3.77.21 \(\int \frac {-e^e+e^{2/x} (-4+2 x+x^2) \log ^2(2)+e^{\frac {1}{x}} (-4+6 x+2 x^2) \log ^2(2) \log (5)+(4 x+x^2) \log ^2(2) \log ^2(5)}{8+8 x+2 x^2} \, dx\)

Optimal. Leaf size=31 \[ \frac {e^e+x^2 \log ^2(2) \left (e^{\frac {1}{x}}+\log (5)\right )^2}{2 (2+x)} \]

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Rubi [B]  time = 1.16, antiderivative size = 128, normalized size of antiderivative = 4.13, number of steps used = 40, number of rules used = 11, integrand size = 77, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {27, 12, 6742, 2206, 2210, 2223, 2209, 2222, 2228, 2178, 683} \begin {gather*} \frac {1}{2} x \log ^2(2) \log ^2(5)+e^{\frac {1}{x}} x \log ^2(2) \log (5)+\frac {1}{2} e^{2/x} x \log ^2(2)+\frac {e^e+4 \log ^2(2) \log ^2(5)}{2 (x+2)}-2 e^{\frac {1}{x}} \log ^2(2) \log (5)+\frac {4 e^{\frac {1}{x}} \log ^2(2) \log (5)}{x+2}-e^{2/x} \log ^2(2)+\frac {2 e^{2/x} \log ^2(2)}{x+2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-E^E + E^(2/x)*(-4 + 2*x + x^2)*Log[2]^2 + E^x^(-1)*(-4 + 6*x + 2*x^2)*Log[2]^2*Log[5] + (4*x + x^2)*Log[
2]^2*Log[5]^2)/(8 + 8*x + 2*x^2),x]

[Out]

-(E^(2/x)*Log[2]^2) + (E^(2/x)*x*Log[2]^2)/2 + (2*E^(2/x)*Log[2]^2)/(2 + x) - 2*E^x^(-1)*Log[2]^2*Log[5] + E^x
^(-1)*x*Log[2]^2*Log[5] + (4*E^x^(-1)*Log[2]^2*Log[5])/(2 + x) + (x*Log[2]^2*Log[5]^2)/2 + (E^E + 4*Log[2]^2*L
og[5]^2)/(2*(2 + 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 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 683

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d +
e*x)^m*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[2*c*d - b*e,
0] && IGtQ[p, 0] &&  !(EqQ[m, 3] && NeQ[p, 1])

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 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 2222

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

Rule 2223

Int[(F_)^((a_.) + (b_.)/((c_.) + (d_.)*(x_)))*((e_.) + (f_.)*(x_))^(m_), x_Symbol] :> Simp[((e + f*x)^(m + 1)*
F^(a + b/(c + d*x)))/(f*(m + 1)), x] + Dist[(b*d*Log[F])/(f*(m + 1)), Int[((e + f*x)^(m + 1)*F^(a + b/(c + d*x
)))/(c + d*x)^2, x], x] /; FreeQ[{F, a, b, c, d, e, f}, x] && NeQ[d*e - c*f, 0] && ILtQ[m, -1]

