3.31 \(\int \frac{\sin (a+\sqrt{-\frac{1}{n^2}} \log (c x^n))}{x^2} \, dx\)

Optimal. Leaf size=86 \[ \frac{\sqrt{-\frac{1}{n^2}} n e^{a \sqrt{-\frac{1}{n^2}} n} \left (c x^n\right )^{-1/n}}{4 x}+\frac{\sqrt{-\frac{1}{n^2}} n e^{-a \sqrt{-\frac{1}{n^2}} n} \log (x) \left (c x^n\right )^{\frac{1}{n}}}{2 x} \]

[Out]

(E^(a*Sqrt[-n^(-2)]*n)*Sqrt[-n^(-2)]*n)/(4*x*(c*x^n)^n^(-1)) + (Sqrt[-n^(-2)]*n*(c*x^n)^n^(-1)*Log[x])/(2*E^(a
*Sqrt[-n^(-2)]*n)*x)

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Rubi [A]  time = 0.0611679, antiderivative size = 86, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.087, Rules used = {4493, 4489} \[ \frac{\sqrt{-\frac{1}{n^2}} n e^{a \sqrt{-\frac{1}{n^2}} n} \left (c x^n\right )^{-1/n}}{4 x}+\frac{\sqrt{-\frac{1}{n^2}} n e^{-a \sqrt{-\frac{1}{n^2}} n} \log (x) \left (c x^n\right )^{\frac{1}{n}}}{2 x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(E^(a*Sqrt[-n^(-2)]*n)*Sqrt[-n^(-2)]*n)/(4*x*(c*x^n)^n^(-1)) + (Sqrt[-n^(-2)]*n*(c*x^n)^n^(-1)*Log[x])/(2*E^(a
*Sqrt[-n^(-2)]*n)*x)

Rule 4493

Int[((e_.)*(x_))^(m_.)*Sin[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*(d_.)]^(p_.), x_Symbol] :> Dist[(e*x)^(m + 1)
/(e*n*(c*x^n)^((m + 1)/n)), Subst[Int[x^((m + 1)/n - 1)*Sin[d*(a + b*Log[x])]^p, x], x, c*x^n], x] /; FreeQ[{a
, b, c, d, e, m, n, p}, x] && (NeQ[c, 1] || NeQ[n, 1])

Rule 4489

Int[((e_.)*(x_))^(m_.)*Sin[((a_.) + Log[x_]*(b_.))*(d_.)]^(p_.), x_Symbol] :> Dist[(m + 1)^p/(2^p*b^p*d^p*p^p)
, Int[ExpandIntegrand[(e*x)^m*(E^((a*b*d^2*p)/(m + 1))/x^((m + 1)/p) - x^((m + 1)/p)/E^((a*b*d^2*p)/(m + 1)))^
p, x], x], x] /; FreeQ[{a, b, d, e, m}, x] && IGtQ[p, 0] && EqQ[b^2*d^2*p^2 + (m + 1)^2, 0]

Rubi steps

\begin{align*} \int \frac{\sin \left (a+\sqrt{-\frac{1}{n^2}} \log \left (c x^n\right )\right )}{x^2} \, dx &=\frac{\left (c x^n\right )^{\frac{1}{n}} \operatorname{Subst}\left (\int x^{-1-\frac{1}{n}} \sin \left (a+\sqrt{-\frac{1}{n^2}} \log (x)\right ) \, dx,x,c x^n\right )}{n x}\\ &=\frac{\left (\sqrt{-\frac{1}{n^2}} \left (c x^n\right )^{\frac{1}{n}}\right ) \operatorname{Subst}\left (\int \left (\frac{e^{-a \sqrt{-\frac{1}{n^2}} n}}{x}-e^{a \sqrt{-\frac{1}{n^2}} n} x^{-\frac{2+n}{n}}\right ) \, dx,x,c x^n\right )}{2 x}\\ &=\frac{e^{a \sqrt{-\frac{1}{n^2}} n} \sqrt{-\frac{1}{n^2}} n \left (c x^n\right )^{-1/n}}{4 x}+\frac{e^{-a \sqrt{-\frac{1}{n^2}} n} \sqrt{-\frac{1}{n^2}} n \left (c x^n\right )^{\frac{1}{n}} \log (x)}{2 x}\\ \end{align*}

