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

Optimal. Leaf size=140 \[ \frac{c^2 x (11 b B-15 A c)}{8 b^5 \left (b+c x^2\right )}+\frac{c^2 x (b B-A c)}{4 b^4 \left (b+c x^2\right )^2}+\frac{7 c^{3/2} (5 b B-9 A c) \tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b}}\right )}{8 b^{11/2}}-\frac{b B-3 A c}{3 b^4 x^3}+\frac{3 c (b B-2 A c)}{b^5 x}-\frac{A}{5 b^3 x^5} \]

[Out]

-A/(5*b^3*x^5) - (b*B - 3*A*c)/(3*b^4*x^3) + (3*c*(b*B - 2*A*c))/(b^5*x) + (c^2*(b*B - A*c)*x)/(4*b^4*(b + c*x
^2)^2) + (c^2*(11*b*B - 15*A*c)*x)/(8*b^5*(b + c*x^2)) + (7*c^(3/2)*(5*b*B - 9*A*c)*ArcTan[(Sqrt[c]*x)/Sqrt[b]
])/(8*b^(11/2))

________________________________________________________________________________________

Rubi [A]  time = 0.331962, antiderivative size = 140, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.238, Rules used = {1593, 456, 1805, 1802, 205} \[ \frac{c^2 x (11 b B-15 A c)}{8 b^5 \left (b+c x^2\right )}+\frac{c^2 x (b B-A c)}{4 b^4 \left (b+c x^2\right )^2}+\frac{7 c^{3/2} (5 b B-9 A c) \tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b}}\right )}{8 b^{11/2}}-\frac{b B-3 A c}{3 b^4 x^3}+\frac{3 c (b B-2 A c)}{b^5 x}-\frac{A}{5 b^3 x^5} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*x^2)/(b*x^2 + c*x^4)^3,x]

[Out]

-A/(5*b^3*x^5) - (b*B - 3*A*c)/(3*b^4*x^3) + (3*c*(b*B - 2*A*c))/(b^5*x) + (c^2*(b*B - A*c)*x)/(4*b^4*(b + c*x
^2)^2) + (c^2*(11*b*B - 15*A*c)*x)/(8*b^5*(b + c*x^2)) + (7*c^(3/2)*(5*b*B - 9*A*c)*ArcTan[(Sqrt[c]*x)/Sqrt[b]
])/(8*b^(11/2))

Rule 1593

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, 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 1805

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[(c*x)^m*Pq,
a + b*x^2, x], f = Coeff[PolynomialRemainder[(c*x)^m*Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[
(c*x)^m*Pq, a + b*x^2, x], x, 1]}, Simp[((a*g - b*f*x)*(a + b*x^2)^(p + 1))/(2*a*b*(p + 1)), x] + Dist[1/(2*a*
(p + 1)), Int[(c*x)^m*(a + b*x^2)^(p + 1)*ExpandToSum[(2*a*(p + 1)*Q)/(c*x)^m + (f*(2*p + 3))/(c*x)^m, x], x],
x]] /; FreeQ[{a, b, c}, x] && PolyQ[Pq, x] && LtQ[p, -1] && ILtQ[m, 0]

