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

Optimal. Leaf size=41 \[ \frac{b B-A c}{2 c^2 \left (b+c x^2\right )}+\frac{B \log \left (b+c x^2\right )}{2 c^2} \]

[Out]

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

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Rubi [A]  time = 0.0467391, antiderivative size = 41, 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, 444, 43} \[ \frac{b B-A c}{2 c^2 \left (b+c x^2\right )}+\frac{B \log \left (b+c x^2\right )}{2 c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 444

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[(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] && EqQ[m
- n + 1, 0]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

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

Mathematica [A]  time = 0.0122716, size = 41, normalized size = 1. \[ \frac{b B-A c}{2 c^2 \left (b+c x^2\right )}+\frac{B \log \left (b+c x^2\right )}{2 c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.007, size = 47, normalized size = 1.2 \begin{align*}{\frac{B\ln \left ( c{x}^{2}+b \right ) }{2\,{c}^{2}}}-{\frac{A}{2\,c \left ( c{x}^{2}+b \right ) }}+{\frac{Bb}{2\,{c}^{2} \left ( c{x}^{2}+b \right ) }} \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)^2,x)

[Out]

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

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Maxima [A]  time = 2.76769, size = 54, normalized size = 1.32 \begin{align*} \frac{B b - A c}{2 \,{\left (c^{3} x^{2} + b c^{2}\right )}} + \frac{B \log \left (c x^{2} + b\right )}{2 \, c^{2}} \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)^2,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 0.789347, size = 92, normalized size = 2.24 \begin{align*} \frac{B b - A c +{\left (B c x^{2} + B b\right )} \log \left (c x^{2} + b\right )}{2 \,{\left (c^{3} x^{2} + b c^{2}\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)^2,x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 0.505632, size = 36, normalized size = 0.88 \begin{align*} \frac{B \log{\left (b + c x^{2} \right )}}{2 c^{2}} + \frac{- A c + B b}{2 b c^{2} + 2 c^{3} x^{2}} \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)**2,x)

[Out]

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

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Giac [A]  time = 1.26088, size = 50, normalized size = 1.22 \begin{align*} \frac{B \log \left ({\left | c x^{2} + b \right |}\right )}{2 \, c^{2}} - \frac{B x^{2} + A}{2 \,{\left (c x^{2} + b\right )} 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)^2,x, algorithm="giac")

[Out]

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