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

Optimal. Leaf size=78 \[ \frac {2 a \tanh ^{-1}\left (\frac {b+2 c x^2}{\sqrt {b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{3/2}}+\frac {x^2 \left (2 a+b x^2\right )}{2 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )} \]

[Out]

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

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Rubi [A]  time = 0.07, antiderivative size = 78, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {1114, 722, 618, 206} \[ \frac {x^2 \left (2 a+b x^2\right )}{2 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}+\frac {2 a \tanh ^{-1}\left (\frac {b+2 c x^2}{\sqrt {b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[x^5/(a + b*x^2 + c*x^4)^2,x]

[Out]

(x^2*(2*a + b*x^2))/(2*(b^2 - 4*a*c)*(a + b*x^2 + c*x^4)) + (2*a*ArcTanh[(b + 2*c*x^2)/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 1114

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

Rubi steps

\begin {align*} \int \frac {x^5}{\left (a+b x^2+c x^4\right )^2} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {x^2}{\left (a+b x+c x^2\right )^2} \, dx,x,x^2\right )\\ &=\frac {x^2 \left (2 a+b x^2\right )}{2 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}-\frac {a \operatorname {Subst}\left (\int \frac {1}{a+b x+c x^2} \, dx,x,x^2\right )}{b^2-4 a c}\\ &=\frac {x^2 \left (2 a+b x^2\right )}{2 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}+\frac {(2 a) \operatorname {Subst}\left (\int \frac {1}{b^2-4 a c-x^2} \, dx,x,b+2 c x^2\right )}{b^2-4 a c}\\ &=\frac {x^2 \left (2 a+b x^2\right )}{2 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}+\frac {2 a \tanh ^{-1}\left (\frac {b+2 c x^2}{\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 = 93, normalized size = 1.19 \[ \frac {2 a \tan ^{-1}\left (\frac {b+2 c x^2}{\sqrt {4 a c-b^2}}\right )}{\left (4 a c-b^2\right )^{3/2}}+\frac {a \left (b-2 c x^2\right )+b^2 x^2}{2 c \left (4 a c-b^2\right ) \left (a+b x^2+c x^4\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.79, size = 407, normalized size = 5.22 \[ \left [-\frac {a b^{3} - 4 \, a^{2} b c + {\left (b^{4} - 6 \, a b^{2} c + 8 \, a^{2} c^{2}\right )} x^{2} + 2 \, {\left (a c^{2} x^{4} + a b c x^{2} + a^{2} c\right )} \sqrt {b^{2} - 4 \, a c} \log \left (\frac {2 \, c^{2} x^{4} + 2 \, b c x^{2} + b^{2} - 2 \, a c - {\left (2 \, c x^{2} + b\right )} \sqrt {b^{2} - 4 \, a c}}{c x^{4} + b x^{2} + a}\right )}{2 \, {\left (a b^{4} c - 8 \, a^{2} b^{2} c^{2} + 16 \, a^{3} c^{3} + {\left (b^{4} c^{2} - 8 \, a b^{2} c^{3} + 16 \, a^{2} c^{4}\right )} x^{4} + {\left (b^{5} c - 8 \, a b^{3} c^{2} + 16 \, a^{2} b c^{3}\right )} x^{2}\right )}}, -\frac {a b^{3} - 4 \, a^{2} b c + {\left (b^{4} - 6 \, a b^{2} c + 8 \, a^{2} c^{2}\right )} x^{2} - 4 \, {\left (a c^{2} x^{4} + a b c x^{2} + a^{2} c\right )} \sqrt {-b^{2} + 4 \, a c} \arctan \left (-\frac {{\left (2 \, c x^{2} + b\right )} \sqrt {-b^{2} + 4 \, a c}}{b^{2} - 4 \, a c}\right )}{2 \, {\left (a b^{4} c - 8 \, a^{2} b^{2} c^{2} + 16 \, a^{3} c^{3} + {\left (b^{4} c^{2} - 8 \, a b^{2} c^{3} + 16 \, a^{2} c^{4}\right )} x^{4} + {\left (b^{5} c - 8 \, a b^{3} c^{2} + 16 \, a^{2} b c^{3}\right )} x^{2}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/2*(a*b^3 - 4*a^2*b*c + (b^4 - 6*a*b^2*c + 8*a^2*c^2)*x^2 + 2*(a*c^2*x^4 + a*b*c*x^2 + a^2*c)*sqrt(b^2 - 4*
a*c)*log((2*c^2*x^4 + 2*b*c*x^2 + b^2 - 2*a*c - (2*c*x^2 + b)*sqrt(b^2 - 4*a*c))/(c*x^4 + b*x^2 + a)))/(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^4 + (b^5*c - 8*a*b^3*c^2 + 16*a^2*b*c^
3)*x^2), -1/2*(a*b^3 - 4*a^2*b*c + (b^4 - 6*a*b^2*c + 8*a^2*c^2)*x^2 - 4*(a*c^2*x^4 + a*b*c*x^2 + a^2*c)*sqrt(
-b^2 + 4*a*c)*arctan(-(2*c*x^2 + b)*sqrt(-b^2 + 4*a*c)/(b^2 - 4*a*c)))/(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^4 + (b^5*c - 8*a*b^3*c^2 + 16*a^2*b*c^3)*x^2)]

