\(\int \frac {1}{3} (-9+e^4 (-24-8 e^2)+78 x+24 e^2 x+(72+24 e^2) \log (x)+(36+12 e^2) \log ^2(x)) \, dx\) [6952]

Optimal result

Integrand size = 48, antiderivative size = 25 \[ \int \frac {1}{3} \left (-9+e^4 \left (-24-8 e^2\right )+78 x+24 e^2 x+\left (72+24 e^2\right ) \log (x)+\left (36+12 e^2\right ) \log ^2(x)\right ) \, dx=x \left (-3+x+4 \left (3+e^2\right ) \left (-\frac {2 e^4}{3}+x+\log ^2(x)\right )\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.64, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {6, 12, 2332, 2333} \[ \int \frac {1}{3} \left (-9+e^4 \left (-24-8 e^2\right )+78 x+24 e^2 x+\left (72+24 e^2\right ) \log (x)+\left (36+12 e^2\right ) \log ^2(x)\right ) \, dx=\left (13+4 e^2\right ) x^2-\frac {1}{3} \left (9+24 e^4+8 e^6\right ) x+4 \left (3+e^2\right ) x \log ^2(x) \]

[In]

Int[(-9 + E^4*(-24 - 8*E^2) + 78*x + 24*E^2*x + (72 + 24*E^2)*Log[x] + (36 + 12*E^2)*Log[x]^2)/3,x]

[Out]

-1/3*((9 + 24*E^4 + 8*E^6)*x) + (13 + 4*E^2)*x^2 + 4*(3 + E^2)*x*Log[x]^2

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, 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 2332

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

Rule 2333

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.), x_Symbol] :> Simp[x*(a + b*Log[c*x^n])^p, x] - Dist[b*n*p, In
t[(a + b*Log[c*x^n])^(p - 1), x], x] /; FreeQ[{a, b, c, n}, x] && GtQ[p, 0] && IntegerQ[2*p]

Rubi steps \begin{align*} \text {integral}& = \int \frac {1}{3} \left (-9+e^4 \left (-24-8 e^2\right )+\left (78+24 e^2\right ) x+\left (72+24 e^2\right ) \log (x)+\left (36+12 e^2\right ) \log ^2(x)\right ) \, dx \\ & = \frac {1}{3} \int \left (-9+e^4 \left (-24-8 e^2\right )+\left (78+24 e^2\right ) x+\left (72+24 e^2\right ) \log (x)+\left (36+12 e^2\right ) \log ^2(x)\right ) \, dx \\ & = -\frac {1}{3} \left (9+24 e^4+8 e^6\right ) x+\left (13+4 e^2\right ) x^2+\left (4 \left (3+e^2\right )\right ) \int \log ^2(x) \, dx+\left (8 \left (3+e^2\right )\right ) \int \log (x) \, dx \\ & = -8 \left (3+e^2\right ) x-\frac {1}{3} \left (9+24 e^4+8 e^6\right ) x+\left (13+4 e^2\right ) x^2+8 \left (3+e^2\right ) x \log (x)+4 \left (3+e^2\right ) x \log ^2(x)-\left (8 \left (3+e^2\right )\right ) \int \log (x) \, dx \\ & = -\frac {1}{3} \left (9+24 e^4+8 e^6\right ) x+\left (13+4 e^2\right ) x^2+4 \left (3+e^2\right ) x \log ^2(x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.92 \[ \int \frac {1}{3} \left (-9+e^4 \left (-24-8 e^2\right )+78 x+24 e^2 x+\left (72+24 e^2\right ) \log (x)+\left (36+12 e^2\right ) \log ^2(x)\right ) \, dx=-3 x-8 e^4 x-\frac {8 e^6 x}{3}+13 x^2+4 e^2 x^2+12 x \log ^2(x)+4 e^2 x \log ^2(x) \]

[In]

Integrate[(-9 + E^4*(-24 - 8*E^2) + 78*x + 24*E^2*x + (72 + 24*E^2)*Log[x] + (36 + 12*E^2)*Log[x]^2)/3,x]

[Out]

-3*x - 8*E^4*x - (8*E^6*x)/3 + 13*x^2 + 4*E^2*x^2 + 12*x*Log[x]^2 + 4*E^2*x*Log[x]^2

