Integrand size = 54, antiderivative size = 93 \[ \int \frac {1}{\left (\left (a+b s^2\right )^2+4 b s \left (a+b s^2\right ) x+2 b \left (a+3 b s^2\right ) x^2+4 b^2 s x^3+b^2 x^4\right )^2} \, dx=\frac {s+x}{6 a \left (a+b (s+x)^2\right )^3}+\frac {5 (s+x)}{24 a^2 \left (a+b (s+x)^2\right )^2}+\frac {5 (s+x)}{16 a^3 \left (a+b (s+x)^2\right )}+\frac {5 \arctan \left (\frac {\sqrt {b} (s+x)}{\sqrt {a}}\right )}{16 a^{7/2} \sqrt {b}} \] Output:
1/6*(s+x)/a/(a+b*(s+x)^2)^3+5/24*(s+x)/a^2/(a+b*(s+x)^2)^2+5/16*(s+x)/a^3/ (a+b*(s+x)^2)+5/16*arctan(b^(1/2)*(s+x)/a^(1/2))/a^(7/2)/b^(1/2)
Time = 0.01 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.82 \[ \int \frac {1}{\left (\left (a+b s^2\right )^2+4 b s \left (a+b s^2\right ) x+2 b \left (a+3 b s^2\right ) x^2+4 b^2 s x^3+b^2 x^4\right )^2} \, dx=\frac {(s+x) \left (33 a^2+40 a b (s+x)^2+15 b^2 (s+x)^4\right )}{48 a^3 \left (a+b (s+x)^2\right )^3}+\frac {5 \arctan \left (\frac {\sqrt {b} (s+x)}{\sqrt {a}}\right )}{16 a^{7/2} \sqrt {b}} \] Input:
Integrate[((a + b*s^2)^2 + 4*b*s*(a + b*s^2)*x + 2*b*(a + 3*b*s^2)*x^2 + 4 *b^2*s*x^3 + b^2*x^4)^(-2),x]
Output:
((s + x)*(33*a^2 + 40*a*b*(s + x)^2 + 15*b^2*(s + x)^4))/(48*a^3*(a + b*(s + x)^2)^3) + (5*ArcTan[(Sqrt[b]*(s + x))/Sqrt[a]])/(16*a^(7/2)*Sqrt[b])
Time = 0.43 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.38, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {2458, 1379, 215, 215, 215, 218}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {1}{\left (2 b x^2 \left (a+3 b s^2\right )+4 b s x \left (a+b s^2\right )+\left (a+b s^2\right )^2+4 b^2 s x^3+b^2 x^4\right )^2} \, dx\) |
\(\Big \downarrow \) 2458 |
\(\displaystyle \int \frac {1}{\left (a^2+2 a b (s+x)^2+b^2 (s+x)^4\right )^2}d(s+x)\) |
\(\Big \downarrow \) 1379 |
\(\displaystyle b^4 \int \frac {1}{\left (b^2 (s+x)^2+a b\right )^4}d(s+x)\) |
\(\Big \downarrow \) 215 |
\(\displaystyle b^4 \left (\frac {5 \int \frac {1}{\left (b^2 (s+x)^2+a b\right )^3}d(s+x)}{6 a b}+\frac {s+x}{6 a b^4 \left (a+b (s+x)^2\right )^3}\right )\) |
\(\Big \downarrow \) 215 |
\(\displaystyle b^4 \left (\frac {5 \left (\frac {3 \int \frac {1}{\left (b^2 (s+x)^2+a b\right )^2}d(s+x)}{4 a b}+\frac {s+x}{4 a b^3 \left (a+b (s+x)^2\right )^2}\right )}{6 a b}+\frac {s+x}{6 a b^4 \left (a+b (s+x)^2\right )^3}\right )\) |
\(\Big \downarrow \) 215 |
\(\displaystyle b^4 \left (\frac {5 \left (\frac {3 \left (\frac {\int \frac {1}{b^2 (s+x)^2+a b}d(s+x)}{2 a b}+\frac {s+x}{2 a b^2 \left (a+b (s+x)^2\right )}\right )}{4 a b}+\frac {s+x}{4 a b^3 \left (a+b (s+x)^2\right )^2}\right )}{6 a b}+\frac {s+x}{6 a b^4 \left (a+b (s+x)^2\right )^3}\right )\) |
\(\Big \downarrow \) 218 |
\(\displaystyle b^4 \left (\frac {5 \left (\frac {3 \left (\frac {\arctan \left (\frac {\sqrt {b} (s+x)}{\sqrt {a}}\right )}{2 a^{3/2} b^{5/2}}+\frac {s+x}{2 a b^2 \left (a+b (s+x)^2\right )}\right )}{4 a b}+\frac {s+x}{4 a b^3 \left (a+b (s+x)^2\right )^2}\right )}{6 a b}+\frac {s+x}{6 a b^4 \left (a+b (s+x)^2\right )^3}\right )\) |
