3.65.10 \(\int \frac {6 e^{2 x} x^2+e^x (2+8 x-8 x^2+16 x^3)+(-2 e^{2 x} x^2+e^x (-2 x+x^2-4 x^3)) \log (\frac {2-x+2 e^x x+4 x^2}{2 x})}{2 x-x^2+2 e^x x^2+4 x^3} \, dx\)

Optimal. Leaf size=33 \[ -e^3-e^x \left (-4+\log \left (e^x+\frac {2-x}{2 x}+2 x\right )\right ) \]

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Rubi [A]  time = 3.45, antiderivative size = 25, normalized size of antiderivative = 0.76, number of steps used = 12, number of rules used = 4, integrand size = 107, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.037, Rules used = {6688, 6742, 2194, 2554} \begin {gather*} 4 e^x-e^x \log \left (2 x+e^x+\frac {1}{x}-\frac {1}{2}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(6*E^(2*x)*x^2 + E^x*(2 + 8*x - 8*x^2 + 16*x^3) + (-2*E^(2*x)*x^2 + E^x*(-2*x + x^2 - 4*x^3))*Log[(2 - x +
2*E^x*x + 4*x^2)/(2*x)])/(2*x - x^2 + 2*E^x*x^2 + 4*x^3),x]

[Out]

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

Rule 2194

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rule 2554

Int[Log[u_]*(v_), x_Symbol] :> With[{w = IntHide[v, x]}, Dist[Log[u], w, x] - Int[SimplifyIntegrand[(w*D[u, x]
)/u, x], x] /; InverseFunctionFreeQ[w, x]] /; InverseFunctionFreeQ[u, x]

Rule 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, 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 \frac {e^x \left (2+8 x+\left (-8+6 e^x\right ) x^2+16 x^3-x \left (2+\left (-1+2 e^x\right ) x+4 x^2\right ) \log \left (-\frac {1}{2}+e^x+\frac {1}{x}+2 x\right )\right )}{x \left (2+\left (-1+2 e^x\right ) x+4 x^2\right )} \, dx\\ &=\int \left (3 e^x+\frac {e^x \left (2+2 x-5 x^2+4 x^3\right )}{x \left (2-x+2 e^x x+4 x^2\right )}-e^x \log \left (-\frac {1}{2}+e^x+\frac {1}{x}+2 x\right )\right ) \, dx\\ &=3 \int e^x \, dx+\int \frac {e^x \left (2+2 x-5 x^2+4 x^3\right )}{x \left (2-x+2 e^x x+4 x^2\right )} \, dx-\int e^x \log \left (-\frac {1}{2}+e^x+\frac {1}{x}+2 x\right ) \, dx\\ &=3 e^x-e^x \log \left (-\frac {1}{2}+e^x+\frac {1}{x}+2 x\right )+\int \frac {e^x \left (2+e^x-\frac {1}{x^2}\right )}{-\frac {1}{2}+e^x+\frac {1}{x}+2 x} \, dx+\int \left (\frac {2 e^x}{2-x+2 e^x x+4 x^2}+\frac {2 e^x}{x \left (2-x+2 e^x x+4 x^2\right )}-\frac {5 e^x x}{2-x+2 e^x x+4 x^2}+\frac {4 e^x x^2}{2-x+2 e^x x+4 x^2}\right ) \, dx\\ &=3 e^x-e^x \log \left (-\frac {1}{2}+e^x+\frac {1}{x}+2 x\right )+2 \int \frac {e^x}{2-x+2 e^x x+4 x^2} \, dx+2 \int \frac {e^x}{x \left (2-x+2 e^x x+4 x^2\right )} \, dx+4 \int \frac {e^x x^2}{2-x+2 e^x x+4 x^2} \, dx-5 \int \frac {e^x x}{2-x+2 e^x x+4 x^2} \, dx+\int \left (e^x-\frac {e^x \left (2+2 x-5 x^2+4 x^3\right )}{x \left (2-x+2 e^x x+4 x^2\right )}\right ) \, dx\\ &=3 e^x-e^x \log \left (-\frac {1}{2}+e^x+\frac {1}{x}+2 x\right )+2 \int \frac {e^x}{2-x+2 e^x x+4 x^2} \, dx+2 \int \frac {e^x}{x \left (2-x+2 e^x x+4 x^2\right )} \, dx+4 \int \frac {e^x x^2}{2-x+2 e^x x+4 x^2} \, dx-5 \int \frac {e^x x}{2-x+2 e^x x+4 x^2} \, dx+\int e^x \, dx-\int \frac {e^x \left (2+2 x-5 x^2+4 x^3\right )}{x \left (2-x+2 e^x x+4 x^2\right )} \, dx\\ &=4 e^x-e^x \log \left (-\frac {1}{2}+e^x+\frac {1}{x}+2 x\right )+2 \int \frac {e^x}{2-x+2 e^x x+4 x^2} \, dx+2 \int \frac {e^x}{x \left (2-x+2 e^x x+4 x^2\right )} \, dx+4 \int \frac {e^x x^2}{2-x+2 e^x x+4 x^2} \, dx-5 \int \frac {e^x x}{2-x+2 e^x x+4 x^2} \, dx-\int \left (\frac {2 e^x}{2-x+2 e^x x+4 x^2}+\frac {2 e^x}{x \left (2-x+2 e^x x+4 x^2\right )}-\frac {5 e^x x}{2-x+2 e^x x+4 x^2}+\frac {4 e^x x^2}{2-x+2 e^x x+4 x^2}\right ) \, dx\\ &=4 e^x-e^x \log \left (-\frac {1}{2}+e^x+\frac {1}{x}+2 x\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 1.16, size = 21, normalized size = 0.64 \begin {gather*} -e^x \left (-4+\log \left (-\frac {1}{2}+e^x+\frac {1}{x}+2 x\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(6*E^(2*x)*x^2 + E^x*(2 + 8*x - 8*x^2 + 16*x^3) + (-2*E^(2*x)*x^2 + E^x*(-2*x + x^2 - 4*x^3))*Log[(2
- x + 2*E^x*x + 4*x^2)/(2*x)])/(2*x - x^2 + 2*E^x*x^2 + 4*x^3),x]

