3.44.83 \(\int \frac {10+10 e^{4 x/5}+e^{3 x/5} (40-40 x)-40 x+60 x^2-40 x^3+10 x^4+e^{2 x/5} (60-120 x+60 x^2)+e^{x/5} (40-120 x+120 x^2-40 x^3)+(10+60 x^2-80 x^3+30 x^4+e^{4 x/5} (-10+8 x)+e^{3 x/5} (-40+24 x-24 x^2)+e^{2 x/5} (-60+24 x+12 x^2+24 x^3)+e^{x/5} (-40+8 x+96 x^2-56 x^3-8 x^4)) \log (x)-10 \log ^2(x)}{5 x^2} \, dx\)

Optimal. Leaf size=24 \[ \frac {2 \log (x) \left (\left (-1-e^{x/5}+x\right )^4+\log (x)\right )}{x} \]

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Rubi [B]  time = 1.78, antiderivative size = 166, normalized size of antiderivative = 6.92, number of steps used = 53, number of rules used = 15, integrand size = 198, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.076, Rules used = {12, 14, 2288, 6742, 2199, 2194, 2177, 2178, 2176, 2554, 43, 2357, 2295, 2304, 2305} \begin {gather*} 2 x^3 \log (x)-8 e^{x/5} x^2 \log (x)-8 x^2 \log (x)+\frac {8 e^{3 x/5} \left (x \log (x)-x^2 \log (x)\right )}{x^2}+\frac {2 \log ^2(x)}{x}+24 e^{x/5} x \log (x)+12 e^{2 x/5} x \log (x)+12 x \log (x)-24 e^{x/5} \log (x)-24 e^{2 x/5} \log (x)-8 \log (x)+\frac {8 e^{x/5} \log (x)}{x}+\frac {12 e^{2 x/5} \log (x)}{x}+\frac {2 e^{4 x/5} \log (x)}{x}+\frac {2 \log (x)}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(10 + 10*E^((4*x)/5) + E^((3*x)/5)*(40 - 40*x) - 40*x + 60*x^2 - 40*x^3 + 10*x^4 + E^((2*x)/5)*(60 - 120*x
+ 60*x^2) + E^(x/5)*(40 - 120*x + 120*x^2 - 40*x^3) + (10 + 60*x^2 - 80*x^3 + 30*x^4 + E^((4*x)/5)*(-10 + 8*x
) + E^((3*x)/5)*(-40 + 24*x - 24*x^2) + E^((2*x)/5)*(-60 + 24*x + 12*x^2 + 24*x^3) + E^(x/5)*(-40 + 8*x + 96*x
^2 - 56*x^3 - 8*x^4))*Log[x] - 10*Log[x]^2)/(5*x^2),x]

[Out]

-8*Log[x] - 24*E^(x/5)*Log[x] - 24*E^((2*x)/5)*Log[x] + (2*Log[x])/x + (8*E^(x/5)*Log[x])/x + (12*E^((2*x)/5)*
Log[x])/x + (2*E^((4*x)/5)*Log[x])/x + 12*x*Log[x] + 24*E^(x/5)*x*Log[x] + 12*E^((2*x)/5)*x*Log[x] - 8*x^2*Log
[x] - 8*E^(x/5)*x^2*Log[x] + 2*x^3*Log[x] + (2*Log[x]^2)/x + (8*E^((3*x)/5)*(x*Log[x] - x^2*Log[x]))/x^2

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 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

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 2177

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

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 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 2199

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

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 2295

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{c, n}, x]

Rule 2304

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*Log[c*x^
n]))/(d*(m + 1)), x] - Simp[(b*n*(d*x)^(m + 1))/(d*(m + 1)^2), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[m, -1
]

Rule 2305

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*Lo
g[c*x^n])^p)/(d*(m + 1)), x] - Dist[(b*n*p)/(m + 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] && GtQ[p, 0]

Rule 2357

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(RFx_), x_Symbol] :> With[{u = ExpandIntegrand[(a + b*Log[c*x^
n])^p, RFx, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[{a, b, c, n}, x] && RationalFunctionQ[RFx, x] && IGtQ[p, 0]

