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

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

________________________________________________________________________________________

Rubi [A]  time = 0.33, antiderivative size = 140, normalized size of antiderivative = 1.00, 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} \begin {gather*} \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} \end {gather*}

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 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 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 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 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]

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 = 0.08, size = 140, normalized size = 1.00 \begin {gather*} \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 {c^2 x (11 b B-15 A c)}{8 b^5 \left (b+c x^2\right )}+\frac {3 c (b B-2 A c)}{b^5 x}+\frac {c^2 x (b B-A c)}{4 b^4 \left (b+c x^2\right )^2}-\frac {b B-3 A c}{3 b^4 x^3}-\frac {A}{5 b^3 x^5} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/5*A/(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))

________________________________________________________________________________________

IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {A+B x^2}{\left (b x^2+c x^4\right )^3} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [A]  time = 0.43, size = 426, normalized size = 3.04 \begin {gather*} \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 {gather*}

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)]

________________________________________________________________________________________

giac [A]  time = 0.16, size = 135, normalized size = 0.96 \begin {gather*} \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 {gather*}

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)

________________________________________________________________________________________

maple [A]  time = 0.07, size = 177, normalized size = 1.26 \begin {gather*} -\frac {15 A \,c^{4} x^{3}}{8 \left (c \,x^{2}+b \right )^{2} b^{5}}+\frac {11 B \,c^{3} x^{3}}{8 \left (c \,x^{2}+b \right )^{2} b^{4}}-\frac {17 A \,c^{3} x}{8 \left (c \,x^{2}+b \right )^{2} b^{4}}+\frac {13 B \,c^{2} x}{8 \left (c \,x^{2}+b \right )^{2} b^{3}}-\frac {63 A \,c^{3} \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{8 \sqrt {b c}\, b^{5}}+\frac {35 B \,c^{2} \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{8 \sqrt {b c}\, b^{4}}-\frac {6 A \,c^{2}}{b^{5} x}+\frac {3 B c}{b^{4} x}+\frac {A c}{b^{4} x^{3}}-\frac {B}{3 b^{3} x^{3}}-\frac {A}{5 b^{3} x^{5}} \end {gather*}

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]

-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(1/(b*c)^(1/2)*c*x)*A+35/8/b^4*c^2/(b*c)^(1/2)*arctan(1/(b*c)^(1/2)*
c*x)*B-1/5*A/b^3/x^5+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

________________________________________________________________________________________

maxima [A]  time = 3.02, size = 154, normalized size = 1.10 \begin {gather*} \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}}{120 \, {\left (b^{5} c^{2} x^{9} + 2 \, b^{6} c x^{7} + b^{7} x^{5}\right )}} + \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}} \end {gather*}

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]

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)/(b^5*c^2*x^9 + 2*b^6*c*x^7 + b^7*x^5) + 7/8*(5*B*b*c^2 - 9*A*c^3)*arct
an(c*x/sqrt(b*c))/(sqrt(b*c)*b^5)

________________________________________________________________________________________

mupad [B]  time = 0.19, size = 135, normalized size = 0.96 \begin {gather*} -\frac {\frac {A}{5\,b}-\frac {x^2\,\left (9\,A\,c-5\,B\,b\right )}{15\,b^2}+\frac {35\,c^2\,x^6\,\left (9\,A\,c-5\,B\,b\right )}{24\,b^4}+\frac {7\,c^3\,x^8\,\left (9\,A\,c-5\,B\,b\right )}{8\,b^5}+\frac {7\,c\,x^4\,\left (9\,A\,c-5\,B\,b\right )}{15\,b^3}}{b^2\,x^5+2\,b\,c\,x^7+c^2\,x^9}-\frac {7\,c^{3/2}\,\mathrm {atan}\left (\frac {\sqrt {c}\,x}{\sqrt {b}}\right )\,\left (9\,A\,c-5\,B\,b\right )}{8\,b^{11/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*x^2)/(b*x^2 + c*x^4)^3,x)

[Out]

- (A/(5*b) - (x^2*(9*A*c - 5*B*b))/(15*b^2) + (35*c^2*x^6*(9*A*c - 5*B*b))/(24*b^4) + (7*c^3*x^8*(9*A*c - 5*B*
b))/(8*b^5) + (7*c*x^4*(9*A*c - 5*B*b))/(15*b^3))/(b^2*x^5 + c^2*x^9 + 2*b*c*x^7) - (7*c^(3/2)*atan((c^(1/2)*x
)/b^(1/2))*(9*A*c - 5*B*b))/(8*b^(11/2))

________________________________________________________________________________________

sympy [A]  time = 0.92, size = 260, normalized size = 1.86 \begin {gather*} - \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 {gather*}

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)

________________________________________________________________________________________