Integrand size = 34, antiderivative size = 708 \[ \int \frac {\left (a+b x^2\right )^{5/2} \sqrt {e+f x^2}}{\left (c+d x^2\right )^{5/2}} \, dx=\frac {(b c-a d)^2 x \sqrt {a+b x^2} \sqrt {e+f x^2}}{3 c d^2 \left (c+d x^2\right )^{3/2}}+\frac {b^2 x \sqrt {a+b x^2} \sqrt {e+f x^2}}{2 d^2 \sqrt {c+d x^2}}-\frac {\sqrt {e} \left (b^2 c^2 (13 d e-15 c f)-2 a b c d (3 d e-5 c f)-2 a^2 d^2 (2 d e-c f)\right ) \sqrt {a+b x^2} \sqrt {\frac {c \left (e+f x^2\right )}{e \left (c+d x^2\right )}} E\left (\arcsin \left (\frac {\sqrt {d e-c f} x}{\sqrt {e} \sqrt {c+d x^2}}\right )|-\frac {(b c-a d) e}{a (d e-c f)}\right )}{6 c^2 d^3 \sqrt {d e-c f} \sqrt {\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}} \sqrt {e+f x^2}}+\frac {\sqrt {e} \left (2 a^3 d^3 f+2 a b^2 c d (4 d e-15 c f)-3 b^3 c^2 (d e-5 c f)-2 a^2 b d^2 (d e-5 c f)\right ) \sqrt {a+b x^2} \sqrt {\frac {c \left (e+f x^2\right )}{e \left (c+d x^2\right )}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {d e-c f} x}{\sqrt {e} \sqrt {c+d x^2}}\right ),-\frac {(b c-a d) e}{a (d e-c f)}\right )}{6 a c d^4 \sqrt {d e-c f} \sqrt {\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}} \sqrt {e+f x^2}}+\frac {b^2 c \sqrt {e} (b d e-5 b c f+5 a d f) \sqrt {a+b x^2} \sqrt {\frac {c \left (e+f x^2\right )}{e \left (c+d x^2\right )}} \operatorname {EllipticPi}\left (\frac {d e}{d e-c f},\arcsin \left (\frac {\sqrt {d e-c f} x}{\sqrt {e} \sqrt {c+d x^2}}\right ),-\frac {(b c-a d) e}{a (d e-c f)}\right )}{2 a d^4 \sqrt {d e-c f} \sqrt {\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}} \sqrt {e+f x^2}} \] Output:
1/3*(-a*d+b*c)^2*x*(b*x^2+a)^(1/2)*(f*x^2+e)^(1/2)/c/d^2/(d*x^2+c)^(3/2)+1 /2*b^2*x*(b*x^2+a)^(1/2)*(f*x^2+e)^(1/2)/d^2/(d*x^2+c)^(1/2)-1/6*e^(1/2)*( b^2*c^2*(-15*c*f+13*d*e)-2*a*b*c*d*(-5*c*f+3*d*e)-2*a^2*d^2*(-c*f+2*d*e))* (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^2/d^3/(-c *f+d*e)^(1/2)/(c*(b*x^2+a)/a/(d*x^2+c))^(1/2)/(f*x^2+e)^(1/2)+1/6*e^(1/2)* (2*a^3*d^3*f+2*a*b^2*c*d*(-15*c*f+4*d*e)-3*b^3*c^2*(-5*c*f+d*e)-2*a^2*b*d^ 2*(-5*c*f+d*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))/a/c/d^4/(-c*f+d*e)^(1/2)/(c*(b*x^2+a)/a/(d*x^2+c))^(1/2)/(f*x^2+e)^(1 /2)+1/2*b^2*c*e^(1/2)*(5*a*d*f-5*b*c*f+b*d*e)*(b*x^2+a)^(1/2)*(c*(f*x^2+e) /e/(d*x^2+c))^(1/2)*EllipticPi((-c*f+d*e)^(1/2)*x/e^(1/2)/(d*x^2+c)^(1/2), d*e/(-c*f+d*e),(-(-a*d+b*c)*e/a/(-c*f+d*e))^(1/2))/a/d^4/(-c*f+d*e)^(1/2)/ (c*(b*x^2+a)/a/(d*x^2+c))^(1/2)/(f*x^2+e)^(1/2)
\[ \int \frac {\left (a+b x^2\right )^{5/2} \sqrt {e+f x^2}}{\left (c+d x^2\right )^{5/2}} \, dx=\int \frac {\left (a+b x^2\right )^{5/2} \sqrt {e+f x^2}}{\left (c+d x^2\right )^{5/2}} \, dx \] Input:
Integrate[((a + b*x^2)^(5/2)*Sqrt[e + f*x^2])/(c + d*x^2)^(5/2),x]
Output:
Integrate[((a + b*x^2)^(5/2)*Sqrt[e + f*x^2])/(c + d*x^2)^(5/2), 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.
