3.5.26 \(\int \frac {-10 x^2-2 e^3 x^2+e^{e^{\frac {5 x+25 x^4+5 e^3 x^4}{5+e^3}}+\frac {5 x+25 x^4+5 e^3 x^4}{5+e^3}} (5 x+100 x^4+20 e^3 x^4)+(e^{e^{\frac {5 x+25 x^4+5 e^3 x^4}{5+e^3}}} (-5-e^3)+5 x^2+e^3 x^2) \log (-e^{e^{\frac {5 x+25 x^4+5 e^3 x^4}{5+e^3}}}+x^2)}{-5 x^4-e^3 x^4+e^{e^{\frac {5 x+25 x^4+5 e^3 x^4}{5+e^3}}} (5 x^2+e^3 x^2)} \, dx\)

Optimal. Leaf size=30 \[ \frac {\log \left (-e^{e^{5 \left (\frac {x}{5+e^3}+x^4\right )}}+x^2\right )}{x} \]

________________________________________________________________________________________

Rubi [A]  time = 24.53, antiderivative size = 29, normalized size of antiderivative = 0.97, number of steps used = 11, number of rules used = 4, integrand size = 234, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.017, Rules used = {6, 6688, 6742, 2551} \begin {gather*} \frac {\log \left (x^2-e^{e^{5 x \left (x^3+\frac {1}{5+e^3}\right )}}\right )}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-10*x^2 - 2*E^3*x^2 + E^(E^((5*x + 25*x^4 + 5*E^3*x^4)/(5 + E^3)) + (5*x + 25*x^4 + 5*E^3*x^4)/(5 + E^3))
*(5*x + 100*x^4 + 20*E^3*x^4) + (E^E^((5*x + 25*x^4 + 5*E^3*x^4)/(5 + E^3))*(-5 - E^3) + 5*x^2 + E^3*x^2)*Log[
-E^E^((5*x + 25*x^4 + 5*E^3*x^4)/(5 + E^3)) + x^2])/(-5*x^4 - E^3*x^4 + E^E^((5*x + 25*x^4 + 5*E^3*x^4)/(5 + E
^3))*(5*x^2 + E^3*x^2)),x]

[Out]

Log[-E^E^(5*x*((5 + E^3)^(-1) + x^3)) + x^2]/x

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, x]