Rule 2228

Int[(F_)^((a_.) + (b_.)/((c_.) + (d_.)*(x_)))/(((e_.) + (f_.)*(x_))*((g_.) + (h_.)*(x_))), x_Symbol] :> -Dist[
d/(f*(d*g - c*h)), Subst[Int[F^(a - (b*h)/(d*g - c*h) + (d*b*x)/(d*g - c*h))/x, x], x, (g + h*x)/(c + d*x)], x
] /; FreeQ[{F, a, b, c, d, e, f}, x] && EqQ[d*e - c*f, 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 \frac {-e^e+e^{2/x} \left (-4+2 x+x^2\right ) \log ^2(2)+e^{\frac {1}{x}} \left (-4+6 x+2 x^2\right ) \log ^2(2) \log (5)+\left (4 x+x^2\right ) \log ^2(2) \log ^2(5)}{2 (2+x)^2} \, dx\\ &=\frac {1}{2} \int \frac {-e^e+e^{2/x} \left (-4+2 x+x^2\right ) \log ^2(2)+e^{\frac {1}{x}} \left (-4+6 x+2 x^2\right ) \log ^2(2) \log (5)+\left (4 x+x^2\right ) \log ^2(2) \log ^2(5)}{(2+x)^2} \, dx\\ &=\frac {1}{2} \int \left (\frac {e^{2/x} \left (-4+2 x+x^2\right ) \log ^2(2)}{(2+x)^2}+\frac {2 e^{\frac {1}{x}} \left (-2+3 x+x^2\right ) \log ^2(2) \log (5)}{(2+x)^2}+\frac {-e^e+4 x \log ^2(2) \log ^2(5)+x^2 \log ^2(2) \log ^2(5)}{(2+x)^2}\right ) \, dx\\ &=\frac {1}{2} \int \frac {-e^e+4 x \log ^2(2) \log ^2(5)+x^2 \log ^2(2) \log ^2(5)}{(2+x)^2} \, dx+\frac {1}{2} \log ^2(2) \int \frac {e^{2/x} \left (-4+2 x+x^2\right )}{(2+x)^2} \, dx+\left (\log ^2(2) \log (5)\right ) \int \frac {e^{\frac {1}{x}} \left (-2+3 x+x^2\right )}{(2+x)^2} \, dx\\ &=\frac {1}{2} \int \left (\log ^2(2) \log ^2(5)-\frac {e^e+4 \log ^2(2) \log ^2(5)}{(2+x)^2}\right ) \, dx+\frac {1}{2} \log ^2(2) \int \left (e^{2/x}-\frac {4 e^{2/x}}{(2+x)^2}-\frac {2 e^{2/x}}{2+x}\right ) \, dx+\left (\log ^2(2) \log (5)\right ) \int \left (e^{\frac {1}{x}}+\frac {e^{\frac {1}{x}}}{-2-x}-\frac {4 e^{\frac {1}{x}}}{(2+x)^2}\right ) \, dx\\ &=\frac {1}{2} x \log ^2(2) \log ^2(5)+\frac {e^e+4 \log ^2(2) \log ^2(5)}{2 (2+x)}+\frac {1}{2} \log ^2(2) \int e^{2/x} \, dx-\log ^2(2) \int \frac {e^{2/x}}{2+x} \, dx-\left (2 \log ^2(2)\right ) \int \frac {e^{2/x}}{(2+x)^2} \, dx+\left (\log ^2(2) \log (5)\right ) \int e^{\frac {1}{x}} \, dx+\left (\log ^2(2) \log (5)\right ) \int \frac {e^{\frac {1}{x}}}{-2-x} \, dx-\left (4 \log ^2(2) \log (5)\right ) \int \frac {e^{\frac {1}{x}}}{(2+x)^2} \, dx\\ &=\frac {1}{2} e^{2/x} x \log ^2(2)+\frac {2 e^{2/x} \log ^2(2)}{2+x}+e^{\frac {1}{x}} x \log ^2(2) \log (5)+\frac {4 e^{\frac {1}{x}} \log ^2(2) \log (5)}{2+x}+\frac {1}{2} x \log ^2(2) \log ^2(5)+\frac {e^e+4 \log ^2(2) \log ^2(5)}{2 (2+x)}+\left (2 \log ^2(2)\right ) \int \frac {e^{2/x}}{x (2+x)} \, dx+\left (4 \log ^2(2)\right ) \int \frac {e^{2/x}}{x^2 (2+x)} \, dx-\left (2 \log ^2(2) \log (5)\right ) \int \frac {e^{\frac {1}{x}}}{(-2-x) x} \, dx+\left (4 \log ^2(2) \log (5)\right ) \int \frac {e^{\frac {1}{x}}}{x^2 (2+x)} \, dx\\ &=\frac {1}{2} e^{2/x} x \log ^2(2)+\frac {2 e^{2/x} \log ^2(2)}{2+x}+e^{\frac {1}{x}} x \log ^2(2) \log (5)+\frac {4 e^{\frac {1}{x}} \log ^2(2) \log (5)}{2+x}+\frac {1}{2} x \log ^2(2) \log ^2(5)+\frac {e^e+4 \log ^2(2) \log ^2(5)}{2 (2+x)}-\log ^2(2) \operatorname {Subst}\left (\int \frac {e^{-1+x}}{x} \, dx,x,\frac {2+x}{x}\right )+\left (4 \log ^2(2)\right ) \int \left (\frac {e^{2/x}}{2 x^2}-\frac {e^{2/x}}{4 x}+\frac {e^{2/x}}{4 (2+x)}\right ) \, dx-\left (\log ^2(2) \log (5)\right ) \operatorname {Subst}\left (\int \frac {e^{-\frac {1}{2}-\frac {x}{2}}}{x} \, dx,x,\frac {-2-x}{x}\right )+\left (4 \log ^2(2) \log (5)\right ) \int \left (\frac {e^{\frac {1}{x}}}{2 x^2}-\frac {e^{\frac {1}{x}}}{4 x}+\frac {e^{\frac {1}{x}}}{4 (2+x)}\right ) \, dx\\ &=\frac {1}{2} e^{2/x} x \log ^2(2)+\frac {2 e^{2/x} \log ^2(2)}{2+x}-\frac {\text {Ei}\left (\frac {2+x}{x}\right ) \log ^2(2)}{e}+e^{\frac {1}{x}} x \log ^2(2) \log (5)+\frac {4 e^{\frac {1}{x}} \log ^2(2) \log (5)}{2+x}-\frac {\text {Ei}\left (\frac {1}{2}+\frac {1}{x}\right ) \log ^2(2) \log (5)}{\sqrt {e}}+\frac {1}{2} x \log ^2(2) \log ^2(5)+\frac {e^e+4 \log ^2(2) \log ^2(5)}{2 (2+x)}-\log ^2(2) \int \frac {e^{2/x}}{x} \, dx+\log ^2(2) \int \frac {e^{2/x}}{2+x} \, dx+\left (2 \log ^2(2)\right ) \int \frac {e^{2/x}}{x^2} \, dx-\left (\log ^2(2) \log (5)\right ) \int \frac {e^{\frac {1}{x}}}{x} \, dx+\left (\log ^2(2) \log (5)\right ) \int \frac {e^{\frac {1}{x}}}{2+x} \, dx+\left (2 \log ^2(2) \log (5)\right ) \int \frac {e^{\frac {1}{x}}}{x^2} \, dx\\ &=-e^{2/x} \log ^2(2)+\frac {1}{2} e^{2/x} x \log ^2(2)+\frac {2 e^{2/x} \log ^2(2)}{2+x}+\text {Ei}\left (\frac {2}{x}\right ) \log ^2(2)-\frac {\text {Ei}\left (\frac {2+x}{x}\right ) \log ^2(2)}{e}-2 e^{\frac {1}{x}} \log ^2(2) \log (5)+e^{\frac {1}{x}} x \log ^2(2) \log (5)+\frac {4 e^{\frac {1}{x}} \log ^2(2) \log (5)}{2+x}-\frac {\text {Ei}\left (\frac {1}{2}+\frac {1}{x}\right ) \log ^2(2) \log (5)}{\sqrt {e}}+\text {Ei}\left (\frac {1}{x}\right ) \log ^2(2) \log (5)+\frac {1}{2} x \log ^2(2) \log ^2(5)+\frac {e^e+4 \log ^2(2) \log ^2(5)}{2 (2+x)}+\log ^2(2) \int \frac {e^{2/x}}{x} \, dx-\left (2 \log ^2(2)\right ) \int \frac {e^{2/x}}{x (2+x)} \, dx+\left (\log ^2(2) \log (5)\right ) \int \frac {e^{\frac {1}{x}}}{x} \, dx-\left (2 \log ^2(2) \log (5)\right ) \int \frac {e^{\frac {1}{x}}}{x (2+x)} \, dx\\ &=-e^{2/x} \log ^2(2)+\frac {1}{2} e^{2/x} x \log ^2(2)+\frac {2 e^{2/x} \log ^2(2)}{2+x}-\frac {\text {Ei}\left (\frac {2+x}{x}\right ) \log ^2(2)}{e}-2 e^{\frac {1}{x}} \log ^2(2) \log (5)+e^{\frac {1}{x}} x \log ^2(2) \log (5)+\frac {4 e^{\frac {1}{x}} \log ^2(2) \log (5)}{2+x}-\frac {\text {Ei}\left (\frac {1}{2}+\frac {1}{x}\right ) \log ^2(2) \log (5)}{\sqrt {e}}+\frac {1}{2} x \log ^2(2) \log ^2(5)+\frac {e^e+4 \log ^2(2) \log ^2(5)}{2 (2+x)}+\log ^2(2) \operatorname {Subst}\left (\int \frac {e^{-1+x}}{x} \, dx,x,\frac {2+x}{x}\right )+\left (\log ^2(2) \log (5)\right ) \operatorname {Subst}\left (\int \frac {e^{-\frac {1}{2}+\frac {x}{2}}}{x} \, dx,x,\frac {2+x}{x}\right )\\ &=-e^{2/x} \log ^2(2)+\frac {1}{2} e^{2/x} x \log ^2(2)+\frac {2 e^{2/x} \log ^2(2)}{2+x}-2 e^{\frac {1}{x}} \log ^2(2) \log (5)+e^{\frac {1}{x}} x \log ^2(2) \log (5)+\frac {4 e^{\frac {1}{x}} \log ^2(2) \log (5)}{2+x}+\frac {1}{2} x \log ^2(2) \log ^2(5)+\frac {e^e+4 \log ^2(2) \log ^2(5)}{2 (2+x)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.31, size = 60, normalized size = 1.94 \begin {gather*} \frac {e^e+e^{2/x} x^2 \log ^2(2)+\left (4+2 x+x^2\right ) \log ^2(2) \log ^2(5)+e^{\frac {1}{x}} x^2 \log ^2(2) \log (25)}{2 (2+x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-E^E + E^(2/x)*(-4 + 2*x + x^2)*Log[2]^2 + E^x^(-1)*(-4 + 6*x + 2*x^2)*Log[2]^2*Log[5] + (4*x + x^2
)*Log[2]^2*Log[5]^2)/(8 + 8*x + 2*x^2),x]

