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

Optimal. Leaf size=53 \[ -\frac {a^2 A}{x}+\frac {1}{5} b x^5 (2 a B+A b)+\frac {1}{2} a x^2 (a B+2 A b)+\frac {1}{8} b^2 B x^8 \]

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Rubi [A]  time = 0.03, antiderivative size = 53, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.050, Rules used = {448} \begin {gather*} -\frac {a^2 A}{x}+\frac {1}{5} b x^5 (2 a B+A b)+\frac {1}{2} a x^2 (a B+2 A b)+\frac {1}{8} b^2 B x^8 \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((a^2*A)/x) + (a*(2*A*b + a*B)*x^2)/2 + (b*(A*b + 2*a*B)*x^5)/5 + (b^2*B*x^8)/8

Rule 448

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

Rubi steps

\begin {align*} \int \frac {\left (a+b x^3\right )^2 \left (A+B x^3\right )}{x^2} \, dx &=\int \left (\frac {a^2 A}{x^2}+a (2 A b+a B) x+b (A b+2 a B) x^4+b^2 B x^7\right ) \, dx\\ &=-\frac {a^2 A}{x}+\frac {1}{2} a (2 A b+a B) x^2+\frac {1}{5} b (A b+2 a B) x^5+\frac {1}{8} b^2 B x^8\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 53, normalized size = 1.00 \begin {gather*} -\frac {a^2 A}{x}+\frac {1}{5} b x^5 (2 a B+A b)+\frac {1}{2} a x^2 (a B+2 A b)+\frac {1}{8} b^2 B x^8 \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((a^2*A)/x) + (a*(2*A*b + a*B)*x^2)/2 + (b*(A*b + 2*a*B)*x^5)/5 + (b^2*B*x^8)/8

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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^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 0.74, size = 53, normalized size = 1.00 \begin {gather*} \frac {5 \, B b^{2} x^{9} + 8 \, {\left (2 \, B a b + A b^{2}\right )} x^{6} + 20 \, {\left (B a^{2} + 2 \, A a b\right )} x^{3} - 40 \, A a^{2}}{40 \, 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^2,x, algorithm="fricas")

[Out]

1/40*(5*B*b^2*x^9 + 8*(2*B*a*b + A*b^2)*x^6 + 20*(B*a^2 + 2*A*a*b)*x^3 - 40*A*a^2)/x

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giac [A]  time = 0.15, size = 52, normalized size = 0.98 \begin {gather*} \frac {1}{8} \, B b^{2} x^{8} + \frac {2}{5} \, B a b x^{5} + \frac {1}{5} \, A b^{2} x^{5} + \frac {1}{2} \, B a^{2} x^{2} + A a b x^{2} - \frac {A a^{2}}{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^2,x, algorithm="giac")

[Out]

1/8*B*b^2*x^8 + 2/5*B*a*b*x^5 + 1/5*A*b^2*x^5 + 1/2*B*a^2*x^2 + A*a*b*x^2 - A*a^2/x

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maple [A]  time = 0.04, size = 53, normalized size = 1.00 \begin {gather*} \frac {B \,b^{2} x^{8}}{8}+\frac {A \,b^{2} x^{5}}{5}+\frac {2 B a b \,x^{5}}{5}+A a b \,x^{2}+\frac {B \,a^{2} x^{2}}{2}-\frac {A \,a^{2}}{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^2,x)

[Out]

1/8*b^2*B*x^8+1/5*A*x^5*b^2+2/5*B*x^5*a*b+A*a*b*x^2+1/2*B*a^2*x^2-A*a^2/x

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maxima [A]  time = 0.59, size = 51, normalized size = 0.96 \begin {gather*} \frac {1}{8} \, B b^{2} x^{8} + \frac {1}{5} \, {\left (2 \, B a b + A b^{2}\right )} x^{5} + \frac {1}{2} \, {\left (B a^{2} + 2 \, A a b\right )} x^{2} - \frac {A a^{2}}{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^2,x, algorithm="maxima")

[Out]

1/8*B*b^2*x^8 + 1/5*(2*B*a*b + A*b^2)*x^5 + 1/2*(B*a^2 + 2*A*a*b)*x^2 - A*a^2/x

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mupad [B]  time = 0.05, size = 50, normalized size = 0.94 \begin {gather*} x^2\,\left (\frac {B\,a^2}{2}+A\,b\,a\right )+x^5\,\left (\frac {A\,b^2}{5}+\frac {2\,B\,a\,b}{5}\right )-\frac {A\,a^2}{x}+\frac {B\,b^2\,x^8}{8} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x^2*((B*a^2)/2 + A*a*b) + x^5*((A*b^2)/5 + (2*B*a*b)/5) - (A*a^2)/x + (B*b^2*x^8)/8

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sympy [A]  time = 0.15, size = 49, normalized size = 0.92 \begin {gather*} - \frac {A a^{2}}{x} + \frac {B b^{2} x^{8}}{8} + x^{5} \left (\frac {A b^{2}}{5} + \frac {2 B a b}{5}\right ) + x^{2} \left (A a b + \frac {B a^{2}}{2}\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**2,x)

[Out]

-A*a**2/x + B*b**2*x**8/8 + x**5*(A*b**2/5 + 2*B*a*b/5) + x**2*(A*a*b + B*a**2/2)

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