Rule 2554

Int[Log[u_]*(v_), x_Symbol] :> With[{w = IntHide[v, x]}, Dist[Log[u], w, x] - Int[SimplifyIntegrand[(w*D[u, x]
)/u, x], x] /; InverseFunctionFreeQ[w, x]] /; InverseFunctionFreeQ[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} &=\frac {1}{5} \int \frac {10+10 e^{4 x/5}+e^{3 x/5} (40-40 x)-40 x+60 x^2-40 x^3+10 x^4+e^{2 x/5} \left (60-120 x+60 x^2\right )+e^{x/5} \left (40-120 x+120 x^2-40 x^3\right )+\left (10+60 x^2-80 x^3+30 x^4+e^{4 x/5} (-10+8 x)+e^{3 x/5} \left (-40+24 x-24 x^2\right )+e^{2 x/5} \left (-60+24 x+12 x^2+24 x^3\right )+e^{x/5} \left (-40+8 x+96 x^2-56 x^3-8 x^4\right )\right ) \log (x)-10 \log ^2(x)}{x^2} \, dx\\ &=\frac {1}{5} \int \left (\frac {2 e^{4 x/5} (5-5 \log (x)+4 x \log (x))}{x^2}-\frac {8 e^{x/5} (-1+x)^2 \left (-5+5 x+5 \log (x)+9 x \log (x)+x^2 \log (x)\right )}{x^2}+\frac {12 e^{2 x/5} (-1+x) \left (-5+5 x+5 \log (x)+3 x \log (x)+2 x^2 \log (x)\right )}{x^2}-\frac {8 e^{3 x/5} \left (-5+5 x+5 \log (x)-3 x \log (x)+3 x^2 \log (x)\right )}{x^2}+\frac {10 \left (1-4 x+6 x^2-4 x^3+x^4+\log (x)+6 x^2 \log (x)-8 x^3 \log (x)+3 x^4 \log (x)-\log ^2(x)\right )}{x^2}\right ) \, dx\\ &=\frac {2}{5} \int \frac {e^{4 x/5} (5-5 \log (x)+4 x \log (x))}{x^2} \, dx-\frac {8}{5} \int \frac {e^{x/5} (-1+x)^2 \left (-5+5 x+5 \log (x)+9 x \log (x)+x^2 \log (x)\right )}{x^2} \, dx-\frac {8}{5} \int \frac {e^{3 x/5} \left (-5+5 x+5 \log (x)-3 x \log (x)+3 x^2 \log (x)\right )}{x^2} \, dx+2 \int \frac {1-4 x+6 x^2-4 x^3+x^4+\log (x)+6 x^2 \log (x)-8 x^3 \log (x)+3 x^4 \log (x)-\log ^2(x)}{x^2} \, dx+\frac {12}{5} \int \frac {e^{2 x/5} (-1+x) \left (-5+5 x+5 \log (x)+3 x \log (x)+2 x^2 \log (x)\right )}{x^2} \, dx\\ &=\frac {2 e^{4 x/5} \log (x)}{x}+\frac {8 e^{3 x/5} \left (x \log (x)-x^2 \log (x)\right )}{x^2}-\frac {8}{5} \int \left (\frac {5 e^{x/5} (-1+x)^3}{x^2}+\frac {e^{x/5} (-1+x)^2 \left (5+9 x+x^2\right ) \log (x)}{x^2}\right ) \, dx+2 \int \left (\frac {(-1+x)^4}{x^2}+\frac {\left (1+6 x^2-8 x^3+3 x^4\right ) \log (x)}{x^2}-\frac {\log ^2(x)}{x^2}\right ) \, dx+\frac {12}{5} \int \left (\frac {5 e^{2 x/5} (-1+x)^2}{x^2}+\frac {e^{2 x/5} (-1+x) \left (5+3 x+2 x^2\right ) \log (x)}{x^2}\right ) \, dx\\ &=\frac {2 e^{4 x/5} \log (x)}{x}+\frac {8 e^{3 x/5} \left (x \log (x)-x^2 \log (x)\right )}{x^2}-\frac {8}{5} \int \frac {e^{x/5} (-1+x)^2 \left (5+9 x+x^2\right ) \log (x)}{x^2} \, dx+2 \int \frac {(-1+x)^4}{x^2} \, dx+2 \int \frac {\left (1+6 x^2-8 x^3+3 x^4\right ) \log (x)}{x^2} \, dx-2 \int \frac {\log ^2(x)}{x^2} \, dx+\frac {12}{5} \int \frac {e^{2 x/5} (-1+x) \left (5+3 x+2 x^2\right ) \log (x)}{x^2} \, dx-8 \int \frac {e^{x/5} (-1+x)^3}{x^2} \, dx+12 \int \frac {e^{2 x/5} (-1+x)^2}{x^2} \, dx\\ &=-24 e^{x/5} \log (x)-24 e^{2 x/5} \log (x)+\frac {8 e^{x/5} \log (x)}{x}+\frac {12 e^{2 x/5} \log (x)}{x}+\frac {2 e^{4 x/5} \log (x)}{x}+24 e^{x/5} x \log (x)+12 e^{2 x/5} x \log (x)-8 e^{x/5} x^2 \log (x)+\frac {2 \log ^2(x)}{x}+\frac {8 e^{3 x/5} \left (x \log (x)-x^2 \log (x)\right )}{x^2}+\frac {8}{5} \int \frac {5 e^{x/5} (-1+x)^3}{x^2} \, dx+2 \int \left (6+\frac {1}{x^2}-\frac {4}{x}-4 x+x^2\right ) \, dx+2 \int \left (6 \log (x)+\frac {\log (x)}{x^2}-8 x \log (x)+3 x^2 \log (x)\right ) \, dx-\frac {12}{5} \int \frac {5 e^{2 x/5} (1-x)^2}{x^2} \, dx-4 \int \frac {\log (x)}{x^2} \, dx-8 \int \left (-3 e^{x/5}-\frac {e^{x/5}}{x^2}+\frac {3 e^{x/5}}{x}+e^{x/5} x\right ) \, dx+12 \int \left (e^{2 x/5}+\frac {e^{2 x/5}}{x^2}-\frac {2 e^{2 x/5}}{x}\right ) \, dx\\ &=\frac {2}{x}+12 x-4 x^2+\frac {2 x^3}{3}-8 \log (x)-24 e^{x/5} \log (x)-24 e^{2 x/5} \log (x)+\frac {4 \log (x)}{x}+\frac {8 e^{x/5} \log (x)}{x}+\frac {12 e^{2 x/5} \log (x)}{x}+\frac {2 e^{4 x/5} \log (x)}{x}+24 e^{x/5} x \log (x)+12 e^{2 x/5} x \log (x)-8 e^{x/5} x^2 \log (x)+\frac {2 \log ^2(x)}{x}+\frac {8 e^{3 x/5} \left (x \log (x)-x^2 \log (x)\right )}{x^2}+2 \int \frac {\log (x)}{x^2} \, dx+6 \int x^2 \log (x) \, dx+8 \int \frac {e^{x/5}}{x^2} \, dx+8 \int \frac {e^{x/5} (-1+x)^3}{x^2} \, dx-8 \int e^{x/5} x \, dx+12 \int e^{2 x/5} \, dx+12 \int \frac {e^{2 x/5}}{x^2} \, dx-12 \int \frac {e^{2 x/5} (1-x)^2}{x^2} \, dx+12 \int \log (x) \, dx-16 \int x \log (x) \, dx+24 \int e^{x/5} \, dx-24 \int \frac {e^{x/5}}{x} \, dx-24 \int \frac {e^{2 x/5}}{x} \, dx\\ &=120 e^{x/5}+30 e^{2 x/5}-\frac {8 e^{x/5}}{x}-\frac {12 e^{2 x/5}}{x}-40 e^{x/5} x-24 \text {Ei}\left (\frac {x}{5}\right )-24 \text {Ei}\left (\frac {2 x}{5}\right )-8 \log (x)-24 e^{x/5} \log (x)-24 e^{2 x/5} \log (x)+\frac {2 \log (x)}{x}+\frac {8 e^{x/5} \log (x)}{x}+\frac {12 e^{2 x/5} \log (x)}{x}+\frac {2 e^{4 x/5} \log (x)}{x}+12 x \log (x)+24 e^{x/5} x \log (x)+12 e^{2 x/5} x \log (x)-8 x^2 \log (x)-8 e^{x/5} x^2 \log (x)+2 x^3 \log (x)+\frac {2 \log ^2(x)}{x}+\frac {8 e^{3 x/5} \left (x \log (x)-x^2 \log (x)\right )}{x^2}+\frac {8}{5} \int \frac {e^{x/5}}{x} \, dx+\frac {24}{5} \int \frac {e^{2 x/5}}{x} \, dx+8 \int \left (-3 e^{x/5}-\frac {e^{x/5}}{x^2}+\frac {3 e^{x/5}}{x}+e^{x/5} x\right ) \, dx-12 \int \left (e^{2 x/5}+\frac {e^{2 x/5}}{x^2}-\frac {2 e^{2 x/5}}{x}\right ) \, dx+40 \int e^{x/5} \, dx\\ &=320 e^{x/5}+30 e^{2 x/5}-\frac {8 e^{x/5}}{x}-\frac {12 e^{2 x/5}}{x}-40 e^{x/5} x-\frac {112 \text {Ei}\left (\frac {x}{5}\right )}{5}-\frac {96 \text {Ei}\left (\frac {2 x}{5}\right )}{5}-8 \log (x)-24 e^{x/5} \log (x)-24 e^{2 x/5} \log (x)+\frac {2 \log (x)}{x}+\frac {8 e^{x/5} \log (x)}{x}+\frac {12 e^{2 x/5} \log (x)}{x}+\frac {2 e^{4 x/5} \log (x)}{x}+12 x \log (x)+24 e^{x/5} x \log (x)+12 e^{2 x/5} x \log (x)-8 x^2 \log (x)-8 e^{x/5} x^2 \log (x)+2 x^3 \log (x)+\frac {2 \log ^2(x)}{x}+\frac {8 e^{3 x/5} \left (x \log (x)-x^2 \log (x)\right )}{x^2}-8 \int \frac {e^{x/5}}{x^2} \, dx+8 \int e^{x/5} x \, dx-12 \int e^{2 x/5} \, dx-12 \int \frac {e^{2 x/5}}{x^2} \, dx-24 \int e^{x/5} \, dx+24 \int \frac {e^{x/5}}{x} \, dx+24 \int \frac {e^{2 x/5}}{x} \, dx\\ &=200 e^{x/5}+\frac {8 \text {Ei}\left (\frac {x}{5}\right )}{5}+\frac {24 \text {Ei}\left (\frac {2 x}{5}\right )}{5}-8 \log (x)-24 e^{x/5} \log (x)-24 e^{2 x/5} \log (x)+\frac {2 \log (x)}{x}+\frac {8 e^{x/5} \log (x)}{x}+\frac {12 e^{2 x/5} \log (x)}{x}+\frac {2 e^{4 x/5} \log (x)}{x}+12 x \log (x)+24 e^{x/5} x \log (x)+12 e^{2 x/5} x \log (x)-8 x^2 \log (x)-8 e^{x/5} x^2 \log (x)+2 x^3 \log (x)+\frac {2 \log ^2(x)}{x}+\frac {8 e^{3 x/5} \left (x \log (x)-x^2 \log (x)\right )}{x^2}-\frac {8}{5} \int \frac {e^{x/5}}{x} \, dx-\frac {24}{5} \int \frac {e^{2 x/5}}{x} \, dx-40 \int e^{x/5} \, dx\\ &=-8 \log (x)-24 e^{x/5} \log (x)-24 e^{2 x/5} \log (x)+\frac {2 \log (x)}{x}+\frac {8 e^{x/5} \log (x)}{x}+\frac {12 e^{2 x/5} \log (x)}{x}+\frac {2 e^{4 x/5} \log (x)}{x}+12 x \log (x)+24 e^{x/5} x \log (x)+12 e^{2 x/5} x \log (x)-8 x^2 \log (x)-8 e^{x/5} x^2 \log (x)+2 x^3 \log (x)+\frac {2 \log ^2(x)}{x}+\frac {8 e^{3 x/5} \left (x \log (x)-x^2 \log (x)\right )}{x^2}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.17, size = 24, normalized size = 1.00 \begin {gather*} \frac {2 \log (x) \left (\left (1+e^{x/5}-x\right )^4+\log (x)\right )}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(10 + 10*E^((4*x)/5) + E^((3*x)/5)*(40 - 40*x) - 40*x + 60*x^2 - 40*x^3 + 10*x^4 + E^((2*x)/5)*(60 -
120*x + 60*x^2) + E^(x/5)*(40 - 120*x + 120*x^2 - 40*x^3) + (10 + 60*x^2 - 80*x^3 + 30*x^4 + E^((4*x)/5)*(-10
+ 8*x) + E^((3*x)/5)*(-40 + 24*x - 24*x^2) + E^((2*x)/5)*(-60 + 24*x + 12*x^2 + 24*x^3) + E^(x/5)*(-40 + 8*x
+ 96*x^2 - 56*x^3 - 8*x^4))*Log[x] - 10*Log[x]^2)/(5*x^2),x]

