3.33.5 \(\int \frac {e^{-4+e^x} (16-16 x)+e^{-4+e^x} (e^x (16 x-16 x^2)+4 e^x x \log (-x^4)) \log (4-4 x+\log (-x^4))}{4 x-4 x^2+x \log (-x^4)} \, dx\)

Optimal. Leaf size=21 \[ 4 e^{-4+e^x} \log \left (4-4 x+\log \left (-x^4\right )\right ) \]

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Rubi [A]  time = 0.22, antiderivative size = 21, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 80, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.038, Rules used = {6688, 12, 2288} \begin {gather*} 4 e^{e^x-4} \log \left (\log \left (-x^4\right )-4 x+4\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [A]  time = 0.06, size = 21, normalized size = 1.00 \begin {gather*} 4 e^{-4+e^x} \log \left (4-4 x+\log \left (-x^4\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.53, size = 19, normalized size = 0.90 \begin {gather*} 4 \, e^{\left (e^{x} - 4\right )} \log \left (-4 \, x + \log \left (-x^{4}\right ) + 4\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.36, size = 19, normalized size = 0.90 \begin {gather*} 4 \, e^{\left (e^{x} - 4\right )} \log \left (-4 \, x + \log \left (-x^{4}\right ) + 4\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.19, size = 158, normalized size = 7.52

method	result	size
risch	\(4 \,{\mathrm e}^{{\mathrm e}^{x}-4} \ln \left (i \pi +4 \ln \relax (x )-\frac {i \pi \,\mathrm {csgn}\left (i x^{2}\right ) \left (-\mathrm {csgn}\left (i x^{2}\right )+\mathrm {csgn}\left (i x \right )\right )^{2}}{2}-\frac {i \pi \,\mathrm {csgn}\left (i x^{3}\right ) \left (-\mathrm {csgn}\left (i x^{3}\right )+\mathrm {csgn}\left (i x^{2}\right )\right ) \left (-\mathrm {csgn}\left (i x^{3}\right )+\mathrm {csgn}\left (i x \right )\right )}{2}-\frac {i \pi \,\mathrm {csgn}\left (i x^{4}\right ) \left (-\mathrm {csgn}\left (i x^{4}\right )+\mathrm {csgn}\left (i x^{3}\right )\right ) \left (-\mathrm {csgn}\left (i x^{4}\right )+\mathrm {csgn}\left (i x \right )\right )}{2}+i \pi \mathrm {csgn}\left (i x^{4}\right )^{2} \left (\mathrm {csgn}\left (i x^{4}\right )-1\right )-4 x +4\right )\)	\(158\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [C]  time = 0.47, size = 20, normalized size = 0.95 \begin {gather*} 4 \, e^{\left (e^{x} - 4\right )} \log \left (i \, \pi - 4 \, x + 4 \, \log \relax (x) + 4\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4*e^(e^x - 4)*log(I*pi - 4*x + 4*log(x) + 4)

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mupad [B]  time = 2.16, size = 19, normalized size = 0.90 \begin {gather*} 4\,\ln \left (\ln \left (-x^4\right )-4\,x+4\right )\,{\mathrm {e}}^{{\mathrm {e}}^x-4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(exp(x) - 4)*(16*x - 16) - log(log(-x^4) - 4*x + 4)*exp(exp(x) - 4)*(exp(x)*(16*x - 16*x^2) + 4*x*exp
(x)*log(-x^4)))/(4*x + x*log(-x^4) - 4*x^2),x)

[Out]

4*log(log(-x^4) - 4*x + 4)*exp(exp(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(((4*x*exp(x)*ln(-x**4)+(-16*x**2+16*x)*exp(x))*exp(exp(x)-4)*ln(ln(-x**4)-4*x+4)+(-16*x+16)*exp(exp(
x)-4))/(x*ln(-x**4)-4*x**2+4*x),x)

[Out]

Timed out

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