3.414 \(\int \frac{x}{c+\frac{a}{x^2}+\frac{b}{x}} \, dx\)

Optimal. Leaf size=89 \[ \frac{\left (b^2-a c\right ) \log \left (a+b x+c x^2\right )}{2 c^3}+\frac{b \left (b^2-3 a c\right ) \tanh ^{-1}\left (\frac{b+2 c x}{\sqrt{b^2-4 a c}}\right )}{c^3 \sqrt{b^2-4 a c}}-\frac{b x}{c^2}+\frac{x^2}{2 c} \]

[Out]

-((b*x)/c^2) + x^2/(2*c) + (b*(b^2 - 3*a*c)*ArcTanh[(b + 2*c*x)/Sqrt[b^2 - 4*a*c]])/(c^3*Sqrt[b^2 - 4*a*c]) +
((b^2 - a*c)*Log[a + b*x + c*x^2])/(2*c^3)

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Rubi [A]  time = 0.0861381, antiderivative size = 89, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.375, Rules used = {1354, 701, 634, 618, 206, 628} \[ \frac{\left (b^2-a c\right ) \log \left (a+b x+c x^2\right )}{2 c^3}+\frac{b \left (b^2-3 a c\right ) \tanh ^{-1}\left (\frac{b+2 c x}{\sqrt{b^2-4 a c}}\right )}{c^3 \sqrt{b^2-4 a c}}-\frac{b x}{c^2}+\frac{x^2}{2 c} \]

Antiderivative was successfully verified.

[In]

Int[x/(c + a/x^2 + b/x),x]

[Out]

-((b*x)/c^2) + x^2/(2*c) + (b*(b^2 - 3*a*c)*ArcTanh[(b + 2*c*x)/Sqrt[b^2 - 4*a*c]])/(c^3*Sqrt[b^2 - 4*a*c]) +
((b^2 - a*c)*Log[a + b*x + c*x^2])/(2*c^3)

Rule 1354

Int[(x_)^(m_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[x^(m + 2*n*p)*(c + b/x^n +
a/x^(2*n))^p, x] /; FreeQ[{a, b, c, m, n}, x] && EqQ[n2, 2*n] && ILtQ[p, 0] && NegQ[n]

Rule 701

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

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 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[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{x}{c+\frac{a}{x^2}+\frac{b}{x}} \, dx &=\int \frac{x^3}{a+b x+c x^2} \, dx\\ &=\int \left (-\frac{b}{c^2}+\frac{x}{c}+\frac{a b+\left (b^2-a c\right ) x}{c^2 \left (a+b x+c x^2\right )}\right ) \, dx\\ &=-\frac{b x}{c^2}+\frac{x^2}{2 c}+\frac{\int \frac{a b+\left (b^2-a c\right ) x}{a+b x+c x^2} \, dx}{c^2}\\ &=-\frac{b x}{c^2}+\frac{x^2}{2 c}-\frac{\left (b \left (b^2-3 a c\right )\right ) \int \frac{1}{a+b x+c x^2} \, dx}{2 c^3}+\frac{\left (b^2-a c\right ) \int \frac{b+2 c x}{a+b x+c x^2} \, dx}{2 c^3}\\ &=-\frac{b x}{c^2}+\frac{x^2}{2 c}+\frac{\left (b^2-a c\right ) \log \left (a+b x+c x^2\right )}{2 c^3}+\frac{\left (b \left (b^2-3 a c\right )\right ) \operatorname{Subst}\left (\int \frac{1}{b^2-4 a c-x^2} \, dx,x,b+2 c x\right )}{c^3}\\ &=-\frac{b x}{c^2}+\frac{x^2}{2 c}+\frac{b \left (b^2-3 a c\right ) \tanh ^{-1}\left (\frac{b+2 c x}{\sqrt{b^2-4 a c}}\right )}{c^3 \sqrt{b^2-4 a c}}+\frac{\left (b^2-a c\right ) \log \left (a+b x+c x^2\right )}{2 c^3}\\ \end{align*}

Mathematica [A]  time = 0.107658, size = 84, normalized size = 0.94 \[ \frac{\left (b^2-a c\right ) \log (a+x (b+c x))-\frac{2 b \left (b^2-3 a c\right ) \tan ^{-1}\left (\frac{b+2 c x}{\sqrt{4 a c-b^2}}\right )}{\sqrt{4 a c-b^2}}+c x (c x-2 b)}{2 c^3} \]

Antiderivative was successfully verified.