[Out]

(E^E + E^(2/x)*x^2*Log[2]^2 + (4 + 2*x + x^2)*Log[2]^2*Log[5]^2 + E^x^(-1)*x^2*Log[2]^2*Log[25])/(2*(2 + x))

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fricas [B]  time = 1.25, size = 57, normalized size = 1.84 \begin {gather*} \frac {2 \, x^{2} e^{\frac {1}{x}} \log \relax (5) \log \relax (2)^{2} + x^{2} e^{\frac {2}{x}} \log \relax (2)^{2} + {\left (x^{2} + 2 \, x + 4\right )} \log \relax (5)^{2} \log \relax (2)^{2} + e^{e}}{2 \, {\left (x + 2\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-exp(exp(1))+(x^2+2*x-4)*log(2)^2*exp(1/x)^2+(2*x^2+6*x-4)*log(2)^2*log(5)*exp(1/x)+(x^2+4*x)*log(2
)^2*log(5)^2)/(2*x^2+8*x+8),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.18, size = 56, normalized size = 1.81 \begin {gather*} \frac {4 \, e^{\frac {1}{x}} \log \relax (5) \log \relax (2)^{2} + 2 \, \log \relax (5)^{2} \log \relax (2)^{2} + 2 \, e^{\frac {2}{x}} \log \relax (2)^{2} - \frac {e^{e}}{x}}{4 \, {\left (\frac {1}{x} + \frac {2}{x^{2}}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-exp(exp(1))+(x^2+2*x-4)*log(2)^2*exp(1/x)^2+(2*x^2+6*x-4)*log(2)^2*log(5)*exp(1/x)+(x^2+4*x)*log(2
)^2*log(5)^2)/(2*x^2+8*x+8),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.25, size = 55, normalized size = 1.77