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 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 {-10 x^2-2 e^3 x^2+\exp \left (e^{\frac {5 x+25 x^4+5 e^3 x^4}{5+e^3}}+\frac {5 x+25 x^4+5 e^3 x^4}{5+e^3}\right ) \left (5 x+100 x^4+20 e^3 x^4\right )+\left (e^{e^{\frac {5 x+25 x^4+5 e^3 x^4}{5+e^3}}} \left (-5-e^3\right )+5 x^2+e^3 x^2\right ) \log \left (-e^{e^{\frac {5 x+25 x^4+5 e^3 x^4}{5+e^3}}}+x^2\right )}{\left (-5-e^3\right ) x^4+e^{e^{\frac {5 x+25 x^4+5 e^3 x^4}{5+e^3}}} \left (5 x^2+e^3 x^2\right )} \, dx\\ &=\int \frac {\left (-10-2 e^3\right ) x^2+\exp \left (e^{\frac {5 x+25 x^4+5 e^3 x^4}{5+e^3}}+\frac {5 x+25 x^4+5 e^3 x^4}{5+e^3}\right ) \left (5 x+100 x^4+20 e^3 x^4\right )+\left (e^{e^{\frac {5 x+25 x^4+5 e^3 x^4}{5+e^3}}} \left (-5-e^3\right )+5 x^2+e^3 x^2\right ) \log \left (-e^{e^{\frac {5 x+25 x^4+5 e^3 x^4}{5+e^3}}}+x^2\right )}{\left (-5-e^3\right ) x^4+e^{e^{\frac {5 x+25 x^4+5 e^3 x^4}{5+e^3}}} \left (5 x^2+e^3 x^2\right )} \, dx\\ &=\int \left (-\frac {2}{e^{e^{5 x \left (\frac {1}{5+e^3}+x^3\right )}}-x^2}+\frac {5 \exp \left (e^{5 x \left (\frac {1}{5+e^3}+x^3\right )}+5 x \left (\frac {1}{5+e^3}+x^3\right )\right ) \left (1+4 \left (5+e^3\right ) x^3\right )}{\left (5+e^3\right ) x \left (e^{e^{5 x \left (\frac {1}{5+e^3}+x^3\right )}}-x^2\right )}-\frac {\log \left (-e^{e^{5 x \left (\frac {1}{5+e^3}+x^3\right )}}+x^2\right )}{x^2}\right ) \, dx\\ &=-\left (2 \int \frac {1}{e^{e^{5 x \left (\frac {1}{5+e^3}+x^3\right )}}-x^2} \, dx\right )+\frac {5 \int \frac {\exp \left (e^{5 x \left (\frac {1}{5+e^3}+x^3\right )}+5 x \left (\frac {1}{5+e^3}+x^3\right )\right ) \left (1+4 \left (5+e^3\right ) x^3\right )}{x \left (e^{e^{5 x \left (\frac {1}{5+e^3}+x^3\right )}}-x^2\right )} \, dx}{5+e^3}-\int \frac {\log \left (-e^{e^{5 x \left (\frac {1}{5+e^3}+x^3\right )}}+x^2\right )}{x^2} \, dx\\ &=\frac {\log \left (-e^{e^{5 x \left (\frac {1}{5+e^3}+x^3\right )}}+x^2\right )}{x}-2 \int \frac {1}{e^{e^{5 x \left (\frac {1}{5+e^3}+x^3\right )}}-x^2} \, dx+\frac {5 \int \left (-\frac {\exp \left (e^{5 x \left (\frac {1}{5+e^3}+x^3\right )}+5 x \left (\frac {1}{5+e^3}+x^3\right )\right )}{x \left (-e^{e^{5 x \left (\frac {1}{5+e^3}+x^3\right )}}+x^2\right )}-\frac {4 \exp \left (e^{5 x \left (\frac {1}{5+e^3}+x^3\right )}+5 x \left (\frac {1}{5+e^3}+x^3\right )\right ) \left (5+e^3\right ) x^2}{-e^{e^{5 x \left (\frac {1}{5+e^3}+x^3\right )}}+x^2}\right ) \, dx}{5+e^3}-\int \frac {-2 x+\frac {5 \exp \left (e^{5 x \left (\frac {1}{5+e^3}+x^3\right )}+5 x \left (\frac {1}{5+e^3}+x^3\right )\right ) \left (1+4 \left (5+e^3\right ) x^3\right )}{5+e^3}}{e^{e^{5 x \left (\frac {1}{5+e^3}+x^3\right )}} x-x^3} \, dx\\ &=\frac {\log \left (-e^{e^{5 x \left (\frac {1}{5+e^3}+x^3\right )}}+x^2\right )}{x}-2 \int \frac {1}{e^{e^{5 x \left (\frac {1}{5+e^3}+x^3\right )}}-x^2} \, dx-20 \int \frac {\exp \left (e^{5 x \left (\frac {1}{5+e^3}+x^3\right )}+5 x \left (\frac {1}{5+e^3}+x^3\right )\right ) x^2}{-e^{e^{5 x \left (\frac {1}{5+e^3}+x^3\right )}}+x^2} \, dx-\frac {5 \int \frac {\exp \left (e^{5 x \left (\frac {1}{5+e^3}+x^3\right )}+5 x \left (\frac {1}{5+e^3}+x^3\right )\right )}{x \left (-e^{e^{5 x \left (\frac {1}{5+e^3}+x^3\right )}}+x^2\right )} \, dx}{5+e^3}-\int \left (-\frac {2}{e^{e^{5 x \left (\frac {1}{5+e^3}+x^3\right )}}-x^2}+\frac {5 \exp \left (e^{5 x \left (\frac {1}{5+e^3}+x^3\right )}+5 x \left (\frac {1}{5+e^3}+x^3\right )\right ) \left (1+4 \left (5+e^3\right ) x^3\right )}{\left (5+e^3\right ) x \left (e^{e^{5 x \left (\frac {1}{5+e^3}+x^3\right )}}-x^2\right )}\right ) \, dx\\ &=\frac {\log \left (-e^{e^{5 x \left (\frac {1}{5+e^3}+x^3\right )}}+x^2\right )}{x}-20 \int \frac {\exp \left (e^{5 x \left (\frac {1}{5+e^3}+x^3\right )}+5 x \left (\frac {1}{5+e^3}+x^3\right )\right ) x^2}{-e^{e^{5 x \left (\frac {1}{5+e^3}+x^3\right )}}+x^2} \, dx-\frac {5 \int \frac {\exp \left (e^{5 x \left (\frac {1}{5+e^3}+x^3\right )}+5 x \left (\frac {1}{5+e^3}+x^3\right )\right )}{x \left (-e^{e^{5 x \left (\frac {1}{5+e^3}+x^3\right )}}+x^2\right )} \, dx}{5+e^3}-\frac {5 \int \frac {\exp \left (e^{5 x \left (\frac {1}{5+e^3}+x^3\right )}+5 x \left (\frac {1}{5+e^3}+x^3\right )\right ) \left (1+4 \left (5+e^3\right ) x^3\right )}{x \left (e^{e^{5 x \left (\frac {1}{5+e^3}+x^3\right )}}-x^2\right )} \, dx}{5+e^3}\\ &=\frac {\log \left (-e^{e^{5 x \left (\frac {1}{5+e^3}+x^3\right )}}+x^2\right )}{x}-20 \int \frac {\exp \left (e^{5 x \left (\frac {1}{5+e^3}+x^3\right )}+5 x \left (\frac {1}{5+e^3}+x^3\right )\right ) x^2}{-e^{e^{5 x \left (\frac {1}{5+e^3}+x^3\right )}}+x^2} \, dx-\frac {5 \int \frac {\exp \left (e^{5 x \left (\frac {1}{5+e^3}+x^3\right )}+5 x \left (\frac {1}{5+e^3}+x^3\right )\right )}{x \left (-e^{e^{5 x \left (\frac {1}{5+e^3}+x^3\right )}}+x^2\right )} \, dx}{5+e^3}-\frac {5 \int \left (-\frac {\exp \left (e^{5 x \left (\frac {1}{5+e^3}+x^3\right )}+5 x \left (\frac {1}{5+e^3}+x^3\right )\right )}{x \left (-e^{e^{5 x \left (\frac {1}{5+e^3}+x^3\right )}}+x^2\right )}-\frac {4 \exp \left (e^{5 x \left (\frac {1}{5+e^3}+x^3\right )}+5 x \left (\frac {1}{5+e^3}+x^3\right )\right ) \left (5+e^3\right ) x^2}{-e^{e^{5 x \left (\frac {1}{5+e^3}+x^3\right )}}+x^2}\right ) \, dx}{5+e^3}\\ &=\frac {\log \left (-e^{e^{5 x \left (\frac {1}{5+e^3}+x^3\right )}}+x^2\right )}{x}\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.72, size = 31, normalized size = 1.03 \begin {gather*} \frac {\log \left (-e^{e^{\frac {5 x}{5+e^3}+5 x^4}}+x^2\right )}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-10*x^2 - 2*E^3*x^2 + E^(E^((5*x + 25*x^4 + 5*E^3*x^4)/(5 + E^3)) + (5*x + 25*x^4 + 5*E^3*x^4)/(5 +
E^3))*(5*x + 100*x^4 + 20*E^3*x^4) + (E^E^((5*x + 25*x^4 + 5*E^3*x^4)/(5 + E^3))*(-5 - E^3) + 5*x^2 + E^3*x^2
)*Log[-E^E^((5*x + 25*x^4 + 5*E^3*x^4)/(5 + E^3)) + x^2])/(-5*x^4 - E^3*x^4 + E^E^((5*x + 25*x^4 + 5*E^3*x^4)/
(5 + E^3))*(5*x^2 + E^3*x^2)),x]

