3.5.35 \(\int \frac {1}{(c+\frac {a}{x^2}+\frac {b}{x})^3 x^4} \, dx\) [435]

Optimal. Leaf size=115 \[ \frac {x (2 a+b x)}{2 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^2}+\frac {3 a b+\left (b^2+2 a c\right ) x}{\left (b^2-4 a c\right )^2 \left (a+b x+c x^2\right )}-\frac {2 \left (b^2+2 a c\right ) \tanh ^{-1}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{5/2}} \]

[Out]

1/2*x*(b*x+2*a)/(-4*a*c+b^2)/(c*x^2+b*x+a)^2+(3*a*b+(2*a*c+b^2)*x)/(-4*a*c+b^2)^2/(c*x^2+b*x+a)-2*(2*a*c+b^2)*
arctanh((2*c*x+b)/(-4*a*c+b^2)^(1/2))/(-4*a*c+b^2)^(5/2)

________________________________________________________________________________________

Rubi [A]
time = 0.04, antiderivative size = 115, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.278, Rules used = {1368, 752, 652, 632, 212} \begin {gather*} \frac {x (2 a+b x)}{2 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^2}+\frac {x \left (2 a c+b^2\right )+3 a b}{\left (b^2-4 a c\right )^2 \left (a+b x+c x^2\right )}-\frac {2 \left (2 a c+b^2\right ) \tanh ^{-1}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((c + a/x^2 + b/x)^3*x^4),x]

[Out]

(x*(2*a + b*x))/(2*(b^2 - 4*a*c)*(a + b*x + c*x^2)^2) + (3*a*b + (b^2 + 2*a*c)*x)/((b^2 - 4*a*c)^2*(a + b*x +
c*x^2)) - (2*(b^2 + 2*a*c)*ArcTanh[(b + 2*c*x)/Sqrt[b^2 - 4*a*c]])/(b^2 - 4*a*c)^(5/2)

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 632

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 652

Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((b*d - 2*a*e + (2*c*d -
b*e)*x)/((p + 1)*(b^2 - 4*a*c)))*(a + b*x + c*x^2)^(p + 1), x] - Dist[(2*p + 3)*((2*c*d - b*e)/((p + 1)*(b^2 -
4*a*c))), Int[(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[2*c*d - b*e, 0] && NeQ[b^
2 - 4*a*c, 0] && LtQ[p, -1] && NeQ[p, -3/2]

Rule 752

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(d + e*x)^(m - 1)*(d
*b - 2*a*e + (2*c*d - b*e)*x)*((a + b*x + c*x^2)^(p + 1)/((p + 1)*(b^2 - 4*a*c))), x] + Dist[1/((p + 1)*(b^2 -
4*a*c)), Int[(d + e*x)^(m - 2)*Simp[e*(2*a*e*(m - 1) + b*d*(2*p - m + 4)) - 2*c*d^2*(2*p + 3) + e*(b*e - 2*d*
c)*(m + 2*p + 2)*x, x]*(a + b*x + c*x^2)^(p + 1), 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] && LtQ[p, -1] && GtQ[m, 1] && IntQuadraticQ[a, b, c, d,
e, m, p, x]

Rule 1368

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]

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]
time = 0.09, size = 131, normalized size = 1.14 \begin {gather*} \frac {1}{2} \left (\frac {b^2 x+a (b-2 c x)}{c \left (-b^2+4 a c\right ) (a+x (b+c x))^2}+\frac {\left (b^2+2 a c\right ) (b+2 c x)}{c \left (b^2-4 a c\right )^2 (a+x (b+c x))}+\frac {4 \left (b^2+2 a c\right ) \tan ^{-1}\left (\frac {b+2 c x}{\sqrt {-b^2+4 a c}}\right )}{\left (-b^2+4 a c\right )^{5/2}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((b^2*x + a*(b - 2*c*x))/(c*(-b^2 + 4*a*c)*(a + x*(b + c*x))^2) + ((b^2 + 2*a*c)*(b + 2*c*x))/(c*(b^2 - 4*a*c)
^2*(a + x*(b + c*x))) + (4*(b^2 + 2*a*c)*ArcTan[(b + 2*c*x)/Sqrt[-b^2 + 4*a*c]])/(-b^2 + 4*a*c)^(5/2))/2

