3.437 \(\int \frac {1}{(c+\frac {a}{x^2}+\frac {b}{x})^3 x^6} \, dx\)

Optimal. Leaf size=103 \[ -\frac {12 c^2 \tanh ^{-1}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{5/2}}+\frac {3 c (b+2 c x)}{\left (b^2-4 a c\right )^2 \left (a+b x+c x^2\right )}+\frac {-b-2 c x}{2 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^2} \]

[Out]

1/2*(-2*c*x-b)/(-4*a*c+b^2)/(c*x^2+b*x+a)^2+3*c*(2*c*x+b)/(-4*a*c+b^2)^2/(c*x^2+b*x+a)-12*c^2*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 = 0.04, antiderivative size = 101, normalized size of antiderivative = 0.98, number of steps used = 5, number of rules used = 4, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {1354, 614, 618, 206} \[ -\frac {12 c^2 \tanh ^{-1}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{5/2}}+\frac {3 c (b+2 c x)}{\left (b^2-4 a c\right )^2 \left (a+b x+c x^2\right )}-\frac {b+2 c x}{2 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[1/((c + a/x^2 + b/x)^3*x^6),x]

[Out]

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

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 614

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

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 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]

Rubi steps

\begin {align*} \int \frac {1}{\left (c+\frac {a}{x^2}+\frac {b}{x}\right )^3 x^6} \, dx &=\int \frac {1}{\left (a+b x+c x^2\right )^3} \, dx\\ &=-\frac {b+2 c x}{2 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^2}-\frac {(3 c) \int \frac {1}{\left (a+b x+c x^2\right )^2} \, dx}{b^2-4 a c}\\ &=-\frac {b+2 c x}{2 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^2}+\frac {3 c (b+2 c x)}{\left (b^2-4 a c\right )^2 \left (a+b x+c x^2\right )}+\frac {\left (6 c^2\right ) \int \frac {1}{a+b x+c x^2} \, dx}{\left (b^2-4 a c\right )^2}\\ &=-\frac {b+2 c x}{2 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^2}+\frac {3 c (b+2 c x)}{\left (b^2-4 a c\right )^2 \left (a+b x+c x^2\right )}-\frac {\left (12 c^2\right ) \operatorname {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 {b+2 c x}{2 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^2}+\frac {3 c (b+2 c x)}{\left (b^2-4 a c\right )^2 \left (a+b x+c x^2\right )}-\frac {12 c^2 \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 = 0.10, size = 97, normalized size = 0.94 \[ \frac {\frac {24 c^2 \tan ^{-1}\left (\frac {b+2 c x}{\sqrt {4 a c-b^2}}\right )}{\sqrt {4 a c-b^2}}-\frac {(b+2 c x) \left (-2 c \left (5 a+3 c x^2\right )+b^2-6 b c x\right )}{(a+x (b+c x))^2}}{2 \left (b^2-4 a c\right )^2} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((c + a/x^2 + b/x)^3*x^6),x]

[Out]

(-(((b + 2*c*x)*(b^2 - 6*b*c*x - 2*c*(5*a + 3*c*x^2)))/(a + x*(b + c*x))^2) + (24*c^2*ArcTan[(b + 2*c*x)/Sqrt[
-b^2 + 4*a*c]])/Sqrt[-b^2 + 4*a*c])/(2*(b^2 - 4*a*c)^2)

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fricas [B]  time = 0.98, size = 785, normalized size = 7.62 \[ \left [-\frac {b^{5} - 14 \, a b^{3} c + 40 \, a^{2} b c^{2} - 12 \, {\left (b^{2} c^{3} - 4 \, a c^{4}\right )} x^{3} - 18 \, {\left (b^{3} c^{2} - 4 \, a b c^{3}\right )} x^{2} - 12 \, {\left (c^{4} x^{4} + 2 \, b c^{3} x^{3} + 2 \, a b c^{2} x + a^{2} c^{2} + {\left (b^{2} c^{2} + 2 \, a c^{3}\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 ) - 4 \, {\left (b^{4} c + a b^{2} c^{2} - 20 \, a^{2} c^{3}\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 {b^{5} - 14 \, a b^{3} c + 40 \, a^{2} b c^{2} - 12 \, {\left (b^{2} c^{3} - 4 \, a c^{4}\right )} x^{3} - 18 \, {\left (b^{3} c^{2} - 4 \, a b c^{3}\right )} x^{2} + 24 \, {\left (c^{4} x^{4} + 2 \, b c^{3} x^{3} + 2 \, a b c^{2} x + a^{2} c^{2} + {\left (b^{2} c^{2} + 2 \, a c^{3}\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 ) - 4 \, {\left (b^{4} c + a b^{2} c^{2} - 20 \, a^{2} c^{3}\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 ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c+a/x^2+b/x)^3/x^6,x, algorithm="fricas")