method	result	size
norman	\(\frac {\ln \relax (2)^{2} \ln \relax (5) {\mathrm e}^{\frac {1}{x}} x^{2}+\frac {x^{2} \ln \relax (2)^{2} \ln \relax (5)^{2}}{2}+\frac {\ln \relax (2)^{2} {\mathrm e}^{\frac {2}{x}} x^{2}}{2}+\frac {{\mathrm e}^{{\mathrm e}}}{2}}{2+x}\)	\(55\)
risch	\(\frac {x \ln \relax (2)^{2} \ln \relax (5)^{2}}{2}+\frac {2 \ln \relax (2)^{2} \ln \relax (5)^{2}}{2+x}+\frac {{\mathrm e}^{{\mathrm e}}}{2 x +4}+\frac {\ln \relax (2)^{2} {\mathrm e}^{\frac {2}{x}} x^{2}}{2 x +4}+\frac {\ln \relax (2)^{2} \ln \relax (5) x^{2} {\mathrm e}^{\frac {1}{x}}}{2+x}\)	\(77\)
derivativedivides	\(\frac {\ln \relax (2)^{2} {\mathrm e}^{\frac {2}{x}} x}{2}-\frac {\ln \relax (2)^{2} \ln \relax (5)^{2}}{\frac {2}{x}+1}+\frac {x \ln \relax (2)^{2} \ln \relax (5)^{2}}{2}-\frac {{\mathrm e}^{{\mathrm e}}}{4 \left (\frac {2}{x}+1\right )}-\frac {\ln \relax (2)^{2} {\mathrm e}^{\frac {2}{x}}}{2 \left (\frac {1}{x}+\frac {1}{2}\right )}+\ln \relax (2)^{2} \ln \relax (5) {\mathrm e}^{\frac {1}{x}} x -\frac {\ln \relax (2)^{2} \ln \relax (5) {\mathrm e}^{\frac {1}{x}}}{\frac {1}{x}+\frac {1}{2}}\)	\(109\)
default	\(\frac {\ln \relax (2)^{2} {\mathrm e}^{\frac {2}{x}} x}{2}-\frac {\ln \relax (2)^{2} \ln \relax (5)^{2}}{\frac {2}{x}+1}+\frac {x \ln \relax (2)^{2} \ln \relax (5)^{2}}{2}-\frac {{\mathrm e}^{{\mathrm e}}}{4 \left (\frac {2}{x}+1\right )}-\frac {\ln \relax (2)^{2} {\mathrm e}^{\frac {2}{x}}}{2 \left (\frac {1}{x}+\frac {1}{2}\right )}+\ln \relax (2)^{2} \ln \relax (5) {\mathrm e}^{\frac {1}{x}} x -\frac {\ln \relax (2)^{2} \ln \relax (5) {\mathrm e}^{\frac {1}{x}}}{\frac {1}{x}+\frac {1}{2}}\)	\(109\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-exp(exp(1))+(x^2+2*x-4)*ln(2)^2*exp(1/x)^2+(2*x^2+6*x-4)*ln(2)^2*ln(5)*exp(1/x)+(x^2+4*x)*ln(2)^2*ln(5)^
2)/(2*x^2+8*x+8),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {1}{2} \, {\left (x - \frac {4}{x + 2} - 4 \, \log \left (x + 2\right )\right )} \log \relax (5)^{2} \log \relax (2)^{2} + 2 \, {\left (\frac {2}{x + 2} + \log \left (x + 2\right )\right )} \log \relax (5)^{2} \log \relax (2)^{2} - 4 \, \int \frac {e^{\frac {2}{x}}}{x^{3} + 4 \, x^{2} + 4 \, x}\,{d x} \log \relax (2)^{2} + \frac {{\left (x^{3} \log \relax (2)^{2} - 4 \, x \log \relax (2)^{2}\right )} e^{\frac {2}{x}} + 2 \, {\left (x^{3} \log \relax (5) \log \relax (2)^{2} - 6 \, x \log \relax (5) \log \relax (2)^{2}\right )} e^{\frac {1}{x}}}{2 \, {\left (x^{2} + 4 \, x + 4\right )}} + \frac {e^{e}}{2 \, {\left (x + 2\right )}} + \frac {1}{2} \, \int \frac {4 \, {\left (x \log \relax (5) \log \relax (2)^{2} - 6 \, \log \relax (5) \log \relax (2)^{2}\right )} e^{\frac {1}{x}}}{x^{4} + 6 \, x^{3} + 12 \, x^{2} + 8 \, x}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-exp(exp(1))+(x^2+2*x-4)*log(2)^2*exp(1/x)^2+(2*x^2+6*x-4)*log(2)^2*log(5)*exp(1/x)+(x^2+4*x)*log(2
)^2*log(5)^2)/(2*x^2+8*x+8),x, algorithm="maxima")

