3.2300 \(\int \frac{a+b x}{(1+x)^3 (1-x+x^2)^3} \, dx\)

Optimal. Leaf size=101 \[ \frac{x (a+b x)}{6 \left (x^3+1\right )^2}+\frac{x (5 a+4 b x)}{18 \left (x^3+1\right )}-\frac{1}{54} (5 a-2 b) \log \left (x^2-x+1\right )+\frac{1}{27} (5 a-2 b) \log (x+1)-\frac{(5 a+2 b) \tan ^{-1}\left (\frac{1-2 x}{\sqrt{3}}\right )}{9 \sqrt{3}} \]

[Out]

(x*(a + b*x))/(6*(1 + x^3)^2) + (x*(5*a + 4*b*x))/(18*(1 + x^3)) - ((5*a + 2*b)*ArcTan[(1 - 2*x)/Sqrt[3]])/(9*
Sqrt[3]) + ((5*a - 2*b)*Log[1 + x])/27 - ((5*a - 2*b)*Log[1 - x + x^2])/54

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Rubi [A]  time = 0.097348, antiderivative size = 101, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 8, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.381, Rules used = {809, 1855, 1860, 31, 634, 618, 204, 628} \[ \frac{x (a+b x)}{6 \left (x^3+1\right )^2}+\frac{x (5 a+4 b x)}{18 \left (x^3+1\right )}-\frac{1}{54} (5 a-2 b) \log \left (x^2-x+1\right )+\frac{1}{27} (5 a-2 b) \log (x+1)-\frac{(5 a+2 b) \tan ^{-1}\left (\frac{1-2 x}{\sqrt{3}}\right )}{9 \sqrt{3}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x)/((1 + x)^3*(1 - x + x^2)^3),x]

[Out]

(x*(a + b*x))/(6*(1 + x^3)^2) + (x*(5*a + 4*b*x))/(18*(1 + x^3)) - ((5*a + 2*b)*ArcTan[(1 - 2*x)/Sqrt[3]])/(9*
Sqrt[3]) + ((5*a - 2*b)*Log[1 + x])/27 - ((5*a - 2*b)*Log[1 - x + x^2])/54

Rule 809

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[
((d + e*x)^FracPart[p]*(a + b*x + c*x^2)^FracPart[p])/(a*d + c*e*x^3)^FracPart[p], Int[(f + g*x)*(a*d + c*e*x^
3)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && EqQ[m, p] && EqQ[b*d + a*e, 0] && EqQ[c*d + b*e, 0]

Rule 1855

Int[(Pq_)*((a_) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> -Simp[(x*Pq*(a + b*x^n)^(p + 1))/(a*n*(p + 1)), x] + Di
st[1/(a*n*(p + 1)), Int[ExpandToSum[n*(p + 1)*Pq + D[x*Pq, x], x]*(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b},
x] && PolyQ[Pq, x] && IGtQ[n, 0] && LtQ[p, -1] && LtQ[Expon[Pq, x], n - 1]

Rule 1860

Int[((A_) + (B_.)*(x_))/((a_) + (b_.)*(x_)^3), x_Symbol] :> With[{r = Numerator[Rt[a/b, 3]], s = Denominator[R
t[a/b, 3]]}, -Dist[(r*(B*r - A*s))/(3*a*s), Int[1/(r + s*x), x], x] + Dist[r/(3*a*s), Int[(r*(B*r + 2*A*s) + s
*(B*r - A*s)*x)/(r^2 - r*s*x + s^2*x^2), x], x]] /; FreeQ[{a, b, A, B}, x] && NeQ[a*B^3 - b*A^3, 0] && PosQ[a/
b]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

