3.80 \(\int \frac{x^6}{a x+b x^3+c x^5} \, dx\)

Optimal. Leaf size=81 \[ -\frac{\left (b^2-2 a c\right ) \tanh ^{-1}\left (\frac{b+2 c x^2}{\sqrt{b^2-4 a c}}\right )}{2 c^2 \sqrt{b^2-4 a c}}-\frac{b \log \left (a+b x^2+c x^4\right )}{4 c^2}+\frac{x^2}{2 c} \]

[Out]

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

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Rubi [A]  time = 0.0872205, antiderivative size = 81, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.35, Rules used = {1585, 1114, 703, 634, 618, 206, 628} \[ -\frac{\left (b^2-2 a c\right ) \tanh ^{-1}\left (\frac{b+2 c x^2}{\sqrt{b^2-4 a c}}\right )}{2 c^2 \sqrt{b^2-4 a c}}-\frac{b \log \left (a+b x^2+c x^4\right )}{4 c^2}+\frac{x^2}{2 c} \]

Antiderivative was successfully verified.

[In]

Int[x^6/(a*x + b*x^3 + c*x^5),x]

[Out]

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

Rule 1585

Int[(u_.)*(x_)^(m_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.))^(n_.), x_Symbol] :> Int[u*x^(m +
n*p)*(a + b*x^(q - p) + c*x^(r - p))^n, x] /; FreeQ[{a, b, c, m, p, q, r}, x] && IntegerQ[n] && PosQ[q - p] &
& PosQ[r - p]

Rule 1114

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

Rule 703

Int[((d_.) + (e_.)*(x_))^(m_)/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(e*(d + e*x)^(m - 1))/(c*
(m - 1)), x] + Dist[1/c, Int[((d + e*x)^(m - 2)*Simp[c*d^2 - a*e^2 + e*(2*c*d - b*e)*x, x])/(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] && GtQ[m, 1]

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^6}{a x+b x^3+c x^5} \, dx &=\int \frac{x^5}{a+b x^2+c x^4} \, dx\\ &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{x^2}{a+b x+c x^2} \, dx,x,x^2\right )\\ &=\frac{x^2}{2 c}+\frac{\operatorname{Subst}\left (\int \frac{-a-b x}{a+b x+c x^2} \, dx,x,x^2\right )}{2 c}\\ &=\frac{x^2}{2 c}-\frac{b \operatorname{Subst}\left (\int \frac{b+2 c x}{a+b x+c x^2} \, dx,x,x^2\right )}{4 c^2}+\frac{\left (b^2-2 a c\right ) \operatorname{Subst}\left (\int \frac{1}{a+b x+c x^2} \, dx,x,x^2\right )}{4 c^2}\\ &=\frac{x^2}{2 c}-\frac{b \log \left (a+b x^2+c x^4\right )}{4 c^2}-\frac{\left (b^2-2 a c\right ) \operatorname{Subst}\left (\int \frac{1}{b^2-4 a c-x^2} \, dx,x,b+2 c x^2\right )}{2 c^2}\\ &=\frac{x^2}{2 c}-\frac{\left (b^2-2 a c\right ) \tanh ^{-1}\left (\frac{b+2 c x^2}{\sqrt{b^2-4 a c}}\right )}{2 c^2 \sqrt{b^2-4 a c}}-\frac{b \log \left (a+b x^2+c x^4\right )}{4 c^2}\\ \end{align*}

Mathematica [A]  time = 0.0484484, size = 78, normalized size = 0.96 \[ \frac{\frac{2 \left (b^2-2 a c\right ) \tan ^{-1}\left (\frac{b+2 c x^2}{\sqrt{4 a c-b^2}}\right )}{\sqrt{4 a c-b^2}}-b \log \left (a+b x^2+c x^4\right )+2 c x^2}{4 c^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x^6/(a*x + b*x^3 + c*x^5),x]

[Out]

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

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Maple [A]  time = 0.003, size = 111, normalized size = 1.4 \begin{align*}{\frac{{x}^{2}}{2\,c}}-{\frac{b\ln \left ( c{x}^{4}+b{x}^{2}+a \right ) }{4\,{c}^{2}}}-{\frac{a}{c}\arctan \left ({(2\,c{x}^{2}+b){\frac{1}{\sqrt{4\,ac-{b}^{2}}}}} \right ){\frac{1}{\sqrt{4\,ac-{b}^{2}}}}}+{\frac{{b}^{2}}{2\,{c}^{2}}\arctan \left ({(2\,c{x}^{2}+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^6/(c*x^5+b*x^3+a*x),x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{x^{2}}{2 \, c} - \frac{\frac{b \log \left (c x^{4} + b x^{2} + a\right )}{4 \, c} - \frac{{\left (b^{2} - 2 \, a c\right )} \arctan \left (\frac{2 \, c x^{2} + b}{\sqrt{-b^{2} + 4 \, a c}}\right )}{2 \, \sqrt{-b^{2} + 4 \, a c} c}}{c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^6/(c*x^5+b*x^3+a*x),x, algorithm="maxima")

[Out]

