3.63 \(\int \frac{x^7 (A+B x^2)}{(b x^2+c x^4)^2} \, dx\)

Optimal. Leaf size=61 \[ -\frac{b (b B-A c)}{2 c^3 \left (b+c x^2\right )}-\frac{(2 b B-A c) \log \left (b+c x^2\right )}{2 c^3}+\frac{B x^2}{2 c^2} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0697961, antiderivative size = 61, 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} \[ -\frac{b (b B-A c)}{2 c^3 \left (b+c x^2\right )}-\frac{(2 b B-A c) \log \left (b+c x^2\right )}{2 c^3}+\frac{B x^2}{2 c^2} \]

Antiderivative was successfully verified.

[In]

Int[(x^7*(A + B*x^2))/(b*x^2 + c*x^4)^2,x]

[Out]

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

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

Mathematica [A]  time = 0.0360475, size = 50, normalized size = 0.82 \[ \frac{\frac{b (A c-b B)}{b+c x^2}+(A c-2 b B) \log \left (b+c x^2\right )+B c x^2}{2 c^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^7*(A + B*x^2))/(b*x^2 + c*x^4)^2,x]

[Out]

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

________________________________________________________________________________________

Maple [A]  time = 0.008, size = 74, normalized size = 1.2 \begin{align*}{\frac{B{x}^{2}}{2\,{c}^{2}}}+{\frac{\ln \left ( c{x}^{2}+b \right ) A}{2\,{c}^{2}}}-{\frac{\ln \left ( c{x}^{2}+b \right ) Bb}{{c}^{3}}}+{\frac{Ab}{2\,{c}^{2} \left ( c{x}^{2}+b \right ) }}-{\frac{B{b}^{2}}{2\,{c}^{3} \left ( c{x}^{2}+b \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^7*(B*x^2+A)/(c*x^4+b*x^2)^2,x)

[Out]

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

________________________________________________________________________________________

Maxima [A]  time = 1.26542, size = 81, normalized size = 1.33 \begin{align*} \frac{B x^{2}}{2 \, c^{2}} - \frac{B b^{2} - A b c}{2 \,{\left (c^{4} x^{2} + b c^{3}\right )}} - \frac{{\left (2 \, B b - A c\right )} \log \left (c x^{2} + b\right )}{2 \, c^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^7*(B*x^2+A)/(c*x^4+b*x^2)^2,x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

Fricas [A]  time = 0.689542, size = 165, normalized size = 2.7 \begin{align*} \frac{B c^{2} x^{4} + B b c x^{2} - B b^{2} + A b c -{\left (2 \, B b^{2} - A b c +{\left (2 \, B b c - A c^{2}\right )} x^{2}\right )} \log \left (c x^{2} + b\right )}{2 \,{\left (c^{4} x^{2} + b c^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^7*(B*x^2+A)/(c*x^4+b*x^2)^2,x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

Sympy [A]  time = 0.715123, size = 56, normalized size = 0.92 \begin{align*} \frac{B x^{2}}{2 c^{2}} - \frac{- A b c + B b^{2}}{2 b c^{3} + 2 c^{4} x^{2}} - \frac{\left (- A c + 2 B b\right ) \log{\left (b + c x^{2} \right )}}{2 c^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**7*(B*x**2+A)/(c*x**4+b*x**2)**2,x)

[Out]

B*x**2/(2*c**2) - (-A*b*c + B*b**2)/(2*b*c**3 + 2*c**4*x**2) - (-A*c + 2*B*b)*log(b + c*x**2)/(2*c**3)

________________________________________________________________________________________

Giac [A]  time = 1.29789, size = 95, normalized size = 1.56 \begin{align*} \frac{B x^{2}}{2 \, c^{2}} - \frac{{\left (2 \, B b - A c\right )} \log \left ({\left | c x^{2} + b \right |}\right )}{2 \, c^{3}} + \frac{2 \, B b c x^{2} - A c^{2} x^{2} + B b^{2}}{2 \,{\left (c x^{2} + b\right )} c^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^7*(B*x^2+A)/(c*x^4+b*x^2)^2,x, algorithm="giac")

[Out]

1/2*B*x^2/c^2 - 1/2*(2*B*b - A*c)*log(abs(c*x^2 + b))/c^3 + 1/2*(2*B*b*c*x^2 - A*c^2*x^2 + B*b^2)/((c*x^2 + b)
*c^3)