3.1.21 \(\int \frac {(a+b x^3)^2 (A+B x^3)}{x^8} \, dx\)

Optimal. Leaf size=53 \[ -\frac {a^2 A}{7 x^7}-\frac {a (a B+2 A b)}{4 x^4}-\frac {b (2 a B+A b)}{x}+\frac {1}{2} b^2 B x^2 \]

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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}{7 x^7}-\frac {a (a B+2 A b)}{4 x^4}-\frac {b (2 a B+A b)}{x}+\frac {1}{2} b^2 B x^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((a + b*x^3)^2*(A + B*x^3))/x^8,x]

[Out]

-(a^2*A)/(7*x^7) - (a*(2*A*b + a*B))/(4*x^4) - (b*(A*b + 2*a*B))/x + (b^2*B*x^2)/2

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

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Mathematica [A]  time = 0.02, size = 54, normalized size = 1.02 \begin {gather*} -\frac {a^2 \left (4 A+7 B x^3\right )+14 a b x^3 \left (A+4 B x^3\right )-14 b^2 x^6 \left (B x^3-2 A\right )}{28 x^7} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x^3)^2*(A + B*x^3))/x^8,x]

[Out]

-1/28*(-14*b^2*x^6*(-2*A + B*x^3) + 14*a*b*x^3*(A + 4*B*x^3) + a^2*(4*A + 7*B*x^3))/x^7

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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^8} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[((a + b*x^3)^2*(A + B*x^3))/x^8,x]

[Out]

IntegrateAlgebraic[((a + b*x^3)^2*(A + B*x^3))/x^8, x]

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fricas [A]  time = 0.80, size = 53, normalized size = 1.00 \begin {gather*} \frac {14 \, B b^{2} x^{9} - 28 \, {\left (2 \, B a b + A b^{2}\right )} x^{6} - 7 \, {\left (B a^{2} + 2 \, A a b\right )} x^{3} - 4 \, A a^{2}}{28 \, x^{7}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a)^2*(B*x^3+A)/x^8,x, algorithm="fricas")

[Out]

1/28*(14*B*b^2*x^9 - 28*(2*B*a*b + A*b^2)*x^6 - 7*(B*a^2 + 2*A*a*b)*x^3 - 4*A*a^2)/x^7

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giac [A]  time = 0.15, size = 56, normalized size = 1.06 \begin {gather*} \frac {1}{2} \, B b^{2} x^{2} - \frac {56 \, B a b x^{6} + 28 \, A b^{2} x^{6} + 7 \, B a^{2} x^{3} + 14 \, A a b x^{3} + 4 \, A a^{2}}{28 \, x^{7}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a)^2*(B*x^3+A)/x^8,x, algorithm="giac")

[Out]

1/2*B*b^2*x^2 - 1/28*(56*B*a*b*x^6 + 28*A*b^2*x^6 + 7*B*a^2*x^3 + 14*A*a*b*x^3 + 4*A*a^2)/x^7

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maple [A]  time = 0.04, size = 48, normalized size = 0.91 \begin {gather*} \frac {B \,b^{2} x^{2}}{2}-\frac {\left (A b +2 B a \right ) b}{x}-\frac {\left (2 A b +B a \right ) a}{4 x^{4}}-\frac {A \,a^{2}}{7 x^{7}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^3+a)^2*(B*x^3+A)/x^8,x)

[Out]

-1/7*a^2*A/x^7-1/4*a*(2*A*b+B*a)/x^4-b*(A*b+2*B*a)/x+1/2*b^2*B*x^2

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maxima [A]  time = 0.57, size = 54, normalized size = 1.02 \begin {gather*} \frac {1}{2} \, B b^{2} x^{2} - \frac {28 \, {\left (2 \, B a b + A b^{2}\right )} x^{6} + 7 \, {\left (B a^{2} + 2 \, A a b\right )} x^{3} + 4 \, A a^{2}}{28 \, x^{7}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a)^2*(B*x^3+A)/x^8,x, algorithm="maxima")

[Out]

1/2*B*b^2*x^2 - 1/28*(28*(2*B*a*b + A*b^2)*x^6 + 7*(B*a^2 + 2*A*a*b)*x^3 + 4*A*a^2)/x^7

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mupad [B]  time = 0.05, size = 53, normalized size = 1.00 \begin {gather*} \frac {B\,b^2\,x^2}{2}-\frac {x^3\,\left (\frac {B\,a^2}{4}+\frac {A\,b\,a}{2}\right )+x^6\,\left (A\,b^2+2\,B\,a\,b\right )+\frac {A\,a^2}{7}}{x^7} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*x^3)*(a + b*x^3)^2)/x^8,x)

[Out]

(B*b^2*x^2)/2 - (x^3*((B*a^2)/4 + (A*a*b)/2) + x^6*(A*b^2 + 2*B*a*b) + (A*a^2)/7)/x^7

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sympy [A]  time = 0.87, size = 58, normalized size = 1.09 \begin {gather*} \frac {B b^{2} x^{2}}{2} + \frac {- 4 A a^{2} + x^{6} \left (- 28 A b^{2} - 56 B a b\right ) + x^{3} \left (- 14 A a b - 7 B a^{2}\right )}{28 x^{7}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**3+a)**2*(B*x**3+A)/x**8,x)

[Out]

B*b**2*x**2/2 + (-4*A*a**2 + x**6*(-28*A*b**2 - 56*B*a*b) + x**3*(-14*A*a*b - 7*B*a**2))/(28*x**7)

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