3.36.85 \(\int \frac {-64 e^4 x+(5+16 e^4 x) \log (3)}{20 \log (3)} \, dx\)

Optimal. Leaf size=23 \[ \frac {x}{4}+\frac {2}{5} e^4 x \left (x-\frac {4 x}{\log (3)}\right ) \]

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Rubi [A]  time = 0.01, antiderivative size = 32, normalized size of antiderivative = 1.39, number of steps used = 2, number of rules used = 1, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.038, Rules used = {12} \begin {gather*} \frac {\left (16 e^4 x+5\right )^2}{640 e^4}-\frac {8 e^4 x^2}{5 \log (3)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-64*E^4*x + (5 + 16*E^4*x)*Log[3])/(20*Log[3]),x]

[Out]

(5 + 16*E^4*x)^2/(640*E^4) - (8*E^4*x^2)/(5*Log[3])

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 {\int \left (-64 e^4 x+\left (5+16 e^4 x\right ) \log (3)\right ) \, dx}{20 \log (3)}\\ &=\frac {\left (5+16 e^4 x\right )^2}{640 e^4}-\frac {8 e^4 x^2}{5 \log (3)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 25, normalized size = 1.09 \begin {gather*} \frac {8 e^4 x^2 (-4+\log (3))+x \log (243)}{20 \log (3)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-64*E^4*x + (5 + 16*E^4*x)*Log[3])/(20*Log[3]),x]

[Out]

(8*E^4*x^2*(-4 + Log[3]) + x*Log[243])/(20*Log[3])

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fricas [A]  time = 0.91, size = 29, normalized size = 1.26 \begin {gather*} -\frac {32 \, x^{2} e^{4} - {\left (8 \, x^{2} e^{4} + 5 \, x\right )} \log \relax (3)}{20 \, \log \relax (3)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/20*((16*x*exp(4)+5)*log(3)-64*x*exp(4))/log(3),x, algorithm="fricas")

[Out]

-1/20*(32*x^2*e^4 - (8*x^2*e^4 + 5*x)*log(3))/log(3)

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giac [A]  time = 0.14, size = 29, normalized size = 1.26 \begin {gather*} -\frac {32 \, x^{2} e^{4} - {\left (8 \, x^{2} e^{4} + 5 \, x\right )} \log \relax (3)}{20 \, \log \relax (3)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/20*((16*x*exp(4)+5)*log(3)-64*x*exp(4))/log(3),x, algorithm="giac")

[Out]

-1/20*(32*x^2*e^4 - (8*x^2*e^4 + 5*x)*log(3))/log(3)

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maple [A]  time = 0.03, size = 20, normalized size = 0.87

method	result	size
norman	\(\frac {x}{4}+\frac {2 \,{\mathrm e}^{4} \left (-4+\ln \relax (3)\right ) x^{2}}{5 \ln \relax (3)}\)	\(20\)
risch	\(\frac {2 x^{2} {\mathrm e}^{4}}{5}-\frac {8 x^{2} {\mathrm e}^{4}}{5 \ln \relax (3)}+\frac {x}{4}\)	\(23\)
gosper	\(\frac {x \left (8 \,{\mathrm e}^{4} x \ln \relax (3)-32 x \,{\mathrm e}^{4}+5 \ln \relax (3)\right )}{20 \ln \relax (3)}\)	\(25\)
default	\(\frac {\ln \relax (3) \left (8 x^{2} {\mathrm e}^{4}+5 x \right )-32 x^{2} {\mathrm e}^{4}}{20 \ln \relax (3)}\)	\(29\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/20*((16*x*exp(4)+5)*ln(3)-64*x*exp(4))/ln(3),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.42, size = 29, normalized size = 1.26 \begin {gather*} -\frac {32 \, x^{2} e^{4} - {\left (8 \, x^{2} e^{4} + 5 \, x\right )} \log \relax (3)}{20 \, \log \relax (3)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/20*((16*x*exp(4)+5)*log(3)-64*x*exp(4))/log(3),x, algorithm="maxima")

[Out]

-1/20*(32*x^2*e^4 - (8*x^2*e^4 + 5*x)*log(3))/log(3)

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mupad [B]  time = 2.11, size = 24, normalized size = 1.04 \begin {gather*} \frac {x}{4}-\frac {x^2\,\left (64\,{\mathrm {e}}^4-16\,{\mathrm {e}}^4\,\ln \relax (3)\right )}{40\,\ln \relax (3)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-((16*x*exp(4))/5 - (log(3)*(16*x*exp(4) + 5))/20)/log(3),x)

[Out]

x/4 - (x^2*(64*exp(4) - 16*exp(4)*log(3)))/(40*log(3))

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sympy [A]  time = 0.06, size = 24, normalized size = 1.04 \begin {gather*} \frac {x^{2} \left (- 8 e^{4} + 2 e^{4} \log {\relax (3 )}\right )}{5 \log {\relax (3 )}} + \frac {x}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/20*((16*x*exp(4)+5)*ln(3)-64*x*exp(4))/ln(3),x)

[Out]

x**2*(-8*exp(4) + 2*exp(4)*log(3))/(5*log(3)) + x/4

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