3.90.16 \(\int \frac {-5 x^2+e (5 x^2+10 x^3) \log (3)+(-4 x+10 e^4 x-10 x^2+e (10 x^2+10 x^3) \log (3)) \log (\frac {1}{5} (2-5 e^4+5 x+e (-5 x-5 x^2) \log (3)))}{-2+5 e^4-5 x+e (5 x+5 x^2) \log (3)} \, dx\)

Optimal. Leaf size=25 \[ x^2 \log \left (\frac {2}{5}+x-e \left (e^3+\left (x+x^2\right ) \log (3)\right )\right ) \]

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Rubi [A]  time = 1.16, antiderivative size = 34, normalized size of antiderivative = 1.36, number of steps used = 18, number of rules used = 10, integrand size = 106, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.094, Rules used = {6741, 6688, 14, 800, 634, 618, 204, 628, 2525, 12} \begin {gather*} x^2 \log \left (-e x^2 \log (3)+x (1-e \log (3))+\frac {1}{5} \left (2-5 e^4\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-5*x^2 + E*(5*x^2 + 10*x^3)*Log[3] + (-4*x + 10*E^4*x - 10*x^2 + E*(10*x^2 + 10*x^3)*Log[3])*Log[(2 - 5*E
^4 + 5*x + E*(-5*x - 5*x^2)*Log[3])/5])/(-2 + 5*E^4 - 5*x + E*(5*x + 5*x^2)*Log[3]),x]

[Out]

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

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 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 628

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

Rule 634

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[(2*c*d - b*e)/(2*c), Int[1/(a +
b*x + c*x^2), x], x] + Dist[e/(2*c), Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] &
& NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] &&  !NiceSqrtQ[b^2 - 4*a*c]

Rule 800

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Int[Exp
andIntegrand[((d + e*x)^m*(f + g*x))/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[b^2 -
4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegerQ[m]

Rule 2525

Int[((a_.) + Log[(c_.)*(RFx_)^(p_.)]*(b_.))^(n_.)*((d_.) + (e_.)*(x_))^(m_.), x_Symbol] :> Simp[((d + e*x)^(m
+ 1)*(a + b*Log[c*RFx^p])^n)/(e*(m + 1)), x] - Dist[(b*n*p)/(e*(m + 1)), Int[SimplifyIntegrand[((d + e*x)^(m +
1)*(a + b*Log[c*RFx^p])^(n - 1)*D[RFx, x])/RFx, x], x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && RationalFunc
tionQ[RFx, x] && IGtQ[n, 0] && (EqQ[n, 1] || IntegerQ[m]) && NeQ[m, -1]

