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

Optimal. Leaf size=90 \[ -\frac {\sqrt {c} (3 b B-5 A c) \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{2 b^{7/2}}-\frac {c x (b B-A c)}{2 b^3 \left (b+c x^2\right )}-\frac {b B-2 A c}{b^3 x}-\frac {A}{3 b^2 x^3} \]

[Out]

-1/3*A/b^2/x^3+(2*A*c-B*b)/b^3/x-1/2*c*(-A*c+B*b)*x/b^3/(c*x^2+b)-1/2*(-5*A*c+3*B*b)*arctan(x*c^(1/2)/b^(1/2))
*c^(1/2)/b^(7/2)

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Rubi [A]  time = 0.12, antiderivative size = 90, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {1593, 456, 1261, 205} \[ -\frac {c x (b B-A c)}{2 b^3 \left (b+c x^2\right )}-\frac {b B-2 A c}{b^3 x}-\frac {\sqrt {c} (3 b B-5 A c) \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{2 b^{7/2}}-\frac {A}{3 b^2 x^3} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*x^2)/(b*x^2 + c*x^4)^2,x]

[Out]

-A/(3*b^2*x^3) - (b*B - 2*A*c)/(b^3*x) - (c*(b*B - A*c)*x)/(2*b^3*(b + c*x^2)) - (Sqrt[c]*(3*b*B - 5*A*c)*ArcT
an[(Sqrt[c]*x)/Sqrt[b]])/(2*b^(7/2))

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 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 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 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]

Rubi steps

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

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Mathematica [A]  time = 0.08, size = 90, normalized size = 1.00 \[ -\frac {\sqrt {c} (3 b B-5 A c) \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{2 b^{7/2}}-\frac {c x (b B-A c)}{2 b^3 \left (b+c x^2\right )}+\frac {2 A c-b B}{b^3 x}-\frac {A}{3 b^2 x^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/3*A/(b^2*x^3) + (-(b*B) + 2*A*c)/(b^3*x) - (c*(b*B - A*c)*x)/(2*b^3*(b + c*x^2)) - (Sqrt[c]*(3*b*B - 5*A*c)
*ArcTan[(Sqrt[c]*x)/Sqrt[b]])/(2*b^(7/2))

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fricas [A]  time = 0.89, size = 250, normalized size = 2.78 \[ \left [-\frac {6 \, {\left (3 \, B b c - 5 \, A c^{2}\right )} x^{4} + 4 \, A b^{2} + 4 \, {\left (3 \, B b^{2} - 5 \, A b c\right )} x^{2} + 3 \, {\left ({\left (3 \, B b c - 5 \, A c^{2}\right )} x^{5} + {\left (3 \, B b^{2} - 5 \, A b c\right )} x^{3}\right )} \sqrt {-\frac {c}{b}} \log \left (\frac {c x^{2} + 2 \, b x \sqrt {-\frac {c}{b}} - b}{c x^{2} + b}\right )}{12 \, {\left (b^{3} c x^{5} + b^{4} x^{3}\right )}}, -\frac {3 \, {\left (3 \, B b c - 5 \, A c^{2}\right )} x^{4} + 2 \, A b^{2} + 2 \, {\left (3 \, B b^{2} - 5 \, A b c\right )} x^{2} + 3 \, {\left ({\left (3 \, B b c - 5 \, A c^{2}\right )} x^{5} + {\left (3 \, B b^{2} - 5 \, A b c\right )} x^{3}\right )} \sqrt {\frac {c}{b}} \arctan \left (x \sqrt {\frac {c}{b}}\right )}{6 \, {\left (b^{3} c x^{5} + b^{4} x^{3}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/12*(6*(3*B*b*c - 5*A*c^2)*x^4 + 4*A*b^2 + 4*(3*B*b^2 - 5*A*b*c)*x^2 + 3*((3*B*b*c - 5*A*c^2)*x^5 + (3*B*b^
2 - 5*A*b*c)*x^3)*sqrt(-c/b)*log((c*x^2 + 2*b*x*sqrt(-c/b) - b)/(c*x^2 + b)))/(b^3*c*x^5 + b^4*x^3), -1/6*(3*(
3*B*b*c - 5*A*c^2)*x^4 + 2*A*b^2 + 2*(3*B*b^2 - 5*A*b*c)*x^2 + 3*((3*B*b*c - 5*A*c^2)*x^5 + (3*B*b^2 - 5*A*b*c
)*x^3)*sqrt(c/b)*arctan(x*sqrt(c/b)))/(b^3*c*x^5 + b^4*x^3)]

