3.52 \(\int \sin ^3(a+\frac{1}{6} i \log (c x^2)) \, dx\)

Optimal. Leaf size=98 \[ -\frac{i e^{-3 i a} c x^3}{16 \sqrt{c x^2}}+\frac{9}{32} i e^{-i a} x \sqrt [6]{c x^2}-\frac{9 i e^{i a} x}{16 \sqrt [6]{c x^2}}+\frac{i e^{3 i a} x \log (x)}{8 \sqrt{c x^2}} \]

[Out]

((-I/16)*c*x^3)/(E^((3*I)*a)*Sqrt[c*x^2]) - (((9*I)/16)*E^(I*a)*x)/(c*x^2)^(1/6) + (((9*I)/32)*x*(c*x^2)^(1/6)
)/E^(I*a) + ((I/8)*E^((3*I)*a)*x*Log[x])/Sqrt[c*x^2]

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Rubi [A]  time = 0.0609948, antiderivative size = 98, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.118, Rules used = {4483, 4489} \[ -\frac{i e^{-3 i a} c x^3}{16 \sqrt{c x^2}}+\frac{9}{32} i e^{-i a} x \sqrt [6]{c x^2}-\frac{9 i e^{i a} x}{16 \sqrt [6]{c x^2}}+\frac{i e^{3 i a} x \log (x)}{8 \sqrt{c x^2}} \]

Antiderivative was successfully verified.

[In]

Int[Sin[a + (I/6)*Log[c*x^2]]^3,x]

[Out]

((-I/16)*c*x^3)/(E^((3*I)*a)*Sqrt[c*x^2]) - (((9*I)/16)*E^(I*a)*x)/(c*x^2)^(1/6) + (((9*I)/32)*x*(c*x^2)^(1/6)
)/E^(I*a) + ((I/8)*E^((3*I)*a)*x*Log[x])/Sqrt[c*x^2]

Rule 4483

Int[Sin[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*(d_.)]^(p_.), x_Symbol] :> Dist[x/(n*(c*x^n)^(1/n)), Subst[Int[x
^(1/n - 1)*Sin[d*(a + b*Log[x])]^p, x], x, c*x^n], x] /; FreeQ[{a, b, c, d, 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 \sin ^3\left (a+\frac{1}{6} i \log \left (c x^2\right )\right ) \, dx &=\frac{x \operatorname{Subst}\left (\int \frac{\sin ^3\left (a+\frac{1}{6} i \log (x)\right )}{\sqrt{x}} \, dx,x,c x^2\right )}{2 \sqrt{c x^2}}\\ &=\frac{(i x) \operatorname{Subst}\left (\int \left (-e^{-3 i a}+\frac{e^{3 i a}}{x}-\frac{3 e^{i a}}{x^{2/3}}+\frac{3 e^{-i a}}{\sqrt [3]{x}}\right ) \, dx,x,c x^2\right )}{16 \sqrt{c x^2}}\\ &=-\frac{i c e^{-3 i a} x^3}{16 \sqrt{c x^2}}-\frac{9 i e^{i a} x}{16 \sqrt [6]{c x^2}}+\frac{9}{32} i e^{-i a} x \sqrt [6]{c x^2}+\frac{i e^{3 i a} x \log (x)}{8 \sqrt{c x^2}}\\ \end{align*}

Mathematica [A]  time = 0.121187, size = 103, normalized size = 1.05 \[ \frac{x \left (-2 c x^2 \sin (3 a)+9 \sin (a) \left (c x^2\right )^{2/3}+18 \sin (a) \sqrt [3]{c x^2}+9 i \cos (a) \sqrt [3]{c x^2} \left (\sqrt [3]{c x^2}-2\right )-2 i \cos (3 a) \left (c x^2-2 \log (x)\right )-4 \sin (3 a) \log (x)\right )}{32 \sqrt{c x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[a + (I/6)*Log[c*x^2]]^3,x]

[Out]

(x*((9*I)*(c*x^2)^(1/3)*(-2 + (c*x^2)^(1/3))*Cos[a] - (2*I)*Cos[3*a]*(c*x^2 - 2*Log[x]) + 18*(c*x^2)^(1/3)*Sin
[a] + 9*(c*x^2)^(2/3)*Sin[a] - 2*c*x^2*Sin[3*a] - 4*Log[x]*Sin[3*a]))/(32*Sqrt[c*x^2])

