\(\int \frac {1}{4} (-8 x-12 x^2+e^4 (-4 x-3 x^2)+e^4 (-4-4 x) \log (5)) \, dx\) [4871]

Optimal result

Integrand size = 37, antiderivative size = 26 \[ \int \frac {1}{4} \left (-8 x-12 x^2+e^4 \left (-4 x-3 x^2\right )+e^4 (-4-4 x) \log (5)\right ) \, dx=2-x \left (x+x^2+\frac {1}{4} e^4 (2+x) (x+2 \log (5))\right ) \]

[Out]

2-x*(x+x^2+1/4*(2+x)*exp(4)*(x+2*ln(5)))

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.73, number of steps used = 3, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.027, Rules used = {12} \[ \int \frac {1}{4} \left (-8 x-12 x^2+e^4 \left (-4 x-3 x^2\right )+e^4 (-4-4 x) \log (5)\right ) \, dx=-\frac {1}{4} e^4 x^3-x^3-\frac {e^4 x^2}{2}-x^2-\frac {1}{2} e^4 (x+1)^2 \log (5) \]

[In]

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

[Out]

-x^2 - (E^4*x^2)/2 - x^3 - (E^4*x^3)/4 - (E^4*(1 + x)^2*Log[5])/2

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}{4} \int \left (-8 x-12 x^2+e^4 \left (-4 x-3 x^2\right )+e^4 (-4-4 x) \log (5)\right ) \, dx \\ & = -x^2-x^3-\frac {1}{2} e^4 (1+x)^2 \log (5)+\frac {1}{4} e^4 \int \left (-4 x-3 x^2\right ) \, dx \\ & = -x^2-\frac {e^4 x^2}{2}-x^3-\frac {e^4 x^3}{4}-\frac {1}{2} e^4 (1+x)^2 \log (5) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.46 \[ \int \frac {1}{4} \left (-8 x-12 x^2+e^4 \left (-4 x-3 x^2\right )+e^4 (-4-4 x) \log (5)\right ) \, dx=-\frac {1}{4} \left (4+e^4\right ) x^3-e^4 x \log (5)-\frac {1}{2} x^2 \left (2+e^4 (1+\log (5))\right ) \]

[In]

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

[Out]

-1/4*((4 + E^4)*x^3) - E^4*x*Log[5] - (x^2*(2 + E^4*(1 + Log[5])))/2

Maple [A] (verified)

Time = 0.04 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.35

method	result	size
norman	\(\left (-\frac {{\mathrm e}^{4}}{4}-1\right ) x^{3}+\left (-\frac {{\mathrm e}^{4} \ln \left (5\right )}{2}-\frac {{\mathrm e}^{4}}{2}-1\right ) x^{2}-{\mathrm e}^{4} \ln \left (5\right ) x\)	\(35\)
gosper	\(-\frac {x \left (2 \,{\mathrm e}^{4} \ln \left (5\right ) x +x^{2} {\mathrm e}^{4}+4 \,{\mathrm e}^{4} \ln \left (5\right )+2 x \,{\mathrm e}^{4}+4 x^{2}+4 x \right )}{4}\)	\(37\)
default	\(-\frac {{\mathrm e}^{4} \ln \left (5\right ) x^{2}}{2}-\frac {x^{3} {\mathrm e}^{4}}{4}-{\mathrm e}^{4} \ln \left (5\right ) x -\frac {x^{2} {\mathrm e}^{4}}{2}-x^{3}-x^{2}\)	\(42\)
risch	\(-\frac {{\mathrm e}^{4} \ln \left (5\right ) x^{2}}{2}-\frac {x^{3} {\mathrm e}^{4}}{4}-{\mathrm e}^{4} \ln \left (5\right ) x -\frac {x^{2} {\mathrm e}^{4}}{2}-x^{3}-x^{2}\)	\(42\)
parallelrisch	\(-\frac {{\mathrm e}^{4} \ln \left (5\right ) x^{2}}{2}-\frac {x^{3} {\mathrm e}^{4}}{4}-{\mathrm e}^{4} \ln \left (5\right ) x -\frac {x^{2} {\mathrm e}^{4}}{2}-x^{3}-x^{2}\)	\(42\)
parts	\(-\frac {{\mathrm e}^{4} \ln \left (5\right ) x^{2}}{2}-\frac {x^{3} {\mathrm e}^{4}}{4}-{\mathrm e}^{4} \ln \left (5\right ) x -\frac {x^{2} {\mathrm e}^{4}}{2}-x^{3}-x^{2}\)	\(42\)

