3.186 \(\int \frac{(a+b x^2)^2}{x^2 (c+d x^2)^2} \, dx\)

Optimal. Leaf size=106 \[ -\frac{x \left (3 a^2 d^2-2 a b c d+b^2 c^2\right )}{2 c^2 d \left (c+d x^2\right )}-\frac{a^2}{c x \left (c+d x^2\right )}+\frac{(b c-a d) (3 a d+b c) \tan ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c}}\right )}{2 c^{5/2} d^{3/2}} \]

[Out]

-(a^2/(c*x*(c + d*x^2))) - ((b^2*c^2 - 2*a*b*c*d + 3*a^2*d^2)*x)/(2*c^2*d*(c + d*x^2)) + ((b*c - a*d)*(b*c + 3
*a*d)*ArcTan[(Sqrt[d]*x)/Sqrt[c]])/(2*c^(5/2)*d^(3/2))

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Rubi [A]  time = 0.0756784, antiderivative size = 106, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.136, Rules used = {462, 385, 205} \[ -\frac{x \left (3 a^2 d^2-2 a b c d+b^2 c^2\right )}{2 c^2 d \left (c+d x^2\right )}-\frac{a^2}{c x \left (c+d x^2\right )}+\frac{(b c-a d) (3 a d+b c) \tan ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c}}\right )}{2 c^{5/2} d^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2)^2/(x^2*(c + d*x^2)^2),x]

[Out]

-(a^2/(c*x*(c + d*x^2))) - ((b^2*c^2 - 2*a*b*c*d + 3*a^2*d^2)*x)/(2*c^2*d*(c + d*x^2)) + ((b*c - a*d)*(b*c + 3
*a*d)*ArcTan[(Sqrt[d]*x)/Sqrt[c]])/(2*c^(5/2)*d^(3/2))

Rule 462

Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^2, x_Symbol] :> Simp[(c^2*(e*x)^(
m + 1)*(a + b*x^n)^(p + 1))/(a*e*(m + 1)), x] - Dist[1/(a*e^n*(m + 1)), Int[(e*x)^(m + n)*(a + b*x^n)^p*Simp[b
*c^2*n*(p + 1) + c*(b*c - 2*a*d)*(m + 1) - a*(m + 1)*d^2*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, p}, x] && Ne
Q[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[m, -1] && GtQ[n, 0]

Rule 385

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> -Simp[((b*c - a*d)*x*(a + b*x^n)^(p +
1))/(a*b*n*(p + 1)), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(a*b*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /
; FreeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && (LtQ[p, -1] || ILtQ[1/n + p, 0])

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rubi steps

\begin{align*} \int \frac{\left (a+b x^2\right )^2}{x^2 \left (c+d x^2\right )^2} \, dx &=-\frac{a^2}{c x \left (c+d x^2\right )}+\frac{\int \frac{a (2 b c-3 a d)+b^2 c x^2}{\left (c+d x^2\right )^2} \, dx}{c}\\ &=-\frac{a^2}{c x \left (c+d x^2\right )}-\frac{\left (b^2 c^2-2 a b c d+3 a^2 d^2\right ) x}{2 c^2 d \left (c+d x^2\right )}+\frac{((b c-a d) (b c+3 a d)) \int \frac{1}{c+d x^2} \, dx}{2 c^2 d}\\ &=-\frac{a^2}{c x \left (c+d x^2\right )}-\frac{\left (b^2 c^2-2 a b c d+3 a^2 d^2\right ) x}{2 c^2 d \left (c+d x^2\right )}+\frac{(b c-a d) (b c+3 a d) \tan ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c}}\right )}{2 c^{5/2} d^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.0601066, size = 91, normalized size = 0.86 \[ \frac{\left (-3 a^2 d^2+2 a b c d+b^2 c^2\right ) \tan ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c}}\right )}{2 c^{5/2} d^{3/2}}-\frac{a^2}{c^2 x}-\frac{x (b c-a d)^2}{2 c^2 d \left (c+d x^2\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^2)^2/(x^2*(c + d*x^2)^2),x]

