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

Optimal. Leaf size=103 \[ -\frac{(b c-a d) (a d+3 b c) \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{2 a^{5/2} b^{3/2}}-\frac{x \left (\frac{3 b c^2}{a}+\frac{a d^2}{b}-2 c d\right )}{2 a \left (a+b x^2\right )}-\frac{c^2}{a x \left (a+b x^2\right )} \]

[Out]

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

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Rubi [A]  time = 0.0777147, antiderivative size = 103, 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} \[ -\frac{(b c-a d) (a d+3 b c) \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{2 a^{5/2} b^{3/2}}-\frac{x \left (\frac{3 b c^2}{a}+\frac{a d^2}{b}-2 c d\right )}{2 a \left (a+b x^2\right )}-\frac{c^2}{a x \left (a+b x^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.0687163, size = 91, normalized size = 0.88 \[ \frac{\left (a^2 d^2+2 a b c d-3 b^2 c^2\right ) \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{2 a^{5/2} b^{3/2}}-\frac{x (a d-b c)^2}{2 a^2 b \left (a+b x^2\right )}-\frac{c^2}{a^2 x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.01, size = 131, normalized size = 1.3 \begin{align*} -{\frac{{c}^{2}}{{a}^{2}x}}-{\frac{x{d}^{2}}{2\,b \left ( b{x}^{2}+a \right ) }}+{\frac{cxd}{a \left ( b{x}^{2}+a \right ) }}-{\frac{bx{c}^{2}}{2\,{a}^{2} \left ( b{x}^{2}+a \right ) }}+{\frac{{d}^{2}}{2\,b}\arctan \left ({bx{\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}}+{\frac{cd}{a}\arctan \left ({bx{\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}}-{\frac{3\,b{c}^{2}}{2\,{a}^{2}}\arctan \left ({bx{\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.5931, size = 625, normalized size = 6.07 \begin{align*} \left [-\frac{4 \, a^{2} b^{2} c^{2} + 2 \,{\left (3 \, a b^{3} c^{2} - 2 \, a^{2} b^{2} c d + a^{3} b d^{2}\right )} x^{2} -{\left ({\left (3 \, b^{3} c^{2} - 2 \, a b^{2} c d - a^{2} b d^{2}\right )} x^{3} +{\left (3 \, a b^{2} c^{2} - 2 \, a^{2} b c d - a^{3} d^{2}\right )} x\right )} \sqrt{-a b} \log \left (\frac{b x^{2} - 2 \, \sqrt{-a b} x - a}{b x^{2} + a}\right )}{4 \,{\left (a^{3} b^{3} x^{3} + a^{4} b^{2} x\right )}}, -\frac{2 \, a^{2} b^{2} c^{2} +{\left (3 \, a b^{3} c^{2} - 2 \, a^{2} b^{2} c d + a^{3} b d^{2}\right )} x^{2} +{\left ({\left (3 \, b^{3} c^{2} - 2 \, a b^{2} c d - a^{2} b d^{2}\right )} x^{3} +{\left (3 \, a b^{2} c^{2} - 2 \, a^{2} b c d - a^{3} d^{2}\right )} x\right )} \sqrt{a b} \arctan \left (\frac{\sqrt{a b} x}{a}\right )}{2 \,{\left (a^{3} b^{3} x^{3} + a^{4} b^{2} x\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [B]  time = 1.01676, size = 238, normalized size = 2.31 \begin{align*} - \frac{\sqrt{- \frac{1}{a^{5} b^{3}}} \left (a d - b c\right ) \left (a d + 3 b c\right ) \log{\left (- \frac{a^{3} b \sqrt{- \frac{1}{a^{5} b^{3}}} \left (a d - b c\right ) \left (a d + 3 b c\right )}{a^{2} d^{2} + 2 a b c d - 3 b^{2} c^{2}} + x \right )}}{4} + \frac{\sqrt{- \frac{1}{a^{5} b^{3}}} \left (a d - b c\right ) \left (a d + 3 b c\right ) \log{\left (\frac{a^{3} b \sqrt{- \frac{1}{a^{5} b^{3}}} \left (a d - b c\right ) \left (a d + 3 b c\right )}{a^{2} d^{2} + 2 a b c d - 3 b^{2} c^{2}} + x \right )}}{4} - \frac{2 a b c^{2} + x^{2} \left (a^{2} d^{2} - 2 a b c d + 3 b^{2} c^{2}\right )}{2 a^{3} b x + 2 a^{2} b^{2} x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.15639, size = 139, normalized size = 1.35 \begin{align*} -\frac{{\left (3 \, b^{2} c^{2} - 2 \, a b c d - a^{2} d^{2}\right )} \arctan \left (\frac{b x}{\sqrt{a b}}\right )}{2 \, \sqrt{a b} a^{2} b} - \frac{3 \, b^{2} c^{2} x^{2} - 2 \, a b c d x^{2} + a^{2} d^{2} x^{2} + 2 \, a b c^{2}}{2 \,{\left (b x^{3} + a x\right )} a^{2} b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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