Maple [A] (verified)

Time = 0.14 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.52

method	result	size
norman	\(\left (4 \,{\mathrm e}^{2}+13\right ) x^{2}+\left (-\frac {8 \,{\mathrm e}^{2} {\mathrm e}^{4}}{3}-8 \,{\mathrm e}^{4}-3\right ) x +\left (4 \,{\mathrm e}^{2}+12\right ) x \ln \left (x \right )^{2}\)	\(38\)
risch	\(4 x \,{\mathrm e}^{2} \ln \left (x \right )^{2}+12 x \ln \left (x \right )^{2}-3 x -\frac {8 x \,{\mathrm e}^{6}}{3}-8 x \,{\mathrm e}^{4}+4 x^{2} {\mathrm e}^{2}+13 x^{2}\)	\(43\)
parallelrisch	\(4 x \,{\mathrm e}^{2} \ln \left (x \right )^{2}+4 x^{2} {\mathrm e}^{2}+12 x \ln \left (x \right )^{2}+13 x^{2}+\left (-3+\frac {\left (-8 \,{\mathrm e}^{2}-24\right ) {\mathrm e}^{4}}{3}\right ) x\)	\(44\)
parts	\(-3 x +\left (24+8 \,{\mathrm e}^{2}\right ) \left (x \ln \left (x \right )-x \right )+\left (4 \,{\mathrm e}^{2}+12\right ) \left (x \ln \left (x \right )^{2}-2 x \ln \left (x \right )+2 x \right )+13 x^{2}+4 x^{2} {\mathrm e}^{2}-\frac {8 \,{\mathrm e}^{2} {\mathrm e}^{4} x}{3}-8 x \,{\mathrm e}^{4}\)	\(66\)
default	\(-3 x +\frac {8 \left (-{\mathrm e}^{2}-3\right ) {\mathrm e}^{4} x}{3}+\frac {\left (12 \,{\mathrm e}^{2}+36\right ) \left (x \ln \left (x \right )^{2}-2 x \ln \left (x \right )+2 x \right )}{3}+\frac {\left (24 \,{\mathrm e}^{2}+72\right ) \left (x \ln \left (x \right )-x \right )}{3}+13 x^{2}+4 x^{2} {\mathrm e}^{2}\)	\(67\)

[In]

int(1/3*(12*exp(2)+36)*ln(x)^2+1/3*(24*exp(2)+72)*ln(x)+1/3*(-8*exp(2)-24)*exp(4)+8*exp(2)*x+26*x-3,x,method=_
RETURNVERBOSE)

[Out]

(4*exp(2)+13)*x^2+(-8/3*exp(2)*exp(4)-8*exp(4)-3)*x+(4*exp(2)+12)*x*ln(x)^2

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.60 \[ \int \frac {1}{3} \left (-9+e^4 \left (-24-8 e^2\right )+78 x+24 e^2 x+\left (72+24 e^2\right ) \log (x)+\left (36+12 e^2\right ) \log ^2(x)\right ) \, dx=4 \, x^{2} e^{2} + 4 \, {\left (x e^{2} + 3 \, x\right )} \log \left (x\right )^{2} + 13 \, x^{2} - \frac {8}{3} \, x e^{6} - 8 \, x e^{4} - 3 \, x \]

[In]

integrate(1/3*(12*exp(2)+36)*log(x)^2+1/3*(24*exp(2)+72)*log(x)+1/3*(-8*exp(2)-24)*exp(4)+8*exp(2)*x+26*x-3,x,
algorithm="fricas")

[Out]

4*x^2*e^2 + 4*(x*e^2 + 3*x)*log(x)^2 + 13*x^2 - 8/3*x*e^6 - 8*x*e^4 - 3*x

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.64 \[ \int \frac {1}{3} \left (-9+e^4 \left (-24-8 e^2\right )+78 x+24 e^2 x+\left (72+24 e^2\right ) \log (x)+\left (36+12 e^2\right ) \log ^2(x)\right ) \, dx=x^{2} \cdot \left (13 + 4 e^{2}\right ) + x \left (- \frac {8 e^{6}}{3} - 8 e^{4} - 3\right ) + \left (12 x + 4 x e^{2}\right ) \log {\left (x \right )}^{2} \]