________________________________________________________________________________________

Maple [A]
time = 0.05, size = 210, normalized size = 1.83

method	result	size
default	\(\frac {\frac {c \left (2 a c +b^{2}\right ) x^{3}}{16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}}+\frac {3 b \left (2 a c +b^{2}\right ) x^{2}}{2 \left (16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}\right )}-\frac {a \left (2 a c -5 b^{2}\right ) x}{16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}}+\frac {3 a^{2} b}{16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}}}{\left (c \,x^{2}+b x +a \right )^{2}}+\frac {2 \left (2 a c +b^{2}\right ) \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\left (16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}\right ) \sqrt {4 a c -b^{2}}}\)	\(210\)
risch	\(\frac {\frac {c \left (2 a c +b^{2}\right ) x^{3}}{16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}}+\frac {3 b \left (2 a c +b^{2}\right ) x^{2}}{2 \left (16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}\right )}-\frac {a \left (2 a c -5 b^{2}\right ) x}{16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}}+\frac {3 a^{2} b}{16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}}}{\left (c \,x^{2}+b x +a \right )^{2}}-\frac {2 \ln \left (\left (32 a^{2} c^{3}-16 a \,b^{2} c^{2}+2 b^{4} c \right ) x +\left (-4 a c +b^{2}\right )^{\frac {5}{2}}+16 a^{2} b \,c^{2}-8 a \,b^{3} c +b^{5}\right ) a c}{\left (-4 a c +b^{2}\right )^{\frac {5}{2}}}-\frac {\ln \left (\left (32 a^{2} c^{3}-16 a \,b^{2} c^{2}+2 b^{4} c \right ) x +\left (-4 a c +b^{2}\right )^{\frac {5}{2}}+16 a^{2} b \,c^{2}-8 a \,b^{3} c +b^{5}\right ) b^{2}}{\left (-4 a c +b^{2}\right )^{\frac {5}{2}}}+\frac {2 \ln \left (\left (-32 a^{2} c^{3}+16 a \,b^{2} c^{2}-2 b^{4} c \right ) x +\left (-4 a c +b^{2}\right )^{\frac {5}{2}}-16 a^{2} b \,c^{2}+8 a \,b^{3} c -b^{5}\right ) a c}{\left (-4 a c +b^{2}\right )^{\frac {5}{2}}}+\frac {\ln \left (\left (-32 a^{2} c^{3}+16 a \,b^{2} c^{2}-2 b^{4} c \right ) x +\left (-4 a c +b^{2}\right )^{\frac {5}{2}}-16 a^{2} b \,c^{2}+8 a \,b^{3} c -b^{5}\right ) b^{2}}{\left (-4 a c +b^{2}\right )^{\frac {5}{2}}}\)	\(436\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(c+a/x^2+b/x)^3/x^4,x,method=_RETURNVERBOSE)

[Out]

(c*(2*a*c+b^2)/(16*a^2*c^2-8*a*b^2*c+b^4)*x^3+3/2*b*(2*a*c+b^2)/(16*a^2*c^2-8*a*b^2*c+b^4)*x^2-a*(2*a*c-5*b^2)
/(16*a^2*c^2-8*a*b^2*c+b^4)*x+3*a^2*b/(16*a^2*c^2-8*a*b^2*c+b^4))/(c*x^2+b*x+a)^2+2*(2*a*c+b^2)/(16*a^2*c^2-8*
a*b^2*c+b^4)/(4*a*c-b^2)^(1/2)*arctan((2*c*x+b)/(4*a*c-b^2)^(1/2))

________________________________________________________________________________________

Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*a*c-b^2>0)', see `assume?` f
or more deta

