3.9.95 \(\int \frac {1}{9} (-17+8 x+12 x^2+e^{2 x} (1+2 x)+e^{2 x^2} (1+4 x^2)+e^x (2+10 x+4 x^2)+e^{x^2} (-2-8 x-4 x^2-8 x^3+e^x (-2-2 x-4 x^2))) \, dx\)

Optimal. Leaf size=25 \[ x \left (-2+\frac {1}{9} \left (1+e^x-e^{x^2}+2 x\right )^2\right ) \]

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Rubi [B]  time = 0.39, antiderivative size = 120, normalized size of antiderivative = 4.80, number of steps used = 26, number of rules used = 10, integrand size = 89, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.112, Rules used = {12, 2176, 2194, 2226, 2204, 2212, 2196, 6742, 2209, 2288} \begin {gather*} \frac {4 x^3}{9}-\frac {4}{9} e^{x^2} x^2+\frac {4 e^x x^2}{9}+\frac {4 x^2}{9}-\frac {2 e^{x^2} x}{9}+\frac {1}{9} e^{2 x^2} x-\frac {2 e^{x^2+x} \left (2 x^2+x\right )}{9 (2 x+1)}+\frac {2 e^x x}{9}-\frac {17 x}{9}-\frac {e^{2 x}}{18}+\frac {1}{18} e^{2 x} (2 x+1) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-17 + 8*x + 12*x^2 + E^(2*x)*(1 + 2*x) + E^(2*x^2)*(1 + 4*x^2) + E^x*(2 + 10*x + 4*x^2) + E^x^2*(-2 - 8*x
- 4*x^2 - 8*x^3 + E^x*(-2 - 2*x - 4*x^2)))/9,x]

[Out]

-1/18*E^(2*x) - (17*x)/9 + (2*E^x*x)/9 - (2*E^x^2*x)/9 + (E^(2*x^2)*x)/9 + (4*x^2)/9 + (4*E^x*x^2)/9 - (4*E^x^
2*x^2)/9 + (4*x^3)/9 + (E^(2*x)*(1 + 2*x))/18 - (2*E^(x + x^2)*(x + 2*x^2))/(9*(1 + 2*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 2176

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^m
*(b*F^(g*(e + f*x)))^n)/(f*g*n*Log[F]), x] - Dist[(d*m)/(f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*(b*F^(g*(e + f*x
)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && GtQ[m, 0] && IntegerQ[2*m] &&  !$UseGamma === True

Rule 2194

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rule 2196

Int[(F_)^((c_.)*(v_))*(u_), x_Symbol] :> Int[ExpandIntegrand[F^(c*ExpandToSum[v, x]), u, x], x] /; FreeQ[{F, c
}, x] && PolynomialQ[u, x] && LinearQ[v, x] &&  !$UseGamma === True

Rule 2204

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 2209

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 2212

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 2226

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]

