3.19 \(\int \frac{(a+b x^2)^2 (A+B x^2)}{x^6} \, dx\)

Optimal. Leaf size=48 \[ -\frac{a^2 A}{5 x^5}-\frac{a (a B+2 A b)}{3 x^3}-\frac{b (2 a B+A b)}{x}+b^2 B x \]

[Out]

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

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Rubi [A]  time = 0.0280537, antiderivative size = 48, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 1, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.05, Rules used = {448} \[ -\frac{a^2 A}{5 x^5}-\frac{a (a B+2 A b)}{3 x^3}-\frac{b (2 a B+A b)}{x}+b^2 B x \]

Antiderivative was successfully verified.

[In]

Int[((a + b*x^2)^2*(A + B*x^2))/x^6,x]

[Out]

-(a^2*A)/(5*x^5) - (a*(2*A*b + a*B))/(3*x^3) - (b*(A*b + 2*a*B))/x + 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^2\right )^2 \left (A+B x^2\right )}{x^6} \, dx &=\int \left (b^2 B+\frac{a^2 A}{x^6}+\frac{a (2 A b+a B)}{x^4}+\frac{b (A b+2 a B)}{x^2}\right ) \, dx\\ &=-\frac{a^2 A}{5 x^5}-\frac{a (2 A b+a B)}{3 x^3}-\frac{b (A b+2 a B)}{x}+b^2 B x\\ \end{align*}

Mathematica [A]  time = 0.0192372, size = 48, normalized size = 1. \[ -\frac{a^2 A}{5 x^5}-\frac{a (a B+2 A b)}{3 x^3}-\frac{b (2 a B+A b)}{x}+b^2 B x \]

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x^2)^2*(A + B*x^2))/x^6,x]

[Out]

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

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Maple [A]  time = 0.005, size = 45, normalized size = 0.9 \begin{align*} -{\frac{A{a}^{2}}{5\,{x}^{5}}}-{\frac{a \left ( 2\,Ab+Ba \right ) }{3\,{x}^{3}}}-{\frac{b \left ( Ab+2\,Ba \right ) }{x}}+{b}^{2}Bx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2+a)^2*(B*x^2+A)/x^6,x)

[Out]

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

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Maxima [A]  time = 0.985484, size = 69, normalized size = 1.44 \begin{align*} B b^{2} x - \frac{15 \,{\left (2 \, B a b + A b^{2}\right )} x^{4} + 3 \, A a^{2} + 5 \,{\left (B a^{2} + 2 \, A a b\right )} x^{2}}{15 \, x^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^2*(B*x^2+A)/x^6,x, algorithm="maxima")

[Out]

B*b^2*x - 1/15*(15*(2*B*a*b + A*b^2)*x^4 + 3*A*a^2 + 5*(B*a^2 + 2*A*a*b)*x^2)/x^5

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Fricas [A]  time = 1.39845, size = 119, normalized size = 2.48 \begin{align*} \frac{15 \, B b^{2} x^{6} - 15 \,{\left (2 \, B a b + A b^{2}\right )} x^{4} - 3 \, A a^{2} - 5 \,{\left (B a^{2} + 2 \, A a b\right )} x^{2}}{15 \, x^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^2*(B*x^2+A)/x^6,x, algorithm="fricas")

[Out]

1/15*(15*B*b^2*x^6 - 15*(2*B*a*b + A*b^2)*x^4 - 3*A*a^2 - 5*(B*a^2 + 2*A*a*b)*x^2)/x^5

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Sympy [A]  time = 0.70299, size = 51, normalized size = 1.06 \begin{align*} B b^{2} x - \frac{3 A a^{2} + x^{4} \left (15 A b^{2} + 30 B a b\right ) + x^{2} \left (10 A a b + 5 B a^{2}\right )}{15 x^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**2+a)**2*(B*x**2+A)/x**6,x)

[Out]

B*b**2*x - (3*A*a**2 + x**4*(15*A*b**2 + 30*B*a*b) + x**2*(10*A*a*b + 5*B*a**2))/(15*x**5)

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Giac [A]  time = 1.13707, size = 72, normalized size = 1.5 \begin{align*} B b^{2} x - \frac{30 \, B a b x^{4} + 15 \, A b^{2} x^{4} + 5 \, B a^{2} x^{2} + 10 \, A a b x^{2} + 3 \, A a^{2}}{15 \, x^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^2*(B*x^2+A)/x^6,x, algorithm="giac")

[Out]

B*b^2*x - 1/15*(30*B*a*b*x^4 + 15*A*b^2*x^4 + 5*B*a^2*x^2 + 10*A*a*b*x^2 + 3*A*a^2)/x^5