3.88.51 \(\int \frac {e^{\frac {-5 x+10 e^{2 x} x+e^x (15+10 x^2)+(4 e^x x+4 x^2) \log ^2(x)}{5 e^{2 x} x+5 e^x x^2}} (5 e^{3 x} x+5 x^2+5 x^3+e^{2 x} (-15-15 x+10 x^2)+e^x (-30 x+10 x^2+5 x^3)+(8 e^{2 x} x+16 e^x x^2+8 x^3) \log (x)+(-4 e^{2 x} x^2-8 e^x x^3-4 x^4) \log ^2(x))}{5 e^{3 x} x+10 e^{2 x} x^2+5 e^x x^3} \, dx\)

Optimal. Leaf size=36 \[ e^{2+e^{-x} \left (\frac {3}{x}+\frac {4}{5} \left (-\frac {5}{e^x+x}+\log ^2(x)\right )\right )} x \]

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Rubi [F]  time = 50.21, 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 {\exp \left (\frac {-5 x+10 e^{2 x} x+e^x \left (15+10 x^2\right )+\left (4 e^x x+4 x^2\right ) \log ^2(x)}{5 e^{2 x} x+5 e^x x^2}\right ) \left (5 e^{3 x} x+5 x^2+5 x^3+e^{2 x} \left (-15-15 x+10 x^2\right )+e^x \left (-30 x+10 x^2+5 x^3\right )+\left (8 e^{2 x} x+16 e^x x^2+8 x^3\right ) \log (x)+\left (-4 e^{2 x} x^2-8 e^x x^3-4 x^4\right ) \log ^2(x)\right )}{5 e^{3 x} x+10 e^{2 x} x^2+5 e^x x^3} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^((-5*x + 10*E^(2*x)*x + E^x*(15 + 10*x^2) + (4*E^x*x + 4*x^2)*Log[x]^2)/(5*E^(2*x)*x + 5*E^x*x^2))*(5*E
^(3*x)*x + 5*x^2 + 5*x^3 + E^(2*x)*(-15 - 15*x + 10*x^2) + E^x*(-30*x + 10*x^2 + 5*x^3) + (8*E^(2*x)*x + 16*E^
x*x^2 + 8*x^3)*Log[x] + (-4*E^(2*x)*x^2 - 8*E^x*x^3 - 4*x^4)*Log[x]^2))/(5*E^(3*x)*x + 10*E^(2*x)*x^2 + 5*E^x*
x^3),x]

[Out]

-3*Defer[Int][E^(((-5*(x + E^(2*x)*(-2 + x)*x + E^x*(-3 - 2*x^2 + x^3)))/(x*(E^x + x)) + 4*Log[x]^2)/(5*E^x)),
x] + Defer[Int][E^(x + ((-5*(x + E^(2*x)*(-2 + x)*x + E^x*(-3 - 2*x^2 + x^3)))/(x*(E^x + x)) + 4*Log[x]^2)/(5
*E^x)), x] - 3*Defer[Int][E^(((-5*(x + E^(2*x)*(-2 + x)*x + E^x*(-3 - 2*x^2 + x^3)))/(x*(E^x + x)) + 4*Log[x]^
2)/(5*E^x))/x, x] + 4*Defer[Int][(E^(((-5*(x + E^(2*x)*(-2 + x)*x + E^x*(-3 - 2*x^2 + x^3)))/(x*(E^x + x)) + 4
*Log[x]^2)/(5*E^x))*x)/(E^x + x)^2, x] - 4*Defer[Int][(E^(((-5*(x + E^(2*x)*(-2 + x)*x + E^x*(-3 - 2*x^2 + x^3
)))/(x*(E^x + x)) + 4*Log[x]^2)/(5*E^x))*x^2)/(E^x + x)^2, x] + 8*Defer[Int][(E^(((-5*(x + E^(2*x)*(-2 + x)*x
+ E^x*(-3 - 2*x^2 + x^3)))/(x*(E^x + x)) + 4*Log[x]^2)/(5*E^x))*x)/(E^x + x), x] + (8*Defer[Int][E^(((-5*(x +
E^(2*x)*(-2 + x)*x + E^x*(-3 - 2*x^2 + x^3)))/(x*(E^x + x)) + 4*Log[x]^2)/(5*E^x))*Log[x], x])/5 - (4*Defer[In
t][E^(((-5*(x + E^(2*x)*(-2 + x)*x + E^x*(-3 - 2*x^2 + x^3)))/(x*(E^x + x)) + 4*Log[x]^2)/(5*E^x))*x*Log[x]^2,
x])/5

