3.21 \(\int \frac {x^6}{(a x^2+b x^3+c x^4)^2} \, dx\)

Optimal. Leaf size=67 \[ \frac {x (2 a+b x)}{\left (b^2-4 a c\right ) \left (a+b x+c x^2\right )}+\frac {4 a \tanh ^{-1}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 67, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {1585, 722, 618, 206} \[ \frac {x (2 a+b x)}{\left (b^2-4 a c\right ) \left (a+b x+c x^2\right )}+\frac {4 a \tanh ^{-1}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[x^6/(a*x^2 + b*x^3 + c*x^4)^2,x]

[Out]

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

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 722

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 1585

Int[(u_.)*(x_)^(m_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.))^(n_.), x_Symbol] :> Int[u*x^(m +
n*p)*(a + b*x^(q - p) + c*x^(r - p))^n, x] /; FreeQ[{a, b, c, m, p, q, r}, x] && IntegerQ[n] && PosQ[q - p] &
& PosQ[r - p]

Rubi steps

\begin {align*} \int \frac {x^6}{\left (a x^2+b x^3+c x^4\right )^2} \, dx &=\int \frac {x^2}{\left (a+b x+c x^2\right )^2} \, dx\\ &=\frac {x (2 a+b x)}{\left (b^2-4 a c\right ) \left (a+b x+c x^2\right )}-\frac {(2 a) \int \frac {1}{a+b x+c x^2} \, dx}{b^2-4 a c}\\ &=\frac {x (2 a+b x)}{\left (b^2-4 a c\right ) \left (a+b x+c x^2\right )}+\frac {(4 a) \operatorname {Subst}\left (\int \frac {1}{b^2-4 a c-x^2} \, dx,x,b+2 c x\right )}{b^2-4 a c}\\ &=\frac {x (2 a+b x)}{\left (b^2-4 a c\right ) \left (a+b x+c x^2\right )}+\frac {4 a \tanh ^{-1}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 0.09, size = 81, normalized size = 1.21 \[ \frac {a (b-2 c x)+b^2 x}{c \left (4 a c-b^2\right ) (a+x (b+c x))}+\frac {4 a \tan ^{-1}\left (\frac {b+2 c x}{\sqrt {4 a c-b^2}}\right )}{\left (4 a c-b^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^6/(a*x^2 + b*x^3 + c*x^4)^2,x]

[Out]

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

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fricas [B]  time = 0.68, size = 387, normalized size = 5.78 \[ \left [-\frac {a b^{3} - 4 \, a^{2} b c + 2 \, {\left (a c^{2} x^{2} + a b c x + a^{2} c\right )} \sqrt {b^{2} - 4 \, a c} \log \left (\frac {2 \, c^{2} x^{2} + 2 \, b c x + b^{2} - 2 \, a c - \sqrt {b^{2} - 4 \, a c} {\left (2 \, c x + b\right )}}{c x^{2} + b x + a}\right ) + {\left (b^{4} - 6 \, a b^{2} c + 8 \, a^{2} c^{2}\right )} x}{a b^{4} c - 8 \, a^{2} b^{2} c^{2} + 16 \, a^{3} c^{3} + {\left (b^{4} c^{2} - 8 \, a b^{2} c^{3} + 16 \, a^{2} c^{4}\right )} x^{2} + {\left (b^{5} c - 8 \, a b^{3} c^{2} + 16 \, a^{2} b c^{3}\right )} x}, -\frac {a b^{3} - 4 \, a^{2} b c - 4 \, {\left (a c^{2} x^{2} + a b c x + a^{2} c\right )} \sqrt {-b^{2} + 4 \, a c} \arctan \left (-\frac {\sqrt {-b^{2} + 4 \, a c} {\left (2 \, c x + b\right )}}{b^{2} - 4 \, a c}\right ) + {\left (b^{4} - 6 \, a b^{2} c + 8 \, a^{2} c^{2}\right )} x}{a b^{4} c - 8 \, a^{2} b^{2} c^{2} + 16 \, a^{3} c^{3} + {\left (b^{4} c^{2} - 8 \, a b^{2} c^{3} + 16 \, a^{2} c^{4}\right )} x^{2} + {\left (b^{5} c - 8 \, a b^{3} c^{2} + 16 \, a^{2} b c^{3}\right )} x}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^6/(c*x^4+b*x^3+a*x^2)^2,x, algorithm="fricas")

[Out]

