\(\int -\frac {\log (2)}{x^4 (i \pi +\log (5))} \, dx\) [6969]

Optimal result

Integrand size = 17, antiderivative size = 19 \[ \int -\frac {\log (2)}{x^4 (i \pi +\log (5))} \, dx=\frac {\log (2)}{3 x^3 (i \pi +\log (5))} \]

[Out]

1/3*ln(2)/(ln(5)+I*Pi)/x^3

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {12, 30} \[ \int -\frac {\log (2)}{x^4 (i \pi +\log (5))} \, dx=\frac {\log (2)}{3 x^3 (\log (5)+i \pi )} \]

[In]

Int[-(Log[2]/(x^4*(I*Pi + Log[5]))),x]

[Out]

Log[2]/(3*x^3*(I*Pi + Log[5]))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rubi steps \begin{align*} \text {integral}& = -\frac {\log (2) \int \frac {1}{x^4} \, dx}{i \pi +\log (5)} \\ & = \frac {\log (2)}{3 x^3 (i \pi +\log (5))} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00 \[ \int -\frac {\log (2)}{x^4 (i \pi +\log (5))} \, dx=\frac {\log (2)}{3 x^3 (i \pi +\log (5))} \]

[In]

Integrate[-(Log[2]/(x^4*(I*Pi + Log[5]))),x]

[Out]

Log[2]/(3*x^3*(I*Pi + Log[5]))

Maple [A] (verified)

Time = 0.11 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.89

method	result	size
gosper	\(\frac {\ln \left (2\right )}{3 \left (\ln \left (5\right )+i \pi \right ) x^{3}}\)	\(17\)
default	\(\frac {\ln \left (2\right )}{3 \left (\ln \left (5\right )+i \pi \right ) x^{3}}\)	\(17\)
risch	\(\frac {\ln \left (2\right )}{3 \left (\ln \left (5\right )+i \pi \right ) x^{3}}\)	\(17\)
parallelrisch	\(\frac {\ln \left (2\right )}{3 \left (\ln \left (5\right )+i \pi \right ) x^{3}}\)	\(17\)
norman	\(\frac {\ln \left (2\right ) \left (-i \pi +\ln \left (5\right )\right )}{3 \left (\ln \left (5\right )^{2}+\pi ^{2}\right ) x^{3}}\)	\(25\)

[In]

int(-ln(2)/x^4/(ln(5)+I*Pi),x,method=_RETURNVERBOSE)

[Out]

1/3*ln(2)/(ln(5)+I*Pi)/x^3

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00 \[ \int -\frac {\log (2)}{x^4 (i \pi +\log (5))} \, dx=\frac {\log \left (2\right )}{3 i \, \pi x^{3} + 3 \, x^{3} \log \left (5\right )} \]

[In]

integrate(-log(2)/x^4/(log(5)+I*pi),x, algorithm="fricas")

[Out]

log(2)/(3*I*pi*x^3 + 3*x^3*log(5))

Sympy [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.74 \[ \int -\frac {\log (2)}{x^4 (i \pi +\log (5))} \, dx=\frac {\log {\left (2 \right )}}{3 x^{3} \left (\log {\left (5 \right )} + i \pi \right )} \]

[In]

integrate(-ln(2)/x**4/(ln(5)+I*pi),x)

[Out]

log(2)/(3*x**3*(log(5) + I*pi))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.79 \[ \int -\frac {\log (2)}{x^4 (i \pi +\log (5))} \, dx=\frac {\log \left (2\right )}{3 \, {\left (i \, \pi + \log \left (5\right )\right )} x^{3}} \]

[In]

integrate(-log(2)/x^4/(log(5)+I*pi),x, algorithm="maxima")

[Out]

1/3*log(2)/((I*pi + log(5))*x^3)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.79 \[ \int -\frac {\log (2)}{x^4 (i \pi +\log (5))} \, dx=\frac {\log \left (2\right )}{3 \, {\left (i \, \pi + \log \left (5\right )\right )} x^{3}} \]

[In]

integrate(-log(2)/x^4/(log(5)+I*pi),x, algorithm="giac")

[Out]

1/3*log(2)/((I*pi + log(5))*x^3)

Mupad [B] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.84 \[ \int -\frac {\log (2)}{x^4 (i \pi +\log (5))} \, dx=\frac {\ln \left (2\right )}{3\,x^3\,\left (\ln \left (5\right )+\Pi \,1{}\mathrm {i}\right )} \]

[In]

int(-log(2)/(x^4*(Pi*1i + log(5))),x)

[Out]

log(2)/(3*x^3*(Pi*1i + log(5)))