[Out]

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

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fricas [A]  time = 0.62, size = 30, normalized size = 0.91 \begin {gather*} -e^{x} \log \left (\frac {4 \, x^{2} + 2 \, x e^{x} - x + 2}{2 \, x}\right ) + 4 \, e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*exp(x)^2*x^2+(-4*x^3+x^2-2*x)*exp(x))*log(1/2*(2*exp(x)*x+4*x^2-x+2)/x)+6*exp(x)^2*x^2+(16*x^3-
8*x^2+8*x+2)*exp(x))/(2*exp(x)*x^2+4*x^3-x^2+2*x),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.25, size = 30, normalized size = 0.91 \begin {gather*} -e^{x} \log \left (\frac {4 \, x^{2} + 2 \, x e^{x} - x + 2}{2 \, x}\right ) + 4 \, e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*exp(x)^2*x^2+(-4*x^3+x^2-2*x)*exp(x))*log(1/2*(2*exp(x)*x+4*x^2-x+2)/x)+6*exp(x)^2*x^2+(16*x^3-
8*x^2+8*x+2)*exp(x))/(2*exp(x)*x^2+4*x^3-x^2+2*x),x, algorithm="giac")

[Out]

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

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maple [C]  time = 0.14, size = 193, normalized size = 5.85

method	result	size
risch	\(-{\mathrm e}^{x} \ln \left (\frac {1}{2}+x^{2}+\left (\frac {{\mathrm e}^{x}}{2}-\frac {1}{4}\right ) x \right )+{\mathrm e}^{x} \ln \relax (x )+\frac {i \pi \,\mathrm {csgn}\left (\frac {i}{x}\right ) \mathrm {csgn}\left (i \left (\frac {1}{2}+x^{2}+\left (\frac {{\mathrm e}^{x}}{2}-\frac {1}{4}\right ) x \right )\right ) \mathrm {csgn}\left (\frac {i \left (\frac {1}{2}+x^{2}+\left (\frac {{\mathrm e}^{x}}{2}-\frac {1}{4}\right ) x \right )}{x}\right ) {\mathrm e}^{x}}{2}-\frac {i \pi \,\mathrm {csgn}\left (\frac {i}{x}\right ) \mathrm {csgn}\left (\frac {i \left (\frac {1}{2}+x^{2}+\left (\frac {{\mathrm e}^{x}}{2}-\frac {1}{4}\right ) x \right )}{x}\right )^{2} {\mathrm e}^{x}}{2}-\frac {i \pi \,\mathrm {csgn}\left (i \left (\frac {1}{2}+x^{2}+\left (\frac {{\mathrm e}^{x}}{2}-\frac {1}{4}\right ) x \right )\right ) \mathrm {csgn}\left (\frac {i \left (\frac {1}{2}+x^{2}+\left (\frac {{\mathrm e}^{x}}{2}-\frac {1}{4}\right ) x \right )}{x}\right )^{2} {\mathrm e}^{x}}{2}+\frac {i \pi \mathrm {csgn}\left (\frac {i \left (\frac {1}{2}+x^{2}+\left (\frac {{\mathrm e}^{x}}{2}-\frac {1}{4}\right ) x \right )}{x}\right )^{3} {\mathrm e}^{x}}{2}-{\mathrm e}^{x} \ln \relax (2)+4 \,{\mathrm e}^{x}\)	\(193\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-2*exp(x)^2*x^2+(-4*x^3+x^2-2*x)*exp(x))*ln(1/2*(2*exp(x)*x+4*x^2-x+2)/x)+6*exp(x)^2*x^2+(16*x^3-8*x^2+8
