3.50.35 \(\int \frac {-18+12 x^4-2 x^8+e^x (6 x^4-2 x^8)+e^x (24 x^4+6 x^5-2 x^9) \log (x)}{9 x-6 x^5+x^9} \, dx\)

Optimal. Leaf size=26 \[ \log \left (\frac {1}{9} x^{-2+\frac {2 e^x x}{\frac {3}{x^3}-x}}\right ) \]

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Rubi [A]  time = 0.64, antiderivative size = 35, normalized size of antiderivative = 1.35, number of steps used = 5, number of rules used = 4, integrand size = 64, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {1594, 28, 6742, 2288} \begin {gather*} \frac {2 e^x x^3 \left (3 x \log (x)-x^5 \log (x)\right )}{\left (3-x^4\right )^2}-2 \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-18 + 12*x^4 - 2*x^8 + E^x*(6*x^4 - 2*x^8) + E^x*(24*x^4 + 6*x^5 - 2*x^9)*Log[x])/(9*x - 6*x^5 + x^9),x]

[Out]

-2*Log[x] + (2*E^x*x^3*(3*x*Log[x] - x^5*Log[x]))/(3 - x^4)^2

Rule 28

Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Dist[1/c^p, Int[u*(b/2 + c*x^n)^(2*
p), x], x] /; FreeQ[{a, b, c, n}, x] && EqQ[n2, 2*n] && 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 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 {-18+12 x^4-2 x^8+e^x \left (6 x^4-2 x^8\right )+e^x \left (24 x^4+6 x^5-2 x^9\right ) \log (x)}{x \left (9-6 x^4+x^8\right )} \, dx\\ &=\int \frac {-18+12 x^4-2 x^8+e^x \left (6 x^4-2 x^8\right )+e^x \left (24 x^4+6 x^5-2 x^9\right ) \log (x)}{x \left (-3+x^4\right )^2} \, dx\\ &=\int \left (-\frac {2}{x}-\frac {2 e^x x^3 \left (-3+x^4-12 \log (x)-3 x \log (x)+x^5 \log (x)\right )}{\left (-3+x^4\right )^2}\right ) \, dx\\ &=-2 \log (x)-2 \int \frac {e^x x^3 \left (-3+x^4-12 \log (x)-3 x \log (x)+x^5 \log (x)\right )}{\left (-3+x^4\right )^2} \, dx\\ &=-2 \log (x)+\frac {2 e^x x^3 \left (3 x \log (x)-x^5 \log (x)\right )}{\left (3-x^4\right )^2}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.13, size = 22, normalized size = 0.85 \begin {gather*} -\frac {2 \left (-3+\left (1+e^x\right ) x^4\right ) \log (x)}{-3+x^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-18 + 12*x^4 - 2*x^8 + E^x*(6*x^4 - 2*x^8) + E^x*(24*x^4 + 6*x^5 - 2*x^9)*Log[x])/(9*x - 6*x^5 + x^
9),x]

[Out]

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

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fricas [A]  time = 0.60, size = 22, normalized size = 0.85 \begin {gather*} -\frac {2 \, {\left (x^{4} e^{x} + x^{4} - 3\right )} \log \relax (x)}{x^{4} - 3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x^9+6*x^5+24*x^4)*exp(x)*log(x)+(-2*x^8+6*x^4)*exp(x)-2*x^8+12*x^4-18)/(x^9-6*x^5+9*x),x, algor
ithm="fricas")

[Out]

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

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giac [A]  time = 0.17, size = 28, normalized size = 1.08 \begin {gather*} -\frac {2 \, {\left (x^{4} e^{x} \log \relax (x) + x^{4} \log \relax (x) - 3 \, \log \relax (x)\right )}}{x^{4} - 3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x^9+6*x^5+24*x^4)*exp(x)*log(x)+(-2*x^8+6*x^4)*exp(x)-2*x^8+12*x^4-18)/(x^9-6*x^5+9*x),x, algor
ithm="giac")

[Out]

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

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maple [A]  time = 0.06, size = 22, normalized size = 0.85

method	result	size
risch	\(-\frac {2 x^{4} {\mathrm e}^{x} \ln \relax (x )}{x^{4}-3}-2 \ln \relax (x )\)	\(22\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-2*x^9+6*x^5+24*x^4)*exp(x)*ln(x)+(-2*x^8+6*x^4)*exp(x)-2*x^8+12*x^4-18)/(x^9-6*x^5+9*x),x,method=_RETUR
NVERBOSE)

[Out]

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

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maxima [A]  time = 0.50, size = 21, normalized size = 0.81 \begin {gather*} -\frac {2 \, x^{4} e^{x} \log \relax (x)}{x^{4} - 3} - 2 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x^9+6*x^5+24*x^4)*exp(x)*log(x)+(-2*x^8+6*x^4)*exp(x)-2*x^8+12*x^4-18)/(x^9-6*x^5+9*x),x, algor
ithm="maxima")

[Out]

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

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mupad [B]  time = 4.44, size = 22, normalized size = 0.85 \begin {gather*} -\frac {2\,\ln \relax (x)\,\left (x^4\,{\mathrm {e}}^x+x^4-3\right )}{x^4-3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(x)*(6*x^4 - 2*x^8) + 12*x^4 - 2*x^8 + exp(x)*log(x)*(24*x^4 + 6*x^5 - 2*x^9) - 18)/(9*x - 6*x^5 + x^9
),x)

[Out]

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

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sympy [A]  time = 0.33, size = 22, normalized size = 0.85 \begin {gather*} - \frac {2 x^{4} e^{x} \log {\relax (x )}}{x^{4} - 3} - 2 \log {\relax (x )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x**9+6*x**5+24*x**4)*exp(x)*ln(x)+(-2*x**8+6*x**4)*exp(x)-2*x**8+12*x**4-18)/(x**9-6*x**5+9*x),
x)

[Out]

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

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