3.38.74 \(\int (-3+30 x-12 x^2+e^{3 x} (1+3 x)) \, dx\)

Optimal. Leaf size=26 \[ -6+e^2-x \left (3-e^{3 x}+x-4 (4-x) x\right ) \]

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Rubi [A]  time = 0.02, antiderivative size = 37, normalized size of antiderivative = 1.42, number of steps used = 3, number of rules used = 2, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {2176, 2194} \begin {gather*} -4 x^3+15 x^2-3 x-\frac {e^{3 x}}{3}+\frac {1}{3} e^{3 x} (3 x+1) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[-3 + 30*x - 12*x^2 + E^(3*x)*(1 + 3*x),x]

[Out]

-1/3*E^(3*x) - 3*x + 15*x^2 - 4*x^3 + (E^(3*x)*(1 + 3*x))/3

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]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=-3 x+15 x^2-4 x^3+\int e^{3 x} (1+3 x) \, dx\\ &=-3 x+15 x^2-4 x^3+\frac {1}{3} e^{3 x} (1+3 x)-\int e^{3 x} \, dx\\ &=-\frac {e^{3 x}}{3}-3 x+15 x^2-4 x^3+\frac {1}{3} e^{3 x} (1+3 x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 21, normalized size = 0.81 \begin {gather*} -3 x+e^{3 x} x+15 x^2-4 x^3 \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[-3 + 30*x - 12*x^2 + E^(3*x)*(1 + 3*x),x]

[Out]

-3*x + E^(3*x)*x + 15*x^2 - 4*x^3

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fricas [A]  time = 0.77, size = 20, normalized size = 0.77 \begin {gather*} -4 \, x^{3} + 15 \, x^{2} + x e^{\left (3 \, x\right )} - 3 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3*x+1)*exp(3*x)-12*x^2+30*x-3,x, algorithm="fricas")

[Out]

-4*x^3 + 15*x^2 + x*e^(3*x) - 3*x

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giac [A]  time = 0.18, size = 20, normalized size = 0.77 \begin {gather*} -4 \, x^{3} + 15 \, x^{2} + x e^{\left (3 \, x\right )} - 3 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3*x+1)*exp(3*x)-12*x^2+30*x-3,x, algorithm="giac")

[Out]

-4*x^3 + 15*x^2 + x*e^(3*x) - 3*x

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maple [A]  time = 0.02, size = 21, normalized size = 0.81

method	result	size
derivativedivides	\(-3 x +15 x^{2}-4 x^{3}+x \,{\mathrm e}^{3 x}\)	\(21\)
default	\(-3 x +15 x^{2}-4 x^{3}+x \,{\mathrm e}^{3 x}\)	\(21\)
norman	\(-3 x +15 x^{2}-4 x^{3}+x \,{\mathrm e}^{3 x}\)	\(21\)
risch	\(-3 x +15 x^{2}-4 x^{3}+x \,{\mathrm e}^{3 x}\)	\(21\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((3*x+1)*exp(3*x)-12*x^2+30*x-3,x,method=_RETURNVERBOSE)

[Out]

-3*x+15*x^2-4*x^3+x*exp(3*x)

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maxima [A]  time = 0.47, size = 20, normalized size = 0.77 \begin {gather*} -4 \, x^{3} + 15 \, x^{2} + x e^{\left (3 \, x\right )} - 3 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3*x+1)*exp(3*x)-12*x^2+30*x-3,x, algorithm="maxima")

[Out]

-4*x^3 + 15*x^2 + x*e^(3*x) - 3*x

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mupad [B]  time = 0.04, size = 16, normalized size = 0.62 \begin {gather*} x\,\left (15\,x+{\mathrm {e}}^{3\,x}-4\,x^2-3\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(30*x + exp(3*x)*(3*x + 1) - 12*x^2 - 3,x)

[Out]

x*(15*x + exp(3*x) - 4*x^2 - 3)

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sympy [A]  time = 0.08, size = 19, normalized size = 0.73 \begin {gather*} - 4 x^{3} + 15 x^{2} + x e^{3 x} - 3 x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3*x+1)*exp(3*x)-12*x**2+30*x-3,x)

[Out]

-4*x**3 + 15*x**2 + x*exp(3*x) - 3*x

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