3.22 \(\int \frac {x^5}{(a x^2+b x^3+c x^4)^2} \, dx\)

Optimal. Leaf size=66 \[ \frac {2 a+b x}{\left (b^2-4 a c\right ) \left (a+b x+c x^2\right )}-\frac {2 b \tanh ^{-1}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{3/2}} \]

[Out]

(b*x+2*a)/(-4*a*c+b^2)/(c*x^2+b*x+a)-2*b*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 = 66, 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, 638, 618, 206} \[ \frac {2 a+b x}{\left (b^2-4 a c\right ) \left (a+b x+c x^2\right )}-\frac {2 b \tanh ^{-1}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[x^5/(a*x^2 + b*x^3 + c*x^4)^2,x]

[Out]

(2*a + b*x)/((b^2 - 4*a*c)*(a + b*x + c*x^2)) - (2*b*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 638

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)*(a + b*x + c*x^2)^(p + 1))/((p + 1)*(b^2 - 4*a*c)), 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 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^5}{\left (a x^2+b x^3+c x^4\right )^2} \, dx &=\int \frac {x}{\left (a+b x+c x^2\right )^2} \, dx\\ &=\frac {2 a+b x}{\left (b^2-4 a c\right ) \left (a+b x+c x^2\right )}+\frac {b \int \frac {1}{a+b x+c x^2} \, dx}{b^2-4 a c}\\ &=\frac {2 a+b x}{\left (b^2-4 a c\right ) \left (a+b x+c x^2\right )}-\frac {(2 b) \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 {2 a+b x}{\left (b^2-4 a c\right ) \left (a+b x+c x^2\right )}-\frac {2 b \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 = 0.07, size = 69, normalized size = 1.05 \[ \frac {2 a+b x}{\left (b^2-4 a c\right ) (a+x (b+c x))}-\frac {2 b \tan ^{-1}\left (\frac {b+2 c x}{\sqrt {4 a c-b^2}}\right )}{\left (4 a c-b^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^5/(a*x^2 + b*x^3 + c*x^4)^2,x]

[Out]

(2*a + b*x)/((b^2 - 4*a*c)*(a + x*(b + c*x))) - (2*b*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 = 0.59, size = 338, normalized size = 5.12 \[ \left [\frac {2 \, a b^{2} - 8 \, a^{2} c - {\left (b c x^{2} + b^{2} x + a b\right )} \sqrt {b^{2} - 4 \, a c} \log \left (\frac {2 \, c^{2} x^{2} + 2 \, b c x + b^{2} - 2 \, a c + \sqrt {b^{2} - 4 \, a c} {\left (2 \, c x + b\right )}}{c x^{2} + b x + a}\right ) + {\left (b^{3} - 4 \, a b c\right )} x}{a b^{4} - 8 \, a^{2} b^{2} c + 16 \, a^{3} c^{2} + {\left (b^{4} c - 8 \, a b^{2} c^{2} + 16 \, a^{2} c^{3}\right )} x^{2} + {\left (b^{5} - 8 \, a b^{3} c + 16 \, a^{2} b c^{2}\right )} x}, \frac {2 \, a b^{2} - 8 \, a^{2} c - 2 \, {\left (b c x^{2} + b^{2} x + a b\right )} \sqrt {-b^{2} + 4 \, a c} \arctan \left (-\frac {\sqrt {-b^{2} + 4 \, a c} {\left (2 \, c x + b\right )}}{b^{2} - 4 \, a c}\right ) + {\left (b^{3} - 4 \, a b c\right )} x}{a b^{4} - 8 \, a^{2} b^{2} c + 16 \, a^{3} c^{2} + {\left (b^{4} c - 8 \, a b^{2} c^{2} + 16 \, a^{2} c^{3}\right )} x^{2} + {\left (b^{5} - 8 \, a b^{3} c + 16 \, a^{2} b c^{2}\right )} x}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5/(c*x^4+b*x^3+a*x^2)^2,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.38, size = 76, normalized size = 1.15 \[ \frac {2 \, b \arctan \left (\frac {2 \, c x + b}{\sqrt {-b^{2} + 4 \, a c}}\right )}{{\left (b^{2} - 4 \, a c\right )} \sqrt {-b^{2} + 4 \, a c}} + \frac {b x + 2 \, a}{{\left (c x^{2} + b x + a\right )} {\left (b^{2} - 4 \, a c\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5/(c*x^4+b*x^3+a*x^2)^2,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.00, size = 70, normalized size = 1.06 \[ -\frac {2 b \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\left (4 a c -b^{2}\right )^{\frac {3}{2}}}+\frac {-b x -2 a}{\left (4 a c -b^{2}\right ) \left (c \,x^{2}+b x +a \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5/(c*x^4+b*x^3+a*x^2)^2,x)

[Out]

(-b*x-2*a)/(4*a*c-b^2)/(c*x^2+b*x+a)-2*b/(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^5/(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 = 2.18, size = 110, normalized size = 1.67 \[ -\frac {\frac {2\,a}{4\,a\,c-b^2}+\frac {b\,x}{4\,a\,c-b^2}}{c\,x^2+b\,x+a}-\frac {2\,b\,\mathrm {atan}\left (\frac {\left (\frac {b^2}{{\left (4\,a\,c-b^2\right )}^{3/2}}+\frac {2\,b\,c\,x}{{\left (4\,a\,c-b^2\right )}^{3/2}}\right )\,\left (4\,a\,c-b^2\right )}{b}\right )}{{\left (4\,a\,c-b^2\right )}^{3/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5/(a*x^2 + b*x^3 + c*x^4)^2,x)

[Out]

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

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sympy [B]  time = 0.56, size = 253, normalized size = 3.83 \[ b \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{3}}} \log {\left (x + \frac {- 16 a^{2} b c^{2} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{3}}} + 8 a b^{3} c \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{3}}} - b^{5} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{3}}} + b^{2}}{2 b c} \right )} - b \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{3}}} \log {\left (x + \frac {16 a^{2} b c^{2} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{3}}} - 8 a b^{3} c \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{3}}} + b^{5} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{3}}} + b^{2}}{2 b c} \right )} + \frac {- 2 a - b x}{4 a^{2} c - a b^{2} + x^{2} \left (4 a c^{2} - b^{2} c\right ) + x \left (4 a b c - b^{3}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**5/(c*x**4+b*x**3+a*x**2)**2,x)

[Out]

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

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