Integrand size = 23, antiderivative size = 86 \[ \int \frac {1}{\left (a+b x^2\right )^{5/4} \sqrt [4]{c+d x^2}} \, dx=\frac {x \sqrt [4]{\frac {a \left (c+d x^2\right )}{c \left (a+b x^2\right )}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {3}{2},\frac {(b c-a d) x^2}{c \left (a+b x^2\right )}\right )}{a \sqrt [4]{a+b x^2} \sqrt [4]{c+d x^2}} \] Output:
x*(a*(d*x^2+c)/c/(b*x^2+a))^(1/4)*hypergeom([1/4, 1/2],[3/2],(-a*d+b*c)*x^ 2/c/(b*x^2+a))/a/(b*x^2+a)^(1/4)/(d*x^2+c)^(1/4)
Time = 2.52 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.03 \[ \int \frac {1}{\left (a+b x^2\right )^{5/4} \sqrt [4]{c+d x^2}} \, dx=\frac {x \sqrt [4]{1+\frac {b x^2}{a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5}{4},\frac {3}{2},\frac {(-b c+a d) x^2}{a \left (c+d x^2\right )}\right )}{a \sqrt [4]{a+b x^2} \sqrt [4]{c+d x^2} \sqrt [4]{1+\frac {d x^2}{c}}} \] Input:
Integrate[1/((a + b*x^2)^(5/4)*(c + d*x^2)^(1/4)),x]
