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

Optimal. Leaf size=26 \[ x+\frac {-1+\frac {e^{e^x x^2}}{x}}{4 \log (\log (5))} \]

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Rubi [B]  time = 0.12, antiderivative size = 55, normalized size of antiderivative = 2.12, number of steps used = 4, number of rules used = 3, integrand size = 43, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.070, Rules used = {12, 14, 2288} \begin {gather*} \frac {e^{e^x x^2} \left (e^x x^3+2 e^x x^2\right )}{4 \left (e^x x^2+2 e^x x\right ) x^2 \log (\log (5))}+x \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int \frac {e^{e^x x^2} \left (-1+e^x x^2 (2+x)\right )+4 x^2 \log (\log (5))}{x^2} \, dx}{4 \log (\log (5))}\\ &=\frac {\int \left (\frac {e^{e^x x^2} \left (-1+2 e^x x^2+e^x x^3\right )}{x^2}+4 \log (\log (5))\right ) \, dx}{4 \log (\log (5))}\\ &=x+\frac {\int \frac {e^{e^x x^2} \left (-1+2 e^x x^2+e^x x^3\right )}{x^2} \, dx}{4 \log (\log (5))}\\ &=x+\frac {e^{e^x x^2} \left (2 e^x x^2+e^x x^3\right )}{4 x^2 \left (2 e^x x+e^x x^2\right ) \log (\log (5))}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.05, size = 23, normalized size = 0.88 \begin {gather*} x+\frac {e^{e^x x^2}}{4 x \log (\log (5))} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

x + E^(E^x*x^2)/(4*x*Log[Log[5]])

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fricas [A]  time = 0.47, size = 27, normalized size = 1.04 \begin {gather*} \frac {4 \, x^{2} \log \left (\log \relax (5)\right ) + e^{\left (e^{\left (x + \log \left (x^{2}\right )\right )}\right )}}{4 \, x \log \left (\log \relax (5)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(4*x^2*log(log(5)) + e^(e^(x + log(x^2))))/(x*log(log(5)))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.25, size = 20, normalized size = 0.77

method	result	size
risch	\(x +\frac {{\mathrm e}^{{\mathrm e}^{x} x^{2}}}{4 \ln \left (\ln \relax (5)\right ) x}\)	\(20\)
norman	\(\frac {x^{2}+\frac {{\mathrm e}^{{\mathrm e}^{\ln \left (x^{2}\right )+x}}}{4 \ln \left (\ln \relax (5)\right )}}{x}\)	\(24\)
default	\(\frac {4 \ln \left (\ln \relax (5)\right ) x^{2}+{\mathrm e}^{{\mathrm e}^{\ln \left (x^{2}\right )+x}}}{4 \ln \left (\ln \relax (5)\right ) x}\)	\(28\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.40, size = 25, normalized size = 0.96 \begin {gather*} \frac {4 \, x \log \left (\log \relax (5)\right ) + \frac {e^{\left (x^{2} e^{x}\right )}}{x}}{4 \, \log \left (\log \relax (5)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(4*x*log(log(5)) + e^(x^2*e^x)/x)/log(log(5))

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mupad [B]  time = 1.97, size = 19, normalized size = 0.73 \begin {gather*} x+\frac {{\mathrm {e}}^{x^2\,{\mathrm {e}}^x}}{4\,x\,\ln \left (\ln \relax (5)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x + exp(x^2*exp(x))/(4*x*log(log(5)))

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sympy [A]  time = 0.18, size = 17, normalized size = 0.65 \begin {gather*} x + \frac {e^{x^{2} e^{x}}}{4 x \log {\left (\log {\relax (5 )} \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x + exp(x**2*exp(x))/(4*x*log(log(5)))

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