3.45.37 \(\int \frac {-2 e^{2+\frac {2 e^2}{x}} \log (5)+e^{2 x} (2 x^3+2 x^4) \log (5)+e^x (4 x^3+8 x^4+2 x^5+e^3 (2 x^2+2 x^3)) \log (5)+(2 x^3+6 x^4+4 x^5+e^3 (2 x^2+4 x^3)) \log (5)+e^{\frac {e^2}{x}} (e^x (-2 e^2 x+2 x^2+2 x^3) \log (5)+(-2 e^5+2 x^2+4 x^3+e^2 (-2 x-2 x^2)) \log (5))}{x^2} \, dx\)

Optimal. Leaf size=26 \[ \left (e^3+e^{\frac {e^2}{x}}+x+x \left (e^x+x\right )\right )^2 \log (5) \]

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Rubi [B]  time = 1.53, antiderivative size = 191, normalized size of antiderivative = 7.35, number of steps used = 38, number of rules used = 12, integrand size = 177, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.068, Rules used = {14, 2196, 2176, 2194, 6688, 6742, 2288, 2209, 629, 2206, 2210, 2214} \begin {gather*} 2 e^x x^3 \log (5)+2 e^{\frac {e^2}{x}} x^2 \log (5)+2 e^x x^2 \log (5)+e^{2 x} x^2 \log (5)+\left (x^2+x+e^3\right )^2 \log (5)-\frac {2 e^{x+\frac {e^2}{x}} \left (e^2-x^2\right ) \log (5)}{\left (1-\frac {e^2}{x^2}\right ) x}+2 e^{\frac {e^2}{x}} x \log (5)-4 e^x x \log (5)+2 \left (2+e^3\right ) e^x x \log (5)+2 e^{\frac {e^2}{x}+3} \log (5)+e^{\frac {2 e^2}{x}} \log (5)+4 e^x \log (5)+2 e^{x+3} \log (5)-2 \left (2+e^3\right ) e^x \log (5) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-2*E^(2 + (2*E^2)/x)*Log[5] + E^(2*x)*(2*x^3 + 2*x^4)*Log[5] + E^x*(4*x^3 + 8*x^4 + 2*x^5 + E^3*(2*x^2 +
2*x^3))*Log[5] + (2*x^3 + 6*x^4 + 4*x^5 + E^3*(2*x^2 + 4*x^3))*Log[5] + E^(E^2/x)*(E^x*(-2*E^2*x + 2*x^2 + 2*x
^3)*Log[5] + (-2*E^5 + 2*x^2 + 4*x^3 + E^2*(-2*x - 2*x^2))*Log[5]))/x^2,x]

[Out]

2*E^(3 + E^2/x)*Log[5] + E^((2*E^2)/x)*Log[5] + 4*E^x*Log[5] + 2*E^(3 + x)*Log[5] - 2*E^x*(2 + E^3)*Log[5] + 2
*E^(E^2/x)*x*Log[5] - 4*E^x*x*Log[5] + 2*E^x*(2 + E^3)*x*Log[5] + 2*E^(E^2/x)*x^2*Log[5] + 2*E^x*x^2*Log[5] +
E^(2*x)*x^2*Log[5] + 2*E^x*x^3*Log[5] - (2*E^(E^2/x + x)*(E^2 - x^2)*Log[5])/((1 - E^2/x^2)*x) + (E^3 + x + x^
2)^2*Log[5]

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 629

Int[((d_) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(d*(a + b*x + c*x^2)^(p +
1))/(b*(p + 1)), x] /; FreeQ[{a, b, c, d, e, p}, x] && EqQ[2*c*d - b*e, 0] && NeQ[p, -1]

Rule 2176

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^m
*(b*F^(g*(e + f*x)))^n)/(f*g*n*Log[F]), x] - Dist[(d*m)/(f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*(b*F^(g*(e + f*x
)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && GtQ[m, 0] && IntegerQ[2*m] &&  !$UseGamma === True

Rule 2194

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rule 2196

Int[(F_)^((c_.)*(v_))*(u_), x_Symbol] :> Int[ExpandIntegrand[F^(c*ExpandToSum[v, x]), u, x], x] /; FreeQ[{F, c
}, x] && PolynomialQ[u, x] && LinearQ[v, x] &&  !$UseGamma === True

