3.68.11 \(\int \frac {2 e^{6+2 x}+16 e^4 x^3+e^{10} (-10+2 x)+e^7 (-40 x+12 x^2)+e^x (e^8 (-8+2 x)+e^5 (8 x+4 x^2))}{e^{10}} \, dx\)

Optimal. Leaf size=18 \[ \left (-5+e^{-2+x}+x+\frac {2 x^2}{e^3}\right )^2 \]

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Rubi [B]  time = 0.11, antiderivative size = 68, normalized size of antiderivative = 3.78, number of steps used = 15, number of rules used = 4, integrand size = 71, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.056, Rules used = {12, 2194, 2196, 2176} \begin {gather*} \frac {4 x^4}{e^6}+\frac {4 x^3}{e^3}+4 e^{x-5} x^2-\frac {20 x^2}{e^3}-2 e^{x-2}+e^{2 x-4}+(5-x)^2-2 e^{x-2} (4-x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(2*E^(6 + 2*x) + 16*E^4*x^3 + E^10*(-10 + 2*x) + E^7*(-40*x + 12*x^2) + E^x*(E^8*(-8 + 2*x) + E^5*(8*x + 4
*x^2)))/E^10,x]

[Out]

-2*E^(-2 + x) + E^(-4 + 2*x) - 2*E^(-2 + x)*(4 - x) + (5 - x)^2 - (20*x^2)/E^3 + 4*E^(-5 + x)*x^2 + (4*x^3)/E^
3 + (4*x^4)/E^6

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 {\int \left (2 e^{6+2 x}+16 e^4 x^3+e^{10} (-10+2 x)+e^7 \left (-40 x+12 x^2\right )+e^x \left (e^8 (-8+2 x)+e^5 \left (8 x+4 x^2\right )\right )\right ) \, dx}{e^{10}}\\ &=(5-x)^2+\frac {4 x^4}{e^6}+\frac {\int e^x \left (e^8 (-8+2 x)+e^5 \left (8 x+4 x^2\right )\right ) \, dx}{e^{10}}+\frac {2 \int e^{6+2 x} \, dx}{e^{10}}+\frac {\int \left (-40 x+12 x^2\right ) \, dx}{e^3}\\ &=e^{-4+2 x}+(5-x)^2-\frac {20 x^2}{e^3}+\frac {4 x^3}{e^3}+\frac {4 x^4}{e^6}+\frac {\int \left (2 e^{8+x} (-4+x)+4 e^{5+x} x (2+x)\right ) \, dx}{e^{10}}\\ &=e^{-4+2 x}+(5-x)^2-\frac {20 x^2}{e^3}+\frac {4 x^3}{e^3}+\frac {4 x^4}{e^6}+\frac {2 \int e^{8+x} (-4+x) \, dx}{e^{10}}+\frac {4 \int e^{5+x} x (2+x) \, dx}{e^{10}}\\ &=e^{-4+2 x}-2 e^{-2+x} (4-x)+(5-x)^2-\frac {20 x^2}{e^3}+\frac {4 x^3}{e^3}+\frac {4 x^4}{e^6}-\frac {2 \int e^{8+x} \, dx}{e^{10}}+\frac {4 \int \left (2 e^{5+x} x+e^{5+x} x^2\right ) \, dx}{e^{10}}\\ &=-2 e^{-2+x}+e^{-4+2 x}-2 e^{-2+x} (4-x)+(5-x)^2-\frac {20 x^2}{e^3}+\frac {4 x^3}{e^3}+\frac {4 x^4}{e^6}+\frac {4 \int e^{5+x} x^2 \, dx}{e^{10}}+\frac {8 \int e^{5+x} x \, dx}{e^{10}}\\ &=-2 e^{-2+x}+e^{-4+2 x}-2 e^{-2+x} (4-x)+(5-x)^2+8 e^{-5+x} x-\frac {20 x^2}{e^3}+4 e^{-5+x} x^2+\frac {4 x^3}{e^3}+\frac {4 x^4}{e^6}-\frac {8 \int e^{5+x} \, dx}{e^{10}}-\frac {8 \int e^{5+x} x \, dx}{e^{10}}\\ &=-8 e^{-5+x}-2 e^{-2+x}+e^{-4+2 x}-2 e^{-2+x} (4-x)+(5-x)^2-\frac {20 x^2}{e^3}+4 e^{-5+x} x^2+\frac {4 x^3}{e^3}+\frac {4 x^4}{e^6}+\frac {8 \int e^{5+x} \, dx}{e^{10}}\\ &=-2 e^{-2+x}+e^{-4+2 x}-2 e^{-2+x} (4-x)+(5-x)^2-\frac {20 x^2}{e^3}+4 e^{-5+x} x^2+\frac {4 x^3}{e^3}+\frac {4 x^4}{e^6}\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.05, size = 79, normalized size = 4.39 \begin {gather*} \frac {2 \left (\frac {1}{2} e^{2+2 x}-5 e^6 x-10 e^3 x^2+\frac {e^6 x^2}{2}+2 e^{1+x} x^2+2 e^3 x^3+2 x^4+e^x \left (-5 e^4+e^4 x\right )\right )}{e^6} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(2*E^(6 + 2*x) + 16*E^4*x^3 + E^10*(-10 + 2*x) + E^7*(-40*x + 12*x^2) + E^x*(E^8*(-8 + 2*x) + E^5*(8
*x + 4*x^2)))/E^10,x]

