3.36.89 \(\int \frac {e^x (4 x^2+3 x^3)+e^x (-4 x^2-4 x^3-3 x^4) \log (x)+(-12-18 x) \log ^2(x)}{(48 x^2+72 x^3+27 x^4) \log ^2(x)} \, dx\)

Optimal. Leaf size=40 \[ \frac {2 x-\frac {2-x \left (\frac {3}{x}-\frac {e^x x}{3 \log (x)}\right )}{4+3 x}}{x} \]

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Rubi [A]  time = 0.92, antiderivative size = 43, normalized size of antiderivative = 1.08, number of steps used = 7, number of rules used = 6, integrand size = 71, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.085, Rules used = {1594, 27, 12, 6742, 74, 2288} \begin {gather*} \frac {1}{x (3 x+4)}-\frac {e^x \left (3 x^2 \log (x)+4 x \log (x)\right )}{3 (3 x+4)^2 \log ^2(x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^x*(4*x^2 + 3*x^3) + E^x*(-4*x^2 - 4*x^3 - 3*x^4)*Log[x] + (-12 - 18*x)*Log[x]^2)/((48*x^2 + 72*x^3 + 27
*x^4)*Log[x]^2),x]

[Out]

1/(x*(4 + 3*x)) - (E^x*(4*x*Log[x] + 3*x^2*Log[x]))/(3*(4 + 3*x)^2*Log[x]^2)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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 74

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(c + d*x)
^(n + 1)*(e + f*x)^(p + 1))/(d*f*(n + p + 2)), x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2, 0] &
& EqQ[a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)), 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 2288

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = (v*y)/(Log[F]*D[u, x])}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, 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 {e^x \left (4 x^2+3 x^3\right )+e^x \left (-4 x^2-4 x^3-3 x^4\right ) \log (x)+(-12-18 x) \log ^2(x)}{x^2 \left (48+72 x+27 x^2\right ) \log ^2(x)} \, dx\\ &=\int \frac {e^x \left (4 x^2+3 x^3\right )+e^x \left (-4 x^2-4 x^3-3 x^4\right ) \log (x)+(-12-18 x) \log ^2(x)}{3 x^2 (4+3 x)^2 \log ^2(x)} \, dx\\ &=\frac {1}{3} \int \frac {e^x \left (4 x^2+3 x^3\right )+e^x \left (-4 x^2-4 x^3-3 x^4\right ) \log (x)+(-12-18 x) \log ^2(x)}{x^2 (4+3 x)^2 \log ^2(x)} \, dx\\ &=\frac {1}{3} \int \left (-\frac {6 (2+3 x)}{x^2 (4+3 x)^2}-\frac {e^x \left (-4-3 x+4 \log (x)+4 x \log (x)+3 x^2 \log (x)\right )}{(4+3 x)^2 \log ^2(x)}\right ) \, dx\\ &=-\left (\frac {1}{3} \int \frac {e^x \left (-4-3 x+4 \log (x)+4 x \log (x)+3 x^2 \log (x)\right )}{(4+3 x)^2 \log ^2(x)} \, dx\right )-2 \int \frac {2+3 x}{x^2 (4+3 x)^2} \, dx\\ &=\frac {1}{x (4+3 x)}-\frac {e^x \left (4 x \log (x)+3 x^2 \log (x)\right )}{3 (4+3 x)^2 \log ^2(x)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.17, size = 31, normalized size = 0.78 \begin {gather*} -\frac {e^x x^2-3 \log (x)}{3 \left (4 x \log (x)+3 x^2 \log (x)\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^x*(4*x^2 + 3*x^3) + E^x*(-4*x^2 - 4*x^3 - 3*x^4)*Log[x] + (-12 - 18*x)*Log[x]^2)/((48*x^2 + 72*x^
3 + 27*x^4)*Log[x]^2),x]

[Out]

