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

Optimal. Leaf size=35 \[ \frac{B x^2}{2 c}-\frac{(b B-A c) \log \left (b+c x^2\right )}{2 c^2} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0431874, antiderivative size = 35, 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} \[ \frac{B x^2}{2 c}-\frac{(b B-A c) \log \left (b+c x^2\right )}{2 c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.0110185, size = 31, normalized size = 0.89 \[ \frac{(A c-b B) \log \left (b+c x^2\right )+B c x^2}{2 c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [A]  time = 0.003, size = 40, normalized size = 1.1 \begin{align*}{\frac{B{x}^{2}}{2\,c}}+{\frac{\ln \left ( c{x}^{2}+b \right ) A}{2\,c}}-{\frac{\ln \left ( c{x}^{2}+b \right ) Bb}{2\,{c}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [A]  time = 1.15695, size = 42, normalized size = 1.2 \begin{align*} \frac{B x^{2}}{2 \, c} - \frac{{\left (B b - A c\right )} \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^3*(B*x^2+A)/(c*x^4+b*x^2),x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

Fricas [A]  time = 0.607129, size = 65, normalized size = 1.86 \begin{align*} \frac{B c x^{2} -{\left (B b - A c\right )} \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^3*(B*x^2+A)/(c*x^4+b*x^2),x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

Sympy [A]  time = 0.388506, size = 27, normalized size = 0.77 \begin{align*} \frac{B x^{2}}{2 c} - \frac{\left (- A c + B b\right ) \log{\left (b + c x^{2} \right )}}{2 c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Giac [A]  time = 1.28704, size = 43, normalized size = 1.23 \begin{align*} \frac{B x^{2}}{2 \, c} - \frac{{\left (B b - A c\right )} \log \left ({\left | c x^{2} + b \right |}\right )}{2 \, c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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