3.14.99 \(\int \frac {e^{-16+10 x-2 x^2} (7-x) (-69+38 x-4 x^2)+e^x (-7-6 x+x^2)}{-63+9 x} \, dx\)

Optimal. Leaf size=33 \[ \frac {1}{9} e^x \left (e^{-(4-x)^2+x-x^2} (7-x)+x\right ) \]

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Rubi [A]  time = 0.36, antiderivative size = 51, normalized size of antiderivative = 1.55, number of steps used = 17, number of rules used = 7, integrand size = 49, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {6742, 2176, 2194, 2234, 2205, 2240, 2241} \begin {gather*} -\frac {1}{9} e^{-2 x^2+10 x-16} x+\frac {7}{9} e^{-2 x^2+10 x-16}-\frac {e^x}{9}+\frac {1}{9} e^x (x+1) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(-16 + 10*x - 2*x^2)*(7 - x)*(-69 + 38*x - 4*x^2) + E^x*(-7 - 6*x + x^2))/(-63 + 9*x),x]

[Out]

-1/9*E^x + (7*E^(-16 + 10*x - 2*x^2))/9 - (E^(-16 + 10*x - 2*x^2)*x)/9 + (E^x*(1 + x))/9

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 2205

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^2), x_Symbol] :> Simp[(F^a*Sqrt[Pi]*Erf[(c + d*x)*Rt[-(b*Log[F]),
2]])/(2*d*Rt[-(b*Log[F]), 2]), x] /; FreeQ[{F, a, b, c, d}, x] && NegQ[b]

Rule 2234

Int[(F_)^((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[F^(a - b^2/(4*c)), Int[F^((b + 2*c*x)^2/(4*c))
, x], x] /; FreeQ[{F, a, b, c}, x]

Rule 2240

Int[(F_)^((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)*((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[(e*F^(a + b*x + c*x^2))/(
2*c*Log[F]), x] - Dist[(b*e - 2*c*d)/(2*c), Int[F^(a + b*x + c*x^2), x], x] /; FreeQ[{F, a, b, c, d, e}, x] &&
NeQ[b*e - 2*c*d, 0]

Rule 2241

Int[(F_)^((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)*((d_.) + (e_.)*(x_))^(m_), x_Symbol] :> Simp[(e*(d + e*x)^(m - 1)
*F^(a + b*x + c*x^2))/(2*c*Log[F]), x] + (-Dist[(b*e - 2*c*d)/(2*c), Int[(d + e*x)^(m - 1)*F^(a + b*x + c*x^2)
, x], x] - Dist[((m - 1)*e^2)/(2*c*Log[F]), Int[(d + e*x)^(m - 2)*F^(a + b*x + c*x^2), x], x]) /; FreeQ[{F, a,
b, c, d, e}, x] && NeQ[b*e - 2*c*d, 0] && GtQ[m, 1]

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (\frac {1}{9} e^x (1+x)+\frac {1}{9} e^{-16+10 x-2 x^2} \left (69-38 x+4 x^2\right )\right ) \, dx\\ &=\frac {1}{9} \int e^x (1+x) \, dx+\frac {1}{9} \int e^{-16+10 x-2 x^2} \left (69-38 x+4 x^2\right ) \, dx\\ &=\frac {1}{9} e^x (1+x)-\frac {\int e^x \, dx}{9}+\frac {1}{9} \int \left (69 e^{-16+10 x-2 x^2}-38 e^{-16+10 x-2 x^2} x+4 e^{-16+10 x-2 x^2} x^2\right ) \, dx\\ &=-\frac {e^x}{9}+\frac {1}{9} e^x (1+x)+\frac {4}{9} \int e^{-16+10 x-2 x^2} x^2 \, dx-\frac {38}{9} \int e^{-16+10 x-2 x^2} x \, dx+\frac {23}{3} \int e^{-16+10 x-2 x^2} \, dx\\ &=-\frac {e^x}{9}+\frac {19}{18} e^{-16+10 x-2 x^2}-\frac {1}{9} e^{-16+10 x-2 x^2} x+\frac {1}{9} e^x (1+x)+\frac {1}{9} \int e^{-16+10 x-2 x^2} \, dx+\frac {10}{9} \int e^{-16+10 x-2 x^2} x \, dx-\frac {95}{9} \int e^{-16+10 x-2 x^2} \, dx+\frac {23 \int e^{-\frac {1}{8} (10-4 x)^2} \, dx}{3 e^{7/2}}\\ &=-\frac {e^x}{9}+\frac {7}{9} e^{-16+10 x-2 x^2}-\frac {1}{9} e^{-16+10 x-2 x^2} x+\frac {1}{9} e^x (1+x)-\frac {23 \sqrt {\frac {\pi }{2}} \text {erf}\left (\frac {5-2 x}{\sqrt {2}}\right )}{6 e^{7/2}}+\frac {25}{9} \int e^{-16+10 x-2 x^2} \, dx+\frac {\int e^{-\frac {1}{8} (10-4 x)^2} \, dx}{9 e^{7/2}}-\frac {95 \int e^{-\frac {1}{8} (10-4 x)^2} \, dx}{9 e^{7/2}}\\ &=-\frac {e^x}{9}+\frac {7}{9} e^{-16+10 x-2 x^2}-\frac {1}{9} e^{-16+10 x-2 x^2} x+\frac {1}{9} e^x (1+x)+\frac {25 \sqrt {\frac {\pi }{2}} \text {erf}\left (\frac {5-2 x}{\sqrt {2}}\right )}{18 e^{7/2}}+\frac {25 \int e^{-\frac {1}{8} (10-4 x)^2} \, dx}{9 e^{7/2}}\\ &=-\frac {e^x}{9}+\frac {7}{9} e^{-16+10 x-2 x^2}-\frac {1}{9} e^{-16+10 x-2 x^2} x+\frac {1}{9} e^x (1+x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.14, size = 36, normalized size = 1.09 \begin {gather*} \frac {1}{9} e^{-16+x-2 x^2} \left (-e^{9 x} (-7+x)+e^{2 \left (8+x^2\right )} x\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(-16 + 10*x - 2*x^2)*(7 - x)*(-69 + 38*x - 4*x^2) + E^x*(-7 - 6*x + x^2))/(-63 + 9*x),x]

