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]

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

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Rubi [A]  time = 0.04, antiderivative size = 48, normalized size of antiderivative = 1.00, 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 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]

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]

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*}

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Mathematica [A]  time = 0.02, 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]

-1/3*(A*b^2)/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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fricas [A]  time = 0.88, size = 52, normalized size = 1.08 \[ \frac {B c^{2} x^{6} + 3 \, {\left (2 \, B b c + A c^{2}\right )} x^{4} - A b^{2} - 3 \, {\left (B b^{2} + 2 \, A b c\right )} x^{2}}{3 \, x^{3}} \]

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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giac [A]  time = 0.14, size = 50, normalized size = 1.04 \[ \frac {1}{3} \, B c^{2} x^{3} + 2 \, B b c x + A c^{2} x - \frac {3 \, B b^{2} x^{2} + 6 \, A b c x^{2} + A b^{2}}{3 \, x^{3}} \]

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

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maple [A]  time = 0.05, size = 46, normalized size = 0.96 \[ \frac {B \,c^{2} x^{3}}{3}+A \,c^{2} x +2 B b c x -\frac {A \,b^{2}}{3 x^{3}}-\frac {\left (2 A c +b B \right ) b}{x} \]

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 = 1.32, size = 50, normalized size = 1.04 \[ \frac {1}{3} \, B c^{2} x^{3} + {\left (2 \, B b c + A c^{2}\right )} x - \frac {A b^{2} + 3 \, {\left (B b^{2} + 2 \, A b c\right )} x^{2}}{3 \, x^{3}} \]

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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mupad [B]  time = 0.05, size = 50, normalized size = 1.04 \[ x\,\left (A\,c^2+2\,B\,b\,c\right )-\frac {x^2\,\left (B\,b^2+2\,A\,c\,b\right )+\frac {A\,b^2}{3}}{x^3}+\frac {B\,c^2\,x^3}{3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*x^2)*(b*x^2 + c*x^4)^2)/x^8,x)

[Out]

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

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sympy [A]  time = 0.26, size = 51, normalized size = 1.06 \[ \frac {B c^{2} x^{3}}{3} + x \left (A c^{2} + 2 B b c\right ) + \frac {- A b^{2} + x^{2} \left (- 6 A b c - 3 B b^{2}\right )}{3 x^{3}} \]

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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