3.48.13 \(\int \frac {1}{256} (e^{2 x} (-9-18 x)+512 x-27 x^2+e^x (-36 x-18 x^2)) \, dx\)

Optimal. Leaf size=21 \[ -e^{16}+x \left (x-\frac {9}{256} \left (e^x+x\right )^2\right ) \]

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Rubi [B]  time = 0.07, antiderivative size = 44, normalized size of antiderivative = 2.10, number of steps used = 12, number of rules used = 5, integrand size = 37, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.135, Rules used = {12, 2176, 2194, 1593, 2196} \begin {gather*} -\frac {9 x^3}{256}-\frac {9 e^x x^2}{128}+x^2+\frac {9 e^{2 x}}{512}-\frac {9}{512} e^{2 x} (2 x+1) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(2*x)*(-9 - 18*x) + 512*x - 27*x^2 + E^x*(-36*x - 18*x^2))/256,x]

[Out]

(9*E^(2*x))/512 + x^2 - (9*E^x*x^2)/128 - (9*x^3)/256 - (9*E^(2*x)*(1 + 2*x))/512

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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} &=\frac {1}{256} \int \left (e^{2 x} (-9-18 x)+512 x-27 x^2+e^x \left (-36 x-18 x^2\right )\right ) \, dx\\ &=x^2-\frac {9 x^3}{256}+\frac {1}{256} \int e^{2 x} (-9-18 x) \, dx+\frac {1}{256} \int e^x \left (-36 x-18 x^2\right ) \, dx\\ &=x^2-\frac {9 x^3}{256}-\frac {9}{512} e^{2 x} (1+2 x)+\frac {1}{256} \int e^x (-36-18 x) x \, dx+\frac {9}{256} \int e^{2 x} \, dx\\ &=\frac {9 e^{2 x}}{512}+x^2-\frac {9 x^3}{256}-\frac {9}{512} e^{2 x} (1+2 x)+\frac {1}{256} \int \left (-36 e^x x-18 e^x x^2\right ) \, dx\\ &=\frac {9 e^{2 x}}{512}+x^2-\frac {9 x^3}{256}-\frac {9}{512} e^{2 x} (1+2 x)-\frac {9}{128} \int e^x x^2 \, dx-\frac {9}{64} \int e^x x \, dx\\ &=\frac {9 e^{2 x}}{512}-\frac {9 e^x x}{64}+x^2-\frac {9 e^x x^2}{128}-\frac {9 x^3}{256}-\frac {9}{512} e^{2 x} (1+2 x)+\frac {9 \int e^x \, dx}{64}+\frac {9}{64} \int e^x x \, dx\\ &=\frac {9 e^x}{64}+\frac {9 e^{2 x}}{512}+x^2-\frac {9 e^x x^2}{128}-\frac {9 x^3}{256}-\frac {9}{512} e^{2 x} (1+2 x)-\frac {9 \int e^x \, dx}{64}\\ &=\frac {9 e^{2 x}}{512}+x^2-\frac {9 e^x x^2}{128}-\frac {9 x^3}{256}-\frac {9}{512} e^{2 x} (1+2 x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 31, normalized size = 1.48 \begin {gather*} \frac {1}{256} \left (-9 e^{2 x} x+256 x^2-18 e^x x^2-9 x^3\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(2*x)*(-9 - 18*x) + 512*x - 27*x^2 + E^x*(-36*x - 18*x^2))/256,x]

[Out]

(-9*E^(2*x)*x + 256*x^2 - 18*E^x*x^2 - 9*x^3)/256

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fricas [A]  time = 0.79, size = 23, normalized size = 1.10 \begin {gather*} -\frac {9}{256} \, x^{3} - \frac {9}{128} \, x^{2} e^{x} + x^{2} - \frac {9}{256} \, x e^{\left (2 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/256*(-18*x-9)*exp(x)^2+1/256*(-18*x^2-36*x)*exp(x)-27/256*x^2+2*x,x, algorithm="fricas")

[Out]

-9/256*x^3 - 9/128*x^2*e^x + x^2 - 9/256*x*e^(2*x)

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giac [A]  time = 0.24, size = 23, normalized size = 1.10 \begin {gather*} -\frac {9}{256} \, x^{3} - \frac {9}{128} \, x^{2} e^{x} + x^{2} - \frac {9}{256} \, x e^{\left (2 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/256*(-18*x-9)*exp(x)^2+1/256*(-18*x^2-36*x)*exp(x)-27/256*x^2+2*x,x, algorithm="giac")

[Out]

-9/256*x^3 - 9/128*x^2*e^x + x^2 - 9/256*x*e^(2*x)

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maple [A]  time = 0.04, size = 24, normalized size = 1.14

method	result	size
default	\(x^{2}-\frac {9 x^{3}}{256}-\frac {9 x \,{\mathrm e}^{2 x}}{256}-\frac {9 \,{\mathrm e}^{x} x^{2}}{128}\)	\(24\)
norman	\(x^{2}-\frac {9 x^{3}}{256}-\frac {9 x \,{\mathrm e}^{2 x}}{256}-\frac {9 \,{\mathrm e}^{x} x^{2}}{128}\)	\(24\)
risch	\(x^{2}-\frac {9 x^{3}}{256}-\frac {9 x \,{\mathrm e}^{2 x}}{256}-\frac {9 \,{\mathrm e}^{x} x^{2}}{128}\)	\(24\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/256*(-18*x-9)*exp(x)^2+1/256*(-18*x^2-36*x)*exp(x)-27/256*x^2+2*x,x,method=_RETURNVERBOSE)

[Out]

x^2-9/256*x^3-9/256*x*exp(x)^2-9/128*exp(x)*x^2

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maxima [A]  time = 0.36, size = 23, normalized size = 1.10 \begin {gather*} -\frac {9}{256} \, x^{3} - \frac {9}{128} \, x^{2} e^{x} + x^{2} - \frac {9}{256} \, x e^{\left (2 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/256*(-18*x-9)*exp(x)^2+1/256*(-18*x^2-36*x)*exp(x)-27/256*x^2+2*x,x, algorithm="maxima")

[Out]

-9/256*x^3 - 9/128*x^2*e^x + x^2 - 9/256*x*e^(2*x)

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mupad [B]  time = 3.16, size = 23, normalized size = 1.10 \begin {gather*} -\frac {x\,\left (9\,{\mathrm {e}}^{2\,x}-256\,x+18\,x\,{\mathrm {e}}^x+9\,x^2\right )}{256} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(2*x - (exp(x)*(36*x + 18*x^2))/256 - (exp(2*x)*(18*x + 9))/256 - (27*x^2)/256,x)

[Out]

-(x*(9*exp(2*x) - 256*x + 18*x*exp(x) + 9*x^2))/256

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sympy [A]  time = 0.11, size = 29, normalized size = 1.38 \begin {gather*} - \frac {9 x^{3}}{256} - \frac {9 x^{2} e^{x}}{128} + x^{2} - \frac {9 x e^{2 x}}{256} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/256*(-18*x-9)*exp(x)**2+1/256*(-18*x**2-36*x)*exp(x)-27/256*x**2+2*x,x)

[Out]

-9*x**3/256 - 9*x**2*exp(x)/128 + x**2 - 9*x*exp(2*x)/256

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