3.33 \(\int \frac{(A+B x^2) (b x^2+c x^4)^3}{x^{10}} \, dx\)

Optimal. Leaf size=69 \[ -\frac{b^2 (3 A c+b B)}{x}-\frac{A b^3}{3 x^3}+\frac{1}{3} c^2 x^3 (A c+3 b B)+3 b c x (A c+b B)+\frac{1}{5} B c^3 x^5 \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0495511, antiderivative size = 69, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.083, Rules used = {1584, 448} \[ -\frac{b^2 (3 A c+b B)}{x}-\frac{A b^3}{3 x^3}+\frac{1}{3} c^2 x^3 (A c+3 b B)+3 b c x (A c+b B)+\frac{1}{5} B c^3 x^5 \]

Antiderivative was successfully verified.

[In]

Int[((A + B*x^2)*(b*x^2 + c*x^4)^3)/x^10,x]

[Out]

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

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 448

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Int[ExpandI
ntegrand[(e*x)^m*(a + b*x^n)^p*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && NeQ[b*c - a*d, 0] &
& IGtQ[p, 0] && IGtQ[q, 0]

Rubi steps

\begin{align*} \int \frac{\left (A+B x^2\right ) \left (b x^2+c x^4\right )^3}{x^{10}} \, dx &=\int \frac{\left (A+B x^2\right ) \left (b+c x^2\right )^3}{x^4} \, dx\\ &=\int \left (3 b c (b B+A c)+\frac{A b^3}{x^4}+\frac{b^2 (b B+3 A c)}{x^2}+c^2 (3 b B+A c) x^2+B c^3 x^4\right ) \, dx\\ &=-\frac{A b^3}{3 x^3}-\frac{b^2 (b B+3 A c)}{x}+3 b c (b B+A c) x+\frac{1}{3} c^2 (3 b B+A c) x^3+\frac{1}{5} B c^3 x^5\\ \end{align*}

Mathematica [A]  time = 0.0228669, size = 71, normalized size = 1.03 \[ \frac{b^3 (-B)-3 A b^2 c}{x}-\frac{A b^3}{3 x^3}+\frac{1}{3} c^2 x^3 (A c+3 b B)+3 b c x (A c+b B)+\frac{1}{5} B c^3 x^5 \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x^2)*(b*x^2 + c*x^4)^3)/x^10,x]

[Out]

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

________________________________________________________________________________________

Maple [A]  time = 0.006, size = 70, normalized size = 1. \begin{align*}{\frac{B{c}^{3}{x}^{5}}{5}}+{\frac{A{x}^{3}{c}^{3}}{3}}+B{x}^{3}b{c}^{2}+3\,Ab{c}^{2}x+3\,B{b}^{2}cx-{\frac{A{b}^{3}}{3\,{x}^{3}}}-{\frac{{b}^{2} \left ( 3\,Ac+Bb \right ) }{x}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x^2+A)*(c*x^4+b*x^2)^3/x^10,x)

[Out]

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

________________________________________________________________________________________

Maxima [A]  time = 1.13104, size = 99, normalized size = 1.43 \begin{align*} \frac{1}{5} \, B c^{3} x^{5} + \frac{1}{3} \,{\left (3 \, B b c^{2} + A c^{3}\right )} x^{3} + 3 \,{\left (B b^{2} c + A b c^{2}\right )} x - \frac{A b^{3} + 3 \,{\left (B b^{3} + 3 \, A b^{2} c\right )} x^{2}}{3 \, x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)*(c*x^4+b*x^2)^3/x^10,x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

Fricas [A]  time = 0.459542, size = 162, normalized size = 2.35 \begin{align*} \frac{3 \, B c^{3} x^{8} + 5 \,{\left (3 \, B b c^{2} + A c^{3}\right )} x^{6} + 45 \,{\left (B b^{2} c + A b c^{2}\right )} x^{4} - 5 \, A b^{3} - 15 \,{\left (B b^{3} + 3 \, A b^{2} c\right )} x^{2}}{15 \, x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)*(c*x^4+b*x^2)^3/x^10,x, algorithm="fricas")

[Out]

1/15*(3*B*c^3*x^8 + 5*(3*B*b*c^2 + A*c^3)*x^6 + 45*(B*b^2*c + A*b*c^2)*x^4 - 5*A*b^3 - 15*(B*b^3 + 3*A*b^2*c)*
x^2)/x^3

________________________________________________________________________________________

Sympy [A]  time = 0.427238, size = 73, normalized size = 1.06 \begin{align*} \frac{B c^{3} x^{5}}{5} + x^{3} \left (\frac{A c^{3}}{3} + B b c^{2}\right ) + x \left (3 A b c^{2} + 3 B b^{2} c\right ) - \frac{A b^{3} + x^{2} \left (9 A b^{2} c + 3 B b^{3}\right )}{3 x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x**2+A)*(c*x**4+b*x**2)**3/x**10,x)

[Out]

B*c**3*x**5/5 + x**3*(A*c**3/3 + B*b*c**2) + x*(3*A*b*c**2 + 3*B*b**2*c) - (A*b**3 + x**2*(9*A*b**2*c + 3*B*b*
*3))/(3*x**3)

________________________________________________________________________________________

Giac [A]  time = 1.18482, size = 100, normalized size = 1.45 \begin{align*} \frac{1}{5} \, B c^{3} x^{5} + B b c^{2} x^{3} + \frac{1}{3} \, A c^{3} x^{3} + 3 \, B b^{2} c x + 3 \, A b c^{2} x - \frac{3 \, B b^{3} x^{2} + 9 \, A b^{2} c x^{2} + A b^{3}}{3 \, x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)*(c*x^4+b*x^2)^3/x^10,x, algorithm="giac")

[Out]

1/5*B*c^3*x^5 + B*b*c^2*x^3 + 1/3*A*c^3*x^3 + 3*B*b^2*c*x + 3*A*b*c^2*x - 1/3*(3*B*b^3*x^2 + 9*A*b^2*c*x^2 + A
*b^3)/x^3