Rule 1802

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*Pq*(a + b*x
^2)^p, x], x] /; FreeQ[{a, b, c, m}, x] && PolyQ[Pq, x] && IGtQ[p, -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{A+B x^2}{\left (b x^2+c x^4\right )^3} \, dx &=\int \frac{A+B x^2}{x^6 \left (b+c x^2\right )^3} \, dx\\ &=\frac{c^2 (b B-A c) x}{4 b^4 \left (b+c x^2\right )^2}-\frac{1}{4} c^2 \int \frac{-\frac{4 A}{b c^2}-\frac{4 (b B-A c) x^2}{b^2 c^2}+\frac{4 (b B-A c) x^4}{b^3 c}-\frac{3 (b B-A c) x^6}{b^4}}{x^6 \left (b+c x^2\right )^2} \, dx\\ &=\frac{c^2 (b B-A c) x}{4 b^4 \left (b+c x^2\right )^2}+\frac{c^2 (11 b B-15 A c) x}{8 b^5 \left (b+c x^2\right )}+\frac{c^2 \int \frac{\frac{8 A}{b c^2}+\frac{8 (b B-2 A c) x^2}{b^2 c^2}-\frac{8 (2 b B-3 A c) x^4}{b^3 c}+\frac{(11 b B-15 A c) x^6}{b^4}}{x^6 \left (b+c x^2\right )} \, dx}{8 b}\\ &=\frac{c^2 (b B-A c) x}{4 b^4 \left (b+c x^2\right )^2}+\frac{c^2 (11 b B-15 A c) x}{8 b^5 \left (b+c x^2\right )}+\frac{c^2 \int \left (\frac{8 A}{b^2 c^2 x^6}+\frac{8 (b B-3 A c)}{b^3 c^2 x^4}-\frac{24 (b B-2 A c)}{b^4 c x^2}+\frac{7 (5 b B-9 A c)}{b^4 \left (b+c x^2\right )}\right ) \, dx}{8 b}\\ &=-\frac{A}{5 b^3 x^5}-\frac{b B-3 A c}{3 b^4 x^3}+\frac{3 c (b B-2 A c)}{b^5 x}+\frac{c^2 (b B-A c) x}{4 b^4 \left (b+c x^2\right )^2}+\frac{c^2 (11 b B-15 A c) x}{8 b^5 \left (b+c x^2\right )}+\frac{\left (7 c^2 (5 b B-9 A c)\right ) \int \frac{1}{b+c x^2} \, dx}{8 b^5}\\ &=-\frac{A}{5 b^3 x^5}-\frac{b B-3 A c}{3 b^4 x^3}+\frac{3 c (b B-2 A c)}{b^5 x}+\frac{c^2 (b B-A c) x}{4 b^4 \left (b+c x^2\right )^2}+\frac{c^2 (11 b B-15 A c) x}{8 b^5 \left (b+c x^2\right )}+\frac{7 c^{3/2} (5 b B-9 A c) \tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b}}\right )}{8 b^{11/2}}\\ \end{align*}

Mathematica [A]  time = 0.0746567, size = 140, normalized size = 1. \[ \frac{c^2 x (11 b B-15 A c)}{8 b^5 \left (b+c x^2\right )}+\frac{c^2 x (b B-A c)}{4 b^4 \left (b+c x^2\right )^2}+\frac{7 c^{3/2} (5 b B-9 A c) \tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b}}\right )}{8 b^{11/2}}-\frac{b B-3 A c}{3 b^4 x^3}+\frac{3 c (b B-2 A c)}{b^5 x}-\frac{A}{5 b^3 x^5} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-A/(5*b^3*x^5) - (b*B - 3*A*c)/(3*b^4*x^3) + (3*c*(b*B - 2*A*c))/(b^5*x) + (c^2*(b*B - A*c)*x)/(4*b^4*(b + c*x
^2)^2) + (c^2*(11*b*B - 15*A*c)*x)/(8*b^5*(b + c*x^2)) + (7*c^(3/2)*(5*b*B - 9*A*c)*ArcTan[(Sqrt[c]*x)/Sqrt[b]
])/(8*b^(11/2))

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Maple [A]  time = 0.014, size = 177, normalized size = 1.3 \begin{align*} -{\frac{A}{5\,{x}^{5}{b}^{3}}}+{\frac{Ac}{{b}^{4}{x}^{3}}}-{\frac{B}{3\,{b}^{3}{x}^{3}}}-6\,{\frac{A{c}^{2}}{{b}^{5}x}}+3\,{\frac{Bc}{{b}^{4}x}}-{\frac{15\,{c}^{4}A{x}^{3}}{8\,{b}^{5} \left ( c{x}^{2}+b \right ) ^{2}}}+{\frac{11\,B{c}^{3}{x}^{3}}{8\,{b}^{4} \left ( c{x}^{2}+b \right ) ^{2}}}-{\frac{17\,A{c}^{3}x}{8\,{b}^{4} \left ( c{x}^{2}+b \right ) ^{2}}}+{\frac{13\,B{c}^{2}x}{8\,{b}^{3} \left ( c{x}^{2}+b \right ) ^{2}}}-{\frac{63\,A{c}^{3}}{8\,{b}^{5}}\arctan \left ({cx{\frac{1}{\sqrt{bc}}}} \right ){\frac{1}{\sqrt{bc}}}}+{\frac{35\,{c}^{2}B}{8\,{b}^{4}}\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((B*x^2+A)/(c*x^4+b*x^2)^3,x)

