3.64.19 \(\int \frac {16 x^2+e (-4 x-32 x^2)+e^2 (-30+4 x+16 x^2)+e^9 (-16 e x+e^2 (-20+16 x))+(24 x^2+e (-6 x-48 x^2)+e^2 (10+6 x+24 x^2)+e^9 (-24 e x+e^2 (-10+24 x))) \log (x)+(12 x^2+e^9 (-12 e x+12 e^2 x)+e (2 x-24 x^2)+e^2 (-2 x+12 x^2)) \log ^2(x)+(2 x^2+e^9 (-2 e x+2 e^2 x)+e (2 x-4 x^2)+e^2 (-2 x+2 x^2)) \log ^3(x)}{8 e^2 x+12 e^2 x \log (x)+6 e^2 x \log ^2(x)+e^2 x \log ^3(x)} \, dx\)

Optimal. Leaf size=22 \[ \left (-1+e^9+x-\frac {x}{e}+\frac {5}{2+\log (x)}\right )^2 \]

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Rubi [B]  time = 1.18, antiderivative size = 74, normalized size of antiderivative = 3.36, number of steps used = 15, number of rules used = 9, integrand size = 238, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.038, Rules used = {6688, 12, 6742, 2302, 30, 2353, 2297, 2299, 2178} \begin {gather*} \frac {(1-e)^2 x^2}{e^2}+\frac {2 \left (1-e-e^9+e^{10}\right ) x}{e}-\frac {10 (1-e) x}{e (\log (x)+2)}-\frac {10 \left (1-e^9\right )}{\log (x)+2}+\frac {25}{(\log (x)+2)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(16*x^2 + E*(-4*x - 32*x^2) + E^2*(-30 + 4*x + 16*x^2) + E^9*(-16*E*x + E^2*(-20 + 16*x)) + (24*x^2 + E*(-
6*x - 48*x^2) + E^2*(10 + 6*x + 24*x^2) + E^9*(-24*E*x + E^2*(-10 + 24*x)))*Log[x] + (12*x^2 + E^9*(-12*E*x +
12*E^2*x) + E*(2*x - 24*x^2) + E^2*(-2*x + 12*x^2))*Log[x]^2 + (2*x^2 + E^9*(-2*E*x + 2*E^2*x) + E*(2*x - 4*x^
2) + E^2*(-2*x + 2*x^2))*Log[x]^3)/(8*E^2*x + 12*E^2*x*Log[x] + 6*E^2*x*Log[x]^2 + E^2*x*Log[x]^3),x]

[Out]

(2*(1 - E - E^9 + E^10)*x)/E + ((1 - E)^2*x^2)/E^2 + 25/(2 + Log[x])^2 - (10*(1 - E^9))/(2 + Log[x]) - (10*(1
- E)*x)/(E*(2 + Log[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 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -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 2297

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_), x_Symbol] :> Simp[(x*(a + b*Log[c*x^n])^(p + 1))/(b*n*(p + 1))
, x] - Dist[1/(b*n*(p + 1)), Int[(a + b*Log[c*x^n])^(p + 1), x], x] /; FreeQ[{a, b, c, n}, x] && LtQ[p, -1] &&
IntegerQ[2*p]

Rule 2299

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_), x_Symbol] :> Dist[1/(n*c^(1/n)), Subst[Int[E^(x/n)*(a + b*x)^p
, x], x, Log[c*x^n]], x] /; FreeQ[{a, b, c, p}, x] && IntegerQ[1/n]

Rule 2302

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/(x_), x_Symbol] :> Dist[1/(b*n), Subst[Int[x^p, x], x, a + b*L
og[c*x^n]], x] /; FreeQ[{a, b, c, n, p}, x]

Rule 2353

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^(r_.))^(q_.), x_Symbol]
:> With[{u = ExpandIntegrand[(a + b*Log[c*x^n])^p, (f*x)^m*(d + e*x^r)^q, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[
{a, b, c, d, e, f, m, n, p, q, r}, x] && IntegerQ[q] && (GtQ[q, 0] || (IGtQ[p, 0] && IntegerQ[m] && IntegerQ[r
]))

