2)),x]

[Out]

(2*(EllipticF[(2*a - Pi + 2*b*Log[c*x^n])/4, 2] - Cos[a + b*Log[c*x^n]]/Sin[a + b*Log[c*x^n]]^(3/2)))/(3*b*n)

Maple [A] (verified)

Time = 0.88 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.93

method	result	size
derivativedivides	\(\frac {\sqrt {\sin \left (a +b \ln \left (c \,x^{n}\right )\right )+1}\, \sqrt {-2 \sin \left (a +b \ln \left (c \,x^{n}\right )\right )+2}\, \sqrt {-\sin \left (a +b \ln \left (c \,x^{n}\right )\right )}\, \operatorname {EllipticF}\left (\sqrt {\sin \left (a +b \ln \left (c \,x^{n}\right )\right )+1}, \frac {\sqrt {2}}{2}\right ) \sin \left (a +b \ln \left (c \,x^{n}\right )\right )-2 {\cos \left (a +b \ln \left (c \,x^{n}\right )\right )}^{2}}{3 n {\sin \left (a +b \ln \left (c \,x^{n}\right )\right )}^{\frac {3}{2}} \cos \left (a +b \ln \left (c \,x^{n}\right )\right ) b}\)	\(131\)
default	\(\frac {\sqrt {\sin \left (a +b \ln \left (c \,x^{n}\right )\right )+1}\, \sqrt {-2 \sin \left (a +b \ln \left (c \,x^{n}\right )\right )+2}\, \sqrt {-\sin \left (a +b \ln \left (c \,x^{n}\right )\right )}\, \operatorname {EllipticF}\left (\sqrt {\sin \left (a +b \ln \left (c \,x^{n}\right )\right )+1}, \frac {\sqrt {2}}{2}\right ) \sin \left (a +b \ln \left (c \,x^{n}\right )\right )-2 {\cos \left (a +b \ln \left (c \,x^{n}\right )\right )}^{2}}{3 n {\sin \left (a +b \ln \left (c \,x^{n}\right )\right )}^{\frac {3}{2}} \cos \left (a +b \ln \left (c \,x^{n}\right )\right ) b}\)	\(131\)

[In]

int(1/x/sin(a+b*ln(c*x^n))^(5/2),x,method=_RETURNVERBOSE)

[Out]

1/3/n/sin(a+b*ln(c*x^n))^(3/2)*((sin(a+b*ln(c*x^n))+1)^(1/2)*(-2*sin(a+b*ln(c*x^n))+2)^(1/2)*(-sin(a+b*ln(c*x^
n)))^(1/2)*EllipticF((sin(a+b*ln(c*x^n))+1)^(1/2),1/2*2^(1/2))*sin(a+b*ln(c*x^n))-2*cos(a+b*ln(c*x^n))^2)/cos(
a+b*ln(c*x^n))/b

Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.09 (sec) , antiderivative size = 177, normalized size of antiderivative = 2.60 \[ \int \frac {1}{x \sin ^{\frac {5}{2}}\left (a+b \log \left (c x^n\right )\right )} \, dx=\frac {{\left (\sqrt {2} \sqrt {-i} \cos \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right )^{2} - \sqrt {2} \sqrt {-i}\right )} {\rm weierstrassPInverse}\left (4, 0, \cos \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right ) + i \, \sin \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right )\right ) + {\left (\sqrt {2} \sqrt {i} \cos \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right )^{2} - \sqrt {2} \sqrt {i}\right )} {\rm weierstrassPInverse}\left (4, 0, \cos \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right ) - i \, \sin \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right )\right ) + 2 \, \cos \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right ) \sqrt {\sin \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right )}}{3 \, {\left (b n \cos \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right )^{2} - b n\right )}} \]

[In]

integrate(1/x/sin(a+b*log(c*x^n))^(5/2),x, algorithm="fricas")

[Out]

1/3*((sqrt(2)*sqrt(-I)*cos(b*n*log(x) + b*log(c) + a)^2 - sqrt(2)*sqrt(-I))*weierstrassPInverse(4, 0, cos(b*n*
log(x) + b*log(c) + a) + I*sin(b*n*log(x) + b*log(c) + a)) + (sqrt(2)*sqrt(I)*cos(b*n*log(x) + b*log(c) + a)^2
- sqrt(2)*sqrt(I))*weierstrassPInverse(4, 0, cos(b*n*log(x) + b*log(c) + a) - I*sin(b*n*log(x) + b*log(c) + a
)) + 2*cos(b*n*log(x) + b*log(c) + a)*sqrt(sin(b*n*log(x) + b*log(c) + a)))/(b*n*cos(b*n*log(x) + b*log(c) + a
)^2 - b*n)

Sympy [F(-1)]

Timed out. \[ \int \frac {1}{x \sin ^{\frac {5}{2}}\left (a+b \log \left (c x^n\right )\right )} \, dx=\text {Timed out} \]

[In]

integrate(1/x/sin(a+b*ln(c*x**n))**(5/2),x)

[Out]

Timed out

Maxima [F]

\[ \int \frac {1}{x \sin ^{\frac {5}{2}}\left (a+b \log \left (c x^n\right )\right )} \, dx=\int { \frac {1}{x \sin \left (b \log \left (c x^{n}\right ) + a\right )^{\frac {5}{2}}} \,d x } \]

[In]

integrate(1/x/sin(a+b*log(c*x^n))^(5/2),x, algorithm="maxima")

[Out]

integrate(1/(x*sin(b*log(c*x^n) + a)^(5/2)), x)

Giac [F(-1)]

Timed out. \[ \int \frac {1}{x \sin ^{\frac {5}{2}}\left (a+b \log \left (c x^n\right )\right )} \, dx=\text {Timed out} \]

[In]

integrate(1/x/sin(a+b*log(c*x^n))^(5/2),x, algorithm="giac")

[Out]

Timed out

Mupad [B] (verification not implemented)

Time = 26.77 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.96 \[ \int \frac {1}{x \sin ^{\frac {5}{2}}\left (a+b \log \left (c x^n\right )\right )} \, dx=-\frac {\cos \left (a+b\,\ln \left (c\,x^n\right )\right )\,{\left ({\sin \left (a+b\,\ln \left (c\,x^n\right )\right )}^2\right )}^{3/4}\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {7}{4};\ \frac {3}{2};\ {\cos \left (a+b\,\ln \left (c\,x^n\right )\right )}^2\right )}{b\,n\,{\sin \left (a+b\,\ln \left (c\,x^n\right )\right )}^{3/2}} \]

[In]

int(1/(x*sin(a + b*log(c*x^n))^(5/2)),x)

[Out]

-(cos(a + b*log(c*x^n))*(sin(a + b*log(c*x^n))^2)^(3/4)*hypergeom([1/2, 7/4], 3/2, cos(a + b*log(c*x^n))^2))/(
b*n*sin(a + b*log(c*x^n))^(3/2))