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {\exp \left (\frac {1}{5} e^{-x} \left (-\frac {5 \left (x+e^{2 x} (-2+x) x+e^x \left (-3-2 x^2+x^3\right )\right )}{x \left (e^x+x\right )}+4 \log ^2(x)\right )\right ) \left (5 \left (e^{3 x} x+x^2 (1+x)+e^x x \left (-6+2 x+x^2\right )+e^{2 x} \left (-3-3 x+2 x^2\right )\right )+8 x \left (e^x+x\right )^2 \log (x)-4 x^2 \left (e^x+x\right )^2 \log ^2(x)\right )}{5 x \left (e^x+x\right )^2} \, dx\\ &=\frac {1}{5} \int \frac {\exp \left (\frac {1}{5} e^{-x} \left (-\frac {5 \left (x+e^{2 x} (-2+x) x+e^x \left (-3-2 x^2+x^3\right )\right )}{x \left (e^x+x\right )}+4 \log ^2(x)\right )\right ) \left (5 \left (e^{3 x} x+x^2 (1+x)+e^x x \left (-6+2 x+x^2\right )+e^{2 x} \left (-3-3 x+2 x^2\right )\right )+8 x \left (e^x+x\right )^2 \log (x)-4 x^2 \left (e^x+x\right )^2 \log ^2(x)\right )}{x \left (e^x+x\right )^2} \, dx\\ &=\frac {1}{5} \int \left (5 \exp \left (x+\frac {1}{5} e^{-x} \left (-\frac {5 \left (x+e^{2 x} (-2+x) x+e^x \left (-3-2 x^2+x^3\right )\right )}{x \left (e^x+x\right )}+4 \log ^2(x)\right )\right )-\frac {20 \exp \left (\frac {1}{5} e^{-x} \left (-\frac {5 \left (x+e^{2 x} (-2+x) x+e^x \left (-3-2 x^2+x^3\right )\right )}{x \left (e^x+x\right )}+4 \log ^2(x)\right )\right ) (-1+x) x}{\left (e^x+x\right )^2}+\frac {40 \exp \left (\frac {1}{5} e^{-x} \left (-\frac {5 \left (x+e^{2 x} (-2+x) x+e^x \left (-3-2 x^2+x^3\right )\right )}{x \left (e^x+x\right )}+4 \log ^2(x)\right )\right ) x}{e^x+x}+\frac {\exp \left (\frac {1}{5} e^{-x} \left (-\frac {5 \left (x+e^{2 x} (-2+x) x+e^x \left (-3-2 x^2+x^3\right )\right )}{x \left (e^x+x\right )}+4 \log ^2(x)\right )\right ) \left (-15-15 x+8 x \log (x)-4 x^2 \log ^2(x)\right )}{x}\right ) \, dx\\ &=\frac {1}{5} \int \frac {\exp \left (\frac {1}{5} e^{-x} \left (-\frac {5 \left (x+e^{2 x} (-2+x) x+e^x \left (-3-2 x^2+x^3\right )\right )}{x \left (e^x+x\right )}+4 \log ^2(x)\right )\right ) \left (-15-15 x+8 x \log (x)-4 x^2 \log ^2(x)\right )}{x} \, dx-4 \int \frac {\exp \left (\frac {1}{5} e^{-x} \left (-\frac {5 \left (x+e^{2 x} (-2+x) x+e^x \left (-3-2 x^2+x^3\right )\right )}{x \left (e^x+x\right )}+4 \log ^2(x)\right )\right ) (-1+x) x}{\left (e^x+x\right )^2} \, dx+8 \int \frac {\exp \left (\frac {1}{5} e^{-x} \left (-\frac {5 \left (x+e^{2 x} (-2+x) x+e^x \left (-3-2 