3.8 \(\int \frac{(a+b x^2) (c+d x^2)}{(e+f x^2)^4} \, dx\)

Optimal. Leaf size=171 \[ \frac{x (a f (5 c f+d e)+b e (c f+d e))}{16 e^3 f^2 \left (e+f x^2\right )}-\frac{x (3 b e (c f+d e)-a f (5 c f+d e))}{24 e^2 f^2 \left (e+f x^2\right )^2}+\frac{\tan ^{-1}\left (\frac{\sqrt{f} x}{\sqrt{e}}\right ) (a f (5 c f+d e)+b e (c f+d e))}{16 e^{7/2} f^{5/2}}-\frac{x \left (a+b x^2\right ) (d e-c f)}{6 e f \left (e+f x^2\right )^3} \]

[Out]

-((d*e - c*f)*x*(a + b*x^2))/(6*e*f*(e + f*x^2)^3) - ((3*b*e*(d*e + c*f) - a*f*(d*e + 5*c*f))*x)/(24*e^2*f^2*(
e + f*x^2)^2) + ((b*e*(d*e + c*f) + a*f*(d*e + 5*c*f))*x)/(16*e^3*f^2*(e + f*x^2)) + ((b*e*(d*e + c*f) + a*f*(
d*e + 5*c*f))*ArcTan[(Sqrt[f]*x)/Sqrt[e]])/(16*e^(7/2)*f^(5/2))

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Rubi [A]  time = 0.163406, antiderivative size = 171, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {526, 385, 199, 205} \[ \frac{x (a f (5 c f+d e)+b e (c f+d e))}{16 e^3 f^2 \left (e+f x^2\right )}-\frac{x (3 b e (c f+d e)-a f (5 c f+d e))}{24 e^2 f^2 \left (e+f x^2\right )^2}+\frac{\tan ^{-1}\left (\frac{\sqrt{f} x}{\sqrt{e}}\right ) (a f (5 c f+d e)+b e (c f+d e))}{16 e^{7/2} f^{5/2}}-\frac{x \left (a+b x^2\right ) (d e-c f)}{6 e f \left (e+f x^2\right )^3} \]

Antiderivative was successfully verified.

[In]

Int[((a + b*x^2)*(c + d*x^2))/(e + f*x^2)^4,x]

[Out]

-((d*e - c*f)*x*(a + b*x^2))/(6*e*f*(e + f*x^2)^3) - ((3*b*e*(d*e + c*f) - a*f*(d*e + 5*c*f))*x)/(24*e^2*f^2*(
e + f*x^2)^2) + ((b*e*(d*e + c*f) + a*f*(d*e + 5*c*f))*x)/(16*e^3*f^2*(e + f*x^2)) + ((b*e*(d*e + c*f) + a*f*(
d*e + 5*c*f))*ArcTan[(Sqrt[f]*x)/Sqrt[e]])/(16*e^(7/2)*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 199

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[(x*(a + b*x^n)^(p + 1))/(a*n*(p + 1)), x] + Dist[(n*(p +
1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (In
tegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[p]
)

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 )^4} \, dx &=-\frac{(d e-c f) x \left (a+b x^2\right )}{6 e f \left (e+f x^2\right )^3}-\frac{\int \frac{-a (d e+5 c f)-3 b (d e+c f) x^2}{\left (e+f x^2\right )^3} \, dx}{6 e f}\\ &=-\frac{(d e-c f) x \left (a+b x^2\right )}{6 e f \left (e+f x^2\right )^3}-\frac{(3 b e (d e+c f)-a f (d e+5 c f)) x}{24 e^2 f^2 \left (e+f x^2\right )^2}+\frac{(b e (d e+c f)+a f (d e+5 c f)) \int \frac{1}{\left (e+f x^2\right )^2} \, dx}{8 e^2 f^2}\\ &=-\frac{(d e-c f) x \left (a+b x^2\right )}{6 e f \left (e+f x^2\right )^3}-\frac{(3 b e (d e+c f)-a f (d e+5 c f)) x}{24 e^2 f^2 \left (e+f x^2\right )^2}+\frac{(b e (d e+c f)+a f (d e+5 c f)) x}{16 e^3 f^2 \left (e+f x^2\right )}+\frac{(b e (d e+c f)+a f (d e+5 c f)) \int \frac{1}{e+f x^2} \, dx}{16 e^3 f^2}\\ &=-\frac{(d e-c f) x \left (a+b x^2\right )}{6 e f \left (e+f x^2\right )^3}-\frac{(3 b e (d e+c f)-a f (d e+5 c f)) x}{24 e^2 f^2 \left (e+f x^2\right )^2}+\frac{(b e (d e+c f)+a f (d e+5 c f)) x}{16 e^3 f^2 \left (e+f x^2\right )}+\frac{(b e (d e+c f)+a f (d e+5 c f)) \tan ^{-1}\left (\frac{\sqrt{f} x}{\sqrt{e}}\right )}{16 e^{7/2} f^{5/2}}\\ \end{align*}

