3.39 \(\int \frac {\sin ^2(a+\sqrt {-\frac {1}{n^2}} \log (c x^n))}{x^3} \, dx\)

Optimal. Leaf size=76 \[ \frac {e^{2 a \sqrt {-\frac {1}{n^2}} n} \left (c x^n\right )^{-2/n}}{16 x^2}-\frac {e^{-2 a \sqrt {-\frac {1}{n^2}} n} \log (x) \left (c x^n\right )^{2/n}}{4 x^2}-\frac {1}{4 x^2} \]

[Out]

-1/4/x^2+1/16*exp(2*a*n*(-1/n^2)^(1/2))/x^2/((c*x^n)^(2/n))-1/4*(c*x^n)^(2/n)*ln(x)/exp(2*a*n*(-1/n^2)^(1/2))/
x^2

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Rubi [A]  time = 0.06, antiderivative size = 76, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {4493, 4489} \[ \frac {e^{2 a \sqrt {-\frac {1}{n^2}} n} \left (c x^n\right )^{-2/n}}{16 x^2}-\frac {e^{-2 a \sqrt {-\frac {1}{n^2}} n} \log (x) \left (c x^n\right )^{2/n}}{4 x^2}-\frac {1}{4 x^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/(4*x^2) + E^(2*a*Sqrt[-n^(-2)]*n)/(16*x^2*(c*x^n)^(2/n)) - ((c*x^n)^(2/n)*Log[x])/(4*E^(2*a*Sqrt[-n^(-2)]*n
)*x^2)

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]

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])

Rubi steps

\begin {align*} \int \frac {\sin ^2\left (a+\sqrt {-\frac {1}{n^2}} \log \left (c x^n\right )\right )}{x^3} \, dx &=\frac {\left (c x^n\right )^{2/n} \operatorname {Subst}\left (\int x^{-1-\frac {2}{n}} \sin ^2\left (a+\sqrt {-\frac {1}{n^2}} \log (x)\right ) \, dx,x,c x^n\right )}{n x^2}\\ &=-\frac {\left (c x^n\right )^{2/n} \operatorname {Subst}\left (\int \left (\frac {e^{-2 a \sqrt {-\frac {1}{n^2}} n}}{x}-2 x^{-\frac {2+n}{n}}+e^{2 a \sqrt {-\frac {1}{n^2}} n} x^{-\frac {4+n}{n}}\right ) \, dx,x,c x^n\right )}{4 n x^2}\\ &=-\frac {1}{4 x^2}+\frac {e^{2 a \sqrt {-\frac {1}{n^2}} n} \left (c x^n\right )^{-2/n}}{16 x^2}-\frac {e^{-2 a \sqrt {-\frac {1}{n^2}} n} \left (c x^n\right )^{2/n} \log (x)}{4 x^2}\\ \end {align*}

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Mathematica [F]  time = 0.17, size = 0, normalized size = 0.00 \[ \int \frac {\sin ^2\left (a+\sqrt {-\frac {1}{n^2}} \log \left (c x^n\right )\right )}{x^3} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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fricas [C]  time = 0.43, size = 65, normalized size = 0.86 \[ -\frac {{\left (4 \, x^{4} \log \relax (x) + 4 \, x^{2} e^{\left (\frac {2 \, {\left (i \, a n - \log \relax (c)\right )}}{n}\right )} - e^{\left (\frac {4 \, {\left (i \, a n - \log \relax (c)\right )}}{n}\right )}\right )} e^{\left (-\frac {2 \, {\left (i \, a n - \log \relax (c)\right )}}{n}\right )}}{16 \, x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(a+log(c*x^n)*(-1/n^2)^(1/2))^2/x^3,x, algorithm="fricas")

[Out]

-1/16*(4*x^4*log(x) + 4*x^2*e^(2*(I*a*n - log(c))/n) - e^(4*(I*a*n - log(c))/n))*e^(-2*(I*a*n - log(c))/n)/x^4

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sin \left (\sqrt {-\frac {1}{n^{2}}} \log \left (c x^{n}\right ) + a\right )^{2}}{x^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(a+log(c*x^n)*(-1/n^2)^(1/2))^2/x^3,x, algorithm="giac")

[Out]

integrate(sin(sqrt(-1/n^2)*log(c*x^n) + a)^2/x^3, x)

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maple [F]  time = 0.07, size = 0, normalized size = 0.00 \[ \int \frac {\sin ^{2}\left (a +\ln \left (c \,x^{n}\right ) \sqrt {-\frac {1}{n^{2}}}\right )}{x^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.37, size = 54, normalized size = 0.71 \[ -\frac {4 \, c^{\frac {4}{n}} x^{6} \cos \left (2 \, a\right ) \log \relax (x) + 4 \, c^{\frac {2}{n}} x^{4} - x^{2} \cos \left (2 \, a\right )}{16 \, c^{\frac {2}{n}} x^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(a+log(c*x^n)*(-1/n^2)^(1/2))^2/x^3,x, algorithm="maxima")

[Out]