Rule 6688

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

Rule 6741

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {5 x^2-e \left (5 x^2+10 x^3\right ) \log (3)-\left (-4 x+10 e^4 x-10 x^2+e \left (10 x^2+10 x^3\right ) \log (3)\right ) \log \left (\frac {1}{5} \left (2-5 e^4+5 x+e \left (-5 x-5 x^2\right ) \log (3)\right )\right )}{2-5 e^4-5 e x^2 \log (3)+5 x (1-e \log (3))} \, dx\\ &=\int x \left (\frac {5 x (-1+e (\log (3)+x \log (9)))}{-2+5 e^4-5 x+5 e x (1+x) \log (3)}+2 \log \left (\frac {2}{5}-e^4+x-e x (1+x) \log (3)\right )\right ) \, dx\\ &=\int \left (\frac {5 x^2 (1-e \log (3)-e x \log (9))}{2-5 e^4-5 e x^2 \log (3)+5 x (1-e \log (3))}+2 x \log \left (\frac {1}{5} \left (2-5 e^4\right )-e x^2 \log (3)+x (1-e \log (3))\right )\right ) \, dx\\ &=2 \int x \log \left (\frac {1}{5} \left (2-5 e^4\right )-e x^2 \log (3)+x (1-e \log (3))\right ) \, dx+5 \int \frac {x^2 (1-e \log (3)-e x \log (9))}{2-5 e^4-5 e x^2 \log (3)+5 x (1-e \log (3))} \, dx\\ &=x^2 \log \left (\frac {1}{5} \left (2-5 e^4\right )-e x^2 \log (3)+x (1-e \log (3))\right )+5 \int \left (\frac {1}{5} \left (-1+\frac {1}{e \log (3)}\right )+\frac {x \log (9)}{5 \log (3)}-\frac {\left (2-5 e^4\right ) \log (3) (1-e \log (3))+x \left (5 e^2 \log ^3(3)-5 e^5 \log (3) \log (9)-e \log (9) \log (27)+\log (243)\right )}{5 e \log ^2(3) \left (2-5 e^4-5 e x^2 \log (3)+5 x (1-e \log (3))\right )}\right ) \, dx-\int \frac {5 x^2 (1-e \log (3)-e x \log (9))}{2-5 e^4-5 e x^2 \log (3)+5 x (1-e \log (3))} \, dx\\ &=-x \left (1-\frac {1}{e \log (3)}\right )+\frac {x^2 \log (9)}{2 \log (3)}+x^2 \log \left (\frac {1}{5} \left (2-5 e^4\right )-e x^2 \log (3)+x (1-e \log (3))\right )-5 \int \frac {x^2 (1-e \log (3)-e x \log (9))}{2-5 e^4-5 e x^2 \log (3)+5 x (1-e \log (3))} \, dx-\frac {\int \frac {\left (2-5 e^4\right ) \log (3) (1-e \log (3))+x \left (5 e^2 \log ^3(3)-5 e^5 \log (3) \log (9)-e \log (9) \log (27)+\log (243)\right )}{2-5 e^4-5 e x^2 \log (3)+5 x (1-e \log (3))} \, dx}{e \log ^2(3)}\\ &=-x \left (1-\frac {1}{e \log (3)}\right )+\frac {x^2 \log (9)}{2 \log (3)}+x^2 \log \left (\frac {1}{5} \left (2-5 e^4\right )-e x^2 \log (3)+x (1-e \log (3))\right )-5 \int \left (\frac {1}{5} \left (-1+\frac {1}{e \log (3)}\right )+\frac {x \log (9)}{5 \log (3)}-\frac {\left (2-5 e^4\right ) \log (3) (1-e \log (3))+x \left (5 e^2 \log ^3(3)-5 e^5 \log (3) \log (9)-e \log (9) \log (27)+\log (243)\right )}{5 e \log ^2(3) \left (2-5 e^4-5 e x^2 \log (3)+5 x (1-e \log (3))\right )}\right ) \, dx+\frac {\left (5 e^2 \log ^3(3)-5 e^5 \log (3) \log (9)-e \log (9) \log (27)+\log (243)\right ) \int \frac {-10 e x \log (3)+5 (1-e \log (3))}{2-5 e^4-5 e x^2 \log (3)+5 x (1-e \log (3))} \, dx}{10 e^2 \log ^3(3)}-\frac {\left ((1-e \log (3)) \left (4 e \log ^2(3)+5 e^2 \log ^3(3)-e \log (9) \log (27)-5 e^5 \log (3) \log (81)+\log (243)\right )\right ) \int \frac {1}{2-5 e^4-5 e x^2 \log (3)+5 x (1-e \log (3))} \, dx}{2 e^2 \log ^3(3)}\\ &=x^2 \log \left (\frac {1}{5} \left (2-5 e^4\right )-e x^2 \log (3)+x (1-e \log (3))\right )+\frac {\left (5 e^2 \log ^3(3)-5 e^5 \log (3) \log (9)-e \log (9) \log (27)+\log (243)\right ) \log \left (2-5 e^4-5 e x^2 \log (3)+5 x (1-e \log (3))\right )}{10 e^2 \log ^3(3)}+\frac {\int \frac {\left (2-5 e^4\right ) \log (3) (1-e \log (3))+x \left (5 e^2 \log ^3(3)-5 e^5 \log (3) \log (9)-e \log (9) \log (27)+\log (243)\right )}{2-5 e^4-5 e x^2 \log (3)+5 x (1-e \log (3))} \, dx}{e \log ^2(3)}+\frac {\left ((1-e \log (3)) \left (4 e \log ^2(3)+5 e^2 \log ^3(3)-e \log (9) \log (27)-5 e^5 \log (3) \log (81)+\log (243)\right )\right ) \operatorname {Subst}\left (\int \frac {1}{-x^2+5 \left (4 e \left (2-5 e^4\right ) \log (3)+5 (1-e \log (3))^2\right )} \, dx,x,-10 e x \log (3)+5 (1-e \log (3))\right )}{e^2 \log ^3(3)}\\ &=-\frac {\tan ^{-1}\left (\sqrt {\frac {5}{-5+20 e^5 \log (3)-5 e^2 \log ^2(3)+e \log (9)}} (1-e (\log (3)+x \log (9)))\right ) (1-e \log (3)) \left (4 e \log ^2(3)+5 e^2 \log ^3(3)-e \log (9) \log (27)-5 e^5 \log (3) \log (81)+\log (243)\right )}{e^2 \log ^3(3) \sqrt {5 \left (-5+20 e^5 \log (3)-5 e^2 \log ^2(3)+e \log (9)\right )}}+x^2 \log \left (\frac {1}{5} \left (2-5 e^4\right )-e x^2 \log (3)+x (1-e \log (3))\right )+\frac {\left (5 e^2 \log ^3(3)-5 e^5 \log (3) \log (9)-e \log (9) \log (27)+\log (243)\right ) \log \left (2-5 e^4-5 e x^2 \log (3)+5 x (1-e \log (3))\right )}{10 e^2 \log ^3(3)}-\frac {\left (5 e^2 \log ^3(3)-5 e^5 \log (3) \log (9)-e \log (9) \log (27)+\log (243)\right ) \int \frac {-10 e x \log (3)+5 (1-e \log (3))}{2-5 e^4-5 e x^2 \log (3)+5 x (1-e \log (3))} \, dx}{10 e^2 \log ^3(3)}+\frac {\left ((1-e \log (3)) \left (4 e \log ^2(3)+5 e^2 \log ^3(3)-e \log (9) \log (27)-5 e^5 \log (3) \log (81)+\log (243)\right )\right ) \int \frac {1}{2-5 e^4-5 e x^2 \log (3)+5 x (1-e \log (3))} \, dx}{2 e^2 \log ^3(3)}\\ &=-\frac {\tan ^{-1}\left (\sqrt {\frac {5}{-5+20 e^5 \log (3)-5 e^2 \log ^2(3)+e \log (9)}} (1-e (\log (3)+x \log (9)))\right ) (1-e \log (3)) \left (4 e \log ^2(3)+5 e^2 \log ^3(3)-e \log (9) \log (27)-5 e^5 \log (3) \log (81)+\log (243)\right )}{e^2 \log ^3(3) \sqrt {5 \left (-5+20 e^5 \log (3)-5 e^2 \log ^2(3)+e \log (9)\right )}}+x^2 \log \left (\frac {1}{5} \left (2-5 e^4\right )-e x^2 \log (3)+x (1-e \log (3))\right )-\frac {\left ((1-e \log (3)) \left (4 e \log ^2(3)+5 e^2 \log ^3(3)-e \log (9) \log (27)-5 e^5 \log (3) \log (81)+\log (243)\right )\right ) \operatorname {Subst}\left (\int \frac {1}{-x^2+5 \left (4 e \left (2-5 e^4\right ) \log (3)+5 (1-e \log (3))^2\right )} \, dx,x,-10 e x \log (3)+5 (1-e \log (3))\right )}{e^2 \log ^3(3)}\\ &=x^2 \log \left (\frac {1}{5} \left (2-5 e^4\right )-e x^2 \log (3)+x (1-e \log (3))\right )\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.88, size = 385, normalized size = 15.40 \begin {gather*} \frac {1}{10} \left (-\frac {5 x^2 \log (9)}{\log (3)}-\frac {2 \tan ^{-1}\left (\sqrt {\frac {5}{-5+20 e^5 \log (3)-5 e^2 \log ^2(3)+e \log (9)}} (1-e (\log (3)+x \log (9)))\right ) (-1+e \log (3)) \sqrt {5 \left (-5-5 e^2 \log ^2(3)+e \log (9)+5 e^5 \log (81)\right )}}{e^2 \log ^2(3)}+\frac {2 \tan ^{-1}\left (\sqrt {\frac {5}{-5+20 e^5 \log (3)-5 e^2 \log ^2(3)+e \log (9)}} (1-e (\log (3)+x \log (9)))\right ) (-1+e \log (3)) \sqrt {5 \left (-5-5 e^2 \log ^2(3)+e \log (9)+5 e^5 \log (81)\right )} \left (-4 e \log ^2(3)-5 e^2 \log ^3(3)+e \log (9) \log (27)+5 e^5 \log (3) \log (81)-\log (243)\right )}{e^2 \log ^3(3) \left (-5+20 e^5 \log (3)-5 e^2 \log ^2(3)+e \log (9)\right )}+\frac {\left (-5 e^2 \log ^3(3)+5 e^5 \log (3) \log (9)+e \log (9) \log (27)-\log (243)\right ) \log \left (2-5 e^4+5 x-5 e x (1+x) \log (3)\right )}{e^2 \log ^3(3)}-\frac {\left (-5+10 e^5 \log (3)-5 e^2 \log ^2(3)+e \log (729)\right ) \log \left (2-5 e^4+5 x-5 e x (1+x) \log (3)\right )}{e^2 \log ^2(3)}+10 x^2 \left (1+\log \left (\frac {2}{5}-e^4+x-e x (1+x) \log (3)\right )\right )\right ) \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[(-5*x^2 + E*(5*x^2 + 10*x^3)*Log[3] + (-4*x + 10*E^4*x - 10*x^2 + E*(10*x^2 + 10*x^3)*Log[3])*Log[(2
- 5*E^4 + 5*x + E*(-5*x - 5*x^2)*Log[3])/5])/(-2 + 5*E^4 - 5*x + E*(5*x + 5*x^2)*Log[3]),x]