Input:
Int[((a + b*s^2)^2 + 4*b*s*(a + b*s^2)*x + 2*b*(a + 3*b*s^2)*x^2 + 4*b^2*s *x^3 + b^2*x^4)^(-2),x]
Output:
b^4*((s + x)/(6*a*b^4*(a + b*(s + x)^2)^3) + (5*((s + x)/(4*a*b^3*(a + b*( s + x)^2)^2) + (3*((s + x)/(2*a*b^2*(a + b*(s + x)^2)) + ArcTan[(Sqrt[b]*( s + x))/Sqrt[a]]/(2*a^(3/2)*b^(5/2))))/(4*a*b)))/(6*a*b))
Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^2)^(p + 1) /(2*a*(p + 1))), x] + Simp[(2*p + 3)/(2*a*(p + 1)) Int[(a + b*x^2)^(p + 1 ), x], x] /; FreeQ[{a, b}, x] && LtQ[p, -1] && (IntegerQ[4*p] || IntegerQ[6 *p])
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
Int[((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[1/ c^p Int[(b/2 + c*x^n)^(2*p), x], x] /; FreeQ[{a, b, c, n, p}, x] && EqQ[n 2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p] && NeQ[p, 1]
Int[(Pn_)^(p_.), x_Symbol] :> With[{S = Coeff[Pn, x, Expon[Pn, x] - 1]/(Exp on[Pn, x]*Coeff[Pn, x, Expon[Pn, x]])}, Subst[Int[ExpandToSum[Pn /. x -> x - S, x]^p, x], x, x + S] /; BinomialQ[Pn /. x -> x - S, x] || (IntegerQ[Exp on[Pn, x]/2] && TrinomialQ[Pn /. x -> x - S, x])] /; FreeQ[p, x] && PolyQ[P n, x] && GtQ[Expon[Pn, x], 2] && NeQ[Coeff[Pn, x, Expon[Pn, x] - 1], 0]
Time = 0.13 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.60
method | result | size |
default | \(\frac {2 b s +2 b x}{12 a b \left (b \,s^{2}+2 b s x +b \,x^{2}+a \right )^{3}}+\frac {\frac {5 \left (2 b s +2 b x \right )}{48 a b \left (b \,s^{2}+2 b s x +b \,x^{2}+a \right )^{2}}+\frac {5 \left (\frac {3 \left (2 b s +2 b x \right )}{16 a b \left (b \,s^{2}+2 b s x +b \,x^{2}+a \right )}+\frac {3 \arctan \left (\frac {2 b s +2 b x}{2 \sqrt {a b}}\right )}{8 a \sqrt {a b}}\right )}{6 a}}{a}\) | \(149\) |
risch | \(\frac {\frac {5 b^{2} x^{5}}{16 a^{3}}+\frac {25 b^{2} s \,x^{4}}{16 a^{3}}+\frac {5 b \left (15 b \,s^{2}+4 a \right ) x^{3}}{24 a^{3}}+\frac {5 b s \left (5 b \,s^{2}+4 a \right ) x^{2}}{8 a^{3}}+\frac {\left (25 b^{2} s^{4}+40 a b \,s^{2}+11 a^{2}\right ) x}{16 a^{3}}+\frac {s \left (15 b^{2} s^{4}+40 a b \,s^{2}+33 a^{2}\right )}{48 a^{3}}}{\left (b^{2} s^{4}+4 b^{2} s^{3} x +6 b^{2} s^{2} x^{2}+4 b^{2} s \,x^{3}+b^{2} x^{4}+2 a b \,s^{2}+4 a b s x +2 a b \,x^{2}+a^{2}\right ) \left (b \,s^{2}+2 b s x +b \,x^{2}+a \right )}-\frac {5 \ln \left (b s +b x +\sqrt {-a b}\right )}{32 \sqrt {-a b}\, a^{3}}+\frac {5 \ln \left (-b s -b x +\sqrt {-a b}\right )}{32 \sqrt {-a b}\, a^{3}}\) | \(260\) |
Input:
int(1/((b*s^2+a)^2+4*b*s*(b*s^2+a)*x+2*b*(3*b*s^2+a)*x^2+4*b^2*s*x^3+b^2*x ^4)^2,x,method=_RETURNVERBOSE)
Output:
1/12*(2*b*s+2*b*x)/a/b/(b*s^2+2*b*s*x+b*x^2+a)^3+5/6/a*(1/8*(2*b*s+2*b*x)/ a/b/(b*s^2+2*b*s*x+b*x^2+a)^2+3/4/a*(1/4*(2*b*s+2*b*x)/a/b/(b*s^2+2*b*s*x+ b*x^2+a)+1/2/a/(a*b)^(1/2)*arctan(1/2*(2*b*s+2*b*x)/(a*b)^(1/2))))
Leaf count of result is larger than twice the leaf count of optimal. 465 vs. \(2 (77) = 154\).
