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

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

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.102417, antiderivative size = 81, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.389, Rules used = {1354, 709, 800, 634, 618, 206, 628} \[ -\frac{\left (b^2-2 a c\right ) \tanh ^{-1}\left (\frac{b+2 c x}{\sqrt{b^2-4 a c}}\right )}{a^2 \sqrt{b^2-4 a c}}+\frac{b \log \left (a+b x+c x^2\right )}{2 a^2}-\frac{b \log (x)}{a^2}-\frac{1}{a x} \]

Antiderivative was successfully verified.

[In]

Int[1/((c + a/x^2 + b/x)*x^4),x]

[Out]

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

Rule 1354

Int[(x_)^(m_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[x^(m + 2*n*p)*(c + b/x^n +
a/x^(2*n))^p, x] /; FreeQ[{a, b, c, m, n}, x] && EqQ[n2, 2*n] && ILtQ[p, 0] && NegQ[n]

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

Mathematica [A]  time = 0.0792688, size = 77, normalized size = 0.95 \[ \frac{\frac{2 \left (b^2-2 a c\right ) \tan ^{-1}\left (\frac{b+2 c x}{\sqrt{4 a c-b^2}}\right )}{\sqrt{4 a c-b^2}}+b \log (a+x (b+c x))-\frac{2 a}{x}-2 b \log (x)}{2 a^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [A]  time = 0.008, size = 112, normalized size = 1.4 \begin{align*} -{\frac{1}{ax}}-{\frac{b\ln \left ( x \right ) }{{a}^{2}}}+{\frac{b\ln \left ( c{x}^{2}+bx+a \right ) }{2\,{a}^{2}}}-2\,{\frac{c}{a\sqrt{4\,ac-{b}^{2}}}\arctan \left ({\frac{2\,cx+b}{\sqrt{4\,ac-{b}^{2}}}} \right ) }+{\frac{{b}^{2}}{{a}^{2}}\arctan \left ({(2\,cx+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(1/(c+a/x^2+b/x)/x^4,x)

[Out]

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

________________________________________________________________________________________

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

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 1.82571, size = 626, normalized size = 7.73 \begin{align*} \left [-\frac{{\left (b^{2} - 2 \, a c\right )} \sqrt{b^{2} - 4 \, a c} x \log \left (\frac{2 \, c^{2} x^{2} + 2 \, b c x + b^{2} - 2 \, a c + \sqrt{b^{2} - 4 \, a c}{\left (2 \, c x + b\right )}}{c x^{2} + b x + a}\right ) + 2 \, a b^{2} - 8 \, a^{2} c -{\left (b^{3} - 4 \, a b c\right )} x \log \left (c x^{2} + b x + a\right ) + 2 \,{\left (b^{3} - 4 \, a b c\right )} x \log \left (x\right )}{2 \,{\left (a^{2} b^{2} - 4 \, a^{3} c\right )} x}, -\frac{2 \,{\left (b^{2} - 2 \, a c\right )} \sqrt{-b^{2} + 4 \, a c} x \arctan \left (-\frac{\sqrt{-b^{2} + 4 \, a c}{\left (2 \, c x + b\right )}}{b^{2} - 4 \, a c}\right ) + 2 \, a b^{2} - 8 \, a^{2} c -{\left (b^{3} - 4 \, a b c\right )} x \log \left (c x^{2} + b x + a\right ) + 2 \,{\left (b^{3} - 4 \, a b c\right )} x \log \left (x\right )}{2 \,{\left (a^{2} b^{2} - 4 \, a^{3} c\right )} x}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/2*((b^2 - 2*a*c)*sqrt(b^2 - 4*a*c)*x*log((2*c^2*x^2 + 2*b*c*x + b^2 - 2*a*c + sqrt(b^2 - 4*a*c)*(2*c*x + b
))/(c*x^2 + b*x + a)) + 2*a*b^2 - 8*a^2*c - (b^3 - 4*a*b*c)*x*log(c*x^2 + b*x + a) + 2*(b^3 - 4*a*b*c)*x*log(x
))/((a^2*b^2 - 4*a^3*c)*x), -1/2*(2*(b^2 - 2*a*c)*sqrt(-b^2 + 4*a*c)*x*arctan(-sqrt(-b^2 + 4*a*c)*(2*c*x + b)/
(b^2 - 4*a*c)) + 2*a*b^2 - 8*a^2*c - (b^3 - 4*a*b*c)*x*log(c*x^2 + b*x + a) + 2*(b^3 - 4*a*b*c)*x*log(x))/((a^
2*b^2 - 4*a^3*c)*x)]

