Integrand size = 26, antiderivative size = 348 \[ \int \frac {\left (a+b x^2\right ) \left (c+d x^2\right )^3}{\left (e+f x^2\right )^4} \, dx=\frac {d \left (b e \left (105 d^2 e^2-10 c d e f-3 c^2 f^2\right )-a f \left (15 d^2 e^2+14 c d e f+15 c^2 f^2\right )\right ) x}{48 e^3 f^4}-\frac {(b e-a f) x \left (c+d x^2\right )^3}{6 e f \left (e+f x^2\right )^3}-\frac {(b e (7 d e-c f)-a f (d e+5 c f)) x \left (c+d x^2\right )^2}{24 e^2 f^2 \left (e+f x^2\right )^2}-\frac {\left (b e \left (35 d^2 e^2-8 c d e f-3 c^2 f^2\right )-a f \left (5 d^2 e^2+4 c d e f+15 c^2 f^2\right )\right ) x \left (c+d x^2\right )}{48 e^3 f^3 \left (e+f x^2\right )}-\frac {\left (b e \left (35 d^3 e^3-15 c d^2 e^2 f-3 c^2 d e f^2-c^3 f^3\right )-a f \left (5 d^3 e^3+3 c d^2 e^2 f+3 c^2 d e f^2+5 c^3 f^3\right )\right ) \arctan \left (\frac {\sqrt {f} x}{\sqrt {e}}\right )}{16 e^{7/2} f^{9/2}} \]
1/48*d*(b*e*(-3*c^2*f^2-10*c*d*e*f+105*d^2*e^2)-a*f*(15*c^2*f^2+14*c*d*e*f +15*d^2*e^2))*x/e^3/f^4-1/6*(-a*f+b*e)*x*(d*x^2+c)^3/e/f/(f*x^2+e)^3-1/24* (b*e*(-c*f+7*d*e)-a*f*(5*c*f+d*e))*x*(d*x^2+c)^2/e^2/f^2/(f*x^2+e)^2-1/48* (b*e*(-3*c^2*f^2-8*c*d*e*f+35*d^2*e^2)-a*f*(15*c^2*f^2+4*c*d*e*f+5*d^2*e^2 ))*x*(d*x^2+c)/e^3/f^3/(f*x^2+e)-1/16*(b*e*(-c^3*f^3-3*c^2*d*e*f^2-15*c*d^ 2*e^2*f+35*d^3*e^3)-a*f*(5*c^3*f^3+3*c^2*d*e*f^2+3*c*d^2*e^2*f+5*d^3*e^3)) *arctan(x*f^(1/2)/e^(1/2))/e^(7/2)/f^(9/2)
Time = 0.13 (sec) , antiderivative size = 295, normalized size of antiderivative = 0.85 \[ \int \frac {\left (a+b x^2\right ) \left (c+d x^2\right )^3}{\left (e+f x^2\right )^4} \, dx=\frac {b d^3 x}{f^4}+\frac {(b e-a f) (d e-c f)^3 x}{6 e f^4 \left (e+f x^2\right )^3}-\frac {(d e-c f)^2 (b e (19 d e-c f)-a f (13 d e+5 c f)) x}{24 e^2 f^4 \left (e+f x^2\right )^2}+\frac {(d e-c f) \left (b e \left (29 d^2 e^2-4 c d e f-c^2 f^2\right )-a f \left (11 d^2 e^2+8 c d e f+5 c^2 f^2\right )\right ) x}{16 e^3 f^4 \left (e+f x^2\right )}-\frac {\left (b e \left (35 d^3 e^3-15 c d^2 e^2 f-3 c^2 d e f^2-c^3 f^3\right )-a f \left (5 d^3 e^3+3 c d^2 e^2 f+3 c^2 d e f^2+5 c^3 f^3\right )\right ) \arctan \left (\frac {\sqrt {f} x}{\sqrt {e}}\right )}{16 e^{7/2} f^{9/2}} \]
(b*d^3*x)/f^4 + ((b*e - a*f)*(d*e - c*f)^3*x)/(6*e*f^4*(e + f*x^2)^3) - (( d*e - c*f)^2*(b*e*(19*d*e - c*f) - a*f*(13*d*e + 5*c*f))*x)/(24*e^2*f^4*(e + f*x^2)^2) + ((d*e - c*f)*(b*e*(29*d^2*e^2 - 4*c*d*e*f - c^2*f^2) - a*f* (11*d^2*e^2 + 8*c*d*e*f + 5*c^2*f^2))*x)/(16*e^3*f^4*(e + f*x^2)) - ((b*e* (35*d^3*e^3 - 15*c*d^2*e^2*f - 3*c^2*d*e*f^2 - c^3*f^3) - a*f*(5*d^3*e^3 + 3*c*d^2*e^2*f + 3*c^2*d*e*f^2 + 5*c^3*f^3))*ArcTan[(Sqrt[f]*x)/Sqrt[e]])/ (16*e^(7/2)*f^(9/2))
Time = 0.60 (sec) , antiderivative size = 374, normalized size of antiderivative = 1.07, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.269, Rules used = {401, 25, 401, 25, 401, 299, 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.
