3.17 \(\int \frac{(a+b x^2)^2 (A+B x^2)}{x^4} \, dx\)

Optimal. Leaf size=48 \[ -\frac{a^2 A}{3 x^3}+b x (2 a B+A b)-\frac{a (a B+2 A b)}{x}+\frac{1}{3} b^2 B x^3 \]

[Out]

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

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Rubi [A]  time = 0.029933, 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} \[ -\frac{a^2 A}{3 x^3}+b x (2 a B+A b)-\frac{a (a B+2 A b)}{x}+\frac{1}{3} b^2 B x^3 \]

Antiderivative was successfully verified.

[In]

Int[((a + b*x^2)^2*(A + B*x^2))/x^4,x]

[Out]

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

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

Mathematica [A]  time = 0.0177868, size = 50, normalized size = 1.04 \[ \frac{a^2 (-B)-2 a A b}{x}-\frac{a^2 A}{3 x^3}+b x (2 a B+A b)+\frac{1}{3} b^2 B x^3 \]

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x^2)^2*(A + B*x^2))/x^4,x]

[Out]

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

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Maple [A]  time = 0.004, size = 46, normalized size = 1. \begin{align*}{\frac{{b}^{2}B{x}^{3}}{3}}+{b}^{2}Ax+2\,abBx-{\frac{A{a}^{2}}{3\,{x}^{3}}}-{\frac{a \left ( 2\,Ab+Ba \right ) }{x}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2+a)^2*(B*x^2+A)/x^4,x)

[Out]

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

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Maxima [A]  time = 0.96608, size = 68, normalized size = 1.42 \begin{align*} \frac{1}{3} \, B b^{2} x^{3} +{\left (2 \, B a b + A b^{2}\right )} x - \frac{A a^{2} + 3 \,{\left (B a^{2} + 2 \, A a b\right )} x^{2}}{3 \, x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^2*(B*x^2+A)/x^4,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.42695, size = 109, normalized size = 2.27 \begin{align*} \frac{B b^{2} x^{6} + 3 \,{\left (2 \, B a b + A b^{2}\right )} x^{4} - A a^{2} - 3 \,{\left (B a^{2} + 2 \, A a b\right )} x^{2}}{3 \, x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^2*(B*x^2+A)/x^4,x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 0.379725, size = 49, normalized size = 1.02 \begin{align*} \frac{B b^{2} x^{3}}{3} + x \left (A b^{2} + 2 B a b\right ) - \frac{A a^{2} + x^{2} \left (6 A a b + 3 B a^{2}\right )}{3 x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**2+a)**2*(B*x**2+A)/x**4,x)

[Out]

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

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Giac [A]  time = 1.12623, size = 68, normalized size = 1.42 \begin{align*} \frac{1}{3} \, B b^{2} x^{3} + 2 \, B a b x + A b^{2} x - \frac{3 \, B a^{2} x^{2} + 6 \, A a b x^{2} + A a^{2}}{3 \, x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^2*(B*x^2+A)/x^4,x, algorithm="giac")

[Out]

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