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

Optimal. Leaf size=117 \[ -\frac{c x (7 b B-11 A c)}{8 b^4 \left (b+c x^2\right )}-\frac{c x (b B-A c)}{4 b^3 \left (b+c x^2\right )^2}-\frac{b B-3 A c}{b^4 x}-\frac{5 \sqrt{c} (3 b B-7 A c) \tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b}}\right )}{8 b^{9/2}}-\frac{A}{3 b^3 x^3} \]

[Out]

-A/(3*b^3*x^3) - (b*B - 3*A*c)/(b^4*x) - (c*(b*B - A*c)*x)/(4*b^3*(b + c*x^2)^2) - (c*(7*b*B - 11*A*c)*x)/(8*b
^4*(b + c*x^2)) - (5*Sqrt[c]*(3*b*B - 7*A*c)*ArcTan[(Sqrt[c]*x)/Sqrt[b]])/(8*b^(9/2))

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Rubi [A]  time = 0.176717, antiderivative size = 117, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.208, Rules used = {1584, 456, 1259, 1261, 205} \[ -\frac{c x (7 b B-11 A c)}{8 b^4 \left (b+c x^2\right )}-\frac{c x (b B-A c)}{4 b^3 \left (b+c x^2\right )^2}-\frac{b B-3 A c}{b^4 x}-\frac{5 \sqrt{c} (3 b B-7 A c) \tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b}}\right )}{8 b^{9/2}}-\frac{A}{3 b^3 x^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-A/(3*b^3*x^3) - (b*B - 3*A*c)/(b^4*x) - (c*(b*B - A*c)*x)/(4*b^3*(b + c*x^2)^2) - (c*(7*b*B - 11*A*c)*x)/(8*b
^4*(b + c*x^2)) - (5*Sqrt[c]*(3*b*B - 7*A*c)*ArcTan[(Sqrt[c]*x)/Sqrt[b]])/(8*b^(9/2))

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]

Rule 456

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[x^m*(a + b*x^2)^(p +
1)*ExpandToSum[2*b*(p + 1)*Together[(b^(m/2)*(c + d*x^2) - (-a)^(m/2 - 1)*(b*c - a*d)*x^(-m + 2))/(a + b*x^2)]
- ((-a)^(m/2 - 1)*(b*c - a*d))/x^m, x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] &
& ILtQ[m/2, 0] && (IntegerQ[p] || EqQ[m + 2*p + 1, 0])

Rule 1259

Int[(x_)^(m_)*((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Simp[((-d)^(
m/2 - 1)*(c*d^2 - b*d*e + a*e^2)^p*x*(d + e*x^2)^(q + 1))/(2*e^(2*p + m/2)*(q + 1)), x] + Dist[(-d)^(m/2 - 1)/
(2*e^(2*p)*(q + 1)), Int[x^m*(d + e*x^2)^(q + 1)*ExpandToSum[Together[(1*(2*(-d)^(-(m/2) + 1)*e^(2*p)*(q + 1)*
(a + b*x^2 + c*x^4)^p - ((c*d^2 - b*d*e + a*e^2)^p/(e^(m/2)*x^m))*(d + e*(2*q + 3)*x^2)))/(d + e*x^2)], x], x]
, x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && IGtQ[p, 0] && ILtQ[q, -1] && ILtQ[m/2, 0]

Rule 1261

Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> In
t[ExpandIntegrand[(f*x)^m*(d + e*x^2)^q*(a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, q}, x] &&
NeQ[b^2 - 4*a*c, 0] && IGtQ[p, 0] && IGtQ[q, -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]

Rubi steps

\begin{align*} \int \frac{x^2 \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^3} \, dx &=\int \frac{A+B x^2}{x^4 \left (b+c x^2\right )^3} \, dx\\ &=-\frac{c (b B-A c) x}{4 b^3 \left (b+c x^2\right )^2}-\frac{1}{4} c \int \frac{-\frac{4 A}{b c}-\frac{4 (b B-A c) x^2}{b^2 c}+\frac{3 (b B-A c) x^4}{b^3}}{x^4 \left (b+c x^2\right )^2} \, dx\\ &=-\frac{c (b B-A c) x}{4 b^3 \left (b+c x^2\right )^2}-\frac{c (7 b B-11 A c) x}{8 b^4 \left (b+c x^2\right )}-\frac{\int \frac{-8 A b c-8 c (b B-2 A c) x^2+\frac{c^2 (7 b B-11 A c) x^4}{b}}{x^4 \left (b+c x^2\right )} \, dx}{8 b^3 c}\\ &=-\frac{c (b B-A c) x}{4 b^3 \left (b+c x^2\right )^2}-\frac{c (7 b B-11 A c) x}{8 b^4 \left (b+c x^2\right )}-\frac{\int \left (-\frac{8 A c}{x^4}-\frac{8 c (b B-3 A c)}{b x^2}+\frac{5 c^2 (3 b B-7 A c)}{b \left (b+c x^2\right )}\right ) \, dx}{8 b^3 c}\\ &=-\frac{A}{3 b^3 x^3}-\frac{b B-3 A c}{b^4 x}-\frac{c (b B-A c) x}{4 b^3 \left (b+c x^2\right )^2}-\frac{c (7 b B-11 A c) x}{8 b^4 \left (b+c x^2\right )}-\frac{(5 c (3 b B-7 A c)) \int \frac{1}{b+c x^2} \, dx}{8 b^4}\\ &=-\frac{A}{3 b^3 x^3}-\frac{b B-3 A c}{b^4 x}-\frac{c (b B-A c) x}{4 b^3 \left (b+c x^2\right )^2}-\frac{c (7 b B-11 A c) x}{8 b^4 \left (b+c x^2\right )}-\frac{5 \sqrt{c} (3 b B-7 A c) \tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b}}\right )}{8 b^{9/2}}\\ \end{align*}