[Out]

((-5*x^2*Log[9])/Log[3] - (2*ArcTan[Sqrt[5/(-5 + 20*E^5*Log[3] - 5*E^2*Log[3]^2 + E*Log[9])]*(1 - E*(Log[3] +
x*Log[9]))]*(-1 + E*Log[3])*Sqrt[5*(-5 - 5*E^2*Log[3]^2 + E*Log[9] + 5*E^5*Log[81])])/(E^2*Log[3]^2) + (2*ArcT
an[Sqrt[5/(-5 + 20*E^5*Log[3] - 5*E^2*Log[3]^2 + E*Log[9])]*(1 - E*(Log[3] + x*Log[9]))]*(-1 + E*Log[3])*Sqrt[
5*(-5 - 5*E^2*Log[3]^2 + E*Log[9] + 5*E^5*Log[81])]*(-4*E*Log[3]^2 - 5*E^2*Log[3]^3 + E*Log[9]*Log[27] + 5*E^5
*Log[3]*Log[81] - Log[243]))/(E^2*Log[3]^3*(-5 + 20*E^5*Log[3] - 5*E^2*Log[3]^2 + E*Log[9])) + ((-5*E^2*Log[3]
^3 + 5*E^5*Log[3]*Log[9] + E*Log[9]*Log[27] - Log[243])*Log[2 - 5*E^4 + 5*x - 5*E*x*(1 + x)*Log[3]])/(E^2*Log[
3]^3) - ((-5 + 10*E^5*Log[3] - 5*E^2*Log[3]^2 + E*Log[729])*Log[2 - 5*E^4 + 5*x - 5*E*x*(1 + x)*Log[3]])/(E^2*
Log[3]^2) + 10*x^2*(1 + Log[2/5 - E^4 + x - E*x*(1 + x)*Log[3]]))/10

