Integrand size = 34, antiderivative size = 533 \[ \int \frac {\sqrt {a+b x^2} \left (e+f x^2\right )^{3/2}}{\left (c+d x^2\right )^{7/2}} \, dx=\frac {(d e-c f) x \sqrt {a+b x^2} \sqrt {e+f x^2}}{5 c d \left (c+d x^2\right )^{5/2}}+\frac {(3 b c (d e+c f)-2 a d (2 d e+c f)) x \sqrt {a+b x^2} \sqrt {e+f x^2}}{15 c^2 d (b c-a d) \left (c+d x^2\right )^{3/2}}+\frac {\sqrt {e} \left (3 b^2 c^2 e^2-a b c e (13 d e-7 c f)+a^2 \left (8 d^2 e^2-3 c d e f-2 c^2 f^2\right )\right ) \sqrt {a+b x^2} \sqrt {\frac {c \left (e+f x^2\right )}{e \left (c+d x^2\right )}} E\left (\arcsin \left (\frac {\sqrt {d e-c f} x}{\sqrt {e} \sqrt {c+d x^2}}\right )|-\frac {(b c-a d) e}{a (d e-c f)}\right )}{15 c^3 (b c-a d)^2 \sqrt {d e-c f} \sqrt {\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}} \sqrt {e+f x^2}}+\frac {2 \sqrt {e} (b e-a f) (3 b c e-2 a d e-a c f) \sqrt {a+b x^2} \sqrt {\frac {c \left (e+f x^2\right )}{e \left (c+d x^2\right )}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {d e-c f} x}{\sqrt {e} \sqrt {c+d x^2}}\right ),-\frac {(b c-a d) e}{a (d e-c f)}\right )}{15 c^2 (b c-a d)^2 \sqrt {d e-c f} \sqrt {\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}} \sqrt {e+f x^2}} \] Output:
1/5*(-c*f+d*e)*x*(b*x^2+a)^(1/2)*(f*x^2+e)^(1/2)/c/d/(d*x^2+c)^(5/2)+1/15* (3*b*c*(c*f+d*e)-2*a*d*(c*f+2*d*e))*x*(b*x^2+a)^(1/2)*(f*x^2+e)^(1/2)/c^2/ d/(-a*d+b*c)/(d*x^2+c)^(3/2)+1/15*e^(1/2)*(3*b^2*c^2*e^2-a*b*c*e*(-7*c*f+1 3*d*e)+a^2*(-2*c^2*f^2-3*c*d*e*f+8*d^2*e^2))*(b*x^2+a)^(1/2)*(c*(f*x^2+e)/ e/(d*x^2+c))^(1/2)*EllipticE((-c*f+d*e)^(1/2)*x/e^(1/2)/(d*x^2+c)^(1/2),(- (-a*d+b*c)*e/a/(-c*f+d*e))^(1/2))/c^3/(-a*d+b*c)^2/(-c*f+d*e)^(1/2)/(c*(b* x^2+a)/a/(d*x^2+c))^(1/2)/(f*x^2+e)^(1/2)+2/15*e^(1/2)*(-a*f+b*e)*(-a*c*f- 2*a*d*e+3*b*c*e)*(b*x^2+a)^(1/2)*(c*(f*x^2+e)/e/(d*x^2+c))^(1/2)*EllipticF ((-c*f+d*e)^(1/2)*x/e^(1/2)/(d*x^2+c)^(1/2),(-(-a*d+b*c)*e/a/(-c*f+d*e))^( 1/2))/c^2/(-a*d+b*c)^2/(-c*f+d*e)^(1/2)/(c*(b*x^2+a)/a/(d*x^2+c))^(1/2)/(f *x^2+e)^(1/2)
\[ \int \frac {\sqrt {a+b x^2} \left (e+f x^2\right )^{3/2}}{\left (c+d x^2\right )^{7/2}} \, dx=\int \frac {\sqrt {a+b x^2} \left (e+f x^2\right )^{3/2}}{\left (c+d x^2\right )^{7/2}} \, dx \] Input:
Integrate[(Sqrt[a + b*x^2]*(e + f*x^2)^(3/2))/(c + d*x^2)^(7/2),x]
Output:
Integrate[(Sqrt[a + b*x^2]*(e + f*x^2)^(3/2))/(c + d*x^2)^(7/2), x]
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.
