3.24.78 \(\int \frac {-1+e^x (-5+x)-x^3+(-4 x^3+e^x x^4) \log (x)}{e^{2 x} (-5+x)+e^{2 x} x^4 \log (x)+(e^x (10-2 x)-2 e^x x^4 \log (x)) \log (-5+x+x^4 \log (x))+(-5+x+x^4 \log (x)) \log ^2(-5+x+x^4 \log (x))} \, dx\)

Optimal. Leaf size=20 \[ 2+\frac {1}{-e^x+\log \left (-5+x+x^4 \log (x)\right )} \]

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Rubi [A]  time = 0.36, antiderivative size = 20, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 107, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.019, Rules used = {6688, 6686} \begin {gather*} -\frac {1}{e^x-\log \left (x^4 \log (x)+x-5\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-1 + E^x*(-5 + x) - x^3 + (-4*x^3 + E^x*x^4)*Log[x])/(E^(2*x)*(-5 + x) + E^(2*x)*x^4*Log[x] + (E^x*(10 -
2*x) - 2*E^x*x^4*Log[x])*Log[-5 + x + x^4*Log[x]] + (-5 + x + x^4*Log[x])*Log[-5 + x + x^4*Log[x]]^2),x]

[Out]

-(E^x - Log[-5 + x + x^4*Log[x]])^(-1)

Rule 6686

Int[(u_)*(y_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[(q*y^(m + 1))/(m + 1), x] /;  !F
alseQ[q]] /; FreeQ[m, x] && NeQ[m, -1]

Rule 6688

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {1-e^x (-5+x)+x^3-x^3 \left (-4+e^x x\right ) \log (x)}{\left (5-x-x^4 \log (x)\right ) \left (e^x-\log \left (-5+x+x^4 \log (x)\right )\right )^2} \, dx\\ &=-\frac {1}{e^x-\log \left (-5+x+x^4 \log (x)\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.06, size = 20, normalized size = 1.00 \begin {gather*} -\frac {1}{e^x-\log \left (-5+x+x^4 \log (x)\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-1 + E^x*(-5 + x) - x^3 + (-4*x^3 + E^x*x^4)*Log[x])/(E^(2*x)*(-5 + x) + E^(2*x)*x^4*Log[x] + (E^x*
(10 - 2*x) - 2*E^x*x^4*Log[x])*Log[-5 + x + x^4*Log[x]] + (-5 + x + x^4*Log[x])*Log[-5 + x + x^4*Log[x]]^2),x]

[Out]

-(E^x - Log[-5 + x + x^4*Log[x]])^(-1)

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fricas [A]  time = 0.50, size = 19, normalized size = 0.95 \begin {gather*} -\frac {1}{e^{x} - \log \left (x^{4} \log \relax (x) + x - 5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((exp(x)*x^4-4*x^3)*log(x)+(x-5)*exp(x)-x^3-1)/((x^4*log(x)+x-5)*log(x^4*log(x)+x-5)^2+(-2*x^4*exp(x
)*log(x)+(-2*x+10)*exp(x))*log(x^4*log(x)+x-5)+x^4*exp(x)^2*log(x)+(x-5)*exp(x)^2),x, algorithm="fricas")

[Out]

-1/(e^x - log(x^4*log(x) + x - 5))

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giac [A]  time = 0.65, size = 19, normalized size = 0.95 \begin {gather*} -\frac {1}{e^{x} - \log \left (x^{4} \log \relax (x) + x - 5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((exp(x)*x^4-4*x^3)*log(x)+(x-5)*exp(x)-x^3-1)/((x^4*log(x)+x-5)*log(x^4*log(x)+x-5)^2+(-2*x^4*exp(x
)*log(x)+(-2*x+10)*exp(x))*log(x^4*log(x)+x-5)+x^4*exp(x)^2*log(x)+(x-5)*exp(x)^2),x, algorithm="giac")

[Out]

-1/(e^x - log(x^4*log(x) + x - 5))

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maple [A]  time = 0.05, size = 20, normalized size = 1.00

method	result	size
risch	\(-\frac {1}{{\mathrm e}^{x}-\ln \left (x^{4} \ln \relax (x )+x -5\right )}\)	\(20\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((exp(x)*x^4-4*x^3)*ln(x)+(x-5)*exp(x)-x^3-1)/((x^4*ln(x)+x-5)*ln(x^4*ln(x)+x-5)^2+(-2*x^4*exp(x)*ln(x)+(-
2*x+10)*exp(x))*ln(x^4*ln(x)+x-5)+x^4*exp(x)^2*ln(x)+(x-5)*exp(x)^2),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.50, size = 19, normalized size = 0.95 \begin {gather*} -\frac {1}{e^{x} - \log \left (x^{4} \log \relax (x) + x - 5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((exp(x)*x^4-4*x^3)*log(x)+(x-5)*exp(x)-x^3-1)/((x^4*log(x)+x-5)*log(x^4*log(x)+x-5)^2+(-2*x^4*exp(x
)*log(x)+(-2*x+10)*exp(x))*log(x^4*log(x)+x-5)+x^4*exp(x)^2*log(x)+(x-5)*exp(x)^2),x, algorithm="maxima")

[Out]

-1/(e^x - log(x^4*log(x) + x - 5))

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mupad [B]  time = 1.73, size = 17, normalized size = 0.85 \begin {gather*} \frac {1}{\ln \left (x+x^4\,\ln \relax (x)-5\right )-{\mathrm {e}}^x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(x)*(x - 5) - x^3 + log(x)*(x^4*exp(x) - 4*x^3) - 1)/(log(x + x^4*log(x) - 5)^2*(x + x^4*log(x) - 5) +
exp(2*x)*(x - 5) - log(x + x^4*log(x) - 5)*(exp(x)*(2*x - 10) + 2*x^4*exp(x)*log(x)) + x^4*exp(2*x)*log(x)),x
)

[Out]

1/(log(x + x^4*log(x) - 5) - exp(x))

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sympy [A]  time = 0.57, size = 17, normalized size = 0.85 \begin {gather*} - \frac {1}{e^{x} - \log {\left (x^{4} \log {\relax (x )} + x - 5 \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((exp(x)*x**4-4*x**3)*ln(x)+(x-5)*exp(x)-x**3-1)/((x**4*ln(x)+x-5)*ln(x**4*ln(x)+x-5)**2+(-2*x**4*ex
p(x)*ln(x)+(-2*x+10)*exp(x))*ln(x**4*ln(x)+x-5)+x**4*exp(x)**2*ln(x)+(x-5)*exp(x)**2),x)

[Out]

-1/(exp(x) - log(x**4*log(x) + x - 5))

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