Rule 2206

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

Rule 2209

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Simp[((e + f*x)^n*
F^(a + b*(c + d*x)^n))/(b*f*n*(c + d*x)^n*Log[F]), x] /; FreeQ[{F, a, b, c, d, e, f, n}, x] && EqQ[m, n - 1] &
& EqQ[d*e - c*f, 0]

Rule 2210

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> Simp[(F^a*ExpIntegralEi[
b*(c + d*x)^n*Log[F]])/(f*n), x] /; FreeQ[{F, a, b, c, d, e, f, n}, x] && EqQ[d*e - c*f, 0]

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 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]

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 \left (2 e^{2 x} x (1+x) \log (5)+\frac {2 e^x \left (-e^{2+\frac {e^2}{x}}+e^3 x+e^{\frac {e^2}{x}} x+e^{\frac {e^2}{x}} x^2+2 \left (1+\frac {e^3}{2}\right ) x^2+4 x^3+x^4\right ) \log (5)}{x}+\frac {2 \left (-e^{5+\frac {e^2}{x}}-e^{2+\frac {2 e^2}{x}}-e^{2+\frac {e^2}{x}} x+e^3 x^2-\left (1-\frac {1}{e^2}\right ) e^{2+\frac {e^2}{x}} x^2+2 e^{\frac {e^2}{x}} x^3+\left (1+2 e^3\right ) x^3+3 x^4+2 x^5\right ) \log (5)}{x^2}\right ) \, dx\\ &=(2 \log (5)) \int e^{2 x} x (1+x) \, dx+(2 \log (5)) \int \frac {e^x \left (-e^{2+\frac {e^2}{x}}+e^3 x+e^{\frac {e^2}{x}} x+e^{\frac {e^2}{x}} x^2+2 \left (1+\frac {e^3}{2}\right ) x^2+4 x^3+x^4\right )}{x} \, dx+(2 \log (5)) \int \frac {-e^{5+\frac {e^2}{x}}-e^{2+\frac {2 e^2}{x}}-e^{2+\frac {e^2}{x}} x+e^3 x^2-\left (1-\frac {1}{e^2}\right ) e^{2+\frac {e^2}{x}} x^2+2 e^{\frac {e^2}{x}} x^3+\left (1+2 e^3\right ) x^3+3 x^4+2 x^5}{x^2} \, dx\\ &=(2 \log (5)) \int \left (e^{2 x} x+e^{2 x} x^2\right ) \, dx+(2 \log (5)) \int \frac {e^x \left (-e^{2+\frac {e^2}{x}}+e^3 x (1+x)+e^{\frac {e^2}{x}} x (1+x)+x^2 \left (2+4 x+x^2\right )\right )}{x} \, dx+(2 \log (5)) \int \left (-\frac {e^{2+\frac {2 e^2}{x}}}{x^2}+(1+2 x) \left (e^3+x+x^2\right )+\frac {e^{\frac {e^2}{x}} \left (-e^5-e^2 x+\left (1-e^2\right ) x^2+2 x^3\right )}{x^2}\right ) \, dx\\ &=-\left ((2 \log (5)) \int \frac {e^{2+\frac {2 e^2}{x}}}{x^2} \, dx\right )+(2 \log (5)) \int e^{2 x} x \, dx+(2 \log (5)) \int e^{2 x} x^2 \, dx+(2 \log (5)) \int (1+2 x) \left (e^3+x+x^2\right ) \, dx+(2 \log (5)) \int \frac {e^{\frac {e^2}{x}} \left (-e^5-e^2 x+\left (1-e^2\right ) x^2+2 x^3\right )}{x^2} \, dx+(2 \log (5)) \int \left (e^{3+x}+2 e^x \left (1+\frac {e^3}{2}\right ) x+4 e^x x^2+e^x x^3+\frac {e^{\frac {e^2}{x}+x} \left (-e^2+x+x^2\right )}{x}\right ) \, dx\\ &=e^{\frac {2 e^2}{x}} \log (5)+e^{2 x} x \log (5)+e^{2 x} x^2 \log (5)+\left (e^3+x+x^2\right )^2 \log (5)-\log (5) \int e^{2 x} \, dx+(2 \log (5)) \int e^{3+x} \, dx-(2 \log (5)) \int e^{2 x} x \, dx+(2 \log (5)) \int e^x x^3 \, dx+(2 \log (5)) \int \left (-e^{2+\frac {e^2}{x}}+e^{\frac {e^2}{x}}-\frac {e^{5+\frac {e^2}{x}}}{x^2}-\frac {e^{2+\frac {e^2}{x}}}{x}+2 e^{\frac {e^2}{x}} x\right ) \, dx+(2 \log (5)) \int \frac {e^{\frac {e^2}{x}+x} \left (-e^2+x+x^2\right )}{x} \, dx+(8 \log (5)) \int e^x x^2 \, dx+\left (2 \left (2+e^3\right ) \log (5)\right ) \int e^x x \, dx\\ &=e^{\frac {2 e^2}{x}} \log (5)-\frac {1}{2} e^{2 x} \log (5)+2 e^{3+x} \log (5)+2 e^x \left (2+e^3\right ) x \log (5)+8 e^x x^2 \log (5)+e^{2 x} x^2 \log (5)+2 e^x x^3 \log (5)-\frac {2 e^{\frac {e^2}{x}+x} \left (e^2-x^2\right ) \log (5)}{\left (1-\frac {e^2}{x^2}\right ) x}+\left (e^3+x+x^2\right )^2 \log (5)+\log (5) \int e^{2 x} \, dx-(2 \log (5)) \int e^{2+\frac {e^2}{x}} \, dx+(2 \log (5)) \int e^{\frac {e^2}{x}} \, dx-(2 \log (5)) \int \frac {e^{5+\frac {e^2}{x}}}{x^2} \, dx-(2 \log (5)) \int \frac {e^{2+\frac {e^2}{x}}}{x} \, dx+(4 \log (5)) \int e^{\frac {e^2}{x}} x \, dx-(6 \log (5)) \int e^x x^2 \, dx-(16 \log (5)) \int e^x x \, dx-\left (2 \left (2+e^3\right ) \log (5)\right ) \int e^x \, dx\\ &=2 e^{3+\frac {e^2}{x}} \log (5)+e^{\frac {2 e^2}{x}} \log (5)+2 e^{3+x} \log (5)-2 e^x \left (2+e^3\right ) \log (5)-2 e^{2+\frac {e^2}{x}} x \log (5)+2 e^{\frac {e^2}{x}} x \log (5)-16 e^x x \log (5)+2 e^x \left (2+e^3\right ) x \log (5)+2 e^{\frac {e^2}{x}} x^2 \log (5)+2 e^x x^2 \log (5)+e^{2 x} x^2 \log (5)+2 e^x x^3 \log (5)-\frac {2 e^{\frac {e^2}{x}+x} \left (e^2-x^2\right ) \log (5)}{\left (1-\frac {e^2}{x^2}\right ) x}+\left (e^3+x+x^2\right )^2 \log (5)+2 e^2 \text {Ei}\left (\frac {e^2}{x}\right ) \log (5)+(12 \log (5)) \int e^x x \, dx+(16 \log (5)) \int e^x \, dx+\left (2 e^2 \log (5)\right ) \int e^{\frac {e^2}{x}} \, dx-\left (2 e^2 \log (5)\right ) \int \frac {e^{2+\frac {e^2}{x}}}{x} \, dx+\left (2 e^2 \log (5)\right ) \int \frac {e^{\frac {e^2}{x}}}{x} \, dx\\ &=2 e^{3+\frac {e^2}{x}} \log (5)+e^{\frac {2 e^2}{x}} \log (5)+16 e^x \log (5)+2 e^{3+x} \log (5)-2 e^x \left (2+e^3\right ) \log (5)+2 e^{\frac {e^2}{x}} x \log (5)-4 e^x x \log (5)+2 e^x \left (2+e^3\right ) x \log (5)+2 e^{\frac {e^2}{x}} x^2 \log (5)+2 e^x x^2 \log (5)+e^{2 x} x^2 \log (5)+2 e^x x^3 \log (5)-\frac {2 e^{\frac {e^2}{x}+x} \left (e^2-x^2\right ) \log (5)}{\left (1-\frac {e^2}{x^2}\right ) x}+\left (e^3+x+x^2\right )^2 \log (5)+2 e^4 \text {Ei}\left (\frac {e^2}{x}\right ) \log (5)-(12 \log (5)) \int e^x \, dx+\left (2 e^4 \log (5)\right ) \int \frac {e^{\frac {e^2}{x}}}{x} \, dx\\ &=2 e^{3+\frac {e^2}{x}} \log (5)+e^{\frac {2 e^2}{x}} \log (5)+4 e^x \log (5)+2 e^{3+x} \log (5)-2 e^x \left (2+e^3\right ) \log (5)+2 e^{\frac {e^2}{x}} x \log (5)-4 e^x x \log (5)+2 e^x \left (2+e^3\right ) x \log (5)+2 e^{\frac {e^2}{x}} x^2 \log (5)+2 e^x x^2 \log (5)+e^{2 x} x^2 \log (5)+2 e^x x^3 \log (5)-\frac {2 e^{\frac {e^2}{x}+x} \left (e^2-x^2\right ) \log (5)}{\left (1-\frac {e^2}{x^2}\right ) x}+\left (e^3+x+x^2\right )^2 \log (5)\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.19, size = 106, normalized size = 4.08 \begin {gather*} 2 \left (e^{3+\frac {e^2}{x}}+\frac {1}{2} e^{\frac {2 e^2}{x}}+e^{3+x} x+e^{\frac {e^2}{x}+x} x+\frac {1}{2} e^{2 x} x^2+e^3 x (1+x)+e^{\frac {e^2}{x}} x (1+x)+e^x x^2 (1+x)+\frac {1}{2} x^2 (1+x)^2\right ) \log (5) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-2*E^(2 + (2*E^2)/x)*Log[5] + E^(2*x)*(2*x^3 + 2*x^4)*Log[5] + E^x*(4*x^3 + 8*x^4 + 2*x^5 + E^3*(2*
x^2 + 2*x^3))*Log[5] + (2*x^3 + 6*x^4 + 4*x^5 + E^3*(2*x^2 + 4*x^3))*Log[5] + E^(E^2/x)*(E^x*(-2*E^2*x + 2*x^2
+ 2*x^3)*Log[5] + (-2*E^5 + 2*x^2 + 4*x^3 + E^2*(-2*x - 2*x^2))*Log[5]))/x^2,x]

