Integrand size = 35, antiderivative size = 859 \[ \int \frac {1}{x^2 \sqrt {a+b x^2} \sqrt {c+d x^2} \left (e+f x^2\right )^3} \, dx=-\frac {\sqrt {a+b x^2} \sqrt {c+d x^2}}{a c e x \left (e+f x^2\right )^2}+\frac {f (a f (4 d e-5 c f)-4 b e (d e-c f)) x \sqrt {a+b x^2} \sqrt {c+d x^2}}{4 a c e^2 (b e-a f) (d e-c f) \left (e+f x^2\right )^2}+\frac {\left (8 b^2 e^2 (d e-c f)^2+a^2 f^2 \left (8 d^2 e^2-26 c d e f+15 c^2 f^2\right )-a b e f \left (16 d^2 e^2-45 c d e f+26 c^2 f^2\right )\right ) x \sqrt {a+b x^2}}{8 a c e^3 (b e-a f)^2 (d e-c f) \sqrt {c+d x^2} \left (e+f x^2\right )}-\frac {\sqrt {d} \left (8 b^2 e^2 (d e-c f)^2+a^2 f^2 \left (8 d^2 e^2-26 c d e f+15 c^2 f^2\right )-a b e f \left (16 d^2 e^2-45 c d e f+26 c^2 f^2\right )\right ) \sqrt {a+b x^2} E\left (\arctan \left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|1-\frac {b c}{a d}\right )}{8 a \sqrt {c} e^3 (b e-a f)^2 (d e-c f)^2 \sqrt {\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}} \sqrt {c+d x^2}}+\frac {\sqrt {c} \sqrt {d} f \left (3 a f \left (8 d^2 e^2-12 c d e f+5 c^2 f^2\right )-b e \left (24 d^2 e^2-37 c d e f+16 c^2 f^2\right )\right ) \sqrt {a+b x^2} \operatorname {EllipticF}\left (\arctan \left (\frac {\sqrt {d} x}{\sqrt {c}}\right ),1-\frac {b c}{a d}\right )}{8 a e^3 (b e-a f) (d e-c f)^3 \sqrt {\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}} \sqrt {c+d x^2}}+\frac {c^{3/2} f^2 \left (3 a^2 f^2 \left (8 d^2 e^2-12 c d e f+5 c^2 f^2\right )-2 a b e f \left (28 d^2 e^2-43 c d e f+18 c^2 f^2\right )+b^2 e^2 \left (35 d^2 e^2-56 c d e f+24 c^2 f^2\right )\right ) \sqrt {a+b x^2} \operatorname {EllipticPi}\left (1-\frac {c f}{d e},\arctan \left (\frac {\sqrt {d} x}{\sqrt {c}}\right ),1-\frac {b c}{a d}\right )}{8 a \sqrt {d} e^4 (b e-a f)^2 (d e-c f)^3 \sqrt {\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}} \sqrt {c+d x^2}} \] Output:
-(b*x^2+a)^(1/2)*(d*x^2+c)^(1/2)/a/c/e/x/(f*x^2+e)^2+1/4*f*(a*f*(-5*c*f+4* d*e)-4*b*e*(-c*f+d*e))*x*(b*x^2+a)^(1/2)*(d*x^2+c)^(1/2)/a/c/e^2/(-a*f+b*e )/(-c*f+d*e)/(f*x^2+e)^2+1/8*(8*b^2*e^2*(-c*f+d*e)^2+a^2*f^2*(15*c^2*f^2-2 6*c*d*e*f+8*d^2*e^2)-a*b*e*f*(26*c^2*f^2-45*c*d*e*f+16*d^2*e^2))*x*(b*x^2+ a)^(1/2)/a/c/e^3/(-a*f+b*e)^2/(-c*f+d*e)/(d*x^2+c)^(1/2)/(f*x^2+e)-1/8*d^( 1/2)*(8*b^2*e^2*(-c*f+d*e)^2+a^2*f^2*(15*c^2*f^2-26*c*d*e*f+8*d^2*e^2)-a*b *e*f*(26*c^2*f^2-45*c*d*e*f+16*d^2*e^2))*(b*x^2+a)^(1/2)*EllipticE(d^(1/2) *x/c^(1/2)/(1+d*x^2/c)^(1/2),(1-b*c/a/d)^(1/2))/a/c^(1/2)/e^3/(-a*f+b*e)^2 /(-c*f+d*e)^2/(c*(b*x^2+a)/a/(d*x^2+c))^(1/2)/(d*x^2+c)^(1/2)+1/8*c^(1/2)* d^(1/2)*f*(3*a*f*(5*c^2*f^2-12*c*d*e*f+8*d^2*e^2)-b*e*(16*c^2*f^2-37*c*d*e *f+24*d^2*e^2))*(b*x^2+a)^(1/2)*InverseJacobiAM(arctan(d^(1/2)*x/c^(1/2)), (1-b*c/a/d)^(1/2))/a/e^3/(-a*f+b*e)/(-c*f+d*e)^3/(c*(b*x^2+a)/a/(d*x^2+c)) ^(1/2)/(d*x^2+c)^(1/2)+1/8*c^(3/2)*f^2*(3*a^2*f^2*(5*c^2*f^2-12*c*d*e*f+8* d^2*e^2)-2*a*b*e*f*(18*c^2*f^2-43*c*d*e*f+28*d^2*e^2)+b^2*e^2*(24*c^2*f^2- 56*c*d*e*f+35*d^2*e^2))*(b*x^2+a)^(1/2)*EllipticPi(d^(1/2)*x/c^(1/2)/(1+d* x^2/c)^(1/2),1-c*f/d/e,(1-b*c/a/d)^(1/2))/a/d^(1/2)/e^4/(-a*f+b*e)^2/(-c*f +d*e)^3/(c*(b*x^2+a)/a/(d*x^2+c))^(1/2)/(d*x^2+c)^(1/2)
Result contains complex when optimal does not.
Time = 5.13 (sec) , antiderivative size = 567, normalized size of antiderivative = 0.66 \[ \int \frac {1}{x^2 \sqrt {a+b x^2} \sqrt {c+d x^2} \left (e+f x^2\right )^3} \, dx=\frac {\sqrt {\frac {b}{a}} \left (-\sqrt {\frac {b}{a}} e \left (a+b x^2\right ) \left (c+d x^2\right ) \left (2 a c e f^3 (-b e+a f) (-d e+c f) x^2+a c f^3 (b e (13 d e-10 c f)+a f (-10 d e+7 c f)) x^2 \left (e+f x^2\right )+8 (b e-a f)^2 (d e-c f)^2 \left (e+f x^2\right )^2\right )+i c x \sqrt {1+\frac {b x^2}{a}} \sqrt {1+\frac {d x^2}{c}} \left (e+f x^2\right )^2 \left (-b e \left (8 b^2 e^2 (d e-c f)^2+a b e f \left (-16 d^2 e^2+45 c d e f-26 c^2 f^2\right )+a^2 f^2 \left (8 d^2 e^2-26 c d e f+15 c^2 f^2\right )\right ) E\left (i \text {arcsinh}\left (\sqrt {\frac {b}{a}} x\right )|\frac {a d}{b c}\right )+b e (d e-c f) \left (a^2 f^2 (16 d e-15 c f)+8 b^2 e^2 (d e-c f)+a b e f (-27 d e+26 c f)\right ) \operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt {\frac {b}{a}} x\right ),\frac {a d}{b c}\right )+a f \left (3 a^2 f^2 \left (8 d^2 e^2-12 c d e f+5 c^2 f^2\right )-2 a b e f \left (28 d^2 e^2-43 c d e f+18 c^2 f^2\right )+b^2 e^2 \left (35 d^2 e^2-56 c d e f+24 c^2 f^2\right )\right ) \operatorname {EllipticPi}\left (\frac {a f}{b e},i \text {arcsinh}\left (\sqrt {\frac {b}{a}} x\right ),\frac {a d}{b c}\right )\right )\right )}{8 b c e^4 (b e-a f)^2 (d e-c f)^2 x \sqrt {a+b x^2} \sqrt {c+d x^2} \left (e+f x^2\right )^2} \] Input:
