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

Optimal. Leaf size=121 \[ \frac {3 c (b B-2 A c) \log \left (b+c x^2\right )}{2 b^5}-\frac {3 c \log (x) (b B-2 A c)}{b^5}-\frac {c (2 b B-3 A c)}{2 b^4 \left (b+c x^2\right )}-\frac {b B-3 A c}{2 b^4 x^2}-\frac {c (b B-A c)}{4 b^3 \left (b+c x^2\right )^2}-\frac {A}{4 b^3 x^4} \]

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Rubi [A]  time = 0.13, antiderivative size = 121, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {1584, 446, 77} \begin {gather*} -\frac {c (2 b B-3 A c)}{2 b^4 \left (b+c x^2\right )}-\frac {b B-3 A c}{2 b^4 x^2}-\frac {c (b B-A c)}{4 b^3 \left (b+c x^2\right )^2}+\frac {3 c (b B-2 A c) \log \left (b+c x^2\right )}{2 b^5}-\frac {3 c \log (x) (b B-2 A c)}{b^5}-\frac {A}{4 b^3 x^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [A]  time = 0.09, size = 108, normalized size = 0.89 \begin {gather*} \frac {\frac {b^2 c (A c-b B)}{\left (b+c x^2\right )^2}-\frac {A b^2}{x^4}+\frac {2 b c (3 A c-2 b B)}{b+c x^2}-\frac {2 b (b B-3 A c)}{x^2}+6 c (b B-2 A c) \log \left (b+c x^2\right )+12 c \log (x) (2 A c-b B)}{4 b^5} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [B]  time = 0.41, size = 229, normalized size = 1.89 \begin {gather*} -\frac {6 \, {\left (B b^{2} c^{2} - 2 \, A b c^{3}\right )} x^{6} + A b^{4} + 9 \, {\left (B b^{3} c - 2 \, A b^{2} c^{2}\right )} x^{4} + 2 \, {\left (B b^{4} - 2 \, A b^{3} c\right )} x^{2} - 6 \, {\left ({\left (B b c^{3} - 2 \, A c^{4}\right )} x^{8} + 2 \, {\left (B b^{2} c^{2} - 2 \, A b c^{3}\right )} x^{6} + {\left (B b^{3} c - 2 \, A b^{2} c^{2}\right )} x^{4}\right )} \log \left (c x^{2} + b\right ) + 12 \, {\left ({\left (B b c^{3} - 2 \, A c^{4}\right )} x^{8} + 2 \, {\left (B b^{2} c^{2} - 2 \, A b c^{3}\right )} x^{6} + {\left (B b^{3} c - 2 \, A b^{2} c^{2}\right )} x^{4}\right )} \log \relax (x)}{4 \, {\left (b^{5} c^{2} x^{8} + 2 \, b^{6} c x^{6} + b^{7} x^{4}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(6*(B*b^2*c^2 - 2*A*b*c^3)*x^6 + A*b^4 + 9*(B*b^3*c - 2*A*b^2*c^2)*x^4 + 2*(B*b^4 - 2*A*b^3*c)*x^2 - 6*((
B*b*c^3 - 2*A*c^4)*x^8 + 2*(B*b^2*c^2 - 2*A*b*c^3)*x^6 + (B*b^3*c - 2*A*b^2*c^2)*x^4)*log(c*x^2 + b) + 12*((B*
b*c^3 - 2*A*c^4)*x^8 + 2*(B*b^2*c^2 - 2*A*b*c^3)*x^6 + (B*b^3*c - 2*A*b^2*c^2)*x^4)*log(x))/(b^5*c^2*x^8 + 2*b
^6*c*x^6 + b^7*x^4)

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giac [A]  time = 0.17, size = 132, normalized size = 1.09 \begin {gather*} -\frac {3 \, {\left (B b c - 2 \, A c^{2}\right )} \log \left ({\left | x \right |}\right )}{b^{5}} + \frac {3 \, {\left (B b c^{2} - 2 \, A c^{3}\right )} \log \left ({\left | c x^{2} + b \right |}\right )}{2 \, b^{5} c} - \frac {6 \, B b c^{2} x^{6} - 12 \, A c^{3} x^{6} + 9 \, B b^{2} c x^{4} - 18 \, A b c^{2} x^{4} + 2 \, B b^{3} x^{2} - 4 \, A b^{2} c x^{2} + A b^{3}}{4 \, {\left (c x^{4} + b x^{2}\right )}^{2} b^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3*(B*b*c - 2*A*c^2)*log(abs(x))/b^5 + 3/2*(B*b*c^2 - 2*A*c^3)*log(abs(c*x^2 + b))/(b^5*c) - 1/4*(6*B*b*c^2*x^
6 - 12*A*c^3*x^6 + 9*B*b^2*c*x^4 - 18*A*b*c^2*x^4 + 2*B*b^3*x^2 - 4*A*b^2*c*x^2 + A*b^3)/((c*x^4 + b*x^2)^2*b^
4)

