3.20.13 \(\int \frac {e^x (-60 x^2+24 x^3) \log ^2(x)+e^x (24 x^2-12 x^3) \log ^3(x)+e^{\frac {5-3 x^3 \log (x)}{3 x^2 \log (x)}} (20 e^x+40 e^x \log (x)+24 e^x x^2 \log ^2(x))}{3 x^5 \log ^2(x)} \, dx\)

Optimal. Leaf size=33 \[ \frac {4 e^x \left (2-e^{-x+\frac {5}{3 x^2 \log (x)}}-\log (x)\right )}{x^2} \]

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Rubi [A]  time = 0.92, antiderivative size = 66, normalized size of antiderivative = 2.00, number of steps used = 5, number of rules used = 3, integrand size = 98, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.031, Rules used = {12, 6742, 2288} \begin {gather*} \frac {4 e^x (2 x-x \log (x))}{x^3}-\frac {4 e^{\frac {5}{3 x^2 \log (x)}} (2 \log (x)+1)}{x^5 \left (\frac {1}{x^3 \log ^2(x)}+\frac {2}{x^3 \log (x)}\right ) \log ^2(x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^x*(-60*x^2 + 24*x^3)*Log[x]^2 + E^x*(24*x^2 - 12*x^3)*Log[x]^3 + E^((5 - 3*x^3*Log[x])/(3*x^2*Log[x]))*
(20*E^x + 40*E^x*Log[x] + 24*E^x*x^2*Log[x]^2))/(3*x^5*Log[x]^2),x]

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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]

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{3} \int \frac {e^x \left (-60 x^2+24 x^3\right ) \log ^2(x)+e^x \left (24 x^2-12 x^3\right ) \log ^3(x)+e^{\frac {5-3 x^3 \log (x)}{3 x^2 \log (x)}} \left (20 e^x+40 e^x \log (x)+24 e^x x^2 \log ^2(x)\right )}{x^5 \log ^2(x)} \, dx\\ &=\frac {1}{3} \int \left (-\frac {12 e^x (5-2 x-2 \log (x)+x \log (x))}{x^3}+\frac {4 e^{\frac {5}{3 x^2 \log (x)}} \left (5+10 \log (x)+6 x^2 \log ^2(x)\right )}{x^5 \log ^2(x)}\right ) \, dx\\ &=\frac {4}{3} \int \frac {e^{\frac {5}{3 x^2 \log (x)}} \left (5+10 \log (x)+6 x^2 \log ^2(x)\right )}{x^5 \log ^2(x)} \, dx-4 \int \frac {e^x (5-2 x-2 \log (x)+x \log (x))}{x^3} \, dx\\ &=-\frac {4 e^{\frac {5}{3 x^2 \log (x)}} (1+2 \log (x))}{x^5 \left (\frac {1}{x^3 \log ^2(x)}+\frac {2}{x^3 \log (x)}\right ) \log ^2(x)}+\frac {4 e^x (2 x-x \log (x))}{x^3}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.08, size = 30, normalized size = 0.91 \begin {gather*} -\frac {4 \left (-2 e^x+e^{\frac {5}{3 x^2 \log (x)}}+e^x \log (x)\right )}{x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^x*(-60*x^2 + 24*x^3)*Log[x]^2 + E^x*(24*x^2 - 12*x^3)*Log[x]^3 + E^((5 - 3*x^3*Log[x])/(3*x^2*Log
[x]))*(20*E^x + 40*E^x*Log[x] + 24*E^x*x^2*Log[x]^2))/(3*x^5*Log[x]^2),x]

[Out]

(-4*(-2*E^x + E^(5/(3*x^2*Log[x])) + E^x*Log[x]))/x^2

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fricas [A]  time = 0.85, size = 36, normalized size = 1.09 \begin {gather*} -\frac {4 \, {\left (e^{x} \log \relax (x) + e^{\left (x - \frac {3 \, x^{3} \log \relax (x) - 5}{3 \, x^{2} \log \relax (x)}\right )} - 2 \, e^{x}\right )}}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*((24*x^2*exp(x)*log(x)^2+40*exp(x)*log(x)+20*exp(x))*exp(1/3*(-3*x^3*log(x)+5)/x^2/log(x))+(-12*
x^3+24*x^2)*exp(x)*log(x)^3+(24*x^3-60*x^2)*exp(x)*log(x)^2)/x^5/log(x)^2,x, algorithm="fricas")