[Out]

[-1/2*(b^5 - 14*a*b^3*c + 40*a^2*b*c^2 - 12*(b^2*c^3 - 4*a*c^4)*x^3 - 18*(b^3*c^2 - 4*a*b*c^3)*x^2 - 12*(c^4*x
^4 + 2*b*c^3*x^3 + 2*a*b*c^2*x + a^2*c^2 + (b^2*c^2 + 2*a*c^3)*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)) - 4*(b^4*c + a*b^2*c^2 - 20*a^2*c^3)*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*(b^5 - 14*a*b^
3*c + 40*a^2*b*c^2 - 12*(b^2*c^3 - 4*a*c^4)*x^3 - 18*(b^3*c^2 - 4*a*b*c^3)*x^2 + 24*(c^4*x^4 + 2*b*c^3*x^3 + 2
*a*b*c^2*x + a^2*c^2 + (b^2*c^2 + 2*a*c^3)*x^2)*sqrt(-b^2 + 4*a*c)*arctan(-sqrt(-b^2 + 4*a*c)*(2*c*x + b)/(b^2
- 4*a*c)) - 4*(b^4*c + a*b^2*c^2 - 20*a^2*c^3)*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)]

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giac [A]  time = 0.34, size = 136, normalized size = 1.32 \[ \frac {12 \, c^{2} \arctan \left (\frac {2 \, c x + b}{\sqrt {-b^{2} + 4 \, a c}}\right )}{{\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} \sqrt {-b^{2} + 4 \, a c}} + \frac {12 \, c^{3} x^{3} + 18 \, b c^{2} x^{2} + 4 \, b^{2} c x + 20 \, a c^{2} x - b^{3} + 10 \, a b c}{2 \, {\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} {\left (c x^{2} + b x + a\right )}^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c+a/x^2+b/x)^3/x^6,x, algorithm="giac")

[Out]

12*c^2*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*(12*c^
3*x^3 + 18*b*c^2*x^2 + 4*b^2*c*x + 20*a*c^2*x - b^3 + 10*a*b*c)/((b^4 - 8*a*b^2*c + 16*a^2*c^2)*(c*x^2 + b*x +
a)^2)

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maple [A]  time = 0.00, size = 129, normalized size = 1.25 \[ \frac {6 c^{2} x}{\left (4 a c -b^{2}\right )^{2} \left (c \,x^{2}+b x +a \right )}+\frac {12 c^{2} \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\left (4 a c -b^{2}\right )^{\frac {5}{2}}}+\frac {3 b c}{\left (4 a c -b^{2}\right )^{2} \left (c \,x^{2}+b x +a \right )}+\frac {2 c x +b}{2 \left (4 a c -b^{2}\right ) \left (c \,x^{2}+b x +a \right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(c+a/x^2+b/x)^3/x^6,x)

[Out]

1/2*(2*c*x+b)/(4*a*c-b^2)/(c*x^2+b*x+a)^2+6*c^2/(4*a*c-b^2)^2/(c*x^2+b*x+a)*x+3*c/(4*a*c-b^2)^2/(c*x^2+b*x+a)*
b+12*c^2/(4*a*c-b^2)^(5/2)*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 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c+a/x^2+b/x)^3/x^6,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 details)Is 4*a*c-b^2 positive or negative?

