3.99.32 \(\int \frac {24+e^x (-24-12 x)+12 x+e^2 (24+12 x)+e^{-3+2 x} (24+12 x)+(-12 x-12 e^2 x+e^x (-12 x-12 x^2)+e^{-3+2 x} (36 x+24 x^2)) \log (x)}{4 x+4 x^2+x^3} \, dx\)

Optimal. Leaf size=26 \[ \frac {12 \left (1+e^2-e^x+e^{-3+2 x}\right ) \log (x)}{2+x} \]

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Rubi [B]  time = 2.22, antiderivative size = 81, normalized size of antiderivative = 3.12, number of steps used = 33, number of rules used = 11, integrand size = 94, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.117, Rules used = {1594, 27, 6742, 44, 36, 29, 31, 2314, 2178, 2197, 2554} \begin {gather*} -\frac {12 e^x \log (x)}{x+2}+\frac {12 e^{2 x-3} \log (x)}{x+2}-\frac {6 \left (1+e^2\right ) x \log (x)}{x+2}+6 e^2 \log (x)+6 \log (x)+6 \left (1+e^2\right ) \log (x+2)-6 e^2 \log (x+2)-6 \log (x+2) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(24 + E^x*(-24 - 12*x) + 12*x + E^2*(24 + 12*x) + E^(-3 + 2*x)*(24 + 12*x) + (-12*x - 12*E^2*x + E^x*(-12*
x - 12*x^2) + E^(-3 + 2*x)*(36*x + 24*x^2))*Log[x])/(4*x + 4*x^2 + x^3),x]

[Out]

6*Log[x] + 6*E^2*Log[x] - (12*E^x*Log[x])/(2 + x) + (12*E^(-3 + 2*x)*Log[x])/(2 + x) - (6*(1 + E^2)*x*Log[x])/
(2 + x) - 6*Log[2 + x] - 6*E^2*Log[2 + x] + 6*(1 + E^2)*Log[2 + x]

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

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 1594

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^
(q - p) + c*x^(r - p))^n, x] /; FreeQ[{a, b, c, p, q, r}, x] && IntegerQ[n] && PosQ[q - p] && PosQ[r - p]

Rule 2178

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(F^(g*(e - (c*f)/d))*ExpIntegral
Ei[(f*g*(c + d*x)*Log[F])/d])/d, x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !$UseGamma === True

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 2314

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_) + (e_.)*(x_)^(r_.))^(q_), x_Symbol] :> Simp[(x*(d + e*x^r)^(q
+ 1)*(a + b*Log[c*x^n]))/d, x] - Dist[(b*n)/d, Int[(d + e*x^r)^(q + 1), x], x] /; FreeQ[{a, b, c, d, e, n, q,
r}, x] && EqQ[r*(q + 1) + 1, 0]