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 \frac {2 \left (-8 e^{10} x+8 x^2+2 e^{11} (-5+4 x)-2 e x (1+8 x)+e^2 \left (-15+2 x+8 x^2\right )+\left (-12 e^{10} x+12 x^2-3 e x (1+8 x)+e^{11} (-5+12 x)+e^2 \left (5+3 x+12 x^2\right )\right ) \log (x)+(-1+e) x \left (6 e^{10}-6 x+e (-1+6 x)\right ) \log ^2(x)+x \left (e-e^{10}+e^{11}+e^2 (-1+x)+x-2 e x\right ) \log ^3(x)\right )}{e^2 x (2+\log (x))^3} \, dx\\ &=\frac {2 \int \frac {-8 e^{10} x+8 x^2+2 e^{11} (-5+4 x)-2 e x (1+8 x)+e^2 \left (-15+2 x+8 x^2\right )+\left (-12 e^{10} x+12 x^2-3 e x (1+8 x)+e^{11} (-5+12 x)+e^2 \left (5+3 x+12 x^2\right )\right ) \log (x)+(-1+e) x \left (6 e^{10}-6 x+e (-1+6 x)\right ) \log ^2(x)+x \left (e-e^{10}+e^{11}+e^2 (-1+x)+x-2 e x\right ) \log ^3(x)}{x (2+\log (x))^3} \, dx}{e^2}\\ &=\frac {2 \int \left (e \left (1+e \left (-1-e^8+e^9\right )\right )+(1+(-2+e) e) x-\frac {25 e^2}{x (2+\log (x))^3}+\frac {5 e \left (e \left (1-e^9\right )+(1-e) x\right )}{x (2+\log (x))^2}+\frac {5 (-1+e) e}{2+\log (x)}\right ) \, dx}{e^2}\\ &=\frac {2 \left (1-e-e^9+e^{10}\right ) x}{e}+\frac {(1-e)^2 x^2}{e^2}-50 \int \frac {1}{x (2+\log (x))^3} \, dx+\frac {10 \int \frac {e \left (1-e^9\right )+(1-e) x}{x (2+\log (x))^2} \, dx}{e}-\frac {(10 (1-e)) \int \frac {1}{2+\log (x)} \, dx}{e}\\ &=\frac {2 \left (1-e-e^9+e^{10}\right ) x}{e}+\frac {(1-e)^2 x^2}{e^2}-50 \operatorname {Subst}\left (\int \frac {1}{x^3} \, dx,x,2+\log (x)\right )+\frac {10 \int \left (\frac {1-e}{(2+\log (x))^2}+\frac {e-e^{10}}{x (2+\log (x))^2}\right ) \, dx}{e}-\frac {(10 (1-e)) \operatorname {Subst}\left (\int \frac {e^x}{2+x} \, dx,x,\log (x)\right )}{e}\\ &=\frac {2 \left (1-e-e^9+e^{10}\right ) x}{e}+\frac {(1-e)^2 x^2}{e^2}-\frac {10 (1-e) \text {Ei}(2+\log (x))}{e^3}+\frac {25}{(2+\log (x))^2}+\frac {(10 (1-e)) \int \frac {1}{(2+\log (x))^2} \, dx}{e}+\left (10 \left (1-e^9\right )\right ) \int \frac {1}{x (2+\log (x))^2} \, dx\\ &=\frac {2 \left (1-e-e^9+e^{10}\right ) x}{e}+\frac {(1-e)^2 x^2}{e^2}-\frac {10 (1-e) \text {Ei}(2+\log (x))}{e^3}+\frac {25}{(2+\log (x))^2}-\frac {10 (1-e) x}{e (2+\log (x))}+\frac {(10 (1-e)) \int \frac {1}{2+\log (x)} \, dx}{e}+\left (10 \left (1-e^9\right )\right ) \operatorname {Subst}\left (\int \frac {1}{x^2} \, dx,x,2+\log (x)\right )\\ &=\frac {2 \left (1-e-e^9+e^{10}\right ) x}{e}+\frac {(1-e)^2 x^2}{e^2}-\frac {10 (1-e) \text {Ei}(2+\log (x))}{e^3}+\frac {25}{(2+\log (x))^2}-\frac {10 \left (1-e^9\right )}{2+\log (x)}-\frac {10 (1-e) x}{e (2+\log (x))}+\frac {(10 (1-e)) \operatorname {Subst}\left (\int \frac {e^x}{2+x} \, dx,x,\log (x)\right )}{e}\\ &=\frac {2 \left (1-e-e^9+e^{10}\right ) x}{e}+\frac {(1-e)^2 x^2}{e^2}+\frac {25}{(2+\log (x))^2}-\frac {10 \left (1-e^9\right )}{2+\log (x)}-\frac {10 (1-e) x}{e (2+\log (x))}\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.12, size = 68, normalized size = 3.09 \begin {gather*} \frac {2 \left (e \left (1-e-e^9+e^{10}\right ) x+\frac {1}{2} (-1+e)^2 x^2+\frac {25 e^2}{2 (2+\log (x))^2}+\frac {5 e \left (e^{10}+e (-1+x)-x\right )}{2+\log (x)}\right )}{e^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(16*x^2 + E*(-4*x - 32*x^2) + E^2*(-30 + 4*x + 16*x^2) + E^9*(-16*E*x + E^2*(-20 + 16*x)) + (24*x^2
+ E*(-6*x - 48*x^2) + E^2*(10 + 6*x + 24*x^2) + E^9*(-24*E*x + E^2*(-10 + 24*x)))*Log[x] + (12*x^2 + E^9*(-12*
E*x + 12*E^2*x) + E*(2*x - 24*x^2) + E^2*(-2*x + 12*x^2))*Log[x]^2 + (2*x^2 + E^9*(-2*E*x + 2*E^2*x) + E*(2*x
- 4*x^2) + E^2*(-2*x + 2*x^2))*Log[x]^3)/(8*E^2*x + 12*E^2*x*Log[x] + 6*E^2*x*Log[x]^2 + E^2*x*Log[x]^3),x]