x^2+x^3\right )\right )}{x \left (e^x+x\right )}+4 \log ^2(x)\right )\right ) x}{e^x+x} \, dx+\int \exp \left (x+\frac {1}{5} e^{-x} \left (-\frac {5 \left (x+e^{2 x} (-2+x) x+e^x \left (-3-2 x^2+x^3\right )\right )}{x \left (e^x+x\right )}+4 \log ^2(x)\right )\right ) \, dx\\ &=\frac {1}{5} \int \left (-\frac {15 \exp \left (\frac {1}{5} e^{-x} \left (-\frac {5 \left (x+e^{2 x} (-2+x) x+e^x \left (-3-2 x^2+x^3\right )\right )}{x \left (e^x+x\right )}+4 \log ^2(x)\right )\right ) (1+x)}{x}+8 \exp \left (\frac {1}{5} e^{-x} \left (-\frac {5 \left (x+e^{2 x} (-2+x) x+e^x \left (-3-2 x^2+x^3\right )\right )}{x \left (e^x+x\right )}+4 \log ^2(x)\right )\right ) \log (x)-4 \exp \left (\frac {1}{5} e^{-x} \left (-\frac {5 \left (x+e^{2 x} (-2+x) x+e^x \left (-3-2 x^2+x^3\right )\right )}{x \left (e^x+x\right )}+4 \log ^2(x)\right )\right ) x \log ^2(x)\right ) \, dx-4 \int \left (-\frac {\exp \left (\frac {1}{5} e^{-x} \left (-\frac {5 \left (x+e^{2 x} (-2+x) x+e^x \left (-3-2 x^2+x^3\right )\right )}{x \left (e^x+x\right )}+4 \log ^2(x)\right )\right ) x}{\left (e^x+x\right )^2}+\frac {\exp \left (\frac {1}{5} e^{-x} \left (-\frac {5 \left (x+e^{2 x} (-2+x) x+e^x \left (-3-2 x^2+x^3\right )\right )}{x \left (e^x+x\right )}+4 \log ^2(x)\right )\right ) x^2}{\left (e^x+x\right )^2}\right ) \, dx+8 \int \frac {\exp \left (\frac {1}{5} e^{-x} \left (-\frac {5 \left (x+e^{2 x} (-2+x) x+e^x \left (-3-2 x^2+x^3\right )\right )}{x \left (e^x+x\right )}+4 \log ^2(x)\right )\right ) x}{e^x+x} \, dx+\int \exp \left (x+\frac {1}{5} e^{-x} \left (-\frac {5 \left (x+e^{2 x} (-2+x) x+e^x \left (-3-2 x^2+x^3\right )\right )}{x \left (e^x+x\right )}+4 \log ^2(x)\right )\right ) \, dx\\ &=-\left (\frac {4}{5} \int \exp \left (\frac {1}{5} e^{-x} \left (-\frac {5 \left (x+e^{2 x} (-2+x) x+e^x \left (-3-2 x^2+x^3\right )\right )}{x \left (e^x+x\right )}+4 \log ^2(x)\right )\right ) x \log ^2(x) \, dx\right )+\frac {8}{5} \int \exp \left (\frac {1}{5} e^{-x} \left (-\frac {5 \left (x+e^{2 x} (-2+x) x+e^x \left (-3-2 x^2+x^3\right )\right )}{x \left (e^x+x\right )}+4 \log ^2(x)\right )\right ) \log (x) \, dx-3 \int \frac {\exp \left (\frac {1}{5} e^{-x} \left (-\frac {5 \left (x+e^{2 x} (-2+x) x+e^x \left (-3-2 x^2+x^3\right )\right )}{x \left (e^x+x\right )}+4 \log ^2(x)\right )\right ) (1+x)}{x} \, dx+4 \int \frac {\exp \left (\frac {1}{5} e^{-x} \left (-\frac {5 \left (x+e^{2 x} (-2+x) x+e^x \left (-3-2 x^2+x^3\right )\right )}{x \left (e^x+x\right )}+4 \log ^2(x)\right )\right ) x}{\left (e^x+x\right )^2} \, dx-4 \int \frac {\exp \left (\frac {1}{5} e^{-x} \left (-\frac {5 \left (x+e^{2 x} (-2+x) x+e^x \left (-3-2 x^2+x^3\right )\right )}{x \left (e^x+x\right )}+4 \log ^2(x)\right )\right ) x^2}{\left (e^x+x\right )^2} \, dx+8 \int \frac {\exp \left (\frac {1}{5} e^{-x} \left (-\frac {5 \left (x+e^{2 x} (-2+x) x+e^x \left (-3-2 x^2+x^3\right )\right )}{x \left (e^x+x\right )}+4 \log ^2(x)\right )\right ) x}{e^x+x} \, dx+\int \exp \left (x+\frac {1}{5} e^{-x} \left (-\frac {5 \left (x+e^{2 x} (-2+x) x+e^x \left (-3-2 x^2+x^3\right )\right )}{x \left (e^x+x\right )}+4 \log ^2(x)\right )\right ) \, dx\\ &=-\left (\frac {4}{5} \int \exp \left (\frac {1}{5} e^{-x} \left (-\frac {5 \left (x+e^{2 x} (-2+x) x+e^x \left (-3-2 x^2+x^3\right )\right )}{x \left (e^x+x\right )}+4 \log ^2(x)\right )\right ) x \log ^2(x) \, dx\right )+\frac {8}{5} \int \exp \left (\frac {1}{5} e^{-x} \left (-\frac {5 \left (x+e^{2 x} (-2+x) x+e^x \left (-3-2 x^2+x^3\right )\right )}{x \left (e^x+x\right )}+4 \log ^2(x)\right )\right ) \log (x) \, dx-3 \int \left (\exp \left (\frac {1}{5} e^{-x} \left (-\frac {5 \left (x+e^{2 x} (-2+x) x+e^x \left (-3-2 x^2+x^3\right )\right )}{x \left (e^x+x\right )}+4 \log ^2(x)\right )\right )+\frac {\exp \left (\frac {1}{5} e^{-x} \left (-\frac {5 \left (x+e^{2 x} (-2+x) x+e^x \left (-3-2 x^2+x^3\right )\right )}{x \left (e^x+x\right )}+4 \log ^2(x)\right )\right )}{x}\right ) \, dx+4 \int \frac {\exp \left (\frac {1}{5} e^{-x} \left (-\frac {5 \left (x+e^{2 x} (-2+x) x+e^x \left (-3-2 x^2+x^3\right )\right )}{x \left (e^x+x\right )}+4 \log ^2(x)\right )\right ) x}{\left (e^x+x\right )^2} \, dx-4 \int \frac {\exp \left (\frac {1}{5} e^{-x} \left (-\frac {5 \left (x+e^{2 x} (-2+x) x+e^x \left (-3-2 x^2+x^3\right )\right )}{x \left (e^x+x\right )}+4 \log ^2(x)\right )\right ) x^2}{\left (e^x+x\right )^2} \, dx+8 \int \frac {\exp \left (\frac {1}{5} e^{-x} \left (-\frac {5 \left (x+e^{2 x} (-2+x) x+e^x \left (-3-2 x^2+x^3\right )\right )}{x \left (e^x+x\right )}+4 \log ^2(x)\right )\right ) x}{e^x+x} \, dx+\int \exp \left (x+\frac {1}{5} e^{-x} \left (-\frac {5 \left (x+e^{2 x} (-2+x) x+e^x \left (-3-2 x^2+x^3\right )\right )}{x \left (e^x+x\right )}+4 \log ^2(x)\right )\right ) \, dx\\ &=-\left (\frac {4}{5} \int \exp \left (\frac {1}{5} e^{-x} \left (-\frac {5 \left (x+e^{2 x} (-2+x) x+e^x \left (-3-2 x^2+x^3\right )\right )}{x \left (e^x+x\right )}+4 \log ^2(x)\right )\right ) x \log ^2(x) \, dx\right )+\frac {8}{5} \int \exp \left (\frac {1}{5} e^{-x} \left (-\frac {5 \left (x+e^{2 x} (-2+x) x+e^x \left (-3-2 x^2+x^3\right )\right )}{x \left (e^x+x\right )}+4 \log ^2(x)\right )\right ) \log (x) \, dx-3 \int \exp \left (\frac {1}{5} e^{-x} \left (-\frac {5 \left (x+e^{2 x} (-2+x) x+e^x \left (-3-2 x^2+x^3\right )\right )}{x \left (e^x+x\right )}+4 \log ^2(x)\right )\right ) \, dx-3 \int \frac {\exp \left (\frac {1}{5} e^{-x} \left (-\frac {5 \left (x+e^{2 x} (-2+x) x+e^x \left (-3-2 x^2+x^3\right )\right )}{x \left (e^x+x\right )}+4 \log ^2(x)\right )\right )}{x} \, dx+4 \int \frac {\exp \left (\frac {1}{5} e^{-x} \left (-\frac {5 \left (x+e^{2 x} (-2+x) x+e^x \left (-3-2 x^2+x^3\right )\right )}{x \left (e^x+x\right )}+4 \log ^2(x)\right )\right ) x}{\left (e^x+x\right )^2} \, dx-4 \int \frac {\exp \left (\frac {1}{5} e^{-x} \left (-\frac {5 \left (x+e^{2 x} (-2+x) x+e^x \left (-3-2 x^2+x^3\right )\right )}{x \left (e^x+x\right )}+4 \log ^2(x)\right )\right ) x^2}{\left (e^x+x\right )^2} \, dx+8 \int \frac {\exp \left (\frac {1}{5} e^{-x} \left (-\frac {5 \left (x+e^{2 x} (-2+x) x+e^x \left (-3-2 x^2+x^3\right )\right )}{x \left (e^x+x\right )}+4 \log ^2(x)\right )\right ) x}{e^x+x} \, dx+\int \exp \left (x+\frac {1}{5} e^{-x} \left (-\frac {5 \left (x+e^{2 x} (-2+x) x+e^x \left (-3-2 x^2+x^3\right )\right )}{x \left (e^x+x\right )}+4 \log ^2(x)\right )\right ) \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.66, size = 50, normalized size = 1.39 \begin {gather*} e^{2+\frac {4}{x^2}-\frac {e^{-x}}{x}-\frac {4}{x^2 \left (1+e^{-x} x\right )}+\frac {4}{5} e^{-x} \log ^2(x)} x \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((-5*x + 10*E^(2*x)*x + E^x*(15 + 10*x^2) + (4*E^x*x + 4*x^2)*Log[x]^2)/(5*E^(2*x)*x + 5*E^x*x^2)
)*(5*E^(3*x)*x + 5*x^2 + 5*x^3 + E^(2*x)*(-15 - 15*x + 10*x^2) + E^x*(-30*x + 10*x^2 + 5*x^3) + (8*E^(2*x)*x +
16*E^x*x^2 + 8*x^3)*Log[x] + (-4*E^(2*x)*x^2 - 8*E^x*x^3 - 4*x^4)*Log[x]^2))/(5*E^(3*x)*x + 10*E^(2*x)*x^2 +
5*E^x*x^3),x]

