\(\int \frac {x^5}{(a x+b x^3+c x^5)^2} \, dx\) [95]

Optimal result

Integrand size = 20, antiderivative size = 75 \[ \int \frac {x^5}{\left (a x+b x^3+c x^5\right )^2} \, dx=\frac {2 a+b x^2}{2 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}-\frac {b \text {arctanh}\left (\frac {b+2 c x^2}{\sqrt {b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{3/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {1599, 1128, 652, 632, 212} \[ \int \frac {x^5}{\left (a x+b x^3+c x^5\right )^2} \, dx=\frac {2 a+b x^2}{2 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}-\frac {b \text {arctanh}\left (\frac {b+2 c x^2}{\sqrt {b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{3/2}} \]

[In]

Int[x^5/(a*x + b*x^3 + c*x^5)^2,x]

[Out]

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

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 652

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)/((p + 1)*(b^2 - 4*a*c)))*(a + b*x + c*x^2)^(p + 1), 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 1128

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]

Rule 1599

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

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.05 \[ \int \frac {x^5}{\left (a x+b x^3+c x^5\right )^2} \, dx=\frac {2 a+b x^2}{2 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}-\frac {b \arctan \left (\frac {b+2 c x^2}{\sqrt {-b^2+4 a c}}\right )}{\left (-b^2+4 a c\right )^{3/2}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.08 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.03

method	result	size
default	\(\frac {-b \,x^{2}-2 a}{2 \left (4 a c -b^{2}\right ) \left (c \,x^{4}+b \,x^{2}+a \right )}-\frac {b \arctan \left (\frac {2 c \,x^{2}+b}{\sqrt {4 a c -b^{2}}}\right )}{\left (4 a c -b^{2}\right )^{\frac {3}{2}}}\)	\(77\)
risch	\(\frac {-\frac {b \,x^{2}}{2 \left (4 a c -b^{2}\right )}-\frac {a}{4 a c -b^{2}}}{c \,x^{4}+b \,x^{2}+a}+\frac {b \ln \left (\left (-\left (-4 a c +b^{2}\right )^{\frac {3}{2}}+4 a b c -b^{3}\right ) x^{2}+8 c \,a^{2}-2 b^{2} a \right )}{2 \left (-4 a c +b^{2}\right )^{\frac {3}{2}}}-\frac {b \ln \left (\left (-\left (-4 a c +b^{2}\right )^{\frac {3}{2}}-4 a b c +b^{3}\right ) x^{2}-8 c \,a^{2}+2 b^{2} a \right )}{2 \left (-4 a c +b^{2}\right )^{\frac {3}{2}}}\)	\(157\)

[In]

int(x^5/(c*x^5+b*x^3+a*x)^2,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 168 vs. \(2 (69) = 138\).

Time = 0.27 (sec) , antiderivative size = 360, normalized size of antiderivative = 4.80 \[ \int \frac {x^5}{\left (a x+b x^3+c x^5\right )^2} \, dx=\left [\frac {2 \, a b^{2} - 8 \, a^{2} c + {\left (b^{3} - 4 \, a b c\right )} x^{2} - {\left (b c x^{4} + b^{2} x^{2} + a b\right )} \sqrt {b^{2} - 4 \, a c} \log \left (\frac {2 \, c^{2} x^{4} + 2 \, b c x^{2} + b^{2} - 2 \, a c + {\left (2 \, c x^{2} + b\right )} \sqrt {b^{2} - 4 \, a c}}{c x^{4} + b x^{2} + a}\right )}{2 \, {\left (a b^{4} - 8 \, a^{2} b^{2} c + 16 \, a^{3} c^{2} + {\left (b^{4} c - 8 \, a b^{2} c^{2} + 16 \, a^{2} c^{3}\right )} x^{4} + {\left (b^{5} - 8 \, a b^{3} c + 16 \, a^{2} b c^{2}\right )} x^{2}\right )}}, \frac {2 \, a b^{2} - 8 \, a^{2} c + {\left (b^{3} - 4 \, a b c\right )} x^{2} - 2 \, {\left (b c x^{4} + b^{2} x^{2} + a b\right )} \sqrt {-b^{2} + 4 \, a c} \arctan \left (-\frac {{\left (2 \, c x^{2} + b\right )} \sqrt {-b^{2} + 4 \, a c}}{b^{2} - 4 \, a c}\right )}{2 \, {\left (a b^{4} - 8 \, a^{2} b^{2} c + 16 \, a^{3} c^{2} + {\left (b^{4} c - 8 \, a b^{2} c^{2} + 16 \, a^{2} c^{3}\right )} x^{4} + {\left (b^{5} - 8 \, a b^{3} c + 16 \, a^{2} b c^{2}\right )} x^{2}\right )}}\right ] \]

