3.75.54 \(\int \frac {e^{225/x} (2 x-x^9)+e^{225/x} (-450+2 x+225 x^8+7 x^9) \log (x)}{4 x-4 x^9+x^{17}} \, dx\)

Optimal. Leaf size=22 \[ 5+\frac {e^{225/x} x \log (x)}{2-x^8} \]

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Rubi [A]  time = 0.53, antiderivative size = 30, normalized size of antiderivative = 1.36, number of steps used = 4, number of rules used = 4, integrand size = 58, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.069, Rules used = {1594, 28, 6741, 2288} \begin {gather*} \frac {e^{225/x} x \left (2 \log (x)-x^8 \log (x)\right )}{\left (2-x^8\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(225/x)*(2*x - x^9) + E^(225/x)*(-450 + 2*x + 225*x^8 + 7*x^9)*Log[x])/(4*x - 4*x^9 + x^17),x]

[Out]

(E^(225/x)*x*(2*Log[x] - x^8*Log[x]))/(2 - x^8)^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 6741

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{225/x} \left (2 x-x^9\right )+e^{225/x} \left (-450+2 x+225 x^8+7 x^9\right ) \log (x)}{x \left (4-4 x^8+x^{16}\right )} \, dx\\ &=\int \frac {e^{225/x} \left (2 x-x^9\right )+e^{225/x} \left (-450+2 x+225 x^8+7 x^9\right ) \log (x)}{x \left (-2+x^8\right )^2} \, dx\\ &=\int \frac {e^{225/x} \left (2 x-x^9-450 \log (x)+2 x \log (x)+225 x^8 \log (x)+7 x^9 \log (x)\right )}{x \left (2-x^8\right )^2} \, dx\\ &=\frac {e^{225/x} x \left (2 \log (x)-x^8 \log (x)\right )}{\left (2-x^8\right )^2}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 1.08, size = 19, normalized size = 0.86 \begin {gather*} -\frac {e^{225/x} x \log (x)}{-2+x^8} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(225/x)*(2*x - x^9) + E^(225/x)*(-450 + 2*x + 225*x^8 + 7*x^9)*Log[x])/(4*x - 4*x^9 + x^17),x]

[Out]

-((E^(225/x)*x*Log[x])/(-2 + x^8))

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fricas [A]  time = 0.81, size = 18, normalized size = 0.82 \begin {gather*} -\frac {x e^{\frac {225}{x}} \log \relax (x)}{x^{8} - 2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((7*x^9+225*x^8+2*x-450)*exp(225/x)*log(x)+(-x^9+2*x)*exp(225/x))/(x^17-4*x^9+4*x),x, algorithm="fri
cas")

[Out]

-x*e^(225/x)*log(x)/(x^8 - 2)

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((7*x^9+225*x^8+2*x-450)*exp(225/x)*log(x)+(-x^9+2*x)*exp(225/x))/(x^17-4*x^9+4*x),x, algorithm="gia
c")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Evaluation time: 7.21Polynomial exponent overflow. Error: Bad Argument Value

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maple [A]  time = 0.02, size = 19, normalized size = 0.86

method	result	size
risch	\(-\frac {x \,{\mathrm e}^{\frac {225}{x}} \ln \relax (x )}{x^{8}-2}\)	\(19\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((7*x^9+225*x^8+2*x-450)*exp(225/x)*ln(x)+(-x^9+2*x)*exp(225/x))/(x^17-4*x^9+4*x),x,method=_RETURNVERBOSE)

[Out]

-x/(x^8-2)*exp(225/x)*ln(x)

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maxima [A]  time = 0.52, size = 18, normalized size = 0.82 \begin {gather*} -\frac {x e^{\frac {225}{x}} \log \relax (x)}{x^{8} - 2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((7*x^9+225*x^8+2*x-450)*exp(225/x)*log(x)+(-x^9+2*x)*exp(225/x))/(x^17-4*x^9+4*x),x, algorithm="max
ima")

[Out]

-x*e^(225/x)*log(x)/(x^8 - 2)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.05 \begin {gather*} \int \frac {{\mathrm {e}}^{225/x}\,\left (2\,x-x^9\right )+{\mathrm {e}}^{225/x}\,\ln \relax (x)\,\left (7\,x^9+225\,x^8+2\,x-450\right )}{x^{17}-4\,x^9+4\,x} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(225/x)*(2*x - x^9) + exp(225/x)*log(x)*(2*x + 225*x^8 + 7*x^9 - 450))/(4*x - 4*x^9 + x^17),x)

[Out]

int((exp(225/x)*(2*x - x^9) + exp(225/x)*log(x)*(2*x + 225*x^8 + 7*x^9 - 450))/(4*x - 4*x^9 + x^17), x)

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sympy [A]  time = 0.35, size = 15, normalized size = 0.68 \begin {gather*} - \frac {x e^{\frac {225}{x}} \log {\relax (x )}}{x^{8} - 2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((7*x**9+225*x**8+2*x-450)*exp(225/x)*ln(x)+(-x**9+2*x)*exp(225/x))/(x**17-4*x**9+4*x),x)

[Out]

-x*exp(225/x)*log(x)/(x**8 - 2)

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