3.57.80 \(\int (8+16 x+e^{2 x} (88+16 e^2+32 x+16 x^2)) \, dx\)

Optimal. Leaf size=18 \[ 8 \left (1+e^{2 x}\right ) \left (5+e^2+x+x^2\right ) \]

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Rubi [B]  time = 0.06, antiderivative size = 48, normalized size of antiderivative = 2.67, number of steps used = 9, number of rules used = 3, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {2196, 2194, 2176} \begin {gather*} 8 e^{2 x} x^2+8 x^2+8 e^{2 x} x+8 x-4 e^{2 x}+4 \left (11+2 e^2\right ) e^{2 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[8 + 16*x + E^(2*x)*(88 + 16*E^2 + 32*x + 16*x^2),x]

[Out]

-4*E^(2*x) + 4*E^(2*x)*(11 + 2*E^2) + 8*x + 8*E^(2*x)*x + 8*x^2 + 8*E^(2*x)*x^2

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} &=8 x+8 x^2+\int e^{2 x} \left (88+16 e^2+32 x+16 x^2\right ) \, dx\\ &=8 x+8 x^2+\int \left (88 e^{2 x} \left (1+\frac {2 e^2}{11}\right )+32 e^{2 x} x+16 e^{2 x} x^2\right ) \, dx\\ &=8 x+8 x^2+16 \int e^{2 x} x^2 \, dx+32 \int e^{2 x} x \, dx+\left (8 \left (11+2 e^2\right )\right ) \int e^{2 x} \, dx\\ &=4 e^{2 x} \left (11+2 e^2\right )+8 x+16 e^{2 x} x+8 x^2+8 e^{2 x} x^2-16 \int e^{2 x} \, dx-16 \int e^{2 x} x \, dx\\ &=-8 e^{2 x}+4 e^{2 x} \left (11+2 e^2\right )+8 x+8 e^{2 x} x+8 x^2+8 e^{2 x} x^2+8 \int e^{2 x} \, dx\\ &=-4 e^{2 x}+4 e^{2 x} \left (11+2 e^2\right )+8 x+8 e^{2 x} x+8 x^2+8 e^{2 x} x^2\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.04, size = 25, normalized size = 1.39 \begin {gather*} 8 x+8 x^2+8 e^{2 x} \left (5+e^2+x+x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[8 + 16*x + E^(2*x)*(88 + 16*E^2 + 32*x + 16*x^2),x]

[Out]

8*x + 8*x^2 + 8*E^(2*x)*(5 + E^2 + x + x^2)

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fricas [A]  time = 0.75, size = 23, normalized size = 1.28 \begin {gather*} 8 \, x^{2} + 8 \, {\left (x^{2} + x + e^{2} + 5\right )} e^{\left (2 \, x\right )} + 8 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((16*exp(1)^2+16*x^2+32*x+88)*exp(x)^2+16*x+8,x, algorithm="fricas")

[Out]

8*x^2 + 8*(x^2 + x + e^2 + 5)*e^(2*x) + 8*x

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giac [A]  time = 0.15, size = 29, normalized size = 1.61 \begin {gather*} 8 \, x^{2} + 8 \, {\left (x^{2} + x + 5\right )} e^{\left (2 \, x\right )} + 8 \, x + 8 \, e^{\left (2 \, x + 2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((16*exp(1)^2+16*x^2+32*x+88)*exp(x)^2+16*x+8,x, algorithm="giac")

[Out]

8*x^2 + 8*(x^2 + x + 5)*e^(2*x) + 8*x + 8*e^(2*x + 2)

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maple [A]  time = 0.04, size = 29, normalized size = 1.61

method	result	size
risch	\(\left (8 \,{\mathrm e}^{2}+8 x^{2}+8 x +40\right ) {\mathrm e}^{2 x}+8 x^{2}+8 x\)	\(29\)
norman	\(\left (40+8 \,{\mathrm e}^{2}\right ) {\mathrm e}^{2 x}+8 x +8 x^{2}+8 x \,{\mathrm e}^{2 x}+8 \,{\mathrm e}^{2 x} x^{2}\)	\(39\)
default	\(8 x +40 \,{\mathrm e}^{2 x}+8 x \,{\mathrm e}^{2 x}+8 \,{\mathrm e}^{2 x} x^{2}+8 \,{\mathrm e}^{2} {\mathrm e}^{2 x}+8 x^{2}\)	\(42\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((16*exp(1)^2+16*x^2+32*x+88)*exp(x)^2+16*x+8,x,method=_RETURNVERBOSE)

[Out]

(8*exp(2)+8*x^2+8*x+40)*exp(2*x)+8*x^2+8*x

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maxima [A]  time = 0.36, size = 23, normalized size = 1.28 \begin {gather*} 8 \, x^{2} + 8 \, {\left (x^{2} + x + e^{2} + 5\right )} e^{\left (2 \, x\right )} + 8 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((16*exp(1)^2+16*x^2+32*x+88)*exp(x)^2+16*x+8,x, algorithm="maxima")

[Out]

8*x^2 + 8*(x^2 + x + e^2 + 5)*e^(2*x) + 8*x

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mupad [B]  time = 3.48, size = 36, normalized size = 2.00 \begin {gather*} 8\,x+8\,x\,{\mathrm {e}}^{2\,x}+8\,x^2\,{\mathrm {e}}^{2\,x}+{\mathrm {e}}^{2\,x}\,\left (8\,{\mathrm {e}}^2+40\right )+8\,x^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(16*x + exp(2*x)*(32*x + 16*exp(2) + 16*x^2 + 88) + 8,x)

[Out]

8*x + 8*x*exp(2*x) + 8*x^2*exp(2*x) + exp(2*x)*(8*exp(2) + 40) + 8*x^2

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sympy [A]  time = 0.11, size = 27, normalized size = 1.50 \begin {gather*} 8 x^{2} + 8 x + \left (8 x^{2} + 8 x + 40 + 8 e^{2}\right ) e^{2 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((16*exp(1)**2+16*x**2+32*x+88)*exp(x)**2+16*x+8,x)

[Out]

8*x**2 + 8*x + (8*x**2 + 8*x + 40 + 8*exp(2))*exp(2*x)

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