\(\int \frac {1+e^{-8+12 e^5+4 x^2} (-x+32 x^2-8 x^3)}{x} \, dx\) [5281]

Optimal result

Integrand size = 35, antiderivative size = 23 \[ \int \frac {1+e^{-8+12 e^5+4 x^2} \left (-x+32 x^2-8 x^3\right )}{x} \, dx=-4-e^{4 \left (-2+3 e^5+x^2\right )} (-4+x)+\log (x) \]

[Out]

ln(x)-exp(12*exp(5)+4*x^2-8)*(x-4)-4

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.83, number of steps used = 8, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {14, 2258, 2235, 2240, 2243} \[ \int \frac {1+e^{-8+12 e^5+4 x^2} \left (-x+32 x^2-8 x^3\right )}{x} \, dx=-e^{4 x^2-4 \left (2-3 e^5\right )} x+4 e^{4 x^2-4 \left (2-3 e^5\right )}+\log (x) \]

[In]

Int[(1 + E^(-8 + 12*E^5 + 4*x^2)*(-x + 32*x^2 - 8*x^3))/x,x]

[Out]

4*E^(-4*(2 - 3*E^5) + 4*x^2) - E^(-4*(2 - 3*E^5) + 4*x^2)*x + Log[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 2235

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^2), x_Symbol] :> Simp[F^a*Sqrt[Pi]*(Erfi[(c + d*x)*Rt[b*Log[F], 2
]]/(2*d*Rt[b*Log[F], 2])), x] /; FreeQ[{F, a, b, c, d}, x] && PosQ[b]

Rule 2240

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

Rule 2243

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(c + d*x)^(m
- n + 1)*(F^(a + b*(c + d*x)^n)/(b*d*n*Log[F])), x] - Dist[(m - n + 1)/(b*n*Log[F]), Int[(c + d*x)^(m - n)*F^(
a + b*(c + d*x)^n), x], x] /; FreeQ[{F, a, b, c, d}, x] && IntegerQ[2*((m + 1)/n)] && LtQ[0, (m + 1)/n, 5] &&
IntegerQ[n] && (LtQ[0, n, m + 1] || LtQ[m, n, 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}& = \int \left (\frac {1}{x}-e^{-8+12 e^5+4 x^2} \left (1-32 x+8 x^2\right )\right ) \, dx \\ & = \log (x)-\int e^{-8+12 e^5+4 x^2} \left (1-32 x+8 x^2\right ) \, dx \\ & = \log (x)-\int \left (e^{-8+12 e^5+4 x^2}-32 e^{-8+12 e^5+4 x^2} x+8 e^{-8+12 e^5+4 x^2} x^2\right ) \, dx \\ & = \log (x)-8 \int e^{-8+12 e^5+4 x^2} x^2 \, dx+32 \int e^{-8+12 e^5+4 x^2} x \, dx-\int e^{-8+12 e^5+4 x^2} \, dx \\ & = 4 e^{-4 \left (2-3 e^5\right )+4 x^2}-e^{-4 \left (2-3 e^5\right )+4 x^2} x-\frac {1}{4} e^{-8+12 e^5} \sqrt {\pi } \text {erfi}(2 x)+\log (x)+\int e^{-8+12 e^5+4 x^2} \, dx \\ & = 4 e^{-4 \left (2-3 e^5\right )+4 x^2}-e^{-4 \left (2-3 e^5\right )+4 x^2} x+\log (x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.27 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.43 \[ \int \frac {1+e^{-8+12 e^5+4 x^2} \left (-x+32 x^2-8 x^3\right )}{x} \, dx=\frac {e^{4 \left (-2+3 e^5+x^2\right )} \left (32 x-8 x^2\right )}{8 x}+\log (x) \]

[In]

Integrate[(1 + E^(-8 + 12*E^5 + 4*x^2)*(-x + 32*x^2 - 8*x^3))/x,x]

[Out]

(E^(4*(-2 + 3*E^5 + x^2))*(32*x - 8*x^2))/(8*x) + Log[x]

Maple [A] (verified)