[Out]

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

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fricas [B]  time = 0.67, size = 89, normalized size = 3.42 \begin {gather*} x^{2} e^{\left (2 \, x\right )} \log \relax (5) + 2 \, {\left (x^{3} + x^{2} + x e^{3}\right )} e^{x} \log \relax (5) + 2 \, {\left (x e^{x} \log \relax (5) + {\left (x^{2} + x + e^{3}\right )} \log \relax (5)\right )} e^{\left (\frac {e^{2}}{x}\right )} + {\left (x^{4} + 2 \, x^{3} + x^{2} + 2 \, {\left (x^{2} + x\right )} e^{3}\right )} \log \relax (5) + e^{\left (\frac {2 \, e^{2}}{x}\right )} \log \relax (5) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*exp(2)*log(5)*exp(exp(2)/x)^2+((-2*exp(2)*x+2*x^3+2*x^2)*log(5)*exp(x)+(-2*exp(2)*exp(3)+(-2*x^2
-2*x)*exp(2)+4*x^3+2*x^2)*log(5))*exp(exp(2)/x)+(2*x^4+2*x^3)*log(5)*exp(x)^2+((2*x^3+2*x^2)*exp(3)+2*x^5+8*x^
4+4*x^3)*log(5)*exp(x)+((4*x^3+2*x^2)*exp(3)+4*x^5+6*x^4+2*x^3)*log(5))/x^2,x, algorithm="fricas")

[Out]

