3.40 \(\int \frac{(A+B x^2) (b x^2+c x^4)^3}{x^{17}} \, dx\)

Optimal. Leaf size=49 \[ -\frac{\left (b+c x^2\right )^4 (5 b B-A c)}{40 b^2 x^8}-\frac{A \left (b+c x^2\right )^4}{10 b x^{10}} \]

[Out]

-(A*(b + c*x^2)^4)/(10*b*x^10) - ((5*b*B - A*c)*(b + c*x^2)^4)/(40*b^2*x^8)

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Rubi [A]  time = 0.0397545, antiderivative size = 49, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {1584, 446, 78, 37} \[ -\frac{\left (b+c x^2\right )^4 (5 b B-A c)}{40 b^2 x^8}-\frac{A \left (b+c x^2\right )^4}{10 b x^{10}} \]

Antiderivative was successfully verified.

[In]

Int[((A + B*x^2)*(b*x^2 + c*x^4)^3)/x^17,x]

[Out]

-(A*(b + c*x^2)^4)/(10*b*x^10) - ((5*b*B - A*c)*(b + c*x^2)^4)/(40*b^2*x^8)

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 446

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> -Simp[((b*e - a*f
)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(f*(p + 1)*(c*f - d*e)), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1)
+ c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f,
n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ
[p, n]))))

Rule 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n +
1))/((b*c - a*d)*(m + 1)), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

Rubi steps

\begin{align*} \int \frac{\left (A+B x^2\right ) \left (b x^2+c x^4\right )^3}{x^{17}} \, dx &=\int \frac{\left (A+B x^2\right ) \left (b+c x^2\right )^3}{x^{11}} \, dx\\ &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{(A+B x) (b+c x)^3}{x^6} \, dx,x,x^2\right )\\ &=-\frac{A \left (b+c x^2\right )^4}{10 b x^{10}}+\frac{(5 b B-A c) \operatorname{Subst}\left (\int \frac{(b+c x)^3}{x^5} \, dx,x,x^2\right )}{10 b}\\ &=-\frac{A \left (b+c x^2\right )^4}{10 b x^{10}}-\frac{(5 b B-A c) \left (b+c x^2\right )^4}{40 b^2 x^8}\\ \end{align*}

Mathematica [A]  time = 0.0194762, size = 78, normalized size = 1.59 \[ -\frac{A \left (15 b^2 c x^2+4 b^3+20 b c^2 x^4+10 c^3 x^6\right )+5 B x^2 \left (4 b^2 c x^2+b^3+6 b c^2 x^4+4 c^3 x^6\right )}{40 x^{10}} \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x^2)*(b*x^2 + c*x^4)^3)/x^17,x]

[Out]

-(5*B*x^2*(b^3 + 4*b^2*c*x^2 + 6*b*c^2*x^4 + 4*c^3*x^6) + A*(4*b^3 + 15*b^2*c*x^2 + 20*b*c^2*x^4 + 10*c^3*x^6)
)/(40*x^10)

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Maple [A]  time = 0.004, size = 66, normalized size = 1.4 \begin{align*} -{\frac{{c}^{2} \left ( Ac+3\,Bb \right ) }{4\,{x}^{4}}}-{\frac{B{c}^{3}}{2\,{x}^{2}}}-{\frac{{b}^{2} \left ( 3\,Ac+Bb \right ) }{8\,{x}^{8}}}-{\frac{A{b}^{3}}{10\,{x}^{10}}}-{\frac{bc \left ( Ac+Bb \right ) }{2\,{x}^{6}}} \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^17,x)

[Out]

-1/4*c^2*(A*c+3*B*b)/x^4-1/2*B*c^3/x^2-1/8*b^2*(3*A*c+B*b)/x^8-1/10*A*b^3/x^10-1/2*b*c*(A*c+B*b)/x^6

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Maxima [A]  time = 1.08587, size = 101, normalized size = 2.06 \begin{align*} -\frac{20 \, B c^{3} x^{8} + 10 \,{\left (3 \, B b c^{2} + A c^{3}\right )} x^{6} + 20 \,{\left (B b^{2} c + A b c^{2}\right )} x^{4} + 4 \, A b^{3} + 5 \,{\left (B b^{3} + 3 \, A b^{2} c\right )} x^{2}}{40 \, x^{10}} \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^17,x, algorithm="maxima")

[Out]

-1/40*(20*B*c^3*x^8 + 10*(3*B*b*c^2 + A*c^3)*x^6 + 20*(B*b^2*c + A*b*c^2)*x^4 + 4*A*b^3 + 5*(B*b^3 + 3*A*b^2*c
)*x^2)/x^10

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Fricas [A]  time = 0.475784, size = 166, normalized size = 3.39 \begin{align*} -\frac{20 \, B c^{3} x^{8} + 10 \,{\left (3 \, B b c^{2} + A c^{3}\right )} x^{6} + 20 \,{\left (B b^{2} c + A b c^{2}\right )} x^{4} + 4 \, A b^{3} + 5 \,{\left (B b^{3} + 3 \, A b^{2} c\right )} x^{2}}{40 \, x^{10}} \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^17,x, algorithm="fricas")

[Out]

-1/40*(20*B*c^3*x^8 + 10*(3*B*b*c^2 + A*c^3)*x^6 + 20*(B*b^2*c + A*b*c^2)*x^4 + 4*A*b^3 + 5*(B*b^3 + 3*A*b^2*c
)*x^2)/x^10

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Sympy [A]  time = 4.43304, size = 80, normalized size = 1.63 \begin{align*} - \frac{4 A b^{3} + 20 B c^{3} x^{8} + x^{6} \left (10 A c^{3} + 30 B b c^{2}\right ) + x^{4} \left (20 A b c^{2} + 20 B b^{2} c\right ) + x^{2} \left (15 A b^{2} c + 5 B b^{3}\right )}{40 x^{10}} \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**17,x)

[Out]

-(4*A*b**3 + 20*B*c**3*x**8 + x**6*(10*A*c**3 + 30*B*b*c**2) + x**4*(20*A*b*c**2 + 20*B*b**2*c) + x**2*(15*A*b
**2*c + 5*B*b**3))/(40*x**10)

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Giac [A]  time = 1.25236, size = 107, normalized size = 2.18 \begin{align*} -\frac{20 \, B c^{3} x^{8} + 30 \, B b c^{2} x^{6} + 10 \, A c^{3} x^{6} + 20 \, B b^{2} c x^{4} + 20 \, A b c^{2} x^{4} + 5 \, B b^{3} x^{2} + 15 \, A b^{2} c x^{2} + 4 \, A b^{3}}{40 \, x^{10}} \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^17,x, algorithm="giac")

[Out]

-1/40*(20*B*c^3*x^8 + 30*B*b*c^2*x^6 + 10*A*c^3*x^6 + 20*B*b^2*c*x^4 + 20*A*b*c^2*x^4 + 5*B*b^3*x^2 + 15*A*b^2
*c*x^2 + 4*A*b^3)/x^10