\(\int \frac {\sqrt {\sin (a+b \log (c x^n))}}{x} \, dx\) [55]

Optimal result

Integrand size = 19, antiderivative size = 29 \[ \int \frac {\sqrt {\sin \left (a+b \log \left (c x^n\right )\right )}}{x} \, dx=\frac {2 E\left (\left .\frac {1}{2} \left (a-\frac {\pi }{2}+b \log \left (c x^n\right )\right )\right |2\right )}{b n} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.053, Rules used = {2719} \[ \int \frac {\sqrt {\sin \left (a+b \log \left (c x^n\right )\right )}}{x} \, dx=\frac {2 E\left (\left .\frac {1}{2} \left (a+b \log \left (c x^n\right )-\frac {\pi }{2}\right )\right |2\right )}{b n} \]

[In]

Int[Sqrt[Sin[a + b*Log[c*x^n]]]/x,x]

[Out]

(2*EllipticE[(a - Pi/2 + b*Log[c*x^n])/2, 2])/(b*n)

Rule 2719

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]

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

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.10 \[ \int \frac {\sqrt {\sin \left (a+b \log \left (c x^n\right )\right )}}{x} \, dx=-\frac {2 E\left (\left .\frac {1}{2} \left (-a+\frac {\pi }{2}-b \log \left (c x^n\right )\right )\right |2\right )}{b n} \]

[In]

Integrate[Sqrt[Sin[a + b*Log[c*x^n]]]/x,x]

[Out]

(-2*EllipticE[(-a + Pi/2 - b*Log[c*x^n])/2, 2])/(b*n)

Maple [A] (verified)

Time = 0.99 (sec) , antiderivative size = 129, normalized size of antiderivative = 4.45

method	result	size
derivativedivides	\(-\frac {\sqrt {\sin \left (a +b \ln \left (c \,x^{n}\right )\right )+1}\, \sqrt {-2 \sin \left (a +b \ln \left (c \,x^{n}\right )\right )+2}\, \sqrt {-\sin \left (a +b \ln \left (c \,x^{n}\right )\right )}\, \left (2 \operatorname {EllipticE}\left (\sqrt {\sin \left (a +b \ln \left (c \,x^{n}\right )\right )+1}, \frac {\sqrt {2}}{2}\right )-\operatorname {EllipticF}\left (\sqrt {\sin \left (a +b \ln \left (c \,x^{n}\right )\right )+1}, \frac {\sqrt {2}}{2}\right )\right )}{n \cos \left (a +b \ln \left (c \,x^{n}\right )\right ) \sqrt {\sin \left (a +b \ln \left (c \,x^{n}\right )\right )}\, b}\)	\(129\)
default	\(-\frac {\sqrt {\sin \left (a +b \ln \left (c \,x^{n}\right )\right )+1}\, \sqrt {-2 \sin \left (a +b \ln \left (c \,x^{n}\right )\right )+2}\, \sqrt {-\sin \left (a +b \ln \left (c \,x^{n}\right )\right )}\, \left (2 \operatorname {EllipticE}\left (\sqrt {\sin \left (a +b \ln \left (c \,x^{n}\right )\right )+1}, \frac {\sqrt {2}}{2}\right )-\operatorname {EllipticF}\left (\sqrt {\sin \left (a +b \ln \left (c \,x^{n}\right )\right )+1}, \frac {\sqrt {2}}{2}\right )\right )}{n \cos \left (a +b \ln \left (c \,x^{n}\right )\right ) \sqrt {\sin \left (a +b \ln \left (c \,x^{n}\right )\right )}\, b}\)	\(129\)

[In]

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

[Out]

-1/n*(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)*(2*EllipticE((si
n(a+b*ln(c*x^n))+1)^(1/2),1/2*2^(1/2))-EllipticF((sin(a+b*ln(c*x^n))+1)^(1/2),1/2*2^(1/2)))/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 = 0.09 (sec) , antiderivative size = 90, normalized size of antiderivative = 3.10 \[ \int \frac {\sqrt {\sin \left (a+b \log \left (c x^n\right )\right )}}{x} \, dx=\frac {i \, \sqrt {2} \sqrt {-i} {\rm weierstrassZeta}\left (4, 0, {\rm weierstrassPInverse}\left (4, 0, \cos \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right ) + i \, \sin \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right )\right )\right ) - i \, \sqrt {2} \sqrt {i} {\rm weierstrassZeta}\left (4, 0, {\rm weierstrassPInverse}\left (4, 0, \cos \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right ) - i \, \sin \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right )\right )\right )}{b n} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 26.55 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.90 \[ \int \frac {\sqrt {\sin \left (a+b \log \left (c x^n\right )\right )}}{x} \, dx=\frac {2\,\mathrm {E}\left (\frac {a}{2}-\frac {\pi }{4}+\frac {b\,\ln \left (c\,x^n\right )}{2}\middle |2\right )}{b\,n} \]

[In]

int(sin(a + b*log(c*x^n))^(1/2)/x,x)

[Out]

(2*ellipticE(a/2 - pi/4 + (b*log(c*x^n))/2, 2))/(b*n)