[Out]

(2*(E*(1 - E - E^9 + E^10)*x + ((-1 + E)^2*x^2)/2 + (25*E^2)/(2*(2 + Log[x])^2) + (5*E*(E^10 + E*(-1 + x) - x)
)/(2 + Log[x])))/E^2

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fricas [B]  time = 0.67, size = 152, normalized size = 6.91 \begin {gather*} \frac {{\left (x^{2} + 2 \, x e^{11} - 2 \, x e^{10} + {\left (x^{2} - 2 \, x\right )} e^{2} - 2 \, {\left (x^{2} - x\right )} e\right )} \log \relax (x)^{2} + 4 \, x^{2} + 4 \, {\left (2 \, x + 5\right )} e^{11} - 8 \, x e^{10} + {\left (4 \, x^{2} + 12 \, x + 5\right )} e^{2} - 4 \, {\left (2 \, x^{2} + 3 \, x\right )} e + 2 \, {\left (2 \, x^{2} + {\left (4 \, x + 5\right )} e^{11} - 4 \, x e^{10} + {\left (2 \, x^{2} + x - 5\right )} e^{2} - {\left (4 \, x^{2} + x\right )} e\right )} \log \relax (x)}{e^{2} \log \relax (x)^{2} + 4 \, e^{2} \log \relax (x) + 4 \, e^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x*exp(1)^2-2*x*exp(1))*exp(9)+(2*x^2-2*x)*exp(1)^2+(-4*x^2+2*x)*exp(1)+2*x^2)*log(x)^3+((12*x*e
xp(1)^2-12*x*exp(1))*exp(9)+(12*x^2-2*x)*exp(1)^2+(-24*x^2+2*x)*exp(1)+12*x^2)*log(x)^2+(((24*x-10)*exp(1)^2-2
4*x*exp(1))*exp(9)+(24*x^2+6*x+10)*exp(1)^2+(-48*x^2-6*x)*exp(1)+24*x^2)*log(x)+((16*x-20)*exp(1)^2-16*x*exp(1
))*exp(9)+(16*x^2+4*x-30)*exp(1)^2+(-32*x^2-4*x)*exp(1)+16*x^2)/(x*exp(1)^2*log(x)^3+6*x*exp(1)^2*log(x)^2+12*
x*exp(1)^2*log(x)+8*x*exp(1)^2),x, algorithm="fricas")

