3.12.10 \(\int \frac {e^x (-1-x)+5 x+10 x \log (x)}{e^{2 x} x^2-10 e^x x^3 \log (x)+25 x^4 \log ^2(x)} \, dx\) [1110]

Optimal. Leaf size=15 \[ \frac {1}{x \left (e^x-5 x \log (x)\right )} \]

[Out]

1/x/(exp(x)-5*x*ln(x))

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Rubi [A]
time = 0.18, antiderivative size = 15, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 50, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.040, Rules used = {6820, 6819} \begin {gather*} \frac {1}{x \left (e^x-5 x \log (x)\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^x*(-1 - x) + 5*x + 10*x*Log[x])/(E^(2*x)*x^2 - 10*E^x*x^3*Log[x] + 25*x^4*Log[x]^2),x]

[Out]

1/(x*(E^x - 5*x*Log[x]))

Rule 6819

Int[(u_)*(y_)^(m_.)*(z_)^(n_.), x_Symbol] :> With[{q = DerivativeDivides[y*z, u*z^(n - m), x]}, Simp[q*y^(m +
1)*(z^(m + 1)/(m + 1)), x] /;  !FalseQ[q]] /; FreeQ[{m, n}, 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 {gather*} \begin {aligned} \text {integral} &=\int \frac {5 x-e^x (1+x)+10 x \log (x)}{x^2 \left (e^x-5 x \log (x)\right )^2} \, dx\\ &=\frac {1}{x \left (e^x-5 x \log (x)\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]
time = 0.08, size = 15, normalized size = 1.00 \begin {gather*} \frac {1}{e^x x-5 x^2 \log (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^x*(-1 - x) + 5*x + 10*x*Log[x])/(E^(2*x)*x^2 - 10*E^x*x^3*Log[x] + 25*x^4*Log[x]^2),x]

[Out]

(E^x*x - 5*x^2*Log[x])^(-1)

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Maple [A]
time = 0.03, size = 18, normalized size = 1.20

method	result	size
risch	\(-\frac {1}{x \left (5 x \ln \left (x \right )-{\mathrm e}^{x}\right )}\)	\(18\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((10*x*ln(x)+(-x-1)*exp(x)+5*x)/(25*x^4*ln(x)^2-10*x^3*exp(x)*ln(x)+exp(x)^2*x^2),x,method=_RETURNVERBOSE)

[Out]

-1/x/(5*x*ln(x)-exp(x))

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Maxima [A]
time = 0.29, size = 17, normalized size = 1.13 \begin {gather*} -\frac {1}{5 \, x^{2} \log \left (x\right ) - x e^{x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((10*x*log(x)+(-1-x)*exp(x)+5*x)/(25*x^4*log(x)^2-10*x^3*exp(x)*log(x)+exp(x)^2*x^2),x, algorithm="ma
xima")

[Out]

-1/(5*x^2*log(x) - x*e^x)

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Fricas [A]
time = 0.34, size = 17, normalized size = 1.13 \begin {gather*} -\frac {1}{5 \, x^{2} \log \left (x\right ) - x e^{x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((10*x*log(x)+(-1-x)*exp(x)+5*x)/(25*x^4*log(x)^2-10*x^3*exp(x)*log(x)+exp(x)^2*x^2),x, algorithm="fr
icas")

[Out]

-1/(5*x^2*log(x) - x*e^x)

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Sympy [A]
time = 0.05, size = 14, normalized size = 0.93 \begin {gather*} \frac {1}{- 5 x^{2} \log {\left (x \right )} + x e^{x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((10*x*ln(x)+(-1-x)*exp(x)+5*x)/(25*x**4*ln(x)**2-10*x**3*exp(x)*ln(x)+exp(x)**2*x**2),x)

[Out]

1/(-5*x**2*log(x) + x*exp(x))

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Giac [A]
time = 0.41, size = 17, normalized size = 1.13 \begin {gather*} -\frac {1}{5 \, x^{2} \log \left (x\right ) - x e^{x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((10*x*log(x)+(-1-x)*exp(x)+5*x)/(25*x^4*log(x)^2-10*x^3*exp(x)*log(x)+exp(x)^2*x^2),x, algorithm="gi
ac")

[Out]

-1/(5*x^2*log(x) - x*e^x)

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Mupad [B]
time = 1.07, size = 17, normalized size = 1.13 \begin {gather*} -\frac {1}{5\,x^2\,\ln \left (x\right )-x\,{\mathrm {e}}^x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((5*x - exp(x)*(x + 1) + 10*x*log(x))/(x^2*exp(2*x) + 25*x^4*log(x)^2 - 10*x^3*exp(x)*log(x)),x)

[Out]

-1/(5*x^2*log(x) - x*exp(x))

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