3.34.16 \(\int \frac {(-2 x^2+e^{e^{e^{x+x^2}}+e^{x+x^2}+x+x^2} (2 x^2+4 x^3)) \log (-e^{e^{e^{x+x^2}}}+x)+(2 e^{e^{e^{x+x^2}}} x-2 x^2) \log ^2(-e^{e^{e^{x+x^2}}}+x)}{27 e^{e^{e^{x+x^2}}}-27 x} \, dx\)

Optimal. Leaf size=25 \[ \frac {1}{27} x^2 \log ^2\left (-e^{e^{e^{x+x^2}}}+x\right ) \]

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Rubi [A]  time = 5.76, antiderivative size = 25, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 6, integrand size = 118, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.051, Rules used = {6688, 12, 14, 6742, 2551, 6687} \begin {gather*} \frac {1}{27} x^2 \log ^2\left (x-e^{e^{e^{x^2+x}}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((-2*x^2 + E^(E^E^(x + x^2) + E^(x + x^2) + x + x^2)*(2*x^2 + 4*x^3))*Log[-E^E^E^(x + x^2) + x] + (2*E^E^E
^(x + x^2)*x - 2*x^2)*Log[-E^E^E^(x + x^2) + x]^2)/(27*E^E^E^(x + x^2) - 27*x),x]

[Out]

(x^2*Log[-E^E^E^(x + x^2) + x]^2)/27

Rule 12

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

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 2551

Int[Log[u_]*((a_.) + (b_.)*(x_))^(m_.), x_Symbol] :> Simp[((a + b*x)^(m + 1)*Log[u])/(b*(m + 1)), x] - Dist[1/
(b*(m + 1)), Int[SimplifyIntegrand[((a + b*x)^(m + 1)*D[u, x])/u, x], x], x] /; FreeQ[{a, b, m}, x] && Inverse
FunctionFreeQ[u, x] && NeQ[m, -1]

Rule 6687

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

Rule 6688

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

Rule 6742

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {2}{27} x \log \left (-e^{e^{e^{x+x^2}}}+x\right ) \left (\frac {x \left (-1+e^{e^{e^{x+x^2}}+e^{x+x^2}+x+x^2} (1+2 x)\right )}{e^{e^{e^{x+x^2}}}-x}+\log \left (-e^{e^{e^{x+x^2}}}+x\right )\right ) \, dx\\ &=\frac {2}{27} \int x \log \left (-e^{e^{e^{x+x^2}}}+x\right ) \left (\frac {x \left (-1+e^{e^{e^{x+x^2}}+e^{x+x^2}+x+x^2} (1+2 x)\right )}{e^{e^{e^{x+x^2}}}-x}+\log \left (-e^{e^{e^{x+x^2}}}+x\right )\right ) \, dx\\ &=\frac {1}{27} x^2 \log ^2\left (-e^{e^{e^{x+x^2}}}+x\right )\\ \end {aligned} \end {gather*}

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Mathematica [F]  time = 0.76, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (-2 x^2+e^{e^{e^{x+x^2}}+e^{x+x^2}+x+x^2} \left (2 x^2+4 x^3\right )\right ) \log \left (-e^{e^{e^{x+x^2}}}+x\right )+\left (2 e^{e^{e^{x+x^2}}} x-2 x^2\right ) \log ^2\left (-e^{e^{e^{x+x^2}}}+x\right )}{27 e^{e^{e^{x+x^2}}}-27 x} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Integrate[((-2*x^2 + E^(E^E^(x + x^2) + E^(x + x^2) + x + x^2)*(2*x^2 + 4*x^3))*Log[-E^E^E^(x + x^2) + x] + (2
*E^E^E^(x + x^2)*x - 2*x^2)*Log[-E^E^E^(x + x^2) + x]^2)/(27*E^E^E^(x + x^2) - 27*x),x]

[Out]

Integrate[((-2*x^2 + E^(E^E^(x + x^2) + E^(x + x^2) + x + x^2)*(2*x^2 + 4*x^3))*Log[-E^E^E^(x + x^2) + x] + (2
*E^E^E^(x + x^2)*x - 2*x^2)*Log[-E^E^E^(x + x^2) + x]^2)/(27*E^E^E^(x + x^2) - 27*x), x]

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fricas [B]  time = 0.60, size = 63, normalized size = 2.52 \begin {gather*} \frac {1}{27} \, x^{2} \log \left ({\left (x e^{\left (x^{2} + x + e^{\left (x^{2} + x\right )}\right )} - e^{\left (x^{2} + x + e^{\left (x^{2} + x\right )} + e^{\left (e^{\left (x^{2} + x\right )}\right )}\right )}\right )} e^{\left (-x^{2} - x - e^{\left (x^{2} + x\right )}\right )}\right )^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x*exp(exp(exp(x^2+x)))-2*x^2)*log(-exp(exp(exp(x^2+x)))+x)^2+((4*x^3+2*x^2)*exp(x^2+x)*exp(exp(x
^2+x))*exp(exp(exp(x^2+x)))-2*x^2)*log(-exp(exp(exp(x^2+x)))+x))/(27*exp(exp(exp(x^2+x)))-27*x),x, algorithm="
fricas")