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fricas [A]  time = 0.43, size = 23, normalized size = 0.92 \begin {gather*} x^{2} \log \left (-{\left (x^{2} + x\right )} e \log \relax (3) + x - e^{4} + \frac {2}{5}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((10*x^3+10*x^2)*exp(1)*log(3)+10*x*exp(1)*exp(3)-10*x^2-4*x)*log(1/5*(-5*x^2-5*x)*exp(1)*log(3)-ex
p(1)*exp(3)+x+2/5)+(10*x^3+5*x^2)*exp(1)*log(3)-5*x^2)/((5*x^2+5*x)*exp(1)*log(3)+5*exp(1)*exp(3)-5*x-2),x, al
gorithm="fricas")

[Out]

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

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giac [A]  time = 0.25, size = 38, normalized size = 1.52 \begin {gather*} -x^{2} \log \relax (5) + x^{2} \log \left (5 \, x^{2} e \log \relax (3) + 5 \, x e \log \relax (3) - 5 \, x + 5 \, e^{4} - 2\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((10*x^3+10*x^2)*exp(1)*log(3)+10*x*exp(1)*exp(3)-10*x^2-4*x)*log(1/5*(-5*x^2-5*x)*exp(1)*log(3)-ex
p(1)*exp(3)+x+2/5)+(10*x^3+5*x^2)*exp(1)*log(3)-5*x^2)/((5*x^2+5*x)*exp(1)*log(3)+5*exp(1)*exp(3)-5*x-2),x, al
gorithm="giac")

[Out]

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

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maple [A]  time = 1.02, size = 28, normalized size = 1.12

method	result	size
risch	\(x^{2} \ln \left (\frac {\left (-5 x^{2}-5 x \right ) {\mathrm e} \ln \relax (3)}{5}-{\mathrm e}^{4}+x +\frac {2}{5}\right )\)	\(28\)
norman	\(x^{2} \ln \left (\frac {\left (-5 x^{2}-5 x \right ) {\mathrm e} \ln \relax (3)}{5}-{\mathrm e} \,{\mathrm e}^{3}+x +\frac {2}{5}\right )\)	\(30\)
default	\(-x^{2} \ln \relax (5)+x^{2} \ln \left (-5 \,{\mathrm e} \ln \relax (3) x^{2}-5 x \,{\mathrm e} \ln \relax (3)-5 \,{\mathrm e} \,{\mathrm e}^{3}+5 x +2\right )\)	\(41\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((10*x^3+10*x^2)*exp(1)*ln(3)+10*x*exp(1)*exp(3)-10*x^2-4*x)*ln(1/5*(-5*x^2-5*x)*exp(1)*ln(3)-exp(1)*exp(
3)+x+2/5)+(10*x^3+5*x^2)*exp(1)*ln(3)-5*x^2)/((5*x^2+5*x)*exp(1)*ln(3)+5*exp(1)*exp(3)-5*x-2),x,method=_RETURN
VERBOSE)

[Out]

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

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maxima [B]  time = 1.95, size = 705, normalized size = 28.20 result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((10*x^3+10*x^2)*exp(1)*log(3)+10*x*exp(1)*exp(3)-10*x^2-4*x)*log(1/5*(-5*x^2-5*x)*exp(1)*log(3)-ex
p(1)*exp(3)+x+2/5)+(10*x^3+5*x^2)*exp(1)*log(3)-5*x^2)/((5*x^2+5*x)*exp(1)*log(3)+5*exp(1)*exp(3)-5*x-2),x, al
gorithm="maxima")

[Out]

