3.48.14 \(\int \frac {e^{3+\frac {e^{2+e^x}}{5}} (4 e^x+e^x (-4+4 x) \log (x)+\frac {1}{5} e^{2+e^x} (8 e^x x^2-4 e^{2 x} x \log (x)))}{4 x^2-4 e^x x \log (x)+e^{2 x} \log ^2(x)} \, dx\)

Optimal. Leaf size=31 \[ \frac {2 e^{3+\frac {e^{2+e^x}}{5}} x}{x-\frac {1}{2} e^x \log (x)} \]

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Rubi [B]  time = 0.28, antiderivative size = 65, normalized size of antiderivative = 2.10, number of steps used = 1, number of rules used = 1, integrand size = 89, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.011, Rules used = {2288} \begin {gather*} \frac {4 e^{-x+\frac {e^{e^x+2}}{5}+3} \left (2 e^x x^2-e^{2 x} x \log (x)\right )}{4 x^2+e^{2 x} \log ^2(x)-4 e^x x \log (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(3 + E^(2 + E^x)/5)*(4*E^x + E^x*(-4 + 4*x)*Log[x] + (E^(2 + E^x)*(8*E^x*x^2 - 4*E^(2*x)*x*Log[x]))/5))
/(4*x^2 - 4*E^x*x*Log[x] + E^(2*x)*Log[x]^2),x]

[Out]

(4*E^(3 + E^(2 + E^x)/5 - x)*(2*E^x*x^2 - E^(2*x)*x*Log[x]))/(4*x^2 - 4*E^x*x*Log[x] + E^(2*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 {4 e^{3+\frac {e^{2+e^x}}{5}-x} \left (2 e^x x^2-e^{2 x} x \log (x)\right )}{4 x^2-4 e^x x \log (x)+e^{2 x} \log ^2(x)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 1.42, size = 30, normalized size = 0.97 \begin {gather*} -\frac {4 e^{3+\frac {e^{2+e^x}}{5}} x}{-2 x+e^x \log (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(3 + E^(2 + E^x)/5)*(4*E^x + E^x*(-4 + 4*x)*Log[x] + (E^(2 + E^x)*(8*E^x*x^2 - 4*E^(2*x)*x*Log[x]
))/5))/(4*x^2 - 4*E^x*x*Log[x] + E^(2*x)*Log[x]^2),x]

[Out]

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

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fricas [A]  time = 0.59, size = 26, normalized size = 0.84 \begin {gather*} -\frac {4 \, x e^{\left (e^{\left (e^{x} - \log \relax (5) + 2\right )} + 3\right )}}{e^{x} \log \relax (x) - 2 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x*exp(x)^2*log(x)+8*exp(x)*x^2)*exp(exp(x)+2-log(5))+(4*x-4)*exp(x)*log(x)+4*exp(x))*exp(exp(ex
p(x)+2-log(5))+3)/(exp(x)^2*log(x)^2-4*x*exp(x)*log(x)+4*x^2),x, algorithm="fricas")

[Out]

-4*x*e^(e^(e^x - log(5) + 2) + 3)/(e^x*log(x) - 2*x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int -\frac {4 \, {\left ({\left (x - 1\right )} e^{x} \log \relax (x) + {\left (2 \, x^{2} e^{x} - x e^{\left (2 \, x\right )} \log \relax (x)\right )} e^{\left (e^{x} - \log \relax (5) + 2\right )} + e^{x}\right )} e^{\left (e^{\left (e^{x} - \log \relax (5) + 2\right )} + 3\right )}}{4 \, x e^{x} \log \relax (x) - e^{\left (2 \, x\right )} \log \relax (x)^{2} - 4 \, x^{2}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x*exp(x)^2*log(x)+8*exp(x)*x^2)*exp(exp(x)+2-log(5))+(4*x-4)*exp(x)*log(x)+4*exp(x))*exp(exp(ex
p(x)+2-log(5))+3)/(exp(x)^2*log(x)^2-4*x*exp(x)*log(x)+4*x^2),x, algorithm="giac")

[Out]

integrate(-4*((x - 1)*e^x*log(x) + (2*x^2*e^x - x*e^(2*x)*log(x))*e^(e^x - log(5) + 2) + e^x)*e^(e^(e^x - log(
5) + 2) + 3)/(4*x*e^x*log(x) - e^(2*x)*log(x)^2 - 4*x^2), x)

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maple [A]  time = 0.04, size = 26, normalized size = 0.84

method	result	size
risch	\(\frac {4 x \,{\mathrm e}^{\frac {{\mathrm e}^{{\mathrm e}^{x}+2}}{5}+3}}{-{\mathrm e}^{x} \ln \relax (x )+2 x}\)	\(26\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-4*x*exp(x)^2*ln(x)+8*exp(x)*x^2)*exp(exp(x)+2-ln(5))+(4*x-4)*exp(x)*ln(x)+4*exp(x))*exp(exp(exp(x)+2-ln
(5))+3)/(exp(x)^2*ln(x)^2-4*x*exp(x)*ln(x)+4*x^2),x,method=_RETURNVERBOSE)

[Out]

4*x/(-exp(x)*ln(x)+2*x)*exp(1/5*exp(exp(x)+2)+3)

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maxima [A]  time = 0.59, size = 24, normalized size = 0.77 \begin {gather*} -\frac {4 \, x e^{\left (\frac {1}{5} \, e^{\left (e^{x} + 2\right )} + 3\right )}}{e^{x} \log \relax (x) - 2 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x*exp(x)^2*log(x)+8*exp(x)*x^2)*exp(exp(x)+2-log(5))+(4*x-4)*exp(x)*log(x)+4*exp(x))*exp(exp(ex
p(x)+2-log(5))+3)/(exp(x)^2*log(x)^2-4*x*exp(x)*log(x)+4*x^2),x, algorithm="maxima")

[Out]

-4*x*e^(1/5*e^(e^x + 2) + 3)/(e^x*log(x) - 2*x)

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mupad [B]  time = 3.42, size = 23, normalized size = 0.74 \begin {gather*} \frac {2\,x\,{\mathrm {e}}^{\frac {{\mathrm {e}}^{{\mathrm {e}}^x}\,{\mathrm {e}}^2}{5}+3}}{x-\frac {{\mathrm {e}}^x\,\ln \relax (x)}{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(exp(exp(x) - log(5) + 2) + 3)*(4*exp(x) + exp(exp(x) - log(5) + 2)*(8*x^2*exp(x) - 4*x*exp(2*x)*log(x
)) + exp(x)*log(x)*(4*x - 4)))/(4*x^2 + exp(2*x)*log(x)^2 - 4*x*exp(x)*log(x)),x)

[Out]

(2*x*exp((exp(exp(x))*exp(2))/5 + 3))/(x - (exp(x)*log(x))/2)

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sympy [A]  time = 0.53, size = 24, normalized size = 0.77 \begin {gather*} \frac {4 x e^{\frac {e^{e^{x} + 2}}{5} + 3}}{2 x - e^{x} \log {\relax (x )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x*exp(x)**2*ln(x)+8*exp(x)*x**2)*exp(exp(x)+2-ln(5))+(4*x-4)*exp(x)*ln(x)+4*exp(x))*exp(exp(exp
(x)+2-ln(5))+3)/(exp(x)**2*ln(x)**2-4*x*exp(x)*ln(x)+4*x**2),x)

[Out]

4*x*exp(exp(exp(x) + 2)/5 + 3)/(2*x - exp(x)*log(x))

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