3.17.36 \(\int \frac {e^2 (-8-18 x-10 x^2-x^3)+e^2 (8+4 x) \log (\frac {4+2 x}{x})}{32+32 x+10 x^2+x^3} \, dx\)

Optimal. Leaf size=27 \[ \frac {e^2 x \left (x-\log \left (3-\frac {-4+x}{x}\right )\right )}{-4-x} \]

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Rubi [A]  time = 0.27, antiderivative size = 50, normalized size of antiderivative = 1.85, number of steps used = 16, number of rules used = 9, integrand size = 55, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.164, Rules used = {6741, 12, 6742, 44, 77, 88, 2463, 514, 72} \begin {gather*} -e^2 x-\frac {16 e^2}{x+4}-\frac {4 e^2 \log \left (\frac {4}{x}+2\right )}{x+4}-e^2 \log (x)+e^2 \log (x+2) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^2*(-8 - 18*x - 10*x^2 - x^3) + E^2*(8 + 4*x)*Log[(4 + 2*x)/x])/(32 + 32*x + 10*x^2 + x^3),x]

[Out]

-(E^2*x) - (16*E^2)/(4 + x) - (4*E^2*Log[2 + 4/x])/(4 + x) - E^2*Log[x] + E^2*Log[2 + x]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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 72

Int[((e_.) + (f_.)*(x_))^(p_.)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Int[ExpandIntegrand[(
e + f*x)^p/((a + b*x)*(c + d*x)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IntegerQ[p]

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 514

Int[(x_)^(m_.)*((c_) + (d_.)*(x_)^(mn_.))^(q_.)*((a_) + (b_.)*(x_)^(n_.))^(p_.), x_Symbol] :> Int[x^(m - n*q)*
(a + b*x^n)^p*(d + c*x^n)^q, x] /; FreeQ[{a, b, c, d, m, n, p}, x] && EqQ[mn, -n] && IntegerQ[q] && (PosQ[n] |
|  !IntegerQ[p])

Rule 2463

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))*((f_.) + (g_.)*(x_))^(r_.), x_Symbol] :> Simp[((
f + g*x)^(r + 1)*(a + b*Log[c*(d + e*x^n)^p]))/(g*(r + 1)), x] - Dist[(b*e*n*p)/(g*(r + 1)), Int[(x^(n - 1)*(f
+ g*x)^(r + 1))/(d + e*x^n), x], x] /; FreeQ[{a, b, c, d, e, f, g, n, p, r}, x] && (IGtQ[r, 0] || RationalQ[n
]) && NeQ[r, -1]

