3.910 \(\int \frac{1}{x^3 (a+b+2 a x^2+a x^4)} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.131711, antiderivative size = 89, 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 = {1114, 709, 800, 634, 618, 204, 628} \[ -\frac{1}{2 x^2 (a+b)}+\frac{a \log \left (a x^4+2 a x^2+a+b\right )}{2 (a+b)^2}+\frac{\sqrt{a} (a-b) \tan ^{-1}\left (\frac{\sqrt{a} \left (x^2+1\right )}{\sqrt{b}}\right )}{2 \sqrt{b} (a+b)^2}-\frac{2 a \log (x)}{(a+b)^2} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^3*(a + b + 2*a*x^2 + a*x^4)),x]

[Out]

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

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 709

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

Rule 800

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Int[Exp
andIntegrand[((d + e*x)^m*(f + g*x))/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[b^2 -
4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegerQ[m]

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

Mathematica [C]  time = 0.0924026, size = 163, normalized size = 1.83 \[ \frac{\left (2 a^{3/2} \sqrt{b}-i a^2+i a b\right ) \log \left (\sqrt{a} x^2+\sqrt{a}-i \sqrt{b}\right )}{4 \sqrt{a} \sqrt{b} (a+b)^2}+\frac{\left (2 a^{3/2} \sqrt{b}+i a^2-i a b\right ) \log \left (\sqrt{a} x^2+\sqrt{a}+i \sqrt{b}\right )}{4 \sqrt{a} \sqrt{b} (a+b)^2}-\frac{1}{2 x^2 (a+b)}-\frac{2 a \log (x)}{(a+b)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^3*(a + b + 2*a*x^2 + a*x^4)),x]

[Out]

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

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Maple [A]  time = 0.051, size = 110, normalized size = 1.2 \begin{align*} -{\frac{1}{ \left ( 2\,a+2\,b \right ){x}^{2}}}-2\,{\frac{a\ln \left ( x \right ) }{ \left ( a+b \right ) ^{2}}}+{\frac{a\ln \left ( a{x}^{4}+2\,a{x}^{2}+a+b \right ) }{2\, \left ( a+b \right ) ^{2}}}+{\frac{{a}^{2}}{2\, \left ( a+b \right ) ^{2}}\arctan \left ({\frac{2\,a{x}^{2}+2\,a}{2}{\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}}-{\frac{ab}{2\, \left ( a+b \right ) ^{2}}\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(1/x^3/(a*x^4+2*a*x^2+a+b),x)

[Out]

-1/2/(a+b)/x^2-2*a*ln(x)/(a+b)^2+1/2*a*ln(a*x^4+2*a*x^2+a+b)/(a+b)^2+1/2/(a+b)^2*a^2/(a*b)^(1/2)*arctan(1/2*(2
*a*x^2+2*a)/(a*b)^(1/2))-1/2/(a+b)^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(1/x^3/(a*x^4+2*a*x^2+a+b),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.5574, size = 490, normalized size = 5.51 \begin{align*} \left [-\frac{{\left (a - b\right )} x^{2} \sqrt{-\frac{a}{b}} \log \left (\frac{a x^{4} + 2 \, a x^{2} - 2 \,{\left (b x^{2} + b\right )} \sqrt{-\frac{a}{b}} + a - b}{a x^{4} + 2 \, a x^{2} + a + b}\right ) - 2 \, a x^{2} \log \left (a x^{4} + 2 \, a x^{2} + a + b\right ) + 8 \, a x^{2} \log \left (x\right ) + 2 \, a + 2 \, b}{4 \,{\left (a^{2} + 2 \, a b + b^{2}\right )} x^{2}}, -\frac{{\left (a - b\right )} x^{2} \sqrt{\frac{a}{b}} \arctan \left (\frac{b \sqrt{\frac{a}{b}}}{a x^{2} + a}\right ) - a x^{2} \log \left (a x^{4} + 2 \, a x^{2} + a + b\right ) + 4 \, a x^{2} \log \left (x\right ) + a + b}{2 \,{\left (a^{2} + 2 \, a b + b^{2}\right )} x^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(a*x^4+2*a*x^2+a+b),x, algorithm="fricas")

[Out]

[-1/4*((a - b)*x^2*sqrt(-a/b)*log((a*x^4 + 2*a*x^2 - 2*(b*x^2 + b)*sqrt(-a/b) + a - b)/(a*x^4 + 2*a*x^2 + a +
b)) - 2*a*x^2*log(a*x^4 + 2*a*x^2 + a + b) + 8*a*x^2*log(x) + 2*a + 2*b)/((a^2 + 2*a*b + b^2)*x^2), -1/2*((a -
b)*x^2*sqrt(a/b)*arctan(b*sqrt(a/b)/(a*x^2 + a)) - a*x^2*log(a*x^4 + 2*a*x^2 + a + b) + 4*a*x^2*log(x) + a +
b)/((a^2 + 2*a*b + b^2)*x^2)]

