3.90.12 \(\int \frac {9-6 x+x^2+e^{2 x} (-24 x^2-20 x^3+8 x^4) \log (2)}{9 x-6 x^2+x^3} \, dx\)

Optimal. Leaf size=22 \[ -\frac {4 e^{2 x} x^2 \log (2)}{3-x}+\log (x) \]

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Rubi [A]  time = 0.53, antiderivative size = 38, normalized size of antiderivative = 1.73, number of steps used = 12, number of rules used = 8, integrand size = 47, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.170, Rules used = {1594, 27, 6742, 2199, 2194, 2177, 2178, 2176} \begin {gather*} 4 e^{2 x} x \log (2)+\log (x)+12 e^{2 x} \log (2)-\frac {36 e^{2 x} \log (2)}{3-x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(9 - 6*x + x^2 + E^(2*x)*(-24*x^2 - 20*x^3 + 8*x^4)*Log[2])/(9*x - 6*x^2 + x^3),x]

[Out]

12*E^(2*x)*Log[2] - (36*E^(2*x)*Log[2])/(3 - x) + 4*E^(2*x)*x*Log[2] + Log[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 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 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 2177

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_), x_Symbol] :> Simp[((c + d*x)^(m
+ 1)*(b*F^(g*(e + f*x)))^n)/(d*(m + 1)), x] - Dist[(f*g*n*Log[F])/(d*(m + 1)), 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] && LtQ[m, -1] && IntegerQ[2*m] &&  !$UseGamma ===
True

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 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 2199

Int[(F_)^((c_.)*(v_))*(u_)^(m_.)*(w_), x_Symbol] :> Int[ExpandIntegrand[F^(c*ExpandToSum[v, x]), w*NormalizePo
werOfLinear[u, x]^m, x], x] /; FreeQ[{F, c}, x] && PolynomialQ[w, x] && LinearQ[v, x] && PowerOfLinearQ[u, x]
&& IntegerQ[m] &&  !$UseGamma === True

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 {9-6 x+x^2+e^{2 x} \left (-24 x^2-20 x^3+8 x^4\right ) \log (2)}{x \left (9-6 x+x^2\right )} \, dx\\ &=\int \frac {9-6 x+x^2+e^{2 x} \left (-24 x^2-20 x^3+8 x^4\right ) \log (2)}{(-3+x)^2 x} \, dx\\ &=\int \left (\frac {1}{x}+\frac {4 e^{2 x} x \left (-6-5 x+2 x^2\right ) \log (2)}{(-3+x)^2}\right ) \, dx\\ &=\log (x)+(4 \log (2)) \int \frac {e^{2 x} x \left (-6-5 x+2 x^2\right )}{(-3+x)^2} \, dx\\ &=\log (x)+(4 \log (2)) \int \left (7 e^{2 x}-\frac {9 e^{2 x}}{(-3+x)^2}+\frac {18 e^{2 x}}{-3+x}+2 e^{2 x} x\right ) \, dx\\ &=\log (x)+(8 \log (2)) \int e^{2 x} x \, dx+(28 \log (2)) \int e^{2 x} \, dx-(36 \log (2)) \int \frac {e^{2 x}}{(-3+x)^2} \, dx+(72 \log (2)) \int \frac {e^{2 x}}{-3+x} \, dx\\ &=14 e^{2 x} \log (2)-\frac {36 e^{2 x} \log (2)}{3-x}+4 e^{2 x} x \log (2)+72 e^6 \text {Ei}(-2 (3-x)) \log (2)+\log (x)-(4 \log (2)) \int e^{2 x} \, dx-(72 \log (2)) \int \frac {e^{2 x}}{-3+x} \, dx\\ &=12 e^{2 x} \log (2)-\frac {36 e^{2 x} \log (2)}{3-x}+4 e^{2 x} x \log (2)+\log (x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.13, size = 19, normalized size = 0.86 \begin {gather*} \frac {e^{2 x} x^2 \log (16)}{-3+x}+\log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(9 - 6*x + x^2 + E^(2*x)*(-24*x^2 - 20*x^3 + 8*x^4)*Log[2])/(9*x - 6*x^2 + x^3),x]

[Out]

(E^(2*x)*x^2*Log[16])/(-3 + x) + Log[x]

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fricas [A]  time = 0.51, size = 24, normalized size = 1.09 \begin {gather*} \frac {4 \, x^{2} e^{\left (2 \, x\right )} \log \relax (2) + {\left (x - 3\right )} \log \relax (x)}{x - 3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((8*x^4-20*x^3-24*x^2)*log(2)*exp(x)^2+x^2-6*x+9)/(x^3-6*x^2+9*x),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.15, size = 26, normalized size = 1.18 \begin {gather*} \frac {4 \, x^{2} e^{\left (2 \, x\right )} \log \relax (2) + x \log \relax (x) - 3 \, \log \relax (x)}{x - 3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((8*x^4-20*x^3-24*x^2)*log(2)*exp(x)^2+x^2-6*x+9)/(x^3-6*x^2+9*x),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.07, size = 20, normalized size = 0.91

method	result	size
norman	\(\frac {4 x^{2} \ln \relax (2) {\mathrm e}^{2 x}}{x -3}+\ln \relax (x )\)	\(20\)
risch	\(\frac {4 x^{2} \ln \relax (2) {\mathrm e}^{2 x}}{x -3}+\ln \relax (x )\)	\(20\)
default	\(\ln \relax (x )+\frac {36 \ln \relax (2) {\mathrm e}^{2 x}}{x -3}+12 \ln \relax (2) {\mathrm e}^{2 x}+4 \ln \relax (2) {\mathrm e}^{2 x} x\)	\(34\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((8*x^4-20*x^3-24*x^2)*ln(2)*exp(x)^2+x^2-6*x+9)/(x^3-6*x^2+9*x),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.48, size = 19, normalized size = 0.86 \begin {gather*} \frac {4 \, x^{2} e^{\left (2 \, x\right )} \log \relax (2)}{x - 3} + \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((8*x^4-20*x^3-24*x^2)*log(2)*exp(x)^2+x^2-6*x+9)/(x^3-6*x^2+9*x),x, algorithm="maxima")

[Out]

4*x^2*e^(2*x)*log(2)/(x - 3) + log(x)

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mupad [B]  time = 6.55, size = 19, normalized size = 0.86 \begin {gather*} \ln \relax (x)+\frac {4\,x^2\,{\mathrm {e}}^{2\,x}\,\ln \relax (2)}{x-3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(6*x - x^2 + exp(2*x)*log(2)*(24*x^2 + 20*x^3 - 8*x^4) - 9)/(9*x - 6*x^2 + x^3),x)

[Out]

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

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sympy [A]  time = 0.15, size = 19, normalized size = 0.86 \begin {gather*} \frac {4 x^{2} e^{2 x} \log {\relax (2 )}}{x - 3} + \log {\relax (x )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((8*x**4-20*x**3-24*x**2)*ln(2)*exp(x)**2+x**2-6*x+9)/(x**3-6*x**2+9*x),x)

[Out]

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

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