Rule 6741

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

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 {e^2 \left (-8-18 x-10 x^2-x^3+8 \log \left (2+\frac {4}{x}\right )+4 x \log \left (2+\frac {4}{x}\right )\right )}{32+32 x+10 x^2+x^3} \, dx\\ &=e^2 \int \frac {-8-18 x-10 x^2-x^3+8 \log \left (2+\frac {4}{x}\right )+4 x \log \left (2+\frac {4}{x}\right )}{32+32 x+10 x^2+x^3} \, dx\\ &=e^2 \int \left (-\frac {8}{(2+x) (4+x)^2}-\frac {18 x}{(2+x) (4+x)^2}-\frac {10 x^2}{(2+x) (4+x)^2}-\frac {x^3}{(2+x) (4+x)^2}+\frac {4 \log \left (2+\frac {4}{x}\right )}{(4+x)^2}\right ) \, dx\\ &=-\left (e^2 \int \frac {x^3}{(2+x) (4+x)^2} \, dx\right )+\left (4 e^2\right ) \int \frac {\log \left (2+\frac {4}{x}\right )}{(4+x)^2} \, dx-\left (8 e^2\right ) \int \frac {1}{(2+x) (4+x)^2} \, dx-\left (10 e^2\right ) \int \frac {x^2}{(2+x) (4+x)^2} \, dx-\left (18 e^2\right ) \int \frac {x}{(2+x) (4+x)^2} \, dx\\ &=-\frac {4 e^2 \log \left (2+\frac {4}{x}\right )}{4+x}-e^2 \int \left (1-\frac {2}{2+x}+\frac {32}{(4+x)^2}-\frac {8}{4+x}\right ) \, dx-\left (8 e^2\right ) \int \left (\frac {1}{4 (2+x)}-\frac {1}{2 (4+x)^2}-\frac {1}{4 (4+x)}\right ) \, dx-\left (10 e^2\right ) \int \left (\frac {1}{2+x}-\frac {8}{(4+x)^2}\right ) \, dx-\left (16 e^2\right ) \int \frac {1}{\left (2+\frac {4}{x}\right ) x^2 (4+x)} \, dx-\left (18 e^2\right ) \int \left (-\frac {1}{2 (2+x)}+\frac {2}{(4+x)^2}+\frac {1}{2 (4+x)}\right ) \, dx\\ &=-e^2 x-\frac {16 e^2}{4+x}-\frac {4 e^2 \log \left (2+\frac {4}{x}\right )}{4+x}-e^2 \log (2+x)+e^2 \log (4+x)-\left (16 e^2\right ) \int \frac {1}{x (4+x) (4+2 x)} \, dx\\ &=-e^2 x-\frac {16 e^2}{4+x}-\frac {4 e^2 \log \left (2+\frac {4}{x}\right )}{4+x}-e^2 \log (2+x)+e^2 \log (4+x)-\left (16 e^2\right ) \int \left (\frac {1}{16 x}-\frac {1}{8 (2+x)}+\frac {1}{16 (4+x)}\right ) \, dx\\ &=-e^2 x-\frac {16 e^2}{4+x}-\frac {4 e^2 \log \left (2+\frac {4}{x}\right )}{4+x}-e^2 \log (x)+e^2 \log (2+x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.06, size = 32, normalized size = 1.19 \begin {gather*} -e^2 \left (x+\frac {4 \left (4+\log \left (2+\frac {4}{x}\right )\right )}{4+x}+\log (x)-\log (2+x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^2*(-8 - 18*x - 10*x^2 - x^3) + E^2*(8 + 4*x)*Log[(4 + 2*x)/x])/(32 + 32*x + 10*x^2 + x^3),x]

[Out]

-(E^2*(x + (4*(4 + Log[2 + 4/x]))/(4 + x) + Log[x] - Log[2 + x]))

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fricas [A]  time = 0.76, size = 32, normalized size = 1.19 \begin {gather*} \frac {x e^{2} \log \left (\frac {2 \, {\left (x + 2\right )}}{x}\right ) - {\left (x^{2} + 4 \, x + 16\right )} e^{2}}{x + 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x+8)*exp(2)*log((2*x+4)/x)+(-x^3-10*x^2-18*x-8)*exp(2))/(x^3+10*x^2+32*x+32),x, algorithm="frica
s")

[Out]

(x*e^2*log(2*(x + 2)/x) - (x^2 + 4*x + 16)*e^2)/(x + 4)

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giac [B]  time = 0.41, size = 59, normalized size = 2.19 \begin {gather*} \frac {\frac {{\left (x + 2\right )} e^{2} \log \left (\frac {2 \, {\left (x + 2\right )}}{x}\right )}{x} - e^{2} \log \left (\frac {2 \, {\left (x + 2\right )}}{x}\right ) - 2 \, e^{2}}{\frac {2 \, {\left (x + 2\right )}^{2}}{x^{2}} - \frac {3 \, {\left (x + 2\right )}}{x} + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x+8)*exp(2)*log((2*x+4)/x)+(-x^3-10*x^2-18*x-8)*exp(2))/(x^3+10*x^2+32*x+32),x, algorithm="giac"
)

[Out]

((x + 2)*e^2*log(2*(x + 2)/x)/x - e^2*log(2*(x + 2)/x) - 2*e^2)/(2*(x + 2)^2/x^2 - 3*(x + 2)/x + 1)

