3.34.33 \(\int \frac {e^{-x} (15-10 x-5 x^2+e^5 (3+x) (42-19 x-16 x^2))}{3+x} \, dx\)

Optimal. Leaf size=23 \[ e^{-x} x \left (5+e^5 \left (16+\frac {3}{x}\right ) (3+x)\right ) \]

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Rubi [B]  time = 0.08, antiderivative size = 73, normalized size of antiderivative = 3.17, number of steps used = 9, number of rules used = 4, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {1586, 2196, 2194, 2176} \begin {gather*} 16 e^{5-x} x^2+32 e^{5-x} x+\left (5+19 e^5\right ) e^{-x} x+32 e^{5-x}-\left (5+42 e^5\right ) e^{-x}+\left (5+19 e^5\right ) e^{-x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(15 - 10*x - 5*x^2 + E^5*(3 + x)*(42 - 19*x - 16*x^2))/(E^x*(3 + x)),x]

[Out]

32*E^(5 - x) + (5 + 19*E^5)/E^x - (5 + 42*E^5)/E^x + 32*E^(5 - x)*x + ((5 + 19*E^5)*x)/E^x + 16*E^(5 - x)*x^2

Rule 1586

Int[(u_.)*(Px_)^(p_.)*(Qx_)^(q_.), x_Symbol] :> Int[u*PolynomialQuotient[Px, Qx, x]^p*Qx^(p + q), x] /; FreeQ[
q, x] && PolyQ[Px, x] && PolyQ[Qx, x] && EqQ[PolynomialRemainder[Px, Qx, x], 0] && IntegerQ[p] && LtQ[p*q, 0]

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} &=\int e^{-x} \left (5+42 e^5+\left (-5-19 e^5\right ) x-16 e^5 x^2\right ) \, dx\\ &=\int \left (5 e^{-x} \left (1+\frac {42 e^5}{5}\right )-e^{-x} \left (5+19 e^5\right ) x-16 e^{5-x} x^2\right ) \, dx\\ &=-\left (16 \int e^{5-x} x^2 \, dx\right )+\left (-5-19 e^5\right ) \int e^{-x} x \, dx+\left (5+42 e^5\right ) \int e^{-x} \, dx\\ &=-e^{-x} \left (5+42 e^5\right )+e^{-x} \left (5+19 e^5\right ) x+16 e^{5-x} x^2-32 \int e^{5-x} x \, dx+\left (-5-19 e^5\right ) \int e^{-x} \, dx\\ &=e^{-x} \left (5+19 e^5\right )-e^{-x} \left (5+42 e^5\right )+32 e^{5-x} x+e^{-x} \left (5+19 e^5\right ) x+16 e^{5-x} x^2-32 \int e^{5-x} \, dx\\ &=32 e^{5-x}+e^{-x} \left (5+19 e^5\right )-e^{-x} \left (5+42 e^5\right )+32 e^{5-x} x+e^{-x} \left (5+19 e^5\right ) x+16 e^{5-x} x^2\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.05, size = 24, normalized size = 1.04 \begin {gather*} e^{-x} \left (5 x+e^5 \left (9+51 x+16 x^2\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(15 - 10*x - 5*x^2 + E^5*(3 + x)*(42 - 19*x - 16*x^2))/(E^x*(3 + x)),x]

[Out]

(5*x + E^5*(9 + 51*x + 16*x^2))/E^x

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fricas [A]  time = 0.51, size = 22, normalized size = 0.96 \begin {gather*} {\left ({\left (16 \, x^{2} + 51 \, x + 9\right )} e^{5} + 5 \, x\right )} e^{\left (-x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-16*x^2-19*x+42)*exp(log(3+x)+5)-5*x^2-10*x+15)/(3+x)/exp(x),x, algorithm="fricas")

[Out]

((16*x^2 + 51*x + 9)*e^5 + 5*x)*e^(-x)

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giac [A]  time = 0.15, size = 36, normalized size = 1.57 \begin {gather*} 16 \, x^{2} e^{\left (-x + 5\right )} + 5 \, x e^{\left (-x\right )} + 51 \, x e^{\left (-x + 5\right )} + 9 \, e^{\left (-x + 5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-16*x^2-19*x+42)*exp(log(3+x)+5)-5*x^2-10*x+15)/(3+x)/exp(x),x, algorithm="giac")

