3.45.15 \(\int (15-6 x+e^x (-5-3 x+x^2)) \, dx\)

Optimal. Leaf size=10 \[ \left (-3+e^x\right ) (-5+x) x \]

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Rubi [B]  time = 0.04, antiderivative size = 22, normalized size of antiderivative = 2.20, number of steps used = 9, number of rules used = 3, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {2196, 2194, 2176} \begin {gather*} e^x x^2-3 x^2-5 e^x x+15 x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[15 - 6*x + E^x*(-5 - 3*x + x^2),x]

[Out]

15*x - 5*E^x*x - 3*x^2 + E^x*x^2

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} &=15 x-3 x^2+\int e^x \left (-5-3 x+x^2\right ) \, dx\\ &=15 x-3 x^2+\int \left (-5 e^x-3 e^x x+e^x x^2\right ) \, dx\\ &=15 x-3 x^2-3 \int e^x x \, dx-5 \int e^x \, dx+\int e^x x^2 \, dx\\ &=-5 e^x+15 x-3 e^x x-3 x^2+e^x x^2-2 \int e^x x \, dx+3 \int e^x \, dx\\ &=-2 e^x+15 x-5 e^x x-3 x^2+e^x x^2+2 \int e^x \, dx\\ &=15 x-5 e^x x-3 x^2+e^x x^2\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 10, normalized size = 1.00 \begin {gather*} \left (-3+e^x\right ) (-5+x) x \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[15 - 6*x + E^x*(-5 - 3*x + x^2),x]

[Out]

(-3 + E^x)*(-5 + x)*x

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fricas [B]  time = 0.55, size = 19, normalized size = 1.90 \begin {gather*} -3 \, x^{2} + {\left (x^{2} - 5 \, x\right )} e^{x} + 15 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2-3*x-5)*exp(x)-6*x+15,x, algorithm="fricas")

[Out]

-3*x^2 + (x^2 - 5*x)*e^x + 15*x

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giac [B]  time = 0.15, size = 19, normalized size = 1.90 \begin {gather*} -3 \, x^{2} + {\left (x^{2} - 5 \, x\right )} e^{x} + 15 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2-3*x-5)*exp(x)-6*x+15,x, algorithm="giac")

[Out]

-3*x^2 + (x^2 - 5*x)*e^x + 15*x

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maple [B]  time = 0.02, size = 20, normalized size = 2.00

method	result	size
risch	\(\left (x^{2}-5 x \right ) {\mathrm e}^{x}-3 x^{2}+15 x\)	\(20\)
default	\(15 x +{\mathrm e}^{x} x^{2}-5 \,{\mathrm e}^{x} x -3 x^{2}\)	\(21\)
norman	\(15 x +{\mathrm e}^{x} x^{2}-5 \,{\mathrm e}^{x} x -3 x^{2}\)	\(21\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2-3*x-5)*exp(x)-6*x+15,x,method=_RETURNVERBOSE)

[Out]

(x^2-5*x)*exp(x)-3*x^2+15*x

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maxima [B]  time = 0.37, size = 19, normalized size = 1.90 \begin {gather*} -3 \, x^{2} + {\left (x^{2} - 5 \, x\right )} e^{x} + 15 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2-3*x-5)*exp(x)-6*x+15,x, algorithm="maxima")

[Out]

-3*x^2 + (x^2 - 5*x)*e^x + 15*x

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mupad [B]  time = 0.07, size = 9, normalized size = 0.90 \begin {gather*} x\,\left ({\mathrm {e}}^x-3\right )\,\left (x-5\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(15 - exp(x)*(3*x - x^2 + 5) - 6*x,x)

[Out]

x*(exp(x) - 3)*(x - 5)

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sympy [B]  time = 0.08, size = 17, normalized size = 1.70 \begin {gather*} - 3 x^{2} + 15 x + \left (x^{2} - 5 x\right ) e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x**2-3*x-5)*exp(x)-6*x+15,x)

[Out]

-3*x**2 + 15*x + (x**2 - 5*x)*exp(x)

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