Mathematica [A]  time = 0.101899, size = 171, normalized size = 1. \[ \frac{x (a f (5 c f+d e)+b e (c f+d e))}{16 e^3 f^2 \left (e+f x^2\right )}+\frac{x (a f (5 c f+d e)+b e (c f-7 d e))}{24 e^2 f^2 \left (e+f x^2\right )^2}+\frac{\tan ^{-1}\left (\frac{\sqrt{f} x}{\sqrt{e}}\right ) (a f (5 c f+d e)+b e (c f+d e))}{16 e^{7/2} f^{5/2}}+\frac{x (b e-a f) (d e-c f)}{6 e f^2 \left (e+f x^2\right )^3} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x^2)*(c + d*x^2))/(e + f*x^2)^4,x]

[Out]

((b*e - a*f)*(d*e - c*f)*x)/(6*e*f^2*(e + f*x^2)^3) + ((b*e*(-7*d*e + c*f) + a*f*(d*e + 5*c*f))*x)/(24*e^2*f^2
*(e + f*x^2)^2) + ((b*e*(d*e + c*f) + a*f*(d*e + 5*c*f))*x)/(16*e^3*f^2*(e + f*x^2)) + ((b*e*(d*e + c*f) + a*f
*(d*e + 5*c*f))*ArcTan[(Sqrt[f]*x)/Sqrt[e]])/(16*e^(7/2)*f^(5/2))

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Maple [A]  time = 0.01, size = 210, normalized size = 1.2 \begin{align*}{\frac{1}{ \left ( f{x}^{2}+e \right ) ^{3}} \left ({\frac{ \left ( 5\,ac{f}^{2}+adef+bcef+bd{e}^{2} \right ){x}^{5}}{16\,{e}^{3}}}+{\frac{ \left ( 5\,ac{f}^{2}+adef+bcef-bd{e}^{2} \right ){x}^{3}}{6\,{e}^{2}f}}+{\frac{ \left ( 11\,ac{f}^{2}-adef-bcef-bd{e}^{2} \right ) x}{16\,e{f}^{2}}} \right ) }+{\frac{5\,ac}{16\,{e}^{3}}\arctan \left ({fx{\frac{1}{\sqrt{ef}}}} \right ){\frac{1}{\sqrt{ef}}}}+{\frac{ad}{16\,{e}^{2}f}\arctan \left ({fx{\frac{1}{\sqrt{ef}}}} \right ){\frac{1}{\sqrt{ef}}}}+{\frac{bc}{16\,{e}^{2}f}\arctan \left ({fx{\frac{1}{\sqrt{ef}}}} \right ){\frac{1}{\sqrt{ef}}}}+{\frac{bd}{16\,e{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)^4,x)

[Out]

