3.906 \(\int \frac{x^5}{a+b+2 a x^2+a x^4} \, dx\)

Optimal. Leaf size=69 \[ \frac{(a-b) \tan ^{-1}\left (\frac{\sqrt{a} \left (x^2+1\right )}{\sqrt{b}}\right )}{2 a^{3/2} \sqrt{b}}-\frac{\log \left (a x^4+2 a x^2+a+b\right )}{2 a}+\frac{x^2}{2 a} \]

[Out]

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

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Rubi [A]  time = 0.0757646, antiderivative size = 69, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.3, Rules used = {1114, 703, 634, 618, 204, 628} \[ \frac{(a-b) \tan ^{-1}\left (\frac{\sqrt{a} \left (x^2+1\right )}{\sqrt{b}}\right )}{2 a^{3/2} \sqrt{b}}-\frac{\log \left (a x^4+2 a x^2+a+b\right )}{2 a}+\frac{x^2}{2 a} \]

Antiderivative was successfully verified.

[In]

Int[x^5/(a + b + 2*a*x^2 + a*x^4),x]

[Out]

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

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

Mathematica [A]  time = 0.0347361, size = 62, normalized size = 0.9 \[ \frac{\sqrt{a} \left (x^2-\log \left (a \left (x^2+1\right )^2+b\right )\right )+\frac{(a-b) \tan ^{-1}\left (\frac{\sqrt{a} \left (x^2+1\right )}{\sqrt{b}}\right )}{\sqrt{b}}}{2 a^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^5/(a + b + 2*a*x^2 + a*x^4),x]

[Out]

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

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Maple [A]  time = 0.046, size = 84, normalized size = 1.2 \begin{align*}{\frac{{x}^{2}}{2\,a}}-{\frac{\ln \left ( a{x}^{4}+2\,a{x}^{2}+a+b \right ) }{2\,a}}+{\frac{1}{2}\arctan \left ({\frac{2\,a{x}^{2}+2\,a}{2}{\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}}-{\frac{b}{2\,a}\arctan \left ({\frac{2\,a{x}^{2}+2\,a}{2}{\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5/(a*x^4+2*a*x^2+a+b),x)

[Out]

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

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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^5/(a*x^4+2*a*x^2+a+b),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.49749, size = 374, normalized size = 5.42 \begin{align*} \left [\frac{2 \, a b x^{2} - 2 \, a b \log \left (a x^{4} + 2 \, a x^{2} + a + b\right ) + \sqrt{-a b}{\left (a - b\right )} \log \left (\frac{a x^{4} + 2 \, a x^{2} + 2 \, \sqrt{-a b}{\left (x^{2} + 1\right )} + a - b}{a x^{4} + 2 \, a x^{2} + a + b}\right )}{4 \, a^{2} b}, \frac{a b x^{2} - a b \log \left (a x^{4} + 2 \, a x^{2} + a + b\right ) - \sqrt{a b}{\left (a - b\right )} \arctan \left (\frac{\sqrt{a b}}{a x^{2} + a}\right )}{2 \, a^{2} b}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5/(a*x^4+2*a*x^2+a+b),x, algorithm="fricas")

[Out]

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

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Sympy [B]  time = 1.11192, size = 144, normalized size = 2.09 \begin{align*} \left (- \frac{1}{2 a} - \frac{\sqrt{- a^{3} b} \left (a - b\right )}{4 a^{3} b}\right ) \log{\left (x^{2} + \frac{4 a b \left (- \frac{1}{2 a} - \frac{\sqrt{- a^{3} b} \left (a - b\right )}{4 a^{3} b}\right ) + a + b}{a - b} \right )} + \left (- \frac{1}{2 a} + \frac{\sqrt{- a^{3} b} \left (a - b\right )}{4 a^{3} b}\right ) \log{\left (x^{2} + \frac{4 a b \left (- \frac{1}{2 a} + \frac{\sqrt{- a^{3} b} \left (a - b\right )}{4 a^{3} b}\right ) + a + b}{a - b} \right )} + \frac{x^{2}}{2 a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**5/(a*x**4+2*a*x**2+a+b),x)

[Out]

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

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Giac [A]  time = 3.5121, size = 78, normalized size = 1.13 \begin{align*} \frac{x^{2}}{2 \, a} + \frac{{\left (a - b\right )} \arctan \left (\frac{a x^{2} + a}{\sqrt{a b}}\right )}{2 \, \sqrt{a b} a} - \frac{\log \left (a x^{4} + 2 \, a x^{2} + a + b\right )}{2 \, a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5/(a*x^4+2*a*x^2+a+b),x, algorithm="giac")

[Out]

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