1/2*x^2/c - integrate((b*x^3 + a*x)/(c*x^4 + b*x^2 + a), x)/c

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Fricas [A]  time = 1.27802, size = 556, normalized size = 6.86 \begin{align*} \left [\frac{2 \,{\left (b^{2} c - 4 \, a c^{2}\right )} x^{2} -{\left (b^{2} - 2 \, a c\right )} \sqrt{b^{2} - 4 \, a c} \log \left (\frac{2 \, c^{2} x^{4} + 2 \, b c x^{2} + b^{2} - 2 \, a c +{\left (2 \, c x^{2} + b\right )} \sqrt{b^{2} - 4 \, a c}}{c x^{4} + b x^{2} + a}\right ) -{\left (b^{3} - 4 \, a b c\right )} \log \left (c x^{4} + b x^{2} + a\right )}{4 \,{\left (b^{2} c^{2} - 4 \, a c^{3}\right )}}, \frac{2 \,{\left (b^{2} c - 4 \, a c^{2}\right )} x^{2} - 2 \,{\left (b^{2} - 2 \, a c\right )} \sqrt{-b^{2} + 4 \, a c} \arctan \left (-\frac{{\left (2 \, c x^{2} + b\right )} \sqrt{-b^{2} + 4 \, a c}}{b^{2} - 4 \, a c}\right ) -{\left (b^{3} - 4 \, a b c\right )} \log \left (c x^{4} + b x^{2} + a\right )}{4 \,{\left (b^{2} c^{2} - 4 \, a c^{3}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^6/(c*x^5+b*x^3+a*x),x, algorithm="fricas")

[Out]

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

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Sympy [B]  time = 1.54081, size = 316, normalized size = 3.9 \begin{align*} \left (- \frac{b}{4 c^{2}} - \frac{\sqrt{- 4 a c + b^{2}} \left (2 a c - b^{2}\right )}{4 c^{2} \left (4 a c - b^{2}\right )}\right ) \log{\left (x^{2} + \frac{- a b - 8 a c^{2} \left (- \frac{b}{4 c^{2}} - \frac{\sqrt{- 4 a c + b^{2}} \left (2 a c - b^{2}\right )}{4 c^{2} \left (4 a c - b^{2}\right )}\right ) + 2 b^{2} c \left (- \frac{b}{4 c^{2}} - \frac{\sqrt{- 4 a c + b^{2}} \left (2 a c - b^{2}\right )}{4 c^{2} \left (4 a c - b^{2}\right )}\right )}{2 a c - b^{2}} \right )} + \left (- \frac{b}{4 c^{2}} + \frac{\sqrt{- 4 a c + b^{2}} \left (2 a c - b^{2}\right )}{4 c^{2} \left (4 a c - b^{2}\right )}\right ) \log{\left (x^{2} + \frac{- a b - 8 a c^{2} \left (- \frac{b}{4 c^{2}} + \frac{\sqrt{- 4 a c + b^{2}} \left (2 a c - b^{2}\right )}{4 c^{2} \left (4 a c - b^{2}\right )}\right ) + 2 b^{2} c \left (- \frac{b}{4 c^{2}} + \frac{\sqrt{- 4 a c + b^{2}} \left (2 a c - b^{2}\right )}{4 c^{2} \left (4 a c - b^{2}\right )}\right )}{2 a c - b^{2}} \right )} + \frac{x^{2}}{2 c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**6/(c*x**5+b*x**3+a*x),x)

[Out]

(-b/(4*c**2) - sqrt(-4*a*c + b**2)*(2*a*c - b**2)/(4*c**2*(4*a*c - b**2)))*log(x**2 + (-a*b - 8*a*c**2*(-b/(4*
c**2) - sqrt(-4*a*c + b**2)*(2*a*c - b**2)/(4*c**2*(4*a*c - b**2))) + 2*b**2*c*(-b/(4*c**2) - sqrt(-4*a*c + b*
*2)*(2*a*c - b**2)/(4*c**2*(4*a*c - b**2))))/(2*a*c - b**2)) + (-b/(4*c**2) + sqrt(-4*a*c + b**2)*(2*a*c - b**
2)/(4*c**2*(4*a*c - b**2)))*log(x**2 + (-a*b - 8*a*c**2*(-b/(4*c**2) + sqrt(-4*a*c + b**2)*(2*a*c - b**2)/(4*c
**2*(4*a*c - b**2))) + 2*b**2*c*(-b/(4*c**2) + sqrt(-4*a*c + b**2)*(2*a*c - b**2)/(4*c**2*(4*a*c - b**2))))/(2
*a*c - b**2)) + x**2/(2*c)

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Giac [A]  time = 1.09547, size = 101, normalized size = 1.25 \begin{align*} \frac{x^{2}}{2 \, c} - \frac{b \log \left (c x^{4} + b x^{2} + a\right )}{4 \, c^{2}} + \frac{{\left (b^{2} - 2 \, a c\right )} \arctan \left (\frac{2 \, c x^{2} + b}{\sqrt{-b^{2} + 4 \, a c}}\right )}{2 \, \sqrt{-b^{2} + 4 \, a c} c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^6/(c*x^5+b*x^3+a*x),x, algorithm="giac")

[Out]

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