(1/16*(5*a*c*f^2+a*d*e*f+b*c*e*f+b*d*e^2)/e^3*x^5+1/6*(5*a*c*f^2+a*d*e*f+b*c*e*f-b*d*e^2)/e^2/f*x^3+1/16*(11*a
*c*f^2-a*d*e*f-b*c*e*f-b*d*e^2)/e/f^2*x)/(f*x^2+e)^3+5/16/e^3/(e*f)^(1/2)*arctan(x*f/(e*f)^(1/2))*a*c+1/16/e^2
/f/(e*f)^(1/2)*arctan(x*f/(e*f)^(1/2))*a*d+1/16/e^2/f/(e*f)^(1/2)*arctan(x*f/(e*f)^(1/2))*b*c+1/16/e/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)^4,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.55886, size = 1355, normalized size = 7.92 \begin{align*} \left [\frac{6 \,{\left (b d e^{3} f^{3} + 5 \, a c e f^{5} +{\left (b c + a d\right )} e^{2} f^{4}\right )} x^{5} - 16 \,{\left (b d e^{4} f^{2} - 5 \, a c e^{2} f^{4} -{\left (b c + a d\right )} e^{3} f^{3}\right )} x^{3} - 3 \,{\left (b d e^{5} + 5 \, a c e^{3} f^{2} +{\left (b d e^{2} f^{3} + 5 \, a c f^{5} +{\left (b c + a d\right )} e f^{4}\right )} x^{6} +{\left (b c + a d\right )} e^{4} f + 3 \,{\left (b d e^{3} f^{2} + 5 \, a c e f^{4} +{\left (b c + a d\right )} e^{2} f^{3}\right )} x^{4} + 3 \,{\left (b d e^{4} f + 5 \, a c e^{2} f^{3} +{\left (b c + a d\right )} e^{3} 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 ) - 6 \,{\left (b d e^{5} f - 11 \, a c e^{3} f^{3} +{\left (b c + a d\right )} e^{4} f^{2}\right )} x}{96 \,{\left (e^{4} f^{6} x^{6} + 3 \, e^{5} f^{5} x^{4} + 3 \, e^{6} f^{4} x^{2} + e^{7} f^{3}\right )}}, \frac{3 \,{\left (b d e^{3} f^{3} + 5 \, a c e f^{5} +{\left (b c + a d\right )} e^{2} f^{4}\right )} x^{5} - 8 \,{\left (b d e^{4} f^{2} - 5 \, a c e^{2} f^{4} -{\left (b c + a d\right )} e^{3} f^{3}\right )} x^{3} + 3 \,{\left (b d e^{5} + 5 \, a c e^{3} f^{2} +{\left (b d e^{2} f^{3} + 5 \, a c f^{5} +{\left (b c + a d\right )} e f^{4}\right )} x^{6} +{\left (b c + a d\right )} e^{4} f + 3 \,{\left (b d e^{3} f^{2} + 5 \, a c e f^{4} +{\left (b c + a d\right )} e^{2} f^{3}\right )} x^{4} + 3 \,{\left (b d e^{4} f + 5 \, a c e^{2} f^{3} +{\left (b c + a d\right )} e^{3} f^{2}\right )} x^{2}\right )} \sqrt{e f} \arctan \left (\frac{\sqrt{e f} x}{e}\right ) - 3 \,{\left (b d e^{5} f - 11 \, a c e^{3} f^{3} +{\left (b c + a d\right )} e^{4} f^{2}\right )} x}{48 \,{\left (e^{4} f^{6} x^{6} + 3 \, e^{5} f^{5} x^{4} + 3 \, e^{6} f^{4} x^{2} + e^{7} 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)^4,x, algorithm="fricas")

[Out]

[1/96*(6*(b*d*e^3*f^3 + 5*a*c*e*f^5 + (b*c + a*d)*e^2*f^4)*x^5 - 16*(b*d*e^4*f^2 - 5*a*c*e^2*f^4 - (b*c + a*d)
*e^3*f^3)*x^3 - 3*(b*d*e^5 + 5*a*c*e^3*f^2 + (b*d*e^2*f^3 + 5*a*c*f^5 + (b*c + a*d)*e*f^4)*x^6 + (b*c + a*d)*e
^4*f + 3*(b*d*e^3*f^2 + 5*a*c*e*f^4 + (b*c + a*d)*e^2*f^3)*x^4 + 3*(b*d*e^4*f + 5*a*c*e^2*f^3 + (b*c + a*d)*e^
3*f^2)*x^2)*sqrt(-e*f)*log((f*x^2 - 2*sqrt(-e*f)*x - e)/(f*x^2 + e)) - 6*(b*d*e^5*f - 11*a*c*e^3*f^3 + (b*c +
a*d)*e^4*f^2)*x)/(e^4*f^6*x^6 + 3*e^5*f^5*x^4 + 3*e^6*f^4*x^2 + e^7*f^3), 1/48*(3*(b*d*e^3*f^3 + 5*a*c*e*f^5 +
(b*c + a*d)*e^2*f^4)*x^5 - 8*(b*d*e^4*f^2 - 5*a*c*e^2*f^4 - (b*c + a*d)*e^3*f^3)*x^3 + 3*(b*d*e^5 + 5*a*c*e^3
*f^2 + (b*d*e^2*f^3 + 5*a*c*f^5 + (b*c + a*d)*e*f^4)*x^6 + (b*c + a*d)*e^4*f + 3*(b*d*e^3*f^2 + 5*a*c*e*f^4 +
(b*c + a*d)*e^2*f^3)*x^4 + 3*(b*d*e^4*f + 5*a*c*e^2*f^3 + (b*c + a*d)*e^3*f^2)*x^2)*sqrt(e*f)*arctan(sqrt(e*f)
*x/e) - 3*(b*d*e^5*f - 11*a*c*e^3*f^3 + (b*c + a*d)*e^4*f^2)*x)/(e^4*f^6*x^6 + 3*e^5*f^5*x^4 + 3*e^6*f^4*x^2 +
e^7*f^3)]

