3.74 \(\int \frac{x^{13} (A+B x^2)}{(b x^2+c x^4)^3} \, dx\)

Optimal. Leaf size=111 \[ \frac{b^2 (4 b B-3 A c)}{2 c^5 \left (b+c x^2\right )}-\frac{b^3 (b B-A c)}{4 c^5 \left (b+c x^2\right )^2}-\frac{x^2 (3 b B-A c)}{2 c^4}+\frac{3 b (2 b B-A c) \log \left (b+c x^2\right )}{2 c^5}+\frac{B x^4}{4 c^3} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.136048, antiderivative size = 111, normalized size of antiderivative = 1., 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} \[ \frac{b^2 (4 b B-3 A c)}{2 c^5 \left (b+c x^2\right )}-\frac{b^3 (b B-A c)}{4 c^5 \left (b+c x^2\right )^2}-\frac{x^2 (3 b B-A c)}{2 c^4}+\frac{3 b (2 b B-A c) \log \left (b+c x^2\right )}{2 c^5}+\frac{B x^4}{4 c^3} \]

Antiderivative was successfully verified.

[In]

Int[(x^13*(A + B*x^2))/(b*x^2 + c*x^4)^3,x]

[Out]

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

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]

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 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]))))

Rubi steps

\begin{align*} \int \frac{x^{13} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^3} \, dx &=\int \frac{x^7 \left (A+B x^2\right )}{\left (b+c x^2\right )^3} \, dx\\ &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{x^3 (A+B x)}{(b+c x)^3} \, dx,x,x^2\right )\\ &=\frac{1}{2} \operatorname{Subst}\left (\int \left (\frac{-3 b B+A c}{c^4}+\frac{B x}{c^3}+\frac{b^3 (b B-A c)}{c^4 (b+c x)^3}-\frac{b^2 (4 b B-3 A c)}{c^4 (b+c x)^2}+\frac{3 b (2 b B-A c)}{c^4 (b+c x)}\right ) \, dx,x,x^2\right )\\ &=-\frac{(3 b B-A c) x^2}{2 c^4}+\frac{B x^4}{4 c^3}-\frac{b^3 (b B-A c)}{4 c^5 \left (b+c x^2\right )^2}+\frac{b^2 (4 b B-3 A c)}{2 c^5 \left (b+c x^2\right )}+\frac{3 b (2 b B-A c) \log \left (b+c x^2\right )}{2 c^5}\\ \end{align*}

Mathematica [A]  time = 0.0628342, size = 94, normalized size = 0.85 \[ \frac{\frac{2 b^2 (4 b B-3 A c)}{b+c x^2}+\frac{b^3 (A c-b B)}{\left (b+c x^2\right )^2}+2 c x^2 (A c-3 b B)+6 b (2 b B-A c) \log \left (b+c x^2\right )+B c^2 x^4}{4 c^5} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^13*(A + B*x^2))/(b*x^2 + c*x^4)^3,x]

[Out]

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

________________________________________________________________________________________

Maple [A]  time = 0.011, size = 134, normalized size = 1.2 \begin{align*}{\frac{B{x}^{4}}{4\,{c}^{3}}}-{\frac{3\,B{x}^{2}b}{2\,{c}^{4}}}+{\frac{A{x}^{2}}{2\,{c}^{3}}}-{\frac{3\,b\ln \left ( c{x}^{2}+b \right ) A}{2\,{c}^{4}}}+3\,{\frac{{b}^{2}\ln \left ( c{x}^{2}+b \right ) B}{{c}^{5}}}-{\frac{3\,A{b}^{2}}{2\,{c}^{4} \left ( c{x}^{2}+b \right ) }}+2\,{\frac{B{b}^{3}}{{c}^{5} \left ( c{x}^{2}+b \right ) }}+{\frac{{b}^{3}A}{4\,{c}^{4} \left ( c{x}^{2}+b \right ) ^{2}}}-{\frac{B{b}^{4}}{4\,{c}^{5} \left ( c{x}^{2}+b \right ) ^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^13*(B*x^2+A)/(c*x^4+b*x^2)^3,x)

[Out]

