3.59.58 \(\int \frac {(18 e^{\frac {1}{x}+2 x} x^2+180 e^{\frac {1}{x}} x^3+e^{\frac {1}{x}+x} (90 x^2+36 x^3)) \log (\frac {1}{3} (e^x+x^2))+(-45 e^{\frac {1}{x}} x^2+e^{\frac {1}{x}+2 x} (-9+9 x^2)+e^{\frac {1}{x}+x} (-45-9 x^2+9 x^4)) \log ^2(\frac {1}{3} (e^x+x^2))}{e^x x^2+x^4} \, dx\)

Optimal. Leaf size=26 \[ 9 e^{\frac {1}{x}} \left (5+e^x\right ) \log ^2\left (\frac {1}{3} \left (e^x+x^2\right )\right ) \]

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Rubi [F]  time = 26.02, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \int \frac {\left (18 e^{\frac {1}{x}+2 x} x^2+180 e^{\frac {1}{x}} x^3+e^{\frac {1}{x}+x} \left (90 x^2+36 x^3\right )\right ) \log \left (\frac {1}{3} \left (e^x+x^2\right )\right )+\left (-45 e^{\frac {1}{x}} x^2+e^{\frac {1}{x}+2 x} \left (-9+9 x^2\right )+e^{\frac {1}{x}+x} \left (-45-9 x^2+9 x^4\right )\right ) \log ^2\left (\frac {1}{3} \left (e^x+x^2\right )\right )}{e^x x^2+x^4} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[((18*E^(x^(-1) + 2*x)*x^2 + 180*E^x^(-1)*x^3 + E^(x^(-1) + x)*(90*x^2 + 36*x^3))*Log[(E^x + x^2)/3] + (-45
*E^x^(-1)*x^2 + E^(x^(-1) + 2*x)*(-9 + 9*x^2) + E^(x^(-1) + x)*(-45 - 9*x^2 + 9*x^4))*Log[(E^x + x^2)/3]^2)/(E
^x*x^2 + x^4),x]

[Out]

