3.60.85 \(\int \frac {1}{3} (18 x^2+e^x (-12 x^2-4 x^3)+e^{e^x (3 x+x^2)} (4+e^x (12 x+20 x^2+4 x^3))) \, dx\)

Optimal. Leaf size=31 \[ x \left (2 x^2+\frac {4}{3} \left (e^{e^x x (3+x)}-e^x x^2\right )\right ) \]

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Rubi [B]  time = 0.14, antiderivative size = 70, normalized size of antiderivative = 2.26, number of steps used = 13, number of rules used = 6, integrand size = 59, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.102, Rules used = {12, 1593, 2196, 2176, 2194, 2288} \begin {gather*} -\frac {4}{3} e^x x^3+2 x^3+\frac {4 e^{e^x \left (x^2+3 x\right )+x} \left (x^3+5 x^2+3 x\right )}{3 \left (e^x \left (x^2+3 x\right )+e^x (2 x+3)\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(18*x^2 + E^x*(-12*x^2 - 4*x^3) + E^(E^x*(3*x + x^2))*(4 + E^x*(12*x + 20*x^2 + 4*x^3)))/3,x]

[Out]

2*x^3 - (4*E^x*x^3)/3 + (4*E^(x + E^x*(3*x + x^2))*(3*x + 5*x^2 + x^3))/(3*(E^x*(3 + 2*x) + E^x*(3*x + x^2)))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 1593

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

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 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]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{3} \int \left (18 x^2+e^x \left (-12 x^2-4 x^3\right )+e^{e^x \left (3 x+x^2\right )} \left (4+e^x \left (12 x+20 x^2+4 x^3\right )\right )\right ) \, dx\\ &=2 x^3+\frac {1}{3} \int e^x \left (-12 x^2-4 x^3\right ) \, dx+\frac {1}{3} \int e^{e^x \left (3 x+x^2\right )} \left (4+e^x \left (12 x+20 x^2+4 x^3\right )\right ) \, dx\\ &=2 x^3+\frac {4 e^{x+e^x \left (3 x+x^2\right )} \left (3 x+5 x^2+x^3\right )}{3 \left (e^x (3+2 x)+e^x \left (3 x+x^2\right )\right )}+\frac {1}{3} \int e^x (-12-4 x) x^2 \, dx\\ &=2 x^3+\frac {4 e^{x+e^x \left (3 x+x^2\right )} \left (3 x+5 x^2+x^3\right )}{3 \left (e^x (3+2 x)+e^x \left (3 x+x^2\right )\right )}+\frac {1}{3} \int \left (-12 e^x x^2-4 e^x x^3\right ) \, dx\\ &=2 x^3+\frac {4 e^{x+e^x \left (3 x+x^2\right )} \left (3 x+5 x^2+x^3\right )}{3 \left (e^x (3+2 x)+e^x \left (3 x+x^2\right )\right )}-\frac {4}{3} \int e^x x^3 \, dx-4 \int e^x x^2 \, dx\\ &=-4 e^x x^2+2 x^3-\frac {4 e^x x^3}{3}+\frac {4 e^{x+e^x \left (3 x+x^2\right )} \left (3 x+5 x^2+x^3\right )}{3 \left (e^x (3+2 x)+e^x \left (3 x+x^2\right )\right )}+4 \int e^x x^2 \, dx+8 \int e^x x \, dx\\ &=8 e^x x+2 x^3-\frac {4 e^x x^3}{3}+\frac {4 e^{x+e^x \left (3 x+x^2\right )} \left (3 x+5 x^2+x^3\right )}{3 \left (e^x (3+2 x)+e^x \left (3 x+x^2\right )\right )}-8 \int e^x \, dx-8 \int e^x x \, dx\\ &=-8 e^x+2 x^3-\frac {4 e^x x^3}{3}+\frac {4 e^{x+e^x \left (3 x+x^2\right )} \left (3 x+5 x^2+x^3\right )}{3 \left (e^x (3+2 x)+e^x \left (3 x+x^2\right )\right )}+8 \int e^x \, dx\\ &=2 x^3-\frac {4 e^x x^3}{3}+\frac {4 e^{x+e^x \left (3 x+x^2\right )} \left (3 x+5 x^2+x^3\right )}{3 \left (e^x (3+2 x)+e^x \left (3 x+x^2\right )\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.09, size = 31, normalized size = 1.00 \begin {gather*} \frac {1}{3} \left (4 e^{e^x x (3+x)} x+6 x^3-4 e^x x^3\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(18*x^2 + E^x*(-12*x^2 - 4*x^3) + E^(E^x*(3*x + x^2))*(4 + E^x*(12*x + 20*x^2 + 4*x^3)))/3,x]

