\(\int \frac {1}{x \sqrt {\sin (a+b \log (c x^n))}} \, dx\) [64]

Optimal result

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

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

[In]

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

[Out]

(2*EllipticF[(a - Pi/2 + b*Log[c*x^n])/2, 2])/(b*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}{\sqrt {\sin (a+b x)}} \, dx,x,\log \left (c x^n\right )\right )}{n} \\ & = \frac {2 \operatorname {EllipticF}\left (\frac {1}{2} \left (a-\frac {\pi }{2}+b \log \left (c x^n\right )\right ),2\right )}{b n} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.10 \[ \int \frac {1}{x \sqrt {\sin \left (a+b \log \left (c x^n\right )\right )}} \, dx=-\frac {2 \operatorname {EllipticF}\left (\frac {1}{2} \left (-a+\frac {\pi }{2}-b \log \left (c x^n\right )\right ),2\right )}{b n} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.87 (sec) , antiderivative size = 102, normalized size of antiderivative = 3.52

method	result	size
derivativedivides	\(\frac {\sqrt {\sin \left (a +b \ln \left (c \,x^{n}\right )\right )+1}\, \sqrt {-2 \sin \left (a +b \ln \left (c \,x^{n}\right )\right )+2}\, \sqrt {-\sin \left (a +b \ln \left (c \,x^{n}\right )\right )}\, \operatorname {EllipticF}\left (\sqrt {\sin \left (a +b \ln \left (c \,x^{n}\right )\right )+1}, \frac {\sqrt {2}}{2}\right )}{n \cos \left (a +b \ln \left (c \,x^{n}\right )\right ) \sqrt {\sin \left (a +b \ln \left (c \,x^{n}\right )\right )}\, b}\)	\(102\)
default	\(\frac {\sqrt {\sin \left (a +b \ln \left (c \,x^{n}\right )\right )+1}\, \sqrt {-2 \sin \left (a +b \ln \left (c \,x^{n}\right )\right )+2}\, \sqrt {-\sin \left (a +b \ln \left (c \,x^{n}\right )\right )}\, \operatorname {EllipticF}\left (\sqrt {\sin \left (a +b \ln \left (c \,x^{n}\right )\right )+1}, \frac {\sqrt {2}}{2}\right )}{n \cos \left (a +b \ln \left (c \,x^{n}\right )\right ) \sqrt {\sin \left (a +b \ln \left (c \,x^{n}\right )\right )}\, b}\)	\(102\)

[In]

int(1/x/sin(a+b*ln(c*x^n))^(1/2),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)*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 = 0.08 (sec) , antiderivative size = 82, normalized size of antiderivative = 2.83 \[ \int \frac {1}{x \sqrt {\sin \left (a+b \log \left (c x^n\right )\right )}} \, dx=\frac {\sqrt {2} \sqrt {-i} {\rm weierstrassPInverse}\left (4, 0, \cos \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right ) + i \, \sin \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right )\right ) + \sqrt {2} \sqrt {i} {\rm weierstrassPInverse}\left (4, 0, \cos \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right ) - i \, \sin \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right )\right )}{b n} \]

[In]

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

[Out]

(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)
))/(b*n)

Sympy [F]

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

[In]

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

[Out]

Integral(1/(x*sqrt(sin(a + b*log(c*x**n)))), x)

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 25.90 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.90 \[ \int \frac {1}{x \sqrt {\sin \left (a+b \log \left (c x^n\right )\right )}} \, dx=-\frac {2\,\mathrm {F}\left (\frac {\pi }{4}-\frac {a}{2}-\frac {b\,\ln \left (c\,x^n\right )}{2}\middle |2\right )}{b\,n} \]

[In]

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

[Out]

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