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maple [A]  time = 0.06, size = 150, normalized size = 1.24 \begin {gather*} \frac {A \,c^{2}}{4 \left (c \,x^{2}+b \right )^{2} b^{3}}-\frac {B c}{4 \left (c \,x^{2}+b \right )^{2} b^{2}}+\frac {3 A \,c^{2}}{2 \left (c \,x^{2}+b \right ) b^{4}}+\frac {6 A \,c^{2} \ln \relax (x )}{b^{5}}-\frac {3 A \,c^{2} \ln \left (c \,x^{2}+b \right )}{b^{5}}-\frac {B c}{\left (c \,x^{2}+b \right ) b^{3}}-\frac {3 B c \ln \relax (x )}{b^{4}}+\frac {3 B c \ln \left (c \,x^{2}+b \right )}{2 b^{4}}+\frac {3 A c}{2 b^{4} x^{2}}-\frac {B}{2 b^{3} x^{2}}-\frac {A}{4 b^{3} x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.43, size = 137, normalized size = 1.13 \begin {gather*} -\frac {6 \, {\left (B b c^{2} - 2 \, A c^{3}\right )} x^{6} + 9 \, {\left (B b^{2} c - 2 \, A b c^{2}\right )} x^{4} + A b^{3} + 2 \, {\left (B b^{3} - 2 \, A b^{2} c\right )} x^{2}}{4 \, {\left (b^{4} c^{2} x^{8} + 2 \, b^{5} c x^{6} + b^{6} x^{4}\right )}} + \frac {3 \, {\left (B b c - 2 \, A c^{2}\right )} \log \left (c x^{2} + b\right )}{2 \, b^{5}} - \frac {3 \, {\left (B b c - 2 \, A c^{2}\right )} \log \left (x^{2}\right )}{2 \, b^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(6*(B*b*c^2 - 2*A*c^3)*x^6 + 9*(B*b^2*c - 2*A*b*c^2)*x^4 + A*b^3 + 2*(B*b^3 - 2*A*b^2*c)*x^2)/(b^4*c^2*x^
8 + 2*b^5*c*x^6 + b^6*x^4) + 3/2*(B*b*c - 2*A*c^2)*log(c*x^2 + b)/b^5 - 3/2*(B*b*c - 2*A*c^2)*log(x^2)/b^5

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mupad [B]  time = 0.17, size = 131, normalized size = 1.08 \begin {gather*} \frac {\frac {x^2\,\left (2\,A\,c-B\,b\right )}{2\,b^2}-\frac {A}{4\,b}+\frac {3\,c^2\,x^6\,\left (2\,A\,c-B\,b\right )}{2\,b^4}+\frac {9\,c\,x^4\,\left (2\,A\,c-B\,b\right )}{4\,b^3}}{b^2\,x^4+2\,b\,c\,x^6+c^2\,x^8}-\frac {\ln \left (c\,x^2+b\right )\,\left (6\,A\,c^2-3\,B\,b\,c\right )}{2\,b^5}+\frac {\ln \relax (x)\,\left (6\,A\,c^2-3\,B\,b\,c\right )}{b^5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((x^2*(2*A*c - B*b))/(2*b^2) - A/(4*b) + (3*c^2*x^6*(2*A*c - B*b))/(2*b^4) + (9*c*x^4*(2*A*c - B*b))/(4*b^3))/
(b^2*x^4 + c^2*x^8 + 2*b*c*x^6) - (log(b + c*x^2)*(6*A*c^2 - 3*B*b*c))/(2*b^5) + (log(x)*(6*A*c^2 - 3*B*b*c))/
b^5

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sympy [A]  time = 1.25, size = 136, normalized size = 1.12 \begin {gather*} \frac {- A b^{3} + x^{6} \left (12 A c^{3} - 6 B b c^{2}\right ) + x^{4} \left (18 A b c^{2} - 9 B b^{2} c\right ) + x^{2} \left (4 A b^{2} c - 2 B b^{3}\right )}{4 b^{6} x^{4} + 8 b^{5} c x^{6} + 4 b^{4} c^{2} x^{8}} - \frac {3 c \left (- 2 A c + B b\right ) \log {\relax (x )}}{b^{5}} + \frac {3 c \left (- 2 A c + B b\right ) \log {\left (\frac {b}{c} + x^{2} \right )}}{2 b^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-A*b**3 + x**6*(12*A*c**3 - 6*B*b*c**2) + x**4*(18*A*b*c**2 - 9*B*b**2*c) + x**2*(4*A*b**2*c - 2*B*b**3))/(4*
b**6*x**4 + 8*b**5*c*x**6 + 4*b**4*c**2*x**8) - 3*c*(-2*A*c + B*b)*log(x)/b**5 + 3*c*(-2*A*c + B*b)*log(b/c +
x**2)/(2*b**5)

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