\(\int \frac {\sin ^{\frac {3}{2}}(a+b \log (c x^n))}{x} \, dx\) [60]

Optimal result

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

[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))*sin(a+b*ln(c*x^n))^(1/2)/b/n

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 = {2715, 2720} \[ \int \frac {\sin ^{\frac {3}{2}}\left (a+b \log \left (c x^n\right )\right )}{x} \, dx=\frac {2 \operatorname {EllipticF}\left (\frac {1}{2} \left (a+b \log \left (c x^n\right )-\frac {\pi }{2}\right ),2\right )}{3 b n}-\frac {2 \sqrt {\sin \left (a+b \log \left (c x^n\right )\right )} \cos \left (a+b \log \left (c x^n\right )\right )}{3 b n} \]

[In]

Int[Sin[a + b*Log[c*x^n]]^(3/2)/x,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]]*Sqrt[Sin[a + b*Log[c*x^n]]])/
(3*b*n)

Rule 2715

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d*x]*((b*Sin[c + d*x])^(n - 1)/(d*n))
, x] + Dist[b^2*((n - 1)/n), Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integ
erQ[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 \sin ^{\frac {3}{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 ) \sqrt {\sin \left (a+b \log \left (c x^n\right )\right )}}{3 b n}+\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 ) \sqrt {\sin \left (a+b \log \left (c x^n\right )\right )}}{3 b n} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.12 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.85 \[ \int \frac {\sin ^{\frac {3}{2}}\left (a+b \log \left (c x^n\right )\right )}{x} \, dx=-\frac {2 \left (\operatorname {EllipticF}\left (\frac {1}{4} \left (-2 a+\pi -2 b \log \left (c x^n\right )\right ),2\right )+\cos \left (a+b \log \left (c x^n\right )\right ) \sqrt {\sin \left (a+b \log \left (c x^n\right )\right )}\right )}{3 b n} \]

[In]

Integrate[Sin[a + b*Log[c*x^n]]^(3/2)/x,x]

[Out]

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

Maple [A] (verified)

Time = 0.95 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.93

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

[In]

int(sin(a+b*ln(c*x^n))^(3/2)/x,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [C] (verification not implemented)

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

Time = 0.10 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.63 \[ \int \frac {\sin ^{\frac {3}{2}}\left (a+b \log \left (c x^n\right )\right )}{x} \, dx=\frac {\sqrt {2} \sqrt {-i} {\rm weierstrassPInverse}\left (4, 0, \cos \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right ) + i \, \sin \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right )\right ) + \sqrt {2} \sqrt {i} {\rm weierstrassPInverse}\left (4, 0, \cos \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right ) - i \, \sin \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right )\right ) - 2 \, \cos \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right ) \sqrt {\sin \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right )}}{3 \, b n} \]

[In]

integrate(sin(a+b*log(c*x^n))^(3/2)/x,x, algorithm="fricas")

[Out]

1/3*(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)*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)

Sympy [F]

\[ \int \frac {\sin ^{\frac {3}{2}}\left (a+b \log \left (c x^n\right )\right )}{x} \, dx=\int \frac {\sin ^{\frac {3}{2}}{\left (a + b \log {\left (c x^{n} \right )} \right )}}{x}\, dx \]

[In]

integrate(sin(a+b*ln(c*x**n))**(3/2)/x,x)

[Out]

Integral(sin(a + b*log(c*x**n))**(3/2)/x, x)

Maxima [F]

\[ \int \frac {\sin ^{\frac {3}{2}}\left (a+b \log \left (c x^n\right )\right )}{x} \, dx=\int { \frac {\sin \left (b \log \left (c x^{n}\right ) + a\right )^{\frac {3}{2}}}{x} \,d x } \]

[In]

integrate(sin(a+b*log(c*x^n))^(3/2)/x,x, algorithm="maxima")

[Out]

integrate(sin(b*log(c*x^n) + a)^(3/2)/x, x)

Giac [F]

\[ \int \frac {\sin ^{\frac {3}{2}}\left (a+b \log \left (c x^n\right )\right )}{x} \, dx=\int { \frac {\sin \left (b \log \left (c x^{n}\right ) + a\right )^{\frac {3}{2}}}{x} \,d x } \]

[In]

integrate(sin(a+b*log(c*x^n))^(3/2)/x,x, algorithm="giac")

[Out]

integrate(sin(b*log(c*x^n) + a)^(3/2)/x, x)

Mupad [B] (verification not implemented)

Time = 26.06 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.96 \[ \int \frac {\sin ^{\frac {3}{2}}\left (a+b \log \left (c x^n\right )\right )}{x} \, dx=-\frac {\cos \left (a+b\,\ln \left (c\,x^n\right )\right )\,{\sin \left (a+b\,\ln \left (c\,x^n\right )\right )}^{5/2}\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {1}{2};\ \frac {3}{2};\ {\cos \left (a+b\,\ln \left (c\,x^n\right )\right )}^2\right )}{b\,n\,{\left ({\sin \left (a+b\,\ln \left (c\,x^n\right )\right )}^2\right )}^{5/4}} \]

[In]

int(sin(a + b*log(c*x^n))^(3/2)/x,x)

[Out]

-(cos(a + b*log(c*x^n))*sin(a + b*log(c*x^n))^(5/2)*hypergeom([-1/4, 1/2], 3/2, cos(a + b*log(c*x^n))^2))/(b*n
*(sin(a + b*log(c*x^n))^2)^(5/4))