Rule 2288

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = (v*y)/(Log[F]*D[u, x])}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, 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} &=\frac {1}{9} \int \left (-17+8 x+12 x^2+e^{2 x} (1+2 x)+e^{2 x^2} \left (1+4 x^2\right )+e^x \left (2+10 x+4 x^2\right )+e^{x^2} \left (-2-8 x-4 x^2-8 x^3+e^x \left (-2-2 x-4 x^2\right )\right )\right ) \, dx\\ &=-\frac {17 x}{9}+\frac {4 x^2}{9}+\frac {4 x^3}{9}+\frac {1}{9} \int e^{2 x} (1+2 x) \, dx+\frac {1}{9} \int e^{2 x^2} \left (1+4 x^2\right ) \, dx+\frac {1}{9} \int e^x \left (2+10 x+4 x^2\right ) \, dx+\frac {1}{9} \int e^{x^2} \left (-2-8 x-4 x^2-8 x^3+e^x \left (-2-2 x-4 x^2\right )\right ) \, dx\\ &=-\frac {17 x}{9}+\frac {4 x^2}{9}+\frac {4 x^3}{9}+\frac {1}{18} e^{2 x} (1+2 x)-\frac {1}{9} \int e^{2 x} \, dx+\frac {1}{9} \int \left (2 e^x+10 e^x x+4 e^x x^2\right ) \, dx+\frac {1}{9} \int \left (e^{2 x^2}+4 e^{2 x^2} x^2\right ) \, dx+\frac {1}{9} \int \left (-2 e^{x^2}-8 e^{x^2} x-4 e^{x^2} x^2-8 e^{x^2} x^3-2 e^{x+x^2} \left (1+x+2 x^2\right )\right ) \, dx\\ &=-\frac {e^{2 x}}{18}-\frac {17 x}{9}+\frac {4 x^2}{9}+\frac {4 x^3}{9}+\frac {1}{18} e^{2 x} (1+2 x)+\frac {1}{9} \int e^{2 x^2} \, dx+\frac {2 \int e^x \, dx}{9}-\frac {2}{9} \int e^{x^2} \, dx-\frac {2}{9} \int e^{x+x^2} \left (1+x+2 x^2\right ) \, dx+\frac {4}{9} \int e^x x^2 \, dx-\frac {4}{9} \int e^{x^2} x^2 \, dx+\frac {4}{9} \int e^{2 x^2} x^2 \, dx-\frac {8}{9} \int e^{x^2} x \, dx-\frac {8}{9} \int e^{x^2} x^3 \, dx+\frac {10}{9} \int e^x x \, dx\\ &=\frac {2 e^x}{9}-\frac {e^{2 x}}{18}-\frac {4 e^{x^2}}{9}-\frac {17 x}{9}+\frac {10 e^x x}{9}-\frac {2 e^{x^2} x}{9}+\frac {1}{9} e^{2 x^2} x+\frac {4 x^2}{9}+\frac {4 e^x x^2}{9}-\frac {4}{9} e^{x^2} x^2+\frac {4 x^3}{9}+\frac {1}{18} e^{2 x} (1+2 x)-\frac {2 e^{x+x^2} \left (x+2 x^2\right )}{9 (1+2 x)}-\frac {1}{9} \sqrt {\pi } \text {erfi}(x)+\frac {1}{18} \sqrt {\frac {\pi }{2}} \text {erfi}\left (\sqrt {2} x\right )-\frac {1}{9} \int e^{2 x^2} \, dx+\frac {2}{9} \int e^{x^2} \, dx-\frac {8}{9} \int e^x x \, dx+\frac {8}{9} \int e^{x^2} x \, dx-\frac {10 \int e^x \, dx}{9}\\ &=-\frac {8 e^x}{9}-\frac {e^{2 x}}{18}-\frac {17 x}{9}+\frac {2 e^x x}{9}-\frac {2 e^{x^2} x}{9}+\frac {1}{9} e^{2 x^2} x+\frac {4 x^2}{9}+\frac {4 e^x x^2}{9}-\frac {4}{9} e^{x^2} x^2+\frac {4 x^3}{9}+\frac {1}{18} e^{2 x} (1+2 x)-\frac {2 e^{x+x^2} \left (x+2 x^2\right )}{9 (1+2 x)}+\frac {8 \int e^x \, dx}{9}\\ &=-\frac {e^{2 x}}{18}-\frac {17 x}{9}+\frac {2 e^x x}{9}-\frac {2 e^{x^2} x}{9}+\frac {1}{9} e^{2 x^2} x+\frac {4 x^2}{9}+\frac {4 e^x x^2}{9}-\frac {4}{9} e^{x^2} x^2+\frac {4 x^3}{9}+\frac {1}{18} e^{2 x} (1+2 x)-\frac {2 e^{x+x^2} \left (x+2 x^2\right )}{9 (1+2 x)}\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.07, size = 57, normalized size = 2.28 \begin {gather*} \frac {1}{9} x \left (-17+e^{2 x}+e^{2 x^2}-2 e^{x+x^2}+4 x+4 x^2-2 e^{x^2} (1+2 x)+e^x (2+4 x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-17 + 8*x + 12*x^2 + E^(2*x)*(1 + 2*x) + E^(2*x^2)*(1 + 4*x^2) + E^x*(2 + 10*x + 4*x^2) + E^x^2*(-2
- 8*x - 4*x^2 - 8*x^3 + E^x*(-2 - 2*x - 4*x^2)))/9,x]

[Out]

