3.65.14 \(\int \frac {e^3 (3-12 x-6 x^2)+e^x (-8 x^4+144 x^5)+(e^3 (3+3 x)+e^x (8 x^3-144 x^4)) \log (x)+e^x (-2 x^2+36 x^3) \log ^2(x)}{8 e^x x^4-8 e^x x^3 \log (x)+2 e^x x^2 \log ^2(x)} \, dx\)

Optimal. Leaf size=31 \[ -x+9 x^2+\frac {3 e^{3-x}}{x (4 x-2 \log (x))} \]

________________________________________________________________________________________

Rubi [A]  time = 1.60, antiderivative size = 47, normalized size of antiderivative = 1.52, number of steps used = 5, number of rules used = 4, integrand size = 111, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.036, Rules used = {6741, 12, 6742, 2288} \begin {gather*} \frac {3 e^{3-x} \left (2 x^2-x \log (x)\right )}{2 x^2 (2 x-\log (x))^2}+\frac {1}{36} (1-18 x)^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^3*(3 - 12*x - 6*x^2) + E^x*(-8*x^4 + 144*x^5) + (E^3*(3 + 3*x) + E^x*(8*x^3 - 144*x^4))*Log[x] + E^x*(-
2*x^2 + 36*x^3)*Log[x]^2)/(8*E^x*x^4 - 8*E^x*x^3*Log[x] + 2*E^x*x^2*Log[x]^2),x]

[Out]

(1 - 18*x)^2/36 + (3*E^(3 - x)*(2*x^2 - x*Log[x]))/(2*x^2*(2*x - Log[x])^2)

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 6741

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

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 \frac {e^{-x} \left (e^3 \left (3-12 x-6 x^2\right )+e^x \left (-8 x^4+144 x^5\right )+\left (e^3 (3+3 x)+e^x \left (8 x^3-144 x^4\right )\right ) \log (x)+e^x \left (-2 x^2+36 x^3\right ) \log ^2(x)\right )}{2 x^2 (2 x-\log (x))^2} \, dx\\ &=\frac {1}{2} \int \frac {e^{-x} \left (e^3 \left (3-12 x-6 x^2\right )+e^x \left (-8 x^4+144 x^5\right )+\left (e^3 (3+3 x)+e^x \left (8 x^3-144 x^4\right )\right ) \log (x)+e^x \left (-2 x^2+36 x^3\right ) \log ^2(x)\right )}{x^2 (2 x-\log (x))^2} \, dx\\ &=\frac {1}{2} \int \left (2 (-1+18 x)-\frac {3 e^{3-x} \left (-1+4 x+2 x^2-\log (x)-x \log (x)\right )}{x^2 (2 x-\log (x))^2}\right ) \, dx\\ &=\frac {1}{36} (1-18 x)^2-\frac {3}{2} \int \frac {e^{3-x} \left (-1+4 x+2 x^2-\log (x)-x \log (x)\right )}{x^2 (2 x-\log (x))^2} \, dx\\ &=\frac {1}{36} (1-18 x)^2+\frac {3 e^{3-x} \left (2 x^2-x \log (x)\right )}{2 x^2 (2 x-\log (x))^2}\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.10, size = 33, normalized size = 1.06 \begin {gather*} \frac {1}{2} \left (-2 x+18 x^2-\frac {3 e^{3-x}}{x (-2 x+\log (x))}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^3*(3 - 12*x - 6*x^2) + E^x*(-8*x^4 + 144*x^5) + (E^3*(3 + 3*x) + E^x*(8*x^3 - 144*x^4))*Log[x] +
E^x*(-2*x^2 + 36*x^3)*Log[x]^2)/(8*E^x*x^4 - 8*E^x*x^3*Log[x] + 2*E^x*x^2*Log[x]^2),x]

[Out]

(-2*x + 18*x^2 - (3*E^(3 - x))/(x*(-2*x + Log[x])))/2

________________________________________________________________________________________

fricas [A]  time = 0.61, size = 56, normalized size = 1.81 \begin {gather*} -\frac {2 \, {\left (9 \, x^{3} - x^{2}\right )} e^{x} \log \relax (x) - 4 \, {\left (9 \, x^{4} - x^{3}\right )} e^{x} - 3 \, e^{3}}{2 \, {\left (2 \, x^{2} e^{x} - x e^{x} \log \relax (x)\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((36*x^3-2*x^2)*exp(x)*log(x)^2+((-144*x^4+8*x^3)*exp(x)+(3*x+3)*exp(3))*log(x)+(144*x^5-8*x^4)*exp(
x)+(-6*x^2-12*x+3)*exp(3))/(2*x^2*exp(x)*log(x)^2-8*x^3*exp(x)*log(x)+8*exp(x)*x^4),x, algorithm="fricas")

[Out]

-1/2*(2*(9*x^3 - x^2)*e^x*log(x) - 4*(9*x^4 - x^3)*e^x - 3*e^3)/(2*x^2*e^x - x*e^x*log(x))