[Out]

1/2*(x - 4/(x + 2) - 4*log(x + 2))*log(5)^2*log(2)^2 + 2*(2/(x + 2) + log(x + 2))*log(5)^2*log(2)^2 - 4*integr
ate(e^(2/x)/(x^3 + 4*x^2 + 4*x), x)*log(2)^2 + 1/2*((x^3*log(2)^2 - 4*x*log(2)^2)*e^(2/x) + 2*(x^3*log(5)*log(
2)^2 - 6*x*log(5)*log(2)^2)*e^(1/x))/(x^2 + 4*x + 4) + 1/2*e^e/(x + 2) + 1/2*integrate(4*(x*log(5)*log(2)^2 -
6*log(5)*log(2)^2)*e^(1/x)/(x^4 + 6*x^3 + 12*x^2 + 8*x), x)

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mupad [B]  time = 4.59, size = 58, normalized size = 1.87 \begin {gather*} \frac {2\,x^2\,{\mathrm {e}}^{2/x}\,{\ln \relax (2)}^2-x\,{\mathrm {e}}^{\mathrm {e}}+2\,x^2\,{\ln \relax (2)}^2\,{\ln \relax (5)}^2+4\,x^2\,{\mathrm {e}}^{1/x}\,{\ln \relax (2)}^2\,\ln \relax (5)}{4\,x+8} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [B]  time = 0.34, size = 100, normalized size = 3.23 \begin {gather*} \frac {x \log {\relax (2 )}^{2} \log {\relax (5 )}^{2}}{2} + \frac {\left (x^{3} \log {\relax (2 )}^{2} + 2 x^{2} \log {\relax (2 )}^{2}\right ) e^{\frac {2}{x}} + \left (2 x^{3} \log {\relax (2 )}^{2} \log {\relax (5 )} + 4 x^{2} \log {\relax (2 )}^{2} \log {\relax (5 )}\right ) e^{\frac {1}{x}}}{2 x^{2} + 8 x + 8} + \frac {4 \log {\relax (2 )}^{2} \log {\relax (5 )}^{2} + e^{e}}{2 x + 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-exp(exp(1))+(x**2+2*x-4)*ln(2)**2*exp(1/x)**2+(2*x**2+6*x-4)*ln(2)**2*ln(5)*exp(1/x)+(x**2+4*x)*ln
(2)**2*ln(5)**2)/(2*x**2+8*x+8),x)

[Out]

x*log(2)**2*log(5)**2/2 + ((x**3*log(2)**2 + 2*x**2*log(2)**2)*exp(2/x) + (2*x**3*log(2)**2*log(5) + 4*x**2*lo
g(2)**2*log(5))*exp(1/x))/(2*x**2 + 8*x + 8) + (4*log(2)**2*log(5)**2 + exp(E))/(2*x + 4)

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