\(\int (6-6 x^2+e^{x^3} (2+6 x^3)-8 x \log (\log (5))-2 \log ^2(\log (5))) \, dx\) [9940]

Optimal result

Integrand size = 33, antiderivative size = 25 \[ \int \left (6-6 x^2+e^{x^3} \left (2+6 x^3\right )-8 x \log (\log (5))-2 \log ^2(\log (5))\right ) \, dx=1-\log (3)+2 x \left (3+e^{x^3}-(x+\log (\log (5)))^2\right ) \]

[Out]

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

Rubi [C] (verified)

Result contains higher order function than in optimal. Order 4 vs. order 3 in optimal.

Time = 0.03 (sec) , antiderivative size = 72, normalized size of antiderivative = 2.88, number of steps used = 5, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {2258, 2239, 2250} \[ \int \left (6-6 x^2+e^{x^3} \left (2+6 x^3\right )-8 x \log (\log (5))-2 \log ^2(\log (5))\right ) \, dx=-2 x^3-\frac {2 x \Gamma \left (\frac {1}{3},-x^3\right )}{3 \sqrt [3]{-x^3}}-4 x^2 \log (\log (5))-\frac {2 x^4 \Gamma \left (\frac {4}{3},-x^3\right )}{\left (-x^3\right )^{4/3}}+2 x \left (3-\log ^2(\log (5))\right ) \]

[In]

Int[6 - 6*x^2 + E^x^3*(2 + 6*x^3) - 8*x*Log[Log[5]] - 2*Log[Log[5]]^2,x]

[Out]

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

Rule 2239

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

Rule 2250

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Simp[(-F^a)*((e +
f*x)^(m + 1)/(f*n*((-b)*(c + d*x)^n*Log[F])^((m + 1)/n)))*Gamma[(m + 1)/n, (-b)*(c + d*x)^n*Log[F]], x] /; Fre
eQ[{F, a, b, c, d, e, f, m, n}, x] && EqQ[d*e - c*f, 0]

Rule 2258

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*(u_), x_Symbol] :> Int[ExpandLinearProduct[F^(a + b*(c + d*
x)^n), u, c, d, x], x] /; FreeQ[{F, a, b, c, d, n}, x] && PolynomialQ[u, x]

Rubi steps \begin{align*} \text {integral}& = -2 x^3-4 x^2 \log (\log (5))+2 x \left (3-\log ^2(\log (5))\right )+\int e^{x^3} \left (2+6 x^3\right ) \, dx \\ & = -2 x^3-4 x^2 \log (\log (5))+2 x \left (3-\log ^2(\log (5))\right )+\int \left (2 e^{x^3}+6 e^{x^3} x^3\right ) \, dx \\ & = -2 x^3-4 x^2 \log (\log (5))+2 x \left (3-\log ^2(\log (5))\right )+2 \int e^{x^3} \, dx+6 \int e^{x^3} x^3 \, dx \\ & = -2 x^3-\frac {2 x \Gamma \left (\frac {1}{3},-x^3\right )}{3 \sqrt [3]{-x^3}}-\frac {2 x^4 \Gamma \left (\frac {4}{3},-x^3\right )}{\left (-x^3\right )^{4/3}}-4 x^2 \log (\log (5))+2 x \left (3-\log ^2(\log (5))\right ) \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 4 vs. order 3 in optimal.

Time = 0.09 (sec) , antiderivative size = 70, normalized size of antiderivative = 2.80 \[ \int \left (6-6 x^2+e^{x^3} \left (2+6 x^3\right )-8 x \log (\log (5))-2 \log ^2(\log (5))\right ) \, dx=-2 x^3-\frac {2 x \Gamma \left (\frac {1}{3},-x^3\right )}{3 \sqrt [3]{-x^3}}-\frac {2 \left (-x^3\right )^{2/3} \Gamma \left (\frac {4}{3},-x^3\right )}{x^2}-4 x^2 \log (\log (5))-2 x \left (-3+\log ^2(\log (5))\right ) \]

[In]

Integrate[6 - 6*x^2 + E^x^3*(2 + 6*x^3) - 8*x*Log[Log[5]] - 2*Log[Log[5]]^2,x]

[Out]

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

Maple [A] (verified)