[Out]

-(a^2/(c^2*x)) - ((b*c - a*d)^2*x)/(2*c^2*d*(c + d*x^2)) + ((b^2*c^2 + 2*a*b*c*d - 3*a^2*d^2)*ArcTan[(Sqrt[d]*
x)/Sqrt[c]])/(2*c^(5/2)*d^(3/2))

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Maple [A]  time = 0.009, size = 131, normalized size = 1.2 \begin{align*} -{\frac{{a}^{2}dx}{2\,{c}^{2} \left ( d{x}^{2}+c \right ) }}+{\frac{abx}{c \left ( d{x}^{2}+c \right ) }}-{\frac{x{b}^{2}}{2\,d \left ( d{x}^{2}+c \right ) }}-{\frac{3\,{a}^{2}d}{2\,{c}^{2}}\arctan \left ({dx{\frac{1}{\sqrt{cd}}}} \right ){\frac{1}{\sqrt{cd}}}}+{\frac{ab}{c}\arctan \left ({dx{\frac{1}{\sqrt{cd}}}} \right ){\frac{1}{\sqrt{cd}}}}+{\frac{{b}^{2}}{2\,d}\arctan \left ({dx{\frac{1}{\sqrt{cd}}}} \right ){\frac{1}{\sqrt{cd}}}}-{\frac{{a}^{2}}{{c}^{2}x}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2+a)^2/x^2/(d*x^2+c)^2,x)

[Out]

-1/2/c^2*d*x/(d*x^2+c)*a^2+1/c*x/(d*x^2+c)*a*b-1/2/d*x/(d*x^2+c)*b^2-3/2/c^2*d/(c*d)^(1/2)*arctan(x*d/(c*d)^(1
/2))*a^2+1/c/(c*d)^(1/2)*arctan(x*d/(c*d)^(1/2))*a*b+1/2/d/(c*d)^(1/2)*arctan(x*d/(c*d)^(1/2))*b^2-a^2/c^2/x

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^2/x^2/(d*x^2+c)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.55426, size = 625, normalized size = 5.9 \begin{align*} \left [-\frac{4 \, a^{2} c^{2} d^{2} + 2 \,{\left (b^{2} c^{3} d - 2 \, a b c^{2} d^{2} + 3 \, a^{2} c d^{3}\right )} x^{2} -{\left ({\left (b^{2} c^{2} d + 2 \, a b c d^{2} - 3 \, a^{2} d^{3}\right )} x^{3} +{\left (b^{2} c^{3} + 2 \, a b c^{2} d - 3 \, a^{2} c d^{2}\right )} x\right )} \sqrt{-c d} \log \left (\frac{d x^{2} + 2 \, \sqrt{-c d} x - c}{d x^{2} + c}\right )}{4 \,{\left (c^{3} d^{3} x^{3} + c^{4} d^{2} x\right )}}, -\frac{2 \, a^{2} c^{2} d^{2} +{\left (b^{2} c^{3} d - 2 \, a b c^{2} d^{2} + 3 \, a^{2} c d^{3}\right )} x^{2} -{\left ({\left (b^{2} c^{2} d + 2 \, a b c d^{2} - 3 \, a^{2} d^{3}\right )} x^{3} +{\left (b^{2} c^{3} + 2 \, a b c^{2} d - 3 \, a^{2} c d^{2}\right )} x\right )} \sqrt{c d} \arctan \left (\frac{\sqrt{c d} x}{c}\right )}{2 \,{\left (c^{3} d^{3} x^{3} + c^{4} d^{2} x\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^2/x^2/(d*x^2+c)^2,x, algorithm="fricas")

[Out]