Time = 0.09 (sec) , antiderivative size = 970, normalized size of antiderivative = 10.43 \[ \int \frac {1}{\left (\left (a+b s^2\right )^2+4 b s \left (a+b s^2\right ) x+2 b \left (a+3 b s^2\right ) x^2+4 b^2 s x^3+b^2 x^4\right )^2} \, dx =\text {Too large to display} \] Input:
integrate(1/((b*s^2+a)^2+4*b*s*(b*s^2+a)*x+2*b*(3*b*s^2+a)*x^2+4*b^2*s*x^3 +b^2*x^4)^2,x, algorithm="fricas")
Output:
[1/96*(30*a*b^3*s^5 + 150*a*b^3*s*x^4 + 30*a*b^3*x^5 + 80*a^2*b^2*s^3 + 66 *a^3*b*s + 20*(15*a*b^3*s^2 + 4*a^2*b^2)*x^3 + 60*(5*a*b^3*s^3 + 4*a^2*b^2 *s)*x^2 - 15*(b^3*s^6 + 6*b^3*s*x^5 + b^3*x^6 + 3*a*b^2*s^4 + 3*a^2*b*s^2 + 3*(5*b^3*s^2 + a*b^2)*x^4 + 4*(5*b^3*s^3 + 3*a*b^2*s)*x^3 + a^3 + 3*(5*b ^3*s^4 + 6*a*b^2*s^2 + a^2*b)*x^2 + 6*(b^3*s^5 + 2*a*b^2*s^3 + a^2*b*s)*x) *sqrt(-a*b)*log((b*s^2 + 2*b*s*x + b*x^2 - 2*sqrt(-a*b)*(s + x) - a)/(b*s^ 2 + 2*b*s*x + b*x^2 + a)) + 6*(25*a*b^3*s^4 + 40*a^2*b^2*s^2 + 11*a^3*b)*x )/(a^4*b^4*s^6 + 6*a^4*b^4*s*x^5 + a^4*b^4*x^6 + 3*a^5*b^3*s^4 + 3*a^6*b^2 *s^2 + a^7*b + 3*(5*a^4*b^4*s^2 + a^5*b^3)*x^4 + 4*(5*a^4*b^4*s^3 + 3*a^5* b^3*s)*x^3 + 3*(5*a^4*b^4*s^4 + 6*a^5*b^3*s^2 + a^6*b^2)*x^2 + 6*(a^4*b^4* s^5 + 2*a^5*b^3*s^3 + a^6*b^2*s)*x), 1/48*(15*a*b^3*s^5 + 75*a*b^3*s*x^4 + 15*a*b^3*x^5 + 40*a^2*b^2*s^3 + 33*a^3*b*s + 10*(15*a*b^3*s^2 + 4*a^2*b^2 )*x^3 + 30*(5*a*b^3*s^3 + 4*a^2*b^2*s)*x^2 + 15*(b^3*s^6 + 6*b^3*s*x^5 + b ^3*x^6 + 3*a*b^2*s^4 + 3*a^2*b*s^2 + 3*(5*b^3*s^2 + a*b^2)*x^4 + 4*(5*b^3* s^3 + 3*a*b^2*s)*x^3 + a^3 + 3*(5*b^3*s^4 + 6*a*b^2*s^2 + a^2*b)*x^2 + 6*( b^3*s^5 + 2*a*b^2*s^3 + a^2*b*s)*x)*sqrt(a*b)*arctan(sqrt(a*b)*(s + x)/a) + 3*(25*a*b^3*s^4 + 40*a^2*b^2*s^2 + 11*a^3*b)*x)/(a^4*b^4*s^6 + 6*a^4*b^4 *s*x^5 + a^4*b^4*x^6 + 3*a^5*b^3*s^4 + 3*a^6*b^2*s^2 + a^7*b + 3*(5*a^4*b^ 4*s^2 + a^5*b^3)*x^4 + 4*(5*a^4*b^4*s^3 + 3*a^5*b^3*s)*x^3 + 3*(5*a^4*b^4* s^4 + 6*a^5*b^3*s^2 + a^6*b^2)*x^2 + 6*(a^4*b^4*s^5 + 2*a^5*b^3*s^3 + a...
