3.38 \(\int \frac{(c+d x^2)^2}{(a+b x^2)^3} \, dx\)

Optimal. Leaf size=116 \[ \frac{3 x \left (\frac{c^2}{a^2}-\frac{d^2}{b^2}\right )}{8 \left (a+b x^2\right )}+\frac{\left (3 a^2 d^2+2 a b c d+3 b^2 c^2\right ) \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{8 a^{5/2} b^{5/2}}+\frac{x \left (c+d x^2\right ) (b c-a d)}{4 a b \left (a+b x^2\right )^2} \]

[Out]

(3*(c^2/a^2 - d^2/b^2)*x)/(8*(a + b*x^2)) + ((b*c - a*d)*x*(c + d*x^2))/(4*a*b*(a + b*x^2)^2) + ((3*b^2*c^2 +
2*a*b*c*d + 3*a^2*d^2)*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(8*a^(5/2)*b^(5/2))

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Rubi [A]  time = 0.0769482, antiderivative size = 116, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.158, Rules used = {413, 385, 205} \[ \frac{3 x \left (\frac{c^2}{a^2}-\frac{d^2}{b^2}\right )}{8 \left (a+b x^2\right )}+\frac{\left (3 a^2 d^2+2 a b c d+3 b^2 c^2\right ) \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{8 a^{5/2} b^{5/2}}+\frac{x \left (c+d x^2\right ) (b c-a d)}{4 a b \left (a+b x^2\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x^2)^2/(a + b*x^2)^3,x]

[Out]

(3*(c^2/a^2 - d^2/b^2)*x)/(8*(a + b*x^2)) + ((b*c - a*d)*x*(c + d*x^2))/(4*a*b*(a + b*x^2)^2) + ((3*b^2*c^2 +
2*a*b*c*d + 3*a^2*d^2)*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(8*a^(5/2)*b^(5/2))

Rule 413

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[((a*d - c*b)*x*(a + b*x^n)^
(p + 1)*(c + d*x^n)^(q - 1))/(a*b*n*(p + 1)), x] - Dist[1/(a*b*n*(p + 1)), Int[(a + b*x^n)^(p + 1)*(c + d*x^n)
^(q - 2)*Simp[c*(a*d - c*b*(n*(p + 1) + 1)) + d*(a*d*(n*(q - 1) + 1) - b*c*(n*(p + q) + 1))*x^n, x], x], x] /;
FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && GtQ[q, 1] && IntBinomialQ[a, b, c, d, n, p, q
, x]

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}{\left (a+b x^2\right )^3} \, dx &=\frac{(b c-a d) x \left (c+d x^2\right )}{4 a b \left (a+b x^2\right )^2}+\frac{\int \frac{c (3 b c+a d)+d (b c+3 a d) x^2}{\left (a+b x^2\right )^2} \, dx}{4 a b}\\ &=\frac{3 \left (\frac{c^2}{a^2}-\frac{d^2}{b^2}\right ) x}{8 \left (a+b x^2\right )}+\frac{(b c-a d) x \left (c+d x^2\right )}{4 a b \left (a+b x^2\right )^2}+\frac{\left (3 b^2 c^2+2 a b c d+3 a^2 d^2\right ) \int \frac{1}{a+b x^2} \, dx}{8 a^2 b^2}\\ &=\frac{3 \left (\frac{c^2}{a^2}-\frac{d^2}{b^2}\right ) x}{8 \left (a+b x^2\right )}+\frac{(b c-a d) x \left (c+d x^2\right )}{4 a b \left (a+b x^2\right )^2}+\frac{\left (3 b^2 c^2+2 a b c d+3 a^2 d^2\right ) \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{8 a^{5/2} b^{5/2}}\\ \end{align*}

Mathematica [A]  time = 0.0945662, size = 124, normalized size = 1.07 \[ \frac{x \left (-a^2 b d \left (2 c+5 d x^2\right )-3 a^3 d^2+a b^2 c \left (5 c+2 d x^2\right )+3 b^3 c^2 x^2\right )}{8 a^2 b^2 \left (a+b x^2\right )^2}+\frac{\left (3 a^2 d^2+2 a b c d+3 b^2 c^2\right ) \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{8 a^{5/2} b^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x^2)^2/(a + b*x^2)^3,x]

[Out]