[Out]

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

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fricas [A]  time = 0.57, size = 56, normalized size = 1.56 \begin {gather*} x e^{\left (\frac {4 \, {\left (x^{2} + x e^{x}\right )} \log \relax (x)^{2} + 10 \, x e^{\left (2 \, x\right )} + 5 \, {\left (2 \, x^{2} + 3\right )} e^{x} - 5 \, x}{5 \, {\left (x^{2} e^{x} + x e^{\left (2 \, x\right )}\right )}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*exp(x)^2*x^2-8*exp(x)*x^3-4*x^4)*log(x)^2+(8*x*exp(x)^2+16*exp(x)*x^2+8*x^3)*log(x)+5*x*exp(x)^
3+(10*x^2-15*x-15)*exp(x)^2+(5*x^3+10*x^2-30*x)*exp(x)+5*x^3+5*x^2)*exp(((4*exp(x)*x+4*x^2)*log(x)^2+10*x*exp(
x)^2+(10*x^2+15)*exp(x)-5*x)/(5*x*exp(x)^2+5*exp(x)*x^2))/(5*x*exp(x)^3+10*exp(x)^2*x^2+5*exp(x)*x^3),x, algor
ithm="fricas")

[Out]

x*e^(1/5*(4*(x^2 + x*e^x)*log(x)^2 + 10*x*e^(2*x) + 5*(2*x^2 + 3)*e^x - 5*x)/(x^2*e^x + x*e^(2*x)))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (5 \, x^{3} - 4 \, {\left (x^{4} + 2 \, x^{3} e^{x} + x^{2} e^{\left (2 \, x\right )}\right )} \log \relax (x)^{2} + 5 \, x^{2} + 5 \, x e^{\left (3 \, x\right )} + 5 \, {\left (2 \, x^{2} - 3 \, x - 3\right )} e^{\left (2 \, x\right )} + 5 \, {\left (x^{3} + 2 \, x^{2} - 6 \, x\right )} e^{x} + 8 \, {\left (x^{3} + 2 \, x^{2} e^{x} + x e^{\left (2 \, x\right )}\right )} \log \relax (x)\right )} e^{\left (\frac {4 \, {\left (x^{2} + x e^{x}\right )} \log \relax (x)^{2} + 10 \, x e^{\left (2 \, x\right )} + 5 \, {\left (2 \, x^{2} + 3\right )} e^{x} - 5 \, x}{5 \, {\left (x^{2} e^{x} + x e^{\left (2 \, x\right )}\right )}}\right )}}{5 \, {\left (x^{3} e^{x} + 2 \, x^{2} e^{\left (2 \, x\right )} + x e^{\left (3 \, x\right )}\right )}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*exp(x)^2*x^2-8*exp(x)*x^3-4*x^4)*log(x)^2+(8*x*exp(x)^2+16*exp(x)*x^2+8*x^3)*log(x)+5*x*exp(x)^
3+(10*x^2-15*x-15)*exp(x)^2+(5*x^3+10*x^2-30*x)*exp(x)+5*x^3+5*x^2)*exp(((4*exp(x)*x+4*x^2)*log(x)^2+10*x*exp(
x)^2+(10*x^2+15)*exp(x)-5*x)/(5*x*exp(x)^2+5*exp(x)*x^2))/(5*x*exp(x)^3+10*exp(x)^2*x^2+5*exp(x)*x^3),x, algor
ithm="giac")

