\(\int \sin (a+\frac {1}{2} i \log (c x^2)) \, dx\) [48]

Optimal result

Integrand size = 15, antiderivative size = 52 \[ \int \sin \left (a+\frac {1}{2} i \log \left (c x^2\right )\right ) \, dx=\frac {i c e^{-i a} x^3}{4 \sqrt {c x^2}}-\frac {i e^{i a} x \log (x)}{2 \sqrt {c x^2}} \]

[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)

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {4571, 4577} \[ \int \sin \left (a+\frac {1}{2} i \log \left (c x^2\right )\right ) \, dx=\frac {i e^{-i a} c x^3}{4 \sqrt {c x^2}}-\frac {i e^{i a} x \log (x)}{2 \sqrt {c x^2}} \]

[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 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*} \text {integral}& = \frac {x \text {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) \text {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*}

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.85 \[ \int \sin \left (a+\frac {1}{2} i \log \left (c x^2\right )\right ) \, dx=\frac {x \left (i \cos (a) \left (c x^2-2 \log (x)\right )+\left (c x^2+2 \log (x)\right ) \sin (a)\right )}{4 \sqrt {c x^2}} \]

[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])

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 105 vs. \(2 (40 ) = 80\).

Time = 1.10 (sec) , antiderivative size = 106, normalized size of antiderivative = 2.04

method	result	size
norman	\(\frac {\frac {i x}{2}-\frac {i x {\tan \left (\frac {a}{2}+\frac {i \ln \left (c \,x^{2}\right )}{4}\right )}^{2}}{2}+\frac {x \ln \left (c \,x^{2}\right ) \tan \left (\frac {a}{2}+\frac {i \ln \left (c \,x^{2}\right )}{4}\right )}{2}-\frac {i x \ln \left (c \,x^{2}\right )}{4}+\frac {i x \ln \left (c \,x^{2}\right ) {\tan \left (\frac {a}{2}+\frac {i \ln \left (c \,x^{2}\right )}{4}\right )}^{2}}{4}}{1+{\tan \left (\frac {a}{2}+\frac {i \ln \left (c \,x^{2}\right )}{4}\right )}^{2}}\)	\(106\)

[In]

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

[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)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.46 \[ \int \sin \left (a+\frac {1}{2} i \log \left (c x^2\right )\right ) \, dx=\frac {{\left (i \, c x^{2} - 2 i \, e^{\left (2 i \, a\right )} \log \left (x\right )\right )} e^{\left (-i \, a\right )}}{4 \, \sqrt {c}} \]

[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)

Sympy [F]

\[ \int \sin \left (a+\frac {1}{2} i \log \left (c x^2\right )\right ) \, dx=\int \sin {\left (a + \frac {i \log {\left (c x^{2} \right )}}{2} \right )}\, dx \]

[In]

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

[Out]

Integral(sin(a + I*log(c*x**2)/2), x)

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.60 \[ \int \sin \left (a+\frac {1}{2} i \log \left (c x^2\right )\right ) \, dx=\frac {c x^{2} {\left (i \, \cos \left (a\right ) + \sin \left (a\right )\right )} - 2 \, {\left (i \, \cos \left (a\right ) - \sin \left (a\right )\right )} \log \left (x\right )}{4 \, \sqrt {c}} \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.46 \[ \int \sin \left (a+\frac {1}{2} i \log \left (c x^2\right )\right ) \, dx=-\frac {-i \, c x^{2} e^{\left (-i \, a\right )} + 2 i \, e^{\left (i \, a\right )} \log \left (x\right )}{4 \, \sqrt {c}} \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \sin \left (a+\frac {1}{2} i \log \left (c x^2\right )\right ) \, dx=\int \sin \left (a+\frac {\ln \left (c\,x^2\right )\,1{}\mathrm {i}}{2}\right ) \,d x \]

[In]

int(sin(a + (log(c*x^2)*1i)/2),x)

[Out]

int(sin(a + (log(c*x^2)*1i)/2), x)