[Out]

Log[-E^E^((5*x)/(5 + E^3) + 5*x^4) + x^2]/x

________________________________________________________________________________________

fricas [B]  time = 0.75, size = 108, normalized size = 3.60 \begin {gather*} \frac {\log \left ({\left (x^{2} e^{\left (\frac {5 \, {\left (x^{4} e^{3} + 5 \, x^{4} + x\right )}}{e^{3} + 5}\right )} - e^{\left (\frac {5 \, x^{4} e^{3} + 25 \, x^{4} + {\left (e^{3} + 5\right )} e^{\left (\frac {5 \, {\left (x^{4} e^{3} + 5 \, x^{4} + x\right )}}{e^{3} + 5}\right )} + 5 \, x}{e^{3} + 5}\right )}\right )} e^{\left (-\frac {5 \, {\left (x^{4} e^{3} + 5 \, x^{4} + x\right )}}{e^{3} + 5}\right )}\right )}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-exp(3)-5)*exp(exp((5*x^4*exp(3)+25*x^4+5*x)/(exp(3)+5)))+x^2*exp(3)+5*x^2)*log(-exp(exp((5*x^4*e
xp(3)+25*x^4+5*x)/(exp(3)+5)))+x^2)+(20*x^4*exp(3)+100*x^4+5*x)*exp((5*x^4*exp(3)+25*x^4+5*x)/(exp(3)+5))*exp(
exp((5*x^4*exp(3)+25*x^4+5*x)/(exp(3)+5)))-2*x^2*exp(3)-10*x^2)/((x^2*exp(3)+5*x^2)*exp(exp((5*x^4*exp(3)+25*x
^4+5*x)/(exp(3)+5)))-x^4*exp(3)-5*x^4),x, algorithm="fricas")