Leaf count of result is larger than twice the leaf count of optimal. 354 vs. \(2 (92) = 184\).
Time = 0.95 (sec) , antiderivative size = 354, normalized size of antiderivative = 3.81 \[ \int \frac {1}{\left (\left (a+b s^2\right )^2+4 b s \left (a+b s^2\right ) x+2 b \left (a+3 b s^2\right ) x^2+4 b^2 s x^3+b^2 x^4\right )^2} \, dx=- \frac {5 \sqrt {- \frac {1}{a^{7} b}} \log {\left (- a^{4} \sqrt {- \frac {1}{a^{7} b}} + s + x \right )}}{32} + \frac {5 \sqrt {- \frac {1}{a^{7} b}} \log {\left (a^{4} \sqrt {- \frac {1}{a^{7} b}} + s + x \right )}}{32} + \frac {33 a^{2} s + 40 a b s^{3} + 15 b^{2} s^{5} + 75 b^{2} s x^{4} + 15 b^{2} x^{5} + x^{3} \cdot \left (40 a b + 150 b^{2} s^{2}\right ) + x^{2} \cdot \left (120 a b s + 150 b^{2} s^{3}\right ) + x \left (33 a^{2} + 120 a b s^{2} + 75 b^{2} s^{4}\right )}{48 a^{6} + 144 a^{5} b s^{2} + 144 a^{4} b^{2} s^{4} + 48 a^{3} b^{3} s^{6} + 288 a^{3} b^{3} s x^{5} + 48 a^{3} b^{3} x^{6} + x^{4} \cdot \left (144 a^{4} b^{2} + 720 a^{3} b^{3} s^{2}\right ) + x^{3} \cdot \left (576 a^{4} b^{2} s + 960 a^{3} b^{3} s^{3}\right ) + x^{2} \cdot \left (144 a^{5} b + 864 a^{4} b^{2} s^{2} + 720 a^{3} b^{3} s^{4}\right ) + x \left (288 a^{5} b s + 576 a^{4} b^{2} s^{3} + 288 a^{3} b^{3} s^{5}\right )} \] Input:
integrate(1/((b*s**2+a)**2+4*b*s*(b*s**2+a)*x+2*b*(3*b*s**2+a)*x**2+4*b**2 *s*x**3+b**2*x**4)**2,x)
Output:
-5*sqrt(-1/(a**7*b))*log(-a**4*sqrt(-1/(a**7*b)) + s + x)/32 + 5*sqrt(-1/( a**7*b))*log(a**4*sqrt(-1/(a**7*b)) + s + x)/32 + (33*a**2*s + 40*a*b*s**3 + 15*b**2*s**5 + 75*b**2*s*x**4 + 15*b**2*x**5 + x**3*(40*a*b + 150*b**2* s**2) + x**2*(120*a*b*s + 150*b**2*s**3) + x*(33*a**2 + 120*a*b*s**2 + 75* b**2*s**4))/(48*a**6 + 144*a**5*b*s**2 + 144*a**4*b**2*s**4 + 48*a**3*b**3 *s**6 + 288*a**3*b**3*s*x**5 + 48*a**3*b**3*x**6 + x**4*(144*a**4*b**2 + 7 20*a**3*b**3*s**2) + x**3*(576*a**4*b**2*s + 960*a**3*b**3*s**3) + x**2*(1 44*a**5*b + 864*a**4*b**2*s**2 + 720*a**3*b**3*s**4) + x*(288*a**5*b*s + 5 76*a**4*b**2*s**3 + 288*a**3*b**3*s**5))
Leaf count of result is larger than twice the leaf count of optimal. 299 vs. \(2 (77) = 154\).