________________________________________________________________________________________

Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 433 vs. \(2 (109) = 218\).
time = 0.37, size = 887, normalized size = 7.71 \begin {gather*} \left [\frac {6 \, a^{2} b^{3} - 24 \, a^{3} b c + 2 \, {\left (b^{4} c - 2 \, a b^{2} c^{2} - 8 \, a^{2} c^{3}\right )} x^{3} + 3 \, {\left (b^{5} - 2 \, a b^{3} c - 8 \, a^{2} b c^{2}\right )} x^{2} + 2 \, {\left ({\left (b^{2} c^{2} + 2 \, a c^{3}\right )} x^{4} + a^{2} b^{2} + 2 \, a^{3} c + 2 \, {\left (b^{3} c + 2 \, a b c^{2}\right )} x^{3} + {\left (b^{4} + 4 \, a b^{2} c + 4 \, a^{2} c^{2}\right )} x^{2} + 2 \, {\left (a b^{3} + 2 \, a^{2} b c\right )} x\right )} \sqrt {b^{2} - 4 \, a c} \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 \, {\left (5 \, a b^{4} - 22 \, a^{2} b^{2} c + 8 \, a^{3} c^{2}\right )} x}{2 \, {\left (a^{2} b^{6} - 12 \, a^{3} b^{4} c + 48 \, a^{4} b^{2} c^{2} - 64 \, a^{5} c^{3} + {\left (b^{6} c^{2} - 12 \, a b^{4} c^{3} + 48 \, a^{2} b^{2} c^{4} - 64 \, a^{3} c^{5}\right )} x^{4} + 2 \, {\left (b^{7} c - 12 \, a b^{5} c^{2} + 48 \, a^{2} b^{3} c^{3} - 64 \, a^{3} b c^{4}\right )} x^{3} + {\left (b^{8} - 10 \, a b^{6} c + 24 \, a^{2} b^{4} c^{2} + 32 \, a^{3} b^{2} c^{3} - 128 \, a^{4} c^{4}\right )} x^{2} + 2 \, {\left (a b^{7} - 12 \, a^{2} b^{5} c + 48 \, a^{3} b^{3} c^{2} - 64 \, a^{4} b c^{3}\right )} x\right )}}, \frac {6 \, a^{2} b^{3} - 24 \, a^{3} b c + 2 \, {\left (b^{4} c - 2 \, a b^{2} c^{2} - 8 \, a^{2} c^{3}\right )} x^{3} + 3 \, {\left (b^{5} - 2 \, a b^{3} c - 8 \, a^{2} b c^{2}\right )} x^{2} - 4 \, {\left ({\left (b^{2} c^{2} + 2 \, a c^{3}\right )} x^{4} + a^{2} b^{2} + 2 \, a^{3} c + 2 \, {\left (b^{3} c + 2 \, a b c^{2}\right )} x^{3} + {\left (b^{4} + 4 \, a b^{2} c + 4 \, a^{2} c^{2}\right )} x^{2} + 2 \, {\left (a b^{3} + 2 \, a^{2} b c\right )} x\right )} \sqrt {-b^{2} + 4 \, a c} \arctan \left (-\frac {\sqrt {-b^{2} + 4 \, a c} {\left (2 \, c x + b\right )}}{b^{2} - 4 \, a c}\right ) + 2 \, {\left (5 \, a b^{4} - 22 \, a^{2} b^{2} c + 8 \, a^{3} c^{2}\right )} x}{2 \, {\left (a^{2} b^{6} - 12 \, a^{3} b^{4} c + 48 \, a^{4} b^{2} c^{2} - 64 \, a^{5} c^{3} + {\left (b^{6} c^{2} - 12 \, a b^{4} c^{3} + 48 \, a^{2} b^{2} c^{4} - 64 \, a^{3} c^{5}\right )} x^{4} + 2 \, {\left (b^{7} c - 12 \, a b^{5} c^{2} + 48 \, a^{2} b^{3} c^{3} - 64 \, a^{3} b c^{4}\right )} x^{3} + {\left (b^{8} - 10 \, a b^{6} c + 24 \, a^{2} b^{4} c^{2} + 32 \, a^{3} b^{2} c^{3} - 128 \, a^{4} c^{4}\right )} x^{2} + 2 \, {\left (a b^{7} - 12 \, a^{2} b^{5} c + 48 \, a^{3} b^{3} c^{2} - 64 \, a^{4} b c^{3}\right )} x\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*(6*a^2*b^3 - 24*a^3*b*c + 2*(b^4*c - 2*a*b^2*c^2 - 8*a^2*c^3)*x^3 + 3*(b^5 - 2*a*b^3*c - 8*a^2*b*c^2)*x^2
+ 2*((b^2*c^2 + 2*a*c^3)*x^4 + a^2*b^2 + 2*a^3*c + 2*(b^3*c + 2*a*b*c^2)*x^3 + (b^4 + 4*a*b^2*c + 4*a^2*c^2)*
x^2 + 2*(a*b^3 + 2*a^2*b*c)*x)*sqrt(b^2 - 4*a*c)*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*(5*a*b^4 - 22*a^2*b^2*c + 8*a^3*c^2)*x)/(a^2*b^6 - 12*a^3*b^4*c + 48*a^4*b^2
*c^2 - 64*a^5*c^3 + (b^6*c^2 - 12*a*b^4*c^3 + 48*a^2*b^2*c^4 - 64*a^3*c^5)*x^4 + 2*(b^7*c - 12*a*b^5*c^2 + 48*
a^2*b^3*c^3 - 64*a^3*b*c^4)*x^3 + (b^8 - 10*a*b^6*c + 24*a^2*b^4*c^2 + 32*a^3*b^2*c^3 - 128*a^4*c^4)*x^2 + 2*(
a*b^7 - 12*a^2*b^5*c + 48*a^3*b^3*c^2 - 64*a^4*b*c^3)*x), 1/2*(6*a^2*b^3 - 24*a^3*b*c + 2*(b^4*c - 2*a*b^2*c^2
- 8*a^2*c^3)*x^3 + 3*(b^5 - 2*a*b^3*c - 8*a^2*b*c^2)*x^2 - 4*((b^2*c^2 + 2*a*c^3)*x^4 + a^2*b^2 + 2*a^3*c + 2
*(b^3*c + 2*a*b*c^2)*x^3 + (b^4 + 4*a*b^2*c + 4*a^2*c^2)*x^2 + 2*(a*b^3 + 2*a^2*b*c)*x)*sqrt(-b^2 + 4*a*c)*arc
tan(-sqrt(-b^2 + 4*a*c)*(2*c*x + b)/(b^2 - 4*a*c)) + 2*(5*a*b^4 - 22*a^2*b^2*c + 8*a^3*c^2)*x)/(a^2*b^6 - 12*a
^3*b^4*c + 48*a^4*b^2*c^2 - 64*a^5*c^3 + (b^6*c^2 - 12*a*b^4*c^3 + 48*a^2*b^2*c^4 - 64*a^3*c^5)*x^4 + 2*(b^7*c
- 12*a*b^5*c^2 + 48*a^2*b^3*c^3 - 64*a^3*b*c^4)*x^3 + (b^8 - 10*a*b^6*c + 24*a^2*b^4*c^2 + 32*a^3*b^2*c^3 - 1
28*a^4*c^4)*x^2 + 2*(a*b^7 - 12*a^2*b^5*c + 48*a^3*b^3*c^2 - 64*a^4*b*c^3)*x)]