-1/3*(E^x*x^2 - 3*Log[x])/(4*x*Log[x] + 3*x^2*Log[x])

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fricas [A]  time = 1.10, size = 28, normalized size = 0.70 \begin {gather*} -\frac {x^{2} e^{x} - 3 \, \log \relax (x)}{3 \, {\left (3 \, x^{2} + 4 \, x\right )} \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-18*x-12)*log(x)^2+(-3*x^4-4*x^3-4*x^2)*exp(x)*log(x)+(3*x^3+4*x^2)*exp(x))/(27*x^4+72*x^3+48*x^2)
/log(x)^2,x, algorithm="fricas")

[Out]

-1/3*(x^2*e^x - 3*log(x))/((3*x^2 + 4*x)*log(x))

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giac [A]  time = 0.24, size = 28, normalized size = 0.70 \begin {gather*} -\frac {x^{2} e^{x} - 3 \, \log \relax (x)}{3 \, {\left (3 \, x^{2} \log \relax (x) + 4 \, x \log \relax (x)\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-18*x-12)*log(x)^2+(-3*x^4-4*x^3-4*x^2)*exp(x)*log(x)+(3*x^3+4*x^2)*exp(x))/(27*x^4+72*x^3+48*x^2)
/log(x)^2,x, algorithm="giac")

[Out]

-1/3*(x^2*e^x - 3*log(x))/(3*x^2*log(x) + 4*x*log(x))

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maple [A]  time = 0.04, size = 29, normalized size = 0.72

method	result	size
risch	\(\frac {1}{\left (4+3 x \right ) x}-\frac {x \,{\mathrm e}^{x}}{3 \left (4+3 x \right ) \ln \relax (x )}\)	\(29\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-18*x-12)*ln(x)^2+(-3*x^4-4*x^3-4*x^2)*exp(x)*ln(x)+(3*x^3+4*x^2)*exp(x))/(27*x^4+72*x^3+48*x^2)/ln(x)^2
,x,method=_RETURNVERBOSE)

[Out]

1/(4+3*x)/x-1/3*x*exp(x)/(4+3*x)/ln(x)

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maxima [A]  time = 0.48, size = 28, normalized size = 0.70 \begin {gather*} -\frac {x^{2} e^{x} - 3 \, \log \relax (x)}{3 \, {\left (3 \, x^{2} + 4 \, x\right )} \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-18*x-12)*log(x)^2+(-3*x^4-4*x^3-4*x^2)*exp(x)*log(x)+(3*x^3+4*x^2)*exp(x))/(27*x^4+72*x^3+48*x^2)
/log(x)^2,x, algorithm="maxima")

[Out]

-1/3*(x^2*e^x - 3*log(x))/((3*x^2 + 4*x)*log(x))

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mupad [B]  time = 2.40, size = 28, normalized size = 0.70 \begin {gather*} \frac {3\,\ln \relax (x)-x^2\,{\mathrm {e}}^x}{3\,x\,\ln \relax (x)\,\left (3\,x+4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(x)^2*(18*x + 12) - exp(x)*(4*x^2 + 3*x^3) + exp(x)*log(x)*(4*x^2 + 4*x^3 + 3*x^4))/(log(x)^2*(48*x^2
+ 72*x^3 + 27*x^4)),x)

[Out]

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

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sympy [A]  time = 0.35, size = 26, normalized size = 0.65 \begin {gather*} - \frac {x e^{x}}{9 x \log {\relax (x )} + 12 \log {\relax (x )}} + \frac {1}{3 x^{2} + 4 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-18*x-12)*ln(x)**2+(-3*x**4-4*x**3-4*x**2)*exp(x)*ln(x)+(3*x**3+4*x**2)*exp(x))/(27*x**4+72*x**3+4
8*x**2)/ln(x)**2,x)

[Out]

-x*exp(x)/(9*x*log(x) + 12*log(x)) + 1/(3*x**2 + 4*x)

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