Time = 0.11 (sec) , antiderivative size = 299, normalized size of antiderivative = 3.22 \[ \int \frac {1}{\left (\left (a+b s^2\right )^2+4 b s \left (a+b s^2\right ) x+2 b \left (a+3 b s^2\right ) x^2+4 b^2 s x^3+b^2 x^4\right )^2} \, dx=\frac {15 \, b^{2} s^{5} + 75 \, b^{2} s x^{4} + 15 \, b^{2} x^{5} + 40 \, a b s^{3} + 10 \, {\left (15 \, b^{2} s^{2} + 4 \, a b\right )} x^{3} + 33 \, a^{2} s + 30 \, {\left (5 \, b^{2} s^{3} + 4 \, a b s\right )} x^{2} + 3 \, {\left (25 \, b^{2} s^{4} + 40 \, a b s^{2} + 11 \, a^{2}\right )} x}{48 \, {\left (a^{3} b^{3} s^{6} + 6 \, a^{3} b^{3} s x^{5} + a^{3} b^{3} x^{6} + 3 \, a^{4} b^{2} s^{4} + 3 \, a^{5} b s^{2} + a^{6} + 3 \, {\left (5 \, a^{3} b^{3} s^{2} + a^{4} b^{2}\right )} x^{4} + 4 \, {\left (5 \, a^{3} b^{3} s^{3} + 3 \, a^{4} b^{2} s\right )} x^{3} + 3 \, {\left (5 \, a^{3} b^{3} s^{4} + 6 \, a^{4} b^{2} s^{2} + a^{5} b\right )} x^{2} + 6 \, {\left (a^{3} b^{3} s^{5} + 2 \, a^{4} b^{2} s^{3} + a^{5} b s\right )} x\right )}} + \frac {5 \, \arctan \left (\frac {b s + b x}{\sqrt {a b}}\right )}{16 \, \sqrt {a b} a^{3}} \] Input:
integrate(1/((b*s^2+a)^2+4*b*s*(b*s^2+a)*x+2*b*(3*b*s^2+a)*x^2+4*b^2*s*x^3 +b^2*x^4)^2,x, algorithm="maxima")
Output:
1/48*(15*b^2*s^5 + 75*b^2*s*x^4 + 15*b^2*x^5 + 40*a*b*s^3 + 10*(15*b^2*s^2 + 4*a*b)*x^3 + 33*a^2*s + 30*(5*b^2*s^3 + 4*a*b*s)*x^2 + 3*(25*b^2*s^4 + 40*a*b*s^2 + 11*a^2)*x)/(a^3*b^3*s^6 + 6*a^3*b^3*s*x^5 + a^3*b^3*x^6 + 3*a ^4*b^2*s^4 + 3*a^5*b*s^2 + a^6 + 3*(5*a^3*b^3*s^2 + a^4*b^2)*x^4 + 4*(5*a^ 3*b^3*s^3 + 3*a^4*b^2*s)*x^3 + 3*(5*a^3*b^3*s^4 + 6*a^4*b^2*s^2 + a^5*b)*x ^2 + 6*(a^3*b^3*s^5 + 2*a^4*b^2*s^3 + a^5*b*s)*x) + 5/16*arctan((b*s + b*x )/sqrt(a*b))/(sqrt(a*b)*a^3)
Time = 0.11 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.59 \[ \int \frac {1}{\left (\left (a+b s^2\right )^2+4 b s \left (a+b s^2\right ) x+2 b \left (a+3 b s^2\right ) x^2+4 b^2 s x^3+b^2 x^4\right )^2} \, dx=\frac {5 \, \arctan \left (\frac {b s + b x}{\sqrt {a b}}\right )}{16 \, \sqrt {a b} a^{3}} + \frac {15 \, b^{2} s^{5} + 75 \, b^{2} s^{4} x + 150 \, b^{2} s^{3} x^{2} + 150 \, b^{2} s^{2} x^{3} + 75 \, b^{2} s x^{4} + 15 \, b^{2} x^{5} + 40 \, a b s^{3} + 120 \, a b s^{2} x + 120 \, a b s x^{2} + 40 \, a b x^{3} + 33 \, a^{2} s + 33 \, a^{2} x}{48 \, {\left (b s^{2} + 2 \, b s x + b x^{2} + a\right )}^{3} a^{3}} \] Input:
