3.1.23 \(\int \frac {-20 e^{2+\frac {5}{-2+e^2 x^4}} x^4+(-4+4 e^2 x^4-e^4 x^8+e^{\frac {5}{-2+e^2 x^4}} (4-4 e^2 x^4+e^4 x^8)) \log (-1+e^{\frac {5}{-2+e^2 x^4}})}{-4+4 e^2 x^4-e^4 x^8+e^{\frac {5}{-2+e^2 x^4}} (4-4 e^2 x^4+e^4 x^8)} \, dx\)

Optimal. Leaf size=23 \[ x \log \left (-1+e^{\frac {5 x}{-2 x+e^2 x^5}}\right ) \]

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Rubi [A]  time = 55.41, antiderivative size = 21, normalized size of antiderivative = 0.91, number of steps used = 84, number of rules used = 5, integrand size = 147, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.034, Rules used = {6688, 6742, 6725, 2548, 12} \begin {gather*} x \log \left (e^{-\frac {5}{2-e^2 x^4}}-1\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-20*E^(2 + 5/(-2 + E^2*x^4))*x^4 + (-4 + 4*E^2*x^4 - E^4*x^8 + E^(5/(-2 + E^2*x^4))*(4 - 4*E^2*x^4 + E^4*
x^8))*Log[-1 + E^(5/(-2 + E^2*x^4))])/(-4 + 4*E^2*x^4 - E^4*x^8 + E^(5/(-2 + E^2*x^4))*(4 - 4*E^2*x^4 + E^4*x^
8)),x]

[Out]

x*Log[-1 + E^(-5/(2 - E^2*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 2548

Int[Log[u_], x_Symbol] :> Simp[x*Log[u], x] - Int[SimplifyIntegrand[(x*D[u, x])/u, x], x] /; InverseFunctionFr
eeQ[u, x]

