3.48.89 \(\int \frac {50 e^{2 x}+100 e^x x+50 x^2+(-100 e^{2 x} x-100 x^2+e^x (-100 x-100 x^2)) \log (x)+(2 x-10 x^2+12 x^3+e^{2 x} (-2-4 x+8 x^2)+e^x (-12 x+8 x^2+8 x^3)) \log ^2(x)+(-4 e^{2 x} x-4 x^2+e^x (-4 x-4 x^2)) \log ^3(x)}{x \log ^2(x)} \, dx\)

Optimal. Leaf size=26 \[ 2 \left (x-\left (e^x+x\right )^2 \left (2-2 x+\frac {25}{\log (x)}+\log (x)\right )\right ) \]

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Rubi [F]  time = 1.54, 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 {50 e^{2 x}+100 e^x x+50 x^2+\left (-100 e^{2 x} x-100 x^2+e^x \left (-100 x-100 x^2\right )\right ) \log (x)+\left (2 x-10 x^2+12 x^3+e^{2 x} \left (-2-4 x+8 x^2\right )+e^x \left (-12 x+8 x^2+8 x^3\right )\right ) \log ^2(x)+\left (-4 e^{2 x} x-4 x^2+e^x \left (-4 x-4 x^2\right )\right ) \log ^3(x)}{x \log ^2(x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(50*E^(2*x) + 100*E^x*x + 50*x^2 + (-100*E^(2*x)*x - 100*x^2 + E^x*(-100*x - 100*x^2))*Log[x] + (2*x - 10*
x^2 + 12*x^3 + E^(2*x)*(-2 - 4*x + 8*x^2) + E^x*(-12*x + 8*x^2 + 8*x^3))*Log[x]^2 + (-4*E^(2*x)*x - 4*x^2 + E^
x*(-4*x - 4*x^2))*Log[x]^3)/(x*Log[x]^2),x]

[Out]

