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

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

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Rubi [A]  time = 0.05, antiderivative size = 21, normalized size of antiderivative = 0.95, number of steps used = 6, number of rules used = 3, integrand size = 41, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.073, Rules used = {12, 14, 2288} \begin {gather*} x^2+\frac {5}{2} e^{x-e^5} \log (x)+\log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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} &=\frac {1}{2} e^{-e^5} \int \frac {5 e^x+e^{e^5} \left (2+4 x^2\right )+5 e^x x \log (x)}{x} \, dx\\ &=\frac {1}{2} e^{-e^5} \int \left (\frac {2 e^{e^5} \left (1+2 x^2\right )}{x}+\frac {5 e^x (1+x \log (x))}{x}\right ) \, dx\\ &=\frac {1}{2} \left (5 e^{-e^5}\right ) \int \frac {e^x (1+x \log (x))}{x} \, dx+\int \frac {1+2 x^2}{x} \, dx\\ &=\frac {5}{2} e^{-e^5+x} \log (x)+\int \left (\frac {1}{x}+2 x\right ) \, dx\\ &=x^2+\log (x)+\frac {5}{2} e^{-e^5+x} \log (x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.05, size = 27, normalized size = 1.23 \begin {gather*} \frac {1}{2} \left (2 x^2+2 \log (x)+5 e^{-e^5+x} \log (x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.54, size = 29, normalized size = 1.32 \begin {gather*} \frac {1}{2} \, {\left (2 \, x^{2} e^{\left (e^{5}\right )} + {\left (5 \, e^{x} + 2 \, e^{\left (e^{5}\right )}\right )} \log \relax (x)\right )} e^{\left (-e^{5}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.26, size = 29, normalized size = 1.32 \begin {gather*} \frac {1}{2} \, {\left (2 \, x^{2} e^{\left (e^{5}\right )} + 5 \, e^{x} \log \relax (x) + 2 \, e^{\left (e^{5}\right )} \log \relax (x)\right )} e^{\left (-e^{5}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.07, size = 18, normalized size = 0.82

method	result	size
norman	\(x^{2}+\ln \relax (x )+\frac {5 \ln \relax (x ) {\mathrm e}^{-{\mathrm e}^{5}} {\mathrm e}^{x}}{2}\)	\(18\)
risch	\(x^{2}+\ln \relax (x )+\frac {5 \ln \relax (x ) {\mathrm e}^{-{\mathrm e}^{5}+x}}{2}\)	\(18\)
default	\(\frac {{\mathrm e}^{-{\mathrm e}^{5}} \left (5 \,{\mathrm e}^{x} \ln \relax (x )+2 x^{2} {\mathrm e}^{{\mathrm e}^{5}}+2 \,{\mathrm e}^{{\mathrm e}^{5}} \ln \relax (x )\right )}{2}\)	\(30\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x^2+ln(x)+5/2*ln(x)/exp(exp(5))*exp(x)

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maxima [A]  time = 0.51, size = 29, normalized size = 1.32 \begin {gather*} \frac {1}{2} \, {\left (2 \, x^{2} e^{\left (e^{5}\right )} + 5 \, e^{x} \log \relax (x) + 2 \, e^{\left (e^{5}\right )} \log \relax (x)\right )} e^{\left (-e^{5}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 2.25, size = 17, normalized size = 0.77 \begin {gather*} \ln \relax (x)+x^2+\frac {5\,{\mathrm {e}}^{-{\mathrm {e}}^5}\,{\mathrm {e}}^x\,\ln \relax (x)}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.34, size = 20, normalized size = 0.91 \begin {gather*} x^{2} + \frac {5 e^{x} \log {\relax (x )}}{2 e^{e^{5}}} + \log {\relax (x )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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