3.5.34 \(\int \frac {1}{(c+\frac {a}{x^2}+\frac {b}{x})^3 x^3} \, dx\) [434]

Optimal. Leaf size=107 \[ -\frac {x^3 (b+2 c x)}{2 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^2}+\frac {3 b x (2 a+b x)}{2 \left (b^2-4 a c\right )^2 \left (a+b x+c x^2\right )}+\frac {6 a b \tanh ^{-1}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{5/2}} \]

[Out]

-1/2*x^3*(2*c*x+b)/(-4*a*c+b^2)/(c*x^2+b*x+a)^2+3/2*b*x*(b*x+2*a)/(-4*a*c+b^2)^2/(c*x^2+b*x+a)+6*a*b*arctanh((
2*c*x+b)/(-4*a*c+b^2)^(1/2))/(-4*a*c+b^2)^(5/2)

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Rubi [A]
time = 0.03, antiderivative size = 107, 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, 742, 736, 632, 212} \begin {gather*} \frac {3 b x (2 a+b x)}{2 \left (b^2-4 a c\right )^2 \left (a+b x+c x^2\right )}-\frac {x^3 (b+2 c x)}{2 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^2}+\frac {6 a b \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^3),x]

[Out]

-1/2*(x^3*(b + 2*c*x))/((b^2 - 4*a*c)*(a + b*x + c*x^2)^2) + (3*b*x*(2*a + b*x))/(2*(b^2 - 4*a*c)^2*(a + b*x +
c*x^2)) + (6*a*b*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 736

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[2*(2*p + 3)*((c*d
^2 - b*d*e + a*e^2)/((p + 1)*(b^2 - 4*a*c))), Int[(d + e*x)^(m - 2)*(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] && EqQ[m +
2*p + 2, 0] && LtQ[p, -1]

Rule 742

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

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^3} \, dx &=\int \frac {x^3}{\left (a+b x+c x^2\right )^3} \, dx\\ &=-\frac {x^3 (b+2 c x)}{2 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^2}+\frac {(3 b) \int \frac {x^2}{\left (a+b x+c x^2\right )^2} \, dx}{2 \left (b^2-4 a c\right )}\\ &=-\frac {x^3 (b+2 c x)}{2 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^2}+\frac {3 b x (2 a+b x)}{2 \left (b^2-4 a c\right )^2 \left (a+b x+c x^2\right )}-\frac {(3 a b) \int \frac {1}{a+b x+c x^2} \, dx}{\left (b^2-4 a c\right )^2}\\ &=-\frac {x^3 (b+2 c x)}{2 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^2}+\frac {3 b x (2 a+b x)}{2 \left (b^2-4 a c\right )^2 \left (a+b x+c x^2\right )}+\frac {(6 a b) \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^3 (b+2 c x)}{2 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^2}+\frac {3 b x (2 a+b x)}{2 \left (b^2-4 a c\right )^2 \left (a+b x+c x^2\right )}+\frac {6 a b \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*}

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Mathematica [A]
time = 0.13, size = 126, normalized size = 1.18 \begin {gather*} -\frac {8 a^3 c+b^4 x^2+a b x \left (2 b^2+b c x+6 c^2 x^2\right )+a^2 \left (b^2+10 b c x+16 c^2 x^2\right )}{2 c \left (b^2-4 a c\right )^2 (a+x (b+c x))^2}-\frac {6 a b \tan ^{-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]

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

[Out]

