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

Optimal. Leaf size=68 \[ -\frac{\left (b+c x^2\right )^5 (2 b B-A c)}{10 c^3}+\frac{b \left (b+c x^2\right )^4 (b B-A c)}{8 c^3}+\frac{B \left (b+c x^2\right )^6}{12 c^3} \]

[Out]

(b*(b*B - A*c)*(b + c*x^2)^4)/(8*c^3) - ((2*b*B - A*c)*(b + c*x^2)^5)/(10*c^3) + (B*(b + c*x^2)^6)/(12*c^3)

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Rubi [A]  time = 0.134347, antiderivative size = 68, 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, 446, 76} \[ -\frac{\left (b+c x^2\right )^5 (2 b B-A c)}{10 c^3}+\frac{b \left (b+c x^2\right )^4 (b B-A c)}{8 c^3}+\frac{B \left (b+c x^2\right )^6}{12 c^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*(b*B - A*c)*(b + c*x^2)^4)/(8*c^3) - ((2*b*B - A*c)*(b + c*x^2)^5)/(10*c^3) + (B*(b + c*x^2)^6)/(12*c^3)

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 76

Int[((d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_))*((e_) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*
x)*(d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, d, e, f, n}, x] && IGtQ[p, 0] && (NeQ[n, -1] || EqQ[p, 1]) && N
eQ[b*e + a*f, 0] && ( !IntegerQ[n] || LtQ[9*p + 5*n, 0] || GeQ[n + p + 1, 0] || (GeQ[n + p + 2, 0] && Rational
Q[a, b, d, e, f])) && (NeQ[n + p + 3, 0] || EqQ[p, 1])

Rubi steps

\begin{align*} \int \frac{\left (A+B x^2\right ) \left (b x^2+c x^4\right )^3}{x^3} \, dx &=\int x^3 \left (A+B x^2\right ) \left (b+c x^2\right )^3 \, dx\\ &=\frac{1}{2} \operatorname{Subst}\left (\int x (A+B x) (b+c x)^3 \, dx,x,x^2\right )\\ &=\frac{1}{2} \operatorname{Subst}\left (\int \left (\frac{b (b B-A c) (b+c x)^3}{c^2}+\frac{(-2 b B+A c) (b+c x)^4}{c^2}+\frac{B (b+c x)^5}{c^2}\right ) \, dx,x,x^2\right )\\ &=\frac{b (b B-A c) \left (b+c x^2\right )^4}{8 c^3}-\frac{(2 b B-A c) \left (b+c x^2\right )^5}{10 c^3}+\frac{B \left (b+c x^2\right )^6}{12 c^3}\\ \end{align*}

Mathematica [A]  time = 0.0186531, size = 69, normalized size = 1.01 \[ \frac{1}{120} x^4 \left (20 b^2 x^2 (3 A c+b B)+30 A b^3+12 c^2 x^6 (A c+3 b B)+45 b c x^4 (A c+b B)+10 B c^3 x^8\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x^4*(30*A*b^3 + 20*b^2*(b*B + 3*A*c)*x^2 + 45*b*c*(b*B + A*c)*x^4 + 12*c^2*(3*b*B + A*c)*x^6 + 10*B*c^3*x^8))
/120

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Maple [A]  time = 0.001, size = 76, normalized size = 1.1 \begin{align*}{\frac{B{c}^{3}{x}^{12}}{12}}+{\frac{ \left ( A{c}^{3}+3\,Bb{c}^{2} \right ){x}^{10}}{10}}+{\frac{ \left ( 3\,Ab{c}^{2}+3\,B{b}^{2}c \right ){x}^{8}}{8}}+{\frac{ \left ( 3\,A{b}^{2}c+B{b}^{3} \right ){x}^{6}}{6}}+{\frac{A{b}^{3}{x}^{4}}{4}} \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^3,x)

[Out]

1/12*B*c^3*x^12+1/10*(A*c^3+3*B*b*c^2)*x^10+1/8*(3*A*b*c^2+3*B*b^2*c)*x^8+1/6*(3*A*b^2*c+B*b^3)*x^6+1/4*A*b^3*
x^4

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Maxima [A]  time = 1.17759, size = 99, normalized size = 1.46 \begin{align*} \frac{1}{12} \, B c^{3} x^{12} + \frac{1}{10} \,{\left (3 \, B b c^{2} + A c^{3}\right )} x^{10} + \frac{3}{8} \,{\left (B b^{2} c + A b c^{2}\right )} x^{8} + \frac{1}{4} \, A b^{3} x^{4} + \frac{1}{6} \,{\left (B b^{3} + 3 \, A b^{2} c\right )} x^{6} \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^3,x, algorithm="maxima")

[Out]

1/12*B*c^3*x^12 + 1/10*(3*B*b*c^2 + A*c^3)*x^10 + 3/8*(B*b^2*c + A*b*c^2)*x^8 + 1/4*A*b^3*x^4 + 1/6*(B*b^3 + 3
*A*b^2*c)*x^6

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Fricas [A]  time = 0.468432, size = 169, normalized size = 2.49 \begin{align*} \frac{1}{12} \, B c^{3} x^{12} + \frac{1}{10} \,{\left (3 \, B b c^{2} + A c^{3}\right )} x^{10} + \frac{3}{8} \,{\left (B b^{2} c + A b c^{2}\right )} x^{8} + \frac{1}{4} \, A b^{3} x^{4} + \frac{1}{6} \,{\left (B b^{3} + 3 \, A b^{2} c\right )} x^{6} \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^3,x, algorithm="fricas")

[Out]

1/12*B*c^3*x^12 + 1/10*(3*B*b*c^2 + A*c^3)*x^10 + 3/8*(B*b^2*c + A*b*c^2)*x^8 + 1/4*A*b^3*x^4 + 1/6*(B*b^3 + 3
*A*b^2*c)*x^6

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Sympy [A]  time = 0.075238, size = 82, normalized size = 1.21 \begin{align*} \frac{A b^{3} x^{4}}{4} + \frac{B c^{3} x^{12}}{12} + x^{10} \left (\frac{A c^{3}}{10} + \frac{3 B b c^{2}}{10}\right ) + x^{8} \left (\frac{3 A b c^{2}}{8} + \frac{3 B b^{2} c}{8}\right ) + x^{6} \left (\frac{A b^{2} c}{2} + \frac{B b^{3}}{6}\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**3,x)

[Out]

A*b**3*x**4/4 + B*c**3*x**12/12 + x**10*(A*c**3/10 + 3*B*b*c**2/10) + x**8*(3*A*b*c**2/8 + 3*B*b**2*c/8) + x**
6*(A*b**2*c/2 + B*b**3/6)

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Giac [A]  time = 1.22454, size = 104, normalized size = 1.53 \begin{align*} \frac{1}{12} \, B c^{3} x^{12} + \frac{3}{10} \, B b c^{2} x^{10} + \frac{1}{10} \, A c^{3} x^{10} + \frac{3}{8} \, B b^{2} c x^{8} + \frac{3}{8} \, A b c^{2} x^{8} + \frac{1}{6} \, B b^{3} x^{6} + \frac{1}{2} \, A b^{2} c x^{6} + \frac{1}{4} \, A b^{3} 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^3,x, algorithm="giac")

[Out]

1/12*B*c^3*x^12 + 3/10*B*b*c^2*x^10 + 1/10*A*c^3*x^10 + 3/8*B*b^2*c*x^8 + 3/8*A*b*c^2*x^8 + 1/6*B*b^3*x^6 + 1/
2*A*b^2*c*x^6 + 1/4*A*b^3*x^4