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Maple [B]  time = 0.081, size = 284, normalized size = 2.9 \begin{align*}{ \left ( -{\frac{23\,i}{40}}x+{\frac{27\,x}{10}\tan \left ({\frac{a}{2}}+{\frac{i}{12}}\ln \left ( c{x}^{2} \right ) \right ) }+{\frac{27\,x}{10} \left ( \tan \left ({\frac{a}{2}}+{\frac{i}{12}}\ln \left ( c{x}^{2} \right ) \right ) \right ) ^{5}}+{\frac{33\,i}{8}}x \left ( \tan \left ({\frac{a}{2}}+{\frac{i}{12}}\ln \left ( c{x}^{2} \right ) \right ) \right ) ^{2}+{\frac{23\,i}{40}}x \left ( \tan \left ({\frac{a}{2}}+{\frac{i}{12}}\ln \left ( c{x}^{2} \right ) \right ) \right ) ^{6}-{\frac{33\,i}{8}}x \left ( \tan \left ({\frac{a}{2}}+{\frac{i}{12}}\ln \left ( c{x}^{2} \right ) \right ) \right ) ^{4}-{\frac{3\,x\ln \left ( c{x}^{2} \right ) }{8}\tan \left ({\frac{a}{2}}+{\frac{i}{12}}\ln \left ( c{x}^{2} \right ) \right ) }+{\frac{5\,x\ln \left ( c{x}^{2} \right ) }{4} \left ( \tan \left ({\frac{a}{2}}+{\frac{i}{12}}\ln \left ( c{x}^{2} \right ) \right ) \right ) ^{3}}-{\frac{3\,x\ln \left ( c{x}^{2} \right ) }{8} \left ( \tan \left ({\frac{a}{2}}+{\frac{i}{12}}\ln \left ( c{x}^{2} \right ) \right ) \right ) ^{5}}+{\frac{i}{16}}x\ln \left ( c{x}^{2} \right ) -{\frac{15\,i}{16}}x\ln \left ( c{x}^{2} \right ) \left ( \tan \left ({\frac{a}{2}}+{\frac{i}{12}}\ln \left ( c{x}^{2} \right ) \right ) \right ) ^{2}+{\frac{15\,i}{16}}x\ln \left ( c{x}^{2} \right ) \left ( \tan \left ({\frac{a}{2}}+{\frac{i}{12}}\ln \left ( c{x}^{2} \right ) \right ) \right ) ^{4}-{\frac{i}{16}}x\ln \left ( c{x}^{2} \right ) \left ( \tan \left ({\frac{a}{2}}+{\frac{i}{12}}\ln \left ( c{x}^{2} \right ) \right ) \right ) ^{6} \right ) \left ( 1+ \left ( \tan \left ({\frac{a}{2}}+{\frac{i}{12}}\ln \left ( c{x}^{2} \right ) \right ) \right ) ^{2} \right ) ^{-3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(a+1/6*I*ln(c*x^2))^3,x)

[Out]

(-23/40*I*x+27/10*x*tan(1/2*a+1/12*I*ln(c*x^2))+27/10*x*tan(1/2*a+1/12*I*ln(c*x^2))^5+33/8*I*x*tan(1/2*a+1/12*
I*ln(c*x^2))^2+23/40*I*x*tan(1/2*a+1/12*I*ln(c*x^2))^6-33/8*I*x*tan(1/2*a+1/12*I*ln(c*x^2))^4-3/8*x*ln(c*x^2)*
tan(1/2*a+1/12*I*ln(c*x^2))+5/4*x*ln(c*x^2)*tan(1/2*a+1/12*I*ln(c*x^2))^3-3/8*x*ln(c*x^2)*tan(1/2*a+1/12*I*ln(
c*x^2))^5+1/16*I*x*ln(c*x^2)-15/16*I*x*ln(c*x^2)*tan(1/2*a+1/12*I*ln(c*x^2))^2+15/16*I*x*ln(c*x^2)*tan(1/2*a+1
/12*I*ln(c*x^2))^4-1/16*I*x*ln(c*x^2)*tan(1/2*a+1/12*I*ln(c*x^2))^6)/(1+tan(1/2*a+1/12*I*ln(c*x^2))^2)^3

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Maxima [A]  time = 1.08594, size = 101, normalized size = 1.03 \begin{align*} -\frac{9 \, c^{\frac{4}{3}} x^{\frac{4}{3}}{\left (-i \, \cos \left (a\right ) - \sin \left (a\right )\right )} + 18 \, c x^{\frac{2}{3}}{\left (i \, \cos \left (a\right ) - \sin \left (a\right )\right )} + 2 \,{\left (c x^{2}{\left (i \, \cos \left (3 \, a\right ) + \sin \left (3 \, a\right )\right )} + 2 \,{\left (-i \, \cos \left (3 \, a\right ) + \sin \left (3 \, a\right )\right )} \log \left (x\right )\right )} c^{\frac{2}{3}}}{32 \, c^{\frac{7}{6}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(a+1/6*I*log(c*x^2))^3,x, algorithm="maxima")

[Out]

-1/32*(9*c^(4/3)*x^(4/3)*(-I*cos(a) - sin(a)) + 18*c*x^(2/3)*(I*cos(a) - sin(a)) + 2*(c*x^2*(I*cos(3*a) + sin(
3*a)) + 2*(-I*cos(3*a) + sin(3*a))*log(x))*c^(2/3))/c^(7/6)

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Fricas [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: UnboundLocalError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(a+1/6*I*log(c*x^2))^3,x, algorithm="fricas")

[Out]

Exception raised: UnboundLocalError

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(a+1/6*I*ln(c*x**2))**3,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sin \left (a + \frac{1}{6} i \, \log \left (c x^{2}\right )\right )^{3}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(a+1/6*I*log(c*x^2))^3,x, algorithm="giac")

[Out]

integrate(sin(a + 1/6*I*log(c*x^2))^3, x)