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

Optimal. Leaf size=97 \[ -\frac {1}{2 x^2 (a-b)}-\frac {\sqrt {a} (a+b) \tanh ^{-1}\left (\frac {\sqrt {a} \left (x^2+1\right )}{\sqrt {b}}\right )}{2 \sqrt {b} (a-b)^2}+\frac {a \log \left (a x^4+2 a x^2+a-b\right )}{2 (a-b)^2}-\frac {2 a \log (x)}{(a-b)^2} \]

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Rubi [A]  time = 0.14, antiderivative size = 97, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.318, Rules used = {1114, 709, 800, 634, 618, 206, 628} \begin {gather*} -\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) \tanh ^{-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} \end {gather*}

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

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 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 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]

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 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 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]

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) \tanh ^{-1}\left (\frac {\sqrt {a} \left (1+x^2\right )}{\sqrt {b}}\right )}{2 (a-b)^2 \sqrt {b}}-\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*}

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Mathematica [A]  time = 0.09, size = 146, normalized size = 1.51 \begin {gather*} \frac {-8 a \sqrt {b} x^2 \log (x)+\sqrt {a} x^2 \left (\sqrt {a}+\sqrt {b}\right )^2 \log \left (\sqrt {a} \left (x^2+1\right )-\sqrt {b}\right )-\left (\sqrt {a}-\sqrt {b}\right ) \left (\left (a x^2-\sqrt {a} \sqrt {b} x^2\right ) \log \left (\sqrt {a} \left (x^2+1\right )+\sqrt {b}\right )+2 \left (\sqrt {a} \sqrt {b}+b\right )\right )}{4 \sqrt {b} x^2 (a-b)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-8*a*Sqrt[b]*x^2*Log[x] + Sqrt[a]*(Sqrt[a] + Sqrt[b])^2*x^2*Log[-Sqrt[b] + Sqrt[a]*(1 + x^2)] - (Sqrt[a] - Sq
rt[b])*(2*(Sqrt[a]*Sqrt[b] + b) + (a*x^2 - Sqrt[a]*Sqrt[b]*x^2)*Log[Sqrt[b] + Sqrt[a]*(1 + x^2)]))/(4*(a - b)^
2*Sqrt[b]*x^2)

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{x^3 \left (a-b+2 a x^2+a x^4\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 0.78, size = 209, normalized size = 2.15 \begin {gather*} \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 \relax (x) - 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 \relax (x) - a + b}{2 \, {\left (a^{2} - 2 \, a b + b^{2}\right )} x^{2}}\right ] \end {gather*}

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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giac [A]  time = 0.28, size = 126, normalized size = 1.30 \begin {gather*} \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 {gather*}

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)

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maple [A]  time = 0.01, size = 122, normalized size = 1.26 \begin {gather*} -\frac {a^{2} \arctanh \left (\frac {2 a \,x^{2}+2 a}{2 \sqrt {a b}}\right )}{2 \left (a -b \right )^{2} \sqrt {a b}}-\frac {a b \arctanh \left (\frac {2 a \,x^{2}+2 a}{2 \sqrt {a b}}\right )}{2 \left (a -b \right )^{2} \sqrt {a b}}-\frac {2 a \ln \relax (x )}{\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 {1}{2 \left (a -b \right ) x^{2}} \end {gather*}

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)*arctanh(1/2*(
2*a*x^2+2*a)/(a*b)^(1/2))-1/2/(a-b)^2*a/(a*b)^(1/2)*arctanh(1/2*(2*a*x^2+2*a)/(a*b)^(1/2))*b

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maxima [A]  time = 2.97, size = 123, normalized size = 1.27 \begin {gather*} \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 )} \log \left (\frac {a x^{2} + a - \sqrt {a b}}{a x^{2} + a + \sqrt {a b}}\right )}{4 \, {\left (a^{2} - 2 \, a b + b^{2}\right )} \sqrt {a b}} - \frac {1}{2 \, {\left (a - b\right )} x^{2}} \end {gather*}

