3.42.71 \(\int \frac {e^{4 x \log (5) \log (e^{3 x}-x)} (-100 x \log (5)+300 e^{3 x} x \log (5)+(100 e^{3 x} \log (5)-100 x \log (5)) \log (e^{3 x}-x))}{e^{3 x}-x} \, dx\)

Optimal. Leaf size=19 \[ 25 e^{4 x \log (5) \log \left (e^{3 x}-x\right )} \]

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Rubi [A]  time = 0.45, antiderivative size = 19, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 71, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.014, Rules used = {6706} \begin {gather*} 25 e^{4 x \log (5) \log \left (e^{3 x}-x\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(4*x*Log[5]*Log[E^(3*x) - x])*(-100*x*Log[5] + 300*E^(3*x)*x*Log[5] + (100*E^(3*x)*Log[5] - 100*x*Log[5
])*Log[E^(3*x) - x]))/(E^(3*x) - x),x]

[Out]

25*E^(4*x*Log[5]*Log[E^(3*x) - x])

Rule 6706

Int[(F_)^(v_)*(u_), x_Symbol] :> With[{q = DerivativeDivides[v, u, x]}, Simp[(q*F^v)/Log[F], x] /;  !FalseQ[q]
] /; FreeQ[F, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=25 e^{4 x \log (5) \log \left (e^{3 x}-x\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.50, size = 25, normalized size = 1.32 \begin {gather*} \frac {4\ 5^{2+4 x \log \left (e^{3 x}-x\right )} \log (5)}{\log (625)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(4*x*Log[5]*Log[E^(3*x) - x])*(-100*x*Log[5] + 300*E^(3*x)*x*Log[5] + (100*E^(3*x)*Log[5] - 100*x
*Log[5])*Log[E^(3*x) - x]))/(E^(3*x) - x),x]

[Out]

(4*5^(2 + 4*x*Log[E^(3*x) - x])*Log[5])/Log[625]

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fricas [A]  time = 1.17, size = 17, normalized size = 0.89 \begin {gather*} 25 \, e^{\left (4 \, x \log \relax (5) \log \left (-x + e^{\left (3 \, x\right )}\right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((100*log(5)*exp(x)^3-100*x*log(5))*log(exp(x)^3-x)+300*x*log(5)*exp(x)^3-100*x*log(5))*exp(2*x*log(
5)*log(exp(x)^3-x))^2/(exp(x)^3-x),x, algorithm="fricas")

[Out]

25*e^(4*x*log(5)*log(-x + e^(3*x)))

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giac [A]  time = 0.34, size = 17, normalized size = 0.89 \begin {gather*} 25 \, e^{\left (4 \, x \log \relax (5) \log \left (-x + e^{\left (3 \, x\right )}\right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((100*log(5)*exp(x)^3-100*x*log(5))*log(exp(x)^3-x)+300*x*log(5)*exp(x)^3-100*x*log(5))*exp(2*x*log(
5)*log(exp(x)^3-x))^2/(exp(x)^3-x),x, algorithm="giac")

[Out]

25*e^(4*x*log(5)*log(-x + e^(3*x)))

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maple [A]  time = 0.04, size = 19, normalized size = 1.00

method	result	size
risch	\(25 \left ({\mathrm e}^{3 x}-x \right )^{4 x \ln \relax (5)}\)	\(19\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((100*ln(5)*exp(x)^3-100*x*ln(5))*ln(exp(x)^3-x)+300*x*ln(5)*exp(x)^3-100*x*ln(5))*exp(2*x*ln(5)*ln(exp(x)
^3-x))^2/(exp(x)^3-x),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.58, size = 17, normalized size = 0.89 \begin {gather*} 25 \, e^{\left (4 \, x \log \relax (5) \log \left (-x + e^{\left (3 \, x\right )}\right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((100*log(5)*exp(x)^3-100*x*log(5))*log(exp(x)^3-x)+300*x*log(5)*exp(x)^3-100*x*log(5))*exp(2*x*log(
5)*log(exp(x)^3-x))^2/(exp(x)^3-x),x, algorithm="maxima")

[Out]

25*e^(4*x*log(5)*log(-x + e^(3*x)))

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mupad [B]  time = 3.19, size = 16, normalized size = 0.84 \begin {gather*} 25\,{\left ({\mathrm {e}}^{3\,x}-x\right )}^{4\,x\,\ln \relax (5)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(4*x*log(5)*log(exp(3*x) - x))*(log(exp(3*x) - x)*(100*exp(3*x)*log(5) - 100*x*log(5)) - 100*x*log(5)
+ 300*x*exp(3*x)*log(5)))/(x - exp(3*x)),x)

[Out]

25*(exp(3*x) - x)^(4*x*log(5))

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sympy [A]  time = 4.73, size = 17, normalized size = 0.89 \begin {gather*} 25 e^{4 x \log {\relax (5 )} \log {\left (- x + e^{3 x} \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((100*ln(5)*exp(x)**3-100*x*ln(5))*ln(exp(x)**3-x)+300*x*ln(5)*exp(x)**3-100*x*ln(5))*exp(2*x*ln(5)*
ln(exp(x)**3-x))**2/(exp(x)**3-x),x)

[Out]

25*exp(4*x*log(5)*log(-x + exp(3*x)))

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