[Out]

(2*(E^(2 + 2*x)/2 - 5*E^6*x - 10*E^3*x^2 + (E^6*x^2)/2 + 2*E^(1 + x)*x^2 + 2*E^3*x^3 + 2*x^4 + E^x*(-5*E^4 + E
^4*x)))/E^6

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fricas [B]  time = 0.66, size = 56, normalized size = 3.11 \begin {gather*} {\left (4 \, x^{4} + {\left (x^{2} - 10 \, x\right )} e^{6} + 4 \, {\left (x^{3} - 5 \, x^{2}\right )} e^{3} + 2 \, {\left (2 \, x^{2} e + {\left (x - 5\right )} e^{4}\right )} e^{x} + e^{\left (2 \, x + 2\right )}\right )} e^{\left (-6\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*exp(3)^2*exp(x)^2+((2*x-8)*exp(2)*exp(3)^2+(4*x^2+8*x)*exp(2)*exp(3))*exp(x)+(2*x-10)*exp(2)^2*ex
p(3)^2+(12*x^2-40*x)*exp(2)^2*exp(3)+16*x^3*exp(2)^2)/exp(2)^2/exp(3)^2,x, algorithm="fricas")

[Out]

(4*x^4 + (x^2 - 10*x)*e^6 + 4*(x^3 - 5*x^2)*e^3 + 2*(2*x^2*e + (x - 5)*e^4)*e^x + e^(2*x + 2))*e^(-6)

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giac [B]  time = 0.23, size = 58, normalized size = 3.22 \begin {gather*} {\left (4 \, x^{4} e^{4} + 4 \, x^{2} e^{\left (x + 5\right )} + {\left (x^{2} - 10 \, x\right )} e^{10} + 4 \, {\left (x^{3} - 5 \, x^{2}\right )} e^{7} + 2 \, {\left (x - 5\right )} e^{\left (x + 8\right )} + e^{\left (2 \, x + 6\right )}\right )} e^{\left (-10\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*exp(3)^2*exp(x)^2+((2*x-8)*exp(2)*exp(3)^2+(4*x^2+8*x)*exp(2)*exp(3))*exp(x)+(2*x-10)*exp(2)^2*ex
p(3)^2+(12*x^2-40*x)*exp(2)^2*exp(3)+16*x^3*exp(2)^2)/exp(2)^2/exp(3)^2,x, algorithm="giac")

[Out]

(4*x^4*e^4 + 4*x^2*e^(x + 5) + (x^2 - 10*x)*e^10 + 4*(x^3 - 5*x^2)*e^7 + 2*(x - 5)*e^(x + 8) + e^(2*x + 6))*e^
(-10)

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maple [B]  time = 0.05, size = 70, normalized size = 3.89

method	result	size
risch	\(4 \,{\mathrm e}^{-6} x^{3} {\mathrm e}^{3}+4 \,{\mathrm e}^{-6} x^{4}-20 \,{\mathrm e}^{-6} x^{2} {\mathrm e}^{3}+{\mathrm e}^{-6} x^{2} {\mathrm e}^{6}-10 \,{\mathrm e}^{-6} x \,{\mathrm e}^{6}+{\mathrm e}^{2 x -4}+\left (2 x \,{\mathrm e}^{8}-10 \,{\mathrm e}^{8}+4 x^{2} {\mathrm e}^{5}\right ) {\mathrm e}^{x -10}\)	\(70\)
norman	\(\left ({\mathrm e}^{-2} {\mathrm e}^{3} {\mathrm e}^{2 x}+{\mathrm e}^{2} \left ({\mathrm e}^{3}-20\right ) x^{2}+4 x^{3} {\mathrm e}^{2}+4 \,{\mathrm e}^{x} x^{2}-10 \,{\mathrm e}^{x} {\mathrm e}^{3}-10 x \,{\mathrm e}^{2} {\mathrm e}^{3}+2 x \,{\mathrm e}^{3} {\mathrm e}^{x}+4 \,{\mathrm e}^{2} {\mathrm e}^{-3} x^{4}\right ) {\mathrm e}^{-2} {\mathrm e}^{-3}\)	\(77\)
default	\({\mathrm e}^{-4} {\mathrm e}^{-6} \left (-8 \,{\mathrm e}^{2} {\mathrm e}^{6} {\mathrm e}^{x}+8 \,{\mathrm e}^{2} {\mathrm e}^{3} \left ({\mathrm e}^{x} x -{\mathrm e}^{x}\right )+4 \,{\mathrm e}^{2} {\mathrm e}^{3} \left ({\mathrm e}^{x} x^{2}-2 \,{\mathrm e}^{x} x +2 \,{\mathrm e}^{x}\right )+2 \,{\mathrm e}^{2} {\mathrm e}^{6} \left ({\mathrm e}^{x} x -{\mathrm e}^{x}\right )+{\mathrm e}^{4} {\mathrm e}^{6} \left (x^{2}-10 x \right )+{\mathrm e}^{4} {\mathrm e}^{3} \left (4 x^{3}-20 x^{2}\right )+4 x^{4} {\mathrm e}^{4}+{\mathrm e}^{6} {\mathrm e}^{2 x}\right )\)	\(127\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*exp(3)^2*exp(x)^2+((2*x-8)*exp(2)*exp(3)^2+(4*x^2+8*x)*exp(2)*exp(3))*exp(x)+(2*x-10)*exp(2)^2*exp(3)^2
+(12*x^2-40*x)*exp(2)^2*exp(3)+16*x^3*exp(2)^2)/exp(2)^2/exp(3)^2,x,method=_RETURNVERBOSE)