________________________________________________________________________________________

Sympy [B]  time = 3.49527, size = 862, normalized size = 10.64 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(b/(2*a**2) - sqrt(-4*a*c + b**2)*(2*a*c - b**2)/(2*a**2*(4*a*c - b**2)))*log(x + (-28*a**6*b*c**2*(b/(2*a**2)
- sqrt(-4*a*c + b**2)*(2*a*c - b**2)/(2*a**2*(4*a*c - b**2)))**2 + 15*a**5*b**3*c*(b/(2*a**2) - sqrt(-4*a*c +
b**2)*(2*a*c - b**2)/(2*a**2*(4*a*c - b**2)))**2 - 4*a**5*c**3*(b/(2*a**2) - sqrt(-4*a*c + b**2)*(2*a*c - b**
2)/(2*a**2*(4*a*c - b**2))) - 2*a**4*b**5*(b/(2*a**2) - sqrt(-4*a*c + b**2)*(2*a*c - b**2)/(2*a**2*(4*a*c - b*
*2)))**2 - 3*a**4*b**2*c**2*(b/(2*a**2) - sqrt(-4*a*c + b**2)*(2*a*c - b**2)/(2*a**2*(4*a*c - b**2))) + a**3*b
**4*c*(b/(2*a**2) - sqrt(-4*a*c + b**2)*(2*a*c - b**2)/(2*a**2*(4*a*c - b**2))) - 4*a**3*b*c**3 + 25*a**2*b**3
*c**2 - 14*a*b**5*c + 2*b**7)/(2*a**3*c**4 + 15*a**2*b**2*c**3 - 12*a*b**4*c**2 + 2*b**6*c)) + (b/(2*a**2) + s
qrt(-4*a*c + b**2)*(2*a*c - b**2)/(2*a**2*(4*a*c - b**2)))*log(x + (-28*a**6*b*c**2*(b/(2*a**2) + sqrt(-4*a*c
+ b**2)*(2*a*c - b**2)/(2*a**2*(4*a*c - b**2)))**2 + 15*a**5*b**3*c*(b/(2*a**2) + sqrt(-4*a*c + b**2)*(2*a*c -
b**2)/(2*a**2*(4*a*c - b**2)))**2 - 4*a**5*c**3*(b/(2*a**2) + sqrt(-4*a*c + b**2)*(2*a*c - b**2)/(2*a**2*(4*a
*c - b**2))) - 2*a**4*b**5*(b/(2*a**2) + sqrt(-4*a*c + b**2)*(2*a*c - b**2)/(2*a**2*(4*a*c - b**2)))**2 - 3*a*
*4*b**2*c**2*(b/(2*a**2) + sqrt(-4*a*c + b**2)*(2*a*c - b**2)/(2*a**2*(4*a*c - b**2))) + a**3*b**4*c*(b/(2*a**
2) + sqrt(-4*a*c + b**2)*(2*a*c - b**2)/(2*a**2*(4*a*c - b**2))) - 4*a**3*b*c**3 + 25*a**2*b**3*c**2 - 14*a*b*
*5*c + 2*b**7)/(2*a**3*c**4 + 15*a**2*b**2*c**3 - 12*a*b**4*c**2 + 2*b**6*c)) - 1/(a*x) - b*log(x)/a**2

________________________________________________________________________________________

Giac [A]  time = 1.11534, size = 107, normalized size = 1.32 \begin{align*} \frac{b \log \left (c x^{2} + b x + a\right )}{2 \, a^{2}} - \frac{b \log \left ({\left | x \right |}\right )}{a^{2}} + \frac{{\left (b^{2} - 2 \, a c\right )} \arctan \left (\frac{2 \, c x + b}{\sqrt{-b^{2} + 4 \, a c}}\right )}{\sqrt{-b^{2} + 4 \, a c} a^{2}} - \frac{1}{a x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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