________________________________________________________________________________________

Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 570 vs. \(2 (107) = 214\).
time = 0.74, size = 570, normalized size = 4.96 \begin {gather*} - \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{5}}} \cdot \left (2 a c + b^{2}\right ) \log {\left (x + \frac {- 64 a^{3} c^{3} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{5}}} \cdot \left (2 a c + b^{2}\right ) + 48 a^{2} b^{2} c^{2} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{5}}} \cdot \left (2 a c + b^{2}\right ) - 12 a b^{4} c \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{5}}} \cdot \left (2 a c + b^{2}\right ) + 2 a b c + b^{6} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{5}}} \cdot \left (2 a c + b^{2}\right ) + b^{3}}{4 a c^{2} + 2 b^{2} c} \right )} + \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{5}}} \cdot \left (2 a c + b^{2}\right ) \log {\left (x + \frac {64 a^{3} c^{3} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{5}}} \cdot \left (2 a c + b^{2}\right ) - 48 a^{2} b^{2} c^{2} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{5}}} \cdot \left (2 a c + b^{2}\right ) + 12 a b^{4} c \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{5}}} \cdot \left (2 a c + b^{2}\right ) + 2 a b c - b^{6} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{5}}} \cdot \left (2 a c + b^{2}\right ) + b^{3}}{4 a c^{2} + 2 b^{2} c} \right )} + \frac {6 a^{2} b + x^{3} \cdot \left (4 a c^{2} + 2 b^{2} c\right ) + x^{2} \cdot \left (6 a b c + 3 b^{3}\right ) + x \left (- 4 a^{2} c + 10 a b^{2}\right )}{32 a^{4} c^{2} - 16 a^{3} b^{2} c + 2 a^{2} b^{4} + x^{4} \cdot \left (32 a^{2} c^{4} - 16 a b^{2} c^{3} + 2 b^{4} c^{2}\right ) + x^{3} \cdot \left (64 a^{2} b c^{3} - 32 a b^{3} c^{2} + 4 b^{5} c\right ) + x^{2} \cdot \left (64 a^{3} c^{3} - 12 a b^{4} c + 2 b^{6}\right ) + x \left (64 a^{3} b c^{2} - 32 a^{2} b^{3} c + 4 a b^{5}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-sqrt(-1/(4*a*c - b**2)**5)*(2*a*c + b**2)*log(x + (-64*a**3*c**3*sqrt(-1/(4*a*c - b**2)**5)*(2*a*c + b**2) +
48*a**2*b**2*c**2*sqrt(-1/(4*a*c - b**2)**5)*(2*a*c + b**2) - 12*a*b**4*c*sqrt(-1/(4*a*c - b**2)**5)*(2*a*c +
b**2) + 2*a*b*c + b**6*sqrt(-1/(4*a*c - b**2)**5)*(2*a*c + b**2) + b**3)/(4*a*c**2 + 2*b**2*c)) + sqrt(-1/(4*a
*c - b**2)**5)*(2*a*c + b**2)*log(x + (64*a**3*c**3*sqrt(-1/(4*a*c - b**2)**5)*(2*a*c + b**2) - 48*a**2*b**2*c
**2*sqrt(-1/(4*a*c - b**2)**5)*(2*a*c + b**2) + 12*a*b**4*c*sqrt(-1/(4*a*c - b**2)**5)*(2*a*c + b**2) + 2*a*b*
c - b**6*sqrt(-1/(4*a*c - b**2)**5)*(2*a*c + b**2) + b**3)/(4*a*c**2 + 2*b**2*c)) + (6*a**2*b + x**3*(4*a*c**2
+ 2*b**2*c) + x**2*(6*a*b*c + 3*b**3) + x*(-4*a**2*c + 10*a*b**2))/(32*a**4*c**2 - 16*a**3*b**2*c + 2*a**2*b*
*4 + x**4*(32*a**2*c**4 - 16*a*b**2*c**3 + 2*b**4*c**2) + x**3*(64*a**2*b*c**3 - 32*a*b**3*c**2 + 4*b**5*c) +
x**2*(64*a**3*c**3 - 12*a*b**4*c + 2*b**6) + x*(64*a**3*b*c**2 - 32*a**2*b**3*c + 4*a*b**5))