[Out]

1/27*x^2*log((x*e^(x^2 + x + e^(x^2 + x)) - e^(x^2 + x + e^(x^2 + x) + e^(e^(x^2 + x))))*e^(-x^2 - x - e^(x^2
+ x)))^2

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giac [B]  time = 0.81, size = 63, normalized size = 2.52 \begin {gather*} \frac {1}{27} \, x^{2} \log \left ({\left (x e^{\left (x^{2} + x + e^{\left (x^{2} + x\right )}\right )} - e^{\left (x^{2} + x + e^{\left (x^{2} + x\right )} + e^{\left (e^{\left (x^{2} + x\right )}\right )}\right )}\right )} e^{\left (-x^{2} - x - e^{\left (x^{2} + x\right )}\right )}\right )^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x*exp(exp(exp(x^2+x)))-2*x^2)*log(-exp(exp(exp(x^2+x)))+x)^2+((4*x^3+2*x^2)*exp(x^2+x)*exp(exp(x
^2+x))*exp(exp(exp(x^2+x)))-2*x^2)*log(-exp(exp(exp(x^2+x)))+x))/(27*exp(exp(exp(x^2+x)))-27*x),x, algorithm="
giac")

[Out]

1/27*x^2*log((x*e^(x^2 + x + e^(x^2 + x)) - e^(x^2 + x + e^(x^2 + x) + e^(e^(x^2 + x))))*e^(-x^2 - x - e^(x^2
+ x)))^2

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maple [A]  time = 0.07, size = 21, normalized size = 0.84

method	result	size
risch	\(\frac {x^{2} \ln \left (-{\mathrm e}^{{\mathrm e}^{{\mathrm e}^{\left (x +1\right ) x}}}+x \right )^{2}}{27}\)	\(21\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((2*x*exp(exp(exp(x^2+x)))-2*x^2)*ln(-exp(exp(exp(x^2+x)))+x)^2+((4*x^3+2*x^2)*exp(x^2+x)*exp(exp(x^2+x))*
exp(exp(exp(x^2+x)))-2*x^2)*ln(-exp(exp(exp(x^2+x)))+x))/(27*exp(exp(exp(x^2+x)))-27*x),x,method=_RETURNVERBOS
E)

[Out]

1/27*x^2*ln(-exp(exp(exp((x+1)*x)))+x)^2

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maxima [A]  time = 0.96, size = 20, normalized size = 0.80 \begin {gather*} \frac {1}{27} \, x^{2} \log \left (x - e^{\left (e^{\left (e^{\left (x^{2} + x\right )}\right )}\right )}\right )^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x*exp(exp(exp(x^2+x)))-2*x^2)*log(-exp(exp(exp(x^2+x)))+x)^2+((4*x^3+2*x^2)*exp(x^2+x)*exp(exp(x
^2+x))*exp(exp(exp(x^2+x)))-2*x^2)*log(-exp(exp(exp(x^2+x)))+x))/(27*exp(exp(exp(x^2+x)))-27*x),x, algorithm="
maxima")

[Out]

1/27*x^2*log(x - e^(e^(e^(x^2 + x))))^2

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mupad [B]  time = 2.17, size = 21, normalized size = 0.84 \begin {gather*} \frac {x^2\,{\ln \left (x-{\mathrm {e}}^{{\mathrm {e}}^{{\mathrm {e}}^{x^2}\,{\mathrm {e}}^x}}\right )}^2}{27} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(x - exp(exp(exp(x + x^2))))*(2*x^2 - exp(x + x^2)*exp(exp(x + x^2))*exp(exp(exp(x + x^2)))*(2*x^2 + 4
*x^3)) - log(x - exp(exp(exp(x + x^2))))^2*(2*x*exp(exp(exp(x + x^2))) - 2*x^2))/(27*x - 27*exp(exp(exp(x + x^
2)))),x)

[Out]

(x^2*log(x - exp(exp(exp(x^2)*exp(x))))^2)/27

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sympy [A]  time = 8.09, size = 19, normalized size = 0.76 \begin {gather*} \frac {x^{2} \log {\left (x - e^{e^{e^{x^{2} + x}}} \right )}^{2}}{27} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x*exp(exp(exp(x**2+x)))-2*x**2)*ln(-exp(exp(exp(x**2+x)))+x)**2+((4*x**3+2*x**2)*exp(x**2+x)*exp
(exp(x**2+x))*exp(exp(exp(x**2+x)))-2*x**2)*ln(-exp(exp(exp(x**2+x)))+x))/(27*exp(exp(exp(x**2+x)))-27*x),x)

[Out]

x**2*log(x - exp(exp(exp(x**2 + x))))**2/27

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