(x*(-17 + E^(2*x) + E^(2*x^2) - 2*E^(x + x^2) + 4*x + 4*x^2 - 2*E^x^2*(1 + 2*x) + E^x*(2 + 4*x)))/9

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fricas [B]  time = 0.61, size = 58, normalized size = 2.32 \begin {gather*} \frac {4}{9} \, x^{3} + \frac {4}{9} \, x^{2} + \frac {1}{9} \, x e^{\left (2 \, x^{2}\right )} - \frac {2}{9} \, {\left (2 \, x^{2} + x e^{x} + x\right )} e^{\left (x^{2}\right )} + \frac {1}{9} \, x e^{\left (2 \, x\right )} + \frac {2}{9} \, {\left (2 \, x^{2} + x\right )} e^{x} - \frac {17}{9} \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/9*(4*x^2+1)*exp(x^2)^2+1/9*((-4*x^2-2*x-2)*exp(x)-8*x^3-4*x^2-8*x-2)*exp(x^2)+1/9*(2*x+1)*exp(x)^2
+1/9*(4*x^2+10*x+2)*exp(x)+4/3*x^2+8/9*x-17/9,x, algorithm="fricas")

[Out]

4/9*x^3 + 4/9*x^2 + 1/9*x*e^(2*x^2) - 2/9*(2*x^2 + x*e^x + x)*e^(x^2) + 1/9*x*e^(2*x) + 2/9*(2*x^2 + x)*e^x -
17/9*x

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giac [B]  time = 0.29, size = 63, normalized size = 2.52 \begin {gather*} \frac {4}{9} \, x^{3} + \frac {4}{9} \, x^{2} + \frac {1}{9} \, x e^{\left (2 \, x^{2}\right )} - \frac {2}{9} \, x e^{\left (x^{2} + x\right )} - \frac {2}{9} \, {\left (2 \, x^{2} + x\right )} e^{\left (x^{2}\right )} + \frac {1}{9} \, x e^{\left (2 \, x\right )} + \frac {2}{9} \, {\left (2 \, x^{2} + x\right )} e^{x} - \frac {17}{9} \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/9*(4*x^2+1)*exp(x^2)^2+1/9*((-4*x^2-2*x-2)*exp(x)-8*x^3-4*x^2-8*x-2)*exp(x^2)+1/9*(2*x+1)*exp(x)^2
+1/9*(4*x^2+10*x+2)*exp(x)+4/3*x^2+8/9*x-17/9,x, algorithm="giac")

[Out]

4/9*x^3 + 4/9*x^2 + 1/9*x*e^(2*x^2) - 2/9*x*e^(x^2 + x) - 2/9*(2*x^2 + x)*e^(x^2) + 1/9*x*e^(2*x) + 2/9*(2*x^2
+ x)*e^x - 17/9*x

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maple [A]  time = 0.06, size = 64, normalized size = 2.56

method	result	size
risch	\(\frac {x \,{\mathrm e}^{2 x^{2}}}{9}+\frac {\left (-4 x^{2}-2 \,{\mathrm e}^{x} x -2 x \right ) {\mathrm e}^{x^{2}}}{9}+\frac {x \,{\mathrm e}^{2 x}}{9}+\frac {\left (4 x^{2}+2 x \right ) {\mathrm e}^{x}}{9}+\frac {4 x^{3}}{9}+\frac {4 x^{2}}{9}-\frac {17 x}{9}\)	\(64\)
default	\(-\frac {17 x}{9}+\frac {4 x^{2}}{9}+\frac {4 x^{3}}{9}+\frac {x \,{\mathrm e}^{2 x}}{9}-\frac {4 x^{2} {\mathrm e}^{x^{2}}}{9}-\frac {2 \,{\mathrm e}^{x^{2}} x}{9}-\frac {2 x \,{\mathrm e}^{x^{2}+x}}{9}+\frac {x \,{\mathrm e}^{2 x^{2}}}{9}+\frac {4 \,{\mathrm e}^{x} x^{2}}{9}+\frac {2 \,{\mathrm e}^{x} x}{9}\)	\(68\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/9*(4*x^2+1)*exp(x^2)^2+1/9*((-4*x^2-2*x-2)*exp(x)-8*x^3-4*x^2-8*x-2)*exp(x^2)+1/9*(2*x+1)*exp(x)^2+1/9*(
4*x^2+10*x+2)*exp(x)+4/3*x^2+8/9*x-17/9,x,method=_RETURNVERBOSE)

