Integrand size = 35, antiderivative size = 879 \[ \int \frac {x^4 \sqrt {a+b x^2} \left (c+d x^2\right )^{3/2}}{\left (e+f x^2\right )^2} \, dx=-\frac {\left (4 a^2 d^2 f^2+2 a b d f (10 d e-7 c f)-b^2 \left (105 d^2 e^2-95 c d e f+6 c^2 f^2\right )\right ) x \sqrt {c+d x^2}}{30 b d f^4 \sqrt {a+b x^2}}-\frac {\left (2 a^2 d f^2 (d e-c f)-a b f \left (37 d^2 e^2-64 c d e f+12 c^2 f^2\right )+b^2 e \left (35 d^2 e^2-62 c d e f+27 c^2 f^2\right )\right ) x \sqrt {a+b x^2} \sqrt {c+d x^2}}{30 b f^3 (b e-a f) (d e-c f)}+\frac {d (b e (7 d e-12 c f)-a f (7 d e-2 c f)) x^3 \sqrt {a+b x^2} \sqrt {c+d x^2}}{10 f^2 (b e-a f) (d e-c f)}-\frac {b d^2 e x^5 \sqrt {a+b x^2} \sqrt {c+d x^2}}{2 f (b e-a f) (d e-c f)}+\frac {e x \left (a+b x^2\right )^{3/2} \left (c+d x^2\right )^{5/2}}{2 (b e-a f) (d e-c f) \left (e+f x^2\right )}+\frac {\sqrt {a} \left (4 a^2 d^2 f^2+2 a b d f (10 d e-7 c f)-b^2 \left (105 d^2 e^2-95 c d e f+6 c^2 f^2\right )\right ) \sqrt {c+d x^2} E\left (\arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )|1-\frac {a d}{b c}\right )}{30 b^{3/2} d f^4 \sqrt {a+b x^2} \sqrt {\frac {a \left (c+d x^2\right )}{c \left (a+b x^2\right )}}}+\frac {a^{3/2} \left (2 a^2 c d f^3-2 a b f \left (45 d^2 e^2-49 c d e f+9 c^2 f^2\right )+b^2 e \left (105 d^2 e^2-130 c d e f+33 c^2 f^2\right )\right ) \sqrt {c+d x^2} \operatorname {EllipticF}\left (\arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ),1-\frac {a d}{b c}\right )}{30 b^{3/2} c f^4 (b e-a f) \sqrt {a+b x^2} \sqrt {\frac {a \left (c+d x^2\right )}{c \left (a+b x^2\right )}}}-\frac {a^{3/2} (d e-c f) (b e (7 d e-4 c f)-3 a f (2 d e-c f)) \sqrt {c+d x^2} \operatorname {EllipticPi}\left (1-\frac {a f}{b e},\arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ),1-\frac {a d}{b c}\right )}{2 \sqrt {b} c f^4 (b e-a f) \sqrt {a+b x^2} \sqrt {\frac {a \left (c+d x^2\right )}{c \left (a+b x^2\right )}}} \] Output:
-1/30*(4*a^2*d^2*f^2+2*a*b*d*f*(-7*c*f+10*d*e)-b^2*(6*c^2*f^2-95*c*d*e*f+1 05*d^2*e^2))*x*(d*x^2+c)^(1/2)/b/d/f^4/(b*x^2+a)^(1/2)-1/30*(2*a^2*d*f^2*( -c*f+d*e)-a*b*f*(12*c^2*f^2-64*c*d*e*f+37*d^2*e^2)+b^2*e*(27*c^2*f^2-62*c* d*e*f+35*d^2*e^2))*x*(b*x^2+a)^(1/2)*(d*x^2+c)^(1/2)/b/f^3/(-a*f+b*e)/(-c* f+d*e)+1/10*d*(b*e*(-12*c*f+7*d*e)-a*f*(-2*c*f+7*d*e))*x^3*(b*x^2+a)^(1/2) *(d*x^2+c)^(1/2)/f^2/(-a*f+b*e)/(-c*f+d*e)-1/2*b*d^2*e*x^5*(b*x^2+a)^(1/2) *(d*x^2+c)^(1/2)/f/(-a*f+b*e)/(-c*f+d*e)+1/2*e*x*(b*x^2+a)^(3/2)*(d*x^2+c) ^(5/2)/(-a*f+b*e)/(-c*f+d*e)/(f*x^2+e)+1/30*a^(1/2)*(4*a^2*d^2*f^2+2*a*b*d *f*(-7*c*f+10*d*e)-b^2*(6*c^2*f^2-95*c*d*e*f+105*d^2*e^2))*(d*x^2+c)^(1/2) *EllipticE(b^(1/2)*x/a^(1/2)/(1+b*x^2/a)^(1/2),(1-a*d/b/c)^(1/2))/b^(3/2)/ d/f^4/(b*x^2+a)^(1/2)/(a*(d*x^2+c)/c/(b*x^2+a))^(1/2)+1/30*a^(3/2)*(2*a^2* c*d*f^3-2*a*b*f*(9*c^2*f^2-49*c*d*e*f+45*d^2*e^2)+b^2*e*(33*c^2*f^2-130*c* d*e*f+105*d^2*e^2))*(d*x^2+c)^(1/2)*InverseJacobiAM(arctan(b^(1/2)*x/a^(1/ 2)),(1-a*d/b/c)^(1/2))/b^(3/2)/c/f^4/(-a*f+b*e)/(b*x^2+a)^(1/2)/(a*(d*x^2+ c)/c/(b*x^2+a))^(1/2)-1/2*a^(3/2)*(-c*f+d*e)*(b*e*(-4*c*f+7*d*e)-3*a*f*(-c *f+2*d*e))*(d*x^2+c)^(1/2)*EllipticPi(b^(1/2)*x/a^(1/2)/(1+b*x^2/a)^(1/2), 1-a*f/b/e,(1-a*d/b/c)^(1/2))/b^(1/2)/c/f^4/(-a*f+b*e)/(b*x^2+a)^(1/2)/(a*( d*x^2+c)/c/(b*x^2+a))^(1/2)
Result contains complex when optimal does not.
Time = 6.33 (sec) , antiderivative size = 514, normalized size of antiderivative = 0.58 \[ \int \frac {x^4 \sqrt {a+b x^2} \left (c+d x^2\right )^{3/2}}{\left (e+f x^2\right )^2} \, dx=\frac {i c f \left (4 a^2 d^2 f^2+2 a b d f (10 d e-7 c f)+b^2 \left (-105 d^2 e^2+95 c d e f-6 c^2 f^2\right )\right ) \sqrt {1+\frac {b x^2}{a}} \sqrt {1+\frac {d x^2}{c}} \left (e+f x^2\right ) E\left (i \text {arcsinh}\left (\sqrt {\frac {b}{a}} x\right )|\frac {a d}{b c}\right )-i \left (2 a^2 c d^2 f^3+2 a b d f \left (45 d^2 e^2-40 c d e f+2 c^2 f^2\right )+b^2 \left (-105 d^3 e^3+60 c d^2 e^2 f+35 c^2 d e f^2-6 c^3 f^3\right )\right ) \sqrt {1+\frac {b x^2}{a}} \sqrt {1+\frac {d x^2}{c}} \left (e+f x^2\right ) \operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt {\frac {b}{a}} x\right ),\frac {a d}{b c}\right )+d \left (\sqrt {\frac {b}{a}} f^2 x \left (a+b x^2\right ) \left (c+d x^2\right ) \left (2 a d f \left (e+f x^2\right )+3 b c f \left (9 e+4 f x^2\right )+b d \left (-35 e^2-14 e f x^2+6 f^2 x^4\right )\right )+15 i b (-d e+c f) (b e (7 d e-4 c f)+3 a f (-2 d e+c f)) \sqrt {1+\frac {b x^2}{a}} \sqrt {1+\frac {d x^2}{c}} \left (e+f x^2\right ) \operatorname {EllipticPi}\left (\frac {a f}{b e},i \text {arcsinh}\left (\sqrt {\frac {b}{a}} x\right ),\frac {a d}{b c}\right )\right )}{30 b \sqrt {\frac {b}{a}} d f^5 \sqrt {a+b x^2} \sqrt {c+d x^2} \left (e+f x^2\right )} \] Input:
Integrate[(x^4*Sqrt[a + b*x^2]*(c + d*x^2)^(3/2))/(e + f*x^2)^2,x]
Output:
(I*c*f*(4*a^2*d^2*f^2 + 2*a*b*d*f*(10*d*e - 7*c*f) + b^2*(-105*d^2*e^2 + 9 5*c*d*e*f - 6*c^2*f^2))*Sqrt[1 + (b*x^2)/a]*Sqrt[1 + (d*x^2)/c]*(e + f*x^2 )*EllipticE[I*ArcSinh[Sqrt[b/a]*x], (a*d)/(b*c)] - I*(2*a^2*c*d^2*f^3 + 2* a*b*d*f*(45*d^2*e^2 - 40*c*d*e*f + 2*c^2*f^2) + b^2*(-105*d^3*e^3 + 60*c*d ^2*e^2*f + 35*c^2*d*e*f^2 - 6*c^3*f^3))*Sqrt[1 + (b*x^2)/a]*Sqrt[1 + (d*x^ 2)/c]*(e + f*x^2)*EllipticF[I*ArcSinh[Sqrt[b/a]*x], (a*d)/(b*c)] + d*(Sqrt [b/a]*f^2*x*(a + b*x^2)*(c + d*x^2)*(2*a*d*f*(e + f*x^2) + 3*b*c*f*(9*e + 4*f*x^2) + b*d*(-35*e^2 - 14*e*f*x^2 + 6*f^2*x^4)) + (15*I)*b*(-(d*e) + c* f)*(b*e*(7*d*e - 4*c*f) + 3*a*f*(-2*d*e + c*f))*Sqrt[1 + (b*x^2)/a]*Sqrt[1 + (d*x^2)/c]*(e + f*x^2)*EllipticPi[(a*f)/(b*e), I*ArcSinh[Sqrt[b/a]*x], (a*d)/(b*c)]))/(30*b*Sqrt[b/a]*d*f^5*Sqrt[a + b*x^2]*Sqrt[c + d*x^2]*(e + f*x^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.
| \(\displaystyle \int \frac {x^4 \sqrt {a+b x^2} \left (c+d x^2\right )^{3/2}}{\left (e+f x^2\right )^2} \, dx\) |
| \(\Big \downarrow \) 450 |
| \(\displaystyle \int \frac {x^4 \sqrt {a+b x^2} \left (c+d x^2\right )^{3/2}}{\left (e+f x^2\right )^2}dx\) |
Input:
Int[(x^4*Sqrt[a + b*x^2]*(c + d*x^2)^(3/2))/(e + f*x^2)^2,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]
Time = 21.30 (sec) , antiderivative size = 1603, normalized size of antiderivative = 1.82
| method | result | size |
| risch | \(\text {Expression too large to display}\) | \(1603\) |
| elliptic | \(\text {Expression too large to display}\) | \(2151\) |
| default | \(\text {Expression too large to display}\) | \(3195\) |
Input:
int(x^4*(b*x^2+a)^(1/2)*(d*x^2+c)^(3/2)/(f*x^2+e)^2,x,method=_RETURNVERBOS E)
Output:
1/15*x*(3*b*d*f*x^2+a*d*f+6*b*c*f-10*b*d*e)*(b*x^2+a)^(1/2)*(d*x^2+c)^(1/2 )/b/f^3-1/15/b/f^3*((a^2*c*d*f^3-9*a*b*c^2*f^3+50*a*b*c*d*e*f^2-45*a*b*d^2 *e^2*f+30*b^2*c^2*e*f^2-90*b^2*c*d*e^2*f+60*b^2*d^2*e^3)/f^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)*El lipticF(x*(-b/a)^(1/2),(-1+(a*d+b*c)/c/b)^(1/2))-1/f*(2*a^2*d^2*f^2-7*a*b* c*d*f^2+10*a*b*d^2*e*f-3*b^2*c^2*f^2+40*b^2*c*d*e*f-45*b^2*d^2*e^2)*c/(-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)/d*(EllipticF(x*(-b/a)^(1/2),(-1+(a*d+b*c)/c/b)^(1/2))-EllipticE(x*( -b/a)^(1/2),(-1+(a*d+b*c)/c/b)^(1/2)))+15*b/f^2*(2*a*c^2*f^3-6*a*c*d*e*f^2 +4*a*d^2*e^2*f-3*b*c^2*e*f^2+8*b*c*d*e^2*f-5*b*d^2*e^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)*Ellipti cPi(x*(-b/a)^(1/2),a*f/b/e,(-1/c*d)^(1/2)/(-b/a)^(1/2))-15*b*e^2*(a*c^2*f^ 3-2*a*c*d*e*f^2+a*d^2*e^2*f-b*c^2*e*f^2+2*b*c*d*e^2*f-b*d^2*e^3)/f^2*(1/2* f^2/(a*c*f^2-a*d*e*f-b*c*e*f+b*d*e^2)/e*x*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1 /2)/(f*x^2+e)-1/2*d*b/(a*c*f^2-a*d*e*f-b*c*e*f+b*d*e^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)*Ellipti cF(x*(-b/a)^(1/2),(-1+(a*d+b*c)/c/b)^(1/2))+1/2*f*b/(a*c*f^2-a*d*e*f-b*c*e *f+b*d*e^2)/e*c/(-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))-1/2*f*b/(a*c*f^2-a*d*e*f-b*c*e*f+b*d*e^2)/e*c/(-b/a)^(1/2)*(1+b*x^2...
