3.95.19 \(\int (48 x+3 x^2+e^x (3 x^2+x^3)) \, dx\)

Optimal. Leaf size=12 \[ x^2 \left (24+x+e^x x\right ) \]

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Rubi [A]  time = 0.08, antiderivative size = 16, normalized size of antiderivative = 1.33, number of steps used = 11, number of rules used = 4, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {1593, 2196, 2176, 2194} \begin {gather*} e^x x^3+x^3+24 x^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Int[48*x + 3*x^2 + E^x*(3*x^2 + x^3),x]

[Out]

24*x^2 + x^3 + E^x*x^3

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=24 x^2+x^3+\int e^x \left (3 x^2+x^3\right ) \, dx\\ &=24 x^2+x^3+\int e^x x^2 (3+x) \, dx\\ &=24 x^2+x^3+\int \left (3 e^x x^2+e^x x^3\right ) \, dx\\ &=24 x^2+x^3+3 \int e^x x^2 \, dx+\int e^x x^3 \, dx\\ &=24 x^2+3 e^x x^2+x^3+e^x x^3-3 \int e^x x^2 \, dx-6 \int e^x x \, dx\\ &=-6 e^x x+24 x^2+x^3+e^x x^3+6 \int e^x \, dx+6 \int e^x x \, dx\\ &=6 e^x+24 x^2+x^3+e^x x^3-6 \int e^x \, dx\\ &=24 x^2+x^3+e^x x^3\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 12, normalized size = 1.00 \begin {gather*} x^2 \left (24+x+e^x x\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[48*x + 3*x^2 + E^x*(3*x^2 + x^3),x]

[Out]

x^2*(24 + x + E^x*x)

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fricas [A]  time = 1.07, size = 15, normalized size = 1.25 \begin {gather*} x^{3} e^{x} + x^{3} + 24 \, x^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^3+3*x^2)*exp(x)+3*x^2+48*x,x, algorithm="fricas")

[Out]

x^3*e^x + x^3 + 24*x^2

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giac [A]  time = 0.17, size = 15, normalized size = 1.25 \begin {gather*} x^{3} e^{x} + x^{3} + 24 \, x^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^3+3*x^2)*exp(x)+3*x^2+48*x,x, algorithm="giac")

[Out]

x^3*e^x + x^3 + 24*x^2

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maple [A]  time = 0.02, size = 16, normalized size = 1.33

method	result	size
default	\({\mathrm e}^{x} x^{3}+24 x^{2}+x^{3}\)	\(16\)
norman	\({\mathrm e}^{x} x^{3}+24 x^{2}+x^{3}\)	\(16\)
risch	\({\mathrm e}^{x} x^{3}+24 x^{2}+x^{3}\)	\(16\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^3+3*x^2)*exp(x)+3*x^2+48*x,x,method=_RETURNVERBOSE)

[Out]

exp(x)*x^3+24*x^2+x^3

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maxima [A]  time = 0.35, size = 15, normalized size = 1.25 \begin {gather*} x^{3} e^{x} + x^{3} + 24 \, x^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^3+3*x^2)*exp(x)+3*x^2+48*x,x, algorithm="maxima")

[Out]

x^3*e^x + x^3 + 24*x^2

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mupad [B]  time = 7.27, size = 11, normalized size = 0.92 \begin {gather*} x^2\,\left (x+x\,{\mathrm {e}}^x+24\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(48*x + exp(x)*(3*x^2 + x^3) + 3*x^2,x)

[Out]

x^2*(x + x*exp(x) + 24)

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sympy [A]  time = 0.08, size = 14, normalized size = 1.17 \begin {gather*} x^{3} e^{x} + x^{3} + 24 x^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x**3+3*x**2)*exp(x)+3*x**2+48*x,x)

[Out]

x**3*exp(x) + x**3 + 24*x**2

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