\(\displaystyle \int \frac {\left (a+b x^2\right )^{5/2} \sqrt {e+f x^2}}{\left (c+d x^2\right )^{5/2}} \, dx\) |
\(\Big \downarrow \) 434 |
\(\displaystyle \int \frac {\left (a+b x^2\right )^{5/2} \sqrt {e+f x^2}}{\left (c+d x^2\right )^{5/2}}dx\) |
Input:
Int[((a + b*x^2)^(5/2)*Sqrt[e + f*x^2])/(c + d*x^2)^(5/2),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]
\[\int \frac {\left (b \,x^{2}+a \right )^{\frac {5}{2}} \sqrt {f \,x^{2}+e}}{\left (x^{2} d +c \right )^{\frac {5}{2}}}d x\]
Input:
int((b*x^2+a)^(5/2)*(f*x^2+e)^(1/2)/(d*x^2+c)^(5/2),x)
Output:
int((b*x^2+a)^(5/2)*(f*x^2+e)^(1/2)/(d*x^2+c)^(5/2),x)
Timed out. \[ \int \frac {\left (a+b x^2\right )^{5/2} \sqrt {e+f x^2}}{\left (c+d x^2\right )^{5/2}} \, dx=\text {Timed out} \] Input:
integrate((b*x^2+a)^(5/2)*(f*x^2+e)^(1/2)/(d*x^2+c)^(5/2),x, algorithm="fr icas")
Output:
Timed out
Timed out. \[ \int \frac {\left (a+b x^2\right )^{5/2} \sqrt {e+f x^2}}{\left (c+d x^2\right )^{5/2}} \, dx=\text {Timed out} \] Input:
integrate((b*x**2+a)**(5/2)*(f*x**2+e)**(1/2)/(d*x**2+c)**(5/2),x)
Output:
Timed out
\[ \int \frac {\left (a+b x^2\right )^{5/2} \sqrt {e+f x^2}}{\left (c+d x^2\right )^{5/2}} \, dx=\int { \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} \sqrt {f x^{2} + e}}{{\left (d x^{2} + c\right )}^{\frac {5}{2}}} \,d x } \] Input:
integrate((b*x^2+a)^(5/2)*(f*x^2+e)^(1/2)/(d*x^2+c)^(5/2),x, algorithm="ma xima")
Output:
integrate((b*x^2 + a)^(5/2)*sqrt(f*x^2 + e)/(d*x^2 + c)^(5/2), x)
\[ \int \frac {\left (a+b x^2\right )^{5/2} \sqrt {e+f x^2}}{\left (c+d x^2\right )^{5/2}} \, dx=\int { \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} \sqrt {f x^{2} + e}}{{\left (d x^{2} + c\right )}^{\frac {5}{2}}} \,d x } \] Input:
integrate((b*x^2+a)^(5/2)*(f*x^2+e)^(1/2)/(d*x^2+c)^(5/2),x, algorithm="gi ac")
Output:
integrate((b*x^2 + a)^(5/2)*sqrt(f*x^2 + e)/(d*x^2 + c)^(5/2), x)
Timed out. \[ \int \frac {\left (a+b x^2\right )^{5/2} \sqrt {e+f x^2}}{\left (c+d x^2\right )^{5/2}} \, dx=\int \frac {{\left (b\,x^2+a\right )}^{5/2}\,\sqrt {f\,x^2+e}}{{\left (d\,x^2+c\right )}^{5/2}} \,d x \] Input:
int(((a + b*x^2)^(5/2)*(e + f*x^2)^(1/2))/(c + d*x^2)^(5/2),x)
Output:
int(((a + b*x^2)^(5/2)*(e + f*x^2)^(1/2))/(c + d*x^2)^(5/2), x)
\[ \int \frac {\left (a+b x^2\right )^{5/2} \sqrt {e+f x^2}}{\left (c+d x^2\right )^{5/2}} \, dx=\int \frac {\left (b \,x^{2}+a \right )^{\frac {5}{2}} \sqrt {f \,x^{2}+e}}{\left (d \,x^{2}+c \right )^{\frac {5}{2}}}d x \] Input:
int((b*x^2+a)^(5/2)*(f*x^2+e)^(1/2)/(d*x^2+c)^(5/2),x)
Output:
int((b*x^2+a)^(5/2)*(f*x^2+e)^(1/2)/(d*x^2+c)^(5/2),x)