3.92.90 \(\int \frac {e^{33-3 e^5+3 e^x+96 x^2} (1+(3 e^x x+192 x^2) \log (x))}{x} \, dx\)

Optimal. Leaf size=22 \[ e^{3 \left (11-e^5+e^x+32 x^2\right )} \log (x) \]

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Rubi [B]  time = 0.12, antiderivative size = 48, normalized size of antiderivative = 2.18, number of steps used = 1, number of rules used = 1, integrand size = 40, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.025, Rules used = {2288} \begin {gather*} \frac {e^{96 x^2+3 e^x+3 \left (11-e^5\right )} \left (64 x^2+e^x x\right ) \log (x)}{x \left (64 x+e^x\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(33 - 3*E^5 + 3*E^x + 96*x^2)*(1 + (3*E^x*x + 192*x^2)*Log[x]))/x,x]

[Out]

(E^(3*E^x + 3*(11 - E^5) + 96*x^2)*(E^x*x + 64*x^2)*Log[x])/(x*(E^x + 64*x))

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 {e^{3 e^x+3 \left (11-e^5\right )+96 x^2} \left (e^x x+64 x^2\right ) \log (x)}{x \left (e^x+64 x\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.04, size = 22, normalized size = 1.00 \begin {gather*} e^{33-3 e^5+3 e^x+96 x^2} \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(33 - 3*E^5 + 3*E^x + 96*x^2)*(1 + (3*E^x*x + 192*x^2)*Log[x]))/x,x]

[Out]

E^(33 - 3*E^5 + 3*E^x + 96*x^2)*Log[x]

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fricas [A]  time = 0.77, size = 19, normalized size = 0.86 \begin {gather*} e^{\left (96 \, x^{2} - 3 \, e^{5} + 3 \, e^{x} + 33\right )} \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((3*exp(x)*x+192*x^2)*log(x)+1)*exp(3*exp(x)-3*exp(5)+96*x^2+33)/x,x, algorithm="fricas")

[Out]

e^(96*x^2 - 3*e^5 + 3*e^x + 33)*log(x)

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giac [A]  time = 0.21, size = 19, normalized size = 0.86 \begin {gather*} e^{\left (96 \, x^{2} - 3 \, e^{5} + 3 \, e^{x} + 33\right )} \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((3*exp(x)*x+192*x^2)*log(x)+1)*exp(3*exp(x)-3*exp(5)+96*x^2+33)/x,x, algorithm="giac")

[Out]

e^(96*x^2 - 3*e^5 + 3*e^x + 33)*log(x)

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maple [A]  time = 0.05, size = 20, normalized size = 0.91

method	result	size
risch	\(\ln \relax (x ) {\mathrm e}^{3 \,{\mathrm e}^{x}-3 \,{\mathrm e}^{5}+96 x^{2}+33}\)	\(20\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((3*exp(x)*x+192*x^2)*ln(x)+1)*exp(3*exp(x)-3*exp(5)+96*x^2+33)/x,x,method=_RETURNVERBOSE)

[Out]

ln(x)*exp(3*exp(x)-3*exp(5)+96*x^2+33)

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maxima [A]  time = 0.53, size = 19, normalized size = 0.86 \begin {gather*} e^{\left (96 \, x^{2} - 3 \, e^{5} + 3 \, e^{x} + 33\right )} \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((3*exp(x)*x+192*x^2)*log(x)+1)*exp(3*exp(x)-3*exp(5)+96*x^2+33)/x,x, algorithm="maxima")

[Out]

e^(96*x^2 - 3*e^5 + 3*e^x + 33)*log(x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.05 \begin {gather*} \int \frac {{\mathrm {e}}^{3\,{\mathrm {e}}^x-3\,{\mathrm {e}}^5+96\,x^2+33}\,\left (\ln \relax (x)\,\left (3\,x\,{\mathrm {e}}^x+192\,x^2\right )+1\right )}{x} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(3*exp(x) - 3*exp(5) + 96*x^2 + 33)*(log(x)*(3*x*exp(x) + 192*x^2) + 1))/x,x)

[Out]

int((exp(3*exp(x) - 3*exp(5) + 96*x^2 + 33)*(log(x)*(3*x*exp(x) + 192*x^2) + 1))/x, x)

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sympy [A]  time = 5.88, size = 20, normalized size = 0.91 \begin {gather*} e^{96 x^{2} + 3 e^{x} - 3 e^{5} + 33} \log {\relax (x )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((3*exp(x)*x+192*x**2)*ln(x)+1)*exp(3*exp(x)-3*exp(5)+96*x**2+33)/x,x)

[Out]

exp(96*x**2 + 3*exp(x) - 3*exp(5) + 33)*log(x)

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