[Out]

log((x^2*e^(5*(x^4*e^3 + 5*x^4 + x)/(e^3 + 5)) - e^((5*x^4*e^3 + 25*x^4 + (e^3 + 5)*e^(5*(x^4*e^3 + 5*x^4 + x)
/(e^3 + 5)) + 5*x)/(e^3 + 5)))*e^(-5*(x^4*e^3 + 5*x^4 + x)/(e^3 + 5)))/x

________________________________________________________________________________________

giac [B]  time = 4.05, size = 129, normalized size = 4.30 \begin {gather*} \frac {\log \left ({\left (x^{2} e^{\left (\frac {5 \, {\left (x^{4} e^{3} + 5 \, x^{4} + x\right )}}{e^{3} + 5}\right )} - e^{\left (\frac {5 \, x^{4} e^{3} + 25 \, x^{4} + 5 \, x + 5 \, e^{\left (\frac {5 \, {\left (x^{4} e^{3} + 5 \, x^{4} + x\right )}}{e^{3} + 5}\right )} + e^{\left (\frac {5 \, {\left (x^{4} e^{3} + 5 \, x^{4} + x\right )}}{e^{3} + 5} + 3\right )}}{e^{3} + 5}\right )}\right )} e^{\left (-\frac {5 \, {\left (x^{4} e^{3} + 5 \, x^{4} + x\right )}}{e^{3} + 5}\right )}\right )}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-exp(3)-5)*exp(exp((5*x^4*exp(3)+25*x^4+5*x)/(exp(3)+5)))+x^2*exp(3)+5*x^2)*log(-exp(exp((5*x^4*e
xp(3)+25*x^4+5*x)/(exp(3)+5)))+x^2)+(20*x^4*exp(3)+100*x^4+5*x)*exp((5*x^4*exp(3)+25*x^4+5*x)/(exp(3)+5))*exp(
exp((5*x^4*exp(3)+25*x^4+5*x)/(exp(3)+5)))-2*x^2*exp(3)-10*x^2)/((x^2*exp(3)+5*x^2)*exp(exp((5*x^4*exp(3)+25*x
^4+5*x)/(exp(3)+5)))-x^4*exp(3)-5*x^4),x, algorithm="giac")

[Out]

log((x^2*e^(5*(x^4*e^3 + 5*x^4 + x)/(e^3 + 5)) - e^((5*x^4*e^3 + 25*x^4 + 5*x + 5*e^(5*(x^4*e^3 + 5*x^4 + x)/(
e^3 + 5)) + e^(5*(x^4*e^3 + 5*x^4 + x)/(e^3 + 5) + 3))/(e^3 + 5)))*e^(-5*(x^4*e^3 + 5*x^4 + x)/(e^3 + 5)))/x

________________________________________________________________________________________

maple [A]  time = 0.36, size = 36, normalized size = 1.20

method	result	size
risch	\(\frac {\ln \left (-{\mathrm e}^{{\mathrm e}^{\frac {5 x \left (x^{3} {\mathrm e}^{3}+5 x^{3}+1\right )}{{\mathrm e}^{3}+5}}}+x^{2}\right )}{x}\)	\(36\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-exp(3)-5)*exp(exp((5*x^4*exp(3)+25*x^4+5*x)/(exp(3)+5)))+x^2*exp(3)+5*x^2)*ln(-exp(exp((5*x^4*exp(3)+2
5*x^4+5*x)/(exp(3)+5)))+x^2)+(20*x^4*exp(3)+100*x^4+5*x)*exp((5*x^4*exp(3)+25*x^4+5*x)/(exp(3)+5))*exp(exp((5*
x^4*exp(3)+25*x^4+5*x)/(exp(3)+5)))-2*x^2*exp(3)-10*x^2)/((x^2*exp(3)+5*x^2)*exp(exp((5*x^4*exp(3)+25*x^4+5*x)
/(exp(3)+5)))-x^4*exp(3)-5*x^4),x,method=_RETURNVERBOSE)