[Out]

((x^2 + 2*x*e^11 - 2*x*e^10 + (x^2 - 2*x)*e^2 - 2*(x^2 - x)*e)*log(x)^2 + 4*x^2 + 4*(2*x + 5)*e^11 - 8*x*e^10
+ (4*x^2 + 12*x + 5)*e^2 - 4*(2*x^2 + 3*x)*e + 2*(2*x^2 + (4*x + 5)*e^11 - 4*x*e^10 + (2*x^2 + x - 5)*e^2 - (4
*x^2 + x)*e)*log(x))/(e^2*log(x)^2 + 4*e^2*log(x) + 4*e^2)

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giac [B]  time = 0.17, size = 205, normalized size = 9.32 \begin {gather*} \frac {x^{2} e^{3} \log \relax (x)^{2} - 2 \, x^{2} e^{2} \log \relax (x)^{2} + x^{2} e \log \relax (x)^{2} + 4 \, x^{2} e^{3} \log \relax (x) - 8 \, x^{2} e^{2} \log \relax (x) + 4 \, x^{2} e \log \relax (x) + 2 \, x e^{12} \log \relax (x)^{2} - 2 \, x e^{11} \log \relax (x)^{2} - 2 \, x e^{3} \log \relax (x)^{2} + 2 \, x e^{2} \log \relax (x)^{2} + 4 \, x^{2} e^{3} - 8 \, x^{2} e^{2} + 4 \, x^{2} e + 8 \, x e^{12} \log \relax (x) - 8 \, x e^{11} \log \relax (x) + 2 \, x e^{3} \log \relax (x) - 2 \, x e^{2} \log \relax (x) + 8 \, x e^{12} - 8 \, x e^{11} + 12 \, x e^{3} - 12 \, x e^{2} + 10 \, e^{12} \log \relax (x) - 10 \, e^{3} \log \relax (x) + 20 \, e^{12} + 5 \, e^{3}}{e^{3} \log \relax (x)^{2} + 4 \, e^{3} \log \relax (x) + 4 \, e^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x*exp(1)^2-2*x*exp(1))*exp(9)+(2*x^2-2*x)*exp(1)^2+(-4*x^2+2*x)*exp(1)+2*x^2)*log(x)^3+((12*x*e
xp(1)^2-12*x*exp(1))*exp(9)+(12*x^2-2*x)*exp(1)^2+(-24*x^2+2*x)*exp(1)+12*x^2)*log(x)^2+(((24*x-10)*exp(1)^2-2
4*x*exp(1))*exp(9)+(24*x^2+6*x+10)*exp(1)^2+(-48*x^2-6*x)*exp(1)+24*x^2)*log(x)+((16*x-20)*exp(1)^2-16*x*exp(1
))*exp(9)+(16*x^2+4*x-30)*exp(1)^2+(-32*x^2-4*x)*exp(1)+16*x^2)/(x*exp(1)^2*log(x)^3+6*x*exp(1)^2*log(x)^2+12*
x*exp(1)^2*log(x)+8*x*exp(1)^2),x, algorithm="giac")

[Out]

(x^2*e^3*log(x)^2 - 2*x^2*e^2*log(x)^2 + x^2*e*log(x)^2 + 4*x^2*e^3*log(x) - 8*x^2*e^2*log(x) + 4*x^2*e*log(x)
+ 2*x*e^12*log(x)^2 - 2*x*e^11*log(x)^2 - 2*x*e^3*log(x)^2 + 2*x*e^2*log(x)^2 + 4*x^2*e^3 - 8*x^2*e^2 + 4*x^2
*e + 8*x*e^12*log(x) - 8*x*e^11*log(x) + 2*x*e^3*log(x) - 2*x*e^2*log(x) + 8*x*e^12 - 8*x*e^11 + 12*x*e^3 - 12
*x*e^2 + 10*e^12*log(x) - 10*e^3*log(x) + 20*e^12 + 5*e^3)/(e^3*log(x)^2 + 4*e^3*log(x) + 4*e^3)

