3.10.24 \(\int \frac {1}{4} (-4+e^{3 x} (16+48 x+4 x^3+3 x^4)) \, dx\)

Optimal. Leaf size=20 \[ -x+e^{3 x} x \left (4+\frac {x^3}{4}\right ) \]

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Rubi [A]  time = 0.16, antiderivative size = 24, normalized size of antiderivative = 1.20, number of steps used = 16, number of rules used = 4, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {12, 2196, 2194, 2176} \begin {gather*} \frac {1}{4} e^{3 x} x^4+4 e^{3 x} x-x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-4 + E^(3*x)*(16 + 48*x + 4*x^3 + 3*x^4))/4,x]

[Out]

-x + 4*E^(3*x)*x + (E^(3*x)*x^4)/4

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{4} \int \left (-4+e^{3 x} \left (16+48 x+4 x^3+3 x^4\right )\right ) \, dx\\ &=-x+\frac {1}{4} \int e^{3 x} \left (16+48 x+4 x^3+3 x^4\right ) \, dx\\ &=-x+\frac {1}{4} \int \left (16 e^{3 x}+48 e^{3 x} x+4 e^{3 x} x^3+3 e^{3 x} x^4\right ) \, dx\\ &=-x+\frac {3}{4} \int e^{3 x} x^4 \, dx+4 \int e^{3 x} \, dx+12 \int e^{3 x} x \, dx+\int e^{3 x} x^3 \, dx\\ &=\frac {4 e^{3 x}}{3}-x+4 e^{3 x} x+\frac {1}{3} e^{3 x} x^3+\frac {1}{4} e^{3 x} x^4-4 \int e^{3 x} \, dx-\int e^{3 x} x^2 \, dx-\int e^{3 x} x^3 \, dx\\ &=-x+4 e^{3 x} x-\frac {1}{3} e^{3 x} x^2+\frac {1}{4} e^{3 x} x^4+\frac {2}{3} \int e^{3 x} x \, dx+\int e^{3 x} x^2 \, dx\\ &=-x+\frac {38}{9} e^{3 x} x+\frac {1}{4} e^{3 x} x^4-\frac {2}{9} \int e^{3 x} \, dx-\frac {2}{3} \int e^{3 x} x \, dx\\ &=-\frac {2 e^{3 x}}{27}-x+4 e^{3 x} x+\frac {1}{4} e^{3 x} x^4+\frac {2}{9} \int e^{3 x} \, dx\\ &=-x+4 e^{3 x} x+\frac {1}{4} e^{3 x} x^4\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 24, normalized size = 1.20 \begin {gather*} -x+\frac {1}{3} e^{3 x} \left (12 x+\frac {3 x^4}{4}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-4 + E^(3*x)*(16 + 48*x + 4*x^3 + 3*x^4))/4,x]

[Out]

-x + (E^(3*x)*(12*x + (3*x^4)/4))/3

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fricas [A]  time = 0.62, size = 17, normalized size = 0.85 \begin {gather*} \frac {1}{4} \, {\left (x^{4} + 16 \, x\right )} e^{\left (3 \, x\right )} - x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(3*x^4+4*x^3+48*x+16)*exp(3*x)-1,x, algorithm="fricas")

[Out]

1/4*(x^4 + 16*x)*e^(3*x) - x

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giac [A]  time = 0.30, size = 17, normalized size = 0.85 \begin {gather*} \frac {1}{4} \, {\left (x^{4} + 16 \, x\right )} e^{\left (3 \, x\right )} - x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(3*x^4+4*x^3+48*x+16)*exp(3*x)-1,x, algorithm="giac")

[Out]

1/4*(x^4 + 16*x)*e^(3*x) - x

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maple [A]  time = 0.02, size = 18, normalized size = 0.90

method	result	size
risch	\(\frac {\left (x^{4}+16 x \right ) {\mathrm e}^{3 x}}{4}-x\)	\(18\)
derivativedivides	\(-x +\frac {{\mathrm e}^{3 x} x^{4}}{4}+4 x \,{\mathrm e}^{3 x}\)	\(21\)
default	\(-x +\frac {{\mathrm e}^{3 x} x^{4}}{4}+4 x \,{\mathrm e}^{3 x}\)	\(21\)
norman	\(-x +\frac {{\mathrm e}^{3 x} x^{4}}{4}+4 x \,{\mathrm e}^{3 x}\)	\(21\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/4*(3*x^4+4*x^3+48*x+16)*exp(3*x)-1,x,method=_RETURNVERBOSE)

[Out]

1/4*(x^4+16*x)*exp(3*x)-x

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maxima [B]  time = 0.39, size = 68, normalized size = 3.40 \begin {gather*} \frac {1}{108} \, {\left (27 \, x^{4} - 36 \, x^{3} + 36 \, x^{2} - 24 \, x + 8\right )} e^{\left (3 \, x\right )} + \frac {1}{27} \, {\left (9 \, x^{3} - 9 \, x^{2} + 6 \, x - 2\right )} e^{\left (3 \, x\right )} + \frac {4}{3} \, {\left (3 \, x - 1\right )} e^{\left (3 \, x\right )} - x + \frac {4}{3} \, e^{\left (3 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(3*x^4+4*x^3+48*x+16)*exp(3*x)-1,x, algorithm="maxima")

[Out]

1/108*(27*x^4 - 36*x^3 + 36*x^2 - 24*x + 8)*e^(3*x) + 1/27*(9*x^3 - 9*x^2 + 6*x - 2)*e^(3*x) + 4/3*(3*x - 1)*e
^(3*x) - x + 4/3*e^(3*x)

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mupad [B]  time = 0.05, size = 19, normalized size = 0.95 \begin {gather*} \frac {x\,\left (16\,{\mathrm {e}}^{3\,x}+x^3\,{\mathrm {e}}^{3\,x}-4\right )}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(3*x)*(48*x + 4*x^3 + 3*x^4 + 16))/4 - 1,x)

[Out]

(x*(16*exp(3*x) + x^3*exp(3*x) - 4))/4

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sympy [A]  time = 0.09, size = 14, normalized size = 0.70 \begin {gather*} - x + \frac {\left (x^{4} + 16 x\right ) e^{3 x}}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(3*x**4+4*x**3+48*x+16)*exp(3*x)-1,x)

[Out]

-x + (x**4 + 16*x)*exp(3*x)/4

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