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{a+b x}{(1+x)^3 \left (1-x+x^2\right )^3} \, dx &=\int \frac{a+b x}{\left (1+x^3\right )^3} \, dx\\ &=\frac{x (a+b x)}{6 \left (1+x^3\right )^2}-\frac{1}{6} \int \frac{-5 a-4 b x}{\left (1+x^3\right )^2} \, dx\\ &=\frac{x (a+b x)}{6 \left (1+x^3\right )^2}+\frac{x (5 a+4 b x)}{18 \left (1+x^3\right )}+\frac{1}{18} \int \frac{10 a+4 b x}{1+x^3} \, dx\\ &=\frac{x (a+b x)}{6 \left (1+x^3\right )^2}+\frac{x (5 a+4 b x)}{18 \left (1+x^3\right )}+\frac{1}{54} \int \frac{20 a+4 b+(-10 a+4 b) x}{1-x+x^2} \, dx+\frac{1}{27} (5 a-2 b) \int \frac{1}{1+x} \, dx\\ &=\frac{x (a+b x)}{6 \left (1+x^3\right )^2}+\frac{x (5 a+4 b x)}{18 \left (1+x^3\right )}+\frac{1}{27} (5 a-2 b) \log (1+x)+\frac{1}{54} (-5 a+2 b) \int \frac{-1+2 x}{1-x+x^2} \, dx+\frac{1}{18} (5 a+2 b) \int \frac{1}{1-x+x^2} \, dx\\ &=\frac{x (a+b x)}{6 \left (1+x^3\right )^2}+\frac{x (5 a+4 b x)}{18 \left (1+x^3\right )}+\frac{1}{27} (5 a-2 b) \log (1+x)-\frac{1}{54} (5 a-2 b) \log \left (1-x+x^2\right )+\frac{1}{9} (-5 a-2 b) \operatorname{Subst}\left (\int \frac{1}{-3-x^2} \, dx,x,-1+2 x\right )\\ &=\frac{x (a+b x)}{6 \left (1+x^3\right )^2}+\frac{x (5 a+4 b x)}{18 \left (1+x^3\right )}-\frac{(5 a+2 b) \tan ^{-1}\left (\frac{1-2 x}{\sqrt{3}}\right )}{9 \sqrt{3}}+\frac{1}{27} (5 a-2 b) \log (1+x)-\frac{1}{54} (5 a-2 b) \log \left (1-x+x^2\right )\\ \end{align*}

Mathematica [A]  time = 0.0685713, size = 94, normalized size = 0.93 \[ \frac{1}{54} \left (\frac{9 x (a+b x)}{\left (x^3+1\right )^2}+\frac{3 x (5 a+4 b x)}{x^3+1}+(2 b-5 a) \log \left (x^2-x+1\right )+2 (5 a-2 b) \log (x+1)+2 \sqrt{3} (5 a+2 b) \tan ^{-1}\left (\frac{2 x-1}{\sqrt{3}}\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x)/((1 + x)^3*(1 - x + x^2)^3),x]

[Out]

((9*x*(a + b*x))/(1 + x^3)^2 + (3*x*(5*a + 4*b*x))/(1 + x^3) + 2*Sqrt[3]*(5*a + 2*b)*ArcTan[(-1 + 2*x)/Sqrt[3]
] + 2*(5*a - 2*b)*Log[1 + x] + (-5*a + 2*b)*Log[1 - x + x^2])/54

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Maple [A]  time = 0.015, size = 154, normalized size = 1.5 \begin{align*} -{\frac{a}{54\, \left ( 1+x \right ) ^{2}}}+{\frac{b}{54\, \left ( 1+x \right ) ^{2}}}-{\frac{a}{9+9\,x}}+{\frac{2\,b}{27+27\,x}}+{\frac{5\,\ln \left ( 1+x \right ) a}{27}}-{\frac{2\,\ln \left ( 1+x \right ) b}{27}}-{\frac{1}{27\, \left ({x}^{2}-x+1 \right ) ^{2}} \left ( \left ( -3\,a-4\,b \right ){x}^{3}+ \left ( a+{\frac{13\,b}{2}} \right ){x}^{2}+ \left ( -a-8\,b \right ) x-{\frac{7\,a}{2}}+{\frac{5\,b}{2}} \right ) }-{\frac{5\,\ln \left ({x}^{2}-x+1 \right ) a}{54}}+{\frac{\ln \left ({x}^{2}-x+1 \right ) b}{27}}+{\frac{5\,\sqrt{3}a}{27}\arctan \left ({\frac{ \left ( -1+2\,x \right ) \sqrt{3}}{3}} \right ) }+{\frac{2\,b\sqrt{3}}{27}\arctan \left ({\frac{ \left ( -1+2\,x \right ) \sqrt{3}}{3}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)/(1+x)^3/(x^2-x+1)^3,x)