-55*E^x^(-1)*x - 55*E^x^(-1)*x^2 - 5*E^x^(-1)*x^3 + 55*ExpIntegralEi[x^(-1)] + 6*Gamma[-4, -x^(-1)] + 105*E^x^
(-1)*x*Log[(E^x + x^2)/3] + 15*E^x^(-1)*x^2*Log[(E^x + x^2)/3] - 6*E^x^(-1)*x^3*Log[(E^x + x^2)/3] - 105*ExpIn
tegralEi[x^(-1)]*Log[(E^x + x^2)/3] + 18*Log[(E^x + x^2)/3]*Defer[Int][E^(x^(-1) + x), x] + 9*Log[3]*Log[(E^x
+ x^2)/3]*Defer[Int][E^(x^(-1) + x)/x^2, x] - 9*Log[(E^x + x^2)/3]*Log[E^x + x^2]*Defer[Int][E^(x^(-1) + x)/x^
2, x] + 180*Log[(E^x + x^2)/3]*Defer[Int][(E^x^(-1)*x)/(E^x + x^2), x] - 210*Defer[Int][(E^x^(-1)*x^2)/(E^x +
x^2), x] - 90*Log[(E^x + x^2)/3]*Defer[Int][(E^x^(-1)*x^2)/(E^x + x^2), x] + 75*Defer[Int][(E^x^(-1)*x^3)/(E^x
+ x^2), x] - 36*Log[(E^x + x^2)/3]*Defer[Int][(E^x^(-1)*x^3)/(E^x + x^2), x] + 27*Defer[Int][(E^x^(-1)*x^4)/(
E^x + x^2), x] + 18*Log[(E^x + x^2)/3]*Defer[Int][(E^x^(-1)*x^4)/(E^x + x^2), x] - 6*Defer[Int][(E^x^(-1)*x^5)
/(E^x + x^2), x] + 105*Defer[Int][ExpIntegralEi[x^(-1)], x] + 210*Defer[Int][(x*ExpIntegralEi[x^(-1)])/(E^x +
x^2), x] - 105*Defer[Int][(x^2*ExpIntegralEi[x^(-1)])/(E^x + x^2), x] + 9*Defer[Int][E^(x^(-1) + x)*Log[(E^x +
x^2)/3]^2, x] - 45*Defer[Int][(E^x^(-1)*Log[(E^x + x^2)/3]^2)/x^2, x] - 18*Defer[Int][Defer[Int][E^(x^(-1) +
x), x], x] - 36*Defer[Int][(x*Defer[Int][E^(x^(-1) + x), x])/(E^x + x^2), x] + 18*Defer[Int][(x^2*Defer[Int][E
^(x^(-1) + x), x])/(E^x + x^2), x] - 9*Log[3]*Defer[Int][Defer[Int][E^(x^(-1) + x)/x^2, x], x] + 9*Log[(E^x +
x^2)/3]*Defer[Int][Defer[Int][E^(x^(-1) + x)/x^2, x], x] + 9*Log[E^x + x^2]*Defer[Int][Defer[Int][E^(x^(-1) +
x)/x^2, x], x] - 18*Log[3]*Defer[Int][(x*Defer[Int][E^(x^(-1) + x)/x^2, x])/(E^x + x^2), x] + 18*Log[(E^x + x^
2)/3]*Defer[Int][(x*Defer[Int][E^(x^(-1) + x)/x^2, x])/(E^x + x^2), x] + 18*Log[E^x + x^2]*Defer[Int][(x*Defer
[Int][E^(x^(-1) + x)/x^2, x])/(E^x + x^2), x] + 9*Log[3]*Defer[Int][(x^2*Defer[Int][E^(x^(-1) + x)/x^2, x])/(E
^x + x^2), x] - 9*Log[(E^x + x^2)/3]*Defer[Int][(x^2*Defer[Int][E^(x^(-1) + x)/x^2, x])/(E^x + x^2), x] - 9*Lo
g[E^x + x^2]*Defer[Int][(x^2*Defer[Int][E^(x^(-1) + x)/x^2, x])/(E^x + x^2), x] - 180*Defer[Int][Defer[Int][(E
^x^(-1)*x)/(E^x + x^2), x], x] - 360*Defer[Int][(x*Defer[Int][(E^x^(-1)*x)/(E^x + x^2), x])/(E^x + x^2), x] +
180*Defer[Int][(x^2*Defer[Int][(E^x^(-1)*x)/(E^x + x^2), x])/(E^x + x^2), x] + 90*Defer[Int][Defer[Int][(E^x^(
-1)*x^2)/(E^x + x^2), x], x] + 180*Defer[Int][(x*Defer[Int][(E^x^(-1)*x^2)/(E^x + x^2), x])/(E^x + x^2), x] -
90*Defer[Int][(x^2*Defer[Int][(E^x^(-1)*x^2)/(E^x + x^2), x])/(E^x + x^2), x] + 36*Defer[Int][Defer[Int][(E^x^
(-1)*x^3)/(E^x + x^2), x], x] + 72*Defer[Int][(x*Defer[Int][(E^x^(-1)*x^3)/(E^x + x^2), x])/(E^x + x^2), x] -
36*Defer[Int][(x^2*Defer[Int][(E^x^(-1)*x^3)/(E^x + x^2), x])/(E^x + x^2), x] - 18*Defer[Int][Defer[Int][(E^x^
(-1)*x^4)/(E^x + x^2), x], x] - 36*Defer[Int][(x*Defer[Int][(E^x^(-1)*x^4)/(E^x + x^2), x])/(E^x + x^2), x] +
18*Defer[Int][(x^2*Defer[Int][(E^x^(-1)*x^4)/(E^x + x^2), x])/(E^x + x^2), x] - 18*Defer[Int][Defer[Int][Defer
