3.58.9 \(\int \frac {(3+12 x^2) \log ^2(x)+e^{\frac {4+4 x \log (x)}{x \log (x)}} (-80 x-80 x \log (x)+60 x^2 \log ^2(x))+e^{\frac {2 (4+4 x \log (x))}{x \log (x)}} (-200 x-200 x \log (x)+75 x^2 \log ^2(x))}{\log ^2(x)} \, dx\)

Optimal. Leaf size=31 \[ x \left (3+\left (2+5 e^{\frac {4 \left (x^2+\frac {x}{\log (x)}\right )}{x^2}}\right )^2 x^2\right ) \]

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Rubi [B]  time = 0.44, antiderivative size = 95, normalized size of antiderivative = 3.06, number of steps used = 5, number of rules used = 2, integrand size = 91, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.022, Rules used = {6742, 2288} \begin {gather*} 4 x^3+\frac {20 x e^{\frac {4}{x \log (x)}+4} (\log (x)+1)}{\left (\frac {1}{x^2 \log ^2(x)}+\frac {1}{x^2 \log (x)}\right ) \log ^2(x)}+\frac {25 x e^{\frac {8}{x \log (x)}+8} (\log (x)+1)}{\left (\frac {1}{x^2 \log ^2(x)}+\frac {1}{x^2 \log (x)}\right ) \log ^2(x)}+3 x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((3 + 12*x^2)*Log[x]^2 + E^((4 + 4*x*Log[x])/(x*Log[x]))*(-80*x - 80*x*Log[x] + 60*x^2*Log[x]^2) + E^((2*(
4 + 4*x*Log[x]))/(x*Log[x]))*(-200*x - 200*x*Log[x] + 75*x^2*Log[x]^2))/Log[x]^2,x]

[Out]

3*x + 4*x^3 + (20*E^(4 + 4/(x*Log[x]))*x*(1 + Log[x]))/((1/(x^2*Log[x]^2) + 1/(x^2*Log[x]))*Log[x]^2) + (25*E^
(8 + 8/(x*Log[x]))*x*(1 + Log[x]))/((1/(x^2*Log[x]^2) + 1/(x^2*Log[x]))*Log[x]^2)

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} &=\int \left (3 \left (1+4 x^2\right )+\frac {25 e^{8+\frac {8}{x \log (x)}} x \left (-8-8 \log (x)+3 x \log ^2(x)\right )}{\log ^2(x)}+\frac {20 e^{4+\frac {4}{x \log (x)}} x \left (-4-4 \log (x)+3 x \log ^2(x)\right )}{\log ^2(x)}\right ) \, dx\\ &=3 \int \left (1+4 x^2\right ) \, dx+20 \int \frac {e^{4+\frac {4}{x \log (x)}} x \left (-4-4 \log (x)+3 x \log ^2(x)\right )}{\log ^2(x)} \, dx+25 \int \frac {e^{8+\frac {8}{x \log (x)}} x \left (-8-8 \log (x)+3 x \log ^2(x)\right )}{\log ^2(x)} \, dx\\ &=3 x+4 x^3+\frac {20 e^{4+\frac {4}{x \log (x)}} x (1+\log (x))}{\left (\frac {1}{x^2 \log ^2(x)}+\frac {1}{x^2 \log (x)}\right ) \log ^2(x)}+\frac {25 e^{8+\frac {8}{x \log (x)}} x (1+\log (x))}{\left (\frac {1}{x^2 \log ^2(x)}+\frac {1}{x^2 \log (x)}\right ) \log ^2(x)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.13, size = 27, normalized size = 0.87 \begin {gather*} x \left (3+\left (2+5 e^{4+\frac {4}{x \log (x)}}\right )^2 x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((3 + 12*x^2)*Log[x]^2 + E^((4 + 4*x*Log[x])/(x*Log[x]))*(-80*x - 80*x*Log[x] + 60*x^2*Log[x]^2) + E
^((2*(4 + 4*x*Log[x]))/(x*Log[x]))*(-200*x - 200*x*Log[x] + 75*x^2*Log[x]^2))/Log[x]^2,x]

[Out]

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

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fricas [A]  time = 0.57, size = 51, normalized size = 1.65 \begin {gather*} 25 \, x^{3} e^{\left (\frac {8 \, {\left (x \log \relax (x) + 1\right )}}{x \log \relax (x)}\right )} + 20 \, x^{3} e^{\left (\frac {4 \, {\left (x \log \relax (x) + 1\right )}}{x \log \relax (x)}\right )} + 4 \, x^{3} + 3 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((75*x^2*log(x)^2-200*x*log(x)-200*x)*exp((4*x*log(x)+4)/x/log(x))^2+(60*x^2*log(x)^2-80*x*log(x)-80
*x)*exp((4*x*log(x)+4)/x/log(x))+(12*x^2+3)*log(x)^2)/log(x)^2,x, algorithm="fricas")

