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

Optimal. Leaf size=26 \[ 2-x \left (x+x^2+\frac {1}{4} e^4 (2+x) (x+2 \log (5))\right ) \]

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Rubi [A]  time = 0.01, antiderivative size = 45, normalized size of antiderivative = 1.73, number of steps used = 3, number of rules used = 1, integrand size = 37, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.027, Rules used = {12} \begin {gather*} -\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) \end {gather*}

Antiderivative was successfully verified.

[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 {gather*} \begin {aligned} \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 {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 38, normalized size = 1.46 \begin {gather*} -\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 ) \end {gather*}

Antiderivative was successfully verified.

[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

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fricas [A]  time = 0.57, size = 37, normalized size = 1.42 \begin {gather*} -x^{3} - \frac {1}{2} \, {\left (x^{2} + 2 \, x\right )} e^{4} \log \relax (5) - x^{2} - \frac {1}{4} \, {\left (x^{3} + 2 \, x^{2}\right )} e^{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(-4*x-4)*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

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giac [A]  time = 1.76, size = 37, normalized size = 1.42 \begin {gather*} -x^{3} - \frac {1}{2} \, {\left (x^{2} + 2 \, x\right )} e^{4} \log \relax (5) - x^{2} - \frac {1}{4} \, {\left (x^{3} + 2 \, x^{2}\right )} e^{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(-4*x-4)*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

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maple [A]  time = 0.04, size = 35, normalized size = 1.35

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/4*(-4*x-4)*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

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maxima [A]  time = 0.37, size = 37, normalized size = 1.42 \begin {gather*} -x^{3} - \frac {1}{2} \, {\left (x^{2} + 2 \, x\right )} e^{4} \log \relax (5) - x^{2} - \frac {1}{4} \, {\left (x^{3} + 2 \, x^{2}\right )} e^{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(-4*x-4)*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

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mupad [B]  time = 0.06, size = 36, normalized size = 1.38 \begin {gather*} \left (-\frac {{\mathrm {e}}^4}{4}-1\right )\,x^3+\left (-\frac {{\mathrm {e}}^4}{2}-\frac {{\mathrm {e}}^4\,\ln \relax (5)}{2}-1\right )\,x^2-{\mathrm {e}}^4\,\ln \relax (5)\,x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [A]  time = 0.09, size = 39, normalized size = 1.50 \begin {gather*} x^{3} \left (- \frac {e^{4}}{4} - 1\right ) + x^{2} \left (- \frac {e^{4} \log {\relax (5 )}}{2} - \frac {e^{4}}{2} - 1\right ) - x e^{4} \log {\relax (5 )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(-4*x-4)*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)

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