[Out]

-1/5*A/x^5/b^3+1/b^4/x^3*A*c-1/3/b^3/x^3*B-6*c^2/b^5/x*A+3*c/b^4/x*B-15/8/b^5*c^4/(c*x^2+b)^2*A*x^3+11/8/b^4*c
^3/(c*x^2+b)^2*B*x^3-17/8/b^4*c^3/(c*x^2+b)^2*A*x+13/8/b^3*c^2/(c*x^2+b)^2*B*x-63/8/b^5*c^3/(b*c)^(1/2)*arctan
(x*c/(b*c)^(1/2))*A+35/8/b^4*c^2/(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((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 = 1.07862, size = 909, normalized size = 6.49 \begin{align*} \left [\frac{210 \,{\left (5 \, B b c^{3} - 9 \, A c^{4}\right )} x^{8} + 350 \,{\left (5 \, B b^{2} c^{2} - 9 \, A b c^{3}\right )} x^{6} - 48 \, A b^{4} + 112 \,{\left (5 \, B b^{3} c - 9 \, A b^{2} c^{2}\right )} x^{4} - 16 \,{\left (5 \, B b^{4} - 9 \, A b^{3} c\right )} x^{2} - 105 \,{\left ({\left (5 \, B b c^{3} - 9 \, A c^{4}\right )} x^{9} + 2 \,{\left (5 \, B b^{2} c^{2} - 9 \, A b c^{3}\right )} x^{7} +{\left (5 \, B b^{3} c - 9 \, A b^{2} c^{2}\right )} x^{5}\right )} \sqrt{-\frac{c}{b}} \log \left (\frac{c x^{2} - 2 \, b x \sqrt{-\frac{c}{b}} - b}{c x^{2} + b}\right )}{240 \,{\left (b^{5} c^{2} x^{9} + 2 \, b^{6} c x^{7} + b^{7} x^{5}\right )}}, \frac{105 \,{\left (5 \, B b c^{3} - 9 \, A c^{4}\right )} x^{8} + 175 \,{\left (5 \, B b^{2} c^{2} - 9 \, A b c^{3}\right )} x^{6} - 24 \, A b^{4} + 56 \,{\left (5 \, B b^{3} c - 9 \, A b^{2} c^{2}\right )} x^{4} - 8 \,{\left (5 \, B b^{4} - 9 \, A b^{3} c\right )} x^{2} + 105 \,{\left ({\left (5 \, B b c^{3} - 9 \, A c^{4}\right )} x^{9} + 2 \,{\left (5 \, B b^{2} c^{2} - 9 \, A b c^{3}\right )} x^{7} +{\left (5 \, B b^{3} c - 9 \, A b^{2} c^{2}\right )} x^{5}\right )} \sqrt{\frac{c}{b}} \arctan \left (x \sqrt{\frac{c}{b}}\right )}{120 \,{\left (b^{5} c^{2} x^{9} + 2 \, b^{6} c x^{7} + b^{7} x^{5}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/240*(210*(5*B*b*c^3 - 9*A*c^4)*x^8 + 350*(5*B*b^2*c^2 - 9*A*b*c^3)*x^6 - 48*A*b^4 + 112*(5*B*b^3*c - 9*A*b^
2*c^2)*x^4 - 16*(5*B*b^4 - 9*A*b^3*c)*x^2 - 105*((5*B*b*c^3 - 9*A*c^4)*x^9 + 2*(5*B*b^2*c^2 - 9*A*b*c^3)*x^7 +
(5*B*b^3*c - 9*A*b^2*c^2)*x^5)*sqrt(-c/b)*log((c*x^2 - 2*b*x*sqrt(-c/b) - b)/(c*x^2 + b)))/(b^5*c^2*x^9 + 2*b
^6*c*x^7 + b^7*x^5), 1/120*(105*(5*B*b*c^3 - 9*A*c^4)*x^8 + 175*(5*B*b^2*c^2 - 9*A*b*c^3)*x^6 - 24*A*b^4 + 56*
(5*B*b^3*c - 9*A*b^2*c^2)*x^4 - 8*(5*B*b^4 - 9*A*b^3*c)*x^2 + 105*((5*B*b*c^3 - 9*A*c^4)*x^9 + 2*(5*B*b^2*c^2
- 9*A*b*c^3)*x^7 + (5*B*b^3*c - 9*A*b^2*c^2)*x^5)*sqrt(c/b)*arctan(x*sqrt(c/b)))/(b^5*c^2*x^9 + 2*b^6*c*x^7 +
b^7*x^5)]

