3.22.70 \(\int \frac {-16 e^{x^2} \log (2)+e^{x^2} (-32 x+32 x^2) \log (2) \log (-1+x)}{(-9+9 x) \log ^2(-1+x)} \, dx\)

Optimal. Leaf size=17 \[ \frac {16 e^{x^2} \log (2)}{9 \log (-1+x)} \]

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Rubi [B]  time = 0.42, antiderivative size = 43, normalized size of antiderivative = 2.53, number of steps used = 3, number of rules used = 3, integrand size = 45, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {6741, 12, 2288} \begin {gather*} \frac {16 e^{x^2} \log (2) \left (x \log (x-1)-x^2 \log (x-1)\right )}{9 (1-x) x \log ^2(x-1)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-16*E^x^2*Log[2] + E^x^2*(-32*x + 32*x^2)*Log[2]*Log[-1 + x])/((-9 + 9*x)*Log[-1 + x]^2),x]

[Out]

(16*E^x^2*Log[2]*(x*Log[-1 + x] - x^2*Log[-1 + x]))/(9*(1 - x)*x*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 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]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {16 e^{x^2} \log (2) \left (1+2 x \log (-1+x)-2 x^2 \log (-1+x)\right )}{(9-9 x) \log ^2(-1+x)} \, dx\\ &=(16 \log (2)) \int \frac {e^{x^2} \left (1+2 x \log (-1+x)-2 x^2 \log (-1+x)\right )}{(9-9 x) \log ^2(-1+x)} \, dx\\ &=\frac {16 e^{x^2} \log (2) \left (x \log (-1+x)-x^2 \log (-1+x)\right )}{9 (1-x) x \log ^2(-1+x)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 17, normalized size = 1.00 \begin {gather*} \frac {16 e^{x^2} \log (2)}{9 \log (-1+x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-16*E^x^2*Log[2] + E^x^2*(-32*x + 32*x^2)*Log[2]*Log[-1 + x])/((-9 + 9*x)*Log[-1 + x]^2),x]

[Out]

(16*E^x^2*Log[2])/(9*Log[-1 + x])

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fricas [A]  time = 0.54, size = 14, normalized size = 0.82 \begin {gather*} \frac {16 \, e^{\left (x^{2}\right )} \log \relax (2)}{9 \, \log \left (x - 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((32*x^2-32*x)*log(2)*exp(x^2)*log(x-1)-16*log(2)*exp(x^2))/(9*x-9)/log(x-1)^2,x, algorithm="fricas"
)

[Out]

16/9*e^(x^2)*log(2)/log(x - 1)

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giac [A]  time = 0.21, size = 14, normalized size = 0.82 \begin {gather*} \frac {16 \, e^{\left (x^{2}\right )} \log \relax (2)}{9 \, \log \left (x - 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((32*x^2-32*x)*log(2)*exp(x^2)*log(x-1)-16*log(2)*exp(x^2))/(9*x-9)/log(x-1)^2,x, algorithm="giac")

[Out]

16/9*e^(x^2)*log(2)/log(x - 1)

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maple [A]  time = 0.36, size = 15, normalized size = 0.88

method	result	size
norman	\(\frac {16 \,{\mathrm e}^{x^{2}} \ln \relax (2)}{9 \ln \left (x -1\right )}\)	\(15\)
risch	\(\frac {16 \,{\mathrm e}^{x^{2}} \ln \relax (2)}{9 \ln \left (x -1\right )}\)	\(15\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((32*x^2-32*x)*ln(2)*exp(x^2)*ln(x-1)-16*ln(2)*exp(x^2))/(9*x-9)/ln(x-1)^2,x,method=_RETURNVERBOSE)

[Out]

16/9*exp(x^2)/ln(x-1)*ln(2)

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maxima [A]  time = 0.58, size = 14, normalized size = 0.82 \begin {gather*} \frac {16 \, e^{\left (x^{2}\right )} \log \relax (2)}{9 \, \log \left (x - 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((32*x^2-32*x)*log(2)*exp(x^2)*log(x-1)-16*log(2)*exp(x^2))/(9*x-9)/log(x-1)^2,x, algorithm="maxima"
)

[Out]

16/9*e^(x^2)*log(2)/log(x - 1)

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mupad [B]  time = 1.46, size = 14, normalized size = 0.82 \begin {gather*} \frac {16\,{\mathrm {e}}^{x^2}\,\ln \relax (2)}{9\,\ln \left (x-1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(16*exp(x^2)*log(2) + log(x - 1)*exp(x^2)*log(2)*(32*x - 32*x^2))/(log(x - 1)^2*(9*x - 9)),x)

[Out]

(16*exp(x^2)*log(2))/(9*log(x - 1))

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sympy [A]  time = 0.33, size = 15, normalized size = 0.88 \begin {gather*} \frac {16 e^{x^{2}} \log {\relax (2 )}}{9 \log {\left (x - 1 \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((32*x**2-32*x)*ln(2)*exp(x**2)*ln(x-1)-16*ln(2)*exp(x**2))/(9*x-9)/ln(x-1)**2,x)

[Out]

16*exp(x**2)*log(2)/(9*log(x - 1))

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