3.11.52 \(\int \frac {e^{\frac {x}{\log (e^x-5 x^2)}} (-e^x x+10 x^2+(e^x-5 x^2) \log (e^x-5 x^2))}{e^{\frac {x}{\log (e^x-5 x^2)}} (e^x-5 x^2) \log ^2(e^x-5 x^2)+(-e^{4+x}+5 e^4 x^2) \log ^2(e^x-5 x^2)} \, dx\)

Optimal. Leaf size=25 \[ 1+\log \left (-e^4+e^{\frac {x}{\log \left (e^x-5 x^2\right )}}\right ) \]

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Rubi [A]  time = 1.63, antiderivative size = 23, normalized size of antiderivative = 0.92, number of steps used = 2, number of rules used = 2, integrand size = 119, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.017, Rules used = {6741, 6684} \begin {gather*} \log \left (e^4-e^{\frac {x}{\log \left (e^x-5 x^2\right )}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(x/Log[E^x - 5*x^2])*(-(E^x*x) + 10*x^2 + (E^x - 5*x^2)*Log[E^x - 5*x^2]))/(E^(x/Log[E^x - 5*x^2])*(E^x
- 5*x^2)*Log[E^x - 5*x^2]^2 + (-E^(4 + x) + 5*E^4*x^2)*Log[E^x - 5*x^2]^2),x]

[Out]

Log[E^4 - E^(x/Log[E^x - 5*x^2])]

Rule 6684

Int[(u_)/(y_), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*Log[RemoveContent[y, x]], x] /;  !Fa
lseQ[q]]

Rule 6741

Int[u_, x_Symbol] :> With[{v = NormalizeIntegrand[u, x]}, Int[v, x] /; v =!= u]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{\frac {x}{\log \left (e^x-5 x^2\right )}} \left (e^x x-10 x^2-\left (e^x-5 x^2\right ) \log \left (e^x-5 x^2\right )\right )}{\left (e^4-e^{\frac {x}{\log \left (e^x-5 x^2\right )}}\right ) \left (e^x-5 x^2\right ) \log ^2\left (e^x-5 x^2\right )} \, dx\\ &=\log \left (e^4-e^{\frac {x}{\log \left (e^x-5 x^2\right )}}\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.06, size = 23, normalized size = 0.92 \begin {gather*} \log \left (-e^4+e^{\frac {x}{\log \left (e^x-5 x^2\right )}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(x/Log[E^x - 5*x^2])*(-(E^x*x) + 10*x^2 + (E^x - 5*x^2)*Log[E^x - 5*x^2]))/(E^(x/Log[E^x - 5*x^2]
)*(E^x - 5*x^2)*Log[E^x - 5*x^2]^2 + (-E^(4 + x) + 5*E^4*x^2)*Log[E^x - 5*x^2]^2),x]

[Out]

Log[-E^4 + E^(x/Log[E^x - 5*x^2])]

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fricas [A]  time = 0.60, size = 30, normalized size = 1.20 \begin {gather*} \log \left (-e^{4} + e^{\left (\frac {x}{\log \left (-{\left (5 \, x^{2} e^{4} - e^{\left (x + 4\right )}\right )} e^{\left (-4\right )}\right )}\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((exp(x)-5*x^2)*log(exp(x)-5*x^2)-exp(x)*x+10*x^2)*exp(x/log(exp(x)-5*x^2))/((exp(x)-5*x^2)*log(exp(
x)-5*x^2)^2*exp(x/log(exp(x)-5*x^2))+(-exp(4)*exp(x)+5*x^2*exp(4))*log(exp(x)-5*x^2)^2),x, algorithm="fricas")

[Out]

