3.1.39 \(\int \frac {(A+B x^2) (b x^2+c x^4)^3}{x^{16}} \, dx\)

Optimal. Leaf size=73 \[ -\frac {A b^3}{9 x^9}-\frac {b^2 (3 A c+b B)}{7 x^7}-\frac {c^2 (A c+3 b B)}{3 x^3}-\frac {3 b c (A c+b B)}{5 x^5}-\frac {B c^3}{x} \]

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Rubi [A]  time = 0.05, antiderivative size = 73, 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} \begin {gather*} -\frac {b^2 (3 A c+b B)}{7 x^7}-\frac {A b^3}{9 x^9}-\frac {c^2 (A c+3 b B)}{3 x^3}-\frac {3 b c (A c+b B)}{5 x^5}-\frac {B c^3}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((A + B*x^2)*(b*x^2 + c*x^4)^3)/x^16,x]

[Out]

-(A*b^3)/(9*x^9) - (b^2*(b*B + 3*A*c))/(7*x^7) - (3*b*c*(b*B + A*c))/(5*x^5) - (c^2*(3*b*B + A*c))/(3*x^3) - (
B*c^3)/x

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 )^3}{x^{16}} \, dx &=\int \frac {\left (A+B x^2\right ) \left (b+c x^2\right )^3}{x^{10}} \, dx\\ &=\int \left (\frac {A b^3}{x^{10}}+\frac {b^2 (b B+3 A c)}{x^8}+\frac {3 b c (b B+A c)}{x^6}+\frac {c^2 (3 b B+A c)}{x^4}+\frac {B c^3}{x^2}\right ) \, dx\\ &=-\frac {A b^3}{9 x^9}-\frac {b^2 (b B+3 A c)}{7 x^7}-\frac {3 b c (b B+A c)}{5 x^5}-\frac {c^2 (3 b B+A c)}{3 x^3}-\frac {B c^3}{x}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 73, normalized size = 1.00 \begin {gather*} -\frac {A b^3}{9 x^9}-\frac {b^2 (3 A c+b B)}{7 x^7}-\frac {c^2 (A c+3 b B)}{3 x^3}-\frac {3 b c (A c+b B)}{5 x^5}-\frac {B c^3}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x^2)*(b*x^2 + c*x^4)^3)/x^16,x]

[Out]

-1/9*(A*b^3)/x^9 - (b^2*(b*B + 3*A*c))/(7*x^7) - (3*b*c*(b*B + A*c))/(5*x^5) - (c^2*(3*b*B + A*c))/(3*x^3) - (
B*c^3)/x

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^3}{x^{16}} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[((A + B*x^2)*(b*x^2 + c*x^4)^3)/x^16,x]

[Out]

IntegrateAlgebraic[((A + B*x^2)*(b*x^2 + c*x^4)^3)/x^16, x]

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fricas [A]  time = 0.39, size = 75, normalized size = 1.03 \begin {gather*} -\frac {315 \, B c^{3} x^{8} + 105 \, {\left (3 \, B b c^{2} + A c^{3}\right )} x^{6} + 189 \, {\left (B b^{2} c + A b c^{2}\right )} x^{4} + 35 \, A b^{3} + 45 \, {\left (B b^{3} + 3 \, A b^{2} c\right )} x^{2}}{315 \, x^{9}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)*(c*x^4+b*x^2)^3/x^16,x, algorithm="fricas")

[Out]

-1/315*(315*B*c^3*x^8 + 105*(3*B*b*c^2 + A*c^3)*x^6 + 189*(B*b^2*c + A*b*c^2)*x^4 + 35*A*b^3 + 45*(B*b^3 + 3*A
*b^2*c)*x^2)/x^9

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giac [A]  time = 0.15, size = 79, normalized size = 1.08 \begin {gather*} -\frac {315 \, B c^{3} x^{8} + 315 \, B b c^{2} x^{6} + 105 \, A c^{3} x^{6} + 189 \, B b^{2} c x^{4} + 189 \, A b c^{2} x^{4} + 45 \, B b^{3} x^{2} + 135 \, A b^{2} c x^{2} + 35 \, A b^{3}}{315 \, x^{9}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)*(c*x^4+b*x^2)^3/x^16,x, algorithm="giac")

