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

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

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Rubi [A]  time = 0.10, antiderivative size = 101, normalized size of antiderivative = 1.00, 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} \begin {gather*} \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}} \end {gather*}

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 31

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

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 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 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]

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 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]

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*}

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Mathematica [A]  time = 0.06, size = 94, normalized size = 0.93 \begin {gather*} \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 ) \end {gather*}

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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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {a+b x}{(1+x)^3 \left (1-x+x^2\right )^3} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 0.41, size = 160, normalized size = 1.58 \begin {gather*} \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 {gather*}

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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giac [A]  time = 0.17, size = 88, normalized size = 0.87 \begin {gather*} \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 {gather*}

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

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maple [A]  time = 0.06, size = 154, normalized size = 1.52 \begin {gather*} \frac {5 \sqrt {3}\, a \arctan \left (\frac {\left (2 x -1\right ) \sqrt {3}}{3}\right )}{27}+\frac {5 a \ln \left (x +1\right )}{27}-\frac {5 a \ln \left (x^{2}-x +1\right )}{54}+\frac {2 \sqrt {3}\, b \arctan \left (\frac {\left (2 x -1\right ) \sqrt {3}}{3}\right )}{27}-\frac {2 b \ln \left (x +1\right )}{27}+\frac {b \ln \left (x^{2}-x +1\right )}{27}-\frac {a}{54 \left (x +1\right )^{2}}-\frac {a}{9 \left (x +1\right )}+\frac {b}{54 \left (x +1\right )^{2}}+\frac {2 b}{27 \left (x +1\right )}-\frac {\left (-3 a -4 b \right ) x^{3}+\left (a +\frac {13 b}{2}\right ) x^{2}-\frac {7 a}{2}+\frac {5 b}{2}+\left (-a -8 b \right ) x}{27 \left (x^{2}-x +1\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.27, size = 92, normalized size = 0.91 \begin {gather*} \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 {gather*}

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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mupad [B]  time = 0.14, size = 114, normalized size = 1.13 \begin {gather*} \ln \left (x+1\right )\,\left (\frac {5\,a}{27}-\frac {2\,b}{27}\right )+\ln \left (x-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (\frac {b}{27}-\frac {5\,a}{54}+\frac {\sqrt {3}\,a\,5{}\mathrm {i}}{54}+\frac {\sqrt {3}\,b\,1{}\mathrm {i}}{27}\right )-\ln \left (x-\frac {1}{2}-\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (\frac {5\,a}{54}-\frac {b}{27}+\frac {\sqrt {3}\,a\,5{}\mathrm {i}}{54}+\frac {\sqrt {3}\,b\,1{}\mathrm {i}}{27}\right )+\frac {\frac {2\,b\,x^5}{9}+\frac {5\,a\,x^4}{18}+\frac {7\,b\,x^2}{18}+\frac {4\,a\,x}{9}}{x^6+2\,x^3+1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(x + (3^(1/2)*1i)/2 - 1/2)*(b/27 - (5*a)/54 + (3^(1/2)*a*5i)/54 + (3^(1/2)*b*1i)/27) - log(x - (3^(1/2)*1i)
/2 - 1/2)*((5*a)/54 - b/27 + (3^(1/2)*a*5i)/54 + (3^(1/2)*b*1i)/27) + log(x + 1)*((5*a)/27 - (2*b)/27) + ((4*a
*x)/9 + (5*a*x^4)/18 + (7*b*x^2)/18 + (2*b*x^5)/9)/(2*x^3 + x^6 + 1)

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sympy [C]  time = 0.71, size = 292, normalized size = 2.89 \begin {gather*} \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 {gather*}

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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