3.103.16 \(\int \frac {8+10 x+e^{9+x} (4+5 x) (-9 x-5 x^2)}{4 x+5 x^2} \, dx\)

Optimal. Leaf size=18 \[ 1-e^{9+x} (4+5 x)+\log \left (x^2\right ) \]

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Rubi [A]  time = 0.19, antiderivative size = 24, normalized size of antiderivative = 1.33, number of steps used = 5, number of rules used = 4, integrand size = 37, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.108, Rules used = {1593, 6688, 2176, 2194} \begin {gather*} -e^{x+9} (5 x+9)+5 e^{x+9}+2 \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(8 + 10*x + E^(9 + x)*(4 + 5*x)*(-9*x - 5*x^2))/(4*x + 5*x^2),x]

[Out]

5*E^(9 + x) - E^(9 + x)*(9 + 5*x) + 2*Log[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 6688

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {8+10 x+e^{9+x} (4+5 x) \left (-9 x-5 x^2\right )}{x (4+5 x)} \, dx\\ &=\int \left (\frac {2}{x}-e^{9+x} (9+5 x)\right ) \, dx\\ &=2 \log (x)-\int e^{9+x} (9+5 x) \, dx\\ &=-e^{9+x} (9+5 x)+2 \log (x)+5 \int e^{9+x} \, dx\\ &=5 e^{9+x}-e^{9+x} (9+5 x)+2 \log (x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 22, normalized size = 1.22 \begin {gather*} -e^x \left (4 e^9+5 e^9 x\right )+2 \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(8 + 10*x + E^(9 + x)*(4 + 5*x)*(-9*x - 5*x^2))/(4*x + 5*x^2),x]

[Out]

-(E^x*(4*E^9 + 5*E^9*x)) + 2*Log[x]

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fricas [A]  time = 1.12, size = 17, normalized size = 0.94 \begin {gather*} -e^{\left (x + \log \left (5 \, x + 4\right ) + 9\right )} + 2 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-5*x^2-9*x)*exp(log(4+5*x)+x+9)+10*x+8)/(5*x^2+4*x),x, algorithm="fricas")

[Out]

-e^(x + log(5*x + 4) + 9) + 2*log(x)

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giac [A]  time = 0.24, size = 18, normalized size = 1.00 \begin {gather*} -5 \, x e^{\left (x + 9\right )} - 4 \, e^{\left (x + 9\right )} + 2 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-5*x^2-9*x)*exp(log(4+5*x)+x+9)+10*x+8)/(5*x^2+4*x),x, algorithm="giac")

[Out]

-5*x*e^(x + 9) - 4*e^(x + 9) + 2*log(x)

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maple [A]  time = 0.08, size = 16, normalized size = 0.89

method	result	size
risch	\(2 \ln \relax (x )+\left (-5 x -4\right ) {\mathrm e}^{x +9}\)	\(16\)
norman	\(-{\mathrm e}^{\ln \left (4+5 x \right )+x +9}+2 \ln \relax (x )\)	\(18\)
default	\(2 \ln \relax (x )-5 \,{\mathrm e}^{x +9} \left (x +9\right )+41 \,{\mathrm e}^{x +9}\)	\(21\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-5*x^2-9*x)*exp(ln(4+5*x)+x+9)+10*x+8)/(5*x^2+4*x),x,method=_RETURNVERBOSE)

[Out]

2*ln(x)+(-5*x-4)*exp(x+9)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {36}{5} \, e^{\frac {41}{5}} E_{1}\left (-x - \frac {4}{5}\right ) - \frac {5 \, {\left (5 \, x^{2} e^{9} + 8 \, x e^{9}\right )} e^{x}}{5 \, x + 4} + \int \frac {20 \, {\left (5 \, x e^{9} + 8 \, e^{9}\right )} e^{x}}{25 \, x^{2} + 40 \, x + 16}\,{d x} + 2 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-5*x^2-9*x)*exp(log(4+5*x)+x+9)+10*x+8)/(5*x^2+4*x),x, algorithm="maxima")

[Out]

36/5*e^(41/5)*exp_integral_e(1, -x - 4/5) - 5*(5*x^2*e^9 + 8*x*e^9)*e^x/(5*x + 4) + integrate(20*(5*x*e^9 + 8*
e^9)*e^x/(25*x^2 + 40*x + 16), x) + 2*log(x)

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mupad [B]  time = 6.85, size = 18, normalized size = 1.00 \begin {gather*} 2\,\ln \relax (x)-4\,{\mathrm {e}}^{x+9}-5\,x\,{\mathrm {e}}^{x+9} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((10*x - exp(x + log(5*x + 4) + 9)*(9*x + 5*x^2) + 8)/(4*x + 5*x^2),x)

[Out]

2*log(x) - 4*exp(x + 9) - 5*x*exp(x + 9)

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sympy [A]  time = 0.11, size = 15, normalized size = 0.83 \begin {gather*} \left (- 5 x - 4\right ) e^{x + 9} + 2 \log {\relax (x )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-5*x**2-9*x)*exp(ln(4+5*x)+x+9)+10*x+8)/(5*x**2+4*x),x)

[Out]

(-5*x - 4)*exp(x + 9) + 2*log(x)

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