[Out]

16*x^2*e^(-x + 5) + 5*x*e^(-x) + 51*x*e^(-x + 5) + 9*e^(-x + 5)

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maple [A]  time = 0.44, size = 26, normalized size = 1.13

method	result	size
norman	\(\left (\left (5+51 \,{\mathrm e}^{5}\right ) x +16 x^{2} {\mathrm e}^{5}+9 \,{\mathrm e}^{5}\right ) {\mathrm e}^{-x}\)	\(26\)
risch	\(\left (16 x^{2} {\mathrm e}^{5}+51 x \,{\mathrm e}^{5}+9 \,{\mathrm e}^{5}+5 x \right ) {\mathrm e}^{-x}\)	\(26\)
default	\(5 x \,{\mathrm e}^{-x}-126 \,{\mathrm e}^{5} {\mathrm e}^{3} \expIntegralEi \left (1, 3+x \right )-15 \,{\mathrm e}^{5} \left (-{\mathrm e}^{-x}+3 \,{\mathrm e}^{3} \expIntegralEi \left (1, 3+x \right )\right )+67 \,{\mathrm e}^{5} \left (x \,{\mathrm e}^{-x}-2 \,{\mathrm e}^{-x}+9 \,{\mathrm e}^{3} \expIntegralEi \left (1, 3+x \right )\right )-16 \,{\mathrm e}^{5} \left (-x^{2} {\mathrm e}^{-x}+x \,{\mathrm e}^{-x}-8 \,{\mathrm e}^{-x}+27 \,{\mathrm e}^{3} \expIntegralEi \left (1, 3+x \right )\right )\)	\(101\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-16*x^2-19*x+42)*exp(ln(3+x)+5)-5*x^2-10*x+15)/(3+x)/exp(x),x,method=_RETURNVERBOSE)

[Out]

((5+51*exp(5))*x+16*x^2*exp(5)+9*exp(5))/exp(x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} -126 \, e^{8} E_{1}\left (x + 3\right ) - 15 \, e^{3} E_{1}\left (x + 3\right ) + \frac {{\left (16 \, x^{3} e^{5} + x^{2} {\left (99 \, e^{5} + 5\right )} + 3 \, x {\left (54 \, e^{5} + 5\right )}\right )} e^{\left (-x\right )}}{x + 3} - \int \frac {3 \, {\left (x {\left (51 \, e^{5} + 5\right )} + 162 \, e^{5} + 15\right )} e^{\left (-x\right )}}{x^{2} + 6 \, x + 9}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-16*x^2-19*x+42)*exp(log(3+x)+5)-5*x^2-10*x+15)/(3+x)/exp(x),x, algorithm="maxima")

[Out]

-126*e^8*exp_integral_e(1, x + 3) - 15*e^3*exp_integral_e(1, x + 3) + (16*x^3*e^5 + x^2*(99*e^5 + 5) + 3*x*(54
*e^5 + 5))*e^(-x)/(x + 3) - integrate(3*(x*(51*e^5 + 5) + 162*e^5 + 15)*e^(-x)/(x^2 + 6*x + 9), x)

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mupad [B]  time = 0.08, size = 25, normalized size = 1.09 \begin {gather*} {\mathrm {e}}^{-x}\,\left (5\,x+9\,{\mathrm {e}}^5+51\,x\,{\mathrm {e}}^5+16\,x^2\,{\mathrm {e}}^5\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(-x)*(10*x + exp(log(x + 3) + 5)*(19*x + 16*x^2 - 42) + 5*x^2 - 15))/(x + 3),x)

[Out]

exp(-x)*(5*x + 9*exp(5) + 51*x*exp(5) + 16*x^2*exp(5))

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sympy [A]  time = 0.15, size = 26, normalized size = 1.13 \begin {gather*} \left (16 x^{2} e^{5} + 5 x + 51 x e^{5} + 9 e^{5}\right ) e^{- x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-16*x**2-19*x+42)*exp(ln(3+x)+5)-5*x**2-10*x+15)/(3+x)/exp(x),x)

[Out]

(16*x**2*exp(5) + 5*x + 51*x*exp(5) + 9*exp(5))*exp(-x)

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