\(\int \frac {1}{5} (10-12 x^2 \log (5)) \, dx\) [7700]

Optimal result

Integrand size = 13, antiderivative size = 13 \[ \int \frac {1}{5} \left (10-12 x^2 \log (5)\right ) \, dx=x \left (2-\frac {4}{5} x^2 \log (5)\right ) \]

[Out]

(-4/5*x^2*ln(5)+2)*x

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 13, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {12} \[ \int \frac {1}{5} \left (10-12 x^2 \log (5)\right ) \, dx=2 x-\frac {4}{5} x^3 \log (5) \]

[In]

Int[(10 - 12*x^2*Log[5])/5,x]

[Out]

2*x - (4*x^3*Log[5])/5

Rule 12

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

Rubi steps \begin{align*} \text {integral}& = \frac {1}{5} \int \left (10-12 x^2 \log (5)\right ) \, dx \\ & = 2 x-\frac {4}{5} x^3 \log (5) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 13, normalized size of antiderivative = 1.00 \[ \int \frac {1}{5} \left (10-12 x^2 \log (5)\right ) \, dx=2 x-\frac {4}{5} x^3 \log (5) \]

[In]

Integrate[(10 - 12*x^2*Log[5])/5,x]

[Out]

2*x - (4*x^3*Log[5])/5

Maple [A] (verified)

Time = 0.04 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.92

method	result	size
default	\(-\frac {4 x^{3} \ln \left (5\right )}{5}+2 x\)	\(12\)
norman	\(-\frac {4 x^{3} \ln \left (5\right )}{5}+2 x\)	\(12\)
risch	\(-\frac {4 x^{3} \ln \left (5\right )}{5}+2 x\)	\(12\)
parallelrisch	\(-\frac {4 x^{3} \ln \left (5\right )}{5}+2 x\)	\(12\)
parts	\(-\frac {4 x^{3} \ln \left (5\right )}{5}+2 x\)	\(12\)
gosper	\(-\frac {2 x \left (2 x^{2} \ln \left (5\right )-5\right )}{5}\)	\(13\)

[In]

int(-12/5*x^2*ln(5)+2,x,method=_RETURNVERBOSE)

[Out]

-4/5*x^3*ln(5)+2*x

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.85 \[ \int \frac {1}{5} \left (10-12 x^2 \log (5)\right ) \, dx=-\frac {4}{5} \, x^{3} \log \left (5\right ) + 2 \, x \]

[In]

integrate(-12/5*x^2*log(5)+2,x, algorithm="fricas")

[Out]

-4/5*x^3*log(5) + 2*x

Sympy [A] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.92 \[ \int \frac {1}{5} \left (10-12 x^2 \log (5)\right ) \, dx=- \frac {4 x^{3} \log {\left (5 \right )}}{5} + 2 x \]

[In]

integrate(-12/5*x**2*ln(5)+2,x)

[Out]

-4*x**3*log(5)/5 + 2*x

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.85 \[ \int \frac {1}{5} \left (10-12 x^2 \log (5)\right ) \, dx=-\frac {4}{5} \, x^{3} \log \left (5\right ) + 2 \, x \]

[In]

integrate(-12/5*x^2*log(5)+2,x, algorithm="maxima")

[Out]

-4/5*x^3*log(5) + 2*x

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.85 \[ \int \frac {1}{5} \left (10-12 x^2 \log (5)\right ) \, dx=-\frac {4}{5} \, x^{3} \log \left (5\right ) + 2 \, x \]

[In]

integrate(-12/5*x^2*log(5)+2,x, algorithm="giac")

[Out]

-4/5*x^3*log(5) + 2*x

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.85 \[ \int \frac {1}{5} \left (10-12 x^2 \log (5)\right ) \, dx=2\,x-\frac {4\,x^3\,\ln \left (5\right )}{5} \]

[In]

int(2 - (12*x^2*log(5))/5,x)

[Out]

2*x - (4*x^3*log(5))/5