Time = 0.03 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96

method	result	size
risch	\(\ln \left (x \right )+\left (-x +4\right ) {\mathrm e}^{12 \,{\mathrm e}^{5}+4 x^{2}-8}\)	\(22\)
norman	\(-x \,{\mathrm e}^{12 \,{\mathrm e}^{5}+4 x^{2}-8}+4 \,{\mathrm e}^{12 \,{\mathrm e}^{5}+4 x^{2}-8}+\ln \left (x \right )\)	\(33\)
parallelrisch	\(-x \,{\mathrm e}^{12 \,{\mathrm e}^{5}+4 x^{2}-8}+4 \,{\mathrm e}^{12 \,{\mathrm e}^{5}+4 x^{2}-8}+\ln \left (x \right )\)	\(33\)
default	\(\ln \left (x \right )-\frac {{\mathrm e}^{12 \,{\mathrm e}^{5}} {\mathrm e}^{-8} \sqrt {\pi }\, \operatorname {erfi}\left (2 x \right )}{4}+4 \,{\mathrm e}^{12 \,{\mathrm e}^{5}} {\mathrm e}^{4 x^{2}} {\mathrm e}^{-8}-8 \,{\mathrm e}^{12 \,{\mathrm e}^{5}} {\mathrm e}^{-8} \left (\frac {x \,{\mathrm e}^{4 x^{2}}}{8}-\frac {\sqrt {\pi }\, \operatorname {erfi}\left (2 x \right )}{32}\right )\)	\(63\)
parts	\(\ln \left (x \right )-\frac {{\mathrm e}^{12 \,{\mathrm e}^{5}} {\mathrm e}^{-8} \sqrt {\pi }\, \operatorname {erfi}\left (2 x \right )}{4}+4 \,{\mathrm e}^{12 \,{\mathrm e}^{5}} {\mathrm e}^{4 x^{2}} {\mathrm e}^{-8}-8 \,{\mathrm e}^{12 \,{\mathrm e}^{5}} {\mathrm e}^{-8} \left (\frac {x \,{\mathrm e}^{4 x^{2}}}{8}-\frac {\sqrt {\pi }\, \operatorname {erfi}\left (2 x \right )}{32}\right )\)	\(63\)

[In]

int(((-8*x^3+32*x^2-x)*exp(12*exp(5)+4*x^2-8)+1)/x,x,method=_RETURNVERBOSE)

[Out]

ln(x)+(-x+4)*exp(12*exp(5)+4*x^2-8)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.87 \[ \int \frac {1+e^{-8+12 e^5+4 x^2} \left (-x+32 x^2-8 x^3\right )}{x} \, dx=-{\left (x - 4\right )} e^{\left (4 \, x^{2} + 12 \, e^{5} - 8\right )} + \log \left (x\right ) \]

[In]

integrate(((-8*x^3+32*x^2-x)*exp(12*exp(5)+4*x^2-8)+1)/x,x, algorithm="fricas")

[Out]

-(x - 4)*e^(4*x^2 + 12*e^5 - 8) + log(x)

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83 \[ \int \frac {1+e^{-8+12 e^5+4 x^2} \left (-x+32 x^2-8 x^3\right )}{x} \, dx=\left (4 - x\right ) e^{4 x^{2} - 8 + 12 e^{5}} + \log {\left (x \right )} \]

[In]

integrate(((-8*x**3+32*x**2-x)*exp(12*exp(5)+4*x**2-8)+1)/x,x)

[Out]

(4 - x)*exp(4*x**2 - 8 + 12*exp(5)) + log(x)

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.39 \[ \int \frac {1+e^{-8+12 e^5+4 x^2} \left (-x+32 x^2-8 x^3\right )}{x} \, dx=-x e^{\left (4 \, x^{2} + 12 \, e^{5} - 8\right )} + 4 \, e^{\left (4 \, x^{2} + 12 \, e^{5} - 8\right )} + \log \left (x\right ) \]

[In]

integrate(((-8*x^3+32*x^2-x)*exp(12*exp(5)+4*x^2-8)+1)/x,x, algorithm="maxima")

[Out]

-x*e^(4*x^2 + 12*e^5 - 8) + 4*e^(4*x^2 + 12*e^5 - 8) + log(x)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.39 \[ \int \frac {1+e^{-8+12 e^5+4 x^2} \left (-x+32 x^2-8 x^3\right )}{x} \, dx=-x e^{\left (4 \, x^{2} + 12 \, e^{5} - 8\right )} + 4 \, e^{\left (4 \, x^{2} + 12 \, e^{5} - 8\right )} + \log \left (x\right ) \]

[In]

integrate(((-8*x^3+32*x^2-x)*exp(12*exp(5)+4*x^2-8)+1)/x,x, algorithm="giac")

[Out]

-x*e^(4*x^2 + 12*e^5 - 8) + 4*e^(4*x^2 + 12*e^5 - 8) + log(x)

Mupad [B] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.48 \[ \int \frac {1+e^{-8+12 e^5+4 x^2} \left (-x+32 x^2-8 x^3\right )}{x} \, dx=\ln \left (x\right )+4\,{\mathrm {e}}^{12\,{\mathrm {e}}^5}\,{\mathrm {e}}^{-8}\,{\mathrm {e}}^{4\,x^2}-x\,{\mathrm {e}}^{12\,{\mathrm {e}}^5}\,{\mathrm {e}}^{-8}\,{\mathrm {e}}^{4\,x^2} \]

[In]

int(-(exp(12*exp(5) + 4*x^2 - 8)*(x - 32*x^2 + 8*x^3) - 1)/x,x)

[Out]

log(x) + 4*exp(12*exp(5))*exp(-8)*exp(4*x^2) - x*exp(12*exp(5))*exp(-8)*exp(4*x^2)