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

Optimal. Leaf size=130 \[ -\frac{x (b e (c f+3 d e)-a f (3 c f+d e))}{8 e^2 f^2 \left (e+f x^2\right )}+\frac{\tan ^{-1}\left (\frac{\sqrt{f} x}{\sqrt{e}}\right ) (a f (3 c f+d e)+b e (c f+3 d e))}{8 e^{5/2} f^{5/2}}-\frac{x \left (a+b x^2\right ) (d e-c f)}{4 e f \left (e+f x^2\right )^2} \]

[Out]

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

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Rubi [A]  time = 0.108164, antiderivative size = 130, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {526, 385, 205} \[ -\frac{x (b e (c f+3 d e)-a f (3 c f+d e))}{8 e^2 f^2 \left (e+f x^2\right )}+\frac{\tan ^{-1}\left (\frac{\sqrt{f} x}{\sqrt{e}}\right ) (a f (3 c f+d e)+b e (c f+3 d e))}{8 e^{5/2} f^{5/2}}-\frac{x \left (a+b x^2\right ) (d e-c f)}{4 e f \left (e+f x^2\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 526

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)), x_Symbol] :> -Simp[
((b*e - a*f)*x*(a + b*x^n)^(p + 1)*(c + d*x^n)^q)/(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 - 1)*Simp[c*(b*e*n*(p + 1) + b*e - a*f) + d*(b*e*n*(p + 1) + (b*e - a*f)*(n*q + 1))*x
^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && LtQ[p, -1] && GtQ[q, 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 ) \left (c+d x^2\right )}{\left (e+f x^2\right )^3} \, dx &=-\frac{(d e-c f) x \left (a+b x^2\right )}{4 e f \left (e+f x^2\right )^2}-\frac{\int \frac{-a (d e+3 c f)-b (3 d e+c f) x^2}{\left (e+f x^2\right )^2} \, dx}{4 e f}\\ &=-\frac{(d e-c f) x \left (a+b x^2\right )}{4 e f \left (e+f x^2\right )^2}-\frac{(b e (3 d e+c f)-a f (d e+3 c f)) x}{8 e^2 f^2 \left (e+f x^2\right )}+\frac{(b e (3 d e+c f)+a f (d e+3 c f)) \int \frac{1}{e+f x^2} \, dx}{8 e^2 f^2}\\ &=-\frac{(d e-c f) x \left (a+b x^2\right )}{4 e f \left (e+f x^2\right )^2}-\frac{(b e (3 d e+c f)-a f (d e+3 c f)) x}{8 e^2 f^2 \left (e+f x^2\right )}+\frac{(b e (3 d e+c f)+a f (d e+3 c f)) \tan ^{-1}\left (\frac{\sqrt{f} x}{\sqrt{e}}\right )}{8 e^{5/2} f^{5/2}}\\ \end{align*}