Rule 6688

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

Rule 6725

Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a + b*x^n), x]}, Int[v, x]
/; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ[n, 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} &=\int \left (-\frac {20 e^{2+\frac {5}{-2+e^2 x^4}} x^4}{\left (-1+e^{\frac {5}{-2+e^2 x^4}}\right ) \left (-2+e^2 x^4\right )^2}+\log \left (-1+e^{\frac {5}{-2+e^2 x^4}}\right )\right ) \, dx\\ &=-\left (20 \int \frac {e^{2+\frac {5}{-2+e^2 x^4}} x^4}{\left (-1+e^{\frac {5}{-2+e^2 x^4}}\right ) \left (-2+e^2 x^4\right )^2} \, dx\right )+\int \log \left (-1+e^{\frac {5}{-2+e^2 x^4}}\right ) \, dx\\ &=x \log \left (-1+e^{-\frac {5}{2-e^2 x^4}}\right )-20 \int \left (\frac {e^{2+\frac {5}{-2+e^2 x^4}} x^4}{5 \left (-1+e^{\frac {1}{-2+e^2 x^4}}\right ) \left (-2+e^2 x^4\right )^2}-\frac {e^{2+\frac {5}{-2+e^2 x^4}} \left (4+3 e^{\frac {1}{-2+e^2 x^4}}+2 e^{\frac {2}{-2+e^2 x^4}}+e^{\frac {3}{-2+e^2 x^4}}\right ) x^4}{5 \left (1+e^{\frac {1}{-2+e^2 x^4}}+e^{\frac {2}{-2+e^2 x^4}}+e^{\frac {3}{-2+e^2 x^4}}+e^{\frac {4}{-2+e^2 x^4}}\right ) \left (-2+e^2 x^4\right )^2}\right ) \, dx-\int -\frac {20 e^{2+\frac {5}{-2+e^2 x^4}} x^4}{\left (-1+e^{\frac {5}{-2+e^2 x^4}}\right ) \left (-2+e^2 x^4\right )^2} \, dx\\ &=x \log \left (-1+e^{-\frac {5}{2-e^2 x^4}}\right )-4 \int \frac {e^{2+\frac {5}{-2+e^2 x^4}} x^4}{\left (-1+e^{\frac {1}{-2+e^2 x^4}}\right ) \left (-2+e^2 x^4\right )^2} \, dx+4 \int \frac {e^{2+\frac {5}{-2+e^2 x^4}} \left (4+3 e^{\frac {1}{-2+e^2 x^4}}+2 e^{\frac {2}{-2+e^2 x^4}}+e^{\frac {3}{-2+e^2 x^4}}\right ) x^4}{\left (1+e^{\frac {1}{-2+e^2 x^4}}+e^{\frac {2}{-2+e^2 x^4}}+e^{\frac {3}{-2+e^2 x^4}}+e^{\frac {4}{-2+e^2 x^4}}\right ) \left (-2+e^2 x^4\right )^2} \, dx+20 \int \frac {e^{2+\frac {5}{-2+e^2 x^4}} x^4}{\left (-1+e^{\frac {5}{-2+e^2 x^4}}\right ) \left (-2+e^2 x^4\right )^2} \, dx\\ &=x \log \left (-1+e^{-\frac {5}{2-e^2 x^4}}\right )-4 \int \left (\frac {2 e^{\frac {5}{-2+e^2 x^4}}}{\left (-1+e^{\frac {1}{-2+e^2 x^4}}\right ) \left (-2+e^2 x^4\right )^2}+\frac {e^{\frac {5}{-2+e^2 x^4}}}{\left (-1+e^{\frac {1}{-2+e^2 x^4}}\right ) \left (-2+e^2 x^4\right )}\right ) \, dx+4 \int \left (\frac {2 e^{\frac {5}{-2+e^2 x^4}} \left (4+3 e^{\frac {1}{-2+e^2 x^4}}+2 e^{\frac {2}{-2+e^2 x^4}}+e^{\frac {3}{-2+e^2 x^4}}\right )}{\left (1+e^{\frac {1}{-2+e^2 x^4}}+e^{\frac {2}{-2+e^2 x^4}}+e^{\frac {3}{-2+e^2 x^4}}+e^{\frac {4}{-2+e^2 x^4}}\right ) \left (-2+e^2 x^4\right )^2}+\frac {e^{\frac {5}{-2+e^2 x^4}} \left (4+3 e^{\frac {1}{-2+e^2 x^4}}+2 e^{\frac {2}{-2+e^2 x^4}}+e^{\frac {3}{-2+e^2 x^4}}\right )}{\left (1+e^{\frac {1}{-2+e^2 x^4}}+e^{\frac {2}{-2+e^2 x^4}}+e^{\frac {3}{-2+e^2 x^4}}+e^{\frac {4}{-2+e^2 x^4}}\right ) \left (-2+e^2 x^4\right )}\right ) \, dx+20 \int \left (\frac {e^{2+\frac {5}{-2+e^2 x^4}} x^4}{5 \left (-1+e^{\frac {1}{-2+e^2 x^4}}\right ) \left (-2+e^2 x^4\right )^2}-\frac {e^{2+\frac {5}{-2+e^2 x^4}} \left (4+3 e^{\frac {1}{-2+e^2 x^4}}+2 e^{\frac {2}{-2+e^2 x^4}}+e^{\frac {3}{-2+e^2 x^4}}\right ) x^4}{5 \left (1+e^{\frac {1}{-2+e^2 x^4}}+e^{\frac {2}{-2+e^2 x^4}}+e^{\frac {3}{-2+e^2 x^4}}+e^{\frac {4}{-2+e^2 x^4}}\right ) \left (-2+e^2 x^4\right )^2}\right ) \, dx\\ &=x \log \left (-1+e^{-\frac {5}{2-e^2 x^4}}\right )+4 \int \frac {e^{2+\frac {5}{-2+e^2 x^4}} x^4}{\left (-1+e^{\frac {1}{-2+e^2 x^4}}\right ) \left (-2+e^2 x^4\right )^2} \, dx-4 \int \frac {e^{2+\frac {5}{-2+e^2 x^4}} \left (4+3 e^{\frac {1}{-2+e^2 x^4}}+2 e^{\frac {2}{-2+e^2 x^4}}+e^{\frac {3}{-2+e^2 x^4}}\right ) x^4}{\left (1+e^{\frac {1}{-2+e^2 x^4}}+e^{\frac {2}{-2+e^2 x^4}}+e^{\frac {3}{-2+e^2 x^4}}+e^{\frac {4}{-2+e^2 x^4}}\right ) \left (-2+e^2 x^4\right )^2} \, dx-4 \int \frac {e^{\frac {5}{-2+e^2 x^4}}}{\left (-1+e^{\frac {1}{-2+e^2 x^4}}\right ) \left (-2+e^2 x^4\right )} \, dx+4 \int \frac {e^{\frac {5}{-2+e^2 x^4}} \left (4+3 e^{\frac {1}{-2+e^2 x^4}}+2 e^{\frac {2}{-2+e^2 x^4}}+e^{\frac {3}{-2+e^2 x^4}}\right )}{\left (1+e^{\frac {1}{-2+e^2 x^4}}+e^{\frac {2}{-2+e^2 x^4}}+e^{\frac {3}{-2+e^2 x^4}}+e^{\frac {4}{-2+e^2 x^4}}\right ) \left (-2+e^2 x^4\right )} \, dx-8 \int \frac {e^{\frac {5}{-2+e^2 x^4}}}{\left (-1+e^{\frac {1}{-2+e^2 x^4}}\right ) \left (-2+e^2 x^4\right )^2} \, dx+8 \int \frac {e^{\frac {5}{-2+e^2 x^4}} \left (4+3 e^{\frac {1}{-2+e^2 x^4}}+2 e^{\frac {2}{-2+e^2 x^4}}+e^{\frac {3}{-2+e^2 x^4}}\right )}{\left (1+e^{\frac {1}{-2+e^2 x^4}}+e^{\frac {2}{-2+e^2 x^4}}+e^{\frac {3}{-2+e^2 x^4}}+e^{\frac {4}{-2+e^2 x^4}}\right ) \left (-2+e^2 x^4\right )^2} \, dx\\ &=\text {Rest of rules removed due to large latex content} \end {aligned} \end {gather*}

