3.83.51 \(\int (1-255 x^4+e^{2 x} (-153 x^2-102 x^3)+e^x (408 x^3+102 x^4)) \, dx\)

Optimal. Leaf size=18 \[ x-51 x \left (e^x x-x^2\right )^2 \]

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Rubi [A]  time = 0.20, antiderivative size = 25, normalized size of antiderivative = 1.39, number of steps used = 23, number of rules used = 4, integrand size = 39, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {1593, 2196, 2176, 2194} \begin {gather*} -51 x^5+102 e^x x^4-51 e^{2 x} x^3+x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1 - 255*x^4 + E^(2*x)*(-153*x^2 - 102*x^3) + E^x*(408*x^3 + 102*x^4),x]

[Out]

x - 51*E^(2*x)*x^3 + 102*E^x*x^4 - 51*x^5

Rule 1593

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

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} &=x-51 x^5+\int e^{2 x} \left (-153 x^2-102 x^3\right ) \, dx+\int e^x \left (408 x^3+102 x^4\right ) \, dx\\ &=x-51 x^5+\int e^{2 x} (-153-102 x) x^2 \, dx+\int e^x x^3 (408+102 x) \, dx\\ &=x-51 x^5+\int \left (-153 e^{2 x} x^2-102 e^{2 x} x^3\right ) \, dx+\int \left (408 e^x x^3+102 e^x x^4\right ) \, dx\\ &=x-51 x^5-102 \int e^{2 x} x^3 \, dx+102 \int e^x x^4 \, dx-153 \int e^{2 x} x^2 \, dx+408 \int e^x x^3 \, dx\\ &=x-\frac {153}{2} e^{2 x} x^2+408 e^x x^3-51 e^{2 x} x^3+102 e^x x^4-51 x^5+153 \int e^{2 x} x \, dx+153 \int e^{2 x} x^2 \, dx-408 \int e^x x^3 \, dx-1224 \int e^x x^2 \, dx\\ &=x+\frac {153}{2} e^{2 x} x-1224 e^x x^2-51 e^{2 x} x^3+102 e^x x^4-51 x^5-\frac {153}{2} \int e^{2 x} \, dx-153 \int e^{2 x} x \, dx+1224 \int e^x x^2 \, dx+2448 \int e^x x \, dx\\ &=-\frac {153 e^{2 x}}{4}+x+2448 e^x x-51 e^{2 x} x^3+102 e^x x^4-51 x^5+\frac {153}{2} \int e^{2 x} \, dx-2448 \int e^x \, dx-2448 \int e^x x \, dx\\ &=-2448 e^x+x-51 e^{2 x} x^3+102 e^x x^4-51 x^5+2448 \int e^x \, dx\\ &=x-51 e^{2 x} x^3+102 e^x x^4-51 x^5\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 25, normalized size = 1.39 \begin {gather*} x-51 e^{2 x} x^3+102 e^x x^4-51 x^5 \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1 - 255*x^4 + E^(2*x)*(-153*x^2 - 102*x^3) + E^x*(408*x^3 + 102*x^4),x]

[Out]

x - 51*E^(2*x)*x^3 + 102*E^x*x^4 - 51*x^5

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fricas [A]  time = 0.62, size = 23, normalized size = 1.28 \begin {gather*} -51 \, x^{5} + 102 \, x^{4} e^{x} - 51 \, x^{3} e^{\left (2 \, x\right )} + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-102*x^3-153*x^2)*exp(x)^2+(102*x^4+408*x^3)*exp(x)-255*x^4+1,x, algorithm="fricas")

[Out]

-51*x^5 + 102*x^4*e^x - 51*x^3*e^(2*x) + x

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giac [A]  time = 0.14, size = 23, normalized size = 1.28 \begin {gather*} -51 \, x^{5} + 102 \, x^{4} e^{x} - 51 \, x^{3} e^{\left (2 \, x\right )} + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-102*x^3-153*x^2)*exp(x)^2+(102*x^4+408*x^3)*exp(x)-255*x^4+1,x, algorithm="giac")

[Out]

-51*x^5 + 102*x^4*e^x - 51*x^3*e^(2*x) + x

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maple [A]  time = 0.03, size = 24, normalized size = 1.33

method	result	size
default	\(x -51 \,{\mathrm e}^{2 x} x^{3}+102 \,{\mathrm e}^{x} x^{4}-51 x^{5}\)	\(24\)
norman	\(x -51 \,{\mathrm e}^{2 x} x^{3}+102 \,{\mathrm e}^{x} x^{4}-51 x^{5}\)	\(24\)
risch	\(x -51 \,{\mathrm e}^{2 x} x^{3}+102 \,{\mathrm e}^{x} x^{4}-51 x^{5}\)	\(24\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-102*x^3-153*x^2)*exp(x)^2+(102*x^4+408*x^3)*exp(x)-255*x^4+1,x,method=_RETURNVERBOSE)

[Out]

x-51*exp(x)^2*x^3+102*exp(x)*x^4-51*x^5

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maxima [A]  time = 0.36, size = 23, normalized size = 1.28 \begin {gather*} -51 \, x^{5} + 102 \, x^{4} e^{x} - 51 \, x^{3} e^{\left (2 \, x\right )} + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-102*x^3-153*x^2)*exp(x)^2+(102*x^4+408*x^3)*exp(x)-255*x^4+1,x, algorithm="maxima")

[Out]

-51*x^5 + 102*x^4*e^x - 51*x^3*e^(2*x) + x

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mupad [B]  time = 4.83, size = 23, normalized size = 1.28 \begin {gather*} x+102\,x^4\,{\mathrm {e}}^x-51\,x^3\,{\mathrm {e}}^{2\,x}-51\,x^5 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(x)*(408*x^3 + 102*x^4) - exp(2*x)*(153*x^2 + 102*x^3) - 255*x^4 + 1,x)

[Out]

x + 102*x^4*exp(x) - 51*x^3*exp(2*x) - 51*x^5

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sympy [A]  time = 0.11, size = 24, normalized size = 1.33 \begin {gather*} - 51 x^{5} + 102 x^{4} e^{x} - 51 x^{3} e^{2 x} + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-102*x**3-153*x**2)*exp(x)**2+(102*x**4+408*x**3)*exp(x)-255*x**4+1,x)

[Out]

-51*x**5 + 102*x**4*exp(x) - 51*x**3*exp(2*x) + x

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