| \(\displaystyle \int \frac {\left (a+b x^2\right ) \left (c+d x^2\right )^3}{\left (e+f x^2\right )^4} \, dx\) |
| \(\Big \downarrow \) 401 |
| \(\displaystyle -\frac {\int -\frac {\left (d x^2+c\right )^2 \left (d (7 b e-a f) x^2+c (b e+5 a f)\right )}{\left (f x^2+e\right )^3}dx}{6 e f}-\frac {x \left (c+d x^2\right )^3 (b e-a f)}{6 e f \left (e+f x^2\right )^3}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\int \frac {\left (d x^2+c\right )^2 \left (d (7 b e-a f) x^2+c (b e+5 a f)\right )}{\left (f x^2+e\right )^3}dx}{6 e f}-\frac {x \left (c+d x^2\right )^3 (b e-a f)}{6 e f \left (e+f x^2\right )^3}\) |
| \(\Big \downarrow \) 401 |
| \(\displaystyle \frac {-\frac {\int -\frac {\left (d x^2+c\right ) \left (d (b e (35 d e-c f)-5 a f (d e+c f)) x^2+c (d e (7 b e-a f)+3 c f (b e+5 a f))\right )}{\left (f x^2+e\right )^2}dx}{4 e f}-\frac {x \left (c+d x^2\right )^2 (b e (7 d e-c f)-a f (5 c f+d e))}{4 e f \left (e+f x^2\right )^2}}{6 e f}-\frac {x \left (c+d x^2\right )^3 (b e-a f)}{6 e f \left (e+f x^2\right )^3}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {\int \frac {\left (d x^2+c\right ) \left (d (b e (35 d e-c f)-5 a f (d e+c f)) x^2+c (d e (7 b e-a f)+3 c f (b e+5 a f))\right )}{\left (f x^2+e\right )^2}dx}{4 e f}-\frac {x \left (c+d x^2\right )^2 (b e (7 d e-c f)-a f (5 c f+d e))}{4 e f \left (e+f x^2\right )^2}}{6 e f}-\frac {x \left (c+d x^2\right )^3 (b e-a f)}{6 e f \left (e+f x^2\right )^3}\) |
| \(\Big \downarrow \) 401 |
| \(\displaystyle \frac {\frac {-\frac {\int \frac {c \left (a f \left (5 d^2 e^2+6 c d f e-15 c^2 f^2\right )-b e \left (35 d^2 e^2+6 c d f e+3 c^2 f^2\right )\right )-d \left (b e \left (105 d^2 e^2-10 c d f e-3 c^2 f^2\right )-a f \left (15 d^2 e^2+14 c d f e+15 c^2 f^2\right )\right ) x^2}{f x^2+e}dx}{2 e f}-\frac {x \left (c+d x^2\right ) \left (b e \left (-3 c^2 f^2-8 c d e f+35 d^2 e^2\right )-a f \left (15 c^2 f^2+4 c d e f+5 d^2 e^2\right )\right )}{2 e f \left (e+f x^2\right )}}{4 e f}-\frac {x \left (c+d x^2\right )^2 (b e (7 d e-c f)-a f (5 c f+d e))}{4 e f \left (e+f x^2\right )^2}}{6 e f}-\frac {x \left (c+d x^2\right )^3 (b e-a f)}{6 e f \left (e+f x^2\right )^3}\) |
| \(\Big \downarrow \) 299 |