[Out]

(E^(-16 + x - 2*x^2)*(-(E^(9*x)*(-7 + x)) + E^(2*(8 + x^2))*x))/9

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fricas [A]  time = 0.68, size = 25, normalized size = 0.76 \begin {gather*} \frac {1}{9} \, x e^{x} + \frac {1}{9} \, e^{\left (-2 \, x^{2} + 10 \, x + \log \left (-x + 7\right ) - 16\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x^2+38*x-69)*exp(x)*exp(log(-x+7)-2*x^2+9*x-16)+(x^2-6*x-7)*exp(x))/(9*x-63),x, algorithm="fric
as")

[Out]

1/9*x*e^x + 1/9*e^(-2*x^2 + 10*x + log(-x + 7) - 16)

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giac [A]  time = 0.21, size = 36, normalized size = 1.09 \begin {gather*} -\frac {1}{9} \, {\left (x e^{\left (-2 \, x^{2} + 10 \, x\right )} - x e^{\left (x + 16\right )} - 7 \, e^{\left (-2 \, x^{2} + 10 \, x\right )}\right )} e^{\left (-16\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x^2+38*x-69)*exp(x)*exp(log(-x+7)-2*x^2+9*x-16)+(x^2-6*x-7)*exp(x))/(9*x-63),x, algorithm="giac
")

[Out]

-1/9*(x*e^(-2*x^2 + 10*x) - x*e^(x + 16) - 7*e^(-2*x^2 + 10*x))*e^(-16)

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maple [A]  time = 0.28, size = 25, normalized size = 0.76

method	result	size
risch	\(\frac {{\mathrm e}^{x} x}{9}+\frac {\left (-x +7\right ) {\mathrm e}^{-2 x^{2}+10 x -16}}{9}\)	\(25\)
norman	\(\frac {{\mathrm e}^{x} x}{9}+\frac {{\mathrm e}^{x} {\mathrm e}^{\ln \left (-x +7\right )-2 x^{2}+9 x -16}}{9}\)	\(28\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-4*x^2+38*x-69)*exp(x)*exp(ln(-x+7)-2*x^2+9*x-16)+(x^2-6*x-7)*exp(x))/(9*x-63),x,method=_RETURNVERBOSE)

[Out]

1/9*exp(x)*x+1/9*(-x+7)*exp(-2*x^2+10*x-16)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} -\frac {1}{9} \, {\left ({\left (x - 7\right )} e^{\left (-2 \, x^{2} + 10 \, x\right )} - x e^{\left (x + 16\right )}\right )} e^{\left (-16\right )} + \frac {7}{9} \, e^{7} E_{1}\left (-x + 7\right ) + \frac {7}{9} \, \int \frac {e^{x}}{x - 7}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x^2+38*x-69)*exp(x)*exp(log(-x+7)-2*x^2+9*x-16)+(x^2-6*x-7)*exp(x))/(9*x-63),x, algorithm="maxi
ma")

[Out]

-1/9*((x - 7)*e^(-2*x^2 + 10*x) - x*e^(x + 16))*e^(-16) + 7/9*e^7*exp_integral_e(1, -x + 7) + 7/9*integrate(e^
x/(x - 7), x)

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mupad [B]  time = 1.10, size = 35, normalized size = 1.06 \begin {gather*} \frac {x\,{\mathrm {e}}^x}{9}+\frac {7\,{\mathrm {e}}^{10\,x}\,{\mathrm {e}}^{-16}\,{\mathrm {e}}^{-2\,x^2}}{9}-\frac {x\,{\mathrm {e}}^{10\,x}\,{\mathrm {e}}^{-16}\,{\mathrm {e}}^{-2\,x^2}}{9} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(x)*(6*x - x^2 + 7) + exp(9*x + log(7 - x) - 2*x^2 - 16)*exp(x)*(4*x^2 - 38*x + 69))/(9*x - 63),x)

[Out]

(x*exp(x))/9 + (7*exp(10*x)*exp(-16)*exp(-2*x^2))/9 - (x*exp(10*x)*exp(-16)*exp(-2*x^2))/9

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sympy [A]  time = 0.23, size = 29, normalized size = 0.88 \begin {gather*} \frac {x e^{x}}{9} + \frac {\left (- x e^{x} + 7 e^{x}\right ) e^{- 2 x^{2} + 9 x - 16}}{9} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x**2+38*x-69)*exp(x)*exp(ln(-x+7)-2*x**2+9*x-16)+(x**2-6*x-7)*exp(x))/(9*x-63),x)

[Out]

x*exp(x)/9 + (-x*exp(x) + 7*exp(x))*exp(-2*x**2 + 9*x - 16)/9

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