3.31 \(\int \frac{(A+B x^2) (b x^2+c x^4)^3}{x^8} \, dx\)

Optimal. Leaf size=65 \[ b^2 x (3 A c+b B)-\frac{A b^3}{x}+\frac{1}{5} c^2 x^5 (A c+3 b B)+b c x^3 (A c+b B)+\frac{1}{7} B c^3 x^7 \]

[Out]

-((A*b^3)/x) + b^2*(b*B + 3*A*c)*x + b*c*(b*B + A*c)*x^3 + (c^2*(3*b*B + A*c)*x^5)/5 + (B*c^3*x^7)/7

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Rubi [A]  time = 0.0454822, antiderivative size = 65, 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} \[ b^2 x (3 A c+b B)-\frac{A b^3}{x}+\frac{1}{5} c^2 x^5 (A c+3 b B)+b c x^3 (A c+b B)+\frac{1}{7} B c^3 x^7 \]

Antiderivative was successfully verified.

[In]

Int[((A + B*x^2)*(b*x^2 + c*x^4)^3)/x^8,x]

[Out]

-((A*b^3)/x) + b^2*(b*B + 3*A*c)*x + b*c*(b*B + A*c)*x^3 + (c^2*(3*b*B + A*c)*x^5)/5 + (B*c^3*x^7)/7

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^8} \, dx &=\int \frac{\left (A+B x^2\right ) \left (b+c x^2\right )^3}{x^2} \, dx\\ &=\int \left (b^2 (b B+3 A c)+\frac{A b^3}{x^2}+3 b c (b B+A c) x^2+c^2 (3 b B+A c) x^4+B c^3 x^6\right ) \, dx\\ &=-\frac{A b^3}{x}+b^2 (b B+3 A c) x+b c (b B+A c) x^3+\frac{1}{5} c^2 (3 b B+A c) x^5+\frac{1}{7} B c^3 x^7\\ \end{align*}

Mathematica [A]  time = 0.0222336, size = 65, normalized size = 1. \[ b^2 x (3 A c+b B)-\frac{A b^3}{x}+\frac{1}{5} c^2 x^5 (A c+3 b B)+b c x^3 (A c+b B)+\frac{1}{7} B c^3 x^7 \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x^2)*(b*x^2 + c*x^4)^3)/x^8,x]

[Out]

-((A*b^3)/x) + b^2*(b*B + 3*A*c)*x + b*c*(b*B + A*c)*x^3 + (c^2*(3*b*B + A*c)*x^5)/5 + (B*c^3*x^7)/7

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Maple [A]  time = 0.004, size = 71, normalized size = 1.1 \begin{align*}{\frac{B{c}^{3}{x}^{7}}{7}}+{\frac{A{x}^{5}{c}^{3}}{5}}+{\frac{3\,B{x}^{5}b{c}^{2}}{5}}+A{x}^{3}b{c}^{2}+B{x}^{3}{b}^{2}c+3\,A{b}^{2}cx+B{b}^{3}x-{\frac{A{b}^{3}}{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^8,x)

[Out]

1/7*B*c^3*x^7+1/5*A*x^5*c^3+3/5*B*x^5*b*c^2+A*x^3*b*c^2+B*x^3*b^2*c+3*A*b^2*c*x+B*b^3*x-A*b^3/x

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Maxima [A]  time = 1.16804, size = 93, normalized size = 1.43 \begin{align*} \frac{1}{7} \, B c^{3} x^{7} + \frac{1}{5} \,{\left (3 \, B b c^{2} + A c^{3}\right )} x^{5} +{\left (B b^{2} c + A b c^{2}\right )} x^{3} - \frac{A b^{3}}{x} +{\left (B b^{3} + 3 \, A b^{2} c\right )} x \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^8,x, algorithm="maxima")

[Out]

1/7*B*c^3*x^7 + 1/5*(3*B*b*c^2 + A*c^3)*x^5 + (B*b^2*c + A*b*c^2)*x^3 - A*b^3/x + (B*b^3 + 3*A*b^2*c)*x

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Fricas [A]  time = 0.481409, size = 161, normalized size = 2.48 \begin{align*} \frac{5 \, B c^{3} x^{8} + 7 \,{\left (3 \, B b c^{2} + A c^{3}\right )} x^{6} + 35 \,{\left (B b^{2} c + A b c^{2}\right )} x^{4} - 35 \, A b^{3} + 35 \,{\left (B b^{3} + 3 \, A b^{2} c\right )} x^{2}}{35 \, x} \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^8,x, algorithm="fricas")

[Out]

1/35*(5*B*c^3*x^8 + 7*(3*B*b*c^2 + A*c^3)*x^6 + 35*(B*b^2*c + A*b*c^2)*x^4 - 35*A*b^3 + 35*(B*b^3 + 3*A*b^2*c)
*x^2)/x

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Sympy [A]  time = 0.313686, size = 68, normalized size = 1.05 \begin{align*} - \frac{A b^{3}}{x} + \frac{B c^{3} x^{7}}{7} + x^{5} \left (\frac{A c^{3}}{5} + \frac{3 B b c^{2}}{5}\right ) + x^{3} \left (A b c^{2} + B b^{2} c\right ) + x \left (3 A b^{2} c + B b^{3}\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)**3/x**8,x)

[Out]

-A*b**3/x + B*c**3*x**7/7 + x**5*(A*c**3/5 + 3*B*b*c**2/5) + x**3*(A*b*c**2 + B*b**2*c) + x*(3*A*b**2*c + B*b*
*3)

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Giac [A]  time = 1.1633, size = 95, normalized size = 1.46 \begin{align*} \frac{1}{7} \, B c^{3} x^{7} + \frac{3}{5} \, B b c^{2} x^{5} + \frac{1}{5} \, A c^{3} x^{5} + B b^{2} c x^{3} + A b c^{2} x^{3} + B b^{3} x + 3 \, A b^{2} c x - \frac{A b^{3}}{x} \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^8,x, algorithm="giac")

[Out]

1/7*B*c^3*x^7 + 3/5*B*b*c^2*x^5 + 1/5*A*c^3*x^5 + B*b^2*c*x^3 + A*b*c^2*x^3 + B*b^3*x + 3*A*b^2*c*x - A*b^3/x