[Out]

(4*E^(E^x*x*(3 + x))*x + 6*x^3 - 4*E^x*x^3)/3

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fricas [A]  time = 0.57, size = 27, normalized size = 0.87 \begin {gather*} -\frac {4}{3} \, x^{3} e^{x} + 2 \, x^{3} + \frac {4}{3} \, x e^{\left ({\left (x^{2} + 3 \, x\right )} e^{x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*((4*x^3+20*x^2+12*x)*exp(x)+4)*exp((x^2+3*x)*exp(x))+1/3*(-4*x^3-12*x^2)*exp(x)+6*x^2,x, algorit
hm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int 6 \, x^{2} + \frac {4}{3} \, {\left ({\left (x^{3} + 5 \, x^{2} + 3 \, x\right )} e^{x} + 1\right )} e^{\left ({\left (x^{2} + 3 \, x\right )} e^{x}\right )} - \frac {4}{3} \, {\left (x^{3} + 3 \, x^{2}\right )} e^{x}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*((4*x^3+20*x^2+12*x)*exp(x)+4)*exp((x^2+3*x)*exp(x))+1/3*(-4*x^3-12*x^2)*exp(x)+6*x^2,x, algorit
hm="giac")

[Out]

integrate(6*x^2 + 4/3*((x^3 + 5*x^2 + 3*x)*e^x + 1)*e^((x^2 + 3*x)*e^x) - 4/3*(x^3 + 3*x^2)*e^x, x)

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maple [A]  time = 0.06, size = 25, normalized size = 0.81

method	result	size
risch	\(2 x^{3}+\frac {4 x \,{\mathrm e}^{{\mathrm e}^{x} \left (3+x \right ) x}}{3}-\frac {4 \,{\mathrm e}^{x} x^{3}}{3}\)	\(25\)
default	\(2 x^{3}+\frac {4 x \,{\mathrm e}^{\left (x^{2}+3 x \right ) {\mathrm e}^{x}}}{3}-\frac {4 \,{\mathrm e}^{x} x^{3}}{3}\)	\(28\)
norman	\(2 x^{3}+\frac {4 x \,{\mathrm e}^{\left (x^{2}+3 x \right ) {\mathrm e}^{x}}}{3}-\frac {4 \,{\mathrm e}^{x} x^{3}}{3}\)	\(28\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/3*((4*x^3+20*x^2+12*x)*exp(x)+4)*exp((x^2+3*x)*exp(x))+1/3*(-4*x^3-12*x^2)*exp(x)+6*x^2,x,method=_RETURN
VERBOSE)

[Out]

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

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maxima [A]  time = 0.35, size = 51, normalized size = 1.65 \begin {gather*} 2 \, x^{3} + \frac {4}{3} \, x e^{\left (x^{2} e^{x} + 3 \, x e^{x}\right )} - \frac {4}{3} \, {\left (x^{3} - 3 \, x^{2} + 6 \, x - 6\right )} e^{x} - 4 \, {\left (x^{2} - 2 \, x + 2\right )} e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*((4*x^3+20*x^2+12*x)*exp(x)+4)*exp((x^2+3*x)*exp(x))+1/3*(-4*x^3-12*x^2)*exp(x)+6*x^2,x, algorit
hm="maxima")

[Out]

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

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mupad [B]  time = 0.15, size = 29, normalized size = 0.94 \begin {gather*} 2\,x^3-\frac {4\,x^3\,{\mathrm {e}}^x}{3}+\frac {4\,x\,{\mathrm {e}}^{3\,x\,{\mathrm {e}}^x}\,{\mathrm {e}}^{x^2\,{\mathrm {e}}^x}}{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(exp(x)*(3*x + x^2))*(exp(x)*(12*x + 20*x^2 + 4*x^3) + 4))/3 - (exp(x)*(12*x^2 + 4*x^3))/3 + 6*x^2,x)

[Out]

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

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sympy [A]  time = 0.24, size = 31, normalized size = 1.00 \begin {gather*} - \frac {4 x^{3} e^{x}}{3} + 2 x^{3} + \frac {4 x e^{\left (x^{2} + 3 x\right ) e^{x}}}{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*((4*x**3+20*x**2+12*x)*exp(x)+4)*exp((x**2+3*x)*exp(x))+1/3*(-4*x**3-12*x**2)*exp(x)+6*x**2,x)

[Out]

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

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