integrate(1/((b*s^2+a)^2+4*b*s*(b*s^2+a)*x+2*b*(3*b*s^2+a)*x^2+4*b^2*s*x^3 +b^2*x^4)^2,x, algorithm="giac")
Output:
5/16*arctan((b*s + b*x)/sqrt(a*b))/(sqrt(a*b)*a^3) + 1/48*(15*b^2*s^5 + 75 *b^2*s^4*x + 150*b^2*s^3*x^2 + 150*b^2*s^2*x^3 + 75*b^2*s*x^4 + 15*b^2*x^5 + 40*a*b*s^3 + 120*a*b*s^2*x + 120*a*b*s*x^2 + 40*a*b*x^3 + 33*a^2*s + 33 *a^2*x)/((b*s^2 + 2*b*s*x + b*x^2 + a)^3*a^3)
Time = 21.62 (sec) , antiderivative size = 297, normalized size of antiderivative = 3.19 \[ \int \frac {1}{\left (\left (a+b s^2\right )^2+4 b s \left (a+b s^2\right ) x+2 b \left (a+3 b s^2\right ) x^2+4 b^2 s x^3+b^2 x^4\right )^2} \, dx=\frac {\frac {33\,a^2\,s+40\,a\,b\,s^3+15\,b^2\,s^5}{48\,a^3}+\frac {5\,x^3\,\left (15\,b^2\,s^2+4\,a\,b\right )}{24\,a^3}+\frac {5\,b^2\,x^5}{16\,a^3}+\frac {5\,x^2\,\left (5\,b^2\,s^3+4\,a\,b\,s\right )}{8\,a^3}+\frac {x\,\left (11\,a^2+40\,a\,b\,s^2+25\,b^2\,s^4\right )}{16\,a^3}+\frac {25\,b^2\,s\,x^4}{16\,a^3}}{x^4\,\left (15\,b^3\,s^2+3\,a\,b^2\right )+x^2\,\left (3\,a^2\,b+18\,a\,b^2\,s^2+15\,b^3\,s^4\right )+x\,\left (6\,a^2\,b\,s+12\,a\,b^2\,s^3+6\,b^3\,s^5\right )+x^3\,\left (20\,b^3\,s^3+12\,a\,b^2\,s\right )+a^3+b^3\,s^6+b^3\,x^6+3\,a^2\,b\,s^2+3\,a\,b^2\,s^4+6\,b^3\,s\,x^5}+\frac {5\,\mathrm {atan}\left (\frac {16\,a^3\,\left (\frac {5\,\sqrt {b}\,s}{16\,a^{7/2}}+\frac {5\,\sqrt {b}\,x}{16\,a^{7/2}}\right )}{5}\right )}{16\,a^{7/2}\,\sqrt {b}} \] Input:
int(1/((a + b*s^2)^2 + b^2*x^4 + 4*b^2*s*x^3 + 2*b*x^2*(a + 3*b*s^2) + 4*b *s*x*(a + b*s^2))^2,x)
Output:
((33*a^2*s + 15*b^2*s^5 + 40*a*b*s^3)/(48*a^3) + (5*x^3*(4*a*b + 15*b^2*s^ 2))/(24*a^3) + (5*b^2*x^5)/(16*a^3) + (5*x^2*(5*b^2*s^3 + 4*a*b*s))/(8*a^3 ) + (x*(11*a^2 + 25*b^2*s^4 + 40*a*b*s^2))/(16*a^3) + (25*b^2*s*x^4)/(16*a ^3))/(x^4*(3*a*b^2 + 15*b^3*s^2) + x^2*(3*a^2*b + 15*b^3*s^4 + 18*a*b^2*s^ 2) + x*(6*b^3*s^5 + 12*a*b^2*s^3 + 6*a^2*b*s) + x^3*(20*b^3*s^3 + 12*a*b^2 *s) + a^3 + b^3*s^6 + b^3*x^6 + 3*a^2*b*s^2 + 3*a*b^2*s^4 + 6*b^3*s*x^5) + (5*atan((16*a^3*((5*b^(1/2)*s)/(16*a^(7/2)) + (5*b^(1/2)*x)/(16*a^(7/2))) )/5))/(16*a^(7/2)*b^(1/2))