Integrate[1/(x^2*Sqrt[a + b*x^2]*Sqrt[c + d*x^2]*(e + f*x^2)^3),x]
Output:
(Sqrt[b/a]*(-(Sqrt[b/a]*e*(a + b*x^2)*(c + d*x^2)*(2*a*c*e*f^3*(-(b*e) + a *f)*(-(d*e) + c*f)*x^2 + a*c*f^3*(b*e*(13*d*e - 10*c*f) + a*f*(-10*d*e + 7 *c*f))*x^2*(e + f*x^2) + 8*(b*e - a*f)^2*(d*e - c*f)^2*(e + f*x^2)^2)) + I *c*x*Sqrt[1 + (b*x^2)/a]*Sqrt[1 + (d*x^2)/c]*(e + f*x^2)^2*(-(b*e*(8*b^2*e ^2*(d*e - c*f)^2 + a*b*e*f*(-16*d^2*e^2 + 45*c*d*e*f - 26*c^2*f^2) + a^2*f ^2*(8*d^2*e^2 - 26*c*d*e*f + 15*c^2*f^2))*EllipticE[I*ArcSinh[Sqrt[b/a]*x] , (a*d)/(b*c)]) + b*e*(d*e - c*f)*(a^2*f^2*(16*d*e - 15*c*f) + 8*b^2*e^2*( d*e - c*f) + a*b*e*f*(-27*d*e + 26*c*f))*EllipticF[I*ArcSinh[Sqrt[b/a]*x], (a*d)/(b*c)] + a*f*(3*a^2*f^2*(8*d^2*e^2 - 12*c*d*e*f + 5*c^2*f^2) - 2*a* b*e*f*(28*d^2*e^2 - 43*c*d*e*f + 18*c^2*f^2) + b^2*e^2*(35*d^2*e^2 - 56*c* d*e*f + 24*c^2*f^2))*EllipticPi[(a*f)/(b*e), I*ArcSinh[Sqrt[b/a]*x], (a*d) /(b*c)])))/(8*b*c*e^4*(b*e - a*f)^2*(d*e - c*f)^2*x*Sqrt[a + b*x^2]*Sqrt[c + d*x^2]*(e + f*x^2)^2)
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.
| \(\displaystyle \int \frac {1}{x^2 \sqrt {a+b x^2} \sqrt {c+d x^2} \left (e+f x^2\right )^3} \, dx\) |
| \(\Big \downarrow \) 450 |
| \(\displaystyle \int \frac {1}{x^2 \sqrt {a+b x^2} \sqrt {c+d x^2} \left (e+f x^2\right )^3}dx\) |
Input:
Int[1/(x^2*Sqrt[a + b*x^2]*Sqrt[c + d*x^2]*(e + f*x^2)^3),x]
Output:
$Aborted
Int[((g_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_)^2)^(p_.)*((c_.) + (d_.)*(x_)^2)^ (q_.)*((e_.) + (f_.)*(x_)^2)^(r_.), x_Symbol] :> Unintegrable[(g*x)^m*(a + b*x^2)^p*(c + d*x^2)^q*(e + f*x^2)^r, x] /; FreeQ[{a, b, c, d, e, f, g, m, p, q, r}, x]
Leaf count of result is larger than twice the leaf count of optimal. \(2868\) vs. \(2(823)=1646\).
Time = 28.13 (sec) , antiderivative size = 2869, normalized size of antiderivative = 3.34
| method | result | size |
| elliptic | \(\text {Expression too large to display}\) | \(2869\) |
| risch | \(\text {Expression too large to display}\) | \(3831\) |
| default | \(\text {Expression too large to display}\) | \(7270\) |
Input:
int(1/x^2/(b*x^2+a)^(1/2)/(d*x^2+c)^(1/2)/(f*x^2+e)^3,x,method=_RETURNVERB OSE)
Output:
