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

Optimal. Leaf size=42 \[ \frac{B \left (b+c x^2\right )^4}{8 c^2}-\frac{\left (b+c x^2\right )^3 (b B-A c)}{6 c^2} \]

[Out]

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

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Rubi [A]  time = 0.0696529, antiderivative size = 42, 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} \[ \frac{B \left (b+c x^2\right )^4}{8 c^2}-\frac{\left (b+c x^2\right )^3 (b B-A c)}{6 c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((b*B - A*c)*(b + c*x^2)^3)/(6*c^2) + (B*(b + c*x^2)^4)/(8*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{\left (A+B x^2\right ) \left (b x^2+c x^4\right )^2}{x^3} \, dx &=\int x \left (A+B x^2\right ) \left (b+c x^2\right )^2 \, dx\\ &=\frac{1}{2} \operatorname{Subst}\left (\int (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) (b+c x)^2}{c}+\frac{B (b+c x)^3}{c}\right ) \, dx,x,x^2\right )\\ &=-\frac{(b B-A c) \left (b+c x^2\right )^3}{6 c^2}+\frac{B \left (b+c x^2\right )^4}{8 c^2}\\ \end{align*}

Mathematica [A]  time = 0.0128354, size = 51, normalized size = 1.21 \[ \frac{1}{24} x^2 \left (12 A b^2+4 c x^4 (A c+2 b B)+6 b x^2 (2 A c+b B)+3 B c^2 x^6\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x^2*(12*A*b^2 + 6*b*(b*B + 2*A*c)*x^2 + 4*c*(2*b*B + A*c)*x^4 + 3*B*c^2*x^6))/24

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Maple [A]  time = 0.001, size = 52, normalized size = 1.2 \begin{align*}{\frac{B{c}^{2}{x}^{8}}{8}}+{\frac{ \left ( A{c}^{2}+2\,Bbc \right ){x}^{6}}{6}}+{\frac{ \left ( 2\,Abc+B{b}^{2} \right ){x}^{4}}{4}}+{\frac{A{b}^{2}{x}^{2}}{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.10798, size = 69, normalized size = 1.64 \begin{align*} \frac{1}{8} \, B c^{2} x^{8} + \frac{1}{6} \,{\left (2 \, B b c + A c^{2}\right )} x^{6} + \frac{1}{2} \, A b^{2} x^{2} + \frac{1}{4} \,{\left (B b^{2} + 2 \, A b c\right )} x^{4} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 0.457673, size = 117, normalized size = 2.79 \begin{align*} \frac{1}{8} \, B c^{2} x^{8} + \frac{1}{6} \,{\left (2 \, B b c + A c^{2}\right )} x^{6} + \frac{1}{2} \, A b^{2} x^{2} + \frac{1}{4} \,{\left (B b^{2} + 2 \, A b c\right )} x^{4} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.069064, size = 53, normalized size = 1.26 \begin{align*} \frac{A b^{2} x^{2}}{2} + \frac{B c^{2} x^{8}}{8} + x^{6} \left (\frac{A c^{2}}{6} + \frac{B b c}{3}\right ) + x^{4} \left (\frac{A b c}{2} + \frac{B b^{2}}{4}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

A*b**2*x**2/2 + B*c**2*x**8/8 + x**6*(A*c**2/6 + B*b*c/3) + x**4*(A*b*c/2 + B*b**2/4)

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Giac [A]  time = 1.26475, size = 72, normalized size = 1.71 \begin{align*} \frac{1}{8} \, B c^{2} x^{8} + \frac{1}{3} \, B b c x^{6} + \frac{1}{6} \, A c^{2} x^{6} + \frac{1}{4} \, B b^{2} x^{4} + \frac{1}{2} \, A b c x^{4} + \frac{1}{2} \, A b^{2} x^{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*B*c^2*x^8 + 1/3*B*b*c*x^6 + 1/6*A*c^2*x^6 + 1/4*B*b^2*x^4 + 1/2*A*b*c*x^4 + 1/2*A*b^2*x^2