[In]

integrate(1/3*(12*exp(2)+36)*ln(x)**2+1/3*(24*exp(2)+72)*ln(x)+1/3*(-8*exp(2)-24)*exp(4)+8*exp(2)*x+26*x-3,x)

[Out]

x**2*(13 + 4*exp(2)) + x*(-8*exp(6)/3 - 8*exp(4) - 3) + (12*x + 4*x*exp(2))*log(x)**2

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 56, normalized size of antiderivative = 2.24 \[ \int \frac {1}{3} \left (-9+e^4 \left (-24-8 e^2\right )+78 x+24 e^2 x+\left (72+24 e^2\right ) \log (x)+\left (36+12 e^2\right ) \log ^2(x)\right ) \, dx=4 \, {\left (\log \left (x\right )^{2} - 2 \, \log \left (x\right ) + 2\right )} x {\left (e^{2} + 3\right )} - \frac {8}{3} \, x {\left (e^{2} + 3\right )} e^{4} + 4 \, x^{2} e^{2} + 13 \, x^{2} + 8 \, {\left (x \log \left (x\right ) - x\right )} {\left (e^{2} + 3\right )} - 3 \, x \]

[In]

integrate(1/3*(12*exp(2)+36)*log(x)^2+1/3*(24*exp(2)+72)*log(x)+1/3*(-8*exp(2)-24)*exp(4)+8*exp(2)*x+26*x-3,x,
algorithm="maxima")

[Out]

4*(log(x)^2 - 2*log(x) + 2)*x*(e^2 + 3) - 8/3*x*(e^2 + 3)*e^4 + 4*x^2*e^2 + 13*x^2 + 8*(x*log(x) - x)*(e^2 + 3
) - 3*x

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 60 vs. \(2 (28) = 56\).

Time = 0.28 (sec) , antiderivative size = 60, normalized size of antiderivative = 2.40 \[ \int \frac {1}{3} \left (-9+e^4 \left (-24-8 e^2\right )+78 x+24 e^2 x+\left (72+24 e^2\right ) \log (x)+\left (36+12 e^2\right ) \log ^2(x)\right ) \, dx=-\frac {8}{3} \, x {\left (e^{2} + 3\right )} e^{4} + 4 \, x^{2} e^{2} + 13 \, x^{2} + 4 \, {\left (x \log \left (x\right )^{2} - 2 \, x \log \left (x\right ) + 2 \, x\right )} {\left (e^{2} + 3\right )} + 8 \, {\left (x \log \left (x\right ) - x\right )} {\left (e^{2} + 3\right )} - 3 \, x \]

[In]

integrate(1/3*(12*exp(2)+36)*log(x)^2+1/3*(24*exp(2)+72)*log(x)+1/3*(-8*exp(2)-24)*exp(4)+8*exp(2)*x+26*x-3,x,
algorithm="giac")

[Out]

-8/3*x*(e^2 + 3)*e^4 + 4*x^2*e^2 + 13*x^2 + 4*(x*log(x)^2 - 2*x*log(x) + 2*x)*(e^2 + 3) + 8*(x*log(x) - x)*(e^
2 + 3) - 3*x

Mupad [B] (verification not implemented)

Time = 14.29 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.44 \[ \int \frac {1}{3} \left (-9+e^4 \left (-24-8 e^2\right )+78 x+24 e^2 x+\left (72+24 e^2\right ) \log (x)+\left (36+12 e^2\right ) \log ^2(x)\right ) \, dx=x^2\,\left (4\,{\mathrm {e}}^2+13\right )-x\,\left (8\,{\mathrm {e}}^4+\frac {8\,{\mathrm {e}}^6}{3}-\frac {{\ln \left (x\right )}^2\,\left (12\,{\mathrm {e}}^2+36\right )}{3}+3\right ) \]

[In]

int(26*x + 8*x*exp(2) + (log(x)*(24*exp(2) + 72))/3 + (log(x)^2*(12*exp(2) + 36))/3 - (exp(4)*(8*exp(2) + 24))
/3 - 3,x)

[Out]

x^2*(4*exp(2) + 13) - x*(8*exp(4) + (8*exp(6))/3 - (log(x)^2*(12*exp(2) + 36))/3 + 3)