3.61.81 \(\int \frac {1}{16} (192+768 x+576 x^2+e^{8+4 x} (3+12 x)+e^{4+2 x} (-48-192 x-96 x^2)) \, dx\)

Optimal. Leaf size=21 \[ 3 \left (-2+\frac {1}{4} e^{4+2 x}-2 x\right )^2 x \]

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Rubi [B]  time = 0.08, antiderivative size = 63, normalized size of antiderivative = 3.00, number of steps used = 12, number of rules used = 4, integrand size = 45, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.089, Rules used = {12, 2176, 2194, 2196} \begin {gather*} 12 x^3-3 e^{2 x+4} x^2+24 x^2-3 e^{2 x+4} x+12 x-\frac {3}{64} e^{4 x+8}+\frac {3}{64} e^{4 x+8} (4 x+1) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(192 + 768*x + 576*x^2 + E^(8 + 4*x)*(3 + 12*x) + E^(4 + 2*x)*(-48 - 192*x - 96*x^2))/16,x]

[Out]

(-3*E^(8 + 4*x))/64 + 12*x - 3*E^(4 + 2*x)*x + 24*x^2 - 3*E^(4 + 2*x)*x^2 + 12*x^3 + (3*E^(8 + 4*x)*(1 + 4*x))
/64

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}{16} \int \left (192+768 x+576 x^2+e^{8+4 x} (3+12 x)+e^{4+2 x} \left (-48-192 x-96 x^2\right )\right ) \, dx\\ &=12 x+24 x^2+12 x^3+\frac {1}{16} \int e^{8+4 x} (3+12 x) \, dx+\frac {1}{16} \int e^{4+2 x} \left (-48-192 x-96 x^2\right ) \, dx\\ &=12 x+24 x^2+12 x^3+\frac {3}{64} e^{8+4 x} (1+4 x)+\frac {1}{16} \int \left (-48 e^{4+2 x}-192 e^{4+2 x} x-96 e^{4+2 x} x^2\right ) \, dx-\frac {3}{16} \int e^{8+4 x} \, dx\\ &=-\frac {3}{64} e^{8+4 x}+12 x+24 x^2+12 x^3+\frac {3}{64} e^{8+4 x} (1+4 x)-3 \int e^{4+2 x} \, dx-6 \int e^{4+2 x} x^2 \, dx-12 \int e^{4+2 x} x \, dx\\ &=-\frac {3}{2} e^{4+2 x}-\frac {3}{64} e^{8+4 x}+12 x-6 e^{4+2 x} x+24 x^2-3 e^{4+2 x} x^2+12 x^3+\frac {3}{64} e^{8+4 x} (1+4 x)+6 \int e^{4+2 x} \, dx+6 \int e^{4+2 x} x \, dx\\ &=\frac {3}{2} e^{4+2 x}-\frac {3}{64} e^{8+4 x}+12 x-3 e^{4+2 x} x+24 x^2-3 e^{4+2 x} x^2+12 x^3+\frac {3}{64} e^{8+4 x} (1+4 x)-3 \int e^{4+2 x} \, dx\\ &=-\frac {3}{64} e^{8+4 x}+12 x-3 e^{4+2 x} x+24 x^2-3 e^{4+2 x} x^2+12 x^3+\frac {3}{64} e^{8+4 x} (1+4 x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.05, size = 20, normalized size = 0.95 \begin {gather*} \frac {3}{16} x \left (e^{4+2 x}-8 (1+x)\right )^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(192 + 768*x + 576*x^2 + E^(8 + 4*x)*(3 + 12*x) + E^(4 + 2*x)*(-48 - 192*x - 96*x^2))/16,x]

[Out]

(3*x*(E^(4 + 2*x) - 8*(1 + x))^2)/16

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fricas [A]  time = 0.61, size = 36, normalized size = 1.71 \begin {gather*} 12 \, x^{3} + 24 \, x^{2} + \frac {3}{16} \, x e^{\left (4 \, x + 8\right )} - 3 \, {\left (x^{2} + x\right )} e^{\left (2 \, x + 4\right )} + 12 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/16*(12*x+3)*exp(2+x)^4+1/16*(-96*x^2-192*x-48)*exp(2+x)^2+36*x^2+48*x+12,x, algorithm="fricas")

[Out]