-4*E^(2*x) + 2*x - 8*E^x*x + 4*E^(2*x)*x - 4*x^2 + 8*E^x*x^2 + 4*x^3 - (50*x^2)/Log[x] - 2*(E^x + x)^2*Log[x]
+ 100*Defer[Int][E^x/Log[x]^2, x] + 50*Defer[Int][E^(2*x)/(x*Log[x]^2), x] - 100*Defer[Int][E^x/Log[x], x] - 1
00*Defer[Int][E^(2*x)/Log[x], x] - 100*Defer[Int][(E^x*x)/Log[x], x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int 2 \left (1-5 x+6 x^2+e^{2 x} \left (-2-\frac {1}{x}+4 x\right )+e^x \left (-6+4 x+4 x^2\right )+\frac {25 \left (e^x+x\right )^2}{x \log ^2(x)}-\frac {50 \left (1+e^x\right ) \left (e^x+x\right )}{\log (x)}-2 \left (1+e^x\right ) \left (e^x+x\right ) \log (x)\right ) \, dx\\ &=2 \int \left (1-5 x+6 x^2+e^{2 x} \left (-2-\frac {1}{x}+4 x\right )+e^x \left (-6+4 x+4 x^2\right )+\frac {25 \left (e^x+x\right )^2}{x \log ^2(x)}-\frac {50 \left (1+e^x\right ) \left (e^x+x\right )}{\log (x)}-2 \left (1+e^x\right ) \left (e^x+x\right ) \log (x)\right ) \, dx\\ &=2 x-5 x^2+4 x^3+2 \int e^{2 x} \left (-2-\frac {1}{x}+4 x\right ) \, dx+2 \int e^x \left (-6+4 x+4 x^2\right ) \, dx-4 \int \left (1+e^x\right ) \left (e^x+x\right ) \log (x) \, dx+50 \int \frac {\left (e^x+x\right )^2}{x \log ^2(x)} \, dx-100 \int \frac {\left (1+e^x\right ) \left (e^x+x\right )}{\log (x)} \, dx\\ &=2 x-5 x^2+4 x^3-2 \left (e^x+x\right )^2 \log (x)+2 \int \left (-2 e^{2 x}-\frac {e^{2 x}}{x}+4 e^{2 x} x\right ) \, dx+2 \int \left (-6 e^x+4 e^x x+4 e^x x^2\right ) \, dx+4 \int \frac {\left (e^x+x\right )^2}{2 x} \, dx+50 \int \left (\frac {2 e^x}{\log ^2(x)}+\frac {e^{2 x}}{x \log ^2(x)}+\frac {x}{\log ^2(x)}\right ) \, dx-100 \int \left (\frac {e^{2 x}}{\log (x)}+\frac {x}{\log (x)}+\frac {e^x (1+x)}{\log (x)}\right ) \, dx\\ &=2 x-5 x^2+4 x^3-2 \left (e^x+x\right )^2 \log (x)-2 \int \frac {e^{2 x}}{x} \, dx+2 \int \frac {\left (e^x+x\right )^2}{x} \, dx-4 \int e^{2 x} \, dx+8 \int e^x x \, dx+8 \int e^{2 x} x \, dx+8 \int e^x x^2 \, dx-12 \int e^x \, dx+50 \int \frac {e^{2 x}}{x \log ^2(x)} \, dx+50 \int \frac {x}{\log ^2(x)} \, dx+100 \int \frac {e^x}{\log ^2(x)} \, dx-100 \int \frac {e^{2 x}}{\log (x)} \, dx-100 \int \frac {x}{\log (x)} \, dx-100 \int \frac {e^x (1+x)}{\log (x)} \, dx\\ &=-12 e^x-2 e^{2 x}+2 x+8 e^x x+4 e^{2 x} x-5 x^2+8 e^x x^2+4 x^3-2 \text {Ei}(2 x)-\frac {50 x^2}{\log (x)}-2 \left (e^x+x\right )^2 \log (x)+2 \int \left (2 e^x+\frac {e^{2 x}}{x}+x\right ) \, dx-4 \int e^{2 x} \, dx-8 \int e^x \, dx-16 \int e^x x \, dx+50 \int \frac {e^{2 x}}{x \log ^2(x)} \, dx-100 \int \left (\frac {e^x}{\log (x)}+\frac {e^x x}{\log (x)}\right ) \, dx+100 \int \frac {e^x}{\log ^2(x)} \, dx-100 \int \frac {e^{2 x}}{\log (x)} \, dx+100 \int \frac {x}{\log (x)} \, dx-100 \operatorname {Subst}\left (\int \frac {e^{2 x}}{x} \, dx,x,\log (x)\right )\\ &=-20 e^x-4 e^{2 x}+2 x-8 e^x x+4 e^{2 x} x-4 x^2+8 e^x x^2+4 x^3-2 \text {Ei}(2 x)-100 \text {Ei}(2 \log (x))-\frac {50 x^2}{\log (x)}-2 \left (e^x+x\right )^2 \log (x)+2 \int \frac {e^{2 x}}{x} \, dx+4 \int e^x \, dx+16 \int e^x \, dx+50 \int \frac {e^{2 x}}{x \log ^2(x)} \, dx+100 \int \frac {e^x}{\log ^2(x)} \, dx-100 \int \frac {e^x}{\log (x)} \, dx-100 \int \frac {e^{2 x}}{\log (x)} \, dx-100 \int \frac {e^x x}{\log (x)} \, dx+100 \operatorname {Subst}\left (\int \frac {e^{2 x}}{x} \, dx,x,\log (x)\right )\\ &=-4 e^{2 x}+2 x-8 e^x x+4 e^{2 x} x-4 x^2+8 e^x x^2+4 x^3-\frac {50 x^2}{\log (x)}-2 \left (e^x+x\right )^2 \log (x)+50 \int \frac {e^{2 x}}{x \log ^2(x)} \, dx+100 \int \frac {e^x}{\log ^2(x)} \, dx-100 \int \frac {e^x}{\log (x)} \, dx-100 \int \frac {e^{2 x}}{\log (x)} \, dx-100 \int \frac {e^x x}{\log (x)} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.30, size = 57, normalized size = 2.19 \begin {gather*} 2 \left (2 e^{2 x} (-1+x)+x+4 e^x (-1+x) x-2 x^2+2 x^3-\frac {25 \left (e^x+x\right )^2}{\log (x)}-\left (e^x+x\right )^2 \log (x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(50*E^(2*x) + 100*E^x*x + 50*x^2 + (-100*E^(2*x)*x - 100*x^2 + E^x*(-100*x - 100*x^2))*Log[x] + (2*x
 - 10*x^2 + 12*x^3 + E^(2*x)*(-2 - 4*x + 8*x^2) + E^x*(-12*x + 8*x^2 + 8*x^3))*Log[x]^2 + (-4*E^(2*x)*x - 4*x^
2 + E^x*(-4*x - 4*x^2))*Log[x]^3)/(x*Log[x]^2),x]