| \(\displaystyle \frac {\frac {-\frac {\frac {3 \left (b e \left (-c^3 f^3-3 c^2 d e f^2-15 c d^2 e^2 f+35 d^3 e^3\right )-a f \left (5 c^3 f^3+3 c^2 d e f^2+3 c d^2 e^2 f+5 d^3 e^3\right )\right ) \int \frac {1}{f x^2+e}dx}{f}-\frac {d x \left (b e \left (-3 c^2 f^2-10 c d e f+105 d^2 e^2\right )-a f \left (15 c^2 f^2+14 c d e f+15 d^2 e^2\right )\right )}{f}}{2 e f}-\frac {x \left (c+d x^2\right ) \left (b e \left (-3 c^2 f^2-8 c d e f+35 d^2 e^2\right )-a f \left (15 c^2 f^2+4 c d e f+5 d^2 e^2\right )\right )}{2 e f \left (e+f x^2\right )}}{4 e f}-\frac {x \left (c+d x^2\right )^2 (b e (7 d e-c f)-a f (5 c f+d e))}{4 e f \left (e+f x^2\right )^2}}{6 e f}-\frac {x \left (c+d x^2\right )^3 (b e-a f)}{6 e f \left (e+f x^2\right )^3}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {\frac {-\frac {\frac {3 \arctan \left (\frac {\sqrt {f} x}{\sqrt {e}}\right ) \left (b e \left (-c^3 f^3-3 c^2 d e f^2-15 c d^2 e^2 f+35 d^3 e^3\right )-a f \left (5 c^3 f^3+3 c^2 d e f^2+3 c d^2 e^2 f+5 d^3 e^3\right )\right )}{\sqrt {e} f^{3/2}}-\frac {d x \left (b e \left (-3 c^2 f^2-10 c d e f+105 d^2 e^2\right )-a f \left (15 c^2 f^2+14 c d e f+15 d^2 e^2\right )\right )}{f}}{2 e f}-\frac {x \left (c+d x^2\right ) \left (b e \left (-3 c^2 f^2-8 c d e f+35 d^2 e^2\right )-a f \left (15 c^2 f^2+4 c d e f+5 d^2 e^2\right )\right )}{2 e f \left (e+f x^2\right )}}{4 e f}-\frac {x \left (c+d x^2\right )^2 (b e (7 d e-c f)-a f (5 c f+d e))}{4 e f \left (e+f x^2\right )^2}}{6 e f}-\frac {x \left (c+d x^2\right )^3 (b e-a f)}{6 e f \left (e+f x^2\right )^3}\) |
-1/6*((b*e - a*f)*x*(c + d*x^2)^3)/(e*f*(e + f*x^2)^3) + (-1/4*((b*e*(7*d* e - c*f) - a*f*(d*e + 5*c*f))*x*(c + d*x^2)^2)/(e*f*(e + f*x^2)^2) + (-1/2 *((b*e*(35*d^2*e^2 - 8*c*d*e*f - 3*c^2*f^2) - a*f*(5*d^2*e^2 + 4*c*d*e*f + 15*c^2*f^2))*x*(c + d*x^2))/(e*f*(e + f*x^2)) - (-((d*(b*e*(105*d^2*e^2 - 10*c*d*e*f - 3*c^2*f^2) - a*f*(15*d^2*e^2 + 14*c*d*e*f + 15*c^2*f^2))*x)/ f) + (3*(b*e*(35*d^3*e^3 - 15*c*d^2*e^2*f - 3*c^2*d*e*f^2 - c^3*f^3) - a*f *(5*d^3*e^3 + 3*c*d^2*e^2*f + 3*c^2*d*e*f^2 + 5*c^3*f^3))*ArcTan[(Sqrt[f]* x)/Sqrt[e]])/(Sqrt[e]*f^(3/2)))/(2*e*f))/(4*e*f))/(6*e*f)
3.1.22.3.1 Defintions of rubi rules used
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_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[d*x *((a + b*x^2)^(p + 1)/(b*(2*p + 3))), x] - Simp[(a*d - b*c*(2*p + 3))/(b*(2 *p + 3)) Int[(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && NeQ[2*p + 3, 0]
