\(\int \sin ^2(a+\frac {1}{4} i \log (c x^2)) \, dx\) [50]

Optimal result

Integrand size = 17, antiderivative size = 53 \[ \int \sin ^2\left (a+\frac {1}{4} i \log \left (c x^2\right )\right ) \, dx=\frac {x}{2}-\frac {c e^{-2 i a} x^3}{8 \sqrt {c x^2}}-\frac {e^{2 i a} x \log (x)}{4 \sqrt {c x^2}} \]

[Out]

1/2*x-1/8*c*x^3/exp(2*I*a)/(c*x^2)^(1/2)-1/4*exp(2*I*a)*x*ln(x)/(c*x^2)^(1/2)

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {4571, 4577} \[ \int \sin ^2\left (a+\frac {1}{4} i \log \left (c x^2\right )\right ) \, dx=-\frac {e^{2 i a} x \log (x)}{4 \sqrt {c x^2}}-\frac {e^{-2 i a} c x^3}{8 \sqrt {c x^2}}+\frac {x}{2} \]

[In]

Int[Sin[a + (I/4)*Log[c*x^2]]^2,x]

[Out]

x/2 - (c*x^3)/(8*E^((2*I)*a)*Sqrt[c*x^2]) - (E^((2*I)*a)*x*Log[x])/(4*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 ^2\left (a+\frac {1}{4} i \log (x)\right )}{\sqrt {x}} \, dx,x,c x^2\right )}{2 \sqrt {c x^2}} \\ & = -\frac {x \text {Subst}\left (\int \left (e^{-2 i a}+\frac {e^{2 i a}}{x}-\frac {2}{\sqrt {x}}\right ) \, dx,x,c x^2\right )}{8 \sqrt {c x^2}} \\ & = \frac {x}{2}-\frac {c e^{-2 i a} x^3}{8 \sqrt {c x^2}}-\frac {e^{2 i a} x \log (x)}{4 \sqrt {c x^2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.13 \[ \int \sin ^2\left (a+\frac {1}{4} i \log \left (c x^2\right )\right ) \, dx=\frac {x \left (4 \sqrt {c x^2}-\cos (2 a) \left (c x^2+2 \log (x)\right )+i \left (c x^2-2 \log (x)\right ) \sin (2 a)\right )}{8 \sqrt {c x^2}} \]

[In]

Integrate[Sin[a + (I/4)*Log[c*x^2]]^2,x]

[Out]

(x*(4*Sqrt[c*x^2] - Cos[2*a]*(c*x^2 + 2*Log[x]) + I*(c*x^2 - 2*Log[x])*Sin[2*a]))/(8*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. 172 vs. \(2 (41 ) = 82\).

Time = 2.57 (sec) , antiderivative size = 173, normalized size of antiderivative = 3.26

method	result	size
norman	\(\frac {\frac {x}{4}+\frac {5 x {\tan \left (\frac {a}{2}+\frac {i \ln \left (c \,x^{2}\right )}{8}\right )}^{2}}{2}+\frac {x {\tan \left (\frac {a}{2}+\frac {i \ln \left (c \,x^{2}\right )}{8}\right )}^{4}}{4}-\frac {x \ln \left (c \,x^{2}\right )}{8}+\frac {3 x \ln \left (c \,x^{2}\right ) {\tan \left (\frac {a}{2}+\frac {i \ln \left (c \,x^{2}\right )}{8}\right )}^{2}}{4}-\frac {x \ln \left (c \,x^{2}\right ) {\tan \left (\frac {a}{2}+\frac {i \ln \left (c \,x^{2}\right )}{8}\right )}^{4}}{8}-\frac {i x \ln \left (c \,x^{2}\right ) \tan \left (\frac {a}{2}+\frac {i \ln \left (c \,x^{2}\right )}{8}\right )}{2}+\frac {i x \ln \left (c \,x^{2}\right ) {\tan \left (\frac {a}{2}+\frac {i \ln \left (c \,x^{2}\right )}{8}\right )}^{3}}{2}}{{\left (1+{\tan \left (\frac {a}{2}+\frac {i \ln \left (c \,x^{2}\right )}{8}\right )}^{2}\right )}^{2}}\)	\(173\)

[In]

int(sin(a+1/4*I*ln(c*x^2))^2,x,method=_RETURNVERBOSE)

[Out]

(1/4*x+5/2*x*tan(1/2*a+1/8*I*ln(c*x^2))^2+1/4*x*tan(1/2*a+1/8*I*ln(c*x^2))^4-1/8*x*ln(c*x^2)+3/4*x*ln(c*x^2)*t
an(1/2*a+1/8*I*ln(c*x^2))^2-1/8*x*ln(c*x^2)*tan(1/2*a+1/8*I*ln(c*x^2))^4-1/2*I*x*ln(c*x^2)*tan(1/2*a+1/8*I*ln(
c*x^2))+1/2*I*x*ln(c*x^2)*tan(1/2*a+1/8*I*ln(c*x^2))^3)/(1+tan(1/2*a+1/8*I*ln(c*x^2))^2)^2

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 145 vs. \(2 (37) = 74\).

