3.1.52 \(\int \sin ^3(a+\frac {1}{6} i \log (c x^2)) \, dx\) [52]

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

[Out]

-9/16*I*exp(I*a)*x/(c*x^2)^(1/6)+9/32*I*x*(c*x^2)^(1/6)/exp(I*a)-1/16*I*c*x^3/exp(3*I*a)/(c*x^2)^(1/2)+1/8*I*e
xp(3*I*a)*x*ln(x)/(c*x^2)^(1/2)

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Rubi [A]
time = 0.04, antiderivative size = 98, normalized size of antiderivative = 1.00, 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 = {4571, 4577} \begin {gather*} \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}}-\frac {i e^{-3 i a} c x^3}{16 \sqrt {c x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sin[a + (I/6)*Log[c*x^2]]^3,x]

[Out]

((-1/16*I)*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 4571

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 4577

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 \text {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) \text {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*}

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Mathematica [A]
time = 0.14, size = 103, normalized size = 1.05 \begin {gather*} \frac {x \left (9 i \sqrt [3]{c x^2} \left (-2+\sqrt [3]{c x^2}\right ) \cos (a)-2 i \cos (3 a) \left (c x^2-2 \log (x)\right )+18 \sqrt [3]{c x^2} \sin (a)+9 \left (c x^2\right )^{2/3} \sin (a)-2 c x^2 \sin (3 a)-4 \log (x) \sin (3 a)\right )}{32 \sqrt {c x^2}} \end {gather*}

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] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 283 vs. \(2 (74 ) = 148\).
time = 0.13, size = 284, normalized size = 2.90

method	result	size
norman	\(\frac {-\frac {23 i x}{40}+\frac {27 x \tan \left (\frac {a}{2}+\frac {i \ln \left (c \,x^{2}\right )}{12}\right )}{10}+\frac {27 x \left (\tan ^{5}\left (\frac {a}{2}+\frac {i \ln \left (c \,x^{2}\right )}{12}\right )\right )}{10}+\frac {33 i x \left (\tan ^{2}\left (\frac {a}{2}+\frac {i \ln \left (c \,x^{2}\right )}{12}\right )\right )}{8}+\frac {23 i x \left (\tan ^{6}\left (\frac {a}{2}+\frac {i \ln \left (c \,x^{2}\right )}{12}\right )\right )}{40}-\frac {33 i x \left (\tan ^{4}\left (\frac {a}{2}+\frac {i \ln \left (c \,x^{2}\right )}{12}\right )\right )}{8}-\frac {3 x \ln \left (c \,x^{2}\right ) \tan \left (\frac {a}{2}+\frac {i \ln \left (c \,x^{2}\right )}{12}\right )}{8}+\frac {5 x \ln \left (c \,x^{2}\right ) \left (\tan ^{3}\left (\frac {a}{2}+\frac {i \ln \left (c \,x^{2}\right )}{12}\right )\right )}{4}-\frac {3 x \ln \left (c \,x^{2}\right ) \left (\tan ^{5}\left (\frac {a}{2}+\frac {i \ln \left (c \,x^{2}\right )}{12}\right )\right )}{8}+\frac {i x \ln \left (c \,x^{2}\right )}{16}-\frac {15 i x \ln \left (c \,x^{2}\right ) \left (\tan ^{2}\left (\frac {a}{2}+\frac {i \ln \left (c \,x^{2}\right )}{12}\right )\right )}{16}+\frac {15 i x \ln \left (c \,x^{2}\right ) \left (\tan ^{4}\left (\frac {a}{2}+\frac {i \ln \left (c \,x^{2}\right )}{12}\right )\right )}{16}-\frac {i x \ln \left (c \,x^{2}\right ) \left (\tan ^{6}\left (\frac {a}{2}+\frac {i \ln \left (c \,x^{2}\right )}{12}\right )\right )}{16}}{\left (1+\tan ^{2}\left (\frac {a}{2}+\frac {i \ln \left (c \,x^{2}\right )}{12}\right )\right )^{3}}\)	\(284\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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 = 0.28, size = 75, normalized size = 0.77 \begin {gather*} -\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 {gather*}

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 [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 204 vs. \(2 (62) = 124\).
time = 6.06, size = 204, normalized size = 2.08 \begin {gather*} \frac {{\left (2 \, c x \sqrt {-\frac {e^{\left (6 i \, a\right )}}{c}} e^{\left (3 i \, a\right )} \log \left (\frac {{\left (\sqrt {c x^{2}} {\left (x^{2} + 1\right )} e^{\left (3 i \, a\right )} - {\left (i \, c x^{3} - i \, c x\right )} \sqrt {-\frac {e^{\left (6 i \, a\right )}}{c}}\right )} e^{\left (-3 i \, a\right )}}{8 \, x^{2}}\right ) - 2 \, c x \sqrt {-\frac {e^{\left (6 i \, a\right )}}{c}} e^{\left (3 i \, a\right )} \log \left (\frac {{\left (\sqrt {c x^{2}} {\left (x^{2} + 1\right )} e^{\left (3 i \, a\right )} - {\left (-i \, c x^{3} + i \, c x\right )} \sqrt {-\frac {e^{\left (6 i \, a\right )}}{c}}\right )} e^{\left (-3 i \, a\right )}}{8 \, x^{2}}\right ) + 9 i \, \left (c x^{2}\right )^{\frac {1}{6}} c x^{2} e^{\left (2 i \, a\right )} - 18 i \, \left (c x^{2}\right )^{\frac {5}{6}} e^{\left (4 i \, a\right )} - 2 \, \sqrt {c x^{2}} {\left (i \, c x^{2} - i \, c\right )}\right )} e^{\left (-3 i \, a\right )}}{32 \, c x} \end {gather*}

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]

1/32*(2*c*x*sqrt(-e^(6*I*a)/c)*e^(3*I*a)*log(1/8*(sqrt(c*x^2)*(x^2 + 1)*e^(3*I*a) - (I*c*x^3 - I*c*x)*sqrt(-e^
(6*I*a)/c))*e^(-3*I*a)/x^2) - 2*c*x*sqrt(-e^(6*I*a)/c)*e^(3*I*a)*log(1/8*(sqrt(c*x^2)*(x^2 + 1)*e^(3*I*a) - (-
I*c*x^3 + I*c*x)*sqrt(-e^(6*I*a)/c))*e^(-3*I*a)/x^2) + 9*I*(c*x^2)^(1/6)*c*x^2*e^(2*I*a) - 18*I*(c*x^2)^(5/6)*
e^(4*I*a) - 2*sqrt(c*x^2)*(I*c*x^2 - I*c))*e^(-3*I*a)/(c*x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \sin ^{3}{\left (a + \frac {i \log {\left (c x^{2} \right )}}{6} \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sin(a + I*log(c*x**2)/6)**3, x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\sin \left (a+\frac {\ln \left (c\,x^2\right )\,1{}\mathrm {i}}{6}\right )}^3 \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(a + (log(c*x^2)*1i)/6)^3,x)

[Out]

int(sin(a + (log(c*x^2)*1i)/6)^3, x)

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