[In]

Integrate[x/(c + a/x^2 + b/x),x]

[Out]

(c*x*(-2*b + c*x) - (2*b*(b^2 - 3*a*c)*ArcTan[(b + 2*c*x)/Sqrt[-b^2 + 4*a*c]])/Sqrt[-b^2 + 4*a*c] + (b^2 - a*c
)*Log[a + x*(b + c*x)])/(2*c^3)

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Maple [A]  time = 0.004, size = 132, normalized size = 1.5 \begin{align*}{\frac{{x}^{2}}{2\,c}}-{\frac{bx}{{c}^{2}}}-{\frac{\ln \left ( c{x}^{2}+bx+a \right ) a}{2\,{c}^{2}}}+{\frac{\ln \left ( c{x}^{2}+bx+a \right ){b}^{2}}{2\,{c}^{3}}}+3\,{\frac{ab}{{c}^{2}\sqrt{4\,ac-{b}^{2}}}\arctan \left ({\frac{2\,cx+b}{\sqrt{4\,ac-{b}^{2}}}} \right ) }-{\frac{{b}^{3}}{{c}^{3}}\arctan \left ({(2\,cx+b){\frac{1}{\sqrt{4\,ac-{b}^{2}}}}} \right ){\frac{1}{\sqrt{4\,ac-{b}^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/(c+a/x^2+b/x),x)

[Out]

1/2*x^2/c-b*x/c^2-1/2/c^2*ln(c*x^2+b*x+a)*a+1/2/c^3*ln(c*x^2+b*x+a)*b^2+3/c^2/(4*a*c-b^2)^(1/2)*arctan((2*c*x+
b)/(4*a*c-b^2)^(1/2))*a*b-1/c^3/(4*a*c-b^2)^(1/2)*arctan((2*c*x+b)/(4*a*c-b^2)^(1/2))*b^3

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(c+a/x^2+b/x),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.72491, size = 653, normalized size = 7.34 \begin{align*} \left [\frac{{\left (b^{2} c^{2} - 4 \, a c^{3}\right )} x^{2} -{\left (b^{3} - 3 \, a b c\right )} \sqrt{b^{2} - 4 \, a c} \log \left (\frac{2 \, c^{2} x^{2} + 2 \, b c x + b^{2} - 2 \, a c - \sqrt{b^{2} - 4 \, a c}{\left (2 \, c x + b\right )}}{c x^{2} + b x + a}\right ) - 2 \,{\left (b^{3} c - 4 \, a b c^{2}\right )} x +{\left (b^{4} - 5 \, a b^{2} c + 4 \, a^{2} c^{2}\right )} \log \left (c x^{2} + b x + a\right )}{2 \,{\left (b^{2} c^{3} - 4 \, a c^{4}\right )}}, \frac{{\left (b^{2} c^{2} - 4 \, a c^{3}\right )} x^{2} + 2 \,{\left (b^{3} - 3 \, a b c\right )} \sqrt{-b^{2} + 4 \, a c} \arctan \left (-\frac{\sqrt{-b^{2} + 4 \, a c}{\left (2 \, c x + b\right )}}{b^{2} - 4 \, a c}\right ) - 2 \,{\left (b^{3} c - 4 \, a b c^{2}\right )} x +{\left (b^{4} - 5 \, a b^{2} c + 4 \, a^{2} c^{2}\right )} \log \left (c x^{2} + b x + a\right )}{2 \,{\left (b^{2} c^{3} - 4 \, a c^{4}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(c+a/x^2+b/x),x, algorithm="fricas")

[Out]

[1/2*((b^2*c^2 - 4*a*c^3)*x^2 - (b^3 - 3*a*b*c)*sqrt(b^2 - 4*a*c)*log((2*c^2*x^2 + 2*b*c*x + b^2 - 2*a*c - sqr
t(b^2 - 4*a*c)*(2*c*x + b))/(c*x^2 + b*x + a)) - 2*(b^3*c - 4*a*b*c^2)*x + (b^4 - 5*a*b^2*c + 4*a^2*c^2)*log(c
*x^2 + b*x + a))/(b^2*c^3 - 4*a*c^4), 1/2*((b^2*c^2 - 4*a*c^3)*x^2 + 2*(b^3 - 3*a*b*c)*sqrt(-b^2 + 4*a*c)*arct
an(-sqrt(-b^2 + 4*a*c)*(2*c*x + b)/(b^2 - 4*a*c)) - 2*(b^3*c - 4*a*b*c^2)*x + (b^4 - 5*a*b^2*c + 4*a^2*c^2)*lo
g(c*x^2 + b*x + a))/(b^2*c^3 - 4*a*c^4)]