((b*x^2+a)*(d*x^2+c))^(1/2)/(b*x^2+a)^(1/2)/(d*x^2+c)^(1/2)*(-15/8/(a*c*f^ 2-a*d*e*f-b*c*e*f+b*d*e^2)^2/e^4*f^5/(-b/a)^(1/2)*(1+b*x^2/a)^(1/2)*(1+d*x ^2/c)^(1/2)/(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*EllipticPi(x*(-b/a)^(1/2), a*f/b/e,(-1/c*d)^(1/2)/(-b/a)^(1/2))*a^2*c^2-3/(a*c*f^2-a*d*e*f-b*c*e*f+b* d*e^2)^2/e^2*f^3/(-b/a)^(1/2)*(1+b*x^2/a)^(1/2)*(1+d*x^2/c)^(1/2)/(b*d*x^4 +a*d*x^2+b*c*x^2+a*c)^(1/2)*EllipticPi(x*(-b/a)^(1/2),a*f/b/e,(-1/c*d)^(1/ 2)/(-b/a)^(1/2))*a^2*d^2-3/(a*c*f^2-a*d*e*f-b*c*e*f+b*d*e^2)^2/e^2*f^3/(-b /a)^(1/2)*(1+b*x^2/a)^(1/2)*(1+d*x^2/c)^(1/2)/(b*d*x^4+a*d*x^2+b*c*x^2+a*c )^(1/2)*EllipticPi(x*(-b/a)^(1/2),a*f/b/e,(-1/c*d)^(1/2)/(-b/a)^(1/2))*b^2 *c^2+9/2/(a*c*f^2-a*d*e*f-b*c*e*f+b*d*e^2)^2/e^3*f^4/(-b/a)^(1/2)*(1+b*x^2 /a)^(1/2)*(1+d*x^2/c)^(1/2)/(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*EllipticPi (x*(-b/a)^(1/2),a*f/b/e,(-1/c*d)^(1/2)/(-b/a)^(1/2))*a^2*c*d+9/2/(a*c*f^2- a*d*e*f-b*c*e*f+b*d*e^2)^2/e^3*f^4/(-b/a)^(1/2)*(1+b*x^2/a)^(1/2)*(1+d*x^2 /c)^(1/2)/(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*EllipticPi(x*(-b/a)^(1/2),a* f/b/e,(-1/c*d)^(1/2)/(-b/a)^(1/2))*a*b*c^2+7/(a*c*f^2-a*d*e*f-b*c*e*f+b*d* e^2)^2*f^2/e/(-b/a)^(1/2)*(1+b*x^2/a)^(1/2)*(1+d*x^2/c)^(1/2)/(b*d*x^4+a*d *x^2+b*c*x^2+a*c)^(1/2)*EllipticPi(x*(-b/a)^(1/2),a*f/b/e,(-1/c*d)^(1/2)/( -b/a)^(1/2))*a*b*d^2+11/8*b^2*d^2*f/(a*c*f^2-a*d*e*f-b*c*e*f+b*d*e^2)^2/(- b/a)^(1/2)*(1+b*x^2/a)^(1/2)*(1+d*x^2/c)^(1/2)/(b*d*x^4+a*d*x^2+b*c*x^2+a* c)^(1/2)*EllipticF(x*(-b/a)^(1/2),(-1+(a*d+b*c)/c/b)^(1/2))-1/(-b/a)^(1...
Timed out. \[ \int \frac {1}{x^2 \sqrt {a+b x^2} \sqrt {c+d x^2} \left (e+f x^2\right )^3} \, dx=\text {Timed out} \] Input:
integrate(1/x^2/(b*x^2+a)^(1/2)/(d*x^2+c)^(1/2)/(f*x^2+e)^3,x, algorithm=" fricas")
Output:
Timed out
Timed out. \[ \int \frac {1}{x^2 \sqrt {a+b x^2} \sqrt {c+d x^2} \left (e+f x^2\right )^3} \, dx=\text {Timed out} \] Input:
integrate(1/x**2/(b*x**2+a)**(1/2)/(d*x**2+c)**(1/2)/(f*x**2+e)**3,x)
Output:
Timed out
\[ \int \frac {1}{x^2 \sqrt {a+b x^2} \sqrt {c+d x^2} \left (e+f x^2\right )^3} \, dx=\int { \frac {1}{\sqrt {b x^{2} + a} \sqrt {d x^{2} + c} {\left (f x^{2} + e\right )}^{3} x^{2}} \,d x } \] Input:
integrate(1/x^2/(b*x^2+a)^(1/2)/(d*x^2+c)^(1/2)/(f*x^2+e)^3,x, algorithm=" maxima")
Output:
integrate(1/(sqrt(b*x^2 + a)*sqrt(d*x^2 + c)*(f*x^2 + e)^3*x^2), x)
