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

Optimal. Leaf size=148 \[ -\frac {c^2 (3 b B-5 A c) \log \left (b+c x^2\right )}{b^6}+\frac {2 c^2 \log (x) (3 b B-5 A c)}{b^6}+\frac {c^2 (3 b B-4 A c)}{2 b^5 \left (b+c x^2\right )}+\frac {3 c (b B-2 A c)}{2 b^5 x^2}+\frac {c^2 (b B-A c)}{4 b^4 \left (b+c x^2\right )^2}-\frac {b B-3 A c}{4 b^4 x^4}-\frac {A}{6 b^3 x^6} \]

[Out]

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

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Rubi [A]  time = 0.17, antiderivative size = 148, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {1584, 446, 77} \[ \frac {c^2 (3 b B-4 A c)}{2 b^5 \left (b+c x^2\right )}+\frac {c^2 (b B-A c)}{4 b^4 \left (b+c x^2\right )^2}-\frac {c^2 (3 b B-5 A c) \log \left (b+c x^2\right )}{b^6}+\frac {2 c^2 \log (x) (3 b B-5 A c)}{b^6}+\frac {3 c (b B-2 A c)}{2 b^5 x^2}-\frac {b B-3 A c}{4 b^4 x^4}-\frac {A}{6 b^3 x^6} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-A/(6*b^3*x^6) - (b*B - 3*A*c)/(4*b^4*x^4) + (3*c*(b*B - 2*A*c))/(2*b^5*x^2) + (c^2*(b*B - A*c))/(4*b^4*(b + c
*x^2)^2) + (c^2*(3*b*B - 4*A*c))/(2*b^5*(b + c*x^2)) + (2*c^2*(3*b*B - 5*A*c)*Log[x])/b^6 - (c^2*(3*b*B - 5*A*
c)*Log[b + c*x^2])/b^6

Rule 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rule 446

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

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 {A+B x^2}{x \left (b x^2+c x^4\right )^3} \, dx &=\int \frac {A+B x^2}{x^7 \left (b+c x^2\right )^3} \, dx\\ &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {A+B x}{x^4 (b+c x)^3} \, dx,x,x^2\right )\\ &=\frac {1}{2} \operatorname {Subst}\left (\int \left (\frac {A}{b^3 x^4}+\frac {b B-3 A c}{b^4 x^3}-\frac {3 c (b B-2 A c)}{b^5 x^2}+\frac {2 c^2 (3 b B-5 A c)}{b^6 x}-\frac {c^3 (b B-A c)}{b^4 (b+c x)^3}-\frac {c^3 (3 b B-4 A c)}{b^5 (b+c x)^2}-\frac {2 c^3 (3 b B-5 A c)}{b^6 (b+c x)}\right ) \, dx,x,x^2\right )\\ &=-\frac {A}{6 b^3 x^6}-\frac {b B-3 A c}{4 b^4 x^4}+\frac {3 c (b B-2 A c)}{2 b^5 x^2}+\frac {c^2 (b B-A c)}{4 b^4 \left (b+c x^2\right )^2}+\frac {c^2 (3 b B-4 A c)}{2 b^5 \left (b+c x^2\right )}+\frac {2 c^2 (3 b B-5 A c) \log (x)}{b^6}-\frac {c^2 (3 b B-5 A c) \log \left (b+c x^2\right )}{b^6}\\ \end {align*}

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Mathematica [A]  time = 0.14, size = 135, normalized size = 0.91 \[ \frac {-\frac {2 A b^3}{x^6}+\frac {3 b^2 c^2 (b B-A c)}{\left (b+c x^2\right )^2}-\frac {3 b^2 (b B-3 A c)}{x^4}+\frac {6 b c^2 (3 b B-4 A c)}{b+c x^2}+12 c^2 (5 A c-3 b B) \log \left (b+c x^2\right )+24 c^2 \log (x) (3 b B-5 A c)+\frac {18 b c (b B-2 A c)}{x^2}}{12 b^6} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-2*A*b^3)/x^6 - (3*b^2*(b*B - 3*A*c))/x^4 + (18*b*c*(b*B - 2*A*c))/x^2 + (3*b^2*c^2*(b*B - A*c))/(b + c*x^2)
^2 + (6*b*c^2*(3*b*B - 4*A*c))/(b + c*x^2) + 24*c^2*(3*b*B - 5*A*c)*Log[x] + 12*c^2*(-3*b*B + 5*A*c)*Log[b + c
*x^2])/(12*b^6)