[Out]

25*x^3*e^(8*(x*log(x) + 1)/(x*log(x))) + 20*x^3*e^(4*(x*log(x) + 1)/(x*log(x))) + 4*x^3 + 3*x

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giac [A]  time = 0.24, size = 51, normalized size = 1.65 \begin {gather*} 25 \, x^{3} e^{\left (\frac {8 \, {\left (x \log \relax (x) + 1\right )}}{x \log \relax (x)}\right )} + 20 \, x^{3} e^{\left (\frac {4 \, {\left (x \log \relax (x) + 1\right )}}{x \log \relax (x)}\right )} + 4 \, x^{3} + 3 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((75*x^2*log(x)^2-200*x*log(x)-200*x)*exp((4*x*log(x)+4)/x/log(x))^2+(60*x^2*log(x)^2-80*x*log(x)-80
*x)*exp((4*x*log(x)+4)/x/log(x))+(12*x^2+3)*log(x)^2)/log(x)^2,x, algorithm="giac")

[Out]

25*x^3*e^(8*(x*log(x) + 1)/(x*log(x))) + 20*x^3*e^(4*(x*log(x) + 1)/(x*log(x))) + 4*x^3 + 3*x

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maple [A]  time = 0.10, size = 52, normalized size = 1.68

method	result	size
default	\(3 x +20 x^{3} {\mathrm e}^{\frac {4 x \ln \relax (x )+4}{x \ln \relax (x )}}+25 x^{3} {\mathrm e}^{\frac {8 x \ln \relax (x )+8}{x \ln \relax (x )}}+4 x^{3}\)	\(52\)
risch	\(3 x +20 x^{3} {\mathrm e}^{\frac {4 x \ln \relax (x )+4}{x \ln \relax (x )}}+25 x^{3} {\mathrm e}^{\frac {8 x \ln \relax (x )+8}{x \ln \relax (x )}}+4 x^{3}\)	\(52\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((75*x^2*ln(x)^2-200*x*ln(x)-200*x)*exp((4*x*ln(x)+4)/x/ln(x))^2+(60*x^2*ln(x)^2-80*x*ln(x)-80*x)*exp((4*x
*ln(x)+4)/x/ln(x))+(12*x^2+3)*ln(x)^2)/ln(x)^2,x,method=_RETURNVERBOSE)

[Out]

3*x+20*x^3*exp(4*(x*ln(x)+1)/ln(x)/x)+25*x^3*exp(8*(x*ln(x)+1)/ln(x)/x)+4*x^3

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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(((75*x^2*log(x)^2-200*x*log(x)-200*x)*exp((4*x*log(x)+4)/x/log(x))^2+(60*x^2*log(x)^2-80*x*log(x)-80
*x)*exp((4*x*log(x)+4)/x/log(x))+(12*x^2+3)*log(x)^2)/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 = 3.73, size = 43, normalized size = 1.39 \begin {gather*} 3\,x+20\,x^3\,{\mathrm {e}}^{\frac {4}{x\,\ln \relax (x)}+4}+25\,x^3\,{\mathrm {e}}^{\frac {8}{x\,\ln \relax (x)}+8}+4\,x^3 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp((4*x*log(x) + 4)/(x*log(x)))*(80*x - 60*x^2*log(x)^2 + 80*x*log(x)) - log(x)^2*(12*x^2 + 3) + exp((2
*(4*x*log(x) + 4))/(x*log(x)))*(200*x - 75*x^2*log(x)^2 + 200*x*log(x)))/log(x)^2,x)

[Out]

3*x + 20*x^3*exp(4/(x*log(x)) + 4) + 25*x^3*exp(8/(x*log(x)) + 8) + 4*x^3

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sympy [A]  time = 2.10, size = 48, normalized size = 1.55 \begin {gather*} 25 x^{3} e^{\frac {2 \left (4 x \log {\relax (x )} + 4\right )}{x \log {\relax (x )}}} + 20 x^{3} e^{\frac {4 x \log {\relax (x )} + 4}{x \log {\relax (x )}}} + 4 x^{3} + 3 x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((75*x**2*ln(x)**2-200*x*ln(x)-200*x)*exp((4*x*ln(x)+4)/x/ln(x))**2+(60*x**2*ln(x)**2-80*x*ln(x)-80*
x)*exp((4*x*ln(x)+4)/x/ln(x))+(12*x**2+3)*ln(x)**2)/ln(x)**2,x)

[Out]

25*x**3*exp(2*(4*x*log(x) + 4)/(x*log(x))) + 20*x**3*exp((4*x*log(x) + 4)/(x*log(x))) + 4*x**3 + 3*x

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