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

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giac [B]  time = 8.86, size = 231, normalized size = 9.24 \begin {gather*} \frac {1}{2} \, \log \left (-2 \, {\left | 5 \, x^{2} - e^{x} \right |}^{\frac {2 \, x}{\pi ^{2} \mathrm {sgn}\left (5 \, x^{2} - e^{x}\right ) + \pi ^{2} + 2 \, \log \left ({\left | 5 \, x^{2} - e^{x} \right |}\right )^{2}}} \cos \left (-\frac {2 \, \pi x \mathrm {sgn}\left (-5 \, x^{2} + e^{x}\right )}{\pi ^{2} \mathrm {sgn}\left (-5 \, x^{2} + e^{x}\right )^{2} - 2 \, \pi ^{2} \mathrm {sgn}\left (-5 \, x^{2} + e^{x}\right ) + \pi ^{2} + 4 \, \log \left ({\left | -5 \, x^{2} + e^{x} \right |}\right )^{2}} + \frac {2 \, \pi x}{\pi ^{2} \mathrm {sgn}\left (-5 \, x^{2} + e^{x}\right )^{2} - 2 \, \pi ^{2} \mathrm {sgn}\left (-5 \, x^{2} + e^{x}\right ) + \pi ^{2} + 4 \, \log \left ({\left | -5 \, x^{2} + e^{x} \right |}\right )^{2}}\right ) e^{4} + {\left | 5 \, x^{2} - e^{x} \right |}^{\frac {4 \, x}{\pi ^{2} \mathrm {sgn}\left (5 \, x^{2} - e^{x}\right ) + \pi ^{2} + 2 \, \log \left ({\left | 5 \, x^{2} - e^{x} \right |}\right )^{2}}} + e^{8}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((exp(x)-5*x^2)*log(exp(x)-5*x^2)-exp(x)*x+10*x^2)*exp(x/log(exp(x)-5*x^2))/((exp(x)-5*x^2)*log(exp(
x)-5*x^2)^2*exp(x/log(exp(x)-5*x^2))+(-exp(4)*exp(x)+5*x^2*exp(4))*log(exp(x)-5*x^2)^2),x, algorithm="giac")

[Out]

1/2*log(-2*abs(5*x^2 - e^x)^(2*x/(pi^2*sgn(5*x^2 - e^x) + pi^2 + 2*log(abs(5*x^2 - e^x))^2))*cos(-2*pi*x*sgn(-
5*x^2 + e^x)/(pi^2*sgn(-5*x^2 + e^x)^2 - 2*pi^2*sgn(-5*x^2 + e^x) + pi^2 + 4*log(abs(-5*x^2 + e^x))^2) + 2*pi*
x/(pi^2*sgn(-5*x^2 + e^x)^2 - 2*pi^2*sgn(-5*x^2 + e^x) + pi^2 + 4*log(abs(-5*x^2 + e^x))^2))*e^4 + abs(5*x^2 -
e^x)^(4*x/(pi^2*sgn(5*x^2 - e^x) + pi^2 + 2*log(abs(5*x^2 - e^x))^2)) + e^8)

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maple [A]  time = 0.05, size = 21, normalized size = 0.84

method	result	size
risch	\(\ln \left (-{\mathrm e}^{4}+{\mathrm e}^{\frac {x}{\ln \left ({\mathrm e}^{x}-5 x^{2}\right )}}\right )\)	\(21\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((exp(x)-5*x^2)*ln(exp(x)-5*x^2)-exp(x)*x+10*x^2)*exp(x/ln(exp(x)-5*x^2))/((exp(x)-5*x^2)*ln(exp(x)-5*x^2)
^2*exp(x/ln(exp(x)-5*x^2))+(-exp(4)*exp(x)+5*x^2*exp(4))*ln(exp(x)-5*x^2)^2),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.52, size = 20, normalized size = 0.80 \begin {gather*} \log \left (-e^{4} + e^{\left (\frac {x}{\log \left (-5 \, x^{2} + e^{x}\right )}\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((exp(x)-5*x^2)*log(exp(x)-5*x^2)-exp(x)*x+10*x^2)*exp(x/log(exp(x)-5*x^2))/((exp(x)-5*x^2)*log(exp(
x)-5*x^2)^2*exp(x/log(exp(x)-5*x^2))+(-exp(4)*exp(x)+5*x^2*exp(4))*log(exp(x)-5*x^2)^2),x, algorithm="maxima")

[Out]

log(-e^4 + e^(x/log(-5*x^2 + e^x)))

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mupad [B]  time = 1.01, size = 20, normalized size = 0.80 \begin {gather*} \ln \left ({\mathrm {e}}^{\frac {x}{\ln \left ({\mathrm {e}}^x-5\,x^2\right )}}-{\mathrm {e}}^4\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(x/log(exp(x) - 5*x^2))*(log(exp(x) - 5*x^2)*(exp(x) - 5*x^2) - x*exp(x) + 10*x^2))/(log(exp(x) - 5*x^
2)^2*(5*x^2*exp(4) - exp(4)*exp(x)) + exp(x/log(exp(x) - 5*x^2))*log(exp(x) - 5*x^2)^2*(exp(x) - 5*x^2)),x)

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((exp(x)-5*x**2)*ln(exp(x)-5*x**2)-exp(x)*x+10*x**2)*exp(x/ln(exp(x)-5*x**2))/((exp(x)-5*x**2)*ln(ex
p(x)-5*x**2)**2*exp(x/ln(exp(x)-5*x**2))+(-exp(4)*exp(x)+5*x**2*exp(4))*ln(exp(x)-5*x**2)**2),x)

[Out]

Timed out

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