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fricas [A]  time = 0.71, size = 267, normalized size = 1.80 \[ \frac {12 \, {\left (3 \, B b^{2} c^{3} - 5 \, A b c^{4}\right )} x^{8} + 18 \, {\left (3 \, B b^{3} c^{2} - 5 \, A b^{2} c^{3}\right )} x^{6} - 2 \, A b^{5} + 4 \, {\left (3 \, B b^{4} c - 5 \, A b^{3} c^{2}\right )} x^{4} - {\left (3 \, B b^{5} - 5 \, A b^{4} c\right )} x^{2} - 12 \, {\left ({\left (3 \, B b c^{4} - 5 \, A c^{5}\right )} x^{10} + 2 \, {\left (3 \, B b^{2} c^{3} - 5 \, A b c^{4}\right )} x^{8} + {\left (3 \, B b^{3} c^{2} - 5 \, A b^{2} c^{3}\right )} x^{6}\right )} \log \left (c x^{2} + b\right ) + 24 \, {\left ({\left (3 \, B b c^{4} - 5 \, A c^{5}\right )} x^{10} + 2 \, {\left (3 \, B b^{2} c^{3} - 5 \, A b c^{4}\right )} x^{8} + {\left (3 \, B b^{3} c^{2} - 5 \, A b^{2} c^{3}\right )} x^{6}\right )} \log \relax (x)}{12 \, {\left (b^{6} c^{2} x^{10} + 2 \, b^{7} c x^{8} + b^{8} x^{6}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(12*(3*B*b^2*c^3 - 5*A*b*c^4)*x^8 + 18*(3*B*b^3*c^2 - 5*A*b^2*c^3)*x^6 - 2*A*b^5 + 4*(3*B*b^4*c - 5*A*b^3
*c^2)*x^4 - (3*B*b^5 - 5*A*b^4*c)*x^2 - 12*((3*B*b*c^4 - 5*A*c^5)*x^10 + 2*(3*B*b^2*c^3 - 5*A*b*c^4)*x^8 + (3*
B*b^3*c^2 - 5*A*b^2*c^3)*x^6)*log(c*x^2 + b) + 24*((3*B*b*c^4 - 5*A*c^5)*x^10 + 2*(3*B*b^2*c^3 - 5*A*b*c^4)*x^
8 + (3*B*b^3*c^2 - 5*A*b^2*c^3)*x^6)*log(x))/(b^6*c^2*x^10 + 2*b^7*c*x^8 + b^8*x^6)

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giac [A]  time = 0.17, size = 201, normalized size = 1.36 \[ \frac {{\left (3 \, B b c^{2} - 5 \, A c^{3}\right )} \log \left (x^{2}\right )}{b^{6}} - \frac {{\left (3 \, B b c^{3} - 5 \, A c^{4}\right )} \log \left ({\left | c x^{2} + b \right |}\right )}{b^{6} c} + \frac {18 \, B b c^{4} x^{4} - 30 \, A c^{5} x^{4} + 42 \, B b^{2} c^{3} x^{2} - 68 \, A b c^{4} x^{2} + 25 \, B b^{3} c^{2} - 39 \, A b^{2} c^{3}}{4 \, {\left (c x^{2} + b\right )}^{2} b^{6}} - \frac {66 \, B b c^{2} x^{6} - 110 \, A c^{3} x^{6} - 18 \, B b^{2} c x^{4} + 36 \, A b c^{2} x^{4} + 3 \, B b^{3} x^{2} - 9 \, A b^{2} c x^{2} + 2 \, A b^{3}}{12 \, b^{6} x^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(3*B*b*c^2 - 5*A*c^3)*log(x^2)/b^6 - (3*B*b*c^3 - 5*A*c^4)*log(abs(c*x^2 + b))/(b^6*c) + 1/4*(18*B*b*c^4*x^4 -
30*A*c^5*x^4 + 42*B*b^2*c^3*x^2 - 68*A*b*c^4*x^2 + 25*B*b^3*c^2 - 39*A*b^2*c^3)/((c*x^2 + b)^2*b^6) - 1/12*(6
6*B*b*c^2*x^6 - 110*A*c^3*x^6 - 18*B*b^2*c*x^4 + 36*A*b*c^2*x^4 + 3*B*b^3*x^2 - 9*A*b^2*c*x^2 + 2*A*b^3)/(b^6*
x^6)

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maple [A]  time = 0.06, size = 180, normalized size = 1.22 \[ -\frac {A \,c^{3}}{4 \left (c \,x^{2}+b \right )^{2} b^{4}}+\frac {B \,c^{2}}{4 \left (c \,x^{2}+b \right )^{2} b^{3}}-\frac {2 A \,c^{3}}{\left (c \,x^{2}+b \right ) b^{5}}-\frac {10 A \,c^{3} \ln \relax (x )}{b^{6}}+\frac {5 A \,c^{3} \ln \left (c \,x^{2}+b \right )}{b^{6}}+\frac {3 B \,c^{2}}{2 \left (c \,x^{2}+b \right ) b^{4}}+\frac {6 B \,c^{2} \ln \relax (x )}{b^{5}}-\frac {3 B \,c^{2} \ln \left (c \,x^{2}+b \right )}{b^{5}}-\frac {3 A \,c^{2}}{b^{5} x^{2}}+\frac {3 B c}{2 b^{4} x^{2}}+\frac {3 A c}{4 b^{4} x^{4}}-\frac {B}{4 b^{3} x^{4}}-\frac {A}{6 b^{3} x^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5/b^6*c^3*ln(c*x^2+b)*A-3/b^5*c^2*ln(c*x^2+b)*B-2/b^5*c^3/(c*x^2+b)*A+3/2/b^4*c^2/(c*x^2+b)*B-1/4/b^4*c^3/(c*x
^2+b)^2*A+1/4/b^3*c^2/(c*x^2+b)^2*B-1/6*A/b^3/x^6+3/4/b^4/x^4*A*c-1/4/b^3/x^4*B-3*c^2/b^5/x^2*A+3/2*c/b^4/x^2*
B-10*c^3/b^6*ln(x)*A+6*c^2/b^5*ln(x)*B