Timed out. \[ \int \frac {x^4 \sqrt {a+b x^2} \left (c+d x^2\right )^{3/2}}{\left (e+f x^2\right )^2} \, dx=\text {Timed out} \] Input:
integrate(x^4*(b*x^2+a)^(1/2)*(d*x^2+c)^(3/2)/(f*x^2+e)^2,x, algorithm="fr icas")
Output:
Timed out
Timed out. \[ \int \frac {x^4 \sqrt {a+b x^2} \left (c+d x^2\right )^{3/2}}{\left (e+f x^2\right )^2} \, dx=\text {Timed out} \] Input:
integrate(x**4*(b*x**2+a)**(1/2)*(d*x**2+c)**(3/2)/(f*x**2+e)**2,x)
Output:
Timed out
\[ \int \frac {x^4 \sqrt {a+b x^2} \left (c+d x^2\right )^{3/2}}{\left (e+f x^2\right )^2} \, dx=\int { \frac {\sqrt {b x^{2} + a} {\left (d x^{2} + c\right )}^{\frac {3}{2}} x^{4}}{{\left (f x^{2} + e\right )}^{2}} \,d x } \] Input:
integrate(x^4*(b*x^2+a)^(1/2)*(d*x^2+c)^(3/2)/(f*x^2+e)^2,x, algorithm="ma xima")
Output:
integrate(sqrt(b*x^2 + a)*(d*x^2 + c)^(3/2)*x^4/(f*x^2 + e)^2, x)
\[ \int \frac {x^4 \sqrt {a+b x^2} \left (c+d x^2\right )^{3/2}}{\left (e+f x^2\right )^2} \, dx=\int { \frac {\sqrt {b x^{2} + a} {\left (d x^{2} + c\right )}^{\frac {3}{2}} x^{4}}{{\left (f x^{2} + e\right )}^{2}} \,d x } \] Input:
integrate(x^4*(b*x^2+a)^(1/2)*(d*x^2+c)^(3/2)/(f*x^2+e)^2,x, algorithm="gi ac")
Output:
integrate(sqrt(b*x^2 + a)*(d*x^2 + c)^(3/2)*x^4/(f*x^2 + e)^2, x)
Timed out. \[ \int \frac {x^4 \sqrt {a+b x^2} \left (c+d x^2\right )^{3/2}}{\left (e+f x^2\right )^2} \, dx=\int \frac {x^4\,\sqrt {b\,x^2+a}\,{\left (d\,x^2+c\right )}^{3/2}}{{\left (f\,x^2+e\right )}^2} \,d x \] Input:
int((x^4*(a + b*x^2)^(1/2)*(c + d*x^2)^(3/2))/(e + f*x^2)^2,x)
Output:
int((x^4*(a + b*x^2)^(1/2)*(c + d*x^2)^(3/2))/(e + f*x^2)^2, x)
\[ \int \frac {x^4 \sqrt {a+b x^2} \left (c+d x^2\right )^{3/2}}{\left (e+f x^2\right )^2} \, dx=\int \frac {x^{4} \sqrt {b \,x^{2}+a}\, \left (d \,x^{2}+c \right )^{\frac {3}{2}}}{\left (f \,x^{2}+e \right )^{2}}d x \] Input:
int(x^4*(b*x^2+a)^(1/2)*(d*x^2+c)^(3/2)/(f*x^2+e)^2,x)
Output:
int(x^4*(b*x^2+a)^(1/2)*(d*x^2+c)^(3/2)/(f*x^2+e)^2,x)