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_.)*((e_) + (f_.)*(x _)^2), x_Symbol] :> Simp[(-(b*e - a*f))*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^ q/(a*b*2*(p + 1))), x] + Simp[1/(a*b*2*(p + 1)) Int[(a + b*x^2)^(p + 1)*( c + d*x^2)^(q - 1)*Simp[c*(b*e*2*(p + 1) + b*e - a*f) + d*(b*e*2*(p + 1) + (b*e - a*f)*(2*q + 1))*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && L tQ[p, -1] && GtQ[q, 0]
Time = 3.37 (sec) , antiderivative size = 417, normalized size of antiderivative = 1.20
| method | result | size |
| default | \(\frac {b \,d^{3} x}{f^{4}}+\frac {\frac {\frac {f^{2} \left (5 a \,c^{3} f^{4}+3 a \,c^{2} d e \,f^{3}+3 a c \,d^{2} e^{2} f^{2}-11 a \,d^{3} e^{3} f +b \,c^{3} e \,f^{3}+3 b \,c^{2} d \,e^{2} f^{2}-33 b c \,d^{2} e^{3} f +29 b \,d^{3} e^{4}\right ) x^{5}}{16 e^{3}}+\frac {f \left (5 a \,c^{3} f^{4}+3 a \,c^{2} d e \,f^{3}-3 a c \,d^{2} e^{2} f^{2}-5 a \,d^{3} e^{3} f +b \,c^{3} e \,f^{3}-3 b \,c^{2} d \,e^{2} f^{2}-15 b c \,d^{2} e^{3} f +17 b \,d^{3} e^{4}\right ) x^{3}}{6 e^{2}}+\frac {\left (11 a \,c^{3} f^{4}-3 a \,c^{2} d e \,f^{3}-3 a c \,d^{2} e^{2} f^{2}-5 a \,d^{3} e^{3} f -b \,c^{3} e \,f^{3}-3 b \,c^{2} d \,e^{2} f^{2}-15 b c \,d^{2} e^{3} f +19 b \,d^{3} e^{4}\right ) x}{16 e}}{\left (f \,x^{2}+e \right )^{3}}+\frac {\left (5 a \,c^{3} f^{4}+3 a \,c^{2} d e \,f^{3}+3 a c \,d^{2} e^{2} f^{2}+5 a \,d^{3} e^{3} f +b \,c^{3} e \,f^{3}+3 b \,c^{2} d \,e^{2} f^{2}+15 b c \,d^{2} e^{3} f -35 b \,d^{3} e^{4}\right ) \arctan \left (\frac {f x}{\sqrt {e f}}\right )}{16 e^{3} \sqrt {e f}}}{f^{4}}\) | \(417\) |
| risch | \(\frac {b \,d^{3} x}{f^{4}}+\frac {\frac {f^{2} \left (5 a \,c^{3} f^{4}+3 a \,c^{2} d e \,f^{3}+3 a c \,d^{2} e^{2} f^{2}-11 a \,d^{3} e^{3} f +b \,c^{3} e \,f^{3}+3 b \,c^{2} d \,e^{2} f^{2}-33 b c \,d^{2} e^{3} f +29 b \,d^{3} e^{4}\right ) x^{5}}{16 e^{3}}+\frac {f \left (5 a \,c^{3} f^{4}+3 a \,c^{2} d e \,f^{3}-3 a c \,d^{2} e^{2} f^{2}-5 a \,d^{3} e^{3} f +b \,c^{3} e \,f^{3}-3 b \,c^{2} d \,e^{2} f^{2}-15 b c \,d^{2} e^{3} f +17 b \,d^{3} e^{4}\right ) x^{3}}{6 e^{2}}+\frac {\left (11 a \,c^{3} f^{4}-3 a \,c^{2} d e \,f^{3}-3 