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maxima [A]  time = 1.43, size = 170, normalized size = 1.15 \[ \frac {12 \, {\left (3 \, B b c^{3} - 5 \, A c^{4}\right )} x^{8} + 18 \, {\left (3 \, B b^{2} c^{2} - 5 \, A b c^{3}\right )} x^{6} - 2 \, A b^{4} + 4 \, {\left (3 \, B b^{3} c - 5 \, A b^{2} c^{2}\right )} x^{4} - {\left (3 \, B b^{4} - 5 \, A b^{3} c\right )} x^{2}}{12 \, {\left (b^{5} c^{2} x^{10} + 2 \, b^{6} c x^{8} + b^{7} x^{6}\right )}} - \frac {{\left (3 \, B b c^{2} - 5 \, A c^{3}\right )} \log \left (c x^{2} + b\right )}{b^{6}} + \frac {{\left (3 \, B b c^{2} - 5 \, A c^{3}\right )} \log \left (x^{2}\right )}{b^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(12*(3*B*b*c^3 - 5*A*c^4)*x^8 + 18*(3*B*b^2*c^2 - 5*A*b*c^3)*x^6 - 2*A*b^4 + 4*(3*B*b^3*c - 5*A*b^2*c^2)*
x^4 - (3*B*b^4 - 5*A*b^3*c)*x^2)/(b^5*c^2*x^10 + 2*b^6*c*x^8 + b^7*x^6) - (3*B*b*c^2 - 5*A*c^3)*log(c*x^2 + b)
/b^6 + (3*B*b*c^2 - 5*A*c^3)*log(x^2)/b^6

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mupad [B]  time = 0.18, size = 155, normalized size = 1.05 \[ \frac {\ln \left (c\,x^2+b\right )\,\left (5\,A\,c^3-3\,B\,b\,c^2\right )}{b^6}-\frac {\frac {A}{6\,b}-\frac {x^2\,\left (5\,A\,c-3\,B\,b\right )}{12\,b^2}+\frac {3\,c^2\,x^6\,\left (5\,A\,c-3\,B\,b\right )}{2\,b^4}+\frac {c^3\,x^8\,\left (5\,A\,c-3\,B\,b\right )}{b^5}+\frac {c\,x^4\,\left (5\,A\,c-3\,B\,b\right )}{3\,b^3}}{b^2\,x^6+2\,b\,c\,x^8+c^2\,x^{10}}-\frac {\ln \relax (x)\,\left (10\,A\,c^3-6\,B\,b\,c^2\right )}{b^6} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(log(b + c*x^2)*(5*A*c^3 - 3*B*b*c^2))/b^6 - (A/(6*b) - (x^2*(5*A*c - 3*B*b))/(12*b^2) + (3*c^2*x^6*(5*A*c - 3
*B*b))/(2*b^4) + (c^3*x^8*(5*A*c - 3*B*b))/b^5 + (c*x^4*(5*A*c - 3*B*b))/(3*b^3))/(b^2*x^6 + c^2*x^10 + 2*b*c*
x^8) - (log(x)*(10*A*c^3 - 6*B*b*c^2))/b^6

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sympy [A]  time = 1.33, size = 165, normalized size = 1.11 \[ \frac {- 2 A b^{4} + x^{8} \left (- 60 A c^{4} + 36 B b c^{3}\right ) + x^{6} \left (- 90 A b c^{3} + 54 B b^{2} c^{2}\right ) + x^{4} \left (- 20 A b^{2} c^{2} + 12 B b^{3} c\right ) + x^{2} \left (5 A b^{3} c - 3 B b^{4}\right )}{12 b^{7} x^{6} + 24 b^{6} c x^{8} + 12 b^{5} c^{2} x^{10}} + \frac {2 c^{2} \left (- 5 A c + 3 B b\right ) \log {\relax (x )}}{b^{6}} - \frac {c^{2} \left (- 5 A c + 3 B b\right ) \log {\left (\frac {b}{c} + x^{2} \right )}}{b^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-2*A*b**4 + x**8*(-60*A*c**4 + 36*B*b*c**3) + x**6*(-90*A*b*c**3 + 54*B*b**2*c**2) + x**4*(-20*A*b**2*c**2 +
12*B*b**3*c) + x**2*(5*A*b**3*c - 3*B*b**4))/(12*b**7*x**6 + 24*b**6*c*x**8 + 12*b**5*c**2*x**10) + 2*c**2*(-5
*A*c + 3*B*b)*log(x)/b**6 - c**2*(-5*A*c + 3*B*b)*log(b/c + x**2)/b**6

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