[Int][E^(x^(-1) + x)/x^2, x], x], x] - 36*Defer[Int][(x*Defer[Int][Defer[Int][E^(x^(-1) + x)/x^2, x], x])/(E^x
+ x^2), x] + 18*Defer[Int][(x^2*Defer[Int][Defer[Int][E^(x^(-1) + x)/x^2, x], x])/(E^x + x^2), x] - 36*Defer[
Int][Defer[Int][(x*Defer[Int][E^(x^(-1) + x)/x^2, x])/(E^x + x^2), x], x] - 72*Defer[Int][(x*Defer[Int][(x*Def
er[Int][E^(x^(-1) + x)/x^2, x])/(E^x + x^2), x])/(E^x + x^2), x] + 36*Defer[Int][(x^2*Defer[Int][(x*Defer[Int]
[E^(x^(-1) + x)/x^2, x])/(E^x + x^2), x])/(E^x + x^2), x] + 18*Defer[Int][Defer[Int][(x^2*Defer[Int][E^(x^(-1)
+ x)/x^2, x])/(E^x + x^2), x], x] + 36*Defer[Int][(x*Defer[Int][(x^2*Defer[Int][E^(x^(-1) + x)/x^2, x])/(E^x
+ x^2), x])/(E^x + x^2), x] - 18*Defer[Int][(x^2*Defer[Int][(x^2*Defer[Int][E^(x^(-1) + x)/x^2, x])/(E^x + x^2
), x])/(E^x + x^2), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {9 e^{\frac {1}{x}} \log \left (\frac {1}{3} \left (e^x+x^2\right )\right ) \left (2 \left (5+e^x\right ) x^2 \left (e^x+2 x\right )+\left (-5 x^2+e^{2 x} \left (-1+x^2\right )+e^x \left (-5-x^2+x^4\right )\right ) \log \left (\frac {1}{3} \left (e^x+x^2\right )\right )\right )}{x^2 \left (e^x+x^2\right )} \, dx\\ &=9 \int \frac {e^{\frac {1}{x}} \log \left (\frac {1}{3} \left (e^x+x^2\right )\right ) \left (2 \left (5+e^x\right ) x^2 \left (e^x+2 x\right )+\left (-5 x^2+e^{2 x} \left (-1+x^2\right )+e^x \left (-5-x^2+x^4\right )\right ) \log \left (\frac {1}{3} \left (e^x+x^2\right )\right )\right )}{x^2 \left (e^x+x^2\right )} \, dx\\ &=9 \int \left (\frac {2 e^{\frac {1}{x}} x \left (10-5 x-2 x^2+x^3\right ) \log \left (\frac {1}{3} \left (e^x+x^2\right )\right )}{e^x+x^2}-\frac {e^{\frac {1}{x}} \log \left (\frac {1}{3} \left (e^x+x^2\right )\right ) \left (-10 x^2-4 x^3+2 x^4+5 \log \left (\frac {1}{3} \left (e^x+x^2\right )\right )\right )}{x^2}+\frac {e^{\frac {1}{x}+x} \log \left (\frac {1}{3} \left (e^x+x^2\right )\right ) \left (2 x^2+\log (3)+x^2 \log \left (\frac {1}{3} \left (e^x+x^2\right )\right )-\log \left (e^x+x^2\right )\right )}{x^2}\right ) \, dx\\ &=-\left (9 \int \frac {e^{\frac {1}{x}} \log \left (\frac {1}{3} \left (e^x+x^2\right )\right ) \left (-10 x^2-4 x^3+2 x^4+5 \log \left (\frac {1}{3} \left (e^x+x^2\right )\right )\right )}{x^2} \, dx\right )+9 \int \frac {e^{\frac {1}{x}+x} \log \left (\frac {1}{3} \left (e^x+x^2\right )\right ) \left (2 x^2+\log (3)+x^2 \log \left (\frac {1}{3} \left (e^x+x^2\right )\right )-\log \left (e^x+x^2\right )\right )}{x^2} \, dx+18 \int \frac {e^{\frac {1}{x}} x \left (10-5 x-2 x^2+x^3\right ) \log \left (\frac {1}{3} \left (e^x+x^2\right )\right )}{e^x+x^2} \, dx\\ &=-\left (9 \int \left (2 e^{\frac {1}{x}} \left (-5-2 x+x^2\right ) \log \left (\frac {1}{3} \left (e^x+x^2\right )\right )+\frac {5 e^{\frac {1}{x}} \log ^2\left (\frac {1}{3} \left (e^x+x^2\right )\right )}{x^2}\right ) \, dx\right )+9 \int \left (\frac {e^{\frac {1}{x}+x} \log \left (\frac {1}{3} \left (e^x+x^2\right )\right ) \left (2 x^2+\log (3)+x^2 \log \left (\frac {1}{3} \left (e^x+x^2\right )\right )\right )}{x^2}-\frac {e^{\frac {1}{x}+x} \log \left (\frac {1}{3} \left (e^x+x^2\right )\right ) \log \left (e^x+x^2\right )}{x^2}\right ) \, dx-18 \int \frac {\left (e^x+2 x\right ) \left (10 \int \frac {e^{\frac {1}{x}} x}{e^x+x^2} \, dx-5 \int \frac {e^{\frac {1}{x}} x^2}{e^x+x^2} \, dx-2 \int \frac {e^{\frac {1}{x}} x^3}{e^x+x^2} \, dx+\int \frac {e^{\frac {1}{x}} x^4}{e^x+x^2} \, dx\right )}{e^x+x^2} \, dx+\left (18 \log \left (\frac {1}{3} \left (e^x+x^2\right )\right )\right ) \int \frac {e^{\frac {1}{x}} x^4}{e^x+x^2} \, dx-\left (36 \log \left (\frac {1}{3} \left (e^x+x^2\right )\right )\right ) \int \frac {e^{\frac {1}{x}} x^3}{e^x+x^2} \, dx-\left (90 \log \left (\frac {1}{3} \left (e^x+x^2\right )\right )\right ) \int \frac {e^{\frac {1}{x}} x^2}{e^x+x^2} \, dx+\left (180 \log \left (\frac {1}{3} \left (e^x+x^2\right )\right )\right ) \int \frac {e^{\frac {1}{x}} x}{e^x+x^2} \, dx\\ &=9 \int \frac {e^{\frac {1}{x}+x} \log \left (\frac {1}{3} \left (e^x+x^2\right )\right ) \left (2 x^2+\log (3)+x^2 \log \left (\frac {1}{3} \left (e^x+x^2\right )\right )\right )}{x^2} \, dx-9 \int \frac {e^{\frac {1}{x}+x} \log \left (\frac {1}{3} \left (e^x+x^2\right )\right ) \log \left (e^x+x^2\right )}{x^2} \, dx-18 \int e^{\frac {1}{x}} \left (-5-2 x+x^2\right ) \log \left (\frac {1}{3} \left (e^x+x^2\right )\right ) \, dx-18 \int \left (10 \int \frac {e^{\frac {1}{x}} x}{e^x+x^2} \, dx-5 \int \frac {e^{\frac {1}{x}} x^2}{e^x+x^2} \, dx-2 \int \frac {e^{\frac {1}{x}} x^3}{e^x+x^2} \, dx+\int \frac {e^{\frac {1}{x}} x^4}{e^x+x^2} \, dx-\frac {(-2+x) x \left (10 \int \frac {e^{\frac {1}{x}} x}{e^x+x^2} \, dx-5 \int \frac {e^{\frac {1}{x}} x^2}{e^x+x^2} \, dx-2 \int \frac {e^{\frac {1}{x}} x^3}{e^x+x^2} \, dx+\int \frac {e^{\frac {1}{x}} x^4}{e^x+x^2} \, dx\right )}{e^x+x^2}\right ) \, dx-45 \int \frac {e^{\frac {1}{x}} \log ^2\left (\frac {1}{3} \left (e^x+x^2\right )\right )}{x^2} \, dx+\left (18 \log \left (\frac {1}{3} \left (e^x+x^2\right )\right )\right ) \int \frac {e^{\frac {1}{x}} x^4}{e^x+x^2} \, dx-\left (36 \log \left (\frac {1}{3} \left (e^x+x^2\right )\right )\right ) \int \frac {e^{\frac {1}{x}} x^3}{e^x+x^2} \, dx-\left (90 \log \left (\frac {1}{3} \left (e^x+x^2\right )\right )\right ) \int \frac {e^{\frac {1}{x}} x^2}{e^x+x^2} \, dx+\left (180 \log \left (\frac {1}{3} \left (e^x+x^2\right )\right )\right ) \int \frac {e^{\frac {1}{x}} x}{e^x+x^2} \, dx\\ &=\text {Rest of rules removed due to large latex content} \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.08, size = 26, normalized size = 1.00 \begin {gather*} 9 e^{\frac {1}{x}} \left (5+e^x\right ) \log ^2\left (\frac {1}{3} \left (e^x+x^2\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((18*E^(x^(-1) + 2*x)*x^2 + 180*E^x^(-1)*x^3 + E^(x^(-1) + x)*(90*x^2 + 36*x^3))*Log[(E^x + x^2)/3]
+ (-45*E^x^(-1)*x^2 + E^(x^(-1) + 2*x)*(-9 + 9*x^2) + E^(x^(-1) + x)*(-45 - 9*x^2 + 9*x^4))*Log[(E^x + x^2)/3]
^2)/(E^x*x^2 + x^4),x]

