3.1.22 \(\int \frac {(a+b x^3)^2 (A+B x^3)}{x^9} \, dx\)

Optimal. Leaf size=50 \[ -\frac {a^2 A}{8 x^8}-\frac {a (a B+2 A b)}{5 x^5}-\frac {b (2 a B+A b)}{2 x^2}+b^2 B x \]

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Rubi [A]  time = 0.03, antiderivative size = 50, 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}{8 x^8}-\frac {a (a B+2 A b)}{5 x^5}-\frac {b (2 a B+A b)}{2 x^2}+b^2 B x \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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^9} \, dx &=\int \left (b^2 B+\frac {a^2 A}{x^9}+\frac {a (2 A b+a B)}{x^6}+\frac {b (A b+2 a B)}{x^3}\right ) \, dx\\ &=-\frac {a^2 A}{8 x^8}-\frac {a (2 A b+a B)}{5 x^5}-\frac {b (A b+2 a B)}{2 x^2}+b^2 B x\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 50, normalized size = 1.00 \begin {gather*} -\frac {a^2 A}{8 x^8}-\frac {a (a B+2 A b)}{5 x^5}-\frac {b (2 a B+A b)}{2 x^2}+b^2 B x \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/8*(a^2*A)/x^8 - (a*(2*A*b + a*B))/(5*x^5) - (b*(A*b + 2*a*B))/(2*x^2) + b^2*B*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^9} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 0.83, size = 53, normalized size = 1.06 \begin {gather*} \frac {40 \, B b^{2} x^{9} - 20 \, {\left (2 \, B a b + A b^{2}\right )} x^{6} - 8 \, {\left (B a^{2} + 2 \, A a b\right )} x^{3} - 5 \, A a^{2}}{40 \, x^{8}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a)^2*(B*x^3+A)/x^9,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.15, size = 53, normalized size = 1.06 \begin {gather*} B b^{2} x - \frac {40 \, B a b x^{6} + 20 \, A b^{2} x^{6} + 8 \, B a^{2} x^{3} + 16 \, A a b x^{3} + 5 \, A a^{2}}{40 \, x^{8}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a)^2*(B*x^3+A)/x^9,x, algorithm="giac")

[Out]

B*b^2*x - 1/40*(40*B*a*b*x^6 + 20*A*b^2*x^6 + 8*B*a^2*x^3 + 16*A*a*b*x^3 + 5*A*a^2)/x^8

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maple [A]  time = 0.05, size = 45, normalized size = 0.90 \begin {gather*} B \,b^{2} x -\frac {\left (A b +2 B a \right ) b}{2 x^{2}}-\frac {\left (2 A b +B a \right ) a}{5 x^{5}}-\frac {A \,a^{2}}{8 x^{8}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^3+a)^2*(B*x^3+A)/x^9,x)

[Out]

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

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maxima [A]  time = 0.49, size = 51, normalized size = 1.02 \begin {gather*} B b^{2} x - \frac {20 \, {\left (2 \, B a b + A b^{2}\right )} x^{6} + 8 \, {\left (B a^{2} + 2 \, A a b\right )} x^{3} + 5 \, A a^{2}}{40 \, x^{8}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a)^2*(B*x^3+A)/x^9,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 2.34, size = 50, normalized size = 1.00 \begin {gather*} B\,b^2\,x-\frac {x^3\,\left (\frac {B\,a^2}{5}+\frac {2\,A\,b\,a}{5}\right )+x^6\,\left (\frac {A\,b^2}{2}+B\,a\,b\right )+\frac {A\,a^2}{8}}{x^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^9,x)

[Out]

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

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sympy [A]  time = 0.99, size = 54, normalized size = 1.08 \begin {gather*} B b^{2} x + \frac {- 5 A a^{2} + x^{6} \left (- 20 A b^{2} - 40 B a b\right ) + x^{3} \left (- 16 A a b - 8 B a^{2}\right )}{40 x^{8}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**3+a)**2*(B*x**3+A)/x**9,x)

[Out]

B*b**2*x + (-5*A*a**2 + x**6*(-20*A*b**2 - 40*B*a*b) + x**3*(-16*A*a*b - 8*B*a**2))/(40*x**8)

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