a c \,d^{2} e^{2} f^{2}-5 a \,d^{3} e^{3} f -b \,c^{3} e \,f^{3}-3 b \,c^{2} d \,e^{2} f^{2}-15 b c \,d^{2} e^{3} f +19 b \,d^{3} e^{4}\right ) x}{16 e}}{f^{4} \left (f \,x^{2}+e \right )^{3}}-\frac {5 \ln \left (f x +\sqrt {-e f}\right ) a \,c^{3}}{32 \sqrt {-e f}\, e^{3}}-\frac {3 \ln \left (f x +\sqrt {-e f}\right ) a \,c^{2} d}{32 f \sqrt {-e f}\, e^{2}}-\frac {3 \ln \left (f x +\sqrt {-e f}\right ) a c \,d^{2}}{32 f^{2} \sqrt {-e f}\, e}-\frac {5 \ln \left (f x +\sqrt {-e f}\right ) a \,d^{3}}{32 f^{3} \sqrt {-e f}}-\frac {\ln \left (f x +\sqrt {-e f}\right ) b \,c^{3}}{32 f \sqrt {-e f}\, e^{2}}-\frac {3 \ln \left (f x +\sqrt {-e f}\right ) b \,c^{2} d}{32 f^{2} \sqrt {-e f}\, e}-\frac {15 \ln \left (f x +\sqrt {-e f}\right ) b c \,d^{2}}{32 f^{3} \sqrt {-e f}}+\frac {35 e \ln \left (f x +\sqrt {-e f}\right ) b \,d^{3}}{32 f^{4} \sqrt {-e f}}+\frac {5 \ln \left (-f x +\sqrt {-e f}\right ) a \,c^{3}}{32 \sqrt {-e f}\, e^{3}}+\frac {3 \ln \left (-f x +\sqrt {-e f}\right ) a \,c^{2} d}{32 f \sqrt {-e f}\, e^{2}}+\frac {3 \ln \left (-f x +\sqrt {-e f}\right ) a c \,d^{2}}{32 f^{2} \sqrt {-e f}\, e}+\frac {5 \ln \left (-f x +\sqrt {-e f}\right ) a \,d^{3}}{32 f^{3} \sqrt {-e f}}+\frac {\ln \left (-f x +\sqrt {-e f}\right ) b \,c^{3}}{32 f \sqrt {-e f}\, e^{2}}+\frac {3 \ln \left (-f x +\sqrt {-e f}\right ) b \,c^{2} d}{32 f^{2} \sqrt {-e f}\, e}+\frac {15 \ln \left (-f x +\sqrt {-e f}\right ) b c \,d^{2}}{32 f^{3} \sqrt {-e f}}-\frac {35 e \ln \left (-f x +\sqrt {-e f}\right ) b \,d^{3}}{32 f^{4} \sqrt {-e f}}\) | \(768\) |
b*d^3/f^4*x+1/f^4*((1/16*f^2*(5*a*c^3*f^4+3*a*c^2*d*e*f^3+3*a*c*d^2*e^2*f^ 2-11*a*d^3*e^3*f+b*c^3*e*f^3+3*b*c^2*d*e^2*f^2-33*b*c*d^2*e^3*f+29*b*d^3*e ^4)/e^3*x^5+1/6*f*(5*a*c^3*f^4+3*a*c^2*d*e*f^3-3*a*c*d^2*e^2*f^2-5*a*d^3*e ^3*f+b*c^3*e*f^3-3*b*c^2*d*e^2*f^2-15*b*c*d^2*e^3*f+17*b*d^3*e^4)/e^2*x^3+ 1/16*(11*a*c^3*f^4-3*a*c^2*d*e*f^3-3*a*c*d^2*e^2*f^2-5*a*d^3*e^3*f-b*c^3*e *f^3-3*b*c^2*d*e^2*f^2-15*b*c*d^2*e^3*f+19*b*d^3*e^4)/e*x)/(f*x^2+e)^3+1/1 6*(5*a*c^3*f^4+3*a*c^2*d*e*f^3+3*a*c*d^2*e^2*f^2+5*a*d^3*e^3*f+b*c^3*e*f^3 +3*b*c^2*d*e^2*f^2+15*b*c*d^2*e^3*f-35*b*d^3*e^4)/e^3/(e*f)^(1/2)*arctan(f *x/(e*f)^(1/2)))
Leaf count of result is larger than twice the leaf count of optimal. 701 vs. \(2 (330) = 660\).
