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

Optimal. Leaf size=48 \[ -\frac{A b^2}{3 x^3}+c x (A c+2 b B)-\frac{b (2 A c+b B)}{x}+\frac{1}{3} B c^2 x^3 \]

[Out]

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

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Rubi [A]  time = 0.0368726, antiderivative size = 48, 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} \[ -\frac{A b^2}{3 x^3}+c x (A c+2 b B)-\frac{b (2 A c+b B)}{x}+\frac{1}{3} B c^2 x^3 \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [A]  time = 0.0185551, size = 50, normalized size = 1.04 \[ \frac{b^2 (-B)-2 A b c}{x}-\frac{A b^2}{3 x^3}+c x (A c+2 b B)+\frac{1}{3} B c^2 x^3 \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.006, size = 46, normalized size = 1. \begin{align*}{\frac{B{c}^{2}{x}^{3}}{3}}+A{c}^{2}x+2\,Bbcx-{\frac{A{b}^{2}}{3\,{x}^{3}}}-{\frac{b \left ( 2\,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)^2/x^8,x)

[Out]

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

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Maxima [A]  time = 1.13744, size = 68, normalized size = 1.42 \begin{align*} \frac{1}{3} \, B c^{2} x^{3} +{\left (2 \, B b c + A c^{2}\right )} x - \frac{A b^{2} + 3 \,{\left (B b^{2} + 2 \, A b 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)^2/x^8,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 0.496618, size = 109, normalized size = 2.27 \begin{align*} \frac{B c^{2} x^{6} + 3 \,{\left (2 \, B b c + A c^{2}\right )} x^{4} - A b^{2} - 3 \,{\left (B b^{2} + 2 \, A b 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)^2/x^8,x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 0.386567, size = 49, normalized size = 1.02 \begin{align*} \frac{B c^{2} x^{3}}{3} + x \left (A c^{2} + 2 B b c\right ) - \frac{A b^{2} + x^{2} \left (6 A b c + 3 B b^{2}\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)**2/x**8,x)

[Out]

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

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Giac [A]  time = 1.30564, size = 68, normalized size = 1.42 \begin{align*} \frac{1}{3} \, B c^{2} x^{3} + 2 \, B b c x + A c^{2} x - \frac{3 \, B b^{2} x^{2} + 6 \, A b c x^{2} + A b^{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)^2/x^8,x, algorithm="giac")

[Out]

1/3*B*c^2*x^3 + 2*B*b*c*x + A*c^2*x - 1/3*(3*B*b^2*x^2 + 6*A*b*c*x^2 + A*b^2)/x^3