[Out]

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

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fricas [B]  time = 0.62, size = 77, normalized size = 3.21 \begin {gather*} \frac {2 \, {\left ({\left (x^{4} - 4 \, x^{3} + 6 \, x^{2} - 4 \, {\left (x - 1\right )} e^{\left (\frac {3}{5} \, x\right )} + 6 \, {\left (x^{2} - 2 \, x + 1\right )} e^{\left (\frac {2}{5} \, x\right )} - 4 \, {\left (x^{3} - 3 \, x^{2} + 3 \, x - 1\right )} e^{\left (\frac {1}{5} \, x\right )} - 4 \, x + e^{\left (\frac {4}{5} \, x\right )} + 1\right )} \log \relax (x) + \log \relax (x)^{2}\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(-10*log(x)^2+((8*x-10)*exp(1/5*x)^4+(-24*x^2+24*x-40)*exp(1/5*x)^3+(24*x^3+12*x^2+24*x-60)*exp(
1/5*x)^2+(-8*x^4-56*x^3+96*x^2+8*x-40)*exp(1/5*x)+30*x^4-80*x^3+60*x^2+10)*log(x)+10*exp(1/5*x)^4+(-40*x+40)*e
xp(1/5*x)^3+(60*x^2-120*x+60)*exp(1/5*x)^2+(-40*x^3+120*x^2-120*x+40)*exp(1/5*x)+10*x^4-40*x^3+60*x^2-40*x+10)
/x^2,x, algorithm="fricas")

[Out]

2*((x^4 - 4*x^3 + 6*x^2 - 4*(x - 1)*e^(3/5*x) + 6*(x^2 - 2*x + 1)*e^(2/5*x) - 4*(x^3 - 3*x^2 + 3*x - 1)*e^(1/5
*x) - 4*x + e^(4/5*x) + 1)*log(x) + log(x)^2)/x

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(-10*log(x)^2+((8*x-10)*exp(1/5*x)^4+(-24*x^2+24*x-40)*exp(1/5*x)^3+(24*x^3+12*x^2+24*x-60)*exp(
1/5*x)^2+(-8*x^4-56*x^3+96*x^2+8*x-40)*exp(1/5*x)+30*x^4-80*x^3+60*x^2+10)*log(x)+10*exp(1/5*x)^4+(-40*x+40)*e
xp(1/5*x)^3+(60*x^2-120*x+60)*exp(1/5*x)^2+(-40*x^3+120*x^2-120*x+40)*exp(1/5*x)+10*x^4-40*x^3+60*x^2-40*x+10)
/x^2,x, algorithm="giac")

[Out]

Timed out

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maple [B]  time = 0.14, size = 107, normalized size = 4.46