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maple [B]  time = 0.13, size = 82, normalized size = 3.73

method	result	size
risch	\(x \left (2 \,{\mathrm e}^{11}-2 \,{\mathrm e}^{10}+{\mathrm e}^{2} x -2 \,{\mathrm e}^{2}-2 x \,{\mathrm e}+2 \,{\mathrm e}+x \right ) {\mathrm e}^{-2}+\frac {5 \left (2 \,{\mathrm e}^{10} \ln \relax (x )+4 \,{\mathrm e}^{10}+2 x \,{\mathrm e} \ln \relax (x )+4 x \,{\mathrm e}-2 \,{\mathrm e} \ln \relax (x )-2 x \ln \relax (x )+{\mathrm e}-4 x \right ) {\mathrm e}^{-1}}{\left (\ln \relax (x )+2\right )^{2}}\)	\(82\)
norman	\(\frac {\left (10 \,{\mathrm e} \left ({\mathrm e}^{9}-1\right ) \ln \relax (x )+\left (8 \,{\mathrm e} \,{\mathrm e}^{9}+12 \,{\mathrm e}-8 \,{\mathrm e}^{9}-12\right ) x +\left (2 \,{\mathrm e} \,{\mathrm e}^{9}-2 \,{\mathrm e}-2 \,{\mathrm e}^{9}+2\right ) x \ln \relax (x )^{2}+\left (8 \,{\mathrm e} \,{\mathrm e}^{9}+2 \,{\mathrm e}-8 \,{\mathrm e}^{9}-2\right ) x \ln \relax (x )+\left ({\mathrm e}^{2}-2 \,{\mathrm e}+1\right ) {\mathrm e}^{-1} x^{2} \ln \relax (x )^{2}+4 \left ({\mathrm e}^{2}-2 \,{\mathrm e}+1\right ) {\mathrm e}^{-1} x^{2}+4 \left ({\mathrm e}^{2}-2 \,{\mathrm e}+1\right ) {\mathrm e}^{-1} x^{2} \ln \relax (x )+5 \,{\mathrm e} \left (4 \,{\mathrm e}^{9}+1\right )\right ) {\mathrm e}^{-1}}{\left (\ln \relax (x )+2\right )^{2}}\)	\(155\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((2*x*exp(1)^2-2*x*exp(1))*exp(9)+(2*x^2-2*x)*exp(1)^2+(-4*x^2+2*x)*exp(1)+2*x^2)*ln(x)^3+((12*x*exp(1)^2
-12*x*exp(1))*exp(9)+(12*x^2-2*x)*exp(1)^2+(-24*x^2+2*x)*exp(1)+12*x^2)*ln(x)^2+(((24*x-10)*exp(1)^2-24*x*exp(
1))*exp(9)+(24*x^2+6*x+10)*exp(1)^2+(-48*x^2-6*x)*exp(1)+24*x^2)*ln(x)+((16*x-20)*exp(1)^2-16*x*exp(1))*exp(9)
+(16*x^2+4*x-30)*exp(1)^2+(-32*x^2-4*x)*exp(1)+16*x^2)/(x*exp(1)^2*ln(x)^3+6*x*exp(1)^2*ln(x)^2+12*x*exp(1)^2*
ln(x)+8*x*exp(1)^2),x,method=_RETURNVERBOSE)

[Out]

x*(2*exp(11)-2*exp(10)+exp(2)*x-2*exp(2)-2*x*exp(1)+2*exp(1)+x)*exp(-2)+5*(2*exp(10)*ln(x)+4*exp(10)+2*x*exp(1
)*ln(x)+4*x*exp(1)-2*exp(1)*ln(x)-2*x*ln(x)+exp(1)-4*x)/(ln(x)+2)^2*exp(-1)

