3.36.81 \(\int \frac {-45-9 x-x^2+e^x (32+12 x+x^2)}{324+153 x+22 x^2+x^3} \, dx\)

Optimal. Leaf size=18 \[ \frac {e^x+x}{9+x}-\log (4+x) \]

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Rubi [A]  time = 0.33, antiderivative size = 23, normalized size of antiderivative = 1.28, number of steps used = 10, number of rules used = 6, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {6688, 6742, 44, 77, 88, 2197} \begin {gather*} \frac {e^x}{x+9}-\frac {9}{x+9}-\log (x+4) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-45 - 9*x - x^2 + E^x*(32 + 12*x + x^2))/(324 + 153*x + 22*x^2 + x^3),x]

[Out]

-9/(9 + x) + E^x/(9 + x) - Log[4 + x]

Rule 44

Int[((a_) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*
x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

Rule 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rule 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rule 2197

Int[(F_)^((c_.)*(v_))*(u_)^(m_.)*(w_), x_Symbol] :> With[{b = Coefficient[v, x, 1], d = Coefficient[u, x, 0],
e = Coefficient[u, x, 1], f = Coefficient[w, x, 0], g = Coefficient[w, x, 1]}, Simp[(g*u^(m + 1)*F^(c*v))/(b*c
*e*Log[F]), x] /; EqQ[e*g*(m + 1) - b*c*(e*f - d*g)*Log[F], 0]] /; FreeQ[{F, c, m}, x] && LinearQ[{u, v, w}, x
]

Rule 6688

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

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 \frac {-45-9 x-x^2+e^x \left (32+12 x+x^2\right )}{(4+x) (9+x)^2} \, dx\\ &=\int \left (-\frac {45}{(4+x) (9+x)^2}-\frac {9 x}{(4+x) (9+x)^2}-\frac {x^2}{(4+x) (9+x)^2}+\frac {e^x (8+x)}{(9+x)^2}\right ) \, dx\\ &=-\left (9 \int \frac {x}{(4+x) (9+x)^2} \, dx\right )-45 \int \frac {1}{(4+x) (9+x)^2} \, dx-\int \frac {x^2}{(4+x) (9+x)^2} \, dx+\int \frac {e^x (8+x)}{(9+x)^2} \, dx\\ &=\frac {e^x}{9+x}-9 \int \left (-\frac {4}{25 (4+x)}+\frac {9}{5 (9+x)^2}+\frac {4}{25 (9+x)}\right ) \, dx-45 \int \left (\frac {1}{25 (4+x)}-\frac {1}{5 (9+x)^2}-\frac {1}{25 (9+x)}\right ) \, dx-\int \left (\frac {16}{25 (4+x)}-\frac {81}{5 (9+x)^2}+\frac {9}{25 (9+x)}\right ) \, dx\\ &=-\frac {9}{9+x}+\frac {e^x}{9+x}-\log (4+x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.10, size = 23, normalized size = 1.28 \begin {gather*} -\frac {9}{9+x}+\frac {e^x}{9+x}-\log (4+x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-45 - 9*x - x^2 + E^x*(32 + 12*x + x^2))/(324 + 153*x + 22*x^2 + x^3),x]

[Out]

-9/(9 + x) + E^x/(9 + x) - Log[4 + x]

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fricas [A]  time = 0.55, size = 21, normalized size = 1.17 \begin {gather*} -\frac {{\left (x + 9\right )} \log \left (x + 4\right ) - e^{x} + 9}{x + 9} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^2+12*x+32)*exp(x)-x^2-9*x-45)/(x^3+22*x^2+153*x+324),x, algorithm="fricas")

[Out]

-((x + 9)*log(x + 4) - e^x + 9)/(x + 9)

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giac [A]  time = 0.14, size = 25, normalized size = 1.39 \begin {gather*} -\frac {x \log \left (x + 4\right ) - e^{x} + 9 \, \log \left (x + 4\right ) + 9}{x + 9} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^2+12*x+32)*exp(x)-x^2-9*x-45)/(x^3+22*x^2+153*x+324),x, algorithm="giac")

[Out]

-(x*log(x + 4) - e^x + 9*log(x + 4) + 9)/(x + 9)

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maple [A]  time = 0.06, size = 18, normalized size = 1.00

method	result	size
norman	\(\frac {-9+{\mathrm e}^{x}}{x +9}-\ln \left (4+x \right )\)	\(18\)
default	\(\frac {{\mathrm e}^{x}}{x +9}-\ln \left (4+x \right )-\frac {9}{x +9}\)	\(23\)
risch	\(\frac {{\mathrm e}^{x}}{x +9}-\ln \left (4+x \right )-\frac {9}{x +9}\)	\(23\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x^2+12*x+32)*exp(x)-x^2-9*x-45)/(x^3+22*x^2+153*x+324),x,method=_RETURNVERBOSE)

[Out]

(-9+exp(x))/(x+9)-ln(4+x)

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maxima [A]  time = 0.44, size = 22, normalized size = 1.22 \begin {gather*} \frac {e^{x}}{x + 9} - \frac {9}{x + 9} - \log \left (x + 4\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^2+12*x+32)*exp(x)-x^2-9*x-45)/(x^3+22*x^2+153*x+324),x, algorithm="maxima")

[Out]

e^x/(x + 9) - 9/(x + 9) - log(x + 4)

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mupad [B]  time = 2.20, size = 17, normalized size = 0.94 \begin {gather*} \frac {x+{\mathrm {e}}^x}{x+9}-\ln \left (x+4\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(9*x - exp(x)*(12*x + x^2 + 32) + x^2 + 45)/(153*x + 22*x^2 + x^3 + 324),x)

[Out]

(x + exp(x))/(x + 9) - log(x + 4)

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sympy [A]  time = 0.14, size = 15, normalized size = 0.83 \begin {gather*} - \log {\left (x + 4 \right )} + \frac {e^{x}}{x + 9} - \frac {9}{x + 9} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x**2+12*x+32)*exp(x)-x**2-9*x-45)/(x**3+22*x**2+153*x+324),x)

[Out]

-log(x + 4) + exp(x)/(x + 9) - 9/(x + 9)

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