[Out]

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

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fricas [B]  time = 0.67, size = 94, normalized size = 3.62 \begin {gather*} 9 \, {\left (e^{\left (\frac {2 \, x^{2} + 1}{x} + \frac {x^{2} + 1}{x}\right )} + 5 \, e^{\left (\frac {2 \, {\left (x^{2} + 1\right )}}{x}\right )}\right )} e^{\left (-\frac {2 \, x^{2} + 1}{x}\right )} \log \left (\frac {1}{3} \, {\left (x^{2} e^{\left (\frac {x^{2} + 1}{x}\right )} + e^{\left (\frac {2 \, x^{2} + 1}{x}\right )}\right )} e^{\left (-\frac {x^{2} + 1}{x}\right )}\right )^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((9*x^2-9)*exp(1/x)*exp(x)^2+(9*x^4-9*x^2-45)*exp(1/x)*exp(x)-45*x^2*exp(1/x))*log(1/3*x^2+1/3*exp(
x))^2+(18*x^2*exp(1/x)*exp(x)^2+(36*x^3+90*x^2)*exp(1/x)*exp(x)+180*x^3*exp(1/x))*log(1/3*x^2+1/3*exp(x)))/(ex
p(x)*x^2+x^4),x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((9*x^2-9)*exp(1/x)*exp(x)^2+(9*x^4-9*x^2-45)*exp(1/x)*exp(x)-45*x^2*exp(1/x))*log(1/3*x^2+1/3*exp(
x))^2+(18*x^2*exp(1/x)*exp(x)^2+(36*x^3+90*x^2)*exp(1/x)*exp(x)+180*x^3*exp(1/x))*log(1/3*x^2+1/3*exp(x)))/(ex
p(x)*x^2+x^4),x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Unable to divide, perhaps due to rounding error%%%{-1,[0,14]%%%}+%%%{4,[0,13]%%%}+%%%{-3,[0,12]%%%}+%%%{-4,
[0,11]%%%}+

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maple [A]  time = 0.32, size = 24, normalized size = 0.92

method	result	size
risch	\(9 \left ({\mathrm e}^{x}+5\right ) {\mathrm e}^{\frac {1}{x}} \ln \left (\frac {x^{2}}{3}+\frac {{\mathrm e}^{x}}{3}\right )^{2}\)	\(24\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((9*x^2-9)*exp(1/x)*exp(x)^2+(9*x^4-9*x^2-45)*exp(1/x)*exp(x)-45*x^2*exp(1/x))*ln(1/3*x^2+1/3*exp(x))^2+(
18*x^2*exp(1/x)*exp(x)^2+(36*x^3+90*x^2)*exp(1/x)*exp(x)+180*x^3*exp(1/x))*ln(1/3*x^2+1/3*exp(x)))/(exp(x)*x^2
+x^4),x,method=_RETURNVERBOSE)

