\(\int \frac {-1+e^x (-5+x)-x^3+(-4 x^3+e^x x^4) \log (x)}{e^{2 x} (-5+x)+e^{2 x} x^4 \log (x)+(e^x (10-2 x)-2 e^x x^4 \log (x)) \log (-5+x+x^4 \log (x))+(-5+x+x^4 \log (x)) \log ^2(-5+x+x^4 \log (x))} \, dx\) [2378]

Optimal result

Integrand size = 107, antiderivative size = 20 \[ \int \frac {-1+e^x (-5+x)-x^3+\left (-4 x^3+e^x x^4\right ) \log (x)}{e^{2 x} (-5+x)+e^{2 x} x^4 \log (x)+\left (e^x (10-2 x)-2 e^x x^4 \log (x)\right ) \log \left (-5+x+x^4 \log (x)\right )+\left (-5+x+x^4 \log (x)\right ) \log ^2\left (-5+x+x^4 \log (x)\right )} \, dx=2+\frac {1}{-e^x+\log \left (-5+x+x^4 \log (x)\right )} \]

[Out]

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

Rubi [A] (verified)

Time = 0.22 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.019, Rules used = {6820, 6818} \[ \int \frac {-1+e^x (-5+x)-x^3+\left (-4 x^3+e^x x^4\right ) \log (x)}{e^{2 x} (-5+x)+e^{2 x} x^4 \log (x)+\left (e^x (10-2 x)-2 e^x x^4 \log (x)\right ) \log \left (-5+x+x^4 \log (x)\right )+\left (-5+x+x^4 \log (x)\right ) \log ^2\left (-5+x+x^4 \log (x)\right )} \, dx=-\frac {1}{e^x-\log \left (x^4 \log (x)+x-5\right )} \]

[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 6818

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 6820

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

Rubi steps \begin{align*} \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{align*}

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int \frac {-1+e^x (-5+x)-x^3+\left (-4 x^3+e^x x^4\right ) \log (x)}{e^{2 x} (-5+x)+e^{2 x} x^4 \log (x)+\left (e^x (10-2 x)-2 e^x x^4 \log (x)\right ) \log \left (-5+x+x^4 \log (x)\right )+\left (-5+x+x^4 \log (x)\right ) \log ^2\left (-5+x+x^4 \log (x)\right )} \, dx=-\frac {1}{e^x-\log \left (-5+x+x^4 \log (x)\right )} \]

[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)

Maple [A] (verified)

Time = 88.37 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00

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

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.95 \[ \int \frac {-1+e^x (-5+x)-x^3+\left (-4 x^3+e^x x^4\right ) \log (x)}{e^{2 x} (-5+x)+e^{2 x} x^4 \log (x)+\left (e^x (10-2 x)-2 e^x x^4 \log (x)\right ) \log \left (-5+x+x^4 \log (x)\right )+\left (-5+x+x^4 \log (x)\right ) \log ^2\left (-5+x+x^4 \log (x)\right )} \, dx=-\frac {1}{e^{x} - \log \left (x^{4} \log \left (x\right ) + x - 5\right )} \]

[In]

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

[Out]

-1/(e^x - log(x^4*log(x) + x - 5))

Sympy [A] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.85 \[ \int \frac {-1+e^x (-5+x)-x^3+\left (-4 x^3+e^x x^4\right ) \log (x)}{e^{2 x} (-5+x)+e^{2 x} x^4 \log (x)+\left (e^x (10-2 x)-2 e^x x^4 \log (x)\right ) \log \left (-5+x+x^4 \log (x)\right )+\left (-5+x+x^4 \log (x)\right ) \log ^2\left (-5+x+x^4 \log (x)\right )} \, dx=- \frac {1}{e^{x} - \log {\left (x^{4} \log {\left (x \right )} + x - 5 \right )}} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.95 \[ \int \frac {-1+e^x (-5+x)-x^3+\left (-4 x^3+e^x x^4\right ) \log (x)}{e^{2 x} (-5+x)+e^{2 x} x^4 \log (x)+\left (e^x (10-2 x)-2 e^x x^4 \log (x)\right ) \log \left (-5+x+x^4 \log (x)\right )+\left (-5+x+x^4 \log (x)\right ) \log ^2\left (-5+x+x^4 \log (x)\right )} \, dx=-\frac {1}{e^{x} - \log \left (x^{4} \log \left (x\right ) + x - 5\right )} \]

[In]

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

[Out]

-1/(e^x - log(x^4*log(x) + x - 5))

Giac [A] (verification not implemented)

none

Time = 0.37 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.95 \[ \int \frac {-1+e^x (-5+x)-x^3+\left (-4 x^3+e^x x^4\right ) \log (x)}{e^{2 x} (-5+x)+e^{2 x} x^4 \log (x)+\left (e^x (10-2 x)-2 e^x x^4 \log (x)\right ) \log \left (-5+x+x^4 \log (x)\right )+\left (-5+x+x^4 \log (x)\right ) \log ^2\left (-5+x+x^4 \log (x)\right )} \, dx=-\frac {1}{e^{x} - \log \left (x^{4} \log \left (x\right ) + x - 5\right )} \]

[In]

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

[Out]

-1/(e^x - log(x^4*log(x) + x - 5))

Mupad [B] (verification not implemented)

Time = 10.76 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.85 \[ \int \frac {-1+e^x (-5+x)-x^3+\left (-4 x^3+e^x x^4\right ) \log (x)}{e^{2 x} (-5+x)+e^{2 x} x^4 \log (x)+\left (e^x (10-2 x)-2 e^x x^4 \log (x)\right ) \log \left (-5+x+x^4 \log (x)\right )+\left (-5+x+x^4 \log (x)\right ) \log ^2\left (-5+x+x^4 \log (x)\right )} \, dx=\frac {1}{\ln \left (x+x^4\,\ln \left (x\right )-5\right )-{\mathrm {e}}^x} \]

[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))