3.83.78 \(\int \frac {-3 e^2-5 e^{2+x} x+e^{2+x} (-5 x-5 x^2) \log (x)}{x} \, dx\)

Optimal. Leaf size=14 \[ e^2 \left (-3-5 e^x x\right ) \log (x) \]

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Rubi [A]  time = 0.02, antiderivative size = 18, normalized size of antiderivative = 1.29, number of steps used = 3, number of rules used = 2, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.057, Rules used = {14, 2288} \begin {gather*} -5 e^{x+2} x \log (x)-3 e^2 \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-3*E^2*Log[x] - 5*E^(2 + x)*x*Log[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 {3 e^2}{x}-5 e^{2+x} (1+\log (x)+x \log (x))\right ) \, dx\\ &=-3 e^2 \log (x)-5 \int e^{2+x} (1+\log (x)+x \log (x)) \, dx\\ &=-3 e^2 \log (x)-5 e^{2+x} x \log (x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 18, normalized size = 1.29 \begin {gather*} -e^2 \left (3 \log (x)+5 e^x x \log (x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-(E^2*(3*Log[x] + 5*E^x*x*Log[x]))

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fricas [A]  time = 0.68, size = 16, normalized size = 1.14 \begin {gather*} -{\left (5 \, x e^{\left (x + 2\right )} + 3 \, e^{2}\right )} \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(5*x*e^(x + 2) + 3*e^2)*log(x)

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giac [A]  time = 0.15, size = 16, normalized size = 1.14 \begin {gather*} -5 \, x e^{\left (x + 2\right )} \log \relax (x) - 3 \, e^{2} \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-5*x*e^(x + 2)*log(x) - 3*e^2*log(x)

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maple [A]  time = 0.06, size = 17, normalized size = 1.21

method	result	size
default	\(-5 x \,{\mathrm e}^{2} {\mathrm e}^{x} \ln \relax (x )-3 \,{\mathrm e}^{2} \ln \relax (x )\)	\(17\)
norman	\(-5 x \,{\mathrm e}^{2} {\mathrm e}^{x} \ln \relax (x )-3 \,{\mathrm e}^{2} \ln \relax (x )\)	\(17\)
risch	\(-5 x \ln \relax (x ) {\mathrm e}^{2+x}-3 \,{\mathrm e}^{2} \ln \relax (x )\)	\(17\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-5*x*exp(2)*exp(x)*ln(x)-3*exp(2)*ln(x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 5.29, size = 13, normalized size = 0.93 \begin {gather*} -{\mathrm {e}}^2\,\ln \relax (x)\,\left (5\,x\,{\mathrm {e}}^x+3\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-exp(2)*log(x)*(5*x*exp(x) + 3)

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sympy [A]  time = 0.31, size = 22, normalized size = 1.57 \begin {gather*} - 5 x e^{2} e^{x} \log {\relax (x )} - 3 e^{2} \log {\relax (x )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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