Mathematica [A]  time = 0.0804119, size = 130, normalized size = 1. \[ \frac{x (a f (3 c f+d e)+b e (c f-5 d e))}{8 e^2 f^2 \left (e+f x^2\right )}+\frac{\tan ^{-1}\left (\frac{\sqrt{f} x}{\sqrt{e}}\right ) (a f (3 c f+d e)+b e (c f+3 d e))}{8 e^{5/2} f^{5/2}}+\frac{x (b e-a f) (d e-c f)}{4 e f^2 \left (e+f x^2\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.009, size = 175, normalized size = 1.4 \begin{align*}{\frac{1}{ \left ( f{x}^{2}+e \right ) ^{2}} \left ({\frac{ \left ( 3\,ac{f}^{2}+adef+bcef-5\,bd{e}^{2} \right ){x}^{3}}{8\,{e}^{2}f}}+{\frac{ \left ( 5\,ac{f}^{2}-adef-bcef-3\,bd{e}^{2} \right ) x}{8\,{f}^{2}e}} \right ) }+{\frac{3\,ac}{8\,{e}^{2}}\arctan \left ({fx{\frac{1}{\sqrt{ef}}}} \right ){\frac{1}{\sqrt{ef}}}}+{\frac{ad}{8\,ef}\arctan \left ({fx{\frac{1}{\sqrt{ef}}}} \right ){\frac{1}{\sqrt{ef}}}}+{\frac{bc}{8\,ef}\arctan \left ({fx{\frac{1}{\sqrt{ef}}}} \right ){\frac{1}{\sqrt{ef}}}}+{\frac{3\,bd}{8\,{f}^{2}}\arctan \left ({fx{\frac{1}{\sqrt{ef}}}} \right ){\frac{1}{\sqrt{ef}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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)*(d*x^2+c)/(f*x^2+e)^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.50555, size = 1002, normalized size = 7.71 \begin{align*} \left [-\frac{2 \,{\left (5 \, b d e^{3} f^{2} - 3 \, a c e f^{4} -{\left (b c + a d\right )} e^{2} f^{3}\right )} x^{3} +{\left (3 \, b d e^{4} + 3 \, a c e^{2} f^{2} +{\left (b c + a d\right )} e^{3} f +{\left (3 \, b d e^{2} f^{2} + 3 \, a c f^{4} +{\left (b c + a d\right )} e f^{3}\right )} x^{4} + 2 \,{\left (3 \, b d e^{3} f + 3 \, a c e f^{3} +{\left (b c + a d\right )} e^{2} f^{2}\right )} x^{2}\right )} \sqrt{-e f} \log \left (\frac{f x^{2} - 2 \, \sqrt{-e f} x - e}{f x^{2} + e}\right ) + 2 \,{\left (3 \, b d e^{4} f - 5 \, a c e^{2} f^{3} +{\left (b c + a d\right )} e^{3} f^{2}\right )} x}{16 \,{\left (e^{3} f^{5} x^{4} + 2 \, e^{4} f^{4} x^{2} + e^{5} f^{3}\right )}}, -\frac{{\left (5 \, b d e^{3} f^{2} - 3 \, a c e f^{4} -{\left (b c + a d\right )} e^{2} f^{3}\right )} x^{3} -{\left (3 \, b d e^{4} + 3 \, a c e^{2} f^{2} +{\left (b c + a d\right )} e^{3} f +{\left (3 \, b d e^{2} f^{2} + 3 \, a c f^{4} +{\left (b c + a d\right )} e f^{3}\right )} x^{4} + 2 \,{\left (3 \, b d e^{3} f + 3 \, a c e f^{3} +{\left (b c + a d\right )} e^{2} f^{2}\right )} x^{2}\right )} \sqrt{e f} \arctan \left (\frac{\sqrt{e f} x}{e}\right ) +{\left (3 \, b d e^{4} f - 5 \, a c e^{2} f^{3} +{\left (b c + a d\right )} e^{3} f^{2}\right )} x}{8 \,{\left (e^{3} f^{5} x^{4} + 2 \, e^{4} f^{4} x^{2} + e^{5} f^{3}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [B]  time = 3.64234, size = 246, normalized size = 1.89 \begin{align*} - \frac{\sqrt{- \frac{1}{e^{5} f^{5}}} \left (3 a c f^{2} + a d e f + b c e f + 3 b d e^{2}\right ) \log{\left (- e^{3} f^{2} \sqrt{- \frac{1}{e^{5} f^{5}}} + x \right )}}{16} + \frac{\sqrt{- \frac{1}{e^{5} f^{5}}} \left (3 a c f^{2} + a d e f + b c e f + 3 b d e^{2}\right ) \log{\left (e^{3} f^{2} \sqrt{- \frac{1}{e^{5} f^{5}}} + x \right )}}{16} + \frac{x^{3} \left (3 a c f^{3} + a d e f^{2} + b c e f^{2} - 5 b d e^{2} f\right ) + x \left (5 a c e f^{2} - a d e^{2} f - b c e^{2} f - 3 b d e^{3}\right )}{8 e^{4} f^{2} + 16 e^{3} f^{3} x^{2} + 8 e^{2} f^{4} x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.15831, size = 182, normalized size = 1.4 \begin{align*} \frac{{\left (3 \, a c f^{2} + b c f e + a d f e + 3 \, b d e^{2}\right )} \arctan \left (\sqrt{f} x e^{\left (-\frac{1}{2}\right )}\right ) e^{\left (-\frac{5}{2}\right )}}{8 \, f^{\frac{5}{2}}} + \frac{{\left (3 \, a c f^{3} x^{3} + b c f^{2} x^{3} e + a d f^{2} x^{3} e - 5 \, b d f x^{3} e^{2} + 5 \, a c f^{2} x e - b c f x e^{2} - a d f x e^{2} - 3 \, b d x e^{3}\right )} e^{\left (-2\right )}}{8 \,{\left (f x^{2} + e\right )}^{2} f^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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