-1/2*(8*a^3*c + b^4*x^2 + a*b*x*(2*b^2 + b*c*x + 6*c^2*x^2) + a^2*(b^2 + 10*b*c*x + 16*c^2*x^2))/(c*(b^2 - 4*a
*c)^2*(a + x*(b + c*x))^2) - (6*a*b*ArcTan[(b + 2*c*x)/Sqrt[-b^2 + 4*a*c]])/(-b^2 + 4*a*c)^(5/2)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(222\) vs. \(2(99)=198\).
time = 0.04, size = 223, normalized size = 2.08

method	result	size
default	\(\frac {-\frac {3 a b c \,x^{3}}{16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}}-\frac {\left (16 a^{2} c^{2}+a \,b^{2} c +b^{4}\right ) x^{2}}{2 c \left (16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}\right )}-\frac {\left (5 a c +b^{2}\right ) a b x}{c \left (16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}\right )}-\frac {a^{2} \left (8 a c +b^{2}\right )}{2 c \left (16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}\right )}}{\left (c \,x^{2}+b x +a \right )^{2}}-\frac {6 a b \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\left (16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}\right ) \sqrt {4 a c -b^{2}}}\)	\(223\)
risch	\(\frac {-\frac {3 a b c \,x^{3}}{16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}}-\frac {\left (16 a^{2} c^{2}+a \,b^{2} c +b^{4}\right ) x^{2}}{2 c \left (16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}\right )}-\frac {\left (5 a c +b^{2}\right ) a b x}{c \left (16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}\right )}-\frac {a^{2} \left (8 a c +b^{2}\right )}{2 c \left (16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}\right )}}{\left (c \,x^{2}+b x +a \right )^{2}}-\frac {3 b a \ln \left (\left (32 a^{2} c^{3}-16 a \,b^{2} c^{2}+2 b^{4} c \right ) x -\left (-4 a c +b^{2}\right )^{\frac {5}{2}}+16 a^{2} b \,c^{2}-8 a \,b^{3} c +b^{5}\right )}{\left (-4 a c +b^{2}\right )^{\frac {5}{2}}}+\frac {3 b a \ln \left (\left (-32 a^{2} c^{3}+16 a \,b^{2} c^{2}-2 b^{4} c \right ) x -\left (-4 a c +b^{2}\right )^{\frac {5}{2}}-16 a^{2} b \,c^{2}+8 a \,b^{3} c -b^{5}\right )}{\left (-4 a c +b^{2}\right )^{\frac {5}{2}}}\)	\(314\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-3*a*b*c/(16*a^2*c^2-8*a*b^2*c+b^4)*x^3-1/2*(16*a^2*c^2+a*b^2*c+b^4)/c/(16*a^2*c^2-8*a*b^2*c+b^4)*x^2-(5*a*c+
b^2)*a*b/c/(16*a^2*c^2-8*a*b^2*c+b^4)*x-1/2*a^2*(8*a*c+b^2)/c/(16*a^2*c^2-8*a*b^2*c+b^4))/(c*x^2+b*x+a)^2-6*a*
b/(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))

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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^3,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