Rule 2554

Int[Log[u_]*(v_), x_Symbol] :> With[{w = IntHide[v, x]}, Dist[Log[u], w, x] - Int[SimplifyIntegrand[(w*D[u, x]
)/u, x], x] /; InverseFunctionFreeQ[w, x]] /; InverseFunctionFreeQ[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 {24+e^x (-24-12 x)+12 x+e^2 (24+12 x)+e^{-3+2 x} (24+12 x)+\left (-12 x-12 e^2 x+e^x \left (-12 x-12 x^2\right )+e^{-3+2 x} \left (36 x+24 x^2\right )\right ) \log (x)}{x \left (4+4 x+x^2\right )} \, dx\\ &=\int \frac {24+e^x (-24-12 x)+12 x+e^2 (24+12 x)+e^{-3+2 x} (24+12 x)+\left (-12 x-12 e^2 x+e^x \left (-12 x-12 x^2\right )+e^{-3+2 x} \left (36 x+24 x^2\right )\right ) \log (x)}{x (2+x)^2} \, dx\\ &=\int \left (\frac {12}{(2+x)^2}+\frac {24}{x (2+x)^2}+\frac {12 e^2}{x (2+x)}-\frac {12 \left (1+e^2\right ) \log (x)}{(2+x)^2}-\frac {12 e^x \left (2+x+x \log (x)+x^2 \log (x)\right )}{x (2+x)^2}+\frac {12 e^{-3+2 x} \left (2+x+3 x \log (x)+2 x^2 \log (x)\right )}{x (2+x)^2}\right ) \, dx\\ &=-\frac {12}{2+x}-12 \int \frac {e^x \left (2+x+x \log (x)+x^2 \log (x)\right )}{x (2+x)^2} \, dx+12 \int \frac {e^{-3+2 x} \left (2+x+3 x \log (x)+2 x^2 \log (x)\right )}{x (2+x)^2} \, dx+24 \int \frac {1}{x (2+x)^2} \, dx+\left (12 e^2\right ) \int \frac {1}{x (2+x)} \, dx-\left (12 \left (1+e^2\right )\right ) \int \frac {\log (x)}{(2+x)^2} \, dx\\ &=-\frac {12}{2+x}-\frac {6 \left (1+e^2\right ) x \log (x)}{2+x}-12 \int \left (\frac {e^x}{x (2+x)}+\frac {e^x (1+x) \log (x)}{(2+x)^2}\right ) \, dx+12 \int \left (\frac {e^{-3+2 x}}{x (2+x)}+\frac {e^{-3+2 x} (3+2 x) \log (x)}{(2+x)^2}\right ) \, dx+24 \int \left (\frac {1}{4 x}-\frac {1}{2 (2+x)^2}-\frac {1}{4 (2+x)}\right ) \, dx+\left (6 e^2\right ) \int \frac {1}{x} \, dx-\left (6 e^2\right ) \int \frac {1}{2+x} \, dx+\left (6 \left (1+e^2\right )\right ) \int \frac {1}{2+x} \, dx\\ &=6 \log (x)+6 e^2 \log (x)-\frac {6 \left (1+e^2\right ) x \log (x)}{2+x}-6 \log (2+x)-6 e^2 \log (2+x)+6 \left (1+e^2\right ) \log (2+x)-12 \int \frac {e^x}{x (2+x)} \, dx+12 \int \frac {e^{-3+2 x}}{x (2+x)} \, dx-12 \int \frac {e^x (1+x) \log (x)}{(2+x)^2} \, dx+12 \int \frac {e^{-3+2 x} (3+2 x) \log (x)}{(2+x)^2} \, dx\\ &=6 \log (x)+6 e^2 \log (x)-\frac {12 e^x \log (x)}{2+x}+\frac {12 e^{-3+2 x} \log (x)}{2+x}-\frac {6 \left (1+e^2\right ) x \log (x)}{2+x}-6 \log (2+x)-6 e^2 \log (2+x)+6 \left (1+e^2\right ) \log (2+x)+12 \int \frac {e^x}{x (2+x)} \, dx-12 \int \frac {e^{-3+2 x}}{x (2+x)} \, dx-12 \int \left (\frac {e^x}{2 x}-\frac {e^x}{2 (2+x)}\right ) \, dx+12 \int \left (\frac {e^{-3+2 x}}{2 x}-\frac {e^{-3+2 x}}{2 (2+x)}\right ) \, dx\\ &=6 \log (x)+6 e^2 \log (x)-\frac {12 e^x \log (x)}{2+x}+\frac {12 e^{-3+2 x} \log (x)}{2+x}-\frac {6 \left (1+e^2\right ) x \log (x)}{2+x}-6 \log (2+x)-6 e^2 \log (2+x)+6 \left (1+e^2\right ) \log (2+x)-6 \int \frac {e^x}{x} \, dx+6 \int \frac {e^{-3+2 x}}{x} \, dx+6 \int \frac {e^x}{2+x} \, dx-6 \int \frac {e^{-3+2 x}}{2+x} \, dx+12 \int \left (\frac {e^x}{2 x}-\frac {e^x}{2 (2+x)}\right ) \, dx-12 \int \left (\frac {e^{-3+2 x}}{2 x}-\frac {e^{-3+2 x}}{2 (2+x)}\right ) \, dx\\ &=-6 \text {Ei}(x)+\frac {6 \text {Ei}(2 x)}{e^3}+\frac {6 \text {Ei}(2+x)}{e^2}-\frac {6 \text {Ei}(2 (2+x))}{e^7}+6 \log (x)+6 e^2 \log (x)-\frac {12 e^x \log (x)}{2+x}+\frac {12 e^{-3+2 x} \log (x)}{2+x}-\frac {6 \left (1+e^2\right ) x \log (x)}{2+x}-6 \log (2+x)-6 e^2 \log (2+x)+6 \left (1+e^2\right ) \log (2+x)+6 \int \frac {e^x}{x} \, dx-6 \int \frac {e^{-3+2 x}}{x} \, dx-6 \int \frac {e^x}{2+x} \, dx+6 \int \frac {e^{-3+2 x}}{2+x} \, dx\\ &=6 \log (x)+6 e^2 \log (x)-\frac {12 e^x \log (x)}{2+x}+\frac {12 e^{-3+2 x} \log (x)}{2+x}-\frac {6 \left (1+e^2\right ) x \log (x)}{2+x}-6 \log (2+x)-6 e^2 \log (2+x)+6 \left (1+e^2\right ) \log (2+x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.45, size = 31, normalized size = 1.19 \begin {gather*} \frac {12 \left (e^3+e^5+e^{2 x}-e^{3+x}\right ) \log (x)}{e^3 (2+x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(24 + E^x*(-24 - 12*x) + 12*x + E^2*(24 + 12*x) + E^(-3 + 2*x)*(24 + 12*x) + (-12*x - 12*E^2*x + E^x
*(-12*x - 12*x^2) + E^(-3 + 2*x)*(36*x + 24*x^2))*Log[x])/(4*x + 4*x^2 + x^3),x]