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

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giac [C]  time = 0.28, size = 238, normalized size = 9.15 \begin {gather*} x^{4} \log \relax (5) - 2 \, x^{2} {\left (\frac {{\rm Ei}\left (\frac {e^{2}}{x}\right ) e^{10}}{x^{2}} - \frac {e^{\left (\frac {e^{2}}{x} + 8\right )}}{x} - e^{\left (\frac {e^{2}}{x} + 6\right )}\right )} e^{\left (-6\right )} \log \relax (5) + 2 \, x^{3} e^{x} \log \relax (5) + 2 \, x^{3} \log \relax (5) + 2 \, x^{2} e^{3} \log \relax (5) + 2 \, x {\left (\frac {{\rm Ei}\left (\frac {e^{2}}{x}\right ) e^{8}}{x} - e^{\left (\frac {e^{2}}{x} + 6\right )}\right )} e^{\left (-4\right )} \log \relax (5) - 2 \, x {\left (\frac {{\rm Ei}\left (\frac {e^{2}}{x}\right ) e^{6}}{x} - e^{\left (\frac {e^{2}}{x} + 4\right )}\right )} e^{\left (-4\right )} \log \relax (5) + x^{2} e^{\left (2 \, x\right )} \log \relax (5) + 2 \, x^{2} e^{x} \log \relax (5) + x^{2} \log \relax (5) + 2 \, x e^{3} \log \relax (5) + 2 \, {\rm Ei}\left (\frac {e^{2}}{x}\right ) e^{2} \log \relax (5) + 2 \, x e^{\left (x + 3\right )} \log \relax (5) + 2 \, x e^{\left (\frac {x^{2} + e^{2}}{x}\right )} \log \relax (5) + e^{\left (\frac {2 \, e^{2}}{x}\right )} \log \relax (5) + 2 \, e^{\left (\frac {e^{2}}{x} + 3\right )} \log \relax (5) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*exp(2)*log(5)*exp(exp(2)/x)^2+((-2*exp(2)*x+2*x^3+2*x^2)*log(5)*exp(x)+(-2*exp(2)*exp(3)+(-2*x^2
-2*x)*exp(2)+4*x^3+2*x^2)*log(5))*exp(exp(2)/x)+(2*x^4+2*x^3)*log(5)*exp(x)^2+((2*x^3+2*x^2)*exp(3)+2*x^5+8*x^
4+4*x^3)*log(5)*exp(x)+((4*x^3+2*x^2)*exp(3)+4*x^5+6*x^4+2*x^3)*log(5))/x^2,x, algorithm="giac")

[Out]

x^4*log(5) - 2*x^2*(Ei(e^2/x)*e^10/x^2 - e^(e^2/x + 8)/x - e^(e^2/x + 6))*e^(-6)*log(5) + 2*x^3*e^x*log(5) + 2
*x^3*log(5) + 2*x^2*e^3*log(5) + 2*x*(Ei(e^2/x)*e^8/x - e^(e^2/x + 6))*e^(-4)*log(5) - 2*x*(Ei(e^2/x)*e^6/x -
e^(e^2/x + 4))*e^(-4)*log(5) + x^2*e^(2*x)*log(5) + 2*x^2*e^x*log(5) + x^2*log(5) + 2*x*e^3*log(5) + 2*Ei(e^2/
x)*e^2*log(5) + 2*x*e^(x + 3)*log(5) + 2*x*e^((x^2 + e^2)/x)*log(5) + e^(2*e^2/x)*log(5) + 2*e^(e^2/x + 3)*log
(5)

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maple [B]  time = 0.13, size = 117, normalized size = 4.50

