3.64.42 \(\int \frac {-16+8 x-32 e^{4 x} x-4 \log (e^{2 e^{4 x}-2 x} x)}{4+12 x+9 x^2+(4 x+6 x^2) \log (e^{2 e^{4 x}-2 x} x)+x^2 \log ^2(e^{2 e^{4 x}-2 x} x)} \, dx\)

Optimal. Leaf size=29 \[ \frac {4}{2-x+x \left (4+\log \left (e^{2 e^{4 x}-2 x} x\right )\right )} \]

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Rubi [A]  time = 0.20, antiderivative size = 27, normalized size of antiderivative = 0.93, number of steps used = 3, number of rules used = 3, integrand size = 92, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.033, Rules used = {6688, 12, 6686} \begin {gather*} \frac {4}{3 x+x \log \left (e^{2 e^{4 x}-2 x} x\right )+2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-16 + 8*x - 32*E^(4*x)*x - 4*Log[E^(2*E^(4*x) - 2*x)*x])/(4 + 12*x + 9*x^2 + (4*x + 6*x^2)*Log[E^(2*E^(4*
x) - 2*x)*x] + x^2*Log[E^(2*E^(4*x) - 2*x)*x]^2),x]

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 6686

Int[(u_)*(y_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[(q*y^(m + 1))/(m + 1), x] /;  !F
alseQ[q]] /; FreeQ[m, x] && NeQ[m, -1]

Rule 6688

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

Rubi steps

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

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Mathematica [A]  time = 0.07, size = 27, normalized size = 0.93 \begin {gather*} \frac {4}{2+3 x+x \log \left (e^{2 e^{4 x}-2 x} x\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-16 + 8*x - 32*E^(4*x)*x - 4*Log[E^(2*E^(4*x) - 2*x)*x])/(4 + 12*x + 9*x^2 + (4*x + 6*x^2)*Log[E^(2
*E^(4*x) - 2*x)*x] + x^2*Log[E^(2*E^(4*x) - 2*x)*x]^2),x]

[Out]

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

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fricas [A]  time = 0.66, size = 25, normalized size = 0.86 \begin {gather*} \frac {4}{x \log \left (x e^{\left (-2 \, x + 2 \, e^{\left (4 \, x\right )}\right )}\right ) + 3 \, x + 2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-4*log(x*exp(exp(4*x))^2/exp(x)^2)-32*x*exp(4*x)+8*x-16)/(x^2*log(x*exp(exp(4*x))^2/exp(x)^2)^2+(6*
x^2+4*x)*log(x*exp(exp(4*x))^2/exp(x)^2)+9*x^2+12*x+4),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.19, size = 26, normalized size = 0.90 \begin {gather*} -\frac {4}{2 \, x^{2} - 2 \, x e^{\left (4 \, x\right )} - x \log \relax (x) - 3 \, x - 2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-4*log(x*exp(exp(4*x))^2/exp(x)^2)-32*x*exp(4*x)+8*x-16)/(x^2*log(x*exp(exp(4*x))^2/exp(x)^2)^2+(6*
x^2+4*x)*log(x*exp(exp(4*x))^2/exp(x)^2)+9*x^2+12*x+4),x, algorithm="giac")

[Out]

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

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maple [C]  time = 0.35, size = 399, normalized size = 13.76

method	result	size
risch	\(\frac {8 i}{\pi x \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i {\mathrm e}^{2 \,{\mathrm e}^{4 x}-2 x}\right ) \mathrm {csgn}\left (i x \,{\mathrm e}^{2 \,{\mathrm e}^{4 x}-2 x}\right )-\pi x \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x \,{\mathrm e}^{2 \,{\mathrm e}^{4 x}-2 x}\right )^{2}+\pi x \mathrm {csgn}\left (i {\mathrm e}^{{\mathrm e}^{4 x}}\right )^{2} \mathrm {csgn}\left (i {\mathrm e}^{2 \,{\mathrm e}^{4 x}}\right )-2 \pi x \,\mathrm {csgn}\left (i {\mathrm e}^{{\mathrm e}^{4 x}}\right ) \mathrm {csgn}\left (i {\mathrm e}^{2 \,{\mathrm e}^{4 x}}\right )^{2}+\pi x \mathrm {csgn}\left (i {\mathrm e}^{2 \,{\mathrm e}^{4 x}}\right )^{3}-\pi x \,\mathrm {csgn}\left (i {\mathrm e}^{2 \,{\mathrm e}^{4 x}}\right ) \mathrm {csgn}\left (i {\mathrm e}^{2 \,{\mathrm e}^{4 x}-2 x}\right )^{2}+\pi x \,\mathrm {csgn}\left (i {\mathrm e}^{2 \,{\mathrm e}^{4 x}}\right ) \mathrm {csgn}\left (i {\mathrm e}^{2 \,{\mathrm e}^{4 x}-2 x}\right ) \mathrm {csgn}\left (i {\mathrm e}^{-2 x}\right )+\pi x \mathrm {csgn}\left (i {\mathrm e}^{2 \,{\mathrm e}^{4 x}-2 x}\right )^{3}-\pi x \mathrm {csgn}\left (i {\mathrm e}^{2 \,{\mathrm e}^{4 x}-2 x}\right )^{2} \mathrm {csgn}\left (i {\mathrm e}^{-2 x}\right )-\pi x \,\mathrm {csgn}\left (i {\mathrm e}^{2 \,{\mathrm e}^{4 x}-2 x}\right ) \mathrm {csgn}\left (i x \,{\mathrm e}^{2 \,{\mathrm e}^{4 x}-2 x}\right )^{2}+\pi x \mathrm {csgn}\left (i x \,{\mathrm e}^{2 \,{\mathrm e}^{4 x}-2 x}\right )^{3}-\pi x \mathrm {csgn}\left (i {\mathrm e}^{x}\right )^{2} \mathrm {csgn}\left (i {\mathrm e}^{2 x}\right )+2 \pi x \,\mathrm {csgn}\left (i {\mathrm e}^{x}\right ) \mathrm {csgn}\left (i {\mathrm e}^{2 x}\right )^{2}-\pi x \mathrm {csgn}\left (i {\mathrm e}^{2 x}\right )^{3}+4 i x \ln \left ({\mathrm e}^{{\mathrm e}^{4 x}}\right )+6 i x -4 i x \ln \left ({\mathrm e}^{x}\right )+2 i x \ln \relax (x )+4 i}\)	\(399\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-4*ln(x*exp(exp(4*x))^2/exp(x)^2)-32*x*exp(4*x)+8*x-16)/(x^2*ln(x*exp(exp(4*x))^2/exp(x)^2)^2+(6*x^2+4*x)
*ln(x*exp(exp(4*x))^2/exp(x)^2)+9*x^2+12*x+4),x,method=_RETURNVERBOSE)