[Out]

-1/315*(315*B*c^3*x^8 + 315*B*b*c^2*x^6 + 105*A*c^3*x^6 + 189*B*b^2*c*x^4 + 189*A*b*c^2*x^4 + 45*B*b^3*x^2 + 1
35*A*b^2*c*x^2 + 35*A*b^3)/x^9

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maple [A]  time = 0.05, size = 66, normalized size = 0.90 \begin {gather*} -\frac {B \,c^{3}}{x}-\frac {\left (A c +3 b B \right ) c^{2}}{3 x^{3}}-\frac {3 \left (A c +b B \right ) b c}{5 x^{5}}-\frac {A \,b^{3}}{9 x^{9}}-\frac {\left (3 A c +b B \right ) b^{2}}{7 x^{7}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x^2+A)*(c*x^4+b*x^2)^3/x^16,x)

[Out]

-1/9*A*b^3/x^9-1/7*b^2*(3*A*c+B*b)/x^7-3/5*b*c*(A*c+B*b)/x^5-1/3*c^2*(A*c+3*B*b)/x^3-B*c^3/x

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maxima [A]  time = 1.38, size = 75, normalized size = 1.03 \begin {gather*} -\frac {315 \, B c^{3} x^{8} + 105 \, {\left (3 \, B b c^{2} + A c^{3}\right )} x^{6} + 189 \, {\left (B b^{2} c + A b c^{2}\right )} x^{4} + 35 \, A b^{3} + 45 \, {\left (B b^{3} + 3 \, A b^{2} c\right )} x^{2}}{315 \, x^{9}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)*(c*x^4+b*x^2)^3/x^16,x, algorithm="maxima")

[Out]

-1/315*(315*B*c^3*x^8 + 105*(3*B*b*c^2 + A*c^3)*x^6 + 189*(B*b^2*c + A*b*c^2)*x^4 + 35*A*b^3 + 45*(B*b^3 + 3*A
*b^2*c)*x^2)/x^9

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mupad [B]  time = 0.07, size = 74, normalized size = 1.01 \begin {gather*} -\frac {x^4\,\left (\frac {3\,B\,b^2\,c}{5}+\frac {3\,A\,b\,c^2}{5}\right )+\frac {A\,b^3}{9}+x^2\,\left (\frac {B\,b^3}{7}+\frac {3\,A\,c\,b^2}{7}\right )+x^6\,\left (\frac {A\,c^3}{3}+B\,b\,c^2\right )+B\,c^3\,x^8}{x^9} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*x^2)*(b*x^2 + c*x^4)^3)/x^16,x)

[Out]

-(x^4*((3*A*b*c^2)/5 + (3*B*b^2*c)/5) + (A*b^3)/9 + x^2*((B*b^3)/7 + (3*A*b^2*c)/7) + x^6*((A*c^3)/3 + B*b*c^2
) + B*c^3*x^8)/x^9

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sympy [A]  time = 2.45, size = 83, normalized size = 1.14 \begin {gather*} \frac {- 35 A b^{3} - 315 B c^{3} x^{8} + x^{6} \left (- 105 A c^{3} - 315 B b c^{2}\right ) + x^{4} \left (- 189 A b c^{2} - 189 B b^{2} c\right ) + x^{2} \left (- 135 A b^{2} c - 45 B b^{3}\right )}{315 x^{9}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x**2+A)*(c*x**4+b*x**2)**3/x**16,x)

[Out]

(-35*A*b**3 - 315*B*c**3*x**8 + x**6*(-105*A*c**3 - 315*B*b*c**2) + x**4*(-189*A*b*c**2 - 189*B*b**2*c) + x**2
*(-135*A*b**2*c - 45*B*b**3))/(315*x**9)

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