Time = 0.29 (sec) , antiderivative size = 1422, normalized size of antiderivative = 4.09 \[ \int \frac {\left (a+b x^2\right ) \left (c+d x^2\right )^3}{\left (e+f x^2\right )^4} \, dx=\text {Too large to display} \]
[1/96*(96*b*d^3*e^4*f^4*x^7 + 6*(77*b*d^3*e^5*f^3 + 5*a*c^3*e*f^7 - 11*(3* b*c*d^2 + a*d^3)*e^4*f^4 + 3*(b*c^2*d + a*c*d^2)*e^3*f^5 + (b*c^3 + 3*a*c^ 2*d)*e^2*f^6)*x^5 + 16*(35*b*d^3*e^6*f^2 + 5*a*c^3*e^2*f^6 - 5*(3*b*c*d^2 + a*d^3)*e^5*f^3 - 3*(b*c^2*d + a*c*d^2)*e^4*f^4 + (b*c^3 + 3*a*c^2*d)*e^3 *f^5)*x^3 + 3*(35*b*d^3*e^7 - 5*a*c^3*e^3*f^4 - 5*(3*b*c*d^2 + a*d^3)*e^6* f - 3*(b*c^2*d + a*c*d^2)*e^5*f^2 - (b*c^3 + 3*a*c^2*d)*e^4*f^3 + (35*b*d^ 3*e^4*f^3 - 5*a*c^3*f^7 - 5*(3*b*c*d^2 + a*d^3)*e^3*f^4 - 3*(b*c^2*d + a*c *d^2)*e^2*f^5 - (b*c^3 + 3*a*c^2*d)*e*f^6)*x^6 + 3*(35*b*d^3*e^5*f^2 - 5*a *c^3*e*f^6 - 5*(3*b*c*d^2 + a*d^3)*e^4*f^3 - 3*(b*c^2*d + a*c*d^2)*e^3*f^4 - (b*c^3 + 3*a*c^2*d)*e^2*f^5)*x^4 + 3*(35*b*d^3*e^6*f - 5*a*c^3*e^2*f^5 - 5*(3*b*c*d^2 + a*d^3)*e^5*f^2 - 3*(b*c^2*d + a*c*d^2)*e^4*f^3 - (b*c^3 + 3*a*c^2*d)*e^3*f^4)*x^2)*sqrt(-e*f)*log((f*x^2 - 2*sqrt(-e*f)*x - e)/(f*x ^2 + e)) + 6*(35*b*d^3*e^7*f + 11*a*c^3*e^3*f^5 - 5*(3*b*c*d^2 + a*d^3)*e^ 6*f^2 - 3*(b*c^2*d + a*c*d^2)*e^5*f^3 - (b*c^3 + 3*a*c^2*d)*e^4*f^4)*x)/(e ^4*f^8*x^6 + 3*e^5*f^7*x^4 + 3*e^6*f^6*x^2 + e^7*f^5), 1/48*(48*b*d^3*e^4* f^4*x^7 + 3*(77*b*d^3*e^5*f^3 + 5*a*c^3*e*f^7 - 11*(3*b*c*d^2 + a*d^3)*e^4 *f^4 + 3*(b*c^2*d + a*c*d^2)*e^3*f^5 + (b*c^3 + 3*a*c^2*d)*e^2*f^6)*x^5 + 8*(35*b*d^3*e^6*f^2 + 5*a*c^3*e^2*f^6 - 5*(3*b*c*d^2 + a*d^3)*e^5*f^3 - 3* (b*c^2*d + a*c*d^2)*e^4*f^4 + (b*c^3 + 3*a*c^2*d)*e^3*f^5)*x^3 - 3*(35*b*d ^3*e^7 - 5*a*c^3*e^3*f^4 - 5*(3*b*c*d^2 + a*d^3)*e^6*f - 3*(b*c^2*d + a...
