3.37 \(\int \frac{(A+B x^2) (b x^2+c x^4)^3}{x^{14}} \, dx\)

Optimal. Leaf size=66 \[ -\frac{b^2 (3 A c+b B)}{5 x^5}-\frac{A b^3}{7 x^7}-\frac{c^2 (A c+3 b B)}{x}-\frac{b c (A c+b B)}{x^3}+B c^3 x \]

[Out]

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

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Rubi [A]  time = 0.0495761, antiderivative size = 66, 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} \[ -\frac{b^2 (3 A c+b B)}{5 x^5}-\frac{A b^3}{7 x^7}-\frac{c^2 (A c+3 b B)}{x}-\frac{b c (A c+b B)}{x^3}+B c^3 x \]

Antiderivative was successfully verified.

[In]

Int[((A + B*x^2)*(b*x^2 + c*x^4)^3)/x^14,x]

[Out]

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

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^{14}} \, dx &=\int \frac{\left (A+B x^2\right ) \left (b+c x^2\right )^3}{x^8} \, dx\\ &=\int \left (B c^3+\frac{A b^3}{x^8}+\frac{b^2 (b B+3 A c)}{x^6}+\frac{3 b c (b B+A c)}{x^4}+\frac{c^2 (3 b B+A c)}{x^2}\right ) \, dx\\ &=-\frac{A b^3}{7 x^7}-\frac{b^2 (b B+3 A c)}{5 x^5}-\frac{b c (b B+A c)}{x^3}-\frac{c^2 (3 b B+A c)}{x}+B c^3 x\\ \end{align*}

Mathematica [A]  time = 0.0274184, size = 66, normalized size = 1. \[ -\frac{b^2 (3 A c+b B)}{5 x^5}-\frac{A b^3}{7 x^7}-\frac{c^2 (A c+3 b B)}{x}-\frac{b c (A c+b B)}{x^3}+B c^3 x \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x^2)*(b*x^2 + c*x^4)^3)/x^14,x]

[Out]

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

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Maple [A]  time = 0.006, size = 63, normalized size = 1. \begin{align*} -{\frac{A{b}^{3}}{7\,{x}^{7}}}-{\frac{{b}^{2} \left ( 3\,Ac+Bb \right ) }{5\,{x}^{5}}}-{\frac{bc \left ( Ac+Bb \right ) }{{x}^{3}}}-{\frac{{c}^{2} \left ( Ac+3\,Bb \right ) }{x}}+B{c}^{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^14,x)

[Out]

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

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Maxima [A]  time = 1.19158, size = 99, normalized size = 1.5 \begin{align*} B c^{3} x - \frac{35 \,{\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} + 5 \, A b^{3} + 7 \,{\left (B b^{3} + 3 \, A b^{2} c\right )} x^{2}}{35 \, x^{7}} \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^14,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 0.483685, size = 163, normalized size = 2.47 \begin{align*} \frac{35 \, B c^{3} x^{8} - 35 \,{\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} - 5 \, A b^{3} - 7 \,{\left (B b^{3} + 3 \, A b^{2} c\right )} x^{2}}{35 \, x^{7}} \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^14,x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 1.60242, size = 75, normalized size = 1.14 \begin{align*} B c^{3} x - \frac{5 A b^{3} + x^{6} \left (35 A c^{3} + 105 B b c^{2}\right ) + x^{4} \left (35 A b c^{2} + 35 B b^{2} c\right ) + x^{2} \left (21 A b^{2} c + 7 B b^{3}\right )}{35 x^{7}} \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**14,x)

[Out]

B*c**3*x - (5*A*b**3 + x**6*(35*A*c**3 + 105*B*b*c**2) + x**4*(35*A*b*c**2 + 35*B*b**2*c) + x**2*(21*A*b**2*c
+ 7*B*b**3))/(35*x**7)

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Giac [A]  time = 1.27154, size = 104, normalized size = 1.58 \begin{align*} B c^{3} x - \frac{105 \, B b c^{2} x^{6} + 35 \, A c^{3} x^{6} + 35 \, B b^{2} c x^{4} + 35 \, A b c^{2} x^{4} + 7 \, B b^{3} x^{2} + 21 \, A b^{2} c x^{2} + 5 \, A b^{3}}{35 \, x^{7}} \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^14,x, algorithm="giac")

[Out]

B*c^3*x - 1/35*(105*B*b*c^2*x^6 + 35*A*c^3*x^6 + 35*B*b^2*c*x^4 + 35*A*b*c^2*x^4 + 7*B*b^3*x^2 + 21*A*b^2*c*x^
2 + 5*A*b^3)/x^7