[Out]

-1/54/(1+x)^2*a+1/54/(1+x)^2*b-1/9/(1+x)*a+2/27/(1+x)*b+5/27*ln(1+x)*a-2/27*ln(1+x)*b-1/27*((-3*a-4*b)*x^3+(a+
13/2*b)*x^2+(-a-8*b)*x-7/2*a+5/2*b)/(x^2-x+1)^2-5/54*ln(x^2-x+1)*a+1/27*ln(x^2-x+1)*b+5/27*3^(1/2)*arctan(1/3*
(-1+2*x)*3^(1/2))*a+2/27*3^(1/2)*arctan(1/3*(-1+2*x)*3^(1/2))*b

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Maxima [A]  time = 1.45901, size = 124, normalized size = 1.23 \begin{align*} \frac{1}{27} \, \sqrt{3}{\left (5 \, a + 2 \, b\right )} \arctan \left (\frac{1}{3} \, \sqrt{3}{\left (2 \, x - 1\right )}\right ) - \frac{1}{54} \,{\left (5 \, a - 2 \, b\right )} \log \left (x^{2} - x + 1\right ) + \frac{1}{27} \,{\left (5 \, a - 2 \, b\right )} \log \left (x + 1\right ) + \frac{4 \, b x^{5} + 5 \, a x^{4} + 7 \, b x^{2} + 8 \, a x}{18 \,{\left (x^{6} + 2 \, x^{3} + 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)/(1+x)^3/(x^2-x+1)^3,x, algorithm="maxima")

[Out]

1/27*sqrt(3)*(5*a + 2*b)*arctan(1/3*sqrt(3)*(2*x - 1)) - 1/54*(5*a - 2*b)*log(x^2 - x + 1) + 1/27*(5*a - 2*b)*
log(x + 1) + 1/18*(4*b*x^5 + 5*a*x^4 + 7*b*x^2 + 8*a*x)/(x^6 + 2*x^3 + 1)

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Fricas [A]  time = 1.28145, size = 394, normalized size = 3.9 \begin{align*} \frac{12 \, b x^{5} + 15 \, a x^{4} + 21 \, b x^{2} + 2 \, \sqrt{3}{\left ({\left (5 \, a + 2 \, b\right )} x^{6} + 2 \,{\left (5 \, a + 2 \, b\right )} x^{3} + 5 \, a + 2 \, b\right )} \arctan \left (\frac{1}{3} \, \sqrt{3}{\left (2 \, x - 1\right )}\right ) + 24 \, a x -{\left ({\left (5 \, a - 2 \, b\right )} x^{6} + 2 \,{\left (5 \, a - 2 \, b\right )} x^{3} + 5 \, a - 2 \, b\right )} \log \left (x^{2} - x + 1\right ) + 2 \,{\left ({\left (5 \, a - 2 \, b\right )} x^{6} + 2 \,{\left (5 \, a - 2 \, b\right )} x^{3} + 5 \, a - 2 \, b\right )} \log \left (x + 1\right )}{54 \,{\left (x^{6} + 2 \, x^{3} + 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)/(1+x)^3/(x^2-x+1)^3,x, algorithm="fricas")

[Out]

1/54*(12*b*x^5 + 15*a*x^4 + 21*b*x^2 + 2*sqrt(3)*((5*a + 2*b)*x^6 + 2*(5*a + 2*b)*x^3 + 5*a + 2*b)*arctan(1/3*
sqrt(3)*(2*x - 1)) + 24*a*x - ((5*a - 2*b)*x^6 + 2*(5*a - 2*b)*x^3 + 5*a - 2*b)*log(x^2 - x + 1) + 2*((5*a - 2
*b)*x^6 + 2*(5*a - 2*b)*x^3 + 5*a - 2*b)*log(x + 1))/(x^6 + 2*x^3 + 1)