Timed out. \[ \int \frac {\left (a+b x^2\right ) \left (c+d x^2\right )^3}{\left (e+f x^2\right )^4} \, dx=\text {Timed out} \]
Exception generated. \[ \int \frac {\left (a+b x^2\right ) \left (c+d x^2\right )^3}{\left (e+f x^2\right )^4} \, dx=\text {Exception raised: ValueError} \]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(e>0)', see `assume?` for more de tails)Is e
Time = 0.32 (sec) , antiderivative size = 475, normalized size of antiderivative = 1.36 \[ \int \frac {\left (a+b x^2\right ) \left (c+d x^2\right )^3}{\left (e+f x^2\right )^4} \, dx=\frac {b d^{3} x}{f^{4}} - \frac {{\left (35 \, b d^{3} e^{4} - 15 \, b c d^{2} e^{3} f - 5 \, a d^{3} e^{3} f - 3 \, b c^{2} d e^{2} f^{2} - 3 \, a c d^{2} e^{2} f^{2} - b c^{3} e f^{3} - 3 \, a c^{2} d e f^{3} - 5 \, a c^{3} f^{4}\right )} \arctan \left (\frac {f x}{\sqrt {e f}}\right )}{16 \, \sqrt {e f} e^{3} f^{4}} + \frac {87 \, b d^{3} e^{4} f^{2} x^{5} - 99 \, b c d^{2} e^{3} f^{3} x^{5} - 33 \, a d^{3} e^{3} f^{3} x^{5} + 9 \, b c^{2} d e^{2} f^{4} x^{5} + 9 \, a c d^{2} e^{2} f^{4} x^{5} + 3 \, b c^{3} e f^{5} x^{5} + 9 \, a c^{2} d e f^{5} x^{5} + 15 \, a c^{3} f^{6} x^{5} + 136 \, b d^{3} e^{5} f x^{3} - 120 \, b c d^{2} e^{4} f^{2} x^{3} - 40 \, a d^{3} e^{4} f^{2} x^{3} - 24 \, b c^{2} d e^{3} f^{3} x^{3} - 24 \, a c d^{2} e^{3} f^{3} x^{3} + 8 \, b c^{3} e^{2} f^{4} x^{3} + 24 \, a c^{2} d e^{2} f^{4} x^{3} + 40 \, a c^{3} e f^{5} x^{3} + 57 \, b d^{3} e^{6} x - 45 \, b c d^{2} e^{5} f x - 15 \, a d^{3} e^{5} f x - 9 \, b c^{2} d e^{4} f^{2} x - 9 \, a c d^{2} e^{4} f^{2} x - 3 \, b c^{3} e^{3} f^{3} x - 9 \, a c^{2} d e^{3} f^{3} x + 33 \, a c^{3} e^{2} f^{4} x}{48 \, {\left (f x^{2} + e\right )}^{3} e^{3} f^{4}} \]
b*d^3*x/f^4 - 1/16*(35*b*d^3*e^4 - 15*b*c*d^2*e^3*f - 5*a*d^3*e^3*f - 3*b* c^2*d*e^2*f^2 - 3*a*c*d^2*e^2*f^2 - b*c^3*e*f^3 - 3*a*c^2*d*e*f^3 - 5*a*c^ 3*f^4)*arctan(f*x/sqrt(e*f))/(sqrt(e*f)*e^3*f^4) + 1/48*(87*b*d^3*e^4*f^2* x^5 - 99*b*c*d^2*e^3*f^3*x^5 - 33*a*d^3*e^3*f^3*x^5 + 9*b*c^2*d*e^2*f^4*x^ 5 + 9*a*c*d^2*e^2*f^4*x^5 + 3*b*c^3*e*f^5*x^5 + 9*a*c^2*d*e*f^5*x^5 + 15*a *c^3*f^6*x^5 + 136*b*d^3*e^5*f*x^3 - 120*b*c*d^2*e^4*f^2*x^3 - 40*a*d^3*e^ 4*f^2*x^3 - 24*b*c^2*d*e^3*f^3*x^3 - 24*a*c*d^2*e^3*f^3*x^3 + 8*b*c^3*e^2* f^4*x^3 + 24*a*c^2*d*e^2*f^4*x^3 + 40*a*c^3*e*f^5*x^3 + 57*b*d^3*e^6*x - 4 5*b*c*d^2*e^5*f*x - 15*a*d^3*e^5*f*x - 9*b*c^2*d*e^4*f^2*x - 9*a*c*d^2*e^4 *f^2*x - 3*b*c^3*e^3*f^3*x - 9*a*c^2*d*e^3*f^3*x + 33*a*c^3*e^2*f^4*x)/((f *x^2 + e)^3*e^3*f^4)