[Out]

integrate(1/5*(5*x^3 - 4*(x^4 + 2*x^3*e^x + x^2*e^(2*x))*log(x)^2 + 5*x^2 + 5*x*e^(3*x) + 5*(2*x^2 - 3*x - 3)*
e^(2*x) + 5*(x^3 + 2*x^2 - 6*x)*e^x + 8*(x^3 + 2*x^2*e^x + x*e^(2*x))*log(x))*e^(1/5*(4*(x^2 + x*e^x)*log(x)^2
+ 10*x*e^(2*x) + 5*(2*x^2 + 3)*e^x - 5*x)/(x^2*e^x + x*e^(2*x)))/(x^3*e^x + 2*x^2*e^(2*x) + x*e^(3*x)), x)

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maple [A]  time = 0.05, size = 60, normalized size = 1.67

method	result	size
risch	\(x \,{\mathrm e}^{\frac {4 x \,{\mathrm e}^{x} \ln \relax (x )^{2}+4 x^{2} \ln \relax (x )^{2}+10 \,{\mathrm e}^{x} x^{2}+10 x \,{\mathrm e}^{2 x}+15 \,{\mathrm e}^{x}-5 x}{5 x \left ({\mathrm e}^{x} x +{\mathrm e}^{2 x}\right )}}\)	\(60\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-4*exp(x)^2*x^2-8*exp(x)*x^3-4*x^4)*ln(x)^2+(8*x*exp(x)^2+16*exp(x)*x^2+8*x^3)*ln(x)+5*x*exp(x)^3+(10*x^
2-15*x-15)*exp(x)^2+(5*x^3+10*x^2-30*x)*exp(x)+5*x^3+5*x^2)*exp(((4*exp(x)*x+4*x^2)*ln(x)^2+10*x*exp(x)^2+(10*
x^2+15)*exp(x)-5*x)/(5*x*exp(x)^2+5*exp(x)*x^2))/(5*x*exp(x)^3+10*exp(x)^2*x^2+5*exp(x)*x^3),x,method=_RETURNV
ERBOSE)

[Out]

x*exp(1/5*(4*x*exp(x)*ln(x)^2+4*x^2*ln(x)^2+10*exp(x)*x^2+10*x*exp(2*x)+15*exp(x)-5*x)/x/(exp(x)*x+exp(2*x)))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {1}{5} \, \int \frac {{\left (5 \, x^{3} - 4 \, {\left (x^{4} + 2 \, x^{3} e^{x} + x^{2} e^{\left (2 \, x\right )}\right )} \log \relax (x)^{2} + 5 \, x^{2} + 5 \, x e^{\left (3 \, x\right )} + 5 \, {\left (2 \, x^{2} - 3 \, x - 3\right )} e^{\left (2 \, x\right )} + 5 \, {\left (x^{3} + 2 \, x^{2} - 6 \, x\right )} e^{x} + 8 \, {\left (x^{3} + 2 \, x^{2} e^{x} + x e^{\left (2 \, x\right )}\right )} \log \relax (x)\right )} e^{\left (\frac {4 \, {\left (x^{2} + x e^{x}\right )} \log \relax (x)^{2} + 10 \, x e^{\left (2 \, x\right )} + 5 \, {\left (2 \, x^{2} + 3\right )} e^{x} - 5 \, x}{5 \, {\left (x^{2} e^{x} + x e^{\left (2 \, x\right )}\right )}}\right )}}{x^{3} e^{x} + 2 \, x^{2} e^{\left (2 \, x\right )} + x e^{\left (3 \, x\right )}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*exp(x)^2*x^2-8*exp(x)*x^3-4*x^4)*log(x)^2+(8*x*exp(x)^2+16*exp(x)*x^2+8*x^3)*log(x)+5*x*exp(x)^
3+(10*x^2-15*x-15)*exp(x)^2+(5*x^3+10*x^2-30*x)*exp(x)+5*x^3+5*x^2)*exp(((4*exp(x)*x+4*x^2)*log(x)^2+10*x*exp(
x)^2+(10*x^2+15)*exp(x)-5*x)/(5*x*exp(x)^2+5*exp(x)*x^2))/(5*x*exp(x)^3+10*exp(x)^2*x^2+5*exp(x)*x^3),x, algor
ithm="maxima")