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Mathematica [A]  time = 4.26, size = 20, normalized size = 0.87 \begin {gather*} x \log \left (-1+e^{\frac {5}{-2+e^2 x^4}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-20*E^(2 + 5/(-2 + E^2*x^4))*x^4 + (-4 + 4*E^2*x^4 - E^4*x^8 + E^(5/(-2 + E^2*x^4))*(4 - 4*E^2*x^4
+ E^4*x^8))*Log[-1 + E^(5/(-2 + E^2*x^4))])/(-4 + 4*E^2*x^4 - E^4*x^8 + E^(5/(-2 + E^2*x^4))*(4 - 4*E^2*x^4 +
E^4*x^8)),x]

[Out]

x*Log[-1 + E^(5/(-2 + E^2*x^4))]

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fricas [A]  time = 0.60, size = 33, normalized size = 1.43 \begin {gather*} x \log \left (-{\left (e^{2} - e^{\left (\frac {2 \, x^{4} e^{2} + 1}{x^{4} e^{2} - 2}\right )}\right )} e^{\left (-2\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x^8*exp(2)^2-4*x^4*exp(2)+4)*exp(5/(x^4*exp(2)-2))-x^8*exp(2)^2+4*x^4*exp(2)-4)*log(exp(5/(x^4*ex
p(2)-2))-1)-20*x^4*exp(2)*exp(5/(x^4*exp(2)-2)))/((x^8*exp(2)^2-4*x^4*exp(2)+4)*exp(5/(x^4*exp(2)-2))-x^8*exp(
2)^2+4*x^4*exp(2)-4),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {20 \, x^{4} e^{\left (\frac {5}{x^{4} e^{2} - 2} + 2\right )} + {\left (x^{8} e^{4} - 4 \, x^{4} e^{2} - {\left (x^{8} e^{4} - 4 \, x^{4} e^{2} + 4\right )} e^{\left (\frac {5}{x^{4} e^{2} - 2}\right )} + 4\right )} \log \left (e^{\left (\frac {5}{x^{4} e^{2} - 2}\right )} - 1\right )}{x^{8} e^{4} - 4 \, x^{4} e^{2} - {\left (x^{8} e^{4} - 4 \, x^{4} e^{2} + 4\right )} e^{\left (\frac {5}{x^{4} e^{2} - 2}\right )} + 4}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x^8*exp(2)^2-4*x^4*exp(2)+4)*exp(5/(x^4*exp(2)-2))-x^8*exp(2)^2+4*x^4*exp(2)-4)*log(exp(5/(x^4*ex
p(2)-2))-1)-20*x^4*exp(2)*exp(5/(x^4*exp(2)-2)))/((x^8*exp(2)^2-4*x^4*exp(2)+4)*exp(5/(x^4*exp(2)-2))-x^8*exp(
2)^2+4*x^4*exp(2)-4),x, algorithm="giac")

