3.66.54 \(\int (8+8 x+e^{2 x} (2 e^{10}+e^5 (-2-4 x)+2 x+2 x^2)+e^x (-4-12 x-4 x^2+e^5 (8+4 x))) \, dx\)

Optimal. Leaf size=23 \[ \left (2+2 x-e^x \left (1-\frac {e^5}{x}\right ) x\right )^2 \]

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Rubi [B]  time = 0.14, antiderivative size = 77, normalized size of antiderivative = 3.35, number of steps used = 21, number of rules used = 3, integrand size = 57, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.053, Rules used = {2196, 2194, 2176} \begin {gather*} -4 e^x x^2+e^{2 x} x^2+4 x^2-4 e^x x+8 x-4 e^{x+5}+e^{2 x+5}+e^{2 x+10}+4 e^{x+5} (x+2)-e^{2 x+5} (2 x+1) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[8 + 8*x + E^(2*x)*(2*E^10 + E^5*(-2 - 4*x) + 2*x + 2*x^2) + E^x*(-4 - 12*x - 4*x^2 + E^5*(8 + 4*x)),x]

[Out]

-4*E^(5 + x) + E^(5 + 2*x) + E^(10 + 2*x) + 8*x - 4*E^x*x + 4*x^2 - 4*E^x*x^2 + E^(2*x)*x^2 + 4*E^(5 + x)*(2 +
x) - E^(5 + 2*x)*(1 + 2*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} &=8 x+4 x^2+\int e^{2 x} \left (2 e^{10}+e^5 (-2-4 x)+2 x+2 x^2\right ) \, dx+\int e^x \left (-4-12 x-4 x^2+e^5 (8+4 x)\right ) \, dx\\ &=8 x+4 x^2+\int \left (-4 e^x-12 e^x x-4 e^x x^2+4 e^{5+x} (2+x)\right ) \, dx+\int \left (2 e^{10+2 x}+2 e^{2 x} x+2 e^{2 x} x^2-2 e^{5+2 x} (1+2 x)\right ) \, dx\\ &=8 x+4 x^2+2 \int e^{10+2 x} \, dx+2 \int e^{2 x} x \, dx+2 \int e^{2 x} x^2 \, dx-2 \int e^{5+2 x} (1+2 x) \, dx-4 \int e^x \, dx-4 \int e^x x^2 \, dx+4 \int e^{5+x} (2+x) \, dx-12 \int e^x x \, dx\\ &=-4 e^x+e^{10+2 x}+8 x-12 e^x x+e^{2 x} x+4 x^2-4 e^x x^2+e^{2 x} x^2+4 e^{5+x} (2+x)-e^{5+2 x} (1+2 x)+2 \int e^{5+2 x} \, dx-2 \int e^{2 x} x \, dx-4 \int e^{5+x} \, dx+8 \int e^x x \, dx+12 \int e^x \, dx-\int e^{2 x} \, dx\\ &=8 e^x-\frac {e^{2 x}}{2}-4 e^{5+x}+e^{5+2 x}+e^{10+2 x}+8 x-4 e^x x+4 x^2-4 e^x x^2+e^{2 x} x^2+4 e^{5+x} (2+x)-e^{5+2 x} (1+2 x)-8 \int e^x \, dx+\int e^{2 x} \, dx\\ &=-4 e^{5+x}+e^{5+2 x}+e^{10+2 x}+8 x-4 e^x x+4 x^2-4 e^x x^2+e^{2 x} x^2+4 e^{5+x} (2+x)-e^{5+2 x} (1+2 x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.13, size = 39, normalized size = 1.70 \begin {gather*} e^{2 x} \left (e^5-x\right )^2+8 x+4 x^2+4 e^x \left (e^5-x\right ) (1+x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[8 + 8*x + E^(2*x)*(2*E^10 + E^5*(-2 - 4*x) + 2*x + 2*x^2) + E^x*(-4 - 12*x - 4*x^2 + E^5*(8 + 4*x)),
x]

[Out]

E^(2*x)*(E^5 - x)^2 + 8*x + 4*x^2 + 4*E^x*(E^5 - x)*(1 + x)

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fricas [B]  time = 0.76, size = 41, normalized size = 1.78 \begin {gather*} 4 \, x^{2} + {\left (x^{2} - 2 \, x e^{5} + e^{10}\right )} e^{\left (2 \, x\right )} - 4 \, {\left (x^{2} - {\left (x + 1\right )} e^{5} + x\right )} e^{x} + 8 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*exp(5)^2+(-4*x-2)*exp(5)+2*x^2+2*x)*exp(x)^2+((4*x+8)*exp(5)-4*x^2-12*x-4)*exp(x)+8*x+8,x, algori
thm="fricas")

[Out]

4*x^2 + (x^2 - 2*x*e^5 + e^10)*e^(2*x) - 4*(x^2 - (x + 1)*e^5 + x)*e^x + 8*x

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giac [B]  time = 0.15, size = 50, normalized size = 2.17 \begin {gather*} x^{2} e^{\left (2 \, x\right )} + 4 \, x^{2} - 2 \, x e^{\left (2 \, x + 5\right )} + 4 \, {\left (x + 1\right )} e^{\left (x + 5\right )} - 4 \, {\left (x^{2} + x\right )} e^{x} + 8 \, x + e^{\left (2 \, x + 10\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*exp(5)^2+(-4*x-2)*exp(5)+2*x^2+2*x)*exp(x)^2+((4*x+8)*exp(5)-4*x^2-12*x-4)*exp(x)+8*x+8,x, algori
thm="giac")