Time = 0.16 (sec) , antiderivative size = 808, normalized size of antiderivative = 8.69 \[ \int \frac {1}{\left (\left (a+b s^2\right )^2+4 b s \left (a+b s^2\right ) x+2 b \left (a+3 b s^2\right ) x^2+4 b^2 s x^3+b^2 x^4\right )^2} \, dx =\text {Too large to display} \] Input:
int(1/((b*s^2+a)^2+4*b*s*(b*s^2+a)*x+2*b*(3*b*s^2+a)*x^2+4*b^2*s*x^3+b^2*x ^4)^2,x)
Output:
(30*sqrt(b)*sqrt(a)*atan((b*s + b*x)/(sqrt(b)*sqrt(a)))*a**3*s + 90*sqrt(b )*sqrt(a)*atan((b*s + b*x)/(sqrt(b)*sqrt(a)))*a**2*b*s**3 + 180*sqrt(b)*sq rt(a)*atan((b*s + b*x)/(sqrt(b)*sqrt(a)))*a**2*b*s**2*x + 90*sqrt(b)*sqrt( a)*atan((b*s + b*x)/(sqrt(b)*sqrt(a)))*a**2*b*s*x**2 + 90*sqrt(b)*sqrt(a)* atan((b*s + b*x)/(sqrt(b)*sqrt(a)))*a*b**2*s**5 + 360*sqrt(b)*sqrt(a)*atan ((b*s + b*x)/(sqrt(b)*sqrt(a)))*a*b**2*s**4*x + 540*sqrt(b)*sqrt(a)*atan(( b*s + b*x)/(sqrt(b)*sqrt(a)))*a*b**2*s**3*x**2 + 360*sqrt(b)*sqrt(a)*atan( (b*s + b*x)/(sqrt(b)*sqrt(a)))*a*b**2*s**2*x**3 + 90*sqrt(b)*sqrt(a)*atan( (b*s + b*x)/(sqrt(b)*sqrt(a)))*a*b**2*s*x**4 + 30*sqrt(b)*sqrt(a)*atan((b* s + b*x)/(sqrt(b)*sqrt(a)))*b**3*s**7 + 180*sqrt(b)*sqrt(a)*atan((b*s + b* x)/(sqrt(b)*sqrt(a)))*b**3*s**6*x + 450*sqrt(b)*sqrt(a)*atan((b*s + b*x)/( sqrt(b)*sqrt(a)))*b**3*s**5*x**2 + 600*sqrt(b)*sqrt(a)*atan((b*s + b*x)/(s qrt(b)*sqrt(a)))*b**3*s**4*x**3 + 450*sqrt(b)*sqrt(a)*atan((b*s + b*x)/(sq rt(b)*sqrt(a)))*b**3*s**3*x**4 + 180*sqrt(b)*sqrt(a)*atan((b*s + b*x)/(sqr t(b)*sqrt(a)))*b**3*s**2*x**5 + 30*sqrt(b)*sqrt(a)*atan((b*s + b*x)/(sqrt( b)*sqrt(a)))*b**3*s*x**6 - 5*a**4 + 51*a**3*b*s**2 + 36*a**3*b*s*x - 15*a* *3*b*x**2 + 65*a**2*b**2*s**4 + 180*a**2*b**2*s**3*x + 150*a**2*b**2*s**2* x**2 + 20*a**2*b**2*s*x**3 - 15*a**2*b**2*x**4 + 25*a*b**3*s**6 + 120*a*b* *3*s**5*x + 225*a*b**3*s**4*x**2 + 200*a*b**3*s**3*x**3 + 75*a*b**3*s**2*x **4 - 5*a*b**3*x**6)/(96*a**4*b*s*(a**3 + 3*a**2*b*s**2 + 6*a**2*b*s*x ...