method	result	size
risch	\(x^{4} \ln \relax (5)+2 x^{2} {\mathrm e}^{3} \ln \relax (5)+2 x^{3} \ln \relax (5)+2 x \,{\mathrm e}^{3} \ln \relax (5)+x^{2} \ln \relax (5)+x^{2} \ln \relax (5) {\mathrm e}^{2 x}+\left (2 x \,{\mathrm e}^{3} \ln \relax (5)+2 x^{3} \ln \relax (5)+2 x^{2} \ln \relax (5)\right ) {\mathrm e}^{x}+\ln \relax (5) {\mathrm e}^{\frac {2 \,{\mathrm e}^{2}}{x}}+\left (2 \,{\mathrm e}^{3} \ln \relax (5)+2 x^{2} \ln \relax (5)+2 x \,{\mathrm e}^{x} \ln \relax (5)+2 x \ln \relax (5)\right ) {\mathrm e}^{\frac {{\mathrm e}^{2}}{x}}\)	\(117\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-2*exp(2)*ln(5)*exp(exp(2)/x)^2+((-2*exp(2)*x+2*x^3+2*x^2)*ln(5)*exp(x)+(-2*exp(2)*exp(3)+(-2*x^2-2*x)*ex
p(2)+4*x^3+2*x^2)*ln(5))*exp(exp(2)/x)+(2*x^4+2*x^3)*ln(5)*exp(x)^2+((2*x^3+2*x^2)*exp(3)+2*x^5+8*x^4+4*x^3)*l
n(5)*exp(x)+((4*x^3+2*x^2)*exp(3)+4*x^5+6*x^4+2*x^3)*ln(5))/x^2,x,method=_RETURNVERBOSE)

[Out]

x^4*ln(5)+2*x^2*exp(3)*ln(5)+2*x^3*ln(5)+2*x*exp(3)*ln(5)+x^2*ln(5)+x^2*ln(5)*exp(2*x)+(2*x*exp(3)*ln(5)+2*x^3
*ln(5)+2*x^2*ln(5))*exp(x)+ln(5)*exp(2*exp(2)/x)+(2*exp(3)*ln(5)+2*x^2*ln(5)+2*x*exp(x)*ln(5)+2*x*ln(5))*exp(e
xp(2)/x)

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maxima [C]  time = 0.53, size = 228, normalized size = 8.77 \begin {gather*} x^{4} \log \relax (5) + 2 \, x^{3} \log \relax (5) + 2 \, x^{2} e^{3} \log \relax (5) + x^{2} \log \relax (5) + 2 \, x e^{3} \log \relax (5) + 2 \, {\rm Ei}\left (\frac {e^{2}}{x}\right ) e^{2} \log \relax (5) + \frac {1}{2} \, {\left (2 \, x^{2} - 2 \, x + 1\right )} e^{\left (2 \, x\right )} \log \relax (5) + \frac {1}{2} \, {\left (2 \, x - 1\right )} e^{\left (2 \, x\right )} \log \relax (5) + 2 \, x e^{\left (x + \frac {e^{2}}{x}\right )} \log \relax (5) + 2 \, {\left (x^{3} - 3 \, x^{2} + 6 \, x - 6\right )} e^{x} \log \relax (5) + 8 \, {\left (x^{2} - 2 \, x + 2\right )} e^{x} \log \relax (5) + 2 \, {\left (x e^{3} - e^{3}\right )} e^{x} \log \relax (5) + 4 \, {\left (x - 1\right )} e^{x} \log \relax (5) + 2 \, e^{4} \Gamma \left (-1, -\frac {e^{2}}{x}\right ) \log \relax (5) - 2 \, e^{2} \Gamma \left (-1, -\frac {e^{2}}{x}\right ) \log \relax (5) + 4 \, e^{4} \Gamma \left (-2, -\frac {e^{2}}{x}\right ) \log \relax (5) + 2 \, e^{\left (x + 3\right )} \log \relax (5) + e^{\left (\frac {2 \, e^{2}}{x}\right )} \log \relax (5) + 2 \, e^{\left (\frac {e^{2}}{x} + 3\right )} \log \relax (5) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*exp(2)*log(5)*exp(exp(2)/x)^2+((-2*exp(2)*x+2*x^3+2*x^2)*log(5)*exp(x)+(-2*exp(2)*exp(3)+(-2*x^2
-2*x)*exp(2)+4*x^3+2*x^2)*log(5))*exp(exp(2)/x)+(2*x^4+2*x^3)*log(5)*exp(x)^2+((2*x^3+2*x^2)*exp(3)+2*x^5+8*x^
4+4*x^3)*log(5)*exp(x)+((4*x^3+2*x^2)*exp(3)+4*x^5+6*x^4+2*x^3)*log(5))/x^2,x, algorithm="maxima")