method	result	size
risch	\(\frac {2 \ln \relax (x )^{2}}{x}+\frac {2 \left (x^{4}-4 \,{\mathrm e}^{\frac {x}{5}} x^{3}+6 \,{\mathrm e}^{\frac {2 x}{5}} x^{2}-4 \,{\mathrm e}^{\frac {3 x}{5}} x +{\mathrm e}^{\frac {4 x}{5}}-4 x^{3}+12 \,{\mathrm e}^{\frac {x}{5}} x^{2}-12 \,{\mathrm e}^{\frac {2 x}{5}} x +4 \,{\mathrm e}^{\frac {3 x}{5}}+6 x^{2}-12 x \,{\mathrm e}^{\frac {x}{5}}+6 \,{\mathrm e}^{\frac {2 x}{5}}+4 \,{\mathrm e}^{\frac {x}{5}}+1\right ) \ln \relax (x )}{x}-8 \ln \relax (x )\)	\(107\)
default	\(\frac {2 \ln \relax (x ) {\mathrm e}^{\frac {4 x}{5}}}{x}+\frac {40 \ln \relax (x ) {\mathrm e}^{\frac {3 x}{5}}-40 \ln \relax (x ) {\mathrm e}^{\frac {3 x}{5}} x}{5 x}+\frac {60 \ln \relax (x ) {\mathrm e}^{\frac {2 x}{5}}-120 \ln \relax (x ) {\mathrm e}^{\frac {2 x}{5}} x +60 \ln \relax (x ) {\mathrm e}^{\frac {2 x}{5}} x^{2}}{5 x}+\frac {40 \ln \relax (x ) {\mathrm e}^{\frac {x}{5}}-120 \ln \relax (x ) {\mathrm e}^{\frac {x}{5}} x +120 \ln \relax (x ) {\mathrm e}^{\frac {x}{5}} x^{2}-40 \ln \relax (x ) {\mathrm e}^{\frac {x}{5}} x^{3}}{5 x}-8 \ln \relax (x )+\frac {2 \ln \relax (x )^{2}}{x}+\frac {2 \ln \relax (x )}{x}+2 x^{3} \ln \relax (x )-8 x^{2} \ln \relax (x )+12 x \ln \relax (x )\)	\(154\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/5*(-10*ln(x)^2+((8*x-10)*exp(1/5*x)^4+(-24*x^2+24*x-40)*exp(1/5*x)^3+(24*x^3+12*x^2+24*x-60)*exp(1/5*x)^
2+(-8*x^4-56*x^3+96*x^2+8*x-40)*exp(1/5*x)+30*x^4-80*x^3+60*x^2+10)*ln(x)+10*exp(1/5*x)^4+(-40*x+40)*exp(1/5*x
)^3+(60*x^2-120*x+60)*exp(1/5*x)^2+(-40*x^3+120*x^2-120*x+40)*exp(1/5*x)+10*x^4-40*x^3+60*x^2-40*x+10)/x^2,x,m
ethod=_RETURNVERBOSE)

[Out]

2*ln(x)^2/x+2*(x^4-4*exp(1/5*x)*x^3+6*exp(2/5*x)*x^2-4*exp(3/5*x)*x+exp(4/5*x)-4*x^3+12*exp(1/5*x)*x^2-12*exp(
2/5*x)*x+4*exp(3/5*x)+6*x^2-12*x*exp(1/5*x)+6*exp(2/5*x)+4*exp(1/5*x)+1)/x*ln(x)-8*ln(x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} 2 \, x^{3} \log \relax (x) - 8 \, x^{2} \log \relax (x) - 40 \, {\left (x - 5\right )} e^{\left (\frac {1}{5} \, x\right )} + 12 \, x \log \relax (x) + 96 \, e^{\left (\frac {1}{5} \, x\right )} \log \relax (x) - \frac {2 \, {\left (4 \, {\left (x - 1\right )} e^{\left (\frac {3}{5} \, x\right )} \log \relax (x) - 6 \, {\left (x^{2} - 2 \, x + 1\right )} e^{\left (\frac {2}{5} \, x\right )} \log \relax (x) + 4 \, {\left (x^{3} - 3 \, x^{2} + 15 \, x - 1\right )} e^{\left (\frac {1}{5} \, x\right )} \log \relax (x) - e^{\left (\frac {4}{5} \, x\right )} \log \relax (x) - \log \relax (x)^{2} - 2 \, \log \relax (x) - 2\right )}}{x} - \frac {2 \, \log \relax (x)}{x} - \frac {4}{x} - 8 \, {\rm Ei}\left (\frac {3}{5} \, x\right ) - 24 \, {\rm Ei}\left (\frac {2}{5} \, x\right ) - 120 \, {\rm Ei}\left (\frac {1}{5} \, x\right ) + 30 \, e^{\left (\frac {2}{5} \, x\right )} + 120 \, e^{\left (\frac {1}{5} \, x\right )} + \frac {8}{5} \, \Gamma \left (-1, -\frac {1}{5} \, x\right ) + \frac {24}{5} \, \Gamma \left (-1, -\frac {2}{5} \, x\right ) + \frac {24}{5} \, \Gamma \left (-1, -\frac {3}{5} \, x\right ) + \frac {8}{5} \, \Gamma \left (-1, -\frac {4}{5} \, x\right ) + \frac {2}{5} \, \int \frac {20 \, {\left (x - 1\right )} e^{\left (\frac {3}{5} \, x\right )}}{x^{2}}\,{d x} - \frac {2}{5} \, \int \frac {30 \, {\left (x^{2} - 2 \, x + 1\right )} e^{\left (\frac {2}{5} \, x\right )}}{x^{2}}\,{d x} + \frac {2}{5} \, \int \frac {20 \, {\left (x^{3} - 3 \, x^{2} + 15 \, x - 1\right )} e^{\left (\frac {1}{5} \, x\right )}}{x^{2}}\,{d x} - 2 \, \int \frac {e^{\left (\frac {4}{5} \, x\right )}}{x^{2}}\,{d x} - 8 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(-10*log(x)^2+((8*x-10)*exp(1/5*x)^4+(-24*x^2+24*x-40)*exp(1/5*x)^3+(24*x^3+12*x^2+24*x-60)*exp(
1/5*x)^2+(-8*x^4-56*x^3+96*x^2+8*x-40)*exp(1/5*x)+30*x^4-80*x^3+60*x^2+10)*log(x)+10*exp(1/5*x)^4+(-40*x+40)*e
xp(1/5*x)^3+(60*x^2-120*x+60)*exp(1/5*x)^2+(-40*x^3+120*x^2-120*x+40)*exp(1/5*x)+10*x^4-40*x^3+60*x^2-40*x+10)
/x^2,x, algorithm="maxima")