[Out]

9*(exp(x)+5)*exp(1/x)*ln(1/3*x^2+1/3*exp(x))^2

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maxima [B]  time = 0.51, size = 63, normalized size = 2.42 \begin {gather*} 9 \, {\left (e^{x} + 5\right )} e^{\frac {1}{x}} \log \left (x^{2} + e^{x}\right )^{2} - 18 \, {\left (e^{x} \log \relax (3) + 5 \, \log \relax (3)\right )} e^{\frac {1}{x}} \log \left (x^{2} + e^{x}\right ) + 9 \, {\left (e^{x} \log \relax (3)^{2} + 5 \, \log \relax (3)^{2}\right )} e^{\frac {1}{x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((9*x^2-9)*exp(1/x)*exp(x)^2+(9*x^4-9*x^2-45)*exp(1/x)*exp(x)-45*x^2*exp(1/x))*log(1/3*x^2+1/3*exp(
x))^2+(18*x^2*exp(1/x)*exp(x)^2+(36*x^3+90*x^2)*exp(1/x)*exp(x)+180*x^3*exp(1/x))*log(1/3*x^2+1/3*exp(x)))/(ex
p(x)*x^2+x^4),x, algorithm="maxima")

[Out]

9*(e^x + 5)*e^(1/x)*log(x^2 + e^x)^2 - 18*(e^x*log(3) + 5*log(3))*e^(1/x)*log(x^2 + e^x) + 9*(e^x*log(3)^2 + 5
*log(3)^2)*e^(1/x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int \frac {\ln \left (\frac {{\mathrm {e}}^x}{3}+\frac {x^2}{3}\right )\,\left ({\mathrm {e}}^{x+\frac {1}{x}}\,\left (36\,x^3+90\,x^2\right )+180\,x^3\,{\mathrm {e}}^{1/x}+18\,x^2\,{\mathrm {e}}^{2\,x+\frac {1}{x}}\right )-{\ln \left (\frac {{\mathrm {e}}^x}{3}+\frac {x^2}{3}\right )}^2\,\left ({\mathrm {e}}^{x+\frac {1}{x}}\,\left (-9\,x^4+9\,x^2+45\right )-{\mathrm {e}}^{2\,x+\frac {1}{x}}\,\left (9\,x^2-9\right )+45\,x^2\,{\mathrm {e}}^{1/x}\right )}{x^2\,{\mathrm {e}}^x+x^4} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(exp(x)/3 + x^2/3)*(180*x^3*exp(1/x) + exp(1/x)*exp(x)*(90*x^2 + 36*x^3) + 18*x^2*exp(2*x)*exp(1/x)) -
log(exp(x)/3 + x^2/3)^2*(45*x^2*exp(1/x) + exp(1/x)*exp(x)*(9*x^2 - 9*x^4 + 45) - exp(2*x)*exp(1/x)*(9*x^2 -
9)))/(x^2*exp(x) + x^4),x)

[Out]

int((log(exp(x)/3 + x^2/3)*(exp(x + 1/x)*(90*x^2 + 36*x^3) + 180*x^3*exp(1/x) + 18*x^2*exp(2*x + 1/x)) - log(e
xp(x)/3 + x^2/3)^2*(exp(x + 1/x)*(9*x^2 - 9*x^4 + 45) - exp(2*x + 1/x)*(9*x^2 - 9) + 45*x^2*exp(1/x)))/(x^2*ex
p(x) + x^4), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((9*x**2-9)*exp(1/x)*exp(x)**2+(9*x**4-9*x**2-45)*exp(1/x)*exp(x)-45*x**2*exp(1/x))*ln(1/3*x**2+1/3
*exp(x))**2+(18*x**2*exp(1/x)*exp(x)**2+(36*x**3+90*x**2)*exp(1/x)*exp(x)+180*x**3*exp(1/x))*ln(1/3*x**2+1/3*e
xp(x)))/(exp(x)*x**2+x**4),x)

[Out]

Timed out

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