Time = 0.25 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.32

method	result	size
default	\(6 x +2 \,{\mathrm e}^{x^{3}} x -2 x^{3}-2 x \ln \left (\ln \left (5\right )\right )^{2}-4 \ln \left (\ln \left (5\right )\right ) x^{2}\)	\(33\)
norman	\(\left (-2 \ln \left (\ln \left (5\right )\right )^{2}+6\right ) x -2 x^{3}+2 \,{\mathrm e}^{x^{3}} x -4 \ln \left (\ln \left (5\right )\right ) x^{2}\)	\(33\)
risch	\(6 x +2 \,{\mathrm e}^{x^{3}} x -2 x^{3}-2 x \ln \left (\ln \left (5\right )\right )^{2}-4 \ln \left (\ln \left (5\right )\right ) x^{2}\)	\(33\)
parallelrisch	\(\left (-2 \ln \left (\ln \left (5\right )\right )^{2}+6\right ) x -2 x^{3}+2 \,{\mathrm e}^{x^{3}} x -4 \ln \left (\ln \left (5\right )\right ) x^{2}\)	\(33\)
parts	\(6 x +2 \,{\mathrm e}^{x^{3}} x -2 x^{3}-2 x \ln \left (\ln \left (5\right )\right )^{2}-4 \ln \left (\ln \left (5\right )\right ) x^{2}\)	\(33\)

[In]

int(-2*ln(ln(5))^2-8*x*ln(ln(5))+(6*x^3+2)*exp(x^3)-6*x^2+6,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.28 \[ \int \left (6-6 x^2+e^{x^3} \left (2+6 x^3\right )-8 x \log (\log (5))-2 \log ^2(\log (5))\right ) \, dx=-2 \, x^{3} - 4 \, x^{2} \log \left (\log \left (5\right )\right ) - 2 \, x \log \left (\log \left (5\right )\right )^{2} + 2 \, x e^{\left (x^{3}\right )} + 6 \, x \]

[In]

integrate(-2*log(log(5))^2-8*x*log(log(5))+(6*x^3+2)*exp(x^3)-6*x^2+6,x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.36 \[ \int \left (6-6 x^2+e^{x^3} \left (2+6 x^3\right )-8 x \log (\log (5))-2 \log ^2(\log (5))\right ) \, dx=- 2 x^{3} - 4 x^{2} \log {\left (\log {\left (5 \right )} \right )} + 2 x e^{x^{3}} + x \left (6 - 2 \log {\left (\log {\left (5 \right )} \right )}^{2}\right ) \]

[In]

integrate(-2*ln(ln(5))**2-8*x*ln(ln(5))+(6*x**3+2)*exp(x**3)-6*x**2+6,x)

[Out]

-2*x**3 - 4*x**2*log(log(5)) + 2*x*exp(x**3) + x*(6 - 2*log(log(5))**2)

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.28 \[ \int \left (6-6 x^2+e^{x^3} \left (2+6 x^3\right )-8 x \log (\log (5))-2 \log ^2(\log (5))\right ) \, dx=-2 \, x^{3} - 4 \, x^{2} \log \left (\log \left (5\right )\right ) - 2 \, x \log \left (\log \left (5\right )\right )^{2} + 2 \, x e^{\left (x^{3}\right )} + 6 \, x \]

[In]

integrate(-2*log(log(5))^2-8*x*log(log(5))+(6*x^3+2)*exp(x^3)-6*x^2+6,x, algorithm="maxima")

[Out]

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

Giac [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 0.27 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.72 \[ \int \left (6-6 x^2+e^{x^3} \left (2+6 x^3\right )-8 x \log (\log (5))-2 \log ^2(\log (5))\right ) \, dx=-2 \, x^{3} - 4 \, x^{2} \log \left (\log \left (5\right )\right ) - 2 \, x \log \left (\log \left (5\right )\right )^{2} + 6 \, x + 2 \, \gamma \left (\frac {4}{3}, -x^{3}\right ) - \frac {2}{3} \, \gamma \left (\frac {1}{3}, -x^{3}\right ) \]

[In]

integrate(-2*log(log(5))^2-8*x*log(log(5))+(6*x^3+2)*exp(x^3)-6*x^2+6,x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int \left (6-6 x^2+e^{x^3} \left (2+6 x^3\right )-8 x \log (\log (5))-2 \log ^2(\log (5))\right ) \, dx=-2\,x\,\left ({\ln \left (\ln \left (5\right )\right )}^2-{\mathrm {e}}^{x^3}+2\,x\,\ln \left (\ln \left (5\right )\right )+x^2-3\right ) \]

[In]

int(exp(x^3)*(6*x^3 + 2) - 2*log(log(5))^2 - 8*x*log(log(5)) - 6*x^2 + 6,x)

[Out]

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