(x*(-3*a^3*d^2 + 3*b^3*c^2*x^2 + a*b^2*c*(5*c + 2*d*x^2) - a^2*b*d*(2*c + 5*d*x^2)))/(8*a^2*b^2*(a + b*x^2)^2)
+ ((3*b^2*c^2 + 2*a*b*c*d + 3*a^2*d^2)*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(8*a^(5/2)*b^(5/2))

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Maple [A]  time = 0.008, size = 147, normalized size = 1.3 \begin{align*}{\frac{1}{ \left ( b{x}^{2}+a \right ) ^{2}} \left ( -{\frac{ \left ( 5\,{a}^{2}{d}^{2}-2\,abcd-3\,{b}^{2}{c}^{2} \right ){x}^{3}}{8\,{a}^{2}b}}-{\frac{ \left ( 3\,{a}^{2}{d}^{2}+2\,abcd-5\,{b}^{2}{c}^{2} \right ) x}{8\,a{b}^{2}}} \right ) }+{\frac{3\,{d}^{2}}{8\,{b}^{2}}\arctan \left ({bx{\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}}+{\frac{cd}{4\,ab}\arctan \left ({bx{\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}}+{\frac{3\,{c}^{2}}{8\,{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/(b*x^2+a)^3,x)

[Out]

(-1/8*(5*a^2*d^2-2*a*b*c*d-3*b^2*c^2)/a^2/b*x^3-1/8*(3*a^2*d^2+2*a*b*c*d-5*b^2*c^2)/a/b^2*x)/(b*x^2+a)^2+3/8/b
^2/(a*b)^(1/2)*arctan(b*x/(a*b)^(1/2))*d^2+1/4/b/a/(a*b)^(1/2)*arctan(b*x/(a*b)^(1/2))*c*d+3/8/a^2/(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/(b*x^2+a)^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.87662, size = 918, normalized size = 7.91 \begin{align*} \left [\frac{2 \,{\left (3 \, a b^{4} c^{2} + 2 \, a^{2} b^{3} c d - 5 \, a^{3} b^{2} d^{2}\right )} x^{3} -{\left (3 \, a^{2} b^{2} c^{2} + 2 \, a^{3} b c d + 3 \, a^{4} d^{2} +{\left (3 \, b^{4} c^{2} + 2 \, a b^{3} c d + 3 \, a^{2} b^{2} d^{2}\right )} x^{4} + 2 \,{\left (3 \, a b^{3} c^{2} + 2 \, a^{2} b^{2} c d + 3 \, a^{3} b d^{2}\right )} x^{2}\right )} \sqrt{-a b} \log \left (\frac{b x^{2} - 2 \, \sqrt{-a b} x - a}{b x^{2} + a}\right ) + 2 \,{\left (5 \, a^{2} b^{3} c^{2} - 2 \, a^{3} b^{2} c d - 3 \, a^{4} b d^{2}\right )} x}{16 \,{\left (a^{3} b^{5} x^{4} + 2 \, a^{4} b^{4} x^{2} + a^{5} b^{3}\right )}}, \frac{{\left (3 \, a b^{4} c^{2} + 2 \, a^{2} b^{3} c d - 5 \, a^{3} b^{2} d^{2}\right )} x^{3} +{\left (3 \, a^{2} b^{2} c^{2} + 2 \, a^{3} b c d + 3 \, a^{4} d^{2} +{\left (3 \, b^{4} c^{2} + 2 \, a b^{3} c d + 3 \, a^{2} b^{2} d^{2}\right )} x^{4} + 2 \,{\left (3 \, a b^{3} c^{2} + 2 \, a^{2} b^{2} c d + 3 \, a^{3} b d^{2}\right )} x^{2}\right )} \sqrt{a b} \arctan \left (\frac{\sqrt{a b} x}{a}\right ) +{\left (5 \, a^{2} b^{3} c^{2} - 2 \, a^{3} b^{2} c d - 3 \, a^{4} b d^{2}\right )} x}{8 \,{\left (a^{3} b^{5} x^{4} + 2 \, a^{4} b^{4} x^{2} + a^{5} b^{3}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x^2+c)^2/(b*x^2+a)^3,x, algorithm="fricas")

[Out]

