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

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

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Rubi [B]  time = 0.74, antiderivative size = 95, normalized size of antiderivative = 3.80, number of steps used = 1, number of rules used = 1, integrand size = 80, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.012, Rules used = {2288} \begin {gather*} -\frac {e^{\frac {1}{5} \left (5 x^4+5 x+5 e^{2 x}+5 e^2+1\right )} \left (4 x^4-\left (4 x^4+2 e^{2 x} x+x\right ) \log (x)+2 e^{2 x} x+x\right )}{\left (4 x^3+2 e^{2 x}+1\right ) \left (x+x \log ^2(x)-2 x \log (x)\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2288

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = (v*y)/(Log[F]*D[u, x])}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

Rubi steps

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

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Mathematica [A]  time = 0.29, size = 25, normalized size = 1.00 \begin {gather*} \frac {e^{\frac {1}{5}+e^2+e^{2 x}+x+x^4}}{-1+\log (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.57, size = 20, normalized size = 0.80 \begin {gather*} \frac {e^{\left (x^{4} + x + e^{2} + e^{\left (2 \, x\right )} + \frac {1}{5}\right )}}{\log \relax (x) - 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

e^(x^4 + x + e^2 + e^(2*x) + 1/5)/(log(x) - 1)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int -\frac {{\left (4 \, x^{4} + 2 \, x e^{\left (2 \, x\right )} - {\left (4 \, x^{4} + 2 \, x e^{\left (2 \, x\right )} + x\right )} \log \relax (x) + x + 1\right )} e^{\left (x^{4} + x + e^{2} + e^{\left (2 \, x\right )} + \frac {1}{5}\right )}}{x \log \relax (x)^{2} - 2 \, x \log \relax (x) + x}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(-(4*x^4 + 2*x*e^(2*x) - (4*x^4 + 2*x*e^(2*x) + x)*log(x) + x + 1)*e^(x^4 + x + e^2 + e^(2*x) + 1/5)/
(x*log(x)^2 - 2*x*log(x) + x), x)

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

method	result	size
risch	\(\frac {{\mathrm e}^{{\mathrm e}^{2 x}+{\mathrm e}^{2}+x^{4}+x +\frac {1}{5}}}{\ln \relax (x )-1}\)	\(21\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.56, size = 20, normalized size = 0.80 \begin {gather*} \frac {e^{\left (x^{4} + x + e^{2} + e^{\left (2 \, x\right )} + \frac {1}{5}\right )}}{\log \relax (x) - 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

e^(x^4 + x + e^2 + e^(2*x) + 1/5)/(log(x) - 1)

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mupad [B]  time = 1.52, size = 23, normalized size = 0.92 \begin {gather*} \frac {{\mathrm {e}}^{x^4}\,{\mathrm {e}}^{1/5}\,{\mathrm {e}}^{{\mathrm {e}}^{2\,x}}\,{\mathrm {e}}^{{\mathrm {e}}^2}\,{\mathrm {e}}^x}{\ln \relax (x)-1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.35, size = 22, normalized size = 0.88 \begin {gather*} \frac {e^{x^{4} + x + e^{2 x} + \frac {1}{5} + e^{2}}}{\log {\relax (x )} - 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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