[Out]

integrate((20*x^4*e^(5/(x^4*e^2 - 2) + 2) + (x^8*e^4 - 4*x^4*e^2 - (x^8*e^4 - 4*x^4*e^2 + 4)*e^(5/(x^4*e^2 - 2
)) + 4)*log(e^(5/(x^4*e^2 - 2)) - 1))/(x^8*e^4 - 4*x^4*e^2 - (x^8*e^4 - 4*x^4*e^2 + 4)*e^(5/(x^4*e^2 - 2)) + 4
), x)

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maple [A]  time = 0.21, size = 19, normalized size = 0.83

method	result	size
risch	\(x \ln \left ({\mathrm e}^{\frac {5}{x^{4} {\mathrm e}^{2}-2}}-1\right )\)	\(19\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((x^8*exp(2)^2-4*x^4*exp(2)+4)*exp(5/(x^4*exp(2)-2))-x^8*exp(2)^2+4*x^4*exp(2)-4)*ln(exp(5/(x^4*exp(2)-2)
)-1)-20*x^4*exp(2)*exp(5/(x^4*exp(2)-2)))/((x^8*exp(2)^2-4*x^4*exp(2)+4)*exp(5/(x^4*exp(2)-2))-x^8*exp(2)^2+4*
x^4*exp(2)-4),x,method=_RETURNVERBOSE)

[Out]

x*ln(exp(5/(x^4*exp(2)-2))-1)

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maxima [B]  time = 0.74, size = 72, normalized size = 3.13 \begin {gather*} x \log \left (e^{\left (\frac {4}{x^{4} e^{2} - 2}\right )} + e^{\left (\frac {3}{x^{4} e^{2} - 2}\right )} + e^{\left (\frac {2}{x^{4} e^{2} - 2}\right )} + e^{\left (\frac {1}{x^{4} e^{2} - 2}\right )} + 1\right ) + x \log \left (e^{\left (\frac {1}{x^{4} e^{2} - 2}\right )} - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x^8*exp(2)^2-4*x^4*exp(2)+4)*exp(5/(x^4*exp(2)-2))-x^8*exp(2)^2+4*x^4*exp(2)-4)*log(exp(5/(x^4*ex
p(2)-2))-1)-20*x^4*exp(2)*exp(5/(x^4*exp(2)-2)))/((x^8*exp(2)^2-4*x^4*exp(2)+4)*exp(5/(x^4*exp(2)-2))-x^8*exp(
2)^2+4*x^4*exp(2)-4),x, algorithm="maxima")

[Out]

x*log(e^(4/(x^4*e^2 - 2)) + e^(3/(x^4*e^2 - 2)) + e^(2/(x^4*e^2 - 2)) + e^(1/(x^4*e^2 - 2)) + 1) + x*log(e^(1/
(x^4*e^2 - 2)) - 1)

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mupad [B]  time = 0.65, size = 18, normalized size = 0.78 \begin {gather*} x\,\ln \left ({\mathrm {e}}^{\frac {5}{x^4\,{\mathrm {e}}^2-2}}-1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(exp(5/(x^4*exp(2) - 2)) - 1)*(exp(5/(x^4*exp(2) - 2))*(x^8*exp(4) - 4*x^4*exp(2) + 4) + 4*x^4*exp(2)
- x^8*exp(4) - 4) - 20*x^4*exp(5/(x^4*exp(2) - 2))*exp(2))/(exp(5/(x^4*exp(2) - 2))*(x^8*exp(4) - 4*x^4*exp(2)
+ 4) + 4*x^4*exp(2) - x^8*exp(4) - 4),x)

[Out]

x*log(exp(5/(x^4*exp(2) - 2)) - 1)

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sympy [A]  time = 0.60, size = 15, normalized size = 0.65 \begin {gather*} x \log {\left (e^{\frac {5}{x^{4} e^{2} - 2}} - 1 \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x**8*exp(2)**2-4*x**4*exp(2)+4)*exp(5/(x**4*exp(2)-2))-x**8*exp(2)**2+4*x**4*exp(2)-4)*ln(exp(5/(
x**4*exp(2)-2))-1)-20*x**4*exp(2)*exp(5/(x**4*exp(2)-2)))/((x**8*exp(2)**2-4*x**4*exp(2)+4)*exp(5/(x**4*exp(2)
-2))-x**8*exp(2)**2+4*x**4*exp(2)-4),x)

[Out]

x*log(exp(5/(x**4*exp(2) - 2)) - 1)

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