[Out]

1/5*integrate((5*x^3 - 4*(x^4 + 2*x^3*e^x + x^2*e^(2*x))*log(x)^2 + 5*x^2 + 5*x*e^(3*x) + 5*(2*x^2 - 3*x - 3)*
e^(2*x) + 5*(x^3 + 2*x^2 - 6*x)*e^x + 8*(x^3 + 2*x^2*e^x + x*e^(2*x))*log(x))*e^(1/5*(4*(x^2 + x*e^x)*log(x)^2
+ 10*x*e^(2*x) + 5*(2*x^2 + 3)*e^x - 5*x)/(x^2*e^x + x*e^(2*x)))/(x^3*e^x + 2*x^2*e^(2*x) + x*e^(3*x)), x)

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mupad [B]  time = 5.79, size = 89, normalized size = 2.47 \begin {gather*} x\,{\mathrm {e}}^{\frac {3}{x\,{\mathrm {e}}^x+x^2}}\,{\mathrm {e}}^{\frac {2\,x}{x+{\mathrm {e}}^x}}\,{\mathrm {e}}^{-\frac {1}{{\mathrm {e}}^{2\,x}+x\,{\mathrm {e}}^x}}\,{\mathrm {e}}^{\frac {4\,x\,{\ln \relax (x)}^2}{5\,{\mathrm {e}}^{2\,x}+5\,x\,{\mathrm {e}}^x}}\,{\mathrm {e}}^{\frac {2\,{\mathrm {e}}^x}{x+{\mathrm {e}}^x}}\,{\mathrm {e}}^{\frac {4\,{\ln \relax (x)}^2}{5\,x+5\,{\mathrm {e}}^x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp((10*x*exp(2*x) - 5*x + exp(x)*(10*x^2 + 15) + log(x)^2*(4*x*exp(x) + 4*x^2))/(5*x*exp(2*x) + 5*x^2*ex
p(x)))*(log(x)*(8*x*exp(2*x) + 16*x^2*exp(x) + 8*x^3) - exp(2*x)*(15*x - 10*x^2 + 15) + 5*x*exp(3*x) - log(x)^
2*(8*x^3*exp(x) + 4*x^2*exp(2*x) + 4*x^4) + 5*x^2 + 5*x^3 + exp(x)*(10*x^2 - 30*x + 5*x^3)))/(5*x*exp(3*x) + 5
*x^3*exp(x) + 10*x^2*exp(2*x)),x)

[Out]

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

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sympy [B]  time = 61.57, size = 58, normalized size = 1.61 \begin {gather*} x e^{\frac {10 x e^{2 x} - 5 x + \left (4 x^{2} + 4 x e^{x}\right ) \log {\relax (x )}^{2} + \left (10 x^{2} + 15\right ) e^{x}}{5 x^{2} e^{x} + 5 x e^{2 x}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*exp(x)**2*x**2-8*exp(x)*x**3-4*x**4)*ln(x)**2+(8*x*exp(x)**2+16*exp(x)*x**2+8*x**3)*ln(x)+5*x*e
xp(x)**3+(10*x**2-15*x-15)*exp(x)**2+(5*x**3+10*x**2-30*x)*exp(x)+5*x**3+5*x**2)*exp(((4*exp(x)*x+4*x**2)*ln(x
)**2+10*x*exp(x)**2+(10*x**2+15)*exp(x)-5*x)/(5*x*exp(x)**2+5*exp(x)*x**2))/(5*x*exp(x)**3+10*exp(x)**2*x**2+5
*exp(x)*x**3),x)

[Out]

x*exp((10*x*exp(2*x) - 5*x + (4*x**2 + 4*x*exp(x))*log(x)**2 + (10*x**2 + 15)*exp(x))/(5*x**2*exp(x) + 5*x*exp
(2*x)))

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