Time = 5.66 (sec) , antiderivative size = 444, normalized size of antiderivative = 1.28 \[ \int \frac {\left (a+b x^2\right ) \left (c+d x^2\right )^3}{\left (e+f x^2\right )^4} \, dx=\frac {\frac {x^3\,\left (b\,c^3\,e\,f^4+5\,a\,c^3\,f^5-3\,b\,c^2\,d\,e^2\,f^3+3\,a\,c^2\,d\,e\,f^4-15\,b\,c\,d^2\,e^3\,f^2-3\,a\,c\,d^2\,e^2\,f^3+17\,b\,d^3\,e^4\,f-5\,a\,d^3\,e^3\,f^2\right )}{6\,e^2}+\frac {x^5\,\left (b\,c^3\,e\,f^5+5\,a\,c^3\,f^6+3\,b\,c^2\,d\,e^2\,f^4+3\,a\,c^2\,d\,e\,f^5-33\,b\,c\,d^2\,e^3\,f^3+3\,a\,c\,d^2\,e^2\,f^4+29\,b\,d^3\,e^4\,f^2-11\,a\,d^3\,e^3\,f^3\right )}{16\,e^3}-\frac {x\,\left (b\,c^3\,e\,f^3-11\,a\,c^3\,f^4+3\,b\,c^2\,d\,e^2\,f^2+3\,a\,c^2\,d\,e\,f^3+15\,b\,c\,d^2\,e^3\,f+3\,a\,c\,d^2\,e^2\,f^2-19\,b\,d^3\,e^4+5\,a\,d^3\,e^3\,f\right )}{16\,e}}{e^3\,f^4+3\,e^2\,f^5\,x^2+3\,e\,f^6\,x^4+f^7\,x^6}+\frac {b\,d^3\,x}{f^4}+\frac {\mathrm {atan}\left (\frac {\sqrt {f}\,x}{\sqrt {e}}\right )\,\left (b\,c^3\,e\,f^3+5\,a\,c^3\,f^4+3\,b\,c^2\,d\,e^2\,f^2+3\,a\,c^2\,d\,e\,f^3+15\,b\,c\,d^2\,e^3\,f+3\,a\,c\,d^2\,e^2\,f^2-35\,b\,d^3\,e^4+5\,a\,d^3\,e^3\,f\right )}{16\,e^{7/2}\,f^{9/2}} \]
((x^3*(5*a*c^3*f^5 - 5*a*d^3*e^3*f^2 + b*c^3*e*f^4 + 17*b*d^3*e^4*f + 3*a* c^2*d*e*f^4 - 3*a*c*d^2*e^2*f^3 - 15*b*c*d^2*e^3*f^2 - 3*b*c^2*d*e^2*f^3)) /(6*e^2) + (x^5*(5*a*c^3*f^6 - 11*a*d^3*e^3*f^3 + 29*b*d^3*e^4*f^2 + b*c^3 *e*f^5 + 3*a*c^2*d*e*f^5 + 3*a*c*d^2*e^2*f^4 - 33*b*c*d^2*e^3*f^3 + 3*b*c^ 2*d*e^2*f^4))/(16*e^3) - (x*(5*a*d^3*e^3*f - 19*b*d^3*e^4 - 11*a*c^3*f^4 + b*c^3*e*f^3 + 3*a*c^2*d*e*f^3 + 15*b*c*d^2*e^3*f + 3*a*c*d^2*e^2*f^2 + 3* b*c^2*d*e^2*f^2))/(16*e))/(e^3*f^4 + f^7*x^6 + 3*e*f^6*x^4 + 3*e^2*f^5*x^2 ) + (b*d^3*x)/f^4 + (atan((f^(1/2)*x)/e^(1/2))*(5*a*c^3*f^4 - 35*b*d^3*e^4 + 5*a*d^3*e^3*f + b*c^3*e*f^3 + 3*a*c^2*d*e*f^3 + 15*b*c*d^2*e^3*f + 3*a* c*d^2*e^2*f^2 + 3*b*c^2*d*e^2*f^2))/(16*e^(7/2)*f^(9/2))