[Out]

-4*(e^x*log(x) + e^(x - 1/3*(3*x^3*log(x) - 5)/(x^2*log(x))) - 2*e^x)/x^2

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giac [A]  time = 3.93, size = 25, normalized size = 0.76 \begin {gather*} -\frac {4 \, {\left (e^{x} \log \relax (x) - 2 \, e^{x} + e^{\left (\frac {5}{3 \, x^{2} \log \relax (x)}\right )}\right )}}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*((24*x^2*exp(x)*log(x)^2+40*exp(x)*log(x)+20*exp(x))*exp(1/3*(-3*x^3*log(x)+5)/x^2/log(x))+(-12*
x^3+24*x^2)*exp(x)*log(x)^3+(24*x^3-60*x^2)*exp(x)*log(x)^2)/x^5/log(x)^2,x, algorithm="giac")

[Out]

-4*(e^x*log(x) - 2*e^x + e^(5/3/(x^2*log(x))))/x^2

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maple [A]  time = 0.04, size = 33, normalized size = 1.00

method	result	size
risch	\(-\frac {4 \,{\mathrm e}^{x} \ln \relax (x )}{x^{2}}+\frac {8 \,{\mathrm e}^{x}}{x^{2}}-\frac {4 \,{\mathrm e}^{\frac {5}{3 x^{2} \ln \relax (x )}}}{x^{2}}\)	\(33\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/3*((24*x^2*exp(x)*ln(x)^2+40*exp(x)*ln(x)+20*exp(x))*exp(1/3*(-3*x^3*ln(x)+5)/x^2/ln(x))+(-12*x^3+24*x^2
)*exp(x)*ln(x)^3+(24*x^3-60*x^2)*exp(x)*ln(x)^2)/x^5/ln(x)^2,x,method=_RETURNVERBOSE)

[Out]

-4/x^2*exp(x)*ln(x)+8*exp(x)/x^2-4/x^2*exp(5/3/x^2/ln(x))

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: RuntimeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*((24*x^2*exp(x)*log(x)^2+40*exp(x)*log(x)+20*exp(x))*exp(1/3*(-3*x^3*log(x)+5)/x^2/log(x))+(-12*
x^3+24*x^2)*exp(x)*log(x)^3+(24*x^3-60*x^2)*exp(x)*log(x)^2)/x^5/log(x)^2,x, algorithm="maxima")

[Out]

Exception raised: RuntimeError >> ECL says: In function CAR, the value of the first argument is  0which is not
of the expected type LIST

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mupad [B]  time = 1.63, size = 32, normalized size = 0.97 \begin {gather*} \frac {8\,{\mathrm {e}}^x}{x^2}-\frac {4\,{\mathrm {e}}^{\frac {5}{3\,x^2\,\ln \relax (x)}}}{x^2}-\frac {4\,{\mathrm {e}}^x\,\ln \relax (x)}{x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((exp(-(x^3*log(x) - 5/3)/(x^2*log(x)))*(20*exp(x) + 40*exp(x)*log(x) + 24*x^2*exp(x)*log(x)^2))/3 + (exp(
x)*log(x)^3*(24*x^2 - 12*x^3))/3 - (exp(x)*log(x)^2*(60*x^2 - 24*x^3))/3)/(x^5*log(x)^2),x)

[Out]

(8*exp(x))/x^2 - (4*exp(5/(3*x^2*log(x))))/x^2 - (4*exp(x)*log(x))/x^2

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sympy [A]  time = 0.54, size = 39, normalized size = 1.18 \begin {gather*} \frac {\left (8 - 4 \log {\relax (x )}\right ) e^{x}}{x^{2}} - \frac {4 e^{x} e^{\frac {- x^{3} \log {\relax (x )} + \frac {5}{3}}{x^{2} \log {\relax (x )}}}}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*((24*x**2*exp(x)*ln(x)**2+40*exp(x)*ln(x)+20*exp(x))*exp(1/3*(-3*x**3*ln(x)+5)/x**2/ln(x))+(-12*
x**3+24*x**2)*exp(x)*ln(x)**3+(24*x**3-60*x**2)*exp(x)*ln(x)**2)/x**5/ln(x)**2,x)

[Out]

(8 - 4*log(x))*exp(x)/x**2 - 4*exp(x)*exp((-x**3*log(x) + 5/3)/(x**2*log(x)))/x**2

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