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giac [A]  time = 0.91, size = 96, normalized size = 1.23 \[ -\frac {2 \, a \arctan \left (\frac {2 \, c x^{2} + b}{\sqrt {-b^{2} + 4 \, a c}}\right )}{{\left (b^{2} - 4 \, a c\right )} \sqrt {-b^{2} + 4 \, a c}} - \frac {b^{2} x^{2} - 2 \, a c x^{2} + a b}{2 \, {\left (c x^{4} + b x^{2} + a\right )} {\left (b^{2} c - 4 \, a c^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*a*arctan((2*c*x^2 + b)/sqrt(-b^2 + 4*a*c))/((b^2 - 4*a*c)*sqrt(-b^2 + 4*a*c)) - 1/2*(b^2*x^2 - 2*a*c*x^2 +
a*b)/((c*x^4 + b*x^2 + a)*(b^2*c - 4*a*c^2))

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maple [A]  time = 0.01, size = 104, normalized size = 1.33 \[ \frac {2 a \arctan \left (\frac {2 c \,x^{2}+b}{\sqrt {4 a c -b^{2}}}\right )}{\left (4 a c -b^{2}\right )^{\frac {3}{2}}}+\frac {\frac {a b}{\left (4 a c -b^{2}\right ) c}-\frac {\left (2 a c -b^{2}\right ) x^{2}}{\left (4 a c -b^{2}\right ) c}}{2 c \,x^{4}+2 b \,x^{2}+2 a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(-(2*a*c-b^2)/c/(4*a*c-b^2)*x^2+a*b/c/(4*a*c-b^2))/(c*x^4+b*x^2+a)+2*a/(4*a*c-b^2)^(3/2)*arctan((2*c*x^2+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^2+a)^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 = 0.18, size = 187, normalized size = 2.40 \[ -\frac {\frac {x^2\,\left (2\,a\,c-b^2\right )}{2\,c\,\left (4\,a\,c-b^2\right )}-\frac {a\,b}{2\,c\,\left (4\,a\,c-b^2\right )}}{c\,x^4+b\,x^2+a}-\frac {2\,a\,\mathrm {atan}\left (\frac {b^3-4\,a\,b\,c}{{\left (4\,a\,c-b^2\right )}^{3/2}}-\frac {x^2\,\left (\frac {4\,a\,c^2}{{\left (4\,a\,c-b^2\right )}^{7/2}}+\frac {4\,a\,\left (b^3\,c^2-4\,a\,b\,c^3\right )\,\left (b^3-4\,a\,b\,c\right )}{{\left (4\,a\,c-b^2\right )}^{13/2}}\right )\,{\left (4\,a\,c-b^2\right )}^4}{8\,a^2\,c^2}\right )}{{\left (4\,a\,c-b^2\right )}^{3/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5/(a + b*x^2 + c*x^4)^2,x)

[Out]

- ((x^2*(2*a*c - b^2))/(2*c*(4*a*c - b^2)) - (a*b)/(2*c*(4*a*c - b^2)))/(a + b*x^2 + c*x^4) - (2*a*atan((b^3 -
4*a*b*c)/(4*a*c - b^2)^(3/2) - (x^2*((4*a*c^2)/(4*a*c - b^2)^(7/2) + (4*a*(b^3*c^2 - 4*a*b*c^3)*(b^3 - 4*a*b*
c))/(4*a*c - b^2)^(13/2))*(4*a*c - b^2)^4)/(8*a^2*c^2)))/(4*a*c - b^2)^(3/2)

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sympy [B]  time = 3.90, size = 282, normalized size = 3.62 \[ - a \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{3}}} \log {\left (x^{2} + \frac {- 16 a^{3} c^{2} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{3}}} + 8 a^{2} b^{2} c \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{3}}} - a b^{4} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{3}}} + a b}{2 a c} \right )} + a \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{3}}} \log {\left (x^{2} + \frac {16 a^{3} c^{2} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{3}}} - 8 a^{2} b^{2} c \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{3}}} + a b^{4} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{3}}} + a b}{2 a c} \right )} + \frac {a b + x^{2} \left (- 2 a c + b^{2}\right )}{8 a^{2} c^{2} - 2 a b^{2} c + x^{4} \left (8 a c^{3} - 2 b^{2} c^{2}\right ) + x^{2} \left (8 a b c^{2} - 2 b^{3} c\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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