3.70.41 \(\int \frac {-25 x+50 x \log (x)+e^{e^{2 x}-x} (1-2 e^{2 x}) \log ^2(x)+2 x \log ^2(x)}{3 \log ^2(x)} \, dx\)

Optimal. Leaf size=31 \[ \frac {1}{3} \left (4-e^{e^{2 x}-x}+x^2+\frac {25 x^2}{\log (x)}\right ) \]

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Rubi [A]  time = 0.40, antiderivative size = 34, normalized size of antiderivative = 1.10, number of steps used = 15, number of rules used = 8, integrand size = 49, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.163, Rules used = {12, 6742, 2282, 2214, 2204, 2306, 2309, 2178} \begin {gather*} \frac {x^2}{3}+\frac {25 x^2}{3 \log (x)}-\frac {1}{3} e^{e^{2 x}-x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-25*x + 50*x*Log[x] + E^(E^(2*x) - x)*(1 - 2*E^(2*x))*Log[x]^2 + 2*x*Log[x]^2)/(3*Log[x]^2),x]

[Out]

-1/3*E^(E^(2*x) - x) + x^2/3 + (25*x^2)/(3*Log[x])

Rule 12

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

Rule 2178

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(F^(g*(e - (c*f)/d))*ExpIntegral
Ei[(f*g*(c + d*x)*Log[F])/d])/d, x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !$UseGamma === True

Rule 2204

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^2), x_Symbol] :> Simp[(F^a*Sqrt[Pi]*Erfi[(c + d*x)*Rt[b*Log[F], 2
]])/(2*d*Rt[b*Log[F], 2]), x] /; FreeQ[{F, a, b, c, d}, x] && PosQ[b]

Rule 2214

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^(m
+ 1)*F^(a + b*(c + d*x)^n))/(d*(m + 1)), x] - Dist[(b*n*Log[F])/(m + 1), Int[(c + d*x)^(m + n)*F^(a + b*(c +
d*x)^n), x], x] /; FreeQ[{F, a, b, c, d}, x] && IntegerQ[(2*(m + 1))/n] && LtQ[-4, (m + 1)/n, 5] && IntegerQ[n
] && ((GtQ[n, 0] && LtQ[m, -1]) || (GtQ[-n, 0] && LeQ[-n, m + 1]))

Rule 2282

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 2306

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*Log
[c*x^n])^(p + 1))/(b*d*n*(p + 1)), x] - Dist[(m + 1)/(b*n*(p + 1)), Int[(d*x)^m*(a + b*Log[c*x^n])^(p + 1), x]
, x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[m, -1] && LtQ[p, -1]

Rule 2309

Int[((a_.) + Log[(c_.)*(x_)]*(b_.))^(p_)*(x_)^(m_.), x_Symbol] :> Dist[1/c^(m + 1), Subst[Int[E^((m + 1)*x)*(a
+ b*x)^p, x], x, Log[c*x]], x] /; FreeQ[{a, b, c, p}, x] && IntegerQ[m]

Rule 6742

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{3} \int \frac {-25 x+50 x \log (x)+e^{e^{2 x}-x} \left (1-2 e^{2 x}\right ) \log ^2(x)+2 x \log ^2(x)}{\log ^2(x)} \, dx\\ &=\frac {1}{3} \int \left (e^{e^{2 x}-x}-2 e^{e^{2 x}+x}+\frac {x \left (-25+50 \log (x)+2 \log ^2(x)\right )}{\log ^2(x)}\right ) \, dx\\ &=\frac {1}{3} \int e^{e^{2 x}-x} \, dx+\frac {1}{3} \int \frac {x \left (-25+50 \log (x)+2 \log ^2(x)\right )}{\log ^2(x)} \, dx-\frac {2}{3} \int e^{e^{2 x}+x} \, dx\\ &=\frac {1}{3} \int \left (2 x-\frac {25 x}{\log ^2(x)}+\frac {50 x}{\log (x)}\right ) \, dx+\frac {1}{3} \operatorname {Subst}\left (\int \frac {e^{x^2}}{x^2} \, dx,x,e^x\right )-\frac {2}{3} \operatorname {Subst}\left (\int e^{x^2} \, dx,x,e^x\right )\\ &=-\frac {1}{3} e^{e^{2 x}-x}+\frac {x^2}{3}-\frac {1}{3} \sqrt {\pi } \text {erfi}\left (e^x\right )+\frac {2}{3} \operatorname {Subst}\left (\int e^{x^2} \, dx,x,e^x\right )-\frac {25}{3} \int \frac {x}{\log ^2(x)} \, dx+\frac {50}{3} \int \frac {x}{\log (x)} \, dx\\ &=-\frac {1}{3} e^{e^{2 x}-x}+\frac {x^2}{3}+\frac {25 x^2}{3 \log (x)}-\frac {50}{3} \int \frac {x}{\log (x)} \, dx+\frac {50}{3} \operatorname {Subst}\left (\int \frac {e^{2 x}}{x} \, dx,x,\log (x)\right )\\ &=-\frac {1}{3} e^{e^{2 x}-x}+\frac {x^2}{3}+\frac {50}{3} \text {Ei}(2 \log (x))+\frac {25 x^2}{3 \log (x)}-\frac {50}{3} \operatorname {Subst}\left (\int \frac {e^{2 x}}{x} \, dx,x,\log (x)\right )\\ &=-\frac {1}{3} e^{e^{2 x}-x}+\frac {x^2}{3}+\frac {25 x^2}{3 \log (x)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.53, size = 30, normalized size = 0.97 \begin {gather*} \frac {1}{3} \left (-e^{e^{2 x}-x}+x^2+\frac {25 x^2}{\log (x)}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-25*x + 50*x*Log[x] + E^(E^(2*x) - x)*(1 - 2*E^(2*x))*Log[x]^2 + 2*x*Log[x]^2)/(3*Log[x]^2),x]

