3.78 \(\int \frac{x^8}{a x+b x^3+c x^5} \, dx\)

Optimal. Leaf size=100 \[ \frac{\left (b^2-a c\right ) \log \left (a+b x^2+c x^4\right )}{4 c^3}+\frac{b \left (b^2-3 a c\right ) \tanh ^{-1}\left (\frac{b+2 c x^2}{\sqrt{b^2-4 a c}}\right )}{2 c^3 \sqrt{b^2-4 a c}}-\frac{b x^2}{2 c^2}+\frac{x^4}{4 c} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.121738, antiderivative size = 100, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.35, Rules used = {1585, 1114, 701, 634, 618, 206, 628} \[ \frac{\left (b^2-a c\right ) \log \left (a+b x^2+c x^4\right )}{4 c^3}+\frac{b \left (b^2-3 a c\right ) \tanh ^{-1}\left (\frac{b+2 c x^2}{\sqrt{b^2-4 a c}}\right )}{2 c^3 \sqrt{b^2-4 a c}}-\frac{b x^2}{2 c^2}+\frac{x^4}{4 c} \]

Antiderivative was successfully verified.

[In]

Int[x^8/(a*x + b*x^3 + c*x^5),x]

[Out]

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

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 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^8}{a x+b x^3+c x^5} \, dx &=\int \frac{x^7}{a+b x^2+c x^4} \, dx\\ &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{x^3}{a+b x+c x^2} \, dx,x,x^2\right )\\ &=\frac{1}{2} \operatorname{Subst}\left (\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,x,x^2\right )\\ &=-\frac{b x^2}{2 c^2}+\frac{x^4}{4 c}+\frac{\operatorname{Subst}\left (\int \frac{a b+\left (b^2-a c\right ) x}{a+b x+c x^2} \, dx,x,x^2\right )}{2 c^2}\\ &=-\frac{b x^2}{2 c^2}+\frac{x^4}{4 c}-\frac{\left (b \left (b^2-3 a c\right )\right ) \operatorname{Subst}\left (\int \frac{1}{a+b x+c x^2} \, dx,x,x^2\right )}{4 c^3}+\frac{\left (b^2-a c\right ) \operatorname{Subst}\left (\int \frac{b+2 c x}{a+b x+c x^2} \, dx,x,x^2\right )}{4 c^3}\\ &=-\frac{b x^2}{2 c^2}+\frac{x^4}{4 c}+\frac{\left (b^2-a c\right ) \log \left (a+b x^2+c x^4\right )}{4 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^2\right )}{2 c^3}\\ &=-\frac{b x^2}{2 c^2}+\frac{x^4}{4 c}+\frac{b \left (b^2-3 a c\right ) \tanh ^{-1}\left (\frac{b+2 c x^2}{\sqrt{b^2-4 a c}}\right )}{2 c^3 \sqrt{b^2-4 a c}}+\frac{\left (b^2-a c\right ) \log \left (a+b x^2+c x^4\right )}{4 c^3}\\ \end{align*}

Mathematica [A]  time = 0.0937252, size = 93, normalized size = 0.93 \[ \frac{\left (b^2-a c\right ) \log \left (a+b x^2+c x^4\right )-\frac{2 b \left (b^2-3 a c\right ) \tan ^{-1}\left (\frac{b+2 c x^2}{\sqrt{4 a c-b^2}}\right )}{\sqrt{4 a c-b^2}}+c x^2 \left (c x^2-2 b\right )}{4 c^3} \]

Antiderivative was successfully verified.

[In]

Integrate[x^8/(a*x + b*x^3 + c*x^5),x]

[Out]

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

________________________________________________________________________________________

Maple [A]  time = 0.005, size = 142, normalized size = 1.4 \begin{align*}{\frac{{x}^{4}}{4\,c}}-{\frac{b{x}^{2}}{2\,{c}^{2}}}-{\frac{\ln \left ( c{x}^{4}+b{x}^{2}+a \right ) a}{4\,{c}^{2}}}+{\frac{\ln \left ( c{x}^{4}+b{x}^{2}+a \right ){b}^{2}}{4\,{c}^{3}}}+{\frac{3\,ab}{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}}}}}-{\frac{{b}^{3}}{2\,{c}^{3}}\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^8/(c*x^5+b*x^3+a*x),x)

[Out]

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

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{c x^{4} - 2 \, b x^{2}}{4 \, c^{2}} - \frac{-\int \frac{{\left (b^{2} - a c\right )} x^{3} + a b x}{c x^{4} + b x^{2} + a}\,{d x}}{c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^8/(c*x^5+b*x^3+a*x),x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