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mupad [B]  time = 1.42, size = 285, normalized size = 2.77 \[ \frac {\frac {6\,c^3\,x^3}{16\,a^2\,c^2-8\,a\,b^2\,c+b^4}-\frac {b^3-10\,a\,b\,c}{2\,\left (16\,a^2\,c^2-8\,a\,b^2\,c+b^4\right )}+\frac {9\,b\,c^2\,x^2}{16\,a^2\,c^2-8\,a\,b^2\,c+b^4}+\frac {2\,c\,x\,\left (b^2+5\,a\,c\right )}{16\,a^2\,c^2-8\,a\,b^2\,c+b^4}}{x^2\,\left (b^2+2\,a\,c\right )+a^2+c^2\,x^4+2\,a\,b\,x+2\,b\,c\,x^3}+\frac {12\,c^2\,\mathrm {atan}\left (\frac {\left (\frac {12\,c^3\,x}{{\left (4\,a\,c-b^2\right )}^{5/2}}+\frac {6\,c^2\,\left (16\,a^2\,b\,c^2-8\,a\,b^3\,c+b^5\right )}{{\left (4\,a\,c-b^2\right )}^{5/2}\,\left (16\,a^2\,c^2-8\,a\,b^2\,c+b^4\right )}\right )\,\left (16\,a^2\,c^2-8\,a\,b^2\,c+b^4\right )}{6\,c^2}\right )}{{\left (4\,a\,c-b^2\right )}^{5/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^6*(c + a/x^2 + b/x)^3),x)

[Out]

((6*c^3*x^3)/(b^4 + 16*a^2*c^2 - 8*a*b^2*c) - (b^3 - 10*a*b*c)/(2*(b^4 + 16*a^2*c^2 - 8*a*b^2*c)) + (9*b*c^2*x
^2)/(b^4 + 16*a^2*c^2 - 8*a*b^2*c) + (2*c*x*(5*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) + (12*c^2*atan((((12*c^3*x)/(4*a*c - b^2)^(5/2) + (6*c^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)))*(b^4 + 16*a^2*c^2 - 8*a*b^2*c))/(6*c
^2)))/(4*a*c - b^2)^(5/2)

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sympy [B]  time = 1.11, size = 474, normalized size = 4.60 \[ - 6 c^{2} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{5}}} \log {\left (x + \frac {- 384 a^{3} c^{5} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{5}}} + 288 a^{2} b^{2} c^{4} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{5}}} - 72 a b^{4} c^{3} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{5}}} + 6 b^{6} c^{2} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{5}}} + 6 b c^{2}}{12 c^{3}} \right )} + 6 c^{2} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{5}}} \log {\left (x + \frac {384 a^{3} c^{5} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{5}}} - 288 a^{2} b^{2} c^{4} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{5}}} + 72 a b^{4} c^{3} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{5}}} - 6 b^{6} c^{2} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{5}}} + 6 b c^{2}}{12 c^{3}} \right )} + \frac {10 a b c - b^{3} + 18 b c^{2} x^{2} + 12 c^{3} x^{3} + x \left (20 a c^{2} + 4 b^{2} c\right )}{32 a^{4} c^{2} - 16 a^{3} b^{2} c + 2 a^{2} b^{4} + x^{4} \left (32 a^{2} c^{4} - 16 a b^{2} c^{3} + 2 b^{4} c^{2}\right ) + x^{3} \left (64 a^{2} b c^{3} - 32 a b^{3} c^{2} + 4 b^{5} c\right ) + x^{2} \left (64 a^{3} c^{3} - 12 a b^{4} c + 2 b^{6}\right ) + x \left (64 a^{3} b c^{2} - 32 a^{2} b^{3} c + 4 a b^{5}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c+a/x**2+b/x)**3/x**6,x)

[Out]

-6*c**2*sqrt(-1/(4*a*c - b**2)**5)*log(x + (-384*a**3*c**5*sqrt(-1/(4*a*c - b**2)**5) + 288*a**2*b**2*c**4*sqr
t(-1/(4*a*c - b**2)**5) - 72*a*b**4*c**3*sqrt(-1/(4*a*c - b**2)**5) + 6*b**6*c**2*sqrt(-1/(4*a*c - b**2)**5) +
6*b*c**2)/(12*c**3)) + 6*c**2*sqrt(-1/(4*a*c - b**2)**5)*log(x + (384*a**3*c**5*sqrt(-1/(4*a*c - b**2)**5) -
288*a**2*b**2*c**4*sqrt(-1/(4*a*c - b**2)**5) + 72*a*b**4*c**3*sqrt(-1/(4*a*c - b**2)**5) - 6*b**6*c**2*sqrt(-
1/(4*a*c - b**2)**5) + 6*b*c**2)/(12*c**3)) + (10*a*b*c - b**3 + 18*b*c**2*x**2 + 12*c**3*x**3 + x*(20*a*c**2
+ 4*b**2*c))/(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))

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