Mathematica [A]  time = 0.0697808, size = 119, normalized size = 1.02 \[ -\frac{x \left (7 b B c-11 A c^2\right )}{8 b^4 \left (b+c x^2\right )}-\frac{c x (b B-A c)}{4 b^3 \left (b+c x^2\right )^2}+\frac{3 A c-b B}{b^4 x}-\frac{5 \sqrt{c} (3 b B-7 A c) \tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b}}\right )}{8 b^{9/2}}-\frac{A}{3 b^3 x^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-A/(3*b^3*x^3) + (-(b*B) + 3*A*c)/(b^4*x) - (c*(b*B - A*c)*x)/(4*b^3*(b + c*x^2)^2) - ((7*b*B*c - 11*A*c^2)*x)
/(8*b^4*(b + c*x^2)) - (5*Sqrt[c]*(3*b*B - 7*A*c)*ArcTan[(Sqrt[c]*x)/Sqrt[b]])/(8*b^(9/2))

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Maple [A]  time = 0.013, size = 152, normalized size = 1.3 \begin{align*} -{\frac{A}{3\,{b}^{3}{x}^{3}}}+3\,{\frac{Ac}{{b}^{4}x}}-{\frac{B}{{b}^{3}x}}+{\frac{11\,A{x}^{3}{c}^{3}}{8\,{b}^{4} \left ( c{x}^{2}+b \right ) ^{2}}}-{\frac{7\,B{c}^{2}{x}^{3}}{8\,{b}^{3} \left ( c{x}^{2}+b \right ) ^{2}}}+{\frac{13\,A{c}^{2}x}{8\,{b}^{3} \left ( c{x}^{2}+b \right ) ^{2}}}-{\frac{9\,Bcx}{8\,{b}^{2} \left ( c{x}^{2}+b \right ) ^{2}}}+{\frac{35\,A{c}^{2}}{8\,{b}^{4}}\arctan \left ({cx{\frac{1}{\sqrt{bc}}}} \right ){\frac{1}{\sqrt{bc}}}}-{\frac{15\,Bc}{8\,{b}^{3}}\arctan \left ({cx{\frac{1}{\sqrt{bc}}}} \right ){\frac{1}{\sqrt{bc}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*A/b^3/x^3+3/b^4/x*A*c-1/b^3/x*B+11/8/b^4*c^3/(c*x^2+b)^2*A*x^3-7/8/b^3*c^2/(c*x^2+b)^2*B*x^3+13/8/b^3*c^2
/(c*x^2+b)^2*A*x-9/8/b^2*c/(c*x^2+b)^2*B*x+35/8/b^4*c^2/(b*c)^(1/2)*arctan(x*c/(b*c)^(1/2))*A-15/8/b^3*c/(b*c)
^(1/2)*arctan(x*c/(b*c)^(1/2))*B