[Out]

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

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fricas [B]  time = 0.62, size = 77, normalized size = 2.96 \begin {gather*} -\frac {2 \, {\left ({\left (x^{2} + 2 \, x e^{x} + e^{\left (2 \, x\right )}\right )} \log \relax (x)^{2} + 25 \, x^{2} + 50 \, x e^{x} - {\left (2 \, x^{3} - 2 \, x^{2} + 2 \, {\left (x - 1\right )} e^{\left (2 \, x\right )} + 4 \, {\left (x^{2} - x\right )} e^{x} + x\right )} \log \relax (x) + 25 \, e^{\left (2 \, x\right )}\right )}}{\log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x*exp(x)^2+(-4*x^2-4*x)*exp(x)-4*x^2)*log(x)^3+((8*x^2-4*x-2)*exp(x)^2+(8*x^3+8*x^2-12*x)*exp(x
)+12*x^3-10*x^2+2*x)*log(x)^2+(-100*x*exp(x)^2+(-100*x^2-100*x)*exp(x)-100*x^2)*log(x)+50*exp(x)^2+100*exp(x)*
x+50*x^2)/x/log(x)^2,x, algorithm="fricas")

[Out]

-2*((x^2 + 2*x*e^x + e^(2*x))*log(x)^2 + 25*x^2 + 50*x*e^x - (2*x^3 - 2*x^2 + 2*(x - 1)*e^(2*x) + 4*(x^2 - x)*
e^x + x)*log(x) + 25*e^(2*x))/log(x)

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giac [B]  time = 0.13, size = 102, normalized size = 3.92 \begin {gather*} \frac {2 \, {\left (2 \, x^{3} \log \relax (x) + 4 \, x^{2} e^{x} \log \relax (x) - x^{2} \log \relax (x)^{2} - 2 \, x e^{x} \log \relax (x)^{2} - 2 \, x^{2} \log \relax (x) + 2 \, x e^{\left (2 \, x\right )} \log \relax (x) - 4 \, x e^{x} \log \relax (x) - e^{\left (2 \, x\right )} \log \relax (x)^{2} - 25 \, x^{2} - 50 \, x e^{x} + x \log \relax (x) - 2 \, e^{\left (2 \, x\right )} \log \relax (x) - 25 \, e^{\left (2 \, x\right )}\right )}}{\log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x*exp(x)^2+(-4*x^2-4*x)*exp(x)-4*x^2)*log(x)^3+((8*x^2-4*x-2)*exp(x)^2+(8*x^3+8*x^2-12*x)*exp(x
)+12*x^3-10*x^2+2*x)*log(x)^2+(-100*x*exp(x)^2+(-100*x^2-100*x)*exp(x)-100*x^2)*log(x)+50*exp(x)^2+100*exp(x)*
x+50*x^2)/x/log(x)^2,x, algorithm="giac")