-1/16*(4*c^(4/n)*x^6*cos(2*a)*log(x) + 4*c^(2/n)*x^4 - x^2*cos(2*a))/(c^(2/n)*x^6)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\sin \left (a+\ln \left (c\,x^n\right )\,\sqrt {-\frac {1}{n^2}}\right )}^2}{x^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(a + log(c*x^n)*(-1/n^2)^(1/2))^2/x^3,x)

[Out]

int(sin(a + log(c*x^n)*(-1/n^2)^(1/2))^2/x^3, x)

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sympy [C]  time = 16.77, size = 462, normalized size = 6.08 \[ \frac {i n \sqrt {\frac {1}{n^{2}}} \log {\relax (x )} \sin {\left (a + i n \sqrt {\frac {1}{n^{2}}} \log {\relax (x )} + i \sqrt {\frac {1}{n^{2}}} \log {\relax (c )} \right )} \cos {\left (a + i n \sqrt {\frac {1}{n^{2}}} \log {\relax (x )} + i \sqrt {\frac {1}{n^{2}}} \log {\relax (c )} \right )}}{2 x^{2}} + \frac {3 i n \sqrt {\frac {1}{n^{2}}} \sin {\left (a + i n \sqrt {\frac {1}{n^{2}}} \log {\relax (x )} + i \sqrt {\frac {1}{n^{2}}} \log {\relax (c )} \right )} \cos {\left (a + i n \sqrt {\frac {1}{n^{2}}} \log {\relax (x )} + i \sqrt {\frac {1}{n^{2}}} \log {\relax (c )} \right )}}{4 x^{2}} + \frac {i \sqrt {\frac {1}{n^{2}}} \log {\relax (c )} \sin {\left (a + i n \sqrt {\frac {1}{n^{2}}} \log {\relax (x )} + i \sqrt {\frac {1}{n^{2}}} \log {\relax (c )} \right )} \cos {\left (a + i n \sqrt {\frac {1}{n^{2}}} \log {\relax (x )} + i \sqrt {\frac {1}{n^{2}}} \log {\relax (c )} \right )}}{2 x^{2}} + \frac {\log {\relax (x )} \sin ^{2}{\left (a + i n \sqrt {\frac {1}{n^{2}}} \log {\relax (x )} + i \sqrt {\frac {1}{n^{2}}} \log {\relax (c )} \right )}}{4 x^{2}} - \frac {\log {\relax (x )} \cos ^{2}{\left (a + i n \sqrt {\frac {1}{n^{2}}} \log {\relax (x )} + i \sqrt {\frac {1}{n^{2}}} \log {\relax (c )} \right )}}{4 x^{2}} - \frac {\cos ^{2}{\left (a + i n \sqrt {\frac {1}{n^{2}}} \log {\relax (x )} + i \sqrt {\frac {1}{n^{2}}} \log {\relax (c )} \right )}}{2 x^{2}} + \frac {\log {\relax (c )} \sin ^{2}{\left (a + i n \sqrt {\frac {1}{n^{2}}} \log {\relax (x )} + i \sqrt {\frac {1}{n^{2}}} \log {\relax (c )} \right )}}{4 n x^{2}} - \frac {\log {\relax (c )} \cos ^{2}{\left (a + i n \sqrt {\frac {1}{n^{2}}} \log {\relax (x )} + i \sqrt {\frac {1}{n^{2}}} \log {\relax (c )} \right )}}{4 n x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(a+ln(c*x**n)*(-1/n**2)**(1/2))**2/x**3,x)

[Out]

I*n*sqrt(n**(-2))*log(x)*sin(a + I*n*sqrt(n**(-2))*log(x) + I*sqrt(n**(-2))*log(c))*cos(a + I*n*sqrt(n**(-2))*
log(x) + I*sqrt(n**(-2))*log(c))/(2*x**2) + 3*I*n*sqrt(n**(-2))*sin(a + I*n*sqrt(n**(-2))*log(x) + I*sqrt(n**(
-2))*log(c))*cos(a + I*n*sqrt(n**(-2))*log(x) + I*sqrt(n**(-2))*log(c))/(4*x**2) + I*sqrt(n**(-2))*log(c)*sin(
a + I*n*sqrt(n**(-2))*log(x) + I*sqrt(n**(-2))*log(c))*cos(a + I*n*sqrt(n**(-2))*log(x) + I*sqrt(n**(-2))*log(
c))/(2*x**2) + log(x)*sin(a + I*n*sqrt(n**(-2))*log(x) + I*sqrt(n**(-2))*log(c))**2/(4*x**2) - log(x)*cos(a +
I*n*sqrt(n**(-2))*log(x) + I*sqrt(n**(-2))*log(c))**2/(4*x**2) - cos(a + I*n*sqrt(n**(-2))*log(x) + I*sqrt(n**
(-2))*log(c))**2/(2*x**2) + log(c)*sin(a + I*n*sqrt(n**(-2))*log(x) + I*sqrt(n**(-2))*log(c))**2/(4*n*x**2) -
log(c)*cos(a + I*n*sqrt(n**(-2))*log(x) + I*sqrt(n**(-2))*log(c))**2/(4*n*x**2)

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