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Sympy [C]  time = 0.757855, size = 292, normalized size = 2.89 \begin{align*} \frac{\left (5 a - 2 b\right ) \log{\left (x + \frac{25 a^{2} \left (5 a - 2 b\right ) + 40 a b^{2} + 2 b \left (5 a - 2 b\right )^{2}}{125 a^{3} + 8 b^{3}} \right )}}{27} + \left (- \frac{5 a}{54} + \frac{b}{27} - \frac{\sqrt{3} i \left (5 a + 2 b\right )}{54}\right ) \log{\left (x + \frac{675 a^{2} \left (- \frac{5 a}{54} + \frac{b}{27} - \frac{\sqrt{3} i \left (5 a + 2 b\right )}{54}\right ) + 40 a b^{2} + 1458 b \left (- \frac{5 a}{54} + \frac{b}{27} - \frac{\sqrt{3} i \left (5 a + 2 b\right )}{54}\right )^{2}}{125 a^{3} + 8 b^{3}} \right )} + \left (- \frac{5 a}{54} + \frac{b}{27} + \frac{\sqrt{3} i \left (5 a + 2 b\right )}{54}\right ) \log{\left (x + \frac{675 a^{2} \left (- \frac{5 a}{54} + \frac{b}{27} + \frac{\sqrt{3} i \left (5 a + 2 b\right )}{54}\right ) + 40 a b^{2} + 1458 b \left (- \frac{5 a}{54} + \frac{b}{27} + \frac{\sqrt{3} i \left (5 a + 2 b\right )}{54}\right )^{2}}{125 a^{3} + 8 b^{3}} \right )} + \frac{5 a x^{4} + 8 a x + 4 b x^{5} + 7 b x^{2}}{18 x^{6} + 36 x^{3} + 18} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)/(1+x)**3/(x**2-x+1)**3,x)

[Out]

(5*a - 2*b)*log(x + (25*a**2*(5*a - 2*b) + 40*a*b**2 + 2*b*(5*a - 2*b)**2)/(125*a**3 + 8*b**3))/27 + (-5*a/54
+ b/27 - sqrt(3)*I*(5*a + 2*b)/54)*log(x + (675*a**2*(-5*a/54 + b/27 - sqrt(3)*I*(5*a + 2*b)/54) + 40*a*b**2 +
1458*b*(-5*a/54 + b/27 - sqrt(3)*I*(5*a + 2*b)/54)**2)/(125*a**3 + 8*b**3)) + (-5*a/54 + b/27 + sqrt(3)*I*(5*
a + 2*b)/54)*log(x + (675*a**2*(-5*a/54 + b/27 + sqrt(3)*I*(5*a + 2*b)/54) + 40*a*b**2 + 1458*b*(-5*a/54 + b/2
7 + sqrt(3)*I*(5*a + 2*b)/54)**2)/(125*a**3 + 8*b**3)) + (5*a*x**4 + 8*a*x + 4*b*x**5 + 7*b*x**2)/(18*x**6 + 3
6*x**3 + 18)

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Giac [A]  time = 1.11403, size = 119, normalized size = 1.18 \begin{align*} \frac{1}{27} \, \sqrt{3}{\left (5 \, a + 2 \, b\right )} \arctan \left (\frac{1}{3} \, \sqrt{3}{\left (2 \, x - 1\right )}\right ) - \frac{1}{54} \,{\left (5 \, a - 2 \, b\right )} \log \left (x^{2} - x + 1\right ) + \frac{1}{27} \,{\left (5 \, a - 2 \, b\right )} \log \left ({\left | x + 1 \right |}\right ) + \frac{4 \, b x^{5} + 5 \, a x^{4} + 7 \, b x^{2} + 8 \, a x}{18 \,{\left (x^{3} + 1\right )}^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)/(1+x)^3/(x^2-x+1)^3,x, algorithm="giac")

[Out]

1/27*sqrt(3)*(5*a + 2*b)*arctan(1/3*sqrt(3)*(2*x - 1)) - 1/54*(5*a - 2*b)*log(x^2 - x + 1) + 1/27*(5*a - 2*b)*
log(abs(x + 1)) + 1/18*(4*b*x^5 + 5*a*x^4 + 7*b*x^2 + 8*a*x)/(x^3 + 1)^2