[Out]

(12*(E^3 + E^5 + E^(2*x) - E^(3 + x))*Log[x])/(E^3*(2 + x))

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fricas [A]  time = 0.95, size = 26, normalized size = 1.00 \begin {gather*} \frac {12 \, {\left (e^{5} + e^{3} + e^{\left (2 \, x\right )} - e^{\left (x + 3\right )}\right )} e^{\left (-3\right )} \log \relax (x)}{x + 2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((24*x^2+36*x)*exp(2*x-3)+(-12*x^2-12*x)*exp(x)-12*exp(2)*x-12*x)*log(x)+(12*x+24)*exp(2*x-3)+(-12*
x-24)*exp(x)+(12*x+24)*exp(2)+12*x+24)/(x^3+4*x^2+4*x),x, algorithm="fricas")

[Out]

12*(e^5 + e^3 + e^(2*x) - e^(x + 3))*e^(-3)*log(x)/(x + 2)

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giac [A]  time = 0.18, size = 39, normalized size = 1.50 \begin {gather*} \frac {12 \, {\left (e^{5} \log \relax (x) + e^{3} \log \relax (x) + e^{\left (2 \, x\right )} \log \relax (x) - e^{\left (x + 3\right )} \log \relax (x)\right )}}{x e^{3} + 2 \, e^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((24*x^2+36*x)*exp(2*x-3)+(-12*x^2-12*x)*exp(x)-12*exp(2)*x-12*x)*log(x)+(12*x+24)*exp(2*x-3)+(-12*
x-24)*exp(x)+(12*x+24)*exp(2)+12*x+24)/(x^3+4*x^2+4*x),x, algorithm="giac")

[Out]

12*(e^5*log(x) + e^3*log(x) + e^(2*x)*log(x) - e^(x + 3)*log(x))/(x*e^3 + 2*e^3)

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maple [A]  time = 0.12, size = 28, normalized size = 1.08

method	result	size
risch	\(\frac {12 \left ({\mathrm e}^{2 x -5}-{\mathrm e}^{x -2}+{\mathrm e}^{-2}+1\right ) {\mathrm e}^{2} \ln \relax (x )}{2+x}\)	\(28\)
default	\(\frac {-24 \,{\mathrm e}^{x} \ln \relax (x )-12 x \,{\mathrm e}^{x} \ln \relax (x )}{\left (2+x \right )^{2}}+\frac {24 \ln \relax (x ) {\mathrm e}^{2 x -3}+12 \ln \relax (x ) {\mathrm e}^{2 x -3} x}{\left (2+x \right )^{2}}+6 \,{\mathrm e}^{2} \ln \relax (x )+6 \ln \relax (x )-\frac {6 \ln \relax (x ) x}{2+x}-\frac {6 \,{\mathrm e}^{2} \ln \relax (x ) x}{2+x}\)	\(82\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((24*x^2+36*x)*exp(2*x-3)+(-12*x^2-12*x)*exp(x)-12*exp(2)*x-12*x)*ln(x)+(12*x+24)*exp(2*x-3)+(-12*x-24)*e
xp(x)+(12*x+24)*exp(2)+12*x+24)/(x^3+4*x^2+4*x),x,method=_RETURNVERBOSE)