[In]

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

[Out]

(-1/4*exp(4)-1)*x^3+(-1/2*exp(4)*ln(5)-1/2*exp(4)-1)*x^2-exp(4)*ln(5)*x

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.42 \[ \int \frac {1}{4} \left (-8 x-12 x^2+e^4 \left (-4 x-3 x^2\right )+e^4 (-4-4 x) \log (5)\right ) \, dx=-x^{3} - \frac {1}{2} \, {\left (x^{2} + 2 \, x\right )} e^{4} \log \left (5\right ) - x^{2} - \frac {1}{4} \, {\left (x^{3} + 2 \, x^{2}\right )} e^{4} \]

[In]

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

[Out]

-x^3 - 1/2*(x^2 + 2*x)*e^4*log(5) - x^2 - 1/4*(x^3 + 2*x^2)*e^4

Sympy [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.50 \[ \int \frac {1}{4} \left (-8 x-12 x^2+e^4 \left (-4 x-3 x^2\right )+e^4 (-4-4 x) \log (5)\right ) \, dx=x^{3} \left (- \frac {e^{4}}{4} - 1\right ) + x^{2} \left (- \frac {e^{4} \log {\left (5 \right )}}{2} - \frac {e^{4}}{2} - 1\right ) - x e^{4} \log {\left (5 \right )} \]

[In]

integrate(1/4*(-4-4*x)*exp(4)*ln(5)+1/4*(-3*x**2-4*x)*exp(4)-3*x**2-2*x,x)

[Out]

x**3*(-exp(4)/4 - 1) + x**2*(-exp(4)*log(5)/2 - exp(4)/2 - 1) - x*exp(4)*log(5)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.42 \[ \int \frac {1}{4} \left (-8 x-12 x^2+e^4 \left (-4 x-3 x^2\right )+e^4 (-4-4 x) \log (5)\right ) \, dx=-x^{3} - \frac {1}{2} \, {\left (x^{2} + 2 \, x\right )} e^{4} \log \left (5\right ) - x^{2} - \frac {1}{4} \, {\left (x^{3} + 2 \, x^{2}\right )} e^{4} \]

[In]

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

[Out]

-x^3 - 1/2*(x^2 + 2*x)*e^4*log(5) - x^2 - 1/4*(x^3 + 2*x^2)*e^4

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.42 \[ \int \frac {1}{4} \left (-8 x-12 x^2+e^4 \left (-4 x-3 x^2\right )+e^4 (-4-4 x) \log (5)\right ) \, dx=-x^{3} - \frac {1}{2} \, {\left (x^{2} + 2 \, x\right )} e^{4} \log \left (5\right ) - x^{2} - \frac {1}{4} \, {\left (x^{3} + 2 \, x^{2}\right )} e^{4} \]

[In]

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

[Out]

-x^3 - 1/2*(x^2 + 2*x)*e^4*log(5) - x^2 - 1/4*(x^3 + 2*x^2)*e^4

Mupad [B] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.38 \[ \int \frac {1}{4} \left (-8 x-12 x^2+e^4 \left (-4 x-3 x^2\right )+e^4 (-4-4 x) \log (5)\right ) \, dx=\left (-\frac {{\mathrm {e}}^4}{4}-1\right )\,x^3+\left (-\frac {{\mathrm {e}}^4}{2}-\frac {{\mathrm {e}}^4\,\ln \left (5\right )}{2}-1\right )\,x^2-{\mathrm {e}}^4\,\ln \left (5\right )\,x \]

[In]

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

[Out]

- x^3*(exp(4)/4 + 1) - x^2*(exp(4)/2 + (exp(4)*log(5))/2 + 1) - x*exp(4)*log(5)