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

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

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Rubi [A]  time = 0.03, antiderivative size = 46, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {446, 80, 43} \begin {gather*} 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 \end {gather*}

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 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])

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 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]]

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*}

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Mathematica [A]  time = 0.04, size = 51, normalized size = 1.11 \begin {gather*} 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 \end {gather*}

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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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (a+b x^3\right )^2 \left (A+B x^3\right )}{x} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 0.77, size = 49, normalized size = 1.07 \begin {gather*} \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 \relax (x) \end {gather*}

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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giac [A]  time = 0.15, size = 52, normalized size = 1.13 \begin {gather*} \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 {gather*}

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))

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maple [A]  time = 0.04, size = 52, normalized size = 1.13 \begin {gather*} \frac {B \,b^{2} x^{9}}{9}+\frac {A \,b^{2} x^{6}}{6}+\frac {B a b \,x^{6}}{3}+\frac {2 A a b \,x^{3}}{3}+\frac {B \,a^{2} x^{3}}{3}+A \,a^{2} \ln \relax (x ) \end {gather*}

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*a^2*x^3+A*a^2*ln(x)

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maxima [A]  time = 0.54, size = 52, normalized size = 1.13 \begin {gather*} \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 {gather*}

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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mupad [B]  time = 0.04, size = 49, normalized size = 1.07 \begin {gather*} x^3\,\left (\frac {B\,a^2}{3}+\frac {2\,A\,b\,a}{3}\right )+x^6\,\left (\frac {A\,b^2}{6}+\frac {B\,a\,b}{3}\right )+\frac {B\,b^2\,x^9}{9}+A\,a^2\,\ln \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*x^3)*(a + b*x^3)^2)/x,x)

[Out]

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

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sympy [A]  time = 0.15, size = 53, normalized size = 1.15 \begin {gather*} A a^{2} \log {\relax (x )} + \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 {gather*}

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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