[Out]

4*exp(-6)*x^3*exp(3)+4*exp(-6)*x^4-20*exp(-6)*x^2*exp(3)+exp(-6)*x^2*exp(6)-10*exp(-6)*x*exp(6)+exp(2*x-4)+(2*
x*exp(8)-10*exp(8)+4*x^2*exp(5))*exp(x-10)

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maxima [B]  time = 0.36, size = 60, normalized size = 3.33 \begin {gather*} {\left (4 \, x^{4} e^{4} + {\left (x^{2} - 10 \, x\right )} e^{10} + 4 \, {\left (x^{3} - 5 \, x^{2}\right )} e^{7} + 2 \, {\left (2 \, x^{2} e^{5} + x e^{8} - 5 \, e^{8}\right )} e^{x} + e^{\left (2 \, x + 6\right )}\right )} e^{\left (-10\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*exp(3)^2*exp(x)^2+((2*x-8)*exp(2)*exp(3)^2+(4*x^2+8*x)*exp(2)*exp(3))*exp(x)+(2*x-10)*exp(2)^2*ex
p(3)^2+(12*x^2-40*x)*exp(2)^2*exp(3)+16*x^3*exp(2)^2)/exp(2)^2/exp(3)^2,x, algorithm="maxima")

[Out]

(4*x^4*e^4 + (x^2 - 10*x)*e^10 + 4*(x^3 - 5*x^2)*e^7 + 2*(2*x^2*e^5 + x*e^8 - 5*e^8)*e^x + e^(2*x + 6))*e^(-10
)

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mupad [B]  time = 0.11, size = 57, normalized size = 3.17 \begin {gather*} {\mathrm {e}}^{2\,x-4}-10\,{\mathrm {e}}^{x-2}-10\,x+2\,x\,{\mathrm {e}}^{x-2}-x^2\,\left (20\,{\mathrm {e}}^{-3}-1\right )+4\,x^2\,{\mathrm {e}}^{x-5}+4\,x^3\,{\mathrm {e}}^{-3}+4\,x^4\,{\mathrm {e}}^{-6} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(-10)*(2*exp(2*x)*exp(6) - exp(7)*(40*x - 12*x^2) + 16*x^3*exp(4) + exp(x)*(exp(5)*(8*x + 4*x^2) + exp(
8)*(2*x - 8)) + exp(10)*(2*x - 10)),x)

[Out]

exp(2*x - 4) - 10*exp(x - 2) - 10*x + 2*x*exp(x - 2) - x^2*(20*exp(-3) - 1) + 4*x^2*exp(x - 5) + 4*x^3*exp(-3)
+ 4*x^4*exp(-6)

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sympy [B]  time = 0.20, size = 66, normalized size = 3.67 \begin {gather*} \frac {4 x^{4}}{e^{6}} + \frac {4 x^{3}}{e^{3}} + \frac {x^{2} \left (-20 + e^{3}\right )}{e^{3}} - 10 x + \frac {\left (4 x^{2} e^{4} + 2 x e^{7} - 10 e^{7}\right ) e^{x} + e^{5} e^{2 x}}{e^{9}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*exp(3)**2*exp(x)**2+((2*x-8)*exp(2)*exp(3)**2+(4*x**2+8*x)*exp(2)*exp(3))*exp(x)+(2*x-10)*exp(2)*
*2*exp(3)**2+(12*x**2-40*x)*exp(2)**2*exp(3)+16*x**3*exp(2)**2)/exp(2)**2/exp(3)**2,x)

[Out]

4*x**4*exp(-6) + 4*x**3*exp(-3) + x**2*(-20 + exp(3))*exp(-3) - 10*x + ((4*x**2*exp(4) + 2*x*exp(7) - 10*exp(7
))*exp(x) + exp(5)*exp(2*x))*exp(-9)

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