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Sympy [B]  time = 0.786597, size = 381, normalized size = 4.28 \begin{align*} - \frac{b x}{c^{2}} + \left (- \frac{b \sqrt{- 4 a c + b^{2}} \left (3 a c - b^{2}\right )}{2 c^{3} \left (4 a c - b^{2}\right )} - \frac{a c - b^{2}}{2 c^{3}}\right ) \log{\left (x + \frac{2 a^{2} c - a b^{2} + 4 a c^{3} \left (- \frac{b \sqrt{- 4 a c + b^{2}} \left (3 a c - b^{2}\right )}{2 c^{3} \left (4 a c - b^{2}\right )} - \frac{a c - b^{2}}{2 c^{3}}\right ) - b^{2} c^{2} \left (- \frac{b \sqrt{- 4 a c + b^{2}} \left (3 a c - b^{2}\right )}{2 c^{3} \left (4 a c - b^{2}\right )} - \frac{a c - b^{2}}{2 c^{3}}\right )}{3 a b c - b^{3}} \right )} + \left (\frac{b \sqrt{- 4 a c + b^{2}} \left (3 a c - b^{2}\right )}{2 c^{3} \left (4 a c - b^{2}\right )} - \frac{a c - b^{2}}{2 c^{3}}\right ) \log{\left (x + \frac{2 a^{2} c - a b^{2} + 4 a c^{3} \left (\frac{b \sqrt{- 4 a c + b^{2}} \left (3 a c - b^{2}\right )}{2 c^{3} \left (4 a c - b^{2}\right )} - \frac{a c - b^{2}}{2 c^{3}}\right ) - b^{2} c^{2} \left (\frac{b \sqrt{- 4 a c + b^{2}} \left (3 a c - b^{2}\right )}{2 c^{3} \left (4 a c - b^{2}\right )} - \frac{a c - b^{2}}{2 c^{3}}\right )}{3 a b c - b^{3}} \right )} + \frac{x^{2}}{2 c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(c+a/x**2+b/x),x)

[Out]

-b*x/c**2 + (-b*sqrt(-4*a*c + b**2)*(3*a*c - b**2)/(2*c**3*(4*a*c - b**2)) - (a*c - b**2)/(2*c**3))*log(x + (2
*a**2*c - a*b**2 + 4*a*c**3*(-b*sqrt(-4*a*c + b**2)*(3*a*c - b**2)/(2*c**3*(4*a*c - b**2)) - (a*c - b**2)/(2*c
**3)) - b**2*c**2*(-b*sqrt(-4*a*c + b**2)*(3*a*c - b**2)/(2*c**3*(4*a*c - b**2)) - (a*c - b**2)/(2*c**3)))/(3*
a*b*c - b**3)) + (b*sqrt(-4*a*c + b**2)*(3*a*c - b**2)/(2*c**3*(4*a*c - b**2)) - (a*c - b**2)/(2*c**3))*log(x
+ (2*a**2*c - a*b**2 + 4*a*c**3*(b*sqrt(-4*a*c + b**2)*(3*a*c - b**2)/(2*c**3*(4*a*c - b**2)) - (a*c - b**2)/(
2*c**3)) - b**2*c**2*(b*sqrt(-4*a*c + b**2)*(3*a*c - b**2)/(2*c**3*(4*a*c - b**2)) - (a*c - b**2)/(2*c**3)))/(
3*a*b*c - b**3)) + x**2/(2*c)

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Giac [A]  time = 1.14168, size = 116, normalized size = 1.3 \begin{align*} \frac{c x^{2} - 2 \, b x}{2 \, c^{2}} + \frac{{\left (b^{2} - a c\right )} \log \left (c x^{2} + b x + a\right )}{2 \, c^{3}} - \frac{{\left (b^{3} - 3 \, a b c\right )} \arctan \left (\frac{2 \, c x + b}{\sqrt{-b^{2} + 4 \, a c}}\right )}{\sqrt{-b^{2} + 4 \, a c} c^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(c+a/x^2+b/x),x, algorithm="giac")

[Out]

1/2*(c*x^2 - 2*b*x)/c^2 + 1/2*(b^2 - a*c)*log(c*x^2 + b*x + a)/c^3 - (b^3 - 3*a*b*c)*arctan((2*c*x + b)/sqrt(-
b^2 + 4*a*c))/(sqrt(-b^2 + 4*a*c)*c^3)