\(\int \frac {-4 x^2+x^2 (i \pi +\log (3))}{60 \log ^2(5)} \, dx\) [2097]

Optimal result

Integrand size = 26, antiderivative size = 20 \[ \int \frac {-4 x^2+x^2 (i \pi +\log (3))}{60 \log ^2(5)} \, dx=\frac {x^3 (-4+i \pi +\log (3))}{180 \log ^2(5)} \]

[Out]

1/180/ln(5)^2*x^3*(ln(3)+I*Pi-4)

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.10, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {6, 12, 30} \[ \int \frac {-4 x^2+x^2 (i \pi +\log (3))}{60 \log ^2(5)} \, dx=-\frac {x^3 (4-i \pi -\log (3))}{180 \log ^2(5)} \]

[In]

Int[(-4*x^2 + x^2*(I*Pi + Log[3]))/(60*Log[5]^2),x]

[Out]

-1/180*(x^3*(4 - I*Pi - Log[3]))/Log[5]^2

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, x]

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}& = \int \frac {x^2 (-4+i \pi +\log (3))}{60 \log ^2(5)} \, dx \\ & = -\frac {(4-i \pi -\log (3)) \int x^2 \, dx}{60 \log ^2(5)} \\ & = -\frac {x^3 (4-i \pi -\log (3))}{180 \log ^2(5)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int \frac {-4 x^2+x^2 (i \pi +\log (3))}{60 \log ^2(5)} \, dx=\frac {x^3 (-4+i \pi +\log (3))}{180 \log ^2(5)} \]

[In]

Integrate[(-4*x^2 + x^2*(I*Pi + Log[3]))/(60*Log[5]^2),x]

[Out]

(x^3*(-4 + I*Pi + Log[3]))/(180*Log[5]^2)

Maple [A] (verified)

Time = 0.08 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.90

method	result	size
gosper	\(\frac {x^{3} \left (\ln \left (3\right )+i \pi -4\right )}{180 \ln \left (5\right )^{2}}\)	\(18\)
norman	\(\frac {x^{3} \left (\ln \left (3\right )+i \pi -4\right )}{180 \ln \left (5\right )^{2}}\)	\(18\)
default	\(\frac {i \left (-i \ln \left (3\right )+\pi +4 i\right ) x^{3}}{180 \ln \left (5\right )^{2}}\)	\(20\)
parallelrisch	\(\frac {\frac {x^{3} \ln \left (3\right )}{3}+\frac {i x^{3} \pi }{3}-\frac {4 x^{3}}{3}}{60 \ln \left (5\right )^{2}}\)	\(27\)
risch	\(\frac {i x^{3} \pi }{180 \ln \left (5\right )^{2}}+\frac {x^{3} \ln \left (3\right )}{180 \ln \left (5\right )^{2}}-\frac {x^{3}}{45 \ln \left (5\right )^{2}}\)	\(33\)

[In]

int(1/60*(x^2*(ln(3)+I*Pi)-4*x^2)/ln(5)^2,x,method=_RETURNVERBOSE)

[Out]

1/180/ln(5)^2*x^3*(ln(3)+I*Pi-4)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.10 \[ \int \frac {-4 x^2+x^2 (i \pi +\log (3))}{60 \log ^2(5)} \, dx=\frac {{\left (i \, \pi - 4\right )} x^{3} + x^{3} \log \left (3\right )}{180 \, \log \left (5\right )^{2}} \]

[In]

integrate(1/60*(x^2*(log(3)+I*pi)-4*x^2)/log(5)^2,x, algorithm="fricas")

[Out]

1/180*((I*pi - 4)*x^3 + x^3*log(3))/log(5)^2

Sympy [A] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.85 \[ \int \frac {-4 x^2+x^2 (i \pi +\log (3))}{60 \log ^2(5)} \, dx=\frac {x^{3} \left (-4 + \log {\left (3 \right )} + i \pi \right )}{180 \log {\left (5 \right )}^{2}} \]

[In]

integrate(1/60*(x**2*(ln(3)+I*pi)-4*x**2)/ln(5)**2,x)

[Out]

x**3*(-4 + log(3) + I*pi)/(180*log(5)**2)

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.20 \[ \int \frac {-4 x^2+x^2 (i \pi +\log (3))}{60 \log ^2(5)} \, dx=-\frac {{\left (-i \, \pi - \log \left (3\right )\right )} x^{3} + 4 \, x^{3}}{180 \, \log \left (5\right )^{2}} \]

[In]

integrate(1/60*(x^2*(log(3)+I*pi)-4*x^2)/log(5)^2,x, algorithm="maxima")

[Out]

-1/180*((-I*pi - log(3))*x^3 + 4*x^3)/log(5)^2

Giac [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.10 \[ \int \frac {-4 x^2+x^2 (i \pi +\log (3))}{60 \log ^2(5)} \, dx=\frac {{\left (i \, \pi + \log \left (3\right )\right )} x^{3} - 4 \, x^{3}}{180 \, \log \left (5\right )^{2}} \]

[In]

integrate(1/60*(x^2*(log(3)+I*pi)-4*x^2)/log(5)^2,x, algorithm="giac")

[Out]

1/180*((I*pi + log(3))*x^3 - 4*x^3)/log(5)^2

Mupad [B] (verification not implemented)

Time = 9.05 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.95 \[ \int \frac {-4 x^2+x^2 (i \pi +\log (3))}{60 \log ^2(5)} \, dx=\frac {x^3\,\left (\Pi -\ln \left (3\right )\,1{}\mathrm {i}+4{}\mathrm {i}\right )\,1{}\mathrm {i}}{180\,{\ln \left (5\right )}^2} \]

[In]

int(((x^2*(Pi*1i + log(3)))/60 - x^2/15)/log(5)^2,x)

[Out]

(x^3*(Pi - log(3)*1i + 4i)*1i)/(180*log(5)^2)