3.79 \(\int \frac {x^8 (A+B x^2)}{(b x^2+c x^4)^3} \, dx\)

Optimal. Leaf size=90 \[ \frac {(A c+3 b B) \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{8 b^{3/2} c^{5/2}}-\frac {x (5 b B-A c)}{8 b c^2 \left (b+c x^2\right )}+\frac {x (b B-A c)}{4 c^2 \left (b+c x^2\right )^2} \]

[Out]

1/4*(-A*c+B*b)*x/c^2/(c*x^2+b)^2-1/8*(-A*c+5*B*b)*x/b/c^2/(c*x^2+b)+1/8*(A*c+3*B*b)*arctan(x*c^(1/2)/b^(1/2))/
b^(3/2)/c^(5/2)

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Rubi [A]  time = 0.08, antiderivative size = 90, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {1584, 455, 385, 205} \[ \frac {(A c+3 b B) \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{8 b^{3/2} c^{5/2}}-\frac {x (5 b B-A c)}{8 b c^2 \left (b+c x^2\right )}+\frac {x (b B-A c)}{4 c^2 \left (b+c x^2\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[(x^8*(A + B*x^2))/(b*x^2 + c*x^4)^3,x]

[Out]

((b*B - A*c)*x)/(4*c^2*(b + c*x^2)^2) - ((5*b*B - A*c)*x)/(8*b*c^2*(b + c*x^2)) + ((3*b*B + A*c)*ArcTan[(Sqrt[
c]*x)/Sqrt[b]])/(8*b^(3/2)*c^(5/2))

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 385

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> -Simp[((b*c - a*d)*x*(a + b*x^n)^(p +
1))/(a*b*n*(p + 1)), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(a*b*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /
; FreeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && (LtQ[p, -1] || ILtQ[1/n + p, 0])

Rule 455

Int[(x_)^(m_)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[((-a)^(m/2 - 1)*(b*c - a*d)*
x*(a + b*x^2)^(p + 1))/(2*b^(m/2 + 1)*(p + 1)), x] + Dist[1/(2*b^(m/2 + 1)*(p + 1)), Int[(a + b*x^2)^(p + 1)*E
xpandToSum[2*b*(p + 1)*x^2*Together[(b^(m/2)*x^(m - 2)*(c + d*x^2) - (-a)^(m/2 - 1)*(b*c - a*d))/(a + b*x^2)]
- (-a)^(m/2 - 1)*(b*c - a*d), x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && IGtQ[
m/2, 0] && (IntegerQ[p] || EqQ[m + 2*p + 1, 0])

Rule 1584

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

Rubi steps

\begin {align*} \int \frac {x^8 \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^3} \, dx &=\int \frac {x^2 \left (A+B x^2\right )}{\left (b+c x^2\right )^3} \, dx\\ &=\frac {(b B-A c) x}{4 c^2 \left (b+c x^2\right )^2}-\frac {\int \frac {b B-A c-4 B c x^2}{\left (b+c x^2\right )^2} \, dx}{4 c^2}\\ &=\frac {(b B-A c) x}{4 c^2 \left (b+c x^2\right )^2}-\frac {(5 b B-A c) x}{8 b c^2 \left (b+c x^2\right )}+\frac {(3 b B+A c) \int \frac {1}{b+c x^2} \, dx}{8 b c^2}\\ &=\frac {(b B-A c) x}{4 c^2 \left (b+c x^2\right )^2}-\frac {(5 b B-A c) x}{8 b c^2 \left (b+c x^2\right )}+\frac {(3 b B+A c) \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{8 b^{3/2} c^{5/2}}\\ \end {align*}

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Mathematica [A]  time = 0.12, size = 83, normalized size = 0.92 \[ \frac {\frac {(A c+3 b B) \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{b^{3/2}}+\frac {\sqrt {c} x \left (-b c \left (A+5 B x^2\right )+A c^2 x^2-3 b^2 B\right )}{b \left (b+c x^2\right )^2}}{8 c^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^8*(A + B*x^2))/(b*x^2 + c*x^4)^3,x]

[Out]

((Sqrt[c]*x*(-3*b^2*B + A*c^2*x^2 - b*c*(A + 5*B*x^2)))/(b*(b + c*x^2)^2) + ((3*b*B + A*c)*ArcTan[(Sqrt[c]*x)/
Sqrt[b]])/b^(3/2))/(8*c^(5/2))

