3.82 \(\int \frac{x^5 (A+B x^2)}{(b x^2+c x^4)^3} \, dx\)

Optimal. Leaf size=68 \[ \frac{A}{2 b^2 \left (b+c x^2\right )}-\frac{A \log \left (b+c x^2\right )}{2 b^3}+\frac{A \log (x)}{b^3}-\frac{b B-A c}{4 b c \left (b+c x^2\right )^2} \]

[Out]

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

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Rubi [A]  time = 0.0728484, antiderivative size = 68, 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} \[ \frac{A}{2 b^2 \left (b+c x^2\right )}-\frac{A \log \left (b+c x^2\right )}{2 b^3}+\frac{A \log (x)}{b^3}-\frac{b B-A c}{4 b c \left (b+c x^2\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.046391, size = 59, normalized size = 0.87 \[ \frac{\frac{b \left (3 A b c+2 A c^2 x^2+b^2 (-B)\right )}{c \left (b+c x^2\right )^2}-2 A \log \left (b+c x^2\right )+4 A \log (x)}{4 b^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.011, size = 68, normalized size = 1. \begin{align*}{\frac{A\ln \left ( x \right ) }{{b}^{3}}}-{\frac{A\ln \left ( c{x}^{2}+b \right ) }{2\,{b}^{3}}}+{\frac{A}{2\,{b}^{2} \left ( c{x}^{2}+b \right ) }}+{\frac{A}{4\,b \left ( c{x}^{2}+b \right ) ^{2}}}-{\frac{B}{4\,c \left ( c{x}^{2}+b \right ) ^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5*(B*x^2+A)/(c*x^4+b*x^2)^3,x)

[Out]

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

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Maxima [A]  time = 1.18785, size = 104, normalized size = 1.53 \begin{align*} \frac{2 \, A c^{2} x^{2} - B b^{2} + 3 \, A b c}{4 \,{\left (b^{2} c^{3} x^{4} + 2 \, b^{3} c^{2} x^{2} + b^{4} c\right )}} - \frac{A \log \left (c x^{2} + b\right )}{2 \, b^{3}} + \frac{A \log \left (x^{2}\right )}{2 \, b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(B*x^2+A)/(c*x^4+b*x^2)^3,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.00913, size = 250, normalized size = 3.68 \begin{align*} \frac{2 \, A b c^{2} x^{2} - B b^{3} + 3 \, A b^{2} c - 2 \,{\left (A c^{3} x^{4} + 2 \, A b c^{2} x^{2} + A b^{2} c\right )} \log \left (c x^{2} + b\right ) + 4 \,{\left (A c^{3} x^{4} + 2 \, A b c^{2} x^{2} + A b^{2} c\right )} \log \left (x\right )}{4 \,{\left (b^{3} c^{3} x^{4} + 2 \, b^{4} c^{2} x^{2} + b^{5} c\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(B*x^2+A)/(c*x^4+b*x^2)^3,x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 0.747113, size = 75, normalized size = 1.1 \begin{align*} \frac{A \log{\left (x \right )}}{b^{3}} - \frac{A \log{\left (\frac{b}{c} + x^{2} \right )}}{2 b^{3}} + \frac{3 A b c + 2 A c^{2} x^{2} - B b^{2}}{4 b^{4} c + 8 b^{3} c^{2} x^{2} + 4 b^{2} c^{3} x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**5*(B*x**2+A)/(c*x**4+b*x**2)**3,x)

[Out]

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

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Giac [A]  time = 1.15348, size = 103, normalized size = 1.51 \begin{align*} \frac{A \log \left (x^{2}\right )}{2 \, b^{3}} - \frac{A \log \left ({\left | c x^{2} + b \right |}\right )}{2 \, b^{3}} + \frac{3 \, A c^{3} x^{4} + 8 \, A b c^{2} x^{2} - B b^{3} + 6 \, A b^{2} c}{4 \,{\left (c x^{2} + b\right )}^{2} b^{3} c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(B*x^2+A)/(c*x^4+b*x^2)^3,x, algorithm="giac")

[Out]

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