________________________________________________________________________________________

Giac [A]
time = 3.23, size = 154, normalized size = 1.34 \begin {gather*} \frac {2 \, {\left (b^{2} + 2 \, a c\right )} \arctan \left (\frac {2 \, c x + b}{\sqrt {-b^{2} + 4 \, a c}}\right )}{{\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} \sqrt {-b^{2} + 4 \, a c}} + \frac {2 \, b^{2} c x^{3} + 4 \, a c^{2} x^{3} + 3 \, b^{3} x^{2} + 6 \, a b c x^{2} + 10 \, a b^{2} x - 4 \, a^{2} c x + 6 \, a^{2} b}{2 \, {\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} {\left (c x^{2} + b x + a\right )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*(b^2 + 2*a*c)*arctan((2*c*x + b)/sqrt(-b^2 + 4*a*c))/((b^4 - 8*a*b^2*c + 16*a^2*c^2)*sqrt(-b^2 + 4*a*c)) + 1
/2*(2*b^2*c*x^3 + 4*a*c^2*x^3 + 3*b^3*x^2 + 6*a*b*c*x^2 + 10*a*b^2*x - 4*a^2*c*x + 6*a^2*b)/((b^4 - 8*a*b^2*c
+ 16*a^2*c^2)*(c*x^2 + b*x + a)^2)

________________________________________________________________________________________

Mupad [B]
time = 1.50, size = 313, normalized size = 2.72 \begin {gather*} \frac {\frac {3\,a^2\,b}{16\,a^2\,c^2-8\,a\,b^2\,c+b^4}-\frac {a\,x\,\left (2\,a\,c-5\,b^2\right )}{16\,a^2\,c^2-8\,a\,b^2\,c+b^4}+\frac {3\,b\,x^2\,\left (b^2+2\,a\,c\right )}{2\,\left (16\,a^2\,c^2-8\,a\,b^2\,c+b^4\right )}+\frac {c\,x^3\,\left (b^2+2\,a\,c\right )}{16\,a^2\,c^2-8\,a\,b^2\,c+b^4}}{x^2\,\left (b^2+2\,a\,c\right )+a^2+c^2\,x^4+2\,a\,b\,x+2\,b\,c\,x^3}+\frac {2\,\mathrm {atan}\left (\frac {\left (\frac {\left (b^2+2\,a\,c\right )\,\left (16\,a^2\,b\,c^2-8\,a\,b^3\,c+b^5\right )}{{\left (4\,a\,c-b^2\right )}^{5/2}\,\left (16\,a^2\,c^2-8\,a\,b^2\,c+b^4\right )}+\frac {2\,c\,x\,\left (b^2+2\,a\,c\right )}{{\left (4\,a\,c-b^2\right )}^{5/2}}\right )\,\left (16\,a^2\,c^2-8\,a\,b^2\,c+b^4\right )}{b^2+2\,a\,c}\right )\,\left (b^2+2\,a\,c\right )}{{\left (4\,a\,c-b^2\right )}^{5/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^4*(c + a/x^2 + b/x)^3),x)

[Out]

((3*a^2*b)/(b^4 + 16*a^2*c^2 - 8*a*b^2*c) - (a*x*(2*a*c - 5*b^2))/(b^4 + 16*a^2*c^2 - 8*a*b^2*c) + (3*b*x^2*(2
*a*c + b^2))/(2*(b^4 + 16*a^2*c^2 - 8*a*b^2*c)) + (c*x^3*(2*a*c + b^2))/(b^4 + 16*a^2*c^2 - 8*a*b^2*c))/(x^2*(
2*a*c + b^2) + a^2 + c^2*x^4 + 2*a*b*x + 2*b*c*x^3) + (2*atan(((((2*a*c + b^2)*(b^5 + 16*a^2*b*c^2 - 8*a*b^3*c
))/((4*a*c - b^2)^(5/2)*(b^4 + 16*a^2*c^2 - 8*a*b^2*c)) + (2*c*x*(2*a*c + b^2))/(4*a*c - b^2)^(5/2))*(b^4 + 16
*a^2*c^2 - 8*a*b^2*c))/(2*a*c + b^2))*(2*a*c + b^2))/(4*a*c - b^2)^(5/2)

________________________________________________________________________________________