[In]

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

[Out]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 269 vs. \(2 (63) = 126\).

Time = 0.69 (sec) , antiderivative size = 269, normalized size of antiderivative = 3.59 \[ \int \frac {x^5}{\left (a x+b x^3+c x^5\right )^2} \, dx=\frac {b \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{3}}} \log {\left (x^{2} + \frac {- 16 a^{2} b c^{2} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{3}}} + 8 a b^{3} c \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{3}}} - b^{5} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{3}}} + b^{2}}{2 b c} \right )}}{2} - \frac {b \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{3}}} \log {\left (x^{2} + \frac {16 a^{2} b c^{2} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{3}}} - 8 a b^{3} c \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{3}}} + b^{5} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{3}}} + b^{2}}{2 b c} \right )}}{2} + \frac {- 2 a - b x^{2}}{8 a^{2} c - 2 a b^{2} + x^{4} \cdot \left (8 a c^{2} - 2 b^{2} c\right ) + x^{2} \cdot \left (8 a b c - 2 b^{3}\right )} \]

[In]

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

[Out]

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

Maxima [F]

\[ \int \frac {x^5}{\left (a x+b x^3+c x^5\right )^2} \, dx=\int { \frac {x^{5}}{{\left (c x^{5} + b x^{3} + a x\right )}^{2}} \,d x } \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.62 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.09 \[ \int \frac {x^5}{\left (a x+b x^3+c x^5\right )^2} \, dx=\frac {b \arctan \left (\frac {2 \, c x^{2} + b}{\sqrt {-b^{2} + 4 \, a c}}\right )}{{\left (b^{2} - 4 \, a c\right )} \sqrt {-b^{2} + 4 \, a c}} + \frac {b x^{2} + 2 \, a}{2 \, {\left (c x^{4} + b x^{2} + a\right )} {\left (b^{2} - 4 \, a c\right )}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 178, normalized size of antiderivative = 2.37 \[ \int \frac {x^5}{\left (a x+b x^3+c x^5\right )^2} \, dx=\frac {b\,\mathrm {atan}\left (\frac {b^3-4\,a\,b\,c}{{\left (4\,a\,c-b^2\right )}^{3/2}}-\frac {x^2\,{\left (4\,a\,c-b^2\right )}^4\,\left (\frac {b^2\,c^2}{a\,{\left (4\,a\,c-b^2\right )}^{7/2}}+\frac {b^2\,\left (2\,b^3\,c^2-8\,a\,b\,c^3\right )\,\left (b^3-4\,a\,b\,c\right )}{2\,a\,{\left (4\,a\,c-b^2\right )}^{13/2}}\right )}{2\,b^2\,c^2}\right )}{{\left (4\,a\,c-b^2\right )}^{3/2}}-\frac {\frac {a}{4\,a\,c-b^2}+\frac {b\,x^2}{2\,\left (4\,a\,c-b^2\right )}}{c\,x^4+b\,x^2+a} \]

[In]

int(x^5/(a*x + b*x^3 + c*x^5)^2,x)

[Out]

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