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fricas [A]  time = 0.99, size = 301, normalized size = 3.34 \[ \left [-\frac {2 \, {\left (5 \, B b^{2} c^{2} - A b c^{3}\right )} x^{3} + {\left ({\left (3 \, B b c^{2} + A c^{3}\right )} x^{4} + 3 \, B b^{3} + A b^{2} c + 2 \, {\left (3 \, B b^{2} c + A b c^{2}\right )} x^{2}\right )} \sqrt {-b c} \log \left (\frac {c x^{2} - 2 \, \sqrt {-b c} x - b}{c x^{2} + b}\right ) + 2 \, {\left (3 \, B b^{3} c + A b^{2} c^{2}\right )} x}{16 \, {\left (b^{2} c^{5} x^{4} + 2 \, b^{3} c^{4} x^{2} + b^{4} c^{3}\right )}}, -\frac {{\left (5 \, B b^{2} c^{2} - A b c^{3}\right )} x^{3} - {\left ({\left (3 \, B b c^{2} + A c^{3}\right )} x^{4} + 3 \, B b^{3} + A b^{2} c + 2 \, {\left (3 \, B b^{2} c + A b c^{2}\right )} x^{2}\right )} \sqrt {b c} \arctan \left (\frac {\sqrt {b c} x}{b}\right ) + {\left (3 \, B b^{3} c + A b^{2} c^{2}\right )} x}{8 \, {\left (b^{2} c^{5} x^{4} + 2 \, b^{3} c^{4} x^{2} + b^{4} c^{3}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^8*(B*x^2+A)/(c*x^4+b*x^2)^3,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.16, size = 78, normalized size = 0.87 \[ \frac {{\left (3 \, B b + A c\right )} \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{8 \, \sqrt {b c} b c^{2}} - \frac {5 \, B b c x^{3} - A c^{2} x^{3} + 3 \, B b^{2} x + A b c x}{8 \, {\left (c x^{2} + b\right )}^{2} b c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^8*(B*x^2+A)/(c*x^4+b*x^2)^3,x, algorithm="giac")

[Out]

1/8*(3*B*b + A*c)*arctan(c*x/sqrt(b*c))/(sqrt(b*c)*b*c^2) - 1/8*(5*B*b*c*x^3 - A*c^2*x^3 + 3*B*b^2*x + A*b*c*x
)/((c*x^2 + b)^2*b*c^2)

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maple [A]  time = 0.06, size = 89, normalized size = 0.99 \[ \frac {A \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{8 \sqrt {b c}\, b c}+\frac {3 B \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{8 \sqrt {b c}\, c^{2}}+\frac {\frac {\left (A c -5 b B \right ) x^{3}}{8 b c}-\frac {\left (A c +3 b B \right ) x}{8 c^{2}}}{\left (c \,x^{2}+b \right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^8*(B*x^2+A)/(c*x^4+b*x^2)^3,x)

[Out]

(1/8*(A*c-5*B*b)/b/c*x^3-1/8*(A*c+3*B*b)/c^2*x)/(c*x^2+b)^2+1/8/c/b/(b*c)^(1/2)*arctan(1/(b*c)^(1/2)*c*x)*A+3/
8/c^2/(b*c)^(1/2)*arctan(1/(b*c)^(1/2)*c*x)*B

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maxima [A]  time = 2.98, size = 92, normalized size = 1.02 \[ -\frac {{\left (5 \, B b c - A c^{2}\right )} x^{3} + {\left (3 \, B b^{2} + A b c\right )} x}{8 \, {\left (b c^{4} x^{4} + 2 \, b^{2} c^{3} x^{2} + b^{3} c^{2}\right )}} + \frac {{\left (3 \, B b + A c\right )} \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{8 \, \sqrt {b c} b c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^8*(B*x^2+A)/(c*x^4+b*x^2)^3,x, algorithm="maxima")

[Out]

-1/8*((5*B*b*c - A*c^2)*x^3 + (3*B*b^2 + A*b*c)*x)/(b*c^4*x^4 + 2*b^2*c^3*x^2 + b^3*c^2) + 1/8*(3*B*b + A*c)*a
rctan(c*x/sqrt(b*c))/(sqrt(b*c)*b*c^2)

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mupad [B]  time = 0.15, size = 82, normalized size = 0.91 \[ \frac {\mathrm {atan}\left (\frac {\sqrt {c}\,x}{\sqrt {b}}\right )\,\left (A\,c+3\,B\,b\right )}{8\,b^{3/2}\,c^{5/2}}-\frac {\frac {x\,\left (A\,c+3\,B\,b\right )}{8\,c^2}-\frac {x^3\,\left (A\,c-5\,B\,b\right )}{8\,b\,c}}{b^2+2\,b\,c\,x^2+c^2\,x^4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^8*(A + B*x^2))/(b*x^2 + c*x^4)^3,x)

[Out]

(atan((c^(1/2)*x)/b^(1/2))*(A*c + 3*B*b))/(8*b^(3/2)*c^(5/2)) - ((x*(A*c + 3*B*b))/(8*c^2) - (x^3*(A*c - 5*B*b
))/(8*b*c))/(b^2 + c^2*x^4 + 2*b*c*x^2)

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sympy [A]  time = 0.76, size = 155, normalized size = 1.72 \[ - \frac {\sqrt {- \frac {1}{b^{3} c^{5}}} \left (A c + 3 B b\right ) \log {\left (- b^{2} c^{2} \sqrt {- \frac {1}{b^{3} c^{5}}} + x \right )}}{16} + \frac {\sqrt {- \frac {1}{b^{3} c^{5}}} \left (A c + 3 B b\right ) \log {\left (b^{2} c^{2} \sqrt {- \frac {1}{b^{3} c^{5}}} + x \right )}}{16} + \frac {x^{3} \left (A c^{2} - 5 B b c\right ) + x \left (- A b c - 3 B b^{2}\right )}{8 b^{3} c^{2} + 16 b^{2} c^{3} x^{2} + 8 b c^{4} x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**8*(B*x**2+A)/(c*x**4+b*x**2)**3,x)

[Out]

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

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