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Sympy [A]  time = 8.42358, size = 313, normalized size = 1.83 \begin{align*} - \frac{\sqrt{- \frac{1}{e^{7} f^{5}}} \left (5 a c f^{2} + a d e f + b c e f + b d e^{2}\right ) \log{\left (- e^{4} f^{2} \sqrt{- \frac{1}{e^{7} f^{5}}} + x \right )}}{32} + \frac{\sqrt{- \frac{1}{e^{7} f^{5}}} \left (5 a c f^{2} + a d e f + b c e f + b d e^{2}\right ) \log{\left (e^{4} f^{2} \sqrt{- \frac{1}{e^{7} f^{5}}} + x \right )}}{32} + \frac{x^{5} \left (15 a c f^{4} + 3 a d e f^{3} + 3 b c e f^{3} + 3 b d e^{2} f^{2}\right ) + x^{3} \left (40 a c e f^{3} + 8 a d e^{2} f^{2} + 8 b c e^{2} f^{2} - 8 b d e^{3} f\right ) + x \left (33 a c e^{2} f^{2} - 3 a d e^{3} f - 3 b c e^{3} f - 3 b d e^{4}\right )}{48 e^{6} f^{2} + 144 e^{5} f^{3} x^{2} + 144 e^{4} f^{4} x^{4} + 48 e^{3} f^{5} x^{6}} \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)**4,x)

[Out]

-sqrt(-1/(e**7*f**5))*(5*a*c*f**2 + a*d*e*f + b*c*e*f + b*d*e**2)*log(-e**4*f**2*sqrt(-1/(e**7*f**5)) + x)/32
+ sqrt(-1/(e**7*f**5))*(5*a*c*f**2 + a*d*e*f + b*c*e*f + b*d*e**2)*log(e**4*f**2*sqrt(-1/(e**7*f**5)) + x)/32
+ (x**5*(15*a*c*f**4 + 3*a*d*e*f**3 + 3*b*c*e*f**3 + 3*b*d*e**2*f**2) + x**3*(40*a*c*e*f**3 + 8*a*d*e**2*f**2
+ 8*b*c*e**2*f**2 - 8*b*d*e**3*f) + x*(33*a*c*e**2*f**2 - 3*a*d*e**3*f - 3*b*c*e**3*f - 3*b*d*e**4))/(48*e**6*
f**2 + 144*e**5*f**3*x**2 + 144*e**4*f**4*x**4 + 48*e**3*f**5*x**6)

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Giac [A]  time = 1.19956, size = 248, normalized size = 1.45 \begin{align*} \frac{{\left (5 \, a c f^{2} + b c f e + a d f e + b d e^{2}\right )} \arctan \left (\sqrt{f} x e^{\left (-\frac{1}{2}\right )}\right ) e^{\left (-\frac{7}{2}\right )}}{16 \, f^{\frac{5}{2}}} + \frac{{\left (15 \, a c f^{4} x^{5} + 3 \, b c f^{3} x^{5} e + 3 \, a d f^{3} x^{5} e + 3 \, b d f^{2} x^{5} e^{2} + 40 \, a c f^{3} x^{3} e + 8 \, b c f^{2} x^{3} e^{2} + 8 \, a d f^{2} x^{3} e^{2} - 8 \, b d f x^{3} e^{3} + 33 \, a c f^{2} x e^{2} - 3 \, b c f x e^{3} - 3 \, a d f x e^{3} - 3 \, b d x e^{4}\right )} e^{\left (-3\right )}}{48 \,{\left (f x^{2} + e\right )}^{3} 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)^4,x, algorithm="giac")

[Out]

1/16*(5*a*c*f^2 + b*c*f*e + a*d*f*e + b*d*e^2)*arctan(sqrt(f)*x*e^(-1/2))*e^(-7/2)/f^(5/2) + 1/48*(15*a*c*f^4*
x^5 + 3*b*c*f^3*x^5*e + 3*a*d*f^3*x^5*e + 3*b*d*f^2*x^5*e^2 + 40*a*c*f^3*x^3*e + 8*b*c*f^2*x^3*e^2 + 8*a*d*f^2
*x^3*e^2 - 8*b*d*f*x^3*e^3 + 33*a*c*f^2*x*e^2 - 3*b*c*f*x*e^3 - 3*a*d*f*x*e^3 - 3*b*d*x*e^4)*e^(-3)/((f*x^2 +
e)^3*f^2)