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giac [A]  time = 0.16, size = 85, normalized size = 0.94 \[ -\frac {{\left (3 \, B b c - 5 \, A c^{2}\right )} \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{2 \, \sqrt {b c} b^{3}} - \frac {B b c x - A c^{2} x}{2 \, {\left (c x^{2} + b\right )} b^{3}} - \frac {3 \, B b x^{2} - 6 \, A c x^{2} + A b}{3 \, b^{3} x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(3*B*b*c - 5*A*c^2)*arctan(c*x/sqrt(b*c))/(sqrt(b*c)*b^3) - 1/2*(B*b*c*x - A*c^2*x)/((c*x^2 + b)*b^3) - 1
/3*(3*B*b*x^2 - 6*A*c*x^2 + A*b)/(b^3*x^3)

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maple [A]  time = 0.07, size = 110, normalized size = 1.22 \[ \frac {A \,c^{2} x}{2 \left (c \,x^{2}+b \right ) b^{3}}+\frac {5 A \,c^{2} \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{2 \sqrt {b c}\, b^{3}}-\frac {B c x}{2 \left (c \,x^{2}+b \right ) b^{2}}-\frac {3 B c \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{2 \sqrt {b c}\, b^{2}}+\frac {2 A c}{b^{3} x}-\frac {B}{b^{2} x}-\frac {A}{3 b^{2} x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 3.00, size = 93, normalized size = 1.03 \[ -\frac {3 \, {\left (3 \, B b c - 5 \, A c^{2}\right )} x^{4} + 2 \, A b^{2} + 2 \, {\left (3 \, B b^{2} - 5 \, A b c\right )} x^{2}}{6 \, {\left (b^{3} c x^{5} + b^{4} x^{3}\right )}} - \frac {{\left (3 \, B b c - 5 \, A c^{2}\right )} \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{2 \, \sqrt {b c} b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(3*(3*B*b*c - 5*A*c^2)*x^4 + 2*A*b^2 + 2*(3*B*b^2 - 5*A*b*c)*x^2)/(b^3*c*x^5 + b^4*x^3) - 1/2*(3*B*b*c -
5*A*c^2)*arctan(c*x/sqrt(b*c))/(sqrt(b*c)*b^3)

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mupad [B]  time = 0.14, size = 83, normalized size = 0.92 \[ \frac {\frac {x^2\,\left (5\,A\,c-3\,B\,b\right )}{3\,b^2}-\frac {A}{3\,b}+\frac {c\,x^4\,\left (5\,A\,c-3\,B\,b\right )}{2\,b^3}}{c\,x^5+b\,x^3}+\frac {\sqrt {c}\,\mathrm {atan}\left (\frac {\sqrt {c}\,x}{\sqrt {b}}\right )\,\left (5\,A\,c-3\,B\,b\right )}{2\,b^{7/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*x^2)/(b*x^2 + c*x^4)^2,x)

[Out]

((x^2*(5*A*c - 3*B*b))/(3*b^2) - A/(3*b) + (c*x^4*(5*A*c - 3*B*b))/(2*b^3))/(b*x^3 + c*x^5) + (c^(1/2)*atan((c
^(1/2)*x)/b^(1/2))*(5*A*c - 3*B*b))/(2*b^(7/2))

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sympy [B]  time = 0.60, size = 184, normalized size = 2.04 \[ \frac {\sqrt {- \frac {c}{b^{7}}} \left (- 5 A c + 3 B b\right ) \log {\left (- \frac {b^{4} \sqrt {- \frac {c}{b^{7}}} \left (- 5 A c + 3 B b\right )}{- 5 A c^{2} + 3 B b c} + x \right )}}{4} - \frac {\sqrt {- \frac {c}{b^{7}}} \left (- 5 A c + 3 B b\right ) \log {\left (\frac {b^{4} \sqrt {- \frac {c}{b^{7}}} \left (- 5 A c + 3 B b\right )}{- 5 A c^{2} + 3 B b c} + x \right )}}{4} + \frac {- 2 A b^{2} + x^{4} \left (15 A c^{2} - 9 B b c\right ) + x^{2} \left (10 A b c - 6 B b^{2}\right )}{6 b^{4} x^{3} + 6 b^{3} c x^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sqrt(-c/b**7)*(-5*A*c + 3*B*b)*log(-b**4*sqrt(-c/b**7)*(-5*A*c + 3*B*b)/(-5*A*c**2 + 3*B*b*c) + x)/4 - sqrt(-c
/b**7)*(-5*A*c + 3*B*b)*log(b**4*sqrt(-c/b**7)*(-5*A*c + 3*B*b)/(-5*A*c**2 + 3*B*b*c) + x)/4 + (-2*A*b**2 + x*
*4*(15*A*c**2 - 9*B*b*c) + x**2*(10*A*b*c - 6*B*b**2))/(6*b**4*x**3 + 6*b**3*c*x**5)

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