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

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

[Out]

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

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Rubi [A]  time = 0.0722065, 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} \[ \frac{3 x \left (\frac{a^2}{c^2}-\frac{b^2}{d^2}\right )}{8 \left (c+d x^2\right )}+\frac{\left (3 a^2 d^2+2 a b c d+3 b^2 c^2\right ) \tan ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c}}\right )}{8 c^{5/2} d^{5/2}}-\frac{x \left (a+b x^2\right ) (b c-a d)}{4 c d \left (c+d x^2\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2)^2/(c + d*x^2)^3,x]

[Out]

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

Mathematica [A]  time = 0.0930836, size = 121, normalized size = 1.04 \[ \frac{x \left (a^2 d^2 \left (5 c+3 d x^2\right )-2 a b c d \left (c-d x^2\right )-b^2 c^2 \left (3 c+5 d x^2\right )\right )}{8 c^2 d^2 \left (c+d x^2\right )^2}+\frac{\left (3 a^2 d^2+2 a b c d+3 b^2 c^2\right ) \tan ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c}}\right )}{8 c^{5/2} d^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.009, size = 147, normalized size = 1.3 \begin{align*}{\frac{1}{ \left ( d{x}^{2}+c \right ) ^{2}} \left ({\frac{ \left ( 3\,{a}^{2}{d}^{2}+2\,abcd-5\,{b}^{2}{c}^{2} \right ){x}^{3}}{8\,{c}^{2}d}}+{\frac{ \left ( 5\,{a}^{2}{d}^{2}-2\,abcd-3\,{b}^{2}{c}^{2} \right ) x}{8\,{d}^{2}c}} \right ) }+{\frac{3\,{a}^{2}}{8\,{c}^{2}}\arctan \left ({dx{\frac{1}{\sqrt{cd}}}} \right ){\frac{1}{\sqrt{cd}}}}+{\frac{ab}{4\,cd}\arctan \left ({dx{\frac{1}{\sqrt{cd}}}} \right ){\frac{1}{\sqrt{cd}}}}+{\frac{3\,{b}^{2}}{8\,{d}^{2}}\arctan \left ({dx{\frac{1}{\sqrt{cd}}}} \right ){\frac{1}{\sqrt{cd}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.75657, size = 921, normalized size = 7.94 \begin{align*} \left [-\frac{2 \,{\left (5 \, b^{2} c^{3} d^{2} - 2 \, a b c^{2} d^{3} - 3 \, a^{2} c d^{4}\right )} x^{3} +{\left (3 \, b^{2} c^{4} + 2 \, a b c^{3} d + 3 \, a^{2} c^{2} d^{2} +{\left (3 \, b^{2} c^{2} d^{2} + 2 \, a b c d^{3} + 3 \, a^{2} d^{4}\right )} x^{4} + 2 \,{\left (3 \, b^{2} c^{3} d + 2 \, a b c^{2} d^{2} + 3 \, a^{2} c d^{3}\right )} x^{2}\right )} \sqrt{-c d} \log \left (\frac{d x^{2} - 2 \, \sqrt{-c d} x - c}{d x^{2} + c}\right ) + 2 \,{\left (3 \, b^{2} c^{4} d + 2 \, a b c^{3} d^{2} - 5 \, a^{2} c^{2} d^{3}\right )} x}{16 \,{\left (c^{3} d^{5} x^{4} + 2 \, c^{4} d^{4} x^{2} + c^{5} d^{3}\right )}}, -\frac{{\left (5 \, b^{2} c^{3} d^{2} - 2 \, a b c^{2} d^{3} - 3 \, a^{2} c d^{4}\right )} x^{3} -{\left (3 \, b^{2} c^{4} + 2 \, a b c^{3} d + 3 \, a^{2} c^{2} d^{2} +{\left (3 \, b^{2} c^{2} d^{2} + 2 \, a b c d^{3} + 3 \, a^{2} d^{4}\right )} x^{4} + 2 \,{\left (3 \, b^{2} c^{3} d + 2 \, a b c^{2} d^{2} + 3 \, a^{2} c d^{3}\right )} x^{2}\right )} \sqrt{c d} \arctan \left (\frac{\sqrt{c d} x}{c}\right ) +{\left (3 \, b^{2} c^{4} d + 2 \, a b c^{3} d^{2} - 5 \, a^{2} c^{2} d^{3}\right )} x}{8 \,{\left (c^{3} d^{5} x^{4} + 2 \, c^{4} d^{4} x^{2} + c^{5} d^{3}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [B]  time = 1.16113, size = 223, normalized size = 1.92 \begin{align*} - \frac{\sqrt{- \frac{1}{c^{5} d^{5}}} \left (3 a^{2} d^{2} + 2 a b c d + 3 b^{2} c^{2}\right ) \log{\left (- c^{3} d^{2} \sqrt{- \frac{1}{c^{5} d^{5}}} + x \right )}}{16} + \frac{\sqrt{- \frac{1}{c^{5} d^{5}}} \left (3 a^{2} d^{2} + 2 a b c d + 3 b^{2} c^{2}\right ) \log{\left (c^{3} d^{2} \sqrt{- \frac{1}{c^{5} d^{5}}} + x \right )}}{16} + \frac{x^{3} \left (3 a^{2} d^{3} + 2 a b c d^{2} - 5 b^{2} c^{2} d\right ) + x \left (5 a^{2} c d^{2} - 2 a b c^{2} d - 3 b^{2} c^{3}\right )}{8 c^{4} d^{2} + 16 c^{3} d^{3} x^{2} + 8 c^{2} d^{4} x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.08346, 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{d x}{\sqrt{c d}}\right )}{8 \, \sqrt{c d} c^{2} d^{2}} - \frac{5 \, b^{2} c^{2} d x^{3} - 2 \, a b c d^{2} x^{3} - 3 \, a^{2} d^{3} x^{3} + 3 \, b^{2} c^{3} x + 2 \, a b c^{2} d x - 5 \, a^{2} c d^{2} x}{8 \,{\left (d x^{2} + c\right )}^{2} c^{2} d^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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