[Out]

2*x^3*log(x) - 8*x^2*log(x) - 40*(x - 5)*e^(1/5*x) + 12*x*log(x) + 96*e^(1/5*x)*log(x) - 2*(4*(x - 1)*e^(3/5*x
)*log(x) - 6*(x^2 - 2*x + 1)*e^(2/5*x)*log(x) + 4*(x^3 - 3*x^2 + 15*x - 1)*e^(1/5*x)*log(x) - e^(4/5*x)*log(x)
- log(x)^2 - 2*log(x) - 2)/x - 2*log(x)/x - 4/x - 8*Ei(3/5*x) - 24*Ei(2/5*x) - 120*Ei(1/5*x) + 30*e^(2/5*x) +
120*e^(1/5*x) + 8/5*gamma(-1, -1/5*x) + 24/5*gamma(-1, -2/5*x) + 24/5*gamma(-1, -3/5*x) + 8/5*gamma(-1, -4/5*
x) + 2/5*integrate(20*(x - 1)*e^(3/5*x)/x^2, x) - 2/5*integrate(30*(x^2 - 2*x + 1)*e^(2/5*x)/x^2, x) + 2/5*int
egrate(20*(x^3 - 3*x^2 + 15*x - 1)*e^(1/5*x)/x^2, x) - 2*integrate(e^(4/5*x)/x^2, x) - 8*log(x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int \frac {2\,{\mathrm {e}}^{\frac {4\,x}{5}}-8\,x+\frac {{\mathrm {e}}^{\frac {2\,x}{5}}\,\left (60\,x^2-120\,x+60\right )}{5}-\frac {{\mathrm {e}}^{x/5}\,\left (40\,x^3-120\,x^2+120\,x-40\right )}{5}-2\,{\ln \relax (x)}^2-\frac {{\mathrm {e}}^{\frac {3\,x}{5}}\,\left (40\,x-40\right )}{5}+12\,x^2-8\,x^3+2\,x^4+\frac {\ln \relax (x)\,\left ({\mathrm {e}}^{\frac {2\,x}{5}}\,\left (24\,x^3+12\,x^2+24\,x-60\right )-{\mathrm {e}}^{\frac {3\,x}{5}}\,\left (24\,x^2-24\,x+40\right )-{\mathrm {e}}^{x/5}\,\left (8\,x^4+56\,x^3-96\,x^2-8\,x+40\right )+{\mathrm {e}}^{\frac {4\,x}{5}}\,\left (8\,x-10\right )+60\,x^2-80\,x^3+30\,x^4+10\right )}{5}+2}{x^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*exp((4*x)/5) - 8*x + (exp((2*x)/5)*(60*x^2 - 120*x + 60))/5 - (exp(x/5)*(120*x - 120*x^2 + 40*x^3 - 40)
)/5 - 2*log(x)^2 - (exp((3*x)/5)*(40*x - 40))/5 + 12*x^2 - 8*x^3 + 2*x^4 + (log(x)*(exp((2*x)/5)*(24*x + 12*x^
2 + 24*x^3 - 60) - exp((3*x)/5)*(24*x^2 - 24*x + 40) - exp(x/5)*(56*x^3 - 96*x^2 - 8*x + 8*x^4 + 40) + exp((4*
x)/5)*(8*x - 10) + 60*x^2 - 80*x^3 + 30*x^4 + 10))/5 + 2)/x^2,x)