[Out]

12*(exp(2*x-5)-exp(x-2)+exp(-2)+1)*exp(2)/(2+x)*ln(x)

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maxima [B]  time = 0.46, size = 95, normalized size = 3.65 \begin {gather*} 6 \, {\left (\frac {2}{x + 2} - \log \left (x + 2\right ) + \log \relax (x)\right )} e^{2} + 6 \, {\left (e^{2} + 1\right )} \log \left (x + 2\right ) - 6 \, {\left (e^{2} + 1\right )} \log \relax (x) + \frac {12 \, {\left ({\left (e^{5} + e^{3}\right )} \log \relax (x) + e^{\left (2 \, x\right )} \log \relax (x) - e^{\left (x + 3\right )} \log \relax (x)\right )}}{x e^{3} + 2 \, e^{3}} - \frac {12 \, e^{2}}{x + 2} - 6 \, \log \left (x + 2\right ) + 6 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((24*x^2+36*x)*exp(2*x-3)+(-12*x^2-12*x)*exp(x)-12*exp(2)*x-12*x)*log(x)+(12*x+24)*exp(2*x-3)+(-12*
x-24)*exp(x)+(12*x+24)*exp(2)+12*x+24)/(x^3+4*x^2+4*x),x, algorithm="maxima")

[Out]

6*(2/(x + 2) - log(x + 2) + log(x))*e^2 + 6*(e^2 + 1)*log(x + 2) - 6*(e^2 + 1)*log(x) + 12*((e^5 + e^3)*log(x)
+ e^(2*x)*log(x) - e^(x + 3)*log(x))/(x*e^3 + 2*e^3) - 12*e^2/(x + 2) - 6*log(x + 2) + 6*log(x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int \frac {12\,x+{\mathrm {e}}^{2\,x-3}\,\left (12\,x+24\right )-{\mathrm {e}}^x\,\left (12\,x+24\right )-\ln \relax (x)\,\left (12\,x+12\,x\,{\mathrm {e}}^2-{\mathrm {e}}^{2\,x-3}\,\left (24\,x^2+36\,x\right )+{\mathrm {e}}^x\,\left (12\,x^2+12\,x\right )\right )+{\mathrm {e}}^2\,\left (12\,x+24\right )+24}{x^3+4\,x^2+4\,x} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((12*x + exp(2*x - 3)*(12*x + 24) - exp(x)*(12*x + 24) - log(x)*(12*x + 12*x*exp(2) - exp(2*x - 3)*(36*x +
24*x^2) + exp(x)*(12*x + 12*x^2)) + exp(2)*(12*x + 24) + 24)/(4*x + 4*x^2 + x^3),x)

[Out]

int((12*x + exp(2*x - 3)*(12*x + 24) - exp(x)*(12*x + 24) - log(x)*(12*x + 12*x*exp(2) - exp(2*x - 3)*(36*x +
24*x^2) + exp(x)*(12*x + 12*x^2)) + exp(2)*(12*x + 24) + 24)/(4*x + 4*x^2 + x^3), x)

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sympy [B]  time = 0.49, size = 71, normalized size = 2.73 \begin {gather*} \frac {\left (12 x \log {\relax (x )} + 24 \log {\relax (x )}\right ) e^{2 x} + \left (- 12 x e^{3} \log {\relax (x )} - 24 e^{3} \log {\relax (x )}\right ) e^{x}}{x^{2} e^{3} + 4 x e^{3} + 4 e^{3}} + \frac {\left (12 + 12 e^{2}\right ) \log {\relax (x )}}{x + 2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((24*x**2+36*x)*exp(2*x-3)+(-12*x**2-12*x)*exp(x)-12*exp(2)*x-12*x)*ln(x)+(12*x+24)*exp(2*x-3)+(-12
*x-24)*exp(x)+(12*x+24)*exp(2)+12*x+24)/(x**3+4*x**2+4*x),x)

[Out]

((12*x*log(x) + 24*log(x))*exp(2*x) + (-12*x*exp(3)*log(x) - 24*exp(3)*log(x))*exp(x))/(x**2*exp(3) + 4*x*exp(
3) + 4*exp(3)) + (12 + 12*exp(2))*log(x)/(x + 2)

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