3.45.76 \(\int \frac {e^{-4+\frac {-2 x+e^4 \log (x+x^2) \log (\log (x))}{e^4 \log (x+x^2)}} ((12 x+24 x^2) \log (x)+(-12 x-12 x^2) \log (x) \log (x+x^2)+e^4 (6+6 x) \log ^2(x+x^2))}{(x+x^2) \log (x) \log ^2(x+x^2)} \, dx\)

Optimal. Leaf size=20 \[ 6 e^{-\frac {2 x}{e^4 \log \left (x+x^2\right )}} \log (x) \]

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Rubi [B]  time = 1.76, antiderivative size = 89, normalized size of antiderivative = 4.45, number of steps used = 4, number of rules used = 4, integrand size = 101, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.040, Rules used = {1593, 6688, 12, 2288} \begin {gather*} \frac {6 e^{-\frac {2 x}{e^4 \log (x (x+1))}-4} \log (x) (2 x-(x+1) \log (x (x+1))+1)}{(x+1) \left (\frac {2 x+1}{e^4 (x+1) \log ^2(x (x+1))}-\frac {1}{e^4 \log (x (x+1))}\right ) \log ^2(x (x+1))} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(-4 + (-2*x + E^4*Log[x + x^2]*Log[Log[x]])/(E^4*Log[x + x^2]))*((12*x + 24*x^2)*Log[x] + (-12*x - 12*x
^2)*Log[x]*Log[x + x^2] + E^4*(6 + 6*x)*Log[x + x^2]^2))/((x + x^2)*Log[x]*Log[x + x^2]^2),x]

[Out]

(6*E^(-4 - (2*x)/(E^4*Log[x*(1 + x)]))*Log[x]*(1 + 2*x - (1 + x)*Log[x*(1 + x)]))/((1 + x)*((1 + 2*x)/(E^4*(1
+ x)*Log[x*(1 + x)]^2) - 1/(E^4*Log[x*(1 + x)]))*Log[x*(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 1593

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

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 \frac {\exp \left (-4+\frac {-2 x+e^4 \log \left (x+x^2\right ) \log (\log (x))}{e^4 \log \left (x+x^2\right )}\right ) \left (\left (12 x+24 x^2\right ) \log (x)+\left (-12 x-12 x^2\right ) \log (x) \log \left (x+x^2\right )+e^4 (6+6 x) \log ^2\left (x+x^2\right )\right )}{x (1+x) \log (x) \log ^2\left (x+x^2\right )} \, dx\\ &=\int \frac {6 e^{-4-\frac {2 x}{e^4 \log (x (1+x))}} \left (e^4 (1+x) \log ^2(x (1+x))-2 x \log (x) (-1-2 x+(1+x) \log (x (1+x)))\right )}{x (1+x) \log ^2(x (1+x))} \, dx\\ &=6 \int \frac {e^{-4-\frac {2 x}{e^4 \log (x (1+x))}} \left (e^4 (1+x) \log ^2(x (1+x))-2 x \log (x) (-1-2 x+(1+x) \log (x (1+x)))\right )}{x (1+x) \log ^2(x (1+x))} \, dx\\ &=\frac {6 e^{-4-\frac {2 x}{e^4 \log (x (1+x))}} \log (x) (1+2 x-(1+x) \log (x (1+x)))}{(1+x) \left (\frac {1+2 x}{e^4 (1+x) \log ^2(x (1+x))}-\frac {1}{e^4 \log (x (1+x))}\right ) \log ^2(x (1+x))}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.05, size = 20, normalized size = 1.00 \begin {gather*} 6 e^{-\frac {2 x}{e^4 \log (x (1+x))}} \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(-4 + (-2*x + E^4*Log[x + x^2]*Log[Log[x]])/(E^4*Log[x + x^2]))*((12*x + 24*x^2)*Log[x] + (-12*x
- 12*x^2)*Log[x]*Log[x + x^2] + E^4*(6 + 6*x)*Log[x + x^2]^2))/((x + x^2)*Log[x]*Log[x + x^2]^2),x]

[Out]

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

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fricas [B]  time = 0.60, size = 42, normalized size = 2.10 \begin {gather*} 6 \, e^{\left (\frac {{\left (e^{4} \log \left (x^{2} + x\right ) \log \left (\log \relax (x)\right ) - 4 \, e^{4} \log \left (x^{2} + x\right ) - 2 \, x\right )} e^{\left (-4\right )}}{\log \left (x^{2} + x\right )} + 4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((6*x+6)*exp(4)*log(x^2+x)^2+(-12*x^2-12*x)*log(x)*log(x^2+x)+(24*x^2+12*x)*log(x))*exp((exp(4)*log(
x^2+x)*log(log(x))-2*x)/exp(4)/log(x^2+x))/(x^2+x)/exp(4)/log(x)/log(x^2+x)^2,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 1.52, size = 20, normalized size = 1.00 \begin {gather*} 6 \, e^{\left (-\frac {2 \, x e^{\left (-4\right )}}{\log \left (x^{2} + x\right )} + \log \left (\log \relax (x)\right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((6*x+6)*exp(4)*log(x^2+x)^2+(-12*x^2-12*x)*log(x)*log(x^2+x)+(24*x^2+12*x)*log(x))*exp((exp(4)*log(
x^2+x)*log(log(x))-2*x)/exp(4)/log(x^2+x))/(x^2+x)/exp(4)/log(x)/log(x^2+x)^2,x, algorithm="giac")

