3.93.57 \(\int e^{3 x+3 x^2} (2 x+3 x^2+6 x^3) \, dx\)

Optimal. Leaf size=12 \[ e^{3 x (1+x)} x^2 \]

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Rubi [B]  time = 0.04, antiderivative size = 27, normalized size of antiderivative = 2.25, number of steps used = 2, number of rules used = 2, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {1594, 2288} \begin {gather*} \frac {e^{3 x^2+3 x} x \left (2 x^2+x\right )}{2 x+1} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[E^(3*x + 3*x^2)*(2*x + 3*x^2 + 6*x^3),x]

[Out]

(E^(3*x + 3*x^2)*x*(x + 2*x^2))/(1 + 2*x)

Rule 1594

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

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} &=\int e^{3 x+3 x^2} x \left (2+3 x+6 x^2\right ) \, dx\\ &=\frac {e^{3 x+3 x^2} x \left (x+2 x^2\right )}{1+2 x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 12, normalized size = 1.00 \begin {gather*} e^{3 x (1+x)} x^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[E^(3*x + 3*x^2)*(2*x + 3*x^2 + 6*x^3),x]

[Out]

E^(3*x*(1 + x))*x^2

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fricas [A]  time = 0.57, size = 14, normalized size = 1.17 \begin {gather*} x^{2} e^{\left (3 \, x^{2} + 3 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x^2*e^(3*x^2 + 3*x)

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giac [B]  time = 0.15, size = 24, normalized size = 2.00 \begin {gather*} \frac {1}{4} \, {\left ({\left (2 \, x + 1\right )}^{2} - 4 \, x - 1\right )} e^{\left (3 \, x^{2} + 3 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.04, size = 12, normalized size = 1.00

method	result	size
risch	\({\mathrm e}^{3 \left (x +1\right ) x} x^{2}\)	\(12\)
gosper	\({\mathrm e}^{3 x^{2}+3 x} x^{2}\)	\(15\)
default	\({\mathrm e}^{3 x^{2}+3 x} x^{2}\)	\(15\)
norman	\({\mathrm e}^{3 x^{2}+3 x} x^{2}\)	\(15\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((6*x^3+3*x^2+2*x)*exp(3*x^2+3*x),x,method=_RETURNVERBOSE)

[Out]

exp(3*(x+1)*x)*x^2

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maxima [C]  time = 0.51, size = 268, normalized size = 22.33 \begin {gather*} \frac {1}{72} \, \sqrt {3} {\left (\frac {36 \, {\left (2 \, x + 1\right )}^{3} \Gamma \left (\frac {3}{2}, -\frac {3}{4} \, {\left (2 \, x + 1\right )}^{2}\right )}{\left (-{\left (2 \, x + 1\right )}^{2}\right )^{\frac {3}{2}}} - \frac {9 \, \sqrt {\pi } {\left (2 \, x + 1\right )} {\left (\operatorname {erf}\left (\frac {1}{2} \, \sqrt {3} \sqrt {-{\left (2 \, x + 1\right )}^{2}}\right ) - 1\right )}}{\sqrt {-{\left (2 \, x + 1\right )}^{2}}} + 18 \, \sqrt {3} e^{\left (\frac {3}{4} \, {\left (2 \, x + 1\right )}^{2}\right )} - 8 \, \sqrt {3} \Gamma \left (2, -\frac {3}{4} \, {\left (2 \, x + 1\right )}^{2}\right )\right )} e^{\left (-\frac {3}{4}\right )} - \frac {1}{24} \, \sqrt {3} {\left (\frac {4 \, {\left (2 \, x + 1\right )}^{3} \Gamma \left (\frac {3}{2}, -\frac {3}{4} \, {\left (2 \, x + 1\right )}^{2}\right )}{\left (-{\left (2 \, x + 1\right )}^{2}\right )^{\frac {3}{2}}} - \frac {3 \, \sqrt {\pi } {\left (2 \, x + 1\right )} {\left (\operatorname {erf}\left (\frac {1}{2} \, \sqrt {3} \sqrt {-{\left (2 \, x + 1\right )}^{2}}\right ) - 1\right )}}{\sqrt {-{\left (2 \, x + 1\right )}^{2}}} + 4 \, \sqrt {3} e^{\left (\frac {3}{4} \, {\left (2 \, x + 1\right )}^{2}\right )}\right )} e^{\left (-\frac {3}{4}\right )} - \frac {1}{18} \, \sqrt {3} {\left (\frac {3 \, \sqrt {\pi } {\left (2 \, x + 1\right )} {\left (\operatorname {erf}\left (\frac {1}{2} \, \sqrt {3} \sqrt {-{\left (2 \, x + 1\right )}^{2}}\right ) - 1\right )}}{\sqrt {-{\left (2 \, x + 1\right )}^{2}}} - 2 \, \sqrt {3} e^{\left (\frac {3}{4} \, {\left (2 \, x + 1\right )}^{2}\right )}\right )} e^{\left (-\frac {3}{4}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/72*sqrt(3)*(36*(2*x + 1)^3*gamma(3/2, -3/4*(2*x + 1)^2)/(-(2*x + 1)^2)^(3/2) - 9*sqrt(pi)*(2*x + 1)*(erf(1/2
*sqrt(3)*sqrt(-(2*x + 1)^2)) - 1)/sqrt(-(2*x + 1)^2) + 18*sqrt(3)*e^(3/4*(2*x + 1)^2) - 8*sqrt(3)*gamma(2, -3/
4*(2*x + 1)^2))*e^(-3/4) - 1/24*sqrt(3)*(4*(2*x + 1)^3*gamma(3/2, -3/4*(2*x + 1)^2)/(-(2*x + 1)^2)^(3/2) - 3*s
qrt(pi)*(2*x + 1)*(erf(1/2*sqrt(3)*sqrt(-(2*x + 1)^2)) - 1)/sqrt(-(2*x + 1)^2) + 4*sqrt(3)*e^(3/4*(2*x + 1)^2)
)*e^(-3/4) - 1/18*sqrt(3)*(3*sqrt(pi)*(2*x + 1)*(erf(1/2*sqrt(3)*sqrt(-(2*x + 1)^2)) - 1)/sqrt(-(2*x + 1)^2) -
2*sqrt(3)*e^(3/4*(2*x + 1)^2))*e^(-3/4)

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mupad [B]  time = 5.64, size = 14, normalized size = 1.17 \begin {gather*} x^2\,{\mathrm {e}}^{3\,x^2+3\,x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(3*x + 3*x^2)*(2*x + 3*x^2 + 6*x^3),x)

[Out]

x^2*exp(3*x + 3*x^2)

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sympy [A]  time = 0.10, size = 12, normalized size = 1.00 \begin {gather*} x^{2} e^{3 x^{2} + 3 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((6*x**3+3*x**2+2*x)*exp(3*x**2+3*x),x)

[Out]

x**2*exp(3*x**2 + 3*x)

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