3.87.11 \(\int \frac {-2+e^{4 x-4 x^2}+e^{4 x-4 x^2} (4-12 x+8 x^2) \log (-1+x)}{(-1+x) \log (4) \log ^2(-1+x)} \, dx\)

Optimal. Leaf size=26 \[ \frac {2-e^{4 x-4 x^2}}{\log (4) \log (-1+x)} \]

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Rubi [B]  time = 0.64, antiderivative size = 70, normalized size of antiderivative = 2.69, number of steps used = 7, number of rules used = 6, integrand size = 55, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.109, Rules used = {12, 6742, 2390, 2302, 30, 2288} \begin {gather*} \frac {2}{\log (4) \log (x-1)}-\frac {e^{4 (1-x) x} \left (2 x^2 \log (x-1)-3 x \log (x-1)+\log (x-1)\right )}{(1-2 x) (1-x) \log (4) \log ^2(x-1)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-2 + E^(4*x - 4*x^2) + E^(4*x - 4*x^2)*(4 - 12*x + 8*x^2)*Log[-1 + x])/((-1 + x)*Log[4]*Log[-1 + x]^2),x]

[Out]

2/(Log[4]*Log[-1 + x]) - (E^(4*(1 - x)*x)*(Log[-1 + x] - 3*x*Log[-1 + x] + 2*x^2*Log[-1 + x]))/((1 - 2*x)*(1 -
x)*Log[4]*Log[-1 + 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 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

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 2302

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/(x_), x_Symbol] :> Dist[1/(b*n), Subst[Int[x^p, x], x, a + b*L
og[c*x^n]], x] /; FreeQ[{a, b, c, n, p}, x]

Rule 2390

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_) + (g_.)*(x_))^(q_.), x_Symbol] :> Dist[1/
e, Subst[Int[((f*x)/d)^q*(a + b*Log[c*x^n])^p, x], x, d + e*x], x] /; FreeQ[{a, b, c, d, e, f, g, n, p, q}, x]
&& EqQ[e*f - d*g, 0]

Rule 6742

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

Rubi steps

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

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Mathematica [A]  time = 0.89, size = 23, normalized size = 0.88 \begin {gather*} \frac {2-e^{-4 (-1+x) x}}{\log (4) \log (-1+x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-2 + E^(4*x - 4*x^2) + E^(4*x - 4*x^2)*(4 - 12*x + 8*x^2)*Log[-1 + x])/((-1 + x)*Log[4]*Log[-1 + x]
^2),x]

[Out]

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

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fricas [A]  time = 0.89, size = 24, normalized size = 0.92 \begin {gather*} -\frac {e^{\left (-4 \, x^{2} + 4 \, x\right )} - 2}{2 \, \log \relax (2) \log \left (x - 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*((8*x^2-12*x+4)*exp(-x^2+x)^4*log(x-1)+exp(-x^2+x)^4-2)/(x-1)/log(2)/log(x-1)^2,x, algorithm="fr
icas")

[Out]

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

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giac [A]  time = 0.16, size = 24, normalized size = 0.92 \begin {gather*} -\frac {e^{\left (-4 \, x^{2} + 4 \, x\right )} - 2}{2 \, \log \relax (2) \log \left (x - 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*((8*x^2-12*x+4)*exp(-x^2+x)^4*log(x-1)+exp(-x^2+x)^4-2)/(x-1)/log(2)/log(x-1)^2,x, algorithm="gi
ac")

[Out]

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

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maple [A]  time = 0.41, size = 22, normalized size = 0.85

method	result	size
risch	\(-\frac {{\mathrm e}^{-4 x \left (x -1\right )}-2}{2 \ln \relax (2) \ln \left (x -1\right )}\)	\(22\)
default	\(\frac {-\frac {{\mathrm e}^{-4 x \left (x -1\right )}}{\ln \left (x -1\right )}+\frac {2}{\ln \left (x -1\right )}}{2 \ln \relax (2)}\)	\(31\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/2*((8*x^2-12*x+4)*exp(-x^2+x)^4*ln(x-1)+exp(-x^2+x)^4-2)/(x-1)/ln(2)/ln(x-1)^2,x,method=_RETURNVERBOSE)

[Out]

-1/2/ln(2)*(exp(-4*x*(x-1))-2)/ln(x-1)

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maxima [A]  time = 0.43, size = 32, normalized size = 1.23 \begin {gather*} -\frac {\frac {e^{\left (-4 \, x^{2} + 4 \, x\right )}}{\log \left (x - 1\right )} - \frac {2}{\log \left (x - 1\right )}}{2 \, \log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*((8*x^2-12*x+4)*exp(-x^2+x)^4*log(x-1)+exp(-x^2+x)^4-2)/(x-1)/log(2)/log(x-1)^2,x, algorithm="ma
xima")

[Out]

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

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mupad [B]  time = 5.66, size = 34, normalized size = 1.31 \begin {gather*} \frac {1}{\ln \left (x-1\right )\,\ln \relax (2)}-\frac {{\mathrm {e}}^{4\,x-4\,x^2}}{2\,\ln \left (x-1\right )\,\ln \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(4*x - 4*x^2)/2 + (log(x - 1)*exp(4*x - 4*x^2)*(8*x^2 - 12*x + 4))/2 - 1)/(log(x - 1)^2*log(2)*(x - 1)
),x)

[Out]

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

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sympy [A]  time = 0.34, size = 29, normalized size = 1.12 \begin {gather*} - \frac {e^{- 4 x^{2} + 4 x}}{2 \log {\relax (2 )} \log {\left (x - 1 \right )}} + \frac {1}{\log {\relax (2 )} \log {\left (x - 1 \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*((8*x**2-12*x+4)*exp(-x**2+x)**4*ln(x-1)+exp(-x**2+x)**4-2)/(x-1)/ln(2)/ln(x-1)**2,x)

[Out]

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

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