[Out]

8*I/(Pi*x*csgn(I*x)*csgn(I*exp(2*exp(4*x)-2*x))*csgn(I*x*exp(2*exp(4*x)-2*x))-Pi*x*csgn(I*x)*csgn(I*x*exp(2*ex
p(4*x)-2*x))^2+Pi*x*csgn(I*exp(exp(4*x)))^2*csgn(I*exp(2*exp(4*x)))-2*Pi*x*csgn(I*exp(exp(4*x)))*csgn(I*exp(2*
exp(4*x)))^2+Pi*x*csgn(I*exp(2*exp(4*x)))^3-Pi*x*csgn(I*exp(2*exp(4*x)))*csgn(I*exp(2*exp(4*x)-2*x))^2+Pi*x*cs
gn(I*exp(2*exp(4*x)))*csgn(I*exp(2*exp(4*x)-2*x))*csgn(I*exp(-2*x))+Pi*x*csgn(I*exp(2*exp(4*x)-2*x))^3-Pi*x*cs
gn(I*exp(2*exp(4*x)-2*x))^2*csgn(I*exp(-2*x))-Pi*x*csgn(I*exp(2*exp(4*x)-2*x))*csgn(I*x*exp(2*exp(4*x)-2*x))^2
+Pi*x*csgn(I*x*exp(2*exp(4*x)-2*x))^3-Pi*x*csgn(I*exp(x))^2*csgn(I*exp(2*x))+2*Pi*x*csgn(I*exp(x))*csgn(I*exp(
2*x))^2-Pi*x*csgn(I*exp(2*x))^3+4*I*x*ln(exp(exp(4*x)))+6*I*x-4*I*x*ln(exp(x))+2*I*x*ln(x)+4*I)

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maxima [A]  time = 0.41, size = 26, normalized size = 0.90 \begin {gather*} -\frac {4}{2 \, x^{2} - 2 \, x e^{\left (4 \, x\right )} - x \log \relax (x) - 3 \, x - 2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-4*log(x*exp(exp(4*x))^2/exp(x)^2)-32*x*exp(4*x)+8*x-16)/(x^2*log(x*exp(exp(4*x))^2/exp(x)^2)^2+(6*
x^2+4*x)*log(x*exp(exp(4*x))^2/exp(x)^2)+9*x^2+12*x+4),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 4.38, size = 23, normalized size = 0.79 \begin {gather*} \frac {4}{x\,\left (2\,{\mathrm {e}}^{4\,x}+\ln \relax (x)+3\right )-2\,x^2+2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(4*log(x*exp(2*exp(4*x))*exp(-2*x)) - 8*x + 32*x*exp(4*x) + 16)/(12*x + x^2*log(x*exp(2*exp(4*x))*exp(-2*
x))^2 + log(x*exp(2*exp(4*x))*exp(-2*x))*(4*x + 6*x^2) + 9*x^2 + 4),x)

[Out]

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

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sympy [A]  time = 0.27, size = 24, normalized size = 0.83 \begin {gather*} \frac {4}{x \log {\left (x e^{- 2 x} e^{2 e^{4 x}} \right )} + 3 x + 2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-4*ln(x*exp(exp(4*x))**2/exp(x)**2)-32*x*exp(4*x)+8*x-16)/(x**2*ln(x*exp(exp(4*x))**2/exp(x)**2)**2
+(6*x**2+4*x)*ln(x*exp(exp(4*x))**2/exp(x)**2)+9*x**2+12*x+4),x)

[Out]

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

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