3.28 \(\int \frac{(A+B x^2) (b x^2+c x^4)^3}{x^5} \, dx\)

Optimal. Leaf size=42 \[ \frac{B \left (b+c x^2\right )^5}{10 c^2}-\frac{\left (b+c x^2\right )^4 (b B-A c)}{8 c^2} \]

[Out]

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

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Rubi [A]  time = 0.0706387, antiderivative size = 42, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {1584, 444, 43} \[ \frac{B \left (b+c x^2\right )^5}{10 c^2}-\frac{\left (b+c x^2\right )^4 (b B-A c)}{8 c^2} \]

Antiderivative was successfully verified.

[In]

Int[((A + B*x^2)*(b*x^2 + c*x^4)^3)/x^5,x]

[Out]

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

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 444

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[(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] && EqQ[m
- n + 1, 0]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

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

Mathematica [A]  time = 0.018608, size = 69, normalized size = 1.64 \[ \frac{1}{40} x^2 \left (10 b^2 x^2 (3 A c+b B)+20 A b^3+5 c^2 x^6 (A c+3 b B)+20 b c x^4 (A c+b B)+4 B c^3 x^8\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x^2)*(b*x^2 + c*x^4)^3)/x^5,x]

[Out]

(x^2*(20*A*b^3 + 10*b^2*(b*B + 3*A*c)*x^2 + 20*b*c*(b*B + A*c)*x^4 + 5*c^2*(3*b*B + A*c)*x^6 + 4*B*c^3*x^8))/4
0

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Maple [A]  time = 0.001, size = 76, normalized size = 1.8 \begin{align*}{\frac{B{c}^{3}{x}^{10}}{10}}+{\frac{ \left ( A{c}^{3}+3\,Bb{c}^{2} \right ){x}^{8}}{8}}+{\frac{ \left ( 3\,Ab{c}^{2}+3\,B{b}^{2}c \right ){x}^{6}}{6}}+{\frac{ \left ( 3\,A{b}^{2}c+B{b}^{3} \right ){x}^{4}}{4}}+{\frac{A{b}^{3}{x}^{2}}{2}} \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^5,x)

[Out]

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

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Maxima [A]  time = 1.13199, size = 99, normalized size = 2.36 \begin{align*} \frac{1}{10} \, B c^{3} x^{10} + \frac{1}{8} \,{\left (3 \, B b c^{2} + A c^{3}\right )} x^{8} + \frac{1}{2} \,{\left (B b^{2} c + A b c^{2}\right )} x^{6} + \frac{1}{2} \, A b^{3} x^{2} + \frac{1}{4} \,{\left (B b^{3} + 3 \, A b^{2} c\right )} x^{4} \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^5,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 0.482166, size = 166, normalized size = 3.95 \begin{align*} \frac{1}{10} \, B c^{3} x^{10} + \frac{1}{8} \,{\left (3 \, B b c^{2} + A c^{3}\right )} x^{8} + \frac{1}{2} \,{\left (B b^{2} c + A b c^{2}\right )} x^{6} + \frac{1}{2} \, A b^{3} x^{2} + \frac{1}{4} \,{\left (B b^{3} + 3 \, A b^{2} c\right )} x^{4} \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^5,x, algorithm="fricas")

[Out]

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

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Sympy [B]  time = 0.075218, size = 80, normalized size = 1.9 \begin{align*} \frac{A b^{3} x^{2}}{2} + \frac{B c^{3} x^{10}}{10} + x^{8} \left (\frac{A c^{3}}{8} + \frac{3 B b c^{2}}{8}\right ) + x^{6} \left (\frac{A b c^{2}}{2} + \frac{B b^{2} c}{2}\right ) + x^{4} \left (\frac{3 A b^{2} c}{4} + \frac{B b^{3}}{4}\right ) \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**5,x)

[Out]

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

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Giac [B]  time = 1.20848, size = 104, normalized size = 2.48 \begin{align*} \frac{1}{10} \, B c^{3} x^{10} + \frac{3}{8} \, B b c^{2} x^{8} + \frac{1}{8} \, A c^{3} x^{8} + \frac{1}{2} \, B b^{2} c x^{6} + \frac{1}{2} \, A b c^{2} x^{6} + \frac{1}{4} \, B b^{3} x^{4} + \frac{3}{4} \, A b^{2} c x^{4} + \frac{1}{2} \, A b^{3} x^{2} \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^5,x, algorithm="giac")

[Out]

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