*x+2)*exp(x))/(2*exp(x)*x^2+4*x^3-x^2+2*x),x,method=_RETURNVERBOSE)

[Out]

-exp(x)*ln(1/2+x^2+(1/2*exp(x)-1/4)*x)+exp(x)*ln(x)+1/2*I*Pi*csgn(I/x)*csgn(I*(1/2+x^2+(1/2*exp(x)-1/4)*x))*cs
gn(I/x*(1/2+x^2+(1/2*exp(x)-1/4)*x))*exp(x)-1/2*I*Pi*csgn(I/x)*csgn(I/x*(1/2+x^2+(1/2*exp(x)-1/4)*x))^2*exp(x)
-1/2*I*Pi*csgn(I*(1/2+x^2+(1/2*exp(x)-1/4)*x))*csgn(I/x*(1/2+x^2+(1/2*exp(x)-1/4)*x))^2*exp(x)+1/2*I*Pi*csgn(I
/x*(1/2+x^2+(1/2*exp(x)-1/4)*x))^3*exp(x)-exp(x)*ln(2)+4*exp(x)

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maxima [A]  time = 0.50, size = 30, normalized size = 0.91 \begin {gather*} {\left (\log \relax (2) + \log \relax (x) + 4\right )} e^{x} - e^{x} \log \left (4 \, x^{2} + 2 \, x e^{x} - x + 2\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*exp(x)^2*x^2+(-4*x^3+x^2-2*x)*exp(x))*log(1/2*(2*exp(x)*x+4*x^2-x+2)/x)+6*exp(x)^2*x^2+(16*x^3-
8*x^2+8*x+2)*exp(x))/(2*exp(x)*x^2+4*x^3-x^2+2*x),x, algorithm="maxima")

[Out]

(log(2) + log(x) + 4)*e^x - e^x*log(4*x^2 + 2*x*e^x - x + 2)

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mupad [B]  time = 4.30, size = 25, normalized size = 0.76 \begin {gather*} -{\mathrm {e}}^x\,\left (\ln \left (\frac {x\,{\mathrm {e}}^x-\frac {x}{2}+2\,x^2+1}{x}\right )-4\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((6*x^2*exp(2*x) - log((x*exp(x) - x/2 + 2*x^2 + 1)/x)*(2*x^2*exp(2*x) + exp(x)*(2*x - x^2 + 4*x^3)) + exp(
x)*(8*x - 8*x^2 + 16*x^3 + 2))/(2*x + 2*x^2*exp(x) - x^2 + 4*x^3),x)

[Out]

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

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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(((-2*exp(x)**2*x**2+(-4*x**3+x**2-2*x)*exp(x))*ln(1/2*(2*exp(x)*x+4*x**2-x+2)/x)+6*exp(x)**2*x**2+(1
6*x**3-8*x**2+8*x+2)*exp(x))/(2*exp(x)*x**2+4*x**3-x**2+2*x),x)

[Out]

Timed out

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