3.12.45 \(\int \frac {1}{2} (2+(2 e^2 x+e (-4 x+3 x^2)) \log (625)) \, dx\)

Optimal. Leaf size=16 \[ x+\frac {1}{2} e x^2 (-2+e+x) \log (625) \]

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Rubi [A]  time = 0.02, antiderivative size = 32, normalized size of antiderivative = 2.00, number of steps used = 4, number of rules used = 1, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.037, Rules used = {12} \begin {gather*} \frac {1}{2} e x^3 \log (625)+\frac {1}{2} e^2 x^2 \log (625)-e x^2 \log (625)+x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(2 + (2*E^2*x + E*(-4*x + 3*x^2))*Log[625])/2,x]

[Out]

x - E*x^2*Log[625] + (E^2*x^2*Log[625])/2 + (E*x^3*Log[625])/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}{2} \int \left (2+\left (2 e^2 x+e \left (-4 x+3 x^2\right )\right ) \log (625)\right ) \, dx\\ &=x+\frac {1}{2} \log (625) \int \left (2 e^2 x+e \left (-4 x+3 x^2\right )\right ) \, dx\\ &=x+\frac {1}{2} e^2 x^2 \log (625)+\frac {1}{2} (e \log (625)) \int \left (-4 x+3 x^2\right ) \, dx\\ &=x-e x^2 \log (625)+\frac {1}{2} e^2 x^2 \log (625)+\frac {1}{2} e x^3 \log (625)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 32, normalized size = 2.00 \begin {gather*} x-e x^2 \log (625)+\frac {1}{2} e^2 x^2 \log (625)+\frac {1}{2} e x^3 \log (625) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(2 + (2*E^2*x + E*(-4*x + 3*x^2))*Log[625])/2,x]

[Out]

x - E*x^2*Log[625] + (E^2*x^2*Log[625])/2 + (E*x^3*Log[625])/2

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fricas [A]  time = 0.76, size = 25, normalized size = 1.56 \begin {gather*} 2 \, {\left (x^{2} e^{2} + {\left (x^{3} - 2 \, x^{2}\right )} e\right )} \log \relax (5) + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.31, size = 25, normalized size = 1.56 \begin {gather*} 2 \, {\left (x^{2} e^{2} + {\left (x^{3} - 2 \, x^{2}\right )} e\right )} \log \relax (5) + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.04, size = 28, normalized size = 1.75

method	result	size
default	\(2 \ln \relax (5) \left (x^{2} {\mathrm e}^{2}+{\mathrm e} \left (x^{3}-2 x^{2}\right )\right )+x\)	\(28\)
gosper	\(x \left (2 x \,{\mathrm e}^{2} \ln \relax (5)+2 \,{\mathrm e} \ln \relax (5) x^{2}-4 x \,{\mathrm e} \ln \relax (5)+1\right )\)	\(30\)
norman	\(2 \,{\mathrm e} \ln \relax (5) x^{3}+\left (2 \,{\mathrm e}^{2} \ln \relax (5)-4 \,{\mathrm e} \ln \relax (5)\right ) x^{2}+x\)	\(31\)
risch	\(2 \,{\mathrm e}^{2} \ln \relax (5) x^{2}+2 \,{\mathrm e} \ln \relax (5) x^{3}-4 \,{\mathrm e} \ln \relax (5) x^{2}+x\)	\(32\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.43, size = 25, normalized size = 1.56 \begin {gather*} 2 \, {\left (x^{2} e^{2} + {\left (x^{3} - 2 \, x^{2}\right )} e\right )} \log \relax (5) + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.75, size = 27, normalized size = 1.69 \begin {gather*} 2\,\mathrm {e}\,\ln \relax (5)\,x^3-\ln \relax (5)\,\left (4\,\mathrm {e}-2\,{\mathrm {e}}^2\right )\,x^2+x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1 - 2*log(5)*(exp(1)*(4*x - 3*x^2) - 2*x*exp(2)),x)

[Out]

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

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sympy [A]  time = 0.06, size = 32, normalized size = 2.00 \begin {gather*} 2 e x^{3} \log {\relax (5 )} + x^{2} \left (- 4 e \log {\relax (5 )} + 2 e^{2} \log {\relax (5 )}\right ) + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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