Time = 0.47 (sec) , antiderivative size = 145, normalized size of antiderivative = 2.74 \[ \int \sin ^2\left (a+\frac {1}{4} i \log \left (c x^2\right )\right ) \, dx=\frac {{\left (4 \, x^{2} e^{\left (2 i \, a\right )} - \frac {x e^{\left (4 i \, a\right )} \log \left (\frac {{\left (\sqrt {c x^{2}} {\left (x^{2} + 1\right )} e^{\left (2 i \, a\right )} + \frac {{\left (c x^{3} - c x\right )} e^{\left (2 i \, a\right )}}{\sqrt {c}}\right )} e^{\left (-2 i \, a\right )}}{8 \, x^{2}}\right )}{\sqrt {c}} + \frac {x e^{\left (4 i \, a\right )} \log \left (\frac {{\left (\sqrt {c x^{2}} {\left (x^{2} + 1\right )} e^{\left (2 i \, a\right )} - \frac {{\left (c x^{3} - c x\right )} e^{\left (2 i \, a\right )}}{\sqrt {c}}\right )} e^{\left (-2 i \, a\right )}}{8 \, x^{2}}\right )}{\sqrt {c}} - \sqrt {c x^{2}} {\left (x^{2} - 1\right )}\right )} e^{\left (-2 i \, a\right )}}{8 \, x} \]

[In]

integrate(sin(a+1/4*I*log(c*x^2))^2,x, algorithm="fricas")

[Out]

1/8*(4*x^2*e^(2*I*a) - x*e^(4*I*a)*log(1/8*(sqrt(c*x^2)*(x^2 + 1)*e^(2*I*a) + (c*x^3 - c*x)*e^(2*I*a)/sqrt(c))
*e^(-2*I*a)/x^2)/sqrt(c) + x*e^(4*I*a)*log(1/8*(sqrt(c*x^2)*(x^2 + 1)*e^(2*I*a) - (c*x^3 - c*x)*e^(2*I*a)/sqrt
(c))*e^(-2*I*a)/x^2)/sqrt(c) - sqrt(c*x^2)*(x^2 - 1))*e^(-2*I*a)/x

Sympy [F]

\[ \int \sin ^2\left (a+\frac {1}{4} i \log \left (c x^2\right )\right ) \, dx=\int \sin ^{2}{\left (a + \frac {i \log {\left (c x^{2} \right )}}{4} \right )}\, dx \]

[In]

integrate(sin(a+1/4*I*ln(c*x**2))**2,x)

[Out]

Integral(sin(a + I*log(c*x**2)/4)**2, x)

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.89 \[ \int \sin ^2\left (a+\frac {1}{4} i \log \left (c x^2\right )\right ) \, dx=\frac {4 \, c x - {\left (c x^{2} {\left (\cos \left (2 \, a\right ) - i \, \sin \left (2 \, a\right )\right )} + 2 \, {\left (\cos \left (2 \, a\right ) + i \, \sin \left (2 \, a\right )\right )} \log \left (x\right )\right )} \sqrt {c}}{8 \, c} \]

[In]

integrate(sin(a+1/4*I*log(c*x^2))^2,x, algorithm="maxima")

[Out]

1/8*(4*c*x - (c*x^2*(cos(2*a) - I*sin(2*a)) + 2*(cos(2*a) + I*sin(2*a))*log(x))*sqrt(c))/c

Giac [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.51 \[ \int \sin ^2\left (a+\frac {1}{4} i \log \left (c x^2\right )\right ) \, dx=\frac {1}{2} \, x - \frac {c x^{2} e^{\left (-2 i \, a\right )} + 2 \, e^{\left (2 i \, a\right )} \log \left (x\right )}{8 \, \sqrt {c}} \]

[In]

integrate(sin(a+1/4*I*log(c*x^2))^2,x, algorithm="giac")

[Out]

1/2*x - 1/8*(c*x^2*e^(-2*I*a) + 2*e^(2*I*a)*log(x))/sqrt(c)

Mupad [F(-1)]

Timed out. \[ \int \sin ^2\left (a+\frac {1}{4} i \log \left (c x^2\right )\right ) \, dx=\int {\sin \left (a+\frac {\ln \left (c\,x^2\right )\,1{}\mathrm {i}}{4}\right )}^2 \,d x \]

[In]

int(sin(a + (log(c*x^2)*1i)/4)^2,x)

[Out]

int(sin(a + (log(c*x^2)*1i)/4)^2, x)