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

Optimal. Leaf size=52 \[ \frac {i e^{-i a} c x^3}{4 \sqrt {c x^2}}-\frac {i e^{i a} x \log (x)}{2 \sqrt {c x^2}} \]

[Out]

1/4*I*c*x^3/exp(I*a)/(c*x^2)^(1/2)-1/2*I*exp(I*a)*x*ln(x)/(c*x^2)^(1/2)

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Rubi [A]  time = 0.04, antiderivative size = 52, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {4483, 4489} \[ \frac {i e^{-i a} c x^3}{4 \sqrt {c x^2}}-\frac {i e^{i a} x \log (x)}{2 \sqrt {c x^2}} \]

Antiderivative was successfully verified.

[In]

Int[Sin[a + (I/2)*Log[c*x^2]],x]

[Out]

((I/4)*c*x^3)/(E^(I*a)*Sqrt[c*x^2]) - ((I/2)*E^(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 \left (a+\frac {1}{2} i \log \left (c x^2\right )\right ) \, dx &=\frac {x \operatorname {Subst}\left (\int \frac {\sin \left (a+\frac {1}{2} 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^{-i a}+\frac {e^{i a}}{x}\right ) \, dx,x,c x^2\right )}{4 \sqrt {c x^2}}\\ &=\frac {i c e^{-i a} x^3}{4 \sqrt {c x^2}}-\frac {i e^{i a} x \log (x)}{2 \sqrt {c x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 44, normalized size = 0.85 \[ \frac {x \left (\sin (a) \left (c x^2+2 \log (x)\right )+i \cos (a) \left (c x^2-2 \log (x)\right )\right )}{4 \sqrt {c x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(I*Cos[a]*(c*x^2 - 2*Log[x]) + (c*x^2 + 2*Log[x])*Sin[a]))/(4*Sqrt[c*x^2])

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fricas [A]  time = 0.44, size = 24, normalized size = 0.46 \[ \frac {{\left (i \, c x^{2} - 2 i \, e^{\left (2 i \, a\right )} \log \relax (x)\right )} e^{\left (-i \, a\right )}}{4 \, \sqrt {c}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.31, size = 29, normalized size = 0.56 \[ -\frac {-i \, c^{\frac {3}{2}} x^{2} e^{\left (-i \, a\right )} + 2 i \, \sqrt {c} e^{\left (i \, a\right )} \log \relax (x)}{4 \, c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(-I*c^(3/2)*x^2*e^(-I*a) + 2*I*sqrt(c)*e^(I*a)*log(x))/c

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maple [B]  time = 0.04, size = 106, normalized size = 2.04 \[ \frac {\frac {i x}{2}-\frac {i x \left (\tan ^{2}\left (\frac {a}{2}+\frac {i \ln \left (c \,x^{2}\right )}{4}\right )\right )}{2}+\frac {x \ln \left (c \,x^{2}\right ) \tan \left (\frac {a}{2}+\frac {i \ln \left (c \,x^{2}\right )}{4}\right )}{2}-\frac {i x \ln \left (c \,x^{2}\right )}{4}+\frac {i x \ln \left (c \,x^{2}\right ) \left (\tan ^{2}\left (\frac {a}{2}+\frac {i \ln \left (c \,x^{2}\right )}{4}\right )\right )}{4}}{1+\tan ^{2}\left (\frac {a}{2}+\frac {i \ln \left (c \,x^{2}\right )}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(1/2*I*x-1/2*I*x*tan(1/2*a+1/4*I*ln(c*x^2))^2+1/2*x*ln(c*x^2)*tan(1/2*a+1/4*I*ln(c*x^2))-1/4*I*x*ln(c*x^2)+1/4
*I*x*ln(c*x^2)*tan(1/2*a+1/4*I*ln(c*x^2))^2)/(1+tan(1/2*a+1/4*I*ln(c*x^2))^2)

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maxima [A]  time = 0.36, size = 31, normalized size = 0.60 \[ \frac {c x^{2} {\left (i \, \cos \relax (a) + \sin \relax (a)\right )} - 2 \, {\left (i \, \cos \relax (a) - \sin \relax (a)\right )} \log \relax (x)}{4 \, \sqrt {c}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(c*x^2*(I*cos(a) + sin(a)) - 2*(I*cos(a) - sin(a))*log(x))/sqrt(c)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \sin \left (a+\frac {\ln \left (c\,x^2\right )\,1{}\mathrm {i}}{2}\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(a + (log(c*x^2)*1i)/2),x)

[Out]

int(sin(a + (log(c*x^2)*1i)/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sin {\left (a + \frac {i \log {\left (c x^{2} \right )}}{2} \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sin(a + I*log(c*x**2)/2), x)

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