12*x^3 + 24*x^2 + 3/16*x*e^(4*x + 8) - 3*(x^2 + x)*e^(2*x + 4) + 12*x

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giac [A]  time = 1.02, size = 36, normalized size = 1.71 \begin {gather*} 12 \, x^{3} + 24 \, x^{2} + \frac {3}{16} \, x e^{\left (4 \, x + 8\right )} - 3 \, {\left (x^{2} + x\right )} e^{\left (2 \, x + 4\right )} + 12 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/16*(12*x+3)*exp(2+x)^4+1/16*(-96*x^2-192*x-48)*exp(2+x)^2+36*x^2+48*x+12,x, algorithm="giac")

[Out]

12*x^3 + 24*x^2 + 3/16*x*e^(4*x + 8) - 3*(x^2 + x)*e^(2*x + 4) + 12*x

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maple [B]  time = 0.11, size = 41, normalized size = 1.95

method	result	size
risch	\(\frac {3 \,{\mathrm e}^{4 x +8} x}{16}+\frac {\left (-48 x^{2}-48 x \right ) {\mathrm e}^{2 x +4}}{16}+12 x^{3}+24 x^{2}+12 x\)	\(41\)
norman	\(12 x +24 x^{2}+12 x^{3}-3 x \,{\mathrm e}^{2 x +4}-3 \,{\mathrm e}^{2 x +4} x^{2}+\frac {3 \,{\mathrm e}^{4 x +8} x}{16}\)	\(44\)
default	\(12 x^{3}+24 x^{2}+12 x -\frac {3 \,{\mathrm e}^{4 x +8}}{8}+\frac {3 \,{\mathrm e}^{4 x +8} \left (2+x \right )}{16}-6 \,{\mathrm e}^{2 x +4}+9 \,{\mathrm e}^{2 x +4} \left (2+x \right )-3 \,{\mathrm e}^{2 x +4} \left (2+x \right )^{2}\)	\(66\)
derivativedivides	\(24 \left (2+x \right )^{2}-168-84 x +12 x^{3}-\frac {3 \,{\mathrm e}^{4 x +8}}{8}+\frac {3 \,{\mathrm e}^{4 x +8} \left (2+x \right )}{16}-6 \,{\mathrm e}^{2 x +4}+9 \,{\mathrm e}^{2 x +4} \left (2+x \right )-3 \,{\mathrm e}^{2 x +4} \left (2+x \right )^{2}\)	\(69\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/16*(12*x+3)*exp(2+x)^4+1/16*(-96*x^2-192*x-48)*exp(2+x)^2+36*x^2+48*x+12,x,method=_RETURNVERBOSE)

[Out]

3/16*exp(4*x+8)*x+1/16*(-48*x^2-48*x)*exp(2*x+4)+12*x^3+24*x^2+12*x

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maxima [B]  time = 0.35, size = 40, normalized size = 1.90 \begin {gather*} 12 \, x^{3} + 24 \, x^{2} - 3 \, {\left (x^{2} e^{4} + x e^{4}\right )} e^{\left (2 \, x\right )} + \frac {3}{16} \, x e^{\left (4 \, x + 8\right )} + 12 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/16*(12*x+3)*exp(2+x)^4+1/16*(-96*x^2-192*x-48)*exp(2+x)^2+36*x^2+48*x+12,x, algorithm="maxima")

[Out]

12*x^3 + 24*x^2 - 3*(x^2*e^4 + x*e^4)*e^(2*x) + 3/16*x*e^(4*x + 8) + 12*x

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mupad [B]  time = 0.16, size = 18, normalized size = 0.86 \begin {gather*} \frac {3\,x\,{\left (8\,x-{\mathrm {e}}^{2\,x+4}+8\right )}^2}{16} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(48*x + (exp(4*x + 8)*(12*x + 3))/16 - (exp(2*x + 4)*(192*x + 96*x^2 + 48))/16 + 36*x^2 + 12,x)

[Out]

(3*x*(8*x - exp(2*x + 4) + 8)^2)/16

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sympy [B]  time = 0.13, size = 42, normalized size = 2.00 \begin {gather*} 12 x^{3} + 24 x^{2} + \frac {3 x e^{4 x + 8}}{16} + 12 x + \frac {\left (- 48 x^{2} - 48 x\right ) e^{2 x + 4}}{16} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/16*(12*x+3)*exp(2+x)**4+1/16*(-96*x**2-192*x-48)*exp(2+x)**2+36*x**2+48*x+12,x)

[Out]

12*x**3 + 24*x**2 + 3*x*exp(4*x + 8)/16 + 12*x + (-48*x**2 - 48*x)*exp(2*x + 4)/16

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