[-(a*b^3 - 4*a^2*b*c + 2*(a*c^2*x^2 + a*b*c*x + a^2*c)*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)) + (b^4 - 6*a*b^2*c + 8*a^2*c^2)*x)/(a*b^4*c - 8*a^2*b^2*
c^2 + 16*a^3*c^3 + (b^4*c^2 - 8*a*b^2*c^3 + 16*a^2*c^4)*x^2 + (b^5*c - 8*a*b^3*c^2 + 16*a^2*b*c^3)*x), -(a*b^3
- 4*a^2*b*c - 4*(a*c^2*x^2 + a*b*c*x + a^2*c)*sqrt(-b^2 + 4*a*c)*arctan(-sqrt(-b^2 + 4*a*c)*(2*c*x + b)/(b^2
- 4*a*c)) + (b^4 - 6*a*b^2*c + 8*a^2*c^2)*x)/(a*b^4*c - 8*a^2*b^2*c^2 + 16*a^3*c^3 + (b^4*c^2 - 8*a*b^2*c^3 +
16*a^2*c^4)*x^2 + (b^5*c - 8*a*b^3*c^2 + 16*a^2*b*c^3)*x)]

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giac [A]  time = 0.49, size = 88, normalized size = 1.31 \[ -\frac {4 \, a \arctan \left (\frac {2 \, c x + b}{\sqrt {-b^{2} + 4 \, a c}}\right )}{{\left (b^{2} - 4 \, a c\right )} \sqrt {-b^{2} + 4 \, a c}} - \frac {b^{2} x - 2 \, a c x + a b}{{\left (b^{2} c - 4 \, a c^{2}\right )} {\left (c x^{2} + b x + a\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^6/(c*x^4+b*x^3+a*x^2)^2,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.01, size = 97, normalized size = 1.45 \[ \frac {4 a \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\left (4 a c -b^{2}\right )^{\frac {3}{2}}}+\frac {\frac {a b}{\left (4 a c -b^{2}\right ) c}-\frac {\left (2 a c -b^{2}\right ) x}{\left (4 a c -b^{2}\right ) c}}{c \,x^{2}+b x +a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^6/(c*x^4+b*x^3+a*x^2)^2,x)

[Out]

(-(2*a*c-b^2)/(4*a*c-b^2)/c*x+1/(4*a*c-b^2)*a*b/c)/(c*x^2+b*x+a)+4*a/(4*a*c-b^2)^(3/2)*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(x^6/(c*x^4+b*x^3+a*x^2)^2,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 = 2.13, size = 135, normalized size = 2.01 \[ -\frac {\frac {x\,\left (2\,a\,c-b^2\right )}{c\,\left (4\,a\,c-b^2\right )}-\frac {a\,b}{c\,\left (4\,a\,c-b^2\right )}}{c\,x^2+b\,x+a}-\frac {4\,a\,\mathrm {atan}\left (\frac {\left (\frac {2\,a\,\left (b^3-4\,a\,b\,c\right )}{{\left (4\,a\,c-b^2\right )}^{5/2}}-\frac {4\,a\,c\,x}{{\left (4\,a\,c-b^2\right )}^{3/2}}\right )\,\left (4\,a\,c-b^2\right )}{2\,a}\right )}{{\left (4\,a\,c-b^2\right )}^{3/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^6/(a*x^2 + b*x^3 + c*x^4)^2,x)

[Out]

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

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sympy [B]  time = 0.60, size = 280, normalized size = 4.18 \[ - 2 a \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{3}}} \log {\left (x + \frac {- 32 a^{3} c^{2} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{3}}} + 16 a^{2} b^{2} c \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{3}}} - 2 a b^{4} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{3}}} + 2 a b}{4 a c} \right )} + 2 a \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{3}}} \log {\left (x + \frac {32 a^{3} c^{2} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{3}}} - 16 a^{2} b^{2} c \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{3}}} + 2 a b^{4} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{3}}} + 2 a b}{4 a c} \right )} + \frac {a b + x \left (- 2 a c + b^{2}\right )}{4 a^{2} c^{2} - a b^{2} c + x^{2} \left (4 a c^{3} - b^{2} c^{2}\right ) + x \left (4 a b c^{2} - b^{3} c\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**6/(c*x**4+b*x**3+a*x**2)**2,x)

[Out]

-2*a*sqrt(-1/(4*a*c - b**2)**3)*log(x + (-32*a**3*c**2*sqrt(-1/(4*a*c - b**2)**3) + 16*a**2*b**2*c*sqrt(-1/(4*
a*c - b**2)**3) - 2*a*b**4*sqrt(-1/(4*a*c - b**2)**3) + 2*a*b)/(4*a*c)) + 2*a*sqrt(-1/(4*a*c - b**2)**3)*log(x
+ (32*a**3*c**2*sqrt(-1/(4*a*c - b**2)**3) - 16*a**2*b**2*c*sqrt(-1/(4*a*c - b**2)**3) + 2*a*b**4*sqrt(-1/(4*
a*c - b**2)**3) + 2*a*b)/(4*a*c)) + (a*b + x*(-2*a*c + b**2))/(4*a**2*c**2 - a*b**2*c + x**2*(4*a*c**3 - b**2*
c**2) + x*(4*a*b*c**2 - b**3*c))

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