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

________________________________________________________________________________________

Maxima [A]  time = 1.153, size = 157, normalized size = 1.41 \begin{align*} \frac{7 \, B b^{4} - 5 \, A b^{3} c + 2 \,{\left (4 \, B b^{3} c - 3 \, A b^{2} c^{2}\right )} x^{2}}{4 \,{\left (c^{7} x^{4} + 2 \, b c^{6} x^{2} + b^{2} c^{5}\right )}} + \frac{B c x^{4} - 2 \,{\left (3 \, B b - A c\right )} x^{2}}{4 \, c^{4}} + \frac{3 \,{\left (2 \, B b^{2} - A b c\right )} \log \left (c x^{2} + b\right )}{2 \, c^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^13*(B*x^2+A)/(c*x^4+b*x^2)^3,x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

Fricas [A]  time = 1.06674, size = 360, normalized size = 3.24 \begin{align*} \frac{B c^{4} x^{8} - 2 \,{\left (2 \, B b c^{3} - A c^{4}\right )} x^{6} + 7 \, B b^{4} - 5 \, A b^{3} c -{\left (11 \, B b^{2} c^{2} - 4 \, A b c^{3}\right )} x^{4} + 2 \,{\left (B b^{3} c - 2 \, A b^{2} c^{2}\right )} x^{2} + 6 \,{\left (2 \, B b^{4} - A b^{3} c +{\left (2 \, B b^{2} c^{2} - A b c^{3}\right )} x^{4} + 2 \,{\left (2 \, B b^{3} c - A b^{2} c^{2}\right )} x^{2}\right )} \log \left (c x^{2} + b\right )}{4 \,{\left (c^{7} x^{4} + 2 \, b c^{6} x^{2} + b^{2} c^{5}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^13*(B*x^2+A)/(c*x^4+b*x^2)^3,x, algorithm="fricas")

[Out]

1/4*(B*c^4*x^8 - 2*(2*B*b*c^3 - A*c^4)*x^6 + 7*B*b^4 - 5*A*b^3*c - (11*B*b^2*c^2 - 4*A*b*c^3)*x^4 + 2*(B*b^3*c
- 2*A*b^2*c^2)*x^2 + 6*(2*B*b^4 - A*b^3*c + (2*B*b^2*c^2 - A*b*c^3)*x^4 + 2*(2*B*b^3*c - A*b^2*c^2)*x^2)*log(
c*x^2 + b))/(c^7*x^4 + 2*b*c^6*x^2 + b^2*c^5)

________________________________________________________________________________________

Sympy [A]  time = 1.64791, size = 116, normalized size = 1.05 \begin{align*} \frac{B x^{4}}{4 c^{3}} + \frac{3 b \left (- A c + 2 B b\right ) \log{\left (b + c x^{2} \right )}}{2 c^{5}} + \frac{- 5 A b^{3} c + 7 B b^{4} + x^{2} \left (- 6 A b^{2} c^{2} + 8 B b^{3} c\right )}{4 b^{2} c^{5} + 8 b c^{6} x^{2} + 4 c^{7} x^{4}} - \frac{x^{2} \left (- A c + 3 B b\right )}{2 c^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**13*(B*x**2+A)/(c*x**4+b*x**2)**3,x)

[Out]

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

________________________________________________________________________________________

Giac [A]  time = 1.13653, size = 178, normalized size = 1.6 \begin{align*} \frac{3 \,{\left (2 \, B b^{2} - A b c\right )} \log \left ({\left | c x^{2} + b \right |}\right )}{2 \, c^{5}} + \frac{B c^{3} x^{4} - 6 \, B b c^{2} x^{2} + 2 \, A c^{3} x^{2}}{4 \, c^{6}} - \frac{18 \, B b^{2} c^{2} x^{4} - 9 \, A b c^{3} x^{4} + 28 \, B b^{3} c x^{2} - 12 \, A b^{2} c^{2} x^{2} + 11 \, B b^{4} - 4 \, A b^{3} c}{4 \,{\left (c x^{2} + b\right )}^{2} c^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^13*(B*x^2+A)/(c*x^4+b*x^2)^3,x, algorithm="giac")

[Out]

3/2*(2*B*b^2 - A*b*c)*log(abs(c*x^2 + b))/c^5 + 1/4*(B*c^3*x^4 - 6*B*b*c^2*x^2 + 2*A*c^3*x^2)/c^6 - 1/4*(18*B*
b^2*c^2*x^4 - 9*A*b*c^3*x^4 + 28*B*b^3*c*x^2 - 12*A*b^2*c^2*x^2 + 11*B*b^4 - 4*A*b^3*c)/((c*x^2 + b)^2*c^5)