3.14 \(\int \frac{(a+b x^3)^2 (A+B x^3)}{x} \, dx\)

Optimal. Leaf size=46 \[ a^2 A \log (x)+\frac{2}{3} a A b x^3+\frac{B \left (a+b x^3\right )^3}{9 b}+\frac{1}{6} A b^2 x^6 \]

[Out]

(2*a*A*b*x^3)/3 + (A*b^2*x^6)/6 + (B*(a + b*x^3)^3)/(9*b) + a^2*A*Log[x]

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Rubi [A]  time = 0.0326931, antiderivative size = 46, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.15, Rules used = {446, 80, 43} \[ a^2 A \log (x)+\frac{2}{3} a A b x^3+\frac{B \left (a+b x^3\right )^3}{9 b}+\frac{1}{6} A b^2 x^6 \]

Antiderivative was successfully verified.

[In]

Int[((a + b*x^3)^2*(A + B*x^3))/x,x]

[Out]

(2*a*A*b*x^3)/3 + (A*b^2*x^6)/6 + (B*(a + b*x^3)^3)/(9*b) + a^2*A*Log[x]

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 80

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(c + d*x)
^(n + 1)*(e + f*x)^(p + 1))/(d*f*(n + p + 2)), x] + Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(
d*f*(n + p + 2)), Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2,
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^3\right )^2 \left (A+B x^3\right )}{x} \, dx &=\frac{1}{3} \operatorname{Subst}\left (\int \frac{(a+b x)^2 (A+B x)}{x} \, dx,x,x^3\right )\\ &=\frac{B \left (a+b x^3\right )^3}{9 b}+\frac{1}{3} A \operatorname{Subst}\left (\int \frac{(a+b x)^2}{x} \, dx,x,x^3\right )\\ &=\frac{B \left (a+b x^3\right )^3}{9 b}+\frac{1}{3} A \operatorname{Subst}\left (\int \left (2 a b+\frac{a^2}{x}+b^2 x\right ) \, dx,x,x^3\right )\\ &=\frac{2}{3} a A b x^3+\frac{1}{6} A b^2 x^6+\frac{B \left (a+b x^3\right )^3}{9 b}+a^2 A \log (x)\\ \end{align*}

Mathematica [A]  time = 0.0142697, size = 51, normalized size = 1.11 \[ a^2 A \log (x)+\frac{1}{6} b x^6 (2 a B+A b)+\frac{1}{3} a x^3 (a B+2 A b)+\frac{1}{9} b^2 B x^9 \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*(2*A*b + a*B)*x^3)/3 + (b*(A*b + 2*a*B)*x^6)/6 + (b^2*B*x^9)/9 + a^2*A*Log[x]

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Maple [A]  time = 0.001, size = 52, normalized size = 1.1 \begin{align*}{\frac{B{b}^{2}{x}^{9}}{9}}+{\frac{A{b}^{2}{x}^{6}}{6}}+{\frac{B{x}^{6}ab}{3}}+{\frac{2\,aAb{x}^{3}}{3}}+{\frac{B{x}^{3}{a}^{2}}{3}}+{a}^{2}A\ln \left ( x \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^3+a)^2*(B*x^3+A)/x,x)

[Out]

1/9*B*b^2*x^9+1/6*A*b^2*x^6+1/3*B*x^6*a*b+2/3*a*A*b*x^3+1/3*B*x^3*a^2+a^2*A*ln(x)

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Maxima [A]  time = 1.21736, size = 70, normalized size = 1.52 \begin{align*} \frac{1}{9} \, B b^{2} x^{9} + \frac{1}{6} \,{\left (2 \, B a b + A b^{2}\right )} x^{6} + \frac{1}{3} \,{\left (B a^{2} + 2 \, A a b\right )} x^{3} + \frac{1}{3} \, A a^{2} \log \left (x^{3}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/9*B*b^2*x^9 + 1/6*(2*B*a*b + A*b^2)*x^6 + 1/3*(B*a^2 + 2*A*a*b)*x^3 + 1/3*A*a^2*log(x^3)

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Fricas [A]  time = 1.44867, size = 116, normalized size = 2.52 \begin{align*} \frac{1}{9} \, B b^{2} x^{9} + \frac{1}{6} \,{\left (2 \, B a b + A b^{2}\right )} x^{6} + \frac{1}{3} \,{\left (B a^{2} + 2 \, A a b\right )} x^{3} + A a^{2} \log \left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/9*B*b^2*x^9 + 1/6*(2*B*a*b + A*b^2)*x^6 + 1/3*(B*a^2 + 2*A*a*b)*x^3 + A*a^2*log(x)

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Sympy [A]  time = 0.282266, size = 53, normalized size = 1.15 \begin{align*} A a^{2} \log{\left (x \right )} + \frac{B b^{2} x^{9}}{9} + x^{6} \left (\frac{A b^{2}}{6} + \frac{B a b}{3}\right ) + x^{3} \left (\frac{2 A a b}{3} + \frac{B a^{2}}{3}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**3+a)**2*(B*x**3+A)/x,x)

[Out]

A*a**2*log(x) + B*b**2*x**9/9 + x**6*(A*b**2/6 + B*a*b/3) + x**3*(2*A*a*b/3 + B*a**2/3)

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Giac [A]  time = 1.11482, size = 70, normalized size = 1.52 \begin{align*} \frac{1}{9} \, B b^{2} x^{9} + \frac{1}{3} \, B a b x^{6} + \frac{1}{6} \, A b^{2} x^{6} + \frac{1}{3} \, B a^{2} x^{3} + \frac{2}{3} \, A a b x^{3} + A a^{2} \log \left ({\left | x \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/9*B*b^2*x^9 + 1/3*B*a*b*x^6 + 1/6*A*b^2*x^6 + 1/3*B*a^2*x^3 + 2/3*A*a*b*x^3 + A*a^2*log(abs(x))