1/2*(2*x*e^(-1)/log(3) - (e*log(3) - 1)*e^(-2)*log(5*x^2*e*log(3) + 5*(e*log(3) - 1)*x + 5*e^4 - 2)/log(3)^2 +
2*(5*e^2*log(3)^2 - 2*(5*e^5 + 3*e)*log(3) + 5)*arctan(5*(2*x*e*log(3) + e*log(3) - 1)/sqrt(-25*e^2*log(3)^2
+ 10*(10*e^5 + e)*log(3) - 25))*e^(-2)/(sqrt(-25*e^2*log(3)^2 + 10*(10*e^5 + e)*log(3) - 25)*log(3)^2))*e*log(
3) + 1/5*(5*(x^2*e*log(3) - 2*(e*log(3) - 1)*x)*e^(-2)/log(3)^2 + (5*e^2*log(3)^2 - (5*e^5 + 8*e)*log(3) + 5)*
e^(-3)*log(5*x^2*e*log(3) + 5*(e*log(3) - 1)*x + 5*e^4 - 2)/log(3)^3 - 10*(5*e^3*log(3)^3 - 3*(5*e^6 + 3*e^2)*
log(3)^2 + 3*(5*e^5 + 3*e)*log(3) - 5)*arctan(5*(2*x*e*log(3) + e*log(3) - 1)/sqrt(-25*e^2*log(3)^2 + 10*(10*e
^5 + e)*log(3) - 25))*e^(-3)/(sqrt(-25*e^2*log(3)^2 + 10*(10*e^5 + e)*log(3) - 25)*log(3)^3))*e*log(3) - x*e^(
-1)/log(3) + 1/2*(e*log(3) - 1)*e^(-2)*log(5*x^2*e*log(3) + 5*(e*log(3) - 1)*x + 5*e^4 - 2)/log(3)^2 + (5*e^3*
log(3)^3 - 20*e^6*log(3)^2 - 7*e^2*log(3)^2 + 20*e^5*log(3) + 7*e*log(3) - 5)*arctan(5*(2*x*e*log(3) + e*log(3
) - 1)/sqrt(-25*e^2*log(3)^2 + 10*(10*e^5 + e)*log(3) - 25))*e^(-2)/(sqrt(-25*e^2*log(3)^2 + 10*(10*e^5 + e)*l
og(3) - 25)*log(3)^2) - (5*e^2*log(3)^2 - 2*(5*e^5 + 3*e)*log(3) + 5)*arctan(5*(2*x*e*log(3) + e*log(3) - 1)/s
qrt(-25*e^2*log(3)^2 + 10*(10*e^5 + e)*log(3) - 25))*e^(-2)/(sqrt(-25*e^2*log(3)^2 + 10*(10*e^5 + e)*log(3) -
25)*log(3)^2) - 1/10*(10*x^2*(log(5) + 1)*e^2*log(3)^2 - 10*(e^2*log(3)^2 - e*log(3))*x - (10*x^2*e^2*log(3)^2
- 5*e^2*log(3)^2 + 10*e^5*log(3) + 6*e*log(3) - 5)*log(-5*x^2*e*log(3) - 5*(e*log(3) - 1)*x - 5*e^4 + 2))*e^(
-2)/log(3)^2

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mupad [B]  time = 8.05, size = 27, normalized size = 1.08 \begin {gather*} x^2\,\ln \left (x-{\mathrm {e}}^4-\frac {\mathrm {e}\,\ln \relax (3)\,\left (5\,x^2+5\,x\right )}{5}+\frac {2}{5}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(x - exp(4) - (exp(1)*log(3)*(5*x + 5*x^2))/5 + 2/5)*(4*x - 10*x*exp(4) + 10*x^2 - exp(1)*log(3)*(10*x
^2 + 10*x^3)) + 5*x^2 - exp(1)*log(3)*(5*x^2 + 10*x^3))/(5*x - 5*exp(4) - exp(1)*log(3)*(5*x + 5*x^2) + 2),x)

[Out]

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

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sympy [A]  time = 0.25, size = 26, normalized size = 1.04 \begin {gather*} x^{2} \log {\left (x + e \left (- x^{2} - x\right ) \log {\relax (3 )} - e^{4} + \frac {2}{5} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((10*x**3+10*x**2)*exp(1)*ln(3)+10*x*exp(1)*exp(3)-10*x**2-4*x)*ln(1/5*(-5*x**2-5*x)*exp(1)*ln(3)-e
xp(1)*exp(3)+x+2/5)+(10*x**3+5*x**2)*exp(1)*ln(3)-5*x**2)/((5*x**2+5*x)*exp(1)*ln(3)+5*exp(1)*exp(3)-5*x-2),x)

[Out]

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

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