\(\displaystyle \int \frac {\sqrt {a+b x^2} \left (e+f x^2\right )^{3/2}}{\left (c+d x^2\right )^{7/2}} \, dx\) |
\(\Big \downarrow \) 434 |
\(\displaystyle \int \frac {\sqrt {a+b x^2} \left (e+f x^2\right )^{3/2}}{\left (c+d x^2\right )^{7/2}}dx\) |
Input:
Int[(Sqrt[a + b*x^2]*(e + f*x^2)^(3/2))/(c + d*x^2)^(7/2),x]
Output:
$Aborted
Int[((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.)*((e_) + (f_.)*( x_)^2)^(r_.), x_Symbol] :> Unintegrable[(a + b*x^2)^p*(c + d*x^2)^q*(e + f* x^2)^r, x] /; FreeQ[{a, b, c, d, e, f, p, q, r}, x]
\[\int \frac {\sqrt {b \,x^{2}+a}\, \left (f \,x^{2}+e \right )^{\frac {3}{2}}}{\left (x^{2} d +c \right )^{\frac {7}{2}}}d x\]
Input:
int((b*x^2+a)^(1/2)*(f*x^2+e)^(3/2)/(d*x^2+c)^(7/2),x)
Output:
int((b*x^2+a)^(1/2)*(f*x^2+e)^(3/2)/(d*x^2+c)^(7/2),x)
\[ \int \frac {\sqrt {a+b x^2} \left (e+f x^2\right )^{3/2}}{\left (c+d x^2\right )^{7/2}} \, dx=\int { \frac {\sqrt {b x^{2} + a} {\left (f x^{2} + e\right )}^{\frac {3}{2}}}{{\left (d x^{2} + c\right )}^{\frac {7}{2}}} \,d x } \] Input:
integrate((b*x^2+a)^(1/2)*(f*x^2+e)^(3/2)/(d*x^2+c)^(7/2),x, algorithm="fr icas")
Output:
integral(sqrt(b*x^2 + a)*sqrt(d*x^2 + c)*(f*x^2 + e)^(3/2)/(d^4*x^8 + 4*c* d^3*x^6 + 6*c^2*d^2*x^4 + 4*c^3*d*x^2 + c^4), x)
Timed out. \[ \int \frac {\sqrt {a+b x^2} \left (e+f x^2\right )^{3/2}}{\left (c+d x^2\right )^{7/2}} \, dx=\text {Timed out} \] Input:
integrate((b*x**2+a)**(1/2)*(f*x**2+e)**(3/2)/(d*x**2+c)**(7/2),x)
Output:
Timed out
\[ \int \frac {\sqrt {a+b x^2} \left (e+f x^2\right )^{3/2}}{\left (c+d x^2\right )^{7/2}} \, dx=\int { \frac {\sqrt {b x^{2} + a} {\left (f x^{2} + e\right )}^{\frac {3}{2}}}{{\left (d x^{2} + c\right )}^{\frac {7}{2}}} \,d x } \] Input:
integrate((b*x^2+a)^(1/2)*(f*x^2+e)^(3/2)/(d*x^2+c)^(7/2),x, algorithm="ma xima")
Output:
integrate(sqrt(b*x^2 + a)*(f*x^2 + e)^(3/2)/(d*x^2 + c)^(7/2), x)
\[ \int \frac {\sqrt {a+b x^2} \left (e+f x^2\right )^{3/2}}{\left (c+d x^2\right )^{7/2}} \, dx=\int { \frac {\sqrt {b x^{2} + a} {\left (f x^{2} + e\right )}^{\frac {3}{2}}}{{\left (d x^{2} + c\right )}^{\frac {7}{2}}} \,d x } \] Input:
integrate((b*x^2+a)^(1/2)*(f*x^2+e)^(3/2)/(d*x^2+c)^(7/2),x, algorithm="gi ac")
Output:
integrate(sqrt(b*x^2 + a)*(f*x^2 + e)^(3/2)/(d*x^2 + c)^(7/2), x)
Timed out. \[ \int \frac {\sqrt {a+b x^2} \left (e+f x^2\right )^{3/2}}{\left (c+d x^2\right )^{7/2}} \, dx=\int \frac {\sqrt {b\,x^2+a}\,{\left (f\,x^2+e\right )}^{3/2}}{{\left (d\,x^2+c\right )}^{7/2}} \,d x \] Input:
int(((a + b*x^2)^(1/2)*(e + f*x^2)^(3/2))/(c + d*x^2)^(7/2),x)
Output:
int(((a + b*x^2)^(1/2)*(e + f*x^2)^(3/2))/(c + d*x^2)^(7/2), x)
\[ \int \frac {\sqrt {a+b x^2} \left (e+f x^2\right )^{3/2}}{\left (c+d x^2\right )^{7/2}} \, dx=\int \frac {\sqrt {b \,x^{2}+a}\, \left (f \,x^{2}+e \right )^{\frac {3}{2}}}{\left (d \,x^{2}+c \right )^{\frac {7}{2}}}d x \] Input:
int((b*x^2+a)^(1/2)*(f*x^2+e)^(3/2)/(d*x^2+c)^(7/2),x)
Output:
int((b*x^2+a)^(1/2)*(f*x^2+e)^(3/2)/(d*x^2+c)^(7/2),x)