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 426 vs. \(2 (99) = 198\).
time = 0.39, size = 872, normalized size = 8.15 \begin {gather*} \left [-\frac {a^{2} b^{4} + 4 \, a^{3} b^{2} c - 32 \, a^{4} c^{2} + 6 \, {\left (a b^{3} c^{2} - 4 \, a^{2} b c^{3}\right )} x^{3} + {\left (b^{6} - 3 \, a b^{4} c + 12 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}\right )} x^{2} - 6 \, {\left (a b c^{3} x^{4} + 2 \, a b^{2} c^{2} x^{3} + 2 \, a^{2} b^{2} c x + a^{3} b c + {\left (a b^{3} c + 2 \, a^{2} b c^{2}\right )} x^{2}\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 (a b^{5} + a^{2} b^{3} c - 20 \, a^{3} b c^{2}\right )} x}{2 \, {\left (a^{2} b^{6} c - 12 \, a^{3} b^{4} c^{2} + 48 \, a^{4} b^{2} c^{3} - 64 \, a^{5} c^{4} + {\left (b^{6} c^{3} - 12 \, a b^{4} c^{4} + 48 \, a^{2} b^{2} c^{5} - 64 \, a^{3} c^{6}\right )} x^{4} + 2 \, {\left (b^{7} c^{2} - 12 \, a b^{5} c^{3} + 48 \, a^{2} b^{3} c^{4} - 64 \, a^{3} b c^{5}\right )} x^{3} + {\left (b^{8} c - 10 \, a b^{6} c^{2} + 24 \, a^{2} b^{4} c^{3} + 32 \, a^{3} b^{2} c^{4} - 128 \, a^{4} c^{5}\right )} x^{2} + 2 \, {\left (a b^{7} c - 12 \, a^{2} b^{5} c^{2} + 48 \, a^{3} b^{3} c^{3} - 64 \, a^{4} b c^{4}\right )} x\right )}}, -\frac {a^{2} b^{4} + 4 \, a^{3} b^{2} c - 32 \, a^{4} c^{2} + 6 \, {\left (a b^{3} c^{2} - 4 \, a^{2} b c^{3}\right )} x^{3} + {\left (b^{6} - 3 \, a b^{4} c + 12 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}\right )} x^{2} - 12 \, {\left (a b c^{3} x^{4} + 2 \, a b^{2} c^{2} x^{3} + 2 \, a^{2} b^{2} c x + a^{3} b c + {\left (a b^{3} c + 2 \, a^{2} b c^{2}\right )} x^{2}\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 (a b^{5} + a^{2} b^{3} c - 20 \, a^{3} b c^{2}\right )} x}{2 \, {\left (a^{2} b^{6} c - 12 \, a^{3} b^{4} c^{2} + 48 \, a^{4} b^{2} c^{3} - 64 \, a^{5} c^{4} + {\left (b^{6} c^{3} - 12 \, a b^{4} c^{4} + 48 \, a^{2} b^{2} c^{5} - 64 \, a^{3} c^{6}\right )} x^{4} + 2 \, {\left (b^{7} c^{2} - 12 \, a b^{5} c^{3} + 48 \, a^{2} b^{3} c^{4} - 64 \, a^{3} b c^{5}\right )} x^{3} + {\left (b^{8} c - 10 \, a b^{6} c^{2} + 24 \, a^{2} b^{4} c^{3} + 32 \, a^{3} b^{2} c^{4} - 128 \, a^{4} c^{5}\right )} x^{2} + 2 \, {\left (a b^{7} c - 12 \, a^{2} b^{5} c^{2} + 48 \, a^{3} b^{3} c^{3} - 64 \, a^{4} b c^{4}\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^3,x, algorithm="fricas")

[Out]