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.04025, size = 782, normalized size = 6.68 \begin{align*} \left [-\frac{30 \,{\left (3 \, B b c^{2} - 7 \, A c^{3}\right )} x^{6} + 50 \,{\left (3 \, B b^{2} c - 7 \, A b c^{2}\right )} x^{4} + 16 \, A b^{3} + 16 \,{\left (3 \, B b^{3} - 7 \, A b^{2} c\right )} x^{2} + 15 \,{\left ({\left (3 \, B b c^{2} - 7 \, A c^{3}\right )} x^{7} + 2 \,{\left (3 \, B b^{2} c - 7 \, A b c^{2}\right )} x^{5} +{\left (3 \, B b^{3} - 7 \, A b^{2} c\right )} x^{3}\right )} \sqrt{-\frac{c}{b}} \log \left (\frac{c x^{2} + 2 \, b x \sqrt{-\frac{c}{b}} - b}{c x^{2} + b}\right )}{48 \,{\left (b^{4} c^{2} x^{7} + 2 \, b^{5} c x^{5} + b^{6} x^{3}\right )}}, -\frac{15 \,{\left (3 \, B b c^{2} - 7 \, A c^{3}\right )} x^{6} + 25 \,{\left (3 \, B b^{2} c - 7 \, A b c^{2}\right )} x^{4} + 8 \, A b^{3} + 8 \,{\left (3 \, B b^{3} - 7 \, A b^{2} c\right )} x^{2} + 15 \,{\left ({\left (3 \, B b c^{2} - 7 \, A c^{3}\right )} x^{7} + 2 \,{\left (3 \, B b^{2} c - 7 \, A b c^{2}\right )} x^{5} +{\left (3 \, B b^{3} - 7 \, A b^{2} c\right )} x^{3}\right )} \sqrt{\frac{c}{b}} \arctan \left (x \sqrt{\frac{c}{b}}\right )}{24 \,{\left (b^{4} c^{2} x^{7} + 2 \, b^{5} c x^{5} + b^{6} x^{3}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/48*(30*(3*B*b*c^2 - 7*A*c^3)*x^6 + 50*(3*B*b^2*c - 7*A*b*c^2)*x^4 + 16*A*b^3 + 16*(3*B*b^3 - 7*A*b^2*c)*x^
2 + 15*((3*B*b*c^2 - 7*A*c^3)*x^7 + 2*(3*B*b^2*c - 7*A*b*c^2)*x^5 + (3*B*b^3 - 7*A*b^2*c)*x^3)*sqrt(-c/b)*log(
(c*x^2 + 2*b*x*sqrt(-c/b) - b)/(c*x^2 + b)))/(b^4*c^2*x^7 + 2*b^5*c*x^5 + b^6*x^3), -1/24*(15*(3*B*b*c^2 - 7*A
*c^3)*x^6 + 25*(3*B*b^2*c - 7*A*b*c^2)*x^4 + 8*A*b^3 + 8*(3*B*b^3 - 7*A*b^2*c)*x^2 + 15*((3*B*b*c^2 - 7*A*c^3)
*x^7 + 2*(3*B*b^2*c - 7*A*b*c^2)*x^5 + (3*B*b^3 - 7*A*b^2*c)*x^3)*sqrt(c/b)*arctan(x*sqrt(c/b)))/(b^4*c^2*x^7
+ 2*b^5*c*x^5 + b^6*x^3)]

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Sympy [B]  time = 1.21224, size = 226, normalized size = 1.93 \begin{align*} \frac{5 \sqrt{- \frac{c}{b^{9}}} \left (- 7 A c + 3 B b\right ) \log{\left (- \frac{5 b^{5} \sqrt{- \frac{c}{b^{9}}} \left (- 7 A c + 3 B b\right )}{- 35 A c^{2} + 15 B b c} + x \right )}}{16} - \frac{5 \sqrt{- \frac{c}{b^{9}}} \left (- 7 A c + 3 B b\right ) \log{\left (\frac{5 b^{5} \sqrt{- \frac{c}{b^{9}}} \left (- 7 A c + 3 B b\right )}{- 35 A c^{2} + 15 B b c} + x \right )}}{16} - \frac{8 A b^{3} + x^{6} \left (- 105 A c^{3} + 45 B b c^{2}\right ) + x^{4} \left (- 175 A b c^{2} + 75 B b^{2} c\right ) + x^{2} \left (- 56 A b^{2} c + 24 B b^{3}\right )}{24 b^{6} x^{3} + 48 b^{5} c x^{5} + 24 b^{4} c^{2} x^{7}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5*sqrt(-c/b**9)*(-7*A*c + 3*B*b)*log(-5*b**5*sqrt(-c/b**9)*(-7*A*c + 3*B*b)/(-35*A*c**2 + 15*B*b*c) + x)/16 -
5*sqrt(-c/b**9)*(-7*A*c + 3*B*b)*log(5*b**5*sqrt(-c/b**9)*(-7*A*c + 3*B*b)/(-35*A*c**2 + 15*B*b*c) + x)/16 - (
8*A*b**3 + x**6*(-105*A*c**3 + 45*B*b*c**2) + x**4*(-175*A*b*c**2 + 75*B*b**2*c) + x**2*(-56*A*b**2*c + 24*B*b
**3))/(24*b**6*x**3 + 48*b**5*c*x**5 + 24*b**4*c**2*x**7)

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Giac [A]  time = 1.17738, size = 146, normalized size = 1.25 \begin{align*} -\frac{5 \,{\left (3 \, B b c - 7 \, A c^{2}\right )} \arctan \left (\frac{c x}{\sqrt{b c}}\right )}{8 \, \sqrt{b c} b^{4}} - \frac{7 \, B b c^{2} x^{3} - 11 \, A c^{3} x^{3} + 9 \, B b^{2} c x - 13 \, A b c^{2} x}{8 \,{\left (c x^{2} + b\right )}^{2} b^{4}} - \frac{3 \, B b x^{2} - 9 \, A c x^{2} + A b}{3 \, b^{4} x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-5/8*(3*B*b*c - 7*A*c^2)*arctan(c*x/sqrt(b*c))/(sqrt(b*c)*b^4) - 1/8*(7*B*b*c^2*x^3 - 11*A*c^3*x^3 + 9*B*b^2*c
*x - 13*A*b*c^2*x)/((c*x^2 + b)^2*b^4) - 1/3*(3*B*b*x^2 - 9*A*c*x^2 + A*b)/(b^4*x^3)