[Out]

2*(2*x^3*log(x) + 4*x^2*e^x*log(x) - x^2*log(x)^2 - 2*x*e^x*log(x)^2 - 2*x^2*log(x) + 2*x*e^(2*x)*log(x) - 4*x
*e^x*log(x) - e^(2*x)*log(x)^2 - 25*x^2 - 50*x*e^x + x*log(x) - 2*e^(2*x)*log(x) - 25*e^(2*x))/log(x)

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maple [B]  time = 0.06, size = 79, normalized size = 3.04




method result size



risch \(\left (-2 x^{2}-4 \,{\mathrm e}^{x} x -2 \,{\mathrm e}^{2 x}\right ) \ln \relax (x )+4 x^{3}+8 \,{\mathrm e}^{x} x^{2}+4 x \,{\mathrm e}^{2 x}-4 x^{2}-8 \,{\mathrm e}^{x} x -4 \,{\mathrm e}^{2 x}+2 x -\frac {50 \left ({\mathrm e}^{2 x}+2 \,{\mathrm e}^{x} x +x^{2}\right )}{\ln \relax (x )}\) \(79\)



Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-4*x*exp(x)^2+(-4*x^2-4*x)*exp(x)-4*x^2)*ln(x)^3+((8*x^2-4*x-2)*exp(x)^2+(8*x^3+8*x^2-12*x)*exp(x)+12*x^
3-10*x^2+2*x)*ln(x)^2+(-100*x*exp(x)^2+(-100*x^2-100*x)*exp(x)-100*x^2)*ln(x)+50*exp(x)^2+100*exp(x)*x+50*x^2)
/x/ln(x)^2,x,method=_RETURNVERBOSE)

[Out]

(-2*x^2-4*exp(x)*x-2*exp(2*x))*ln(x)+4*x^3+8*exp(x)*x^2+4*x*exp(2*x)-4*x^2-8*exp(x)*x-4*exp(2*x)+2*x-50*(exp(2
*x)+2*exp(x)*x+x^2)/ln(x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} 4 \, x^{3} - 5 \, x^{2} + 2 \, {\left (2 \, x - 1\right )} e^{\left (2 \, x\right )} + 8 \, {\left (x^{2} - 2 \, x + 2\right )} e^{x} + 8 \, {\left (x - 1\right )} e^{x} + 2 \, x - \frac {2 \, x^{2} \log \relax (x)^{2} - x^{2} \log \relax (x) + 2 \, e^{\left (2 \, x\right )} \log \relax (x)^{2} + 4 \, {\left (x \log \relax (x)^{2} + 25 \, x - \log \relax (x)\right )} e^{x}}{\log \relax (x)} - 2 \, {\rm Ei}\left (2 \, x\right ) - 2 \, e^{\left (2 \, x\right )} - 12 \, e^{x} + 100 \, \Gamma \left (-1, -2 \, \log \relax (x)\right ) - 100 \, \int \frac {x}{\log \relax (x)}\,{d x} + 2 \, \int -\frac {{\left (50 \, x \log \relax (x) - \log \relax (x)^{2} - 25\right )} e^{\left (2 \, x\right )}}{x \log \relax (x)^{2}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x*exp(x)^2+(-4*x^2-4*x)*exp(x)-4*x^2)*log(x)^3+((8*x^2-4*x-2)*exp(x)^2+(8*x^3+8*x^2-12*x)*exp(x
)+12*x^3-10*x^2+2*x)*log(x)^2+(-100*x*exp(x)^2+(-100*x^2-100*x)*exp(x)-100*x^2)*log(x)+50*exp(x)^2+100*exp(x)*
x+50*x^2)/x/log(x)^2,x, algorithm="maxima")

[Out]