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maxima [B]  time = 0.44, size = 189, normalized size = 8.59 \begin {gather*} \frac {4 \, x^{2} {\left (e^{2} - 2 \, e + 1\right )} + {\left (x^{2} {\left (e^{2} - 2 \, e + 1\right )} + 2 \, x {\left (e^{11} - e^{10} - e^{2} + e\right )}\right )} \log \relax (x)^{2} + 4 \, x {\left (2 \, e^{11} - 2 \, e^{10} + 3 \, e^{2} - 3 \, e\right )} + 2 \, {\left (2 \, x^{2} {\left (e^{2} - 2 \, e + 1\right )} + x {\left (4 \, e^{11} - 4 \, e^{10} + e^{2} - e\right )} + 5 \, e^{11} - 5 \, e^{2}\right )} \log \relax (x) + 10 \, e^{11} - 10 \, e^{2}}{e^{2} \log \relax (x)^{2} + 4 \, e^{2} \log \relax (x) + 4 \, e^{2}} + \frac {10 \, e^{11}}{e^{2} \log \relax (x)^{2} + 4 \, e^{2} \log \relax (x) + 4 \, e^{2}} + \frac {15 \, e^{2}}{e^{2} \log \relax (x)^{2} + 4 \, e^{2} \log \relax (x) + 4 \, e^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x*exp(1)^2-2*x*exp(1))*exp(9)+(2*x^2-2*x)*exp(1)^2+(-4*x^2+2*x)*exp(1)+2*x^2)*log(x)^3+((12*x*e
xp(1)^2-12*x*exp(1))*exp(9)+(12*x^2-2*x)*exp(1)^2+(-24*x^2+2*x)*exp(1)+12*x^2)*log(x)^2+(((24*x-10)*exp(1)^2-2
4*x*exp(1))*exp(9)+(24*x^2+6*x+10)*exp(1)^2+(-48*x^2-6*x)*exp(1)+24*x^2)*log(x)+((16*x-20)*exp(1)^2-16*x*exp(1
))*exp(9)+(16*x^2+4*x-30)*exp(1)^2+(-32*x^2-4*x)*exp(1)+16*x^2)/(x*exp(1)^2*log(x)^3+6*x*exp(1)^2*log(x)^2+12*
x*exp(1)^2*log(x)+8*x*exp(1)^2),x, algorithm="maxima")

[Out]

(4*x^2*(e^2 - 2*e + 1) + (x^2*(e^2 - 2*e + 1) + 2*x*(e^11 - e^10 - e^2 + e))*log(x)^2 + 4*x*(2*e^11 - 2*e^10 +
3*e^2 - 3*e) + 2*(2*x^2*(e^2 - 2*e + 1) + x*(4*e^11 - 4*e^10 + e^2 - e) + 5*e^11 - 5*e^2)*log(x) + 10*e^11 -
10*e^2)/(e^2*log(x)^2 + 4*e^2*log(x) + 4*e^2) + 10*e^11/(e^2*log(x)^2 + 4*e^2*log(x) + 4*e^2) + 15*e^2/(e^2*lo
g(x)^2 + 4*e^2*log(x) + 4*e^2)