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Sympy [A]  time = 1.69213, size = 260, normalized size = 1.86 \begin{align*} - \frac{7 \sqrt{- \frac{c^{3}}{b^{11}}} \left (- 9 A c + 5 B b\right ) \log{\left (- \frac{7 b^{6} \sqrt{- \frac{c^{3}}{b^{11}}} \left (- 9 A c + 5 B b\right )}{- 63 A c^{3} + 35 B b c^{2}} + x \right )}}{16} + \frac{7 \sqrt{- \frac{c^{3}}{b^{11}}} \left (- 9 A c + 5 B b\right ) \log{\left (\frac{7 b^{6} \sqrt{- \frac{c^{3}}{b^{11}}} \left (- 9 A c + 5 B b\right )}{- 63 A c^{3} + 35 B b c^{2}} + x \right )}}{16} + \frac{- 24 A b^{4} + x^{8} \left (- 945 A c^{4} + 525 B b c^{3}\right ) + x^{6} \left (- 1575 A b c^{3} + 875 B b^{2} c^{2}\right ) + x^{4} \left (- 504 A b^{2} c^{2} + 280 B b^{3} c\right ) + x^{2} \left (72 A b^{3} c - 40 B b^{4}\right )}{120 b^{7} x^{5} + 240 b^{6} c x^{7} + 120 b^{5} c^{2} x^{9}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-7*sqrt(-c**3/b**11)*(-9*A*c + 5*B*b)*log(-7*b**6*sqrt(-c**3/b**11)*(-9*A*c + 5*B*b)/(-63*A*c**3 + 35*B*b*c**2
) + x)/16 + 7*sqrt(-c**3/b**11)*(-9*A*c + 5*B*b)*log(7*b**6*sqrt(-c**3/b**11)*(-9*A*c + 5*B*b)/(-63*A*c**3 + 3
5*B*b*c**2) + x)/16 + (-24*A*b**4 + x**8*(-945*A*c**4 + 525*B*b*c**3) + x**6*(-1575*A*b*c**3 + 875*B*b**2*c**2
) + x**4*(-504*A*b**2*c**2 + 280*B*b**3*c) + x**2*(72*A*b**3*c - 40*B*b**4))/(120*b**7*x**5 + 240*b**6*c*x**7
+ 120*b**5*c**2*x**9)

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Giac [A]  time = 1.1575, size = 182, normalized size = 1.3 \begin{align*} \frac{7 \,{\left (5 \, B b c^{2} - 9 \, A c^{3}\right )} \arctan \left (\frac{c x}{\sqrt{b c}}\right )}{8 \, \sqrt{b c} b^{5}} + \frac{11 \, B b c^{3} x^{3} - 15 \, A c^{4} x^{3} + 13 \, B b^{2} c^{2} x - 17 \, A b c^{3} x}{8 \,{\left (c x^{2} + b\right )}^{2} b^{5}} + \frac{45 \, B b c x^{4} - 90 \, A c^{2} x^{4} - 5 \, B b^{2} x^{2} + 15 \, A b c x^{2} - 3 \, A b^{2}}{15 \, b^{5} x^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

7/8*(5*B*b*c^2 - 9*A*c^3)*arctan(c*x/sqrt(b*c))/(sqrt(b*c)*b^5) + 1/8*(11*B*b*c^3*x^3 - 15*A*c^4*x^3 + 13*B*b^
2*c^2*x - 17*A*b*c^3*x)/((c*x^2 + b)^2*b^5) + 1/15*(45*B*b*c*x^4 - 90*A*c^2*x^4 - 5*B*b^2*x^2 + 15*A*b*c*x^2 -
3*A*b^2)/(b^5*x^5)