________________________________________________________________________________________

giac [A]  time = 0.16, size = 48, normalized size = 1.55 \begin {gather*} \frac {36 \, x^{4} - 18 \, x^{3} \log \relax (x) - 4 \, x^{3} + 2 \, x^{2} \log \relax (x) + 3 \, e^{\left (-x + 3\right )}}{2 \, {\left (2 \, x^{2} - x \log \relax (x)\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((36*x^3-2*x^2)*exp(x)*log(x)^2+((-144*x^4+8*x^3)*exp(x)+(3*x+3)*exp(3))*log(x)+(144*x^5-8*x^4)*exp(
x)+(-6*x^2-12*x+3)*exp(3))/(2*x^2*exp(x)*log(x)^2-8*x^3*exp(x)*log(x)+8*exp(x)*x^4),x, algorithm="giac")

[Out]

1/2*(36*x^4 - 18*x^3*log(x) - 4*x^3 + 2*x^2*log(x) + 3*e^(-x + 3))/(2*x^2 - x*log(x))

________________________________________________________________________________________

maple [A]  time = 0.08, size = 31, normalized size = 1.00

method	result	size
risch	\(9 x^{2}-x +\frac {3 \,{\mathrm e}^{3-x}}{2 \left (2 x -\ln \relax (x )\right ) x}\)	\(31\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((36*x^3-2*x^2)*exp(x)*ln(x)^2+((-144*x^4+8*x^3)*exp(x)+(3*x+3)*exp(3))*ln(x)+(144*x^5-8*x^4)*exp(x)+(-6*x
^2-12*x+3)*exp(3))/(2*x^2*exp(x)*ln(x)^2-8*x^3*exp(x)*ln(x)+8*exp(x)*x^4),x,method=_RETURNVERBOSE)

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 0.42, size = 49, normalized size = 1.58 \begin {gather*} \frac {36 \, x^{4} - 4 \, x^{3} - 2 \, {\left (9 \, x^{3} - x^{2}\right )} \log \relax (x) + 3 \, e^{\left (-x + 3\right )}}{2 \, {\left (2 \, x^{2} - x \log \relax (x)\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((36*x^3-2*x^2)*exp(x)*log(x)^2+((-144*x^4+8*x^3)*exp(x)+(3*x+3)*exp(3))*log(x)+(144*x^5-8*x^4)*exp(
x)+(-6*x^2-12*x+3)*exp(3))/(2*x^2*exp(x)*log(x)^2-8*x^3*exp(x)*log(x)+8*exp(x)*x^4),x, algorithm="maxima")

[Out]

1/2*(36*x^4 - 4*x^3 - 2*(9*x^3 - x^2)*log(x) + 3*e^(-x + 3))/(2*x^2 - x*log(x))

________________________________________________________________________________________

mupad [B]  time = 4.49, size = 105, normalized size = 3.39 \begin {gather*} \frac {\frac {3\,{\mathrm {e}}^{-x}\,\left (2\,{\mathrm {e}}^3\,x^2+4\,{\mathrm {e}}^3\,x-{\mathrm {e}}^3\right )}{2\,x\,\left (2\,x-1\right )}-\frac {3\,{\mathrm {e}}^{-x}\,\ln \relax (x)\,\left ({\mathrm {e}}^3+x\,{\mathrm {e}}^3\right )}{2\,x\,\left (2\,x-1\right )}}{2\,x-\ln \relax (x)}-x+9\,x^2+\frac {{\mathrm {e}}^{-x}\,\left (\frac {3\,{\mathrm {e}}^3}{4}+\frac {3\,x\,{\mathrm {e}}^3}{4}\right )}{\frac {x}{2}-x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(x)*(8*x^4 - 144*x^5) + exp(3)*(12*x + 6*x^2 - 3) - log(x)*(exp(x)*(8*x^3 - 144*x^4) + exp(3)*(3*x +
3)) + exp(x)*log(x)^2*(2*x^2 - 36*x^3))/(8*x^4*exp(x) + 2*x^2*exp(x)*log(x)^2 - 8*x^3*exp(x)*log(x)),x)

[Out]

((3*exp(-x)*(4*x*exp(3) - exp(3) + 2*x^2*exp(3)))/(2*x*(2*x - 1)) - (3*exp(-x)*log(x)*(exp(3) + x*exp(3)))/(2*
x*(2*x - 1)))/(2*x - log(x)) - x + 9*x^2 + (exp(-x)*((3*exp(3))/4 + (3*x*exp(3))/4))/(x/2 - x^2)

________________________________________________________________________________________

sympy [A]  time = 0.36, size = 26, normalized size = 0.84 \begin {gather*} 9 x^{2} - x + \frac {3 e^{3} e^{- x}}{4 x^{2} - 2 x \log {\relax (x )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((36*x**3-2*x**2)*exp(x)*ln(x)**2+((-144*x**4+8*x**3)*exp(x)+(3*x+3)*exp(3))*ln(x)+(144*x**5-8*x**4)
*exp(x)+(-6*x**2-12*x+3)*exp(3))/(2*x**2*exp(x)*ln(x)**2-8*x**3*exp(x)*ln(x)+8*exp(x)*x**4),x)

[Out]

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

________________________________________________________________________________________