[Out]

(-E^(E^(2*x) - x) + x^2 + (25*x^2)/Log[x])/3

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fricas [A]  time = 0.66, size = 31, normalized size = 1.00 \begin {gather*} \frac {x^{2} \log \relax (x) + 25 \, x^{2} - e^{\left (-x + e^{\left (2 \, x\right )}\right )} \log \relax (x)}{3 \, \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*((-2*exp(x)^2+1)*log(x)^2*exp(exp(x)^2-x)+2*x*log(x)^2+50*x*log(x)-25*x)/log(x)^2,x, algorithm="
fricas")

[Out]

1/3*(x^2*log(x) + 25*x^2 - e^(-x + e^(2*x))*log(x))/log(x)

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giac [A]  time = 0.25, size = 41, normalized size = 1.32 \begin {gather*} \frac {{\left (x^{2} e^{\left (2 \, x\right )} \log \relax (x) + 25 \, x^{2} e^{\left (2 \, x\right )} - e^{\left (x + e^{\left (2 \, x\right )}\right )} \log \relax (x)\right )} e^{\left (-2 \, x\right )}}{3 \, \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*((-2*exp(x)^2+1)*log(x)^2*exp(exp(x)^2-x)+2*x*log(x)^2+50*x*log(x)-25*x)/log(x)^2,x, algorithm="
giac")

[Out]

1/3*(x^2*e^(2*x)*log(x) + 25*x^2*e^(2*x) - e^(x + e^(2*x))*log(x))*e^(-2*x)/log(x)

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maple [A]  time = 0.04, size = 27, normalized size = 0.87

method	result	size
default	\(\frac {x^{2}}{3}+\frac {25 x^{2}}{3 \ln \relax (x )}-\frac {{\mathrm e}^{{\mathrm e}^{2 x}} {\mathrm e}^{-x}}{3}\)	\(27\)
risch	\(\frac {x^{2}}{3}+\frac {25 x^{2}}{3 \ln \relax (x )}-\frac {{\mathrm e}^{{\mathrm e}^{2 x}-x}}{3}\)	\(27\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/3*((-2*exp(x)^2+1)*ln(x)^2*exp(exp(x)^2-x)+2*x*ln(x)^2+50*x*ln(x)-25*x)/ln(x)^2,x,method=_RETURNVERBOSE)

[Out]

1/3*x^2+25/3*x^2/ln(x)-1/3*exp(exp(x)^2)/exp(x)

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maxima [A]  time = 0.43, size = 33, normalized size = 1.06 \begin {gather*} \frac {1}{3} \, x^{2} + \frac {{\left (25 \, x^{2} e^{x} - e^{\left (e^{\left (2 \, x\right )}\right )} \log \relax (x)\right )} e^{\left (-x\right )}}{3 \, \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*((-2*exp(x)^2+1)*log(x)^2*exp(exp(x)^2-x)+2*x*log(x)^2+50*x*log(x)-25*x)/log(x)^2,x, algorithm="
maxima")

[Out]

1/3*x^2 + 1/3*(25*x^2*e^x - e^(e^(2*x))*log(x))*e^(-x)/log(x)

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mupad [B]  time = 4.21, size = 26, normalized size = 0.84 \begin {gather*} \frac {25\,x^2}{3\,\ln \relax (x)}-\frac {{\mathrm {e}}^{{\mathrm {e}}^{2\,x}-x}}{3}+\frac {x^2}{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-((25*x)/3 - (2*x*log(x)^2)/3 - (50*x*log(x))/3 + (exp(exp(2*x) - x)*log(x)^2*(2*exp(2*x) - 1))/3)/log(x)^
2,x)

[Out]

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

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sympy [A]  time = 0.34, size = 24, normalized size = 0.77 \begin {gather*} \frac {x^{2}}{3} + \frac {25 x^{2}}{3 \log {\relax (x )}} - \frac {e^{- x + e^{2 x}}}{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*((-2*exp(x)**2+1)*ln(x)**2*exp(exp(x)**2-x)+2*x*ln(x)**2+50*x*ln(x)-25*x)/ln(x)**2,x)

[Out]

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

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