3.45.51 \(\int \frac {e^x (-10+5 e^{-x^2})+5 x+e^x (-10 x+e^{-x^2} (5 x-10 x^2)) \log (x)}{x} \, dx\)

Optimal. Leaf size=22 \[ 5 \left (x-e^x \left (2-e^{-x^2}\right ) \log (x)\right ) \]

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Rubi [A]  time = 0.14, antiderivative size = 44, normalized size of antiderivative = 2.00, number of steps used = 6, number of rules used = 2, integrand size = 50, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.040, Rules used = {14, 2288} \begin {gather*} \frac {5 e^{x-x^2} \left (x \log (x)-2 x^2 \log (x)\right )}{(1-2 x) x}+5 x-10 e^x \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

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} &=\int \left (-\frac {5 \left (2 e^x-x+2 e^x x \log (x)\right )}{x}-\frac {5 e^{x-x^2} \left (-1-x \log (x)+2 x^2 \log (x)\right )}{x}\right ) \, dx\\ &=-\left (5 \int \frac {2 e^x-x+2 e^x x \log (x)}{x} \, dx\right )-5 \int \frac {e^{x-x^2} \left (-1-x \log (x)+2 x^2 \log (x)\right )}{x} \, dx\\ &=\frac {5 e^{x-x^2} \left (x \log (x)-2 x^2 \log (x)\right )}{(1-2 x) x}-5 \int \left (-1+\frac {2 e^x (1+x \log (x))}{x}\right ) \, dx\\ &=5 x+\frac {5 e^{x-x^2} \left (x \log (x)-2 x^2 \log (x)\right )}{(1-2 x) x}-10 \int \frac {e^x (1+x \log (x))}{x} \, dx\\ &=5 x-10 e^x \log (x)+\frac {5 e^{x-x^2} \left (x \log (x)-2 x^2 \log (x)\right )}{(1-2 x) x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.33, size = 19, normalized size = 0.86 \begin {gather*} 5 \left (x+e^x \left (-2+e^{-x^2}\right ) \log (x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

5*(x + E^x*(-2 + E^(-x^2))*Log[x])

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fricas [A]  time = 0.49, size = 18, normalized size = 0.82 \begin {gather*} 5 \, {\left (e^{\left (-x^{2}\right )} - 2\right )} e^{x} \log \relax (x) + 5 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5*(e^(-x^2) - 2)*e^x*log(x) + 5*x

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giac [A]  time = 0.14, size = 22, normalized size = 1.00 \begin {gather*} 5 \, e^{\left (-x^{2} + x\right )} \log \relax (x) - 10 \, e^{x} \log \relax (x) + 5 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5*e^(-x^2 + x)*log(x) - 10*e^x*log(x) + 5*x

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maple [A]  time = 0.04, size = 22, normalized size = 1.00

method	result	size
risch	\(\left (5 \,{\mathrm e}^{-x \left (x -1\right )}-10 \,{\mathrm e}^{x}\right ) \ln \relax (x )+5 x\)	\(22\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(5*exp(-x*(x-1))-10*exp(x))*ln(x)+5*x

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} 5 \, e^{\left (-x^{2} + x\right )} \log \relax (x) - 10 \, e^{x} \log \relax (x) + 5 \, x - 10 \, {\rm Ei}\relax (x) + 10 \, \int \frac {e^{x}}{x}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5*e^(-x^2 + x)*log(x) - 10*e^x*log(x) + 5*x - 10*Ei(x) + 10*integrate(e^x/x, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.05 \begin {gather*} \int \frac {5\,x+{\mathrm {e}}^x\,\left (5\,{\mathrm {e}}^{-x^2}-10\right )-{\mathrm {e}}^x\,\ln \relax (x)\,\left (10\,x-{\mathrm {e}}^{-x^2}\,\left (5\,x-10\,x^2\right )\right )}{x} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 12.58, size = 24, normalized size = 1.09 \begin {gather*} 5 x - 10 e^{x} \log {\relax (x )} + 5 e^{x} e^{- x^{2}} \log {\relax (x )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-10*x**2+5*x)*exp(-x**2)-10*x)*exp(x)*ln(x)+(5*exp(-x**2)-10)*exp(x)+5*x)/x,x)

[Out]

5*x - 10*exp(x)*log(x) + 5*exp(x)*exp(-x**2)*log(x)

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