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Sympy [B]  time = 5.18822, size = 386, normalized size = 4.34 \begin{align*} - \frac{2 a \log{\left (x \right )}}{\left (a + b\right )^{2}} + \left (\frac{a}{2 \left (a + b\right )^{2}} - \frac{\sqrt{- a b} \left (a - b\right )}{4 b \left (a^{2} + 2 a b + b^{2}\right )}\right ) \log{\left (x^{2} + \frac{4 a^{2} b \left (\frac{a}{2 \left (a + b\right )^{2}} - \frac{\sqrt{- a b} \left (a - b\right )}{4 b \left (a^{2} + 2 a b + b^{2}\right )}\right ) + a^{2} + 8 a b^{2} \left (\frac{a}{2 \left (a + b\right )^{2}} - \frac{\sqrt{- a b} \left (a - b\right )}{4 b \left (a^{2} + 2 a b + b^{2}\right )}\right ) - 3 a b + 4 b^{3} \left (\frac{a}{2 \left (a + b\right )^{2}} - \frac{\sqrt{- a b} \left (a - b\right )}{4 b \left (a^{2} + 2 a b + b^{2}\right )}\right )}{a^{2} - a b} \right )} + \left (\frac{a}{2 \left (a + b\right )^{2}} + \frac{\sqrt{- a b} \left (a - b\right )}{4 b \left (a^{2} + 2 a b + b^{2}\right )}\right ) \log{\left (x^{2} + \frac{4 a^{2} b \left (\frac{a}{2 \left (a + b\right )^{2}} + \frac{\sqrt{- a b} \left (a - b\right )}{4 b \left (a^{2} + 2 a b + b^{2}\right )}\right ) + a^{2} + 8 a b^{2} \left (\frac{a}{2 \left (a + b\right )^{2}} + \frac{\sqrt{- a b} \left (a - b\right )}{4 b \left (a^{2} + 2 a b + b^{2}\right )}\right ) - 3 a b + 4 b^{3} \left (\frac{a}{2 \left (a + b\right )^{2}} + \frac{\sqrt{- a b} \left (a - b\right )}{4 b \left (a^{2} + 2 a b + b^{2}\right )}\right )}{a^{2} - a b} \right )} - \frac{1}{x^{2} \left (2 a + 2 b\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*a*log(x)/(a + b)**2 + (a/(2*(a + b)**2) - sqrt(-a*b)*(a - b)/(4*b*(a**2 + 2*a*b + b**2)))*log(x**2 + (4*a**
2*b*(a/(2*(a + b)**2) - sqrt(-a*b)*(a - b)/(4*b*(a**2 + 2*a*b + b**2))) + a**2 + 8*a*b**2*(a/(2*(a + b)**2) -
sqrt(-a*b)*(a - b)/(4*b*(a**2 + 2*a*b + b**2))) - 3*a*b + 4*b**3*(a/(2*(a + b)**2) - sqrt(-a*b)*(a - b)/(4*b*(
a**2 + 2*a*b + b**2))))/(a**2 - a*b)) + (a/(2*(a + b)**2) + sqrt(-a*b)*(a - b)/(4*b*(a**2 + 2*a*b + b**2)))*lo
g(x**2 + (4*a**2*b*(a/(2*(a + b)**2) + sqrt(-a*b)*(a - b)/(4*b*(a**2 + 2*a*b + b**2))) + a**2 + 8*a*b**2*(a/(2
*(a + b)**2) + sqrt(-a*b)*(a - b)/(4*b*(a**2 + 2*a*b + b**2))) - 3*a*b + 4*b**3*(a/(2*(a + b)**2) + sqrt(-a*b)
*(a - b)/(4*b*(a**2 + 2*a*b + b**2))))/(a**2 - a*b)) - 1/(x**2*(2*a + 2*b))

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Giac [A]  time = 3.40728, size = 169, normalized size = 1.9 \begin{align*} \frac{a \log \left (a x^{4} + 2 \, a x^{2} + a + b\right )}{2 \,{\left (a^{2} + 2 \, a b + b^{2}\right )}} - \frac{a \log \left (x^{2}\right )}{a^{2} + 2 \, a b + b^{2}} + \frac{{\left (a^{2} - a b\right )} \arctan \left (\frac{a x^{2} + a}{\sqrt{a b}}\right )}{2 \,{\left (a^{2} + 2 \, a b + b^{2}\right )} \sqrt{a b}} + \frac{2 \, a x^{2} - a - b}{2 \,{\left (a^{2} + 2 \, a b + b^{2}\right )} x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(a*x^4+2*a*x^2+a+b),x, algorithm="giac")

[Out]

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