[Out]

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

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maple [F]  time = 0.05, size = 0, normalized size = 0.00 \[\int \frac {\left (\left (6 x +6\right ) {\mathrm e}^{4} \ln \left (x^{2}+x \right )^{2}+\left (-12 x^{2}-12 x \right ) \ln \relax (x ) \ln \left (x^{2}+x \right )+\left (24 x^{2}+12 x \right ) \ln \relax (x )\right ) {\mathrm e}^{\frac {\left ({\mathrm e}^{4} \ln \left (x^{2}+x \right ) \ln \left (\ln \relax (x )\right )-2 x \right ) {\mathrm e}^{-4}}{\ln \left (x^{2}+x \right )}} {\mathrm e}^{-4}}{\left (x^{2}+x \right ) \ln \relax (x ) \ln \left (x^{2}+x \right )^{2}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((6*x+6)*exp(4)*ln(x^2+x)^2+(-12*x^2-12*x)*ln(x)*ln(x^2+x)+(24*x^2+12*x)*ln(x))*exp((exp(4)*ln(x^2+x)*ln(l
n(x))-2*x)/exp(4)/ln(x^2+x))/(x^2+x)/exp(4)/ln(x)/ln(x^2+x)^2,x)

[Out]

int(((6*x+6)*exp(4)*ln(x^2+x)^2+(-12*x^2-12*x)*ln(x)*ln(x^2+x)+(24*x^2+12*x)*ln(x))*exp((exp(4)*ln(x^2+x)*ln(l
n(x))-2*x)/exp(4)/ln(x^2+x))/(x^2+x)/exp(4)/ln(x)/ln(x^2+x)^2,x)

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: RuntimeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((6*x+6)*exp(4)*log(x^2+x)^2+(-12*x^2-12*x)*log(x)*log(x^2+x)+(24*x^2+12*x)*log(x))*exp((exp(4)*log(
x^2+x)*log(log(x))-2*x)/exp(4)/log(x^2+x))/(x^2+x)/exp(4)/log(x)/log(x^2+x)^2,x, algorithm="maxima")

[Out]

Exception raised: RuntimeError >> ECL says: In function CAR, the value of the first argument is  0which is not
of the expected type LIST

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mupad [F]  time = 0.00, size = -1, normalized size = -0.05 \begin {gather*} \int \frac {{\mathrm {e}}^{-\frac {{\mathrm {e}}^{-4}\,\left (2\,x-\ln \left (x^2+x\right )\,\ln \left (\ln \relax (x)\right )\,{\mathrm {e}}^4\right )}{\ln \left (x^2+x\right )}}\,{\mathrm {e}}^{-4}\,\left ({\mathrm {e}}^4\,\left (6\,x+6\right )\,{\ln \left (x^2+x\right )}^2-\ln \relax (x)\,\left (12\,x^2+12\,x\right )\,\ln \left (x^2+x\right )+\ln \relax (x)\,\left (24\,x^2+12\,x\right )\right )}{{\ln \left (x^2+x\right )}^2\,\ln \relax (x)\,\left (x^2+x\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(-(exp(-4)*(2*x - log(x + x^2)*log(log(x))*exp(4)))/log(x + x^2))*exp(-4)*(log(x)*(12*x + 24*x^2) + lo
g(x + x^2)^2*exp(4)*(6*x + 6) - log(x + x^2)*log(x)*(12*x + 12*x^2)))/(log(x + x^2)^2*log(x)*(x + x^2)),x)

[Out]

int((exp(-(exp(-4)*(2*x - log(x + x^2)*log(log(x))*exp(4)))/log(x + x^2))*exp(-4)*(log(x)*(12*x + 24*x^2) + lo
g(x + x^2)^2*exp(4)*(6*x + 6) - log(x + x^2)*log(x)*(12*x + 12*x^2)))/(log(x + x^2)^2*log(x)*(x + x^2)), x)

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sympy [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SympifyError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((6*x+6)*exp(4)*ln(x**2+x)**2+(-12*x**2-12*x)*ln(x)*ln(x**2+x)+(24*x**2+12*x)*ln(x))*exp((exp(4)*ln(
x**2+x)*ln(ln(x))-2*x)/exp(4)/ln(x**2+x))/(x**2+x)/exp(4)/ln(x)/ln(x**2+x)**2,x)

[Out]

Exception raised: SympifyError

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