Fricas [A]  time = 1.329, size = 675, normalized size = 6.75 \begin{align*} \left [\frac{{\left (b^{2} c^{2} - 4 \, a c^{3}\right )} x^{4} - 2 \,{\left (b^{3} c - 4 \, a b c^{2}\right )} x^{2} -{\left (b^{3} - 3 \, a b 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^{4} - 5 \, a b^{2} c + 4 \, a^{2} c^{2}\right )} \log \left (c x^{4} + b x^{2} + a\right )}{4 \,{\left (b^{2} c^{3} - 4 \, a c^{4}\right )}}, \frac{{\left (b^{2} c^{2} - 4 \, a c^{3}\right )} x^{4} - 2 \,{\left (b^{3} c - 4 \, a b c^{2}\right )} x^{2} + 2 \,{\left (b^{3} - 3 \, a b 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^{4} - 5 \, a b^{2} c + 4 \, a^{2} c^{2}\right )} \log \left (c x^{4} + b x^{2} + a\right )}{4 \,{\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^8/(c*x^5+b*x^3+a*x),x, algorithm="fricas")

[Out]

[1/4*((b^2*c^2 - 4*a*c^3)*x^4 - 2*(b^3*c - 4*a*b*c^2)*x^2 - (b^3 - 3*a*b*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^4 - 5*a*b^2*c + 4*a^2*c^
2)*log(c*x^4 + b*x^2 + a))/(b^2*c^3 - 4*a*c^4), 1/4*((b^2*c^2 - 4*a*c^3)*x^4 - 2*(b^3*c - 4*a*b*c^2)*x^2 + 2*(
b^3 - 3*a*b*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^4 - 5*a*b^2*c +
4*a^2*c^2)*log(c*x^4 + b*x^2 + a))/(b^2*c^3 - 4*a*c^4)]

________________________________________________________________________________________

Sympy [B]  time = 1.89168, size = 391, normalized size = 3.91 \begin{align*} - \frac{b x^{2}}{2 c^{2}} + \left (- \frac{b \sqrt{- 4 a c + b^{2}} \left (3 a c - b^{2}\right )}{4 c^{3} \left (4 a c - b^{2}\right )} - \frac{a c - b^{2}}{4 c^{3}}\right ) \log{\left (x^{2} + \frac{2 a^{2} c - a b^{2} + 8 a c^{3} \left (- \frac{b \sqrt{- 4 a c + b^{2}} \left (3 a c - b^{2}\right )}{4 c^{3} \left (4 a c - b^{2}\right )} - \frac{a c - b^{2}}{4 c^{3}}\right ) - 2 b^{2} c^{2} \left (- \frac{b \sqrt{- 4 a c + b^{2}} \left (3 a c - b^{2}\right )}{4 c^{3} \left (4 a c - b^{2}\right )} - \frac{a c - b^{2}}{4 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 )}{4 c^{3} \left (4 a c - b^{2}\right )} - \frac{a c - b^{2}}{4 c^{3}}\right ) \log{\left (x^{2} + \frac{2 a^{2} c - a b^{2} + 8 a c^{3} \left (\frac{b \sqrt{- 4 a c + b^{2}} \left (3 a c - b^{2}\right )}{4 c^{3} \left (4 a c - b^{2}\right )} - \frac{a c - b^{2}}{4 c^{3}}\right ) - 2 b^{2} c^{2} \left (\frac{b \sqrt{- 4 a c + b^{2}} \left (3 a c - b^{2}\right )}{4 c^{3} \left (4 a c - b^{2}\right )} - \frac{a c - b^{2}}{4 c^{3}}\right )}{3 a b c - b^{3}} \right )} + \frac{x^{4}}{4 c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**8/(c*x**5+b*x**3+a*x),x)

[Out]

-b*x**2/(2*c**2) + (-b*sqrt(-4*a*c + b**2)*(3*a*c - b**2)/(4*c**3*(4*a*c - b**2)) - (a*c - b**2)/(4*c**3))*log
(x**2 + (2*a**2*c - a*b**2 + 8*a*c**3*(-b*sqrt(-4*a*c + b**2)*(3*a*c - b**2)/(4*c**3*(4*a*c - b**2)) - (a*c -
b**2)/(4*c**3)) - 2*b**2*c**2*(-b*sqrt(-4*a*c + b**2)*(3*a*c - b**2)/(4*c**3*(4*a*c - b**2)) - (a*c - b**2)/(4
*c**3)))/(3*a*b*c - b**3)) + (b*sqrt(-4*a*c + b**2)*(3*a*c - b**2)/(4*c**3*(4*a*c - b**2)) - (a*c - b**2)/(4*c
**3))*log(x**2 + (2*a**2*c - a*b**2 + 8*a*c**3*(b*sqrt(-4*a*c + b**2)*(3*a*c - b**2)/(4*c**3*(4*a*c - b**2)) -
(a*c - b**2)/(4*c**3)) - 2*b**2*c**2*(b*sqrt(-4*a*c + b**2)*(3*a*c - b**2)/(4*c**3*(4*a*c - b**2)) - (a*c - b
**2)/(4*c**3)))/(3*a*b*c - b**3)) + x**4/(4*c)

________________________________________________________________________________________

Giac [A]  time = 1.09998, size = 124, normalized size = 1.24 \begin{align*} \frac{c x^{4} - 2 \, b x^{2}}{4 \, c^{2}} + \frac{{\left (b^{2} - a c\right )} \log \left (c x^{4} + b x^{2} + a\right )}{4 \, c^{3}} - \frac{{\left (b^{3} - 3 \, a b 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^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^8/(c*x^5+b*x^3+a*x),x, algorithm="giac")

[Out]

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