3.94.82 \(\int (20 x^3+9 e^{10-2 e^x-2 x} (10 x-10 x^2-10 e^x x^2)+3 e^{5-e^x-x} (30 x^2-10 x^3-10 e^x x^3)) \, dx\)

Optimal. Leaf size=25 \[ 5 \left (-3 e^{5-e^x-x} x-x^2\right )^2 \]

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Rubi [B]  time = 0.12, antiderivative size = 70, normalized size of antiderivative = 2.80, number of steps used = 3, number of rules used = 1, integrand size = 70, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.014, Rules used = {2288} \begin {gather*} 5 x^4+\frac {30 e^{-x-e^x+5} \left (e^x x^3+x^3\right )}{e^x+1}+\frac {45 e^{-2 x-2 e^x+10} \left (e^x x^2+x^2\right )}{e^x+1} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[20*x^3 + 9*E^(10 - 2*E^x - 2*x)*(10*x - 10*x^2 - 10*E^x*x^2) + 3*E^(5 - E^x - x)*(30*x^2 - 10*x^3 - 10*E^x
*x^3),x]

[Out]

5*x^4 + (45*E^(10 - 2*E^x - 2*x)*(x^2 + E^x*x^2))/(1 + E^x) + (30*E^(5 - E^x - x)*(x^3 + E^x*x^3))/(1 + E^x)

Rule 2288

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = (v*y)/(Log[F]*D[u, x])}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=5 x^4+3 \int e^{5-e^x-x} \left (30 x^2-10 x^3-10 e^x x^3\right ) \, dx+9 \int e^{10-2 e^x-2 x} \left (10 x-10 x^2-10 e^x x^2\right ) \, dx\\ &=5 x^4+\frac {45 e^{10-2 e^x-2 x} \left (x^2+e^x x^2\right )}{1+e^x}+\frac {30 e^{5-e^x-x} \left (x^3+e^x x^3\right )}{1+e^x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.98, size = 31, normalized size = 1.24 \begin {gather*} 5 e^{-2 \left (e^x+x\right )} x^2 \left (3 e^5+e^{e^x+x} x\right )^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[20*x^3 + 9*E^(10 - 2*E^x - 2*x)*(10*x - 10*x^2 - 10*E^x*x^2) + 3*E^(5 - E^x - x)*(30*x^2 - 10*x^3 -
10*E^x*x^3),x]

[Out]

(5*x^2*(3*E^5 + E^(E^x + x)*x)^2)/E^(2*(E^x + x))

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fricas [A]  time = 0.84, size = 42, normalized size = 1.68 \begin {gather*} 5 \, x^{4} + 10 \, x^{3} e^{\left (-x - e^{x} + \log \relax (3) + 5\right )} + 5 \, x^{2} e^{\left (-2 \, x - 2 \, e^{x} + 2 \, \log \relax (3) + 10\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-10*exp(x)*x^2-10*x^2+10*x)*exp(-exp(x)+log(3)-x+5)^2+(-10*exp(x)*x^3-10*x^3+30*x^2)*exp(-exp(x)+lo
g(3)-x+5)+20*x^3,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.20, size = 36, normalized size = 1.44 \begin {gather*} 5 \, x^{4} + 30 \, x^{3} e^{\left (-x - e^{x} + 5\right )} + 45 \, x^{2} e^{\left (-2 \, x - 2 \, e^{x} + 10\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-10*exp(x)*x^2-10*x^2+10*x)*exp(-exp(x)+log(3)-x+5)^2+(-10*exp(x)*x^3-10*x^3+30*x^2)*exp(-exp(x)+lo
g(3)-x+5)+20*x^3,x, algorithm="giac")

[Out]

5*x^4 + 30*x^3*e^(-x - e^x + 5) + 45*x^2*e^(-2*x - 2*e^x + 10)

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maple [A]  time = 0.18, size = 37, normalized size = 1.48

method	result	size
risch	\(45 \,{\mathrm e}^{-2 \,{\mathrm e}^{x}+10-2 x} x^{2}+30 \,{\mathrm e}^{-{\mathrm e}^{x}+5-x} x^{3}+5 x^{4}\)	\(37\)
default	\(45 \,{\mathrm e}^{-2 \,{\mathrm e}^{x}+10-2 x} x^{2}+10 \,{\mathrm e}^{-{\mathrm e}^{x}+\ln \relax (3)-x +5} x^{3}+5 x^{4}\)	\(43\)
norman	\(45 \,{\mathrm e}^{-2 \,{\mathrm e}^{x}+10-2 x} x^{2}+10 \,{\mathrm e}^{-{\mathrm e}^{x}+\ln \relax (3)-x +5} x^{3}+5 x^{4}\)	\(43\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-10*exp(x)*x^2-10*x^2+10*x)*exp(-exp(x)+ln(3)-x+5)^2+(-10*exp(x)*x^3-10*x^3+30*x^2)*exp(-exp(x)+ln(3)-x+5
)+20*x^3,x,method=_RETURNVERBOSE)

[Out]

45*exp(-2*exp(x)+10-2*x)*x^2+30*exp(-exp(x)+5-x)*x^3+5*x^4

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maxima [A]  time = 0.42, size = 36, normalized size = 1.44 \begin {gather*} 5 \, x^{4} + 30 \, x^{3} e^{\left (-x - e^{x} + 5\right )} + 45 \, x^{2} e^{\left (-2 \, x - 2 \, e^{x} + 10\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-10*exp(x)*x^2-10*x^2+10*x)*exp(-exp(x)+log(3)-x+5)^2+(-10*exp(x)*x^3-10*x^3+30*x^2)*exp(-exp(x)+lo
g(3)-x+5)+20*x^3,x, algorithm="maxima")

[Out]

5*x^4 + 30*x^3*e^(-x - e^x + 5) + 45*x^2*e^(-2*x - 2*e^x + 10)

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mupad [B]  time = 0.19, size = 28, normalized size = 1.12 \begin {gather*} 5\,x^2\,{\mathrm {e}}^{-2\,x-2\,{\mathrm {e}}^x}\,{\left (3\,{\mathrm {e}}^5+x\,{\mathrm {e}}^{x+{\mathrm {e}}^x}\right )}^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(20*x^3 - exp(log(3) - x - exp(x) + 5)*(10*x^3*exp(x) - 30*x^2 + 10*x^3) - exp(2*log(3) - 2*x - 2*exp(x) +
10)*(10*x^2*exp(x) - 10*x + 10*x^2),x)

[Out]

5*x^2*exp(- 2*x - 2*exp(x))*(3*exp(5) + x*exp(x + exp(x)))^2

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sympy [A]  time = 0.21, size = 34, normalized size = 1.36 \begin {gather*} 5 x^{4} + 30 x^{3} e^{- x - e^{x} + 5} + 45 x^{2} e^{- 2 x - 2 e^{x} + 10} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-10*exp(x)*x**2-10*x**2+10*x)*exp(-exp(x)+ln(3)-x+5)**2+(-10*exp(x)*x**3-10*x**3+30*x**2)*exp(-exp(
x)+ln(3)-x+5)+20*x**3,x)

[Out]

5*x**4 + 30*x**3*exp(-x - exp(x) + 5) + 45*x**2*exp(-2*x - 2*exp(x) + 10)

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