[-1/2*(a^2*b^4 + 4*a^3*b^2*c - 32*a^4*c^2 + 6*(a*b^3*c^2 - 4*a^2*b*c^3)*x^3 + (b^6 - 3*a*b^4*c + 12*a^2*b^2*c^
2 - 64*a^3*c^3)*x^2 - 6*(a*b*c^3*x^4 + 2*a*b^2*c^2*x^3 + 2*a^2*b^2*c*x + a^3*b*c + (a*b^3*c + 2*a^2*b*c^2)*x^2
)*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*(a*b^5 + a^2*b^3*c - 20*a^3*b*c^2)*x)/(a^2*b^6*c - 12*a^3*b^4*c^2 + 48*a^4*b^2*c^3 - 64*a^5*c^4 + (b^6*c^
3 - 12*a*b^4*c^4 + 48*a^2*b^2*c^5 - 64*a^3*c^6)*x^4 + 2*(b^7*c^2 - 12*a*b^5*c^3 + 48*a^2*b^3*c^4 - 64*a^3*b*c^
5)*x^3 + (b^8*c - 10*a*b^6*c^2 + 24*a^2*b^4*c^3 + 32*a^3*b^2*c^4 - 128*a^4*c^5)*x^2 + 2*(a*b^7*c - 12*a^2*b^5*
c^2 + 48*a^3*b^3*c^3 - 64*a^4*b*c^4)*x), -1/2*(a^2*b^4 + 4*a^3*b^2*c - 32*a^4*c^2 + 6*(a*b^3*c^2 - 4*a^2*b*c^3
)*x^3 + (b^6 - 3*a*b^4*c + 12*a^2*b^2*c^2 - 64*a^3*c^3)*x^2 - 12*(a*b*c^3*x^4 + 2*a*b^2*c^2*x^3 + 2*a^2*b^2*c*
x + a^3*b*c + (a*b^3*c + 2*a^2*b*c^2)*x^2)*sqrt(-b^2 + 4*a*c)*arctan(-sqrt(-b^2 + 4*a*c)*(2*c*x + b)/(b^2 - 4*
a*c)) + 2*(a*b^5 + a^2*b^3*c - 20*a^3*b*c^2)*x)/(a^2*b^6*c - 12*a^3*b^4*c^2 + 48*a^4*b^2*c^3 - 64*a^5*c^4 + (b
^6*c^3 - 12*a*b^4*c^4 + 48*a^2*b^2*c^5 - 64*a^3*c^6)*x^4 + 2*(b^7*c^2 - 12*a*b^5*c^3 + 48*a^2*b^3*c^4 - 64*a^3
*b*c^5)*x^3 + (b^8*c - 10*a*b^6*c^2 + 24*a^2*b^4*c^3 + 32*a^3*b^2*c^4 - 128*a^4*c^5)*x^2 + 2*(a*b^7*c - 12*a^2
*b^5*c^2 + 48*a^3*b^3*c^3 - 64*a^4*b*c^4)*x)]

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 513 vs. \(2 (102) = 204\).
time = 0.66, size = 513, normalized size = 4.79 \begin {gather*} 3 a b \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{5}}} \log {\left (x + \frac {- 192 a^{4} b c^{3} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{5}}} + 144 a^{3} b^{3} c^{2} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{5}}} - 36 a^{2} b^{5} c \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{5}}} + 3 a b^{7} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{5}}} + 3 a b^{2}}{6 a b c} \right )} - 3 a b \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{5}}} \log {\left (x + \frac {192 a^{4} b c^{3} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{5}}} - 144 a^{3} b^{3} c^{2} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{5}}} + 36 a^{2} b^{5} c \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{5}}} - 3 a b^{7} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{5}}} + 3 a b^{2}}{6 a b c} \right )} + \frac {- 8 a^{3} c - a^{2} b^{2} - 6 a b c^{2} x^{3} + x^{2} \left (- 16 a^{2} c^{2} - a b^{2} c - b^{4}\right ) + x \left (- 10 a^{2} b c - 2 a b^{3}\right )}{32 a^{4} c^{3} - 16 a^{3} b^{2} c^{2} + 2 a^{2} b^{4} c + x^{4} \cdot \left (32 a^{2} c^{5} - 16 a b^{2} c^{4} + 2 b^{4} c^{3}\right ) + x^{3} \cdot \left (64 a^{2} b c^{4} - 32 a b^{3} c^{3} + 4 b^{5} c^{2}\right ) + x^{2} \cdot \left (64 a^{3} c^{4} - 12 a b^{4} c^{2} + 2 b^{6} c\right ) + x \left (64 a^{3} b c^{3} - 32 a^{2} b^{3} c^{2} + 4 a b^{5} c\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3*a*b*sqrt(-1/(4*a*c - b**2)**5)*log(x + (-192*a**4*b*c**3*sqrt(-1/(4*a*c - b**2)**5) + 144*a**3*b**3*c**2*sqr
t(-1/(4*a*c - b**2)**5) - 36*a**2*b**5*c*sqrt(-1/(4*a*c - b**2)**5) + 3*a*b**7*sqrt(-1/(4*a*c - b**2)**5) + 3*
a*b**2)/(6*a*b*c)) - 3*a*b*sqrt(-1/(4*a*c - b**2)**5)*log(x + (192*a**4*b*c**3*sqrt(-1/(4*a*c - b**2)**5) - 14
4*a**3*b**3*c**2*sqrt(-1/(4*a*c - b**2)**5) + 36*a**2*b**5*c*sqrt(-1/(4*a*c - b**2)**5) - 3*a*b**7*sqrt(-1/(4*
a*c - b**2)**5) + 3*a*b**2)/(6*a*b*c)) + (-8*a**3*c - a**2*b**2 - 6*a*b*c**2*x**3 + x**2*(-16*a**2*c**2 - a*b*
*2*c - b**4) + x*(-10*a**2*b*c - 2*a*b**3))/(32*a**4*c**3 - 16*a**3*b**2*c**2 + 2*a**2*b**4*c + x**4*(32*a**2*
c**5 - 16*a*b**2*c**4 + 2*b**4*c**3) + x**3*(64*a**2*b*c**4 - 32*a*b**3*c**3 + 4*b**5*c**2) + x**2*(64*a**3*c*
*4 - 12*a*b**4*c**2 + 2*b**6*c) + x*(64*a**3*b*c**3 - 32*a**2*b**3*c**2 + 4*a*b**5*c))