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mupad [B]  time = 4.91, size = 170, normalized size = 7.73 \begin {gather*} \frac {{\left (\mathrm {e}-1\right )}^2\,x^4\,{\ln \relax (x)}^2+4\,{\left (\mathrm {e}-1\right )}^2\,x^4\,\ln \relax (x)+4\,{\left (\mathrm {e}-1\right )}^2\,x^4+\left (2\,\mathrm {e}-2\,{\mathrm {e}}^2-2\,{\mathrm {e}}^{10}+2\,{\mathrm {e}}^{11}\right )\,x^3\,{\ln \relax (x)}^2+2\,\mathrm {e}\,\left (\mathrm {e}-4\,{\mathrm {e}}^9+4\,{\mathrm {e}}^{10}-1\right )\,x^3\,\ln \relax (x)+4\,\mathrm {e}\,\left (3\,\mathrm {e}-2\,{\mathrm {e}}^9+2\,{\mathrm {e}}^{10}-3\right )\,x^3+\left (-\frac {5\,{\mathrm {e}}^2}{4}-5\,{\mathrm {e}}^{11}\right )\,x^2\,{\ln \relax (x)}^2+\left (-15\,{\mathrm {e}}^2-10\,{\mathrm {e}}^{11}\right )\,x^2\,\ln \relax (x)}{{\mathrm {e}}^2\,x^2\,{\ln \relax (x)}^2+4\,{\mathrm {e}}^2\,x^2\,\ln \relax (x)+4\,{\mathrm {e}}^2\,x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(1)*(4*x + 32*x^2) - exp(2)*(4*x + 16*x^2 - 30) + log(x)^3*(exp(2)*(2*x - 2*x^2) - exp(1)*(2*x - 4*x^
2) + exp(9)*(2*x*exp(1) - 2*x*exp(2)) - 2*x^2) + log(x)^2*(exp(2)*(2*x - 12*x^2) - exp(1)*(2*x - 24*x^2) + exp
(9)*(12*x*exp(1) - 12*x*exp(2)) - 12*x^2) + exp(9)*(16*x*exp(1) - exp(2)*(16*x - 20)) - log(x)*(exp(2)*(6*x +
24*x^2 + 10) - exp(1)*(6*x + 48*x^2) - exp(9)*(24*x*exp(1) - exp(2)*(24*x - 10)) + 24*x^2) - 16*x^2)/(8*x*exp(
2) + 12*x*exp(2)*log(x) + 6*x*exp(2)*log(x)^2 + x*exp(2)*log(x)^3),x)

[Out]

(4*x^4*(exp(1) - 1)^2 - x^2*log(x)*(15*exp(2) + 10*exp(11)) + 4*x^4*log(x)*(exp(1) - 1)^2 + 4*x^3*exp(1)*(3*ex
p(1) - 2*exp(9) + 2*exp(10) - 3) - x^2*log(x)^2*((5*exp(2))/4 + 5*exp(11)) + x^4*log(x)^2*(exp(1) - 1)^2 + x^3
*log(x)^2*(2*exp(1) - 2*exp(2) - 2*exp(10) + 2*exp(11)) + 2*x^3*exp(1)*log(x)*(exp(1) - 4*exp(9) + 4*exp(10) -
1))/(4*x^2*exp(2) + 4*x^2*exp(2)*log(x) + x^2*exp(2)*log(x)^2)

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sympy [B]  time = 0.24, size = 104, normalized size = 4.73 \begin {gather*} \frac {x^{2} \left (- 2 e + 1 + e^{2}\right )}{e^{2}} + \frac {x \left (- 2 e^{9} - 2 e + 2 + 2 e^{10}\right )}{e} + \frac {- 20 x + 20 e x + \left (- 10 x + 10 e x - 10 e + 10 e^{10}\right ) \log {\relax (x )} + 5 e + 20 e^{10}}{e \log {\relax (x )}^{2} + 4 e \log {\relax (x )} + 4 e} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x*exp(1)**2-2*x*exp(1))*exp(9)+(2*x**2-2*x)*exp(1)**2+(-4*x**2+2*x)*exp(1)+2*x**2)*ln(x)**3+((1
2*x*exp(1)**2-12*x*exp(1))*exp(9)+(12*x**2-2*x)*exp(1)**2+(-24*x**2+2*x)*exp(1)+12*x**2)*ln(x)**2+(((24*x-10)*
exp(1)**2-24*x*exp(1))*exp(9)+(24*x**2+6*x+10)*exp(1)**2+(-48*x**2-6*x)*exp(1)+24*x**2)*ln(x)+((16*x-20)*exp(1
)**2-16*x*exp(1))*exp(9)+(16*x**2+4*x-30)*exp(1)**2+(-32*x**2-4*x)*exp(1)+16*x**2)/(x*exp(1)**2*ln(x)**3+6*x*e
xp(1)**2*ln(x)**2+12*x*exp(1)**2*ln(x)+8*x*exp(1)**2),x)

[Out]

x**2*(-2*E + 1 + exp(2))*exp(-2) + x*(-2*exp(9) - 2*E + 2 + 2*exp(10))*exp(-1) + (-20*x + 20*E*x + (-10*x + 10
*E*x - 10*E + 10*exp(10))*log(x) + 5*E + 20*exp(10))/(E*log(x)**2 + 4*E*log(x) + 4*E)

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