3.298 \(\int \frac{-35+70 x-4 x^2+2 x^3}{(26-10 x+x^2) (17-2 x+x^2)} \, dx\)

Optimal. Leaf size=49 \[ \frac{1003 \log \left (x^2-10 x+26\right )}{1025}+\frac{22 \log \left (x^2-2 x+17\right )}{1025}-\frac{15033 \tan ^{-1}(5-x)}{1025}-\frac{4607 \tan ^{-1}\left (\frac{x-1}{4}\right )}{4100} \]

[Out]

(-15033*ArcTan[5 - x])/1025 - (4607*ArcTan[(-1 + x)/4])/4100 + (1003*Log[26 - 10*x + x^2])/1025 + (22*Log[17 -
2*x + x^2])/1025

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Rubi [A]  time = 0.156707, antiderivative size = 49, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 5, integrand size = 36, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.139, Rules used = {6728, 634, 618, 204, 628} \[ \frac{1003 \log \left (x^2-10 x+26\right )}{1025}+\frac{22 \log \left (x^2-2 x+17\right )}{1025}-\frac{15033 \tan ^{-1}(5-x)}{1025}-\frac{4607 \tan ^{-1}\left (\frac{x-1}{4}\right )}{4100} \]

Antiderivative was successfully verified.

[In]

Int[(-35 + 70*x - 4*x^2 + 2*x^3)/((26 - 10*x + x^2)*(17 - 2*x + x^2)),x]

[Out]

(-15033*ArcTan[5 - x])/1025 - (4607*ArcTan[(-1 + x)/4])/4100 + (1003*Log[26 - 10*x + x^2])/1025 + (22*Log[17 -
2*x + x^2])/1025

Rule 6728

Int[(u_)/((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a +
b*x^n + c*x^(2*n)), x]}, Int[v, x] /; SumQ[v]] /; FreeQ[{a, b, c}, x] && EqQ[n2, 2*n] && IGtQ[n, 0]

Rule 634

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[(2*c*d - b*e)/(2*c), Int[1/(a +
b*x + c*x^2), x], x] + Dist[e/(2*c), Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] &
& NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] &&  !NiceSqrtQ[b^2 - 4*a*c]

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rubi steps

\begin{align*} \int \frac{-35+70 x-4 x^2+2 x^3}{\left (26-10 x+x^2\right ) \left (17-2 x+x^2\right )} \, dx &=\int \left (\frac{5003+2006 x}{1025 \left (26-10 x+x^2\right )}+\frac{-4651+44 x}{1025 \left (17-2 x+x^2\right )}\right ) \, dx\\ &=\frac{\int \frac{5003+2006 x}{26-10 x+x^2} \, dx}{1025}+\frac{\int \frac{-4651+44 x}{17-2 x+x^2} \, dx}{1025}\\ &=\frac{22 \int \frac{-2+2 x}{17-2 x+x^2} \, dx}{1025}+\frac{1003 \int \frac{-10+2 x}{26-10 x+x^2} \, dx}{1025}-\frac{4607 \int \frac{1}{17-2 x+x^2} \, dx}{1025}+\frac{15033 \int \frac{1}{26-10 x+x^2} \, dx}{1025}\\ &=\frac{1003 \log \left (26-10 x+x^2\right )}{1025}+\frac{22 \log \left (17-2 x+x^2\right )}{1025}+\frac{9214 \operatorname{Subst}\left (\int \frac{1}{-64-x^2} \, dx,x,-2+2 x\right )}{1025}-\frac{30066 \operatorname{Subst}\left (\int \frac{1}{-4-x^2} \, dx,x,-10+2 x\right )}{1025}\\ &=-\frac{15033 \tan ^{-1}(5-x)}{1025}-\frac{4607 \tan ^{-1}\left (\frac{1}{4} (-1+x)\right )}{4100}+\frac{1003 \log \left (26-10 x+x^2\right )}{1025}+\frac{22 \log \left (17-2 x+x^2\right )}{1025}\\ \end{align*}

Mathematica [A]  time = 0.0135432, size = 49, normalized size = 1. \[ \frac{1003 \log \left (x^2-10 x+26\right )}{1025}+\frac{22 \log \left (x^2-2 x+17\right )}{1025}-\frac{15033 \tan ^{-1}(5-x)}{1025}-\frac{4607 \tan ^{-1}\left (\frac{x-1}{4}\right )}{4100} \]

Antiderivative was successfully verified.