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Giac [A]
time = 3.35, size = 163, normalized size = 1.52 \begin {gather*} -\frac {6 \, a b \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 {6 \, a b c^{2} x^{3} + b^{4} x^{2} + a b^{2} c x^{2} + 16 \, a^{2} c^{2} x^{2} + 2 \, a b^{3} x + 10 \, a^{2} b c x + a^{2} b^{2} + 8 \, a^{3} c}{2 \, {\left (b^{4} c - 8 \, a b^{2} c^{2} + 16 \, a^{2} c^{3}\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^3,x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 1.43, size = 271, normalized size = 2.53 \begin {gather*} -\frac {\frac {x^2\,\left (16\,a^2\,c^2+a\,b^2\,c+b^4\right )}{2\,c\,\left (16\,a^2\,c^2-8\,a\,b^2\,c+b^4\right )}+\frac {a^2\,\left (b^2+8\,a\,c\right )}{2\,c\,\left (16\,a^2\,c^2-8\,a\,b^2\,c+b^4\right )}+\frac {3\,a\,b\,c\,x^3}{16\,a^2\,c^2-8\,a\,b^2\,c+b^4}+\frac {a\,b\,x\,\left (b^2+5\,a\,c\right )}{c\,\left (16\,a^2\,c^2-8\,a\,b^2\,c+b^4\right )}}{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 {6\,a\,b\,\mathrm {atan}\left (\frac {\left (\frac {3\,a\,b^2}{{\left (4\,a\,c-b^2\right )}^{5/2}}+\frac {6\,a\,b\,c\,x}{{\left (4\,a\,c-b^2\right )}^{5/2}}\right )\,\left (16\,a^2\,c^2-8\,a\,b^2\,c+b^4\right )}{3\,a\,b}\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^3*(c + a/x^2 + b/x)^3),x)

[Out]

- ((x^2*(b^4 + 16*a^2*c^2 + a*b^2*c))/(2*c*(b^4 + 16*a^2*c^2 - 8*a*b^2*c)) + (a^2*(8*a*c + b^2))/(2*c*(b^4 + 1
6*a^2*c^2 - 8*a*b^2*c)) + (3*a*b*c*x^3)/(b^4 + 16*a^2*c^2 - 8*a*b^2*c) + (a*b*x*(5*a*c + b^2))/(c*(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) - (6*a*b*atan((((3*a*b^2)/(4*a*
c - b^2)^(5/2) + (6*a*b*c*x)/(4*a*c - b^2)^(5/2))*(b^4 + 16*a^2*c^2 - 8*a*b^2*c))/(3*a*b)))/(4*a*c - b^2)^(5/2
)

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