[1/16*(2*(3*a*b^4*c^2 + 2*a^2*b^3*c*d - 5*a^3*b^2*d^2)*x^3 - (3*a^2*b^2*c^2 + 2*a^3*b*c*d + 3*a^4*d^2 + (3*b^4
*c^2 + 2*a*b^3*c*d + 3*a^2*b^2*d^2)*x^4 + 2*(3*a*b^3*c^2 + 2*a^2*b^2*c*d + 3*a^3*b*d^2)*x^2)*sqrt(-a*b)*log((b
*x^2 - 2*sqrt(-a*b)*x - a)/(b*x^2 + a)) + 2*(5*a^2*b^3*c^2 - 2*a^3*b^2*c*d - 3*a^4*b*d^2)*x)/(a^3*b^5*x^4 + 2*
a^4*b^4*x^2 + a^5*b^3), 1/8*((3*a*b^4*c^2 + 2*a^2*b^3*c*d - 5*a^3*b^2*d^2)*x^3 + (3*a^2*b^2*c^2 + 2*a^3*b*c*d
+ 3*a^4*d^2 + (3*b^4*c^2 + 2*a*b^3*c*d + 3*a^2*b^2*d^2)*x^4 + 2*(3*a*b^3*c^2 + 2*a^2*b^2*c*d + 3*a^3*b*d^2)*x^
2)*sqrt(a*b)*arctan(sqrt(a*b)*x/a) + (5*a^2*b^3*c^2 - 2*a^3*b^2*c*d - 3*a^4*b*d^2)*x)/(a^3*b^5*x^4 + 2*a^4*b^4
*x^2 + a^5*b^3)]

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Sympy [B]  time = 1.19055, size = 223, normalized size = 1.92 \begin{align*} - \frac{\sqrt{- \frac{1}{a^{5} b^{5}}} \left (3 a^{2} d^{2} + 2 a b c d + 3 b^{2} c^{2}\right ) \log{\left (- a^{3} b^{2} \sqrt{- \frac{1}{a^{5} b^{5}}} + x \right )}}{16} + \frac{\sqrt{- \frac{1}{a^{5} b^{5}}} \left (3 a^{2} d^{2} + 2 a b c d + 3 b^{2} c^{2}\right ) \log{\left (a^{3} b^{2} \sqrt{- \frac{1}{a^{5} b^{5}}} + x \right )}}{16} - \frac{x^{3} \left (5 a^{2} b d^{2} - 2 a b^{2} c d - 3 b^{3} c^{2}\right ) + x \left (3 a^{3} d^{2} + 2 a^{2} b c d - 5 a b^{2} c^{2}\right )}{8 a^{4} b^{2} + 16 a^{3} b^{3} x^{2} + 8 a^{2} b^{4} x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x**2+c)**2/(b*x**2+a)**3,x)

[Out]

-sqrt(-1/(a**5*b**5))*(3*a**2*d**2 + 2*a*b*c*d + 3*b**2*c**2)*log(-a**3*b**2*sqrt(-1/(a**5*b**5)) + x)/16 + sq
rt(-1/(a**5*b**5))*(3*a**2*d**2 + 2*a*b*c*d + 3*b**2*c**2)*log(a**3*b**2*sqrt(-1/(a**5*b**5)) + x)/16 - (x**3*
(5*a**2*b*d**2 - 2*a*b**2*c*d - 3*b**3*c**2) + x*(3*a**3*d**2 + 2*a**2*b*c*d - 5*a*b**2*c**2))/(8*a**4*b**2 +
16*a**3*b**3*x**2 + 8*a**2*b**4*x**4)

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Giac [A]  time = 1.18004, size = 170, normalized size = 1.47 \begin{align*} \frac{{\left (3 \, b^{2} c^{2} + 2 \, a b c d + 3 \, a^{2} d^{2}\right )} \arctan \left (\frac{b x}{\sqrt{a b}}\right )}{8 \, \sqrt{a b} a^{2} b^{2}} + \frac{3 \, b^{3} c^{2} x^{3} + 2 \, a b^{2} c d x^{3} - 5 \, a^{2} b d^{2} x^{3} + 5 \, a b^{2} c^{2} x - 2 \, a^{2} b c d x - 3 \, a^{3} d^{2} x}{8 \,{\left (b x^{2} + a\right )}^{2} a^{2} b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x^2+c)^2/(b*x^2+a)^3,x, algorithm="giac")

[Out]

1/8*(3*b^2*c^2 + 2*a*b*c*d + 3*a^2*d^2)*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*a^2*b^2) + 1/8*(3*b^3*c^2*x^3 + 2*a*b
^2*c*d*x^3 - 5*a^2*b*d^2*x^3 + 5*a*b^2*c^2*x - 2*a^2*b*c*d*x - 3*a^3*d^2*x)/((b*x^2 + a)^2*a^2*b^2)