Mathematica [F]  time = 0.0722603, size = 0, normalized size = 0. \[ \int \frac{\sin \left (a+\sqrt{-\frac{1}{n^2}} \log \left (c x^n\right )\right )}{x^2} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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Maple [F]  time = 0.038, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{{x}^{2}}\sin \left ( a+\ln \left ( c{x}^{n} \right ) \sqrt{-{n}^{-2}} \right ) }\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(a+ln(c*x^n)*(-1/n^2)^(1/2))/x^2,x)

[Out]

int(sin(a+ln(c*x^n)*(-1/n^2)^(1/2))/x^2,x)

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Maxima [A]  time = 1.13135, size = 45, normalized size = 0.52 \begin{align*} \frac{2 \, c^{\frac{2}{n}} x^{2} \log \left (x\right ) \sin \left (a\right ) - \sin \left (a\right )}{4 \, c^{\left (\frac{1}{n}\right )} x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(2*c^(2/n)*x^2*log(x)*sin(a) - sin(a))/(c^(1/n)*x^2)

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Fricas [C]  time = 0.472435, size = 107, normalized size = 1.24 \begin{align*} \frac{{\left (2 i \, x^{2} \log \left (x\right ) + i \, e^{\left (\frac{2 \,{\left (i \, a n - \log \left (c\right )\right )}}{n}\right )}\right )} e^{\left (-\frac{i \, a n - \log \left (c\right )}{n}\right )}}{4 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(2*I*x^2*log(x) + I*e^(2*(I*a*n - log(c))/n))*e^(-(I*a*n - log(c))/n)/x^2

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Sympy [C]  time = 14.2811, size = 214, normalized size = 2.49 \begin{align*} \frac{i n \sqrt{\frac{1}{n^{2}}} \log{\left (x \right )} \cos{\left (a + i n \sqrt{\frac{1}{n^{2}}} \log{\left (x \right )} + i \sqrt{\frac{1}{n^{2}}} \log{\left (c \right )} \right )}}{2 x} + \frac{i \sqrt{\frac{1}{n^{2}}} \log{\left (c \right )} \cos{\left (a + i n \sqrt{\frac{1}{n^{2}}} \log{\left (x \right )} + i \sqrt{\frac{1}{n^{2}}} \log{\left (c \right )} \right )}}{2 x} + \frac{\log{\left (x \right )} \sin{\left (a + i n \sqrt{\frac{1}{n^{2}}} \log{\left (x \right )} + i \sqrt{\frac{1}{n^{2}}} \log{\left (c \right )} \right )}}{2 x} - \frac{\sin{\left (a + i n \sqrt{\frac{1}{n^{2}}} \log{\left (x \right )} + i \sqrt{\frac{1}{n^{2}}} \log{\left (c \right )} \right )}}{2 x} + \frac{\log{\left (c \right )} \sin{\left (a + i n \sqrt{\frac{1}{n^{2}}} \log{\left (x \right )} + i \sqrt{\frac{1}{n^{2}}} \log{\left (c \right )} \right )}}{2 n x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

I*n*sqrt(n**(-2))*log(x)*cos(a + I*n*sqrt(n**(-2))*log(x) + I*sqrt(n**(-2))*log(c))/(2*x) + I*sqrt(n**(-2))*lo
g(c)*cos(a + I*n*sqrt(n**(-2))*log(x) + I*sqrt(n**(-2))*log(c))/(2*x) + log(x)*sin(a + I*n*sqrt(n**(-2))*log(x
) + I*sqrt(n**(-2))*log(c))/(2*x) - sin(a + I*n*sqrt(n**(-2))*log(x) + I*sqrt(n**(-2))*log(c))/(2*x) + log(c)*
sin(a + I*n*sqrt(n**(-2))*log(x) + I*sqrt(n**(-2))*log(c))/(2*n*x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sin \left (\sqrt{-\frac{1}{n^{2}}} \log \left (c x^{n}\right ) + a\right )}{x^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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