[Out]

1/9*x*exp(2*x^2)+1/9*(-4*x^2-2*exp(x)*x-2*x)*exp(x^2)+1/9*x*exp(2*x)+1/9*(4*x^2+2*x)*exp(x)+4/9*x^3+4/9*x^2-17
/9*x

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maxima [B]  time = 0.39, size = 58, normalized size = 2.32 \begin {gather*} \frac {4}{9} \, x^{3} + \frac {4}{9} \, x^{2} + \frac {1}{9} \, x e^{\left (2 \, x^{2}\right )} - \frac {2}{9} \, {\left (2 \, x^{2} + x e^{x} + x\right )} e^{\left (x^{2}\right )} + \frac {1}{9} \, x e^{\left (2 \, x\right )} + \frac {2}{9} \, {\left (2 \, x^{2} + x\right )} e^{x} - \frac {17}{9} \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/9*(4*x^2+1)*exp(x^2)^2+1/9*((-4*x^2-2*x-2)*exp(x)-8*x^3-4*x^2-8*x-2)*exp(x^2)+1/9*(2*x+1)*exp(x)^2
+1/9*(4*x^2+10*x+2)*exp(x)+4/3*x^2+8/9*x-17/9,x, algorithm="maxima")

[Out]

4/9*x^3 + 4/9*x^2 + 1/9*x*e^(2*x^2) - 2/9*(2*x^2 + x*e^x + x)*e^(x^2) + 1/9*x*e^(2*x) + 2/9*(2*x^2 + x)*e^x -
17/9*x

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mupad [B]  time = 0.12, size = 53, normalized size = 2.12 \begin {gather*} \frac {x\,\left (4\,x-2\,{\mathrm {e}}^{x^2+x}+{\mathrm {e}}^{2\,x}-2\,{\mathrm {e}}^{x^2}+{\mathrm {e}}^{2\,x^2}+2\,{\mathrm {e}}^x-4\,x\,{\mathrm {e}}^{x^2}+4\,x\,{\mathrm {e}}^x+4\,x^2-17\right )}{9} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((8*x)/9 + (exp(2*x^2)*(4*x^2 + 1))/9 + (exp(x)*(10*x + 4*x^2 + 2))/9 + (exp(2*x)*(2*x + 1))/9 + (4*x^2)/3
- (exp(x^2)*(8*x + exp(x)*(2*x + 4*x^2 + 2) + 4*x^2 + 8*x^3 + 2))/9 - 17/9,x)

[Out]

(x*(4*x - 2*exp(x + x^2) + exp(2*x) - 2*exp(x^2) + exp(2*x^2) + 2*exp(x) - 4*x*exp(x^2) + 4*x*exp(x) + 4*x^2 -
17))/9

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sympy [B]  time = 0.31, size = 73, normalized size = 2.92 \begin {gather*} \frac {4 x^{3}}{9} + \frac {4 x^{2}}{9} + \frac {x e^{2 x}}{9} + \frac {x e^{2 x^{2}}}{9} - \frac {17 x}{9} + \frac {\left (36 x^{2} + 18 x\right ) e^{x}}{81} + \frac {\left (- 36 x^{2} - 18 x e^{x} - 18 x\right ) e^{x^{2}}}{81} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/9*(4*x**2+1)*exp(x**2)**2+1/9*((-4*x**2-2*x-2)*exp(x)-8*x**3-4*x**2-8*x-2)*exp(x**2)+1/9*(2*x+1)*e
xp(x)**2+1/9*(4*x**2+10*x+2)*exp(x)+4/3*x**2+8/9*x-17/9,x)

[Out]

4*x**3/9 + 4*x**2/9 + x*exp(2*x)/9 + x*exp(2*x**2)/9 - 17*x/9 + (36*x**2 + 18*x)*exp(x)/81 + (-36*x**2 - 18*x*
exp(x) - 18*x)*exp(x**2)/81

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