[Out]

1/x*ln(-exp(exp(5*x*(x^3*exp(3)+5*x^3+1)/(exp(3)+5)))+x^2)

________________________________________________________________________________________

maxima [A]  time = 1.02, size = 47, normalized size = 1.57 \begin {gather*} \frac {\log \left (x^{2} - e^{\left (e^{\left (\frac {5 \, x^{4} e^{3}}{e^{3} + 5} + \frac {25 \, x^{4}}{e^{3} + 5} + \frac {5 \, x}{e^{3} + 5}\right )}\right )}\right )}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-exp(3)-5)*exp(exp((5*x^4*exp(3)+25*x^4+5*x)/(exp(3)+5)))+x^2*exp(3)+5*x^2)*log(-exp(exp((5*x^4*e
xp(3)+25*x^4+5*x)/(exp(3)+5)))+x^2)+(20*x^4*exp(3)+100*x^4+5*x)*exp((5*x^4*exp(3)+25*x^4+5*x)/(exp(3)+5))*exp(
exp((5*x^4*exp(3)+25*x^4+5*x)/(exp(3)+5)))-2*x^2*exp(3)-10*x^2)/((x^2*exp(3)+5*x^2)*exp(exp((5*x^4*exp(3)+25*x
^4+5*x)/(exp(3)+5)))-x^4*exp(3)-5*x^4),x, algorithm="maxima")

[Out]

log(x^2 - e^(e^(5*x^4*e^3/(e^3 + 5) + 25*x^4/(e^3 + 5) + 5*x/(e^3 + 5))))/x

________________________________________________________________________________________

mupad [B]  time = 1.03, size = 49, normalized size = 1.63 \begin {gather*} \frac {\ln \left (x^2-{\mathrm {e}}^{{\mathrm {e}}^{\frac {5\,x^4\,{\mathrm {e}}^3}{{\mathrm {e}}^3+5}}\,{\mathrm {e}}^{\frac {5\,x}{{\mathrm {e}}^3+5}}\,{\mathrm {e}}^{\frac {25\,x^4}{{\mathrm {e}}^3+5}}}\right )}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(x^2 - exp(exp((5*x + 5*x^4*exp(3) + 25*x^4)/(exp(3) + 5))))*(x^2*exp(3) - exp(exp((5*x + 5*x^4*exp(3
) + 25*x^4)/(exp(3) + 5)))*(exp(3) + 5) + 5*x^2) - 2*x^2*exp(3) - 10*x^2 + exp(exp((5*x + 5*x^4*exp(3) + 25*x^
4)/(exp(3) + 5)))*exp((5*x + 5*x^4*exp(3) + 25*x^4)/(exp(3) + 5))*(5*x + 20*x^4*exp(3) + 100*x^4))/(x^4*exp(3)
+ 5*x^4 - exp(exp((5*x + 5*x^4*exp(3) + 25*x^4)/(exp(3) + 5)))*(x^2*exp(3) + 5*x^2)),x)

[Out]

log(x^2 - exp(exp((5*x^4*exp(3))/(exp(3) + 5))*exp((5*x)/(exp(3) + 5))*exp((25*x^4)/(exp(3) + 5))))/x

________________________________________________________________________________________

sympy [A]  time = 11.69, size = 31, normalized size = 1.03 \begin {gather*} \frac {\log {\left (x^{2} - e^{e^{\frac {25 x^{4} + 5 x^{4} e^{3} + 5 x}{5 + e^{3}}}} \right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-exp(3)-5)*exp(exp((5*x**4*exp(3)+25*x**4+5*x)/(exp(3)+5)))+x**2*exp(3)+5*x**2)*ln(-exp(exp((5*x*
*4*exp(3)+25*x**4+5*x)/(exp(3)+5)))+x**2)+(20*x**4*exp(3)+100*x**4+5*x)*exp((5*x**4*exp(3)+25*x**4+5*x)/(exp(3
)+5))*exp(exp((5*x**4*exp(3)+25*x**4+5*x)/(exp(3)+5)))-2*x**2*exp(3)-10*x**2)/((x**2*exp(3)+5*x**2)*exp(exp((5
*x**4*exp(3)+25*x**4+5*x)/(exp(3)+5)))-x**4*exp(3)-5*x**4),x)

[Out]

log(x**2 - exp(exp((25*x**4 + 5*x**4*exp(3) + 5*x)/(5 + exp(3)))))/x

________________________________________________________________________________________