\(\int (8 e^4 x-12 x^2-\log (2 e^3)) \log (\log (5)) \, dx\) [1312]

Optimal result

Integrand size = 24, antiderivative size = 23 \[ \int \left (8 e^4 x-12 x^2-\log \left (2 e^3\right )\right ) \log (\log (5)) \, dx=\left (e^4-x\right ) \left (4 x^2+\log \left (2 e^3\right )\right ) \log (\log (5)) \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.30, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.042, Rules used = {12} \[ \int \left (8 e^4 x-12 x^2-\log \left (2 e^3\right )\right ) \log (\log (5)) \, dx=-4 x^3 \log (\log (5))+4 e^4 x^2 \log (\log (5))-x (3+\log (2)) \log (\log (5)) \]

[In]

Int[(8*E^4*x - 12*x^2 - Log[2*E^3])*Log[Log[5]],x]

[Out]

4*E^4*x^2*Log[Log[5]] - 4*x^3*Log[Log[5]] - x*(3 + Log[2])*Log[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]]

Rubi steps \begin{align*} \text {integral}& = \log (\log (5)) \int \left (8 e^4 x-12 x^2-\log \left (2 e^3\right )\right ) \, dx \\ & = 4 e^4 x^2 \log (\log (5))-4 x^3 \log (\log (5))-x (3+\log (2)) \log (\log (5)) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.13 \[ \int \left (8 e^4 x-12 x^2-\log \left (2 e^3\right )\right ) \log (\log (5)) \, dx=\left (-3 x+4 e^4 x^2-4 x^3-x \log (2)\right ) \log (\log (5)) \]

[In]

Integrate[(8*E^4*x - 12*x^2 - Log[2*E^3])*Log[Log[5]],x]

[Out]

(-3*x + 4*E^4*x^2 - 4*x^3 - x*Log[2])*Log[Log[5]]

Maple [A] (verified)

Time = 0.03 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04

method	result	size
gosper	\(x \left (4 x \,{\mathrm e}^{4}-4 x^{2}-\ln \left (2 \,{\mathrm e}^{3}\right )\right ) \ln \left (\ln \left (5\right )\right )\)	\(24\)
default	\(\left (-4 x^{3}+4 x^{2} {\mathrm e}^{4}-\ln \left (2 \,{\mathrm e}^{3}\right ) x \right ) \ln \left (\ln \left (5\right )\right )\)	\(26\)
parallelrisch	\(\left (-4 x^{3}+4 x^{2} {\mathrm e}^{4}-\ln \left (2 \,{\mathrm e}^{3}\right ) x \right ) \ln \left (\ln \left (5\right )\right )\)	\(26\)
risch	\(4 \ln \left (\ln \left (5\right )\right ) {\mathrm e}^{4} x^{2}-4 \ln \left (\ln \left (5\right )\right ) x^{3}-\ln \left (\ln \left (5\right )\right ) x \ln \left (2\right )-3 x \ln \left (\ln \left (5\right )\right )\)	\(34\)
norman	\(\left (-3 \ln \left (\ln \left (5\right )\right )-\ln \left (\ln \left (5\right )\right ) \ln \left (2\right )\right ) x -4 \ln \left (\ln \left (5\right )\right ) x^{3}+4 \ln \left (\ln \left (5\right )\right ) {\mathrm e}^{4} x^{2}\)	\(35\)

[In]

int((-ln(2*exp(3))+8*x*exp(4)-12*x^2)*ln(ln(5)),x,method=_RETURNVERBOSE)

[Out]

x*(4*x*exp(4)-4*x^2-ln(2*exp(3)))*ln(ln(5))

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.09 \[ \int \left (8 e^4 x-12 x^2-\log \left (2 e^3\right )\right ) \log (\log (5)) \, dx=-{\left (4 \, x^{3} - 4 \, x^{2} e^{4} + x \log \left (2\right ) + 3 \, x\right )} \log \left (\log \left (5\right )\right ) \]

[In]

integrate((-log(2*exp(3))+8*x*exp(4)-12*x^2)*log(log(5)),x, algorithm="fricas")

[Out]

-(4*x^3 - 4*x^2*e^4 + x*log(2) + 3*x)*log(log(5))

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 41 vs. \(2 (20) = 40\).

Time = 0.02 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.78 \[ \int \left (8 e^4 x-12 x^2-\log \left (2 e^3\right )\right ) \log (\log (5)) \, dx=- 4 x^{3} \log {\left (\log {\left (5 \right )} \right )} + 4 x^{2} e^{4} \log {\left (\log {\left (5 \right )} \right )} + x \left (- 3 \log {\left (\log {\left (5 \right )} \right )} - \log {\left (2 \right )} \log {\left (\log {\left (5 \right )} \right )}\right ) \]

[In]

integrate((-ln(2*exp(3))+8*x*exp(4)-12*x**2)*ln(ln(5)),x)

[Out]

-4*x**3*log(log(5)) + 4*x**2*exp(4)*log(log(5)) + x*(-3*log(log(5)) - log(2)*log(log(5)))

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.09 \[ \int \left (8 e^4 x-12 x^2-\log \left (2 e^3\right )\right ) \log (\log (5)) \, dx=-{\left (4 \, x^{3} - 4 \, x^{2} e^{4} + x \log \left (2 \, e^{3}\right )\right )} \log \left (\log \left (5\right )\right ) \]

[In]

integrate((-log(2*exp(3))+8*x*exp(4)-12*x^2)*log(log(5)),x, algorithm="maxima")

[Out]

-(4*x^3 - 4*x^2*e^4 + x*log(2*e^3))*log(log(5))

Giac [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.09 \[ \int \left (8 e^4 x-12 x^2-\log \left (2 e^3\right )\right ) \log (\log (5)) \, dx=-{\left (4 \, x^{3} - 4 \, x^{2} e^{4} + x \log \left (2 \, e^{3}\right )\right )} \log \left (\log \left (5\right )\right ) \]

[In]

integrate((-log(2*exp(3))+8*x*exp(4)-12*x^2)*log(log(5)),x, algorithm="giac")

[Out]

-(4*x^3 - 4*x^2*e^4 + x*log(2*e^3))*log(log(5))

Mupad [B] (verification not implemented)

Time = 8.40 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.87 \[ \int \left (8 e^4 x-12 x^2-\log \left (2 e^3\right )\right ) \log (\log (5)) \, dx=-x\,\ln \left (\ln \left (5\right )\right )\,\left (4\,x^2-4\,{\mathrm {e}}^4\,x+\ln \left (2\right )+3\right ) \]

[In]

int(-log(log(5))*(log(2*exp(3)) - 8*x*exp(4) + 12*x^2),x)

[Out]

-x*log(log(5))*(log(2) - 4*x*exp(4) + 4*x^2 + 3)