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]

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

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mupad [B]  time = 4.87, size = 389, normalized size = 4.01 \begin {gather*} \frac {\ln \left (100\,a\,{\left (a\,b\right )}^{7/2}-198\,b\,{\left (a\,b\right )}^{7/2}-a^3\,{\left (a\,b\right )}^{5/2}+100\,b^3\,{\left (a\,b\right )}^{5/2}-b^5\,{\left (a\,b\right )}^{3/2}+a^2\,b^6-100\,a^3\,b^5+198\,a^4\,b^4-100\,a^5\,b^3+a^6\,b^2+a^2\,b^6\,x^2-100\,a^3\,b^5\,x^2+198\,a^4\,b^4\,x^2-100\,a^5\,b^3\,x^2+a^6\,b^2\,x^2\right )\,\left (\frac {a\,\sqrt {a\,b}}{4}+b\,\left (\frac {a}{2}+\frac {\sqrt {a\,b}}{4}\right )\right )}{a^2\,b-2\,a\,b^2+b^3}-\frac {2\,a\,\ln \relax (x)}{a^2-2\,a\,b+b^2}-\frac {\ln \left (198\,b\,{\left (a\,b\right )}^{7/2}-100\,a\,{\left (a\,b\right )}^{7/2}+a^3\,{\left (a\,b\right )}^{5/2}-100\,b^3\,{\left (a\,b\right )}^{5/2}+b^5\,{\left (a\,b\right )}^{3/2}+a^2\,b^6-100\,a^3\,b^5+198\,a^4\,b^4-100\,a^5\,b^3+a^6\,b^2+a^2\,b^6\,x^2-100\,a^3\,b^5\,x^2+198\,a^4\,b^4\,x^2-100\,a^5\,b^3\,x^2+a^6\,b^2\,x^2\right )\,\left (\frac {a\,\sqrt {a\,b}}{4}-b\,\left (\frac {a}{2}-\frac {\sqrt {a\,b}}{4}\right )\right )}{a^2\,b-2\,a\,b^2+b^3}-\frac {1}{2\,x^2\,\left (a-b\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(log(100*a*(a*b)^(7/2) - 198*b*(a*b)^(7/2) - a^3*(a*b)^(5/2) + 100*b^3*(a*b)^(5/2) - b^5*(a*b)^(3/2) + a^2*b^6
- 100*a^3*b^5 + 198*a^4*b^4 - 100*a^5*b^3 + a^6*b^2 + a^2*b^6*x^2 - 100*a^3*b^5*x^2 + 198*a^4*b^4*x^2 - 100*a
^5*b^3*x^2 + a^6*b^2*x^2)*((a*(a*b)^(1/2))/4 + b*(a/2 + (a*b)^(1/2)/4)))/(a^2*b - 2*a*b^2 + b^3) - (2*a*log(x)
)/(a^2 - 2*a*b + b^2) - (log(198*b*(a*b)^(7/2) - 100*a*(a*b)^(7/2) + a^3*(a*b)^(5/2) - 100*b^3*(a*b)^(5/2) + b
^5*(a*b)^(3/2) + a^2*b^6 - 100*a^3*b^5 + 198*a^4*b^4 - 100*a^5*b^3 + a^6*b^2 + a^2*b^6*x^2 - 100*a^3*b^5*x^2 +
198*a^4*b^4*x^2 - 100*a^5*b^3*x^2 + a^6*b^2*x^2)*((a*(a*b)^(1/2))/4 - b*(a/2 - (a*b)^(1/2)/4)))/(a^2*b - 2*a*
b^2 + b^3) - 1/(2*x^2*(a - b))

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sympy [B]  time = 32.93, size = 372, normalized size = 3.84 \begin {gather*} - \frac {2 a \log {\relax (x )}}{\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 {gather*}

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) - s
qrt(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)))*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)) - 1/(x**2*(2*a - 2*b))

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