4*x^3 - 5*x^2 + 2*(2*x - 1)*e^(2*x) + 8*(x^2 - 2*x + 2)*e^x + 8*(x - 1)*e^x + 2*x - (2*x^2*log(x)^2 - x^2*log(
x) + 2*e^(2*x)*log(x)^2 + 4*(x*log(x)^2 + 25*x - log(x))*e^x)/log(x) - 2*Ei(2*x) - 2*e^(2*x) - 12*e^x + 100*ga
mma(-1, -2*log(x)) - 100*integrate(x/log(x), x) + 2*integrate(-(50*x*log(x) - log(x)^2 - 25)*e^(2*x)/(x*log(x)
^2), x)

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mupad [B]  time = 3.60, size = 87, normalized size = 3.35 \begin {gather*} 2\,x-\frac {50\,{\left (x+{\mathrm {e}}^x\right )}^2-100\,x\,\ln \relax (x)\,\left ({\mathrm {e}}^x+1\right )\,\left (x+{\mathrm {e}}^x\right )}{\ln \relax (x)}-\ln \relax (x)\,\left (2\,{\mathrm {e}}^{2\,x}+4\,x\,{\mathrm {e}}^x+2\,x^2\right )-{\mathrm {e}}^x\,\left (92\,x^2+108\,x\right )-{\mathrm {e}}^{2\,x}\,\left (96\,x+4\right )-104\,x^2+4\,x^3 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((50*exp(2*x) + log(x)^2*(2*x - exp(2*x)*(4*x - 8*x^2 + 2) - 10*x^2 + 12*x^3 + exp(x)*(8*x^2 - 12*x + 8*x^3
)) - log(x)*(100*x*exp(2*x) + exp(x)*(100*x + 100*x^2) + 100*x^2) + 100*x*exp(x) + 50*x^2 - log(x)^3*(4*x*exp(
2*x) + exp(x)*(4*x + 4*x^2) + 4*x^2))/(x*log(x)^2),x)

[Out]

2*x - (50*(x + exp(x))^2 - 100*x*log(x)*(exp(x) + 1)*(x + exp(x)))/log(x) - log(x)*(2*exp(2*x) + 4*x*exp(x) +
2*x^2) - exp(x)*(108*x + 92*x^2) - exp(2*x)*(96*x + 4) - 104*x^2 + 4*x^3

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sympy [B]  time = 0.43, size = 104, normalized size = 4.00 \begin {gather*} 4 x^{3} - 2 x^{2} \log {\relax (x )} - 4 x^{2} - \frac {50 x^{2}}{\log {\relax (x )}} + 2 x + \frac {\left (4 x \log {\relax (x )}^{2} - 2 \log {\relax (x )}^{3} - 4 \log {\relax (x )}^{2} - 50 \log {\relax (x )}\right ) e^{2 x} + \left (8 x^{2} \log {\relax (x )}^{2} - 4 x \log {\relax (x )}^{3} - 8 x \log {\relax (x )}^{2} - 100 x \log {\relax (x )}\right ) e^{x}}{\log {\relax (x )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x*exp(x)**2+(-4*x**2-4*x)*exp(x)-4*x**2)*ln(x)**3+((8*x**2-4*x-2)*exp(x)**2+(8*x**3+8*x**2-12*x
)*exp(x)+12*x**3-10*x**2+2*x)*ln(x)**2+(-100*x*exp(x)**2+(-100*x**2-100*x)*exp(x)-100*x**2)*ln(x)+50*exp(x)**2
+100*exp(x)*x+50*x**2)/x/ln(x)**2,x)

[Out]

4*x**3 - 2*x**2*log(x) - 4*x**2 - 50*x**2/log(x) + 2*x + ((4*x*log(x)**2 - 2*log(x)**3 - 4*log(x)**2 - 50*log(
x))*exp(2*x) + (8*x**2*log(x)**2 - 4*x*log(x)**3 - 8*x*log(x)**2 - 100*x*log(x))*exp(x))/log(x)**2

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