[Out]

int((2*exp((4*x)/5) - 8*x + (exp((2*x)/5)*(60*x^2 - 120*x + 60))/5 - (exp(x/5)*(120*x - 120*x^2 + 40*x^3 - 40)
)/5 - 2*log(x)^2 - (exp((3*x)/5)*(40*x - 40))/5 + 12*x^2 - 8*x^3 + 2*x^4 + (log(x)*(exp((2*x)/5)*(24*x + 12*x^
2 + 24*x^3 - 60) - exp((3*x)/5)*(24*x^2 - 24*x + 40) - exp(x/5)*(56*x^3 - 96*x^2 - 8*x + 8*x^4 + 40) + exp((4*
x)/5)*(8*x - 10) + 60*x^2 - 80*x^3 + 30*x^4 + 10))/5 + 2)/x^2, x)

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sympy [B]  time = 0.65, size = 148, normalized size = 6.17 \begin {gather*} - 8 \log {\relax (x )} + \frac {\left (2 x^{4} - 8 x^{3} + 12 x^{2} + 2\right ) \log {\relax (x )}}{x} + \frac {2 \log {\relax (x )}^{2}}{x} + \frac {2 x^{3} e^{\frac {4 x}{5}} \log {\relax (x )} + \left (- 8 x^{4} \log {\relax (x )} + 8 x^{3} \log {\relax (x )}\right ) e^{\frac {3 x}{5}} + \left (12 x^{5} \log {\relax (x )} - 24 x^{4} \log {\relax (x )} + 12 x^{3} \log {\relax (x )}\right ) e^{\frac {2 x}{5}} + \left (- 8 x^{6} \log {\relax (x )} + 24 x^{5} \log {\relax (x )} - 24 x^{4} \log {\relax (x )} + 8 x^{3} \log {\relax (x )}\right ) e^{\frac {x}{5}}}{x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(-10*ln(x)**2+((8*x-10)*exp(1/5*x)**4+(-24*x**2+24*x-40)*exp(1/5*x)**3+(24*x**3+12*x**2+24*x-60)
*exp(1/5*x)**2+(-8*x**4-56*x**3+96*x**2+8*x-40)*exp(1/5*x)+30*x**4-80*x**3+60*x**2+10)*ln(x)+10*exp(1/5*x)**4+
(-40*x+40)*exp(1/5*x)**3+(60*x**2-120*x+60)*exp(1/5*x)**2+(-40*x**3+120*x**2-120*x+40)*exp(1/5*x)+10*x**4-40*x
**3+60*x**2-40*x+10)/x**2,x)

[Out]

-8*log(x) + (2*x**4 - 8*x**3 + 12*x**2 + 2)*log(x)/x + 2*log(x)**2/x + (2*x**3*exp(4*x/5)*log(x) + (-8*x**4*lo
g(x) + 8*x**3*log(x))*exp(3*x/5) + (12*x**5*log(x) - 24*x**4*log(x) + 12*x**3*log(x))*exp(2*x/5) + (-8*x**6*lo
g(x) + 24*x**5*log(x) - 24*x**4*log(x) + 8*x**3*log(x))*exp(x/5))/x**4

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