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maple [A]  time = 0.11, size = 29, normalized size = 1.07

method	result	size
norman	\(\frac {{\mathrm e}^{2} x \ln \left (\frac {2 x +4}{x}\right )-x^{2} {\mathrm e}^{2}}{4+x}\)	\(29\)
derivativedivides	\(\frac {{\mathrm e}^{2} \ln \left (2+\frac {4}{x}\right ) \left (2+\frac {4}{x}\right )}{\frac {4}{x}+1}-{\mathrm e}^{2} x +\frac {4 \,{\mathrm e}^{2}}{\frac {4}{x}+1}-{\mathrm e}^{2} \ln \left (2+\frac {4}{x}\right )\)	\(59\)
default	\(\frac {{\mathrm e}^{2} \ln \left (2+\frac {4}{x}\right ) \left (2+\frac {4}{x}\right )}{\frac {4}{x}+1}-{\mathrm e}^{2} x +\frac {4 \,{\mathrm e}^{2}}{\frac {4}{x}+1}-{\mathrm e}^{2} \ln \left (2+\frac {4}{x}\right )\)	\(59\)
risch	\(-\frac {4 \,{\mathrm e}^{2} \ln \left (\frac {2 x +4}{x}\right )}{4+x}-\frac {{\mathrm e}^{2} \left (\ln \left (-x \right ) x -\ln \left (-x -2\right ) x +x^{2}+4 \ln \left (-x \right )-4 \ln \left (-x -2\right )+4 x +16\right )}{4+x}\)	\(67\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((4*x+8)*exp(2)*ln((2*x+4)/x)+(-x^3-10*x^2-18*x-8)*exp(2))/(x^3+10*x^2+32*x+32),x,method=_RETURNVERBOSE)

[Out]

(exp(2)*x*ln((2*x+4)/x)-x^2*exp(2))/(4+x)

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maxima [B]  time = 0.68, size = 80, normalized size = 2.96 \begin {gather*} -2 \, {\left (\frac {2}{x + 4} - \log \left (x + 4\right ) + \log \left (x + 2\right )\right )} e^{2} - 2 \, e^{2} \log \left (x + 4\right ) - \frac {x^{2} e^{2} + x e^{2} \log \relax (x) + 4 \, x e^{2} + 4 \, {\left (\log \relax (2) + 3\right )} e^{2} - {\left (3 \, x e^{2} + 8 \, e^{2}\right )} \log \left (x + 2\right )}{x + 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x+8)*exp(2)*log((2*x+4)/x)+(-x^3-10*x^2-18*x-8)*exp(2))/(x^3+10*x^2+32*x+32),x, algorithm="maxim
a")

[Out]

-2*(2/(x + 4) - log(x + 4) + log(x + 2))*e^2 - 2*e^2*log(x + 4) - (x^2*e^2 + x*e^2*log(x) + 4*x*e^2 + 4*(log(2
) + 3)*e^2 - (3*x*e^2 + 8*e^2)*log(x + 2))/(x + 4)

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mupad [B]  time = 1.31, size = 23, normalized size = 0.85 \begin {gather*} -\frac {x\,{\mathrm {e}}^2\,\left (x-\ln \left (\frac {2\,\left (x+2\right )}{x}\right )\right )}{x+4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(2)*(18*x + 10*x^2 + x^3 + 8) - exp(2)*log((2*x + 4)/x)*(4*x + 8))/(32*x + 10*x^2 + x^3 + 32),x)

[Out]

-(x*exp(2)*(x - log((2*(x + 2))/x)))/(x + 4)

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sympy [B]  time = 0.33, size = 44, normalized size = 1.63 \begin {gather*} - x e^{2} - e^{2} \log {\relax (x )} + e^{2} \log {\left (x + 2 \right )} - \frac {4 e^{2} \log {\left (\frac {2 x + 4}{x} \right )}}{x + 4} - \frac {16 e^{2}}{x + 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x+8)*exp(2)*ln((2*x+4)/x)+(-x**3-10*x**2-18*x-8)*exp(2))/(x**3+10*x**2+32*x+32),x)

[Out]

-x*exp(2) - exp(2)*log(x) + exp(2)*log(x + 2) - 4*exp(2)*log((2*x + 4)/x)/(x + 4) - 16*exp(2)/(x + 4)

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