[Out]

x^4*log(5) + 2*x^3*log(5) + 2*x^2*e^3*log(5) + x^2*log(5) + 2*x*e^3*log(5) + 2*Ei(e^2/x)*e^2*log(5) + 1/2*(2*x
^2 - 2*x + 1)*e^(2*x)*log(5) + 1/2*(2*x - 1)*e^(2*x)*log(5) + 2*x*e^(x + e^2/x)*log(5) + 2*(x^3 - 3*x^2 + 6*x
- 6)*e^x*log(5) + 8*(x^2 - 2*x + 2)*e^x*log(5) + 2*(x*e^3 - e^3)*e^x*log(5) + 4*(x - 1)*e^x*log(5) + 2*e^4*gam
ma(-1, -e^2/x)*log(5) - 2*e^2*gamma(-1, -e^2/x)*log(5) + 4*e^4*gamma(-2, -e^2/x)*log(5) + 2*e^(x + 3)*log(5) +
e^(2*e^2/x)*log(5) + 2*e^(e^2/x + 3)*log(5)

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mupad [B]  time = 4.26, size = 91, normalized size = 3.50 \begin {gather*} x^2\,\left (\ln \relax (5)+2\,{\mathrm {e}}^3\,\ln \relax (5)\right )+2\,x^3\,\ln \relax (5)+x^4\,\ln \relax (5)+{\mathrm {e}}^{\frac {2\,{\mathrm {e}}^2}{x}}\,\ln \relax (5)+2\,{\mathrm {e}}^{\frac {{\mathrm {e}}^2}{x}}\,\ln \relax (5)\,\left (x+{\mathrm {e}}^3+x\,{\mathrm {e}}^x+x^2\right )+2\,x\,{\mathrm {e}}^3\,\ln \relax (5)+x^2\,{\mathrm {e}}^{2\,x}\,\ln \relax (5)+2\,x\,{\mathrm {e}}^x\,\ln \relax (5)\,\left (x^2+x+{\mathrm {e}}^3\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(5)*(exp(3)*(2*x^2 + 4*x^3) + 2*x^3 + 6*x^4 + 4*x^5) - exp(exp(2)/x)*(log(5)*(2*exp(5) + exp(2)*(2*x +
2*x^2) - 2*x^2 - 4*x^3) - exp(x)*log(5)*(2*x^2 - 2*x*exp(2) + 2*x^3)) - 2*exp((2*exp(2))/x)*exp(2)*log(5) + e
xp(x)*log(5)*(exp(3)*(2*x^2 + 2*x^3) + 4*x^3 + 8*x^4 + 2*x^5) + exp(2*x)*log(5)*(2*x^3 + 2*x^4))/x^2,x)

[Out]

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

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sympy [B]  time = 10.46, size = 134, normalized size = 5.15 \begin {gather*} x^{4} \log {\relax (5 )} + 2 x^{3} \log {\relax (5 )} + x^{2} e^{2 x} \log {\relax (5 )} + x^{2} \left (\log {\relax (5 )} + 2 e^{3} \log {\relax (5 )}\right ) + 2 x e^{3} \log {\relax (5 )} + \left (2 x^{3} \log {\relax (5 )} + 2 x^{2} \log {\relax (5 )} + 2 x e^{3} \log {\relax (5 )}\right ) e^{x} + \left (2 x^{2} \log {\relax (5 )} + 2 x e^{x} \log {\relax (5 )} + 2 x \log {\relax (5 )} + 2 e^{3} \log {\relax (5 )}\right ) e^{\frac {e^{2}}{x}} + e^{\frac {2 e^{2}}{x}} \log {\relax (5 )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*exp(2)*ln(5)*exp(exp(2)/x)**2+((-2*exp(2)*x+2*x**3+2*x**2)*ln(5)*exp(x)+(-2*exp(2)*exp(3)+(-2*x*
*2-2*x)*exp(2)+4*x**3+2*x**2)*ln(5))*exp(exp(2)/x)+(2*x**4+2*x**3)*ln(5)*exp(x)**2+((2*x**3+2*x**2)*exp(3)+2*x
**5+8*x**4+4*x**3)*ln(5)*exp(x)+((4*x**3+2*x**2)*exp(3)+4*x**5+6*x**4+2*x**3)*ln(5))/x**2,x)

[Out]

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

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