[Out]

x^2*e^(2*x) + 4*x^2 - 2*x*e^(2*x + 5) + 4*(x + 1)*e^(x + 5) - 4*(x^2 + x)*e^x + 8*x + e^(2*x + 10)

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maple [B]  time = 0.05, size = 47, normalized size = 2.04

method	result	size
risch	\(\left ({\mathrm e}^{10}-2 x \,{\mathrm e}^{5}+x^{2}\right ) {\mathrm e}^{2 x}+\left (4 x \,{\mathrm e}^{5}-4 x^{2}+4 \,{\mathrm e}^{5}-4 x \right ) {\mathrm e}^{x}+4 x^{2}+8 x\)	\(47\)
norman	\({\mathrm e}^{10} {\mathrm e}^{2 x}+{\mathrm e}^{2 x} x^{2}+\left (4 \,{\mathrm e}^{5}-4\right ) x \,{\mathrm e}^{x}+8 x +4 x^{2}+4 \,{\mathrm e}^{5} {\mathrm e}^{x}-4 \,{\mathrm e}^{x} x^{2}-2 x \,{\mathrm e}^{5} {\mathrm e}^{2 x}\)	\(59\)
default	\(8 x +{\mathrm e}^{2 x} x^{2}-{\mathrm e}^{5} {\mathrm e}^{2 x}+{\mathrm e}^{10} {\mathrm e}^{2 x}-4 \,{\mathrm e}^{5} \left (\frac {x \,{\mathrm e}^{2 x}}{2}-\frac {{\mathrm e}^{2 x}}{4}\right )-4 \,{\mathrm e}^{x} x -4 \,{\mathrm e}^{x} x^{2}+8 \,{\mathrm e}^{5} {\mathrm e}^{x}+4 \,{\mathrm e}^{5} \left ({\mathrm e}^{x} x -{\mathrm e}^{x}\right )+4 x^{2}\)	\(84\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*exp(5)^2+(-4*x-2)*exp(5)+2*x^2+2*x)*exp(x)^2+((4*x+8)*exp(5)-4*x^2-12*x-4)*exp(x)+8*x+8,x,method=_RETUR
NVERBOSE)

[Out]

(exp(10)-2*x*exp(5)+x^2)*exp(2*x)+(4*x*exp(5)-4*x^2+4*exp(5)-4*x)*exp(x)+4*x^2+8*x

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maxima [B]  time = 0.38, size = 44, normalized size = 1.91 \begin {gather*} 4 \, x^{2} + {\left (x^{2} - 2 \, x e^{5} + e^{10}\right )} e^{\left (2 \, x\right )} - 4 \, {\left (x^{2} - x {\left (e^{5} - 1\right )} - e^{5}\right )} e^{x} + 8 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*exp(5)^2+(-4*x-2)*exp(5)+2*x^2+2*x)*exp(x)^2+((4*x+8)*exp(5)-4*x^2-12*x-4)*exp(x)+8*x+8,x, algori
thm="maxima")

[Out]

4*x^2 + (x^2 - 2*x*e^5 + e^10)*e^(2*x) - 4*(x^2 - x*(e^5 - 1) - e^5)*e^x + 8*x

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mupad [B]  time = 0.08, size = 55, normalized size = 2.39 \begin {gather*} 8\,x+4\,{\mathrm {e}}^{x+5}+{\mathrm {e}}^{2\,x+10}-4\,x^2\,{\mathrm {e}}^x-2\,x\,{\mathrm {e}}^{2\,x+5}+x^2\,{\mathrm {e}}^{2\,x}+4\,x^2+x\,{\mathrm {e}}^x\,\left (4\,{\mathrm {e}}^5-4\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(8*x - exp(x)*(12*x + 4*x^2 - exp(5)*(4*x + 8) + 4) + exp(2*x)*(2*x + 2*exp(10) + 2*x^2 - exp(5)*(4*x + 2))
+ 8,x)

[Out]

8*x + 4*exp(x + 5) + exp(2*x + 10) - 4*x^2*exp(x) - 2*x*exp(2*x + 5) + x^2*exp(2*x) + 4*x^2 + x*exp(x)*(4*exp(
5) - 4)

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sympy [B]  time = 0.15, size = 49, normalized size = 2.13 \begin {gather*} 4 x^{2} + 8 x + \left (x^{2} - 2 x e^{5} + e^{10}\right ) e^{2 x} + \left (- 4 x^{2} - 4 x + 4 x e^{5} + 4 e^{5}\right ) e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*exp(5)**2+(-4*x-2)*exp(5)+2*x**2+2*x)*exp(x)**2+((4*x+8)*exp(5)-4*x**2-12*x-4)*exp(x)+8*x+8,x)

[Out]

4*x**2 + 8*x + (x**2 - 2*x*exp(5) + exp(10))*exp(2*x) + (-4*x**2 - 4*x + 4*x*exp(5) + 4*exp(5))*exp(x)

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