\[ \int \frac {1}{x^2 \sqrt {a+b x^2} \sqrt {c+d x^2} \left (e+f x^2\right )^3} \, dx=\int { \frac {1}{\sqrt {b x^{2} + a} \sqrt {d x^{2} + c} {\left (f x^{2} + e\right )}^{3} x^{2}} \,d x } \] Input:
integrate(1/x^2/(b*x^2+a)^(1/2)/(d*x^2+c)^(1/2)/(f*x^2+e)^3,x, algorithm=" giac")
Output:
integrate(1/(sqrt(b*x^2 + a)*sqrt(d*x^2 + c)*(f*x^2 + e)^3*x^2), x)
Timed out. \[ \int \frac {1}{x^2 \sqrt {a+b x^2} \sqrt {c+d x^2} \left (e+f x^2\right )^3} \, dx=\int \frac {1}{x^2\,\sqrt {b\,x^2+a}\,\sqrt {d\,x^2+c}\,{\left (f\,x^2+e\right )}^3} \,d x \] Input:
int(1/(x^2*(a + b*x^2)^(1/2)*(c + d*x^2)^(1/2)*(e + f*x^2)^3),x)
Output:
int(1/(x^2*(a + b*x^2)^(1/2)*(c + d*x^2)^(1/2)*(e + f*x^2)^3), x)
\[ \int \frac {1}{x^2 \sqrt {a+b x^2} \sqrt {c+d x^2} \left (e+f x^2\right )^3} \, dx=\text {too large to display} \] Input:
int(1/x^2/(b*x^2+a)^(1/2)/(d*x^2+c)^(1/2)/(f*x^2+e)^3,x)
Output:
( - sqrt(c + d*x**2)*sqrt(a + b*x**2) - 3*int((sqrt(c + d*x**2)*sqrt(a + b *x**2)*x**4)/(a*c*e**3 + 3*a*c*e**2*f*x**2 + 3*a*c*e*f**2*x**4 + a*c*f**3* x**6 + a*d*e**3*x**2 + 3*a*d*e**2*f*x**4 + 3*a*d*e*f**2*x**6 + a*d*f**3*x* *8 + b*c*e**3*x**2 + 3*b*c*e**2*f*x**4 + 3*b*c*e*f**2*x**6 + b*c*f**3*x**8 + b*d*e**3*x**4 + 3*b*d*e**2*f*x**6 + 3*b*d*e*f**2*x**8 + b*d*f**3*x**10) ,x)*b*d*e**2*f*x - 6*int((sqrt(c + d*x**2)*sqrt(a + b*x**2)*x**4)/(a*c*e** 3 + 3*a*c*e**2*f*x**2 + 3*a*c*e*f**2*x**4 + a*c*f**3*x**6 + a*d*e**3*x**2 + 3*a*d*e**2*f*x**4 + 3*a*d*e*f**2*x**6 + a*d*f**3*x**8 + b*c*e**3*x**2 + 3*b*c*e**2*f*x**4 + 3*b*c*e*f**2*x**6 + b*c*f**3*x**8 + b*d*e**3*x**4 + 3* b*d*e**2*f*x**6 + 3*b*d*e*f**2*x**8 + b*d*f**3*x**10),x)*b*d*e*f**2*x**3 - 3*int((sqrt(c + d*x**2)*sqrt(a + b*x**2)*x**4)/(a*c*e**3 + 3*a*c*e**2*f*x **2 + 3*a*c*e*f**2*x**4 + a*c*f**3*x**6 + a*d*e**3*x**2 + 3*a*d*e**2*f*x** 4 + 3*a*d*e*f**2*x**6 + a*d*f**3*x**8 + b*c*e**3*x**2 + 3*b*c*e**2*f*x**4 + 3*b*c*e*f**2*x**6 + b*c*f**3*x**8 + b*d*e**3*x**4 + 3*b*d*e**2*f*x**6 + 3*b*d*e*f**2*x**8 + b*d*f**3*x**10),x)*b*d*f**3*x**5 - 4*int((sqrt(c + d*x **2)*sqrt(a + b*x**2)*x**2)/(a*c*e**3 + 3*a*c*e**2*f*x**2 + 3*a*c*e*f**2*x **4 + a*c*f**3*x**6 + a*d*e**3*x**2 + 3*a*d*e**2*f*x**4 + 3*a*d*e*f**2*x** 6 + a*d*f**3*x**8 + b*c*e**3*x**2 + 3*b*c*e**2*f*x**4 + 3*b*c*e*f**2*x**6 + b*c*f**3*x**8 + b*d*e**3*x**4 + 3*b*d*e**2*f*x**6 + 3*b*d*e*f**2*x**8 + b*d*f**3*x**10),x)*a*d*e**2*f*x - 8*int((sqrt(c + d*x**2)*sqrt(a + b*x*...