[In]

Integrate[(-35 + 70*x - 4*x^2 + 2*x^3)/((26 - 10*x + x^2)*(17 - 2*x + x^2)),x]

[Out]

(-15033*ArcTan[5 - x])/1025 - (4607*ArcTan[(-1 + x)/4])/4100 + (1003*Log[26 - 10*x + x^2])/1025 + (22*Log[17 -
2*x + x^2])/1025

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Maple [A]  time = 0.007, size = 38, normalized size = 0.8 \begin{align*}{\frac{15033\,\arctan \left ( -5+x \right ) }{1025}}-{\frac{4607}{4100}\arctan \left ({\frac{x}{4}}-{\frac{1}{4}} \right ) }+{\frac{1003\,\ln \left ({x}^{2}-10\,x+26 \right ) }{1025}}+{\frac{22\,\ln \left ({x}^{2}-2\,x+17 \right ) }{1025}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*x^3-4*x^2+70*x-35)/(x^2-10*x+26)/(x^2-2*x+17),x)

[Out]

15033/1025*arctan(-5+x)-4607/4100*arctan(1/4*x-1/4)+1003/1025*ln(x^2-10*x+26)+22/1025*ln(x^2-2*x+17)

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Maxima [A]  time = 1.4725, size = 50, normalized size = 1.02 \begin{align*} \frac{15033}{1025} \, \arctan \left (x - 5\right ) - \frac{4607}{4100} \, \arctan \left (\frac{1}{4} \, x - \frac{1}{4}\right ) + \frac{22}{1025} \, \log \left (x^{2} - 2 \, x + 17\right ) + \frac{1003}{1025} \, \log \left (x^{2} - 10 \, x + 26\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x^3-4*x^2+70*x-35)/(x^2-10*x+26)/(x^2-2*x+17),x, algorithm="maxima")

[Out]

15033/1025*arctan(x - 5) - 4607/4100*arctan(1/4*x - 1/4) + 22/1025*log(x^2 - 2*x + 17) + 1003/1025*log(x^2 - 1
0*x + 26)

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Fricas [A]  time = 1.67226, size = 163, normalized size = 3.33 \begin{align*} \frac{15033}{1025} \, \arctan \left (x - 5\right ) - \frac{4607}{4100} \, \arctan \left (\frac{1}{4} \, x - \frac{1}{4}\right ) + \frac{22}{1025} \, \log \left (x^{2} - 2 \, x + 17\right ) + \frac{1003}{1025} \, \log \left (x^{2} - 10 \, x + 26\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x^3-4*x^2+70*x-35)/(x^2-10*x+26)/(x^2-2*x+17),x, algorithm="fricas")

[Out]

15033/1025*arctan(x - 5) - 4607/4100*arctan(1/4*x - 1/4) + 22/1025*log(x^2 - 2*x + 17) + 1003/1025*log(x^2 - 1
0*x + 26)

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Sympy [A]  time = 0.195851, size = 46, normalized size = 0.94 \begin{align*} \frac{1003 \log{\left (x^{2} - 10 x + 26 \right )}}{1025} + \frac{22 \log{\left (x^{2} - 2 x + 17 \right )}}{1025} - \frac{4607 \operatorname{atan}{\left (\frac{x}{4} - \frac{1}{4} \right )}}{4100} + \frac{15033 \operatorname{atan}{\left (x - 5 \right )}}{1025} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x**3-4*x**2+70*x-35)/(x**2-10*x+26)/(x**2-2*x+17),x)

[Out]

1003*log(x**2 - 10*x + 26)/1025 + 22*log(x**2 - 2*x + 17)/1025 - 4607*atan(x/4 - 1/4)/4100 + 15033*atan(x - 5)
/1025

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Giac [A]  time = 1.12058, size = 50, normalized size = 1.02 \begin{align*} \frac{15033}{1025} \, \arctan \left (x - 5\right ) - \frac{4607}{4100} \, \arctan \left (\frac{1}{4} \, x - \frac{1}{4}\right ) + \frac{22}{1025} \, \log \left (x^{2} - 2 \, x + 17\right ) + \frac{1003}{1025} \, \log \left (x^{2} - 10 \, x + 26\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x^3-4*x^2+70*x-35)/(x^2-10*x+26)/(x^2-2*x+17),x, algorithm="giac")

[Out]

15033/1025*arctan(x - 5) - 4607/4100*arctan(1/4*x - 1/4) + 22/1025*log(x^2 - 2*x + 17) + 1003/1025*log(x^2 - 1
0*x + 26)