[-1/4*(4*a^2*c^2*d^2 + 2*(b^2*c^3*d - 2*a*b*c^2*d^2 + 3*a^2*c*d^3)*x^2 - ((b^2*c^2*d + 2*a*b*c*d^2 - 3*a^2*d^3
)*x^3 + (b^2*c^3 + 2*a*b*c^2*d - 3*a^2*c*d^2)*x)*sqrt(-c*d)*log((d*x^2 + 2*sqrt(-c*d)*x - c)/(d*x^2 + c)))/(c^
3*d^3*x^3 + c^4*d^2*x), -1/2*(2*a^2*c^2*d^2 + (b^2*c^3*d - 2*a*b*c^2*d^2 + 3*a^2*c*d^3)*x^2 - ((b^2*c^2*d + 2*
a*b*c*d^2 - 3*a^2*d^3)*x^3 + (b^2*c^3 + 2*a*b*c^2*d - 3*a^2*c*d^2)*x)*sqrt(c*d)*arctan(sqrt(c*d)*x/c))/(c^3*d^
3*x^3 + c^4*d^2*x)]

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Sympy [B]  time = 1.01355, size = 238, normalized size = 2.25 \begin{align*} \frac{\sqrt{- \frac{1}{c^{5} d^{3}}} \left (a d - b c\right ) \left (3 a d + b c\right ) \log{\left (- \frac{c^{3} d \sqrt{- \frac{1}{c^{5} d^{3}}} \left (a d - b c\right ) \left (3 a d + b c\right )}{3 a^{2} d^{2} - 2 a b c d - b^{2} c^{2}} + x \right )}}{4} - \frac{\sqrt{- \frac{1}{c^{5} d^{3}}} \left (a d - b c\right ) \left (3 a d + b c\right ) \log{\left (\frac{c^{3} d \sqrt{- \frac{1}{c^{5} d^{3}}} \left (a d - b c\right ) \left (3 a d + b c\right )}{3 a^{2} d^{2} - 2 a b c d - b^{2} c^{2}} + x \right )}}{4} - \frac{2 a^{2} c d + x^{2} \left (3 a^{2} d^{2} - 2 a b c d + b^{2} c^{2}\right )}{2 c^{3} d x + 2 c^{2} d^{2} x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**2+a)**2/x**2/(d*x**2+c)**2,x)

[Out]

sqrt(-1/(c**5*d**3))*(a*d - b*c)*(3*a*d + b*c)*log(-c**3*d*sqrt(-1/(c**5*d**3))*(a*d - b*c)*(3*a*d + b*c)/(3*a
**2*d**2 - 2*a*b*c*d - b**2*c**2) + x)/4 - sqrt(-1/(c**5*d**3))*(a*d - b*c)*(3*a*d + b*c)*log(c**3*d*sqrt(-1/(
c**5*d**3))*(a*d - b*c)*(3*a*d + b*c)/(3*a**2*d**2 - 2*a*b*c*d - b**2*c**2) + x)/4 - (2*a**2*c*d + x**2*(3*a**
2*d**2 - 2*a*b*c*d + b**2*c**2))/(2*c**3*d*x + 2*c**2*d**2*x**3)

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Giac [A]  time = 1.1602, size = 138, normalized size = 1.3 \begin{align*} \frac{{\left (b^{2} c^{2} + 2 \, a b c d - 3 \, a^{2} d^{2}\right )} \arctan \left (\frac{d x}{\sqrt{c d}}\right )}{2 \, \sqrt{c d} c^{2} d} - \frac{b^{2} c^{2} x^{2} - 2 \, a b c d x^{2} + 3 \, a^{2} d^{2} x^{2} + 2 \, a^{2} c d}{2 \,{\left (d x^{3} + c x\right )} c^{2} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^2/x^2/(d*x^2+c)^2,x, algorithm="giac")

[Out]

1/2*(b^2*c^2 + 2*a*b*c*d - 3*a^2*d^2)*arctan(d*x/sqrt(c*d))/(sqrt(c*d)*c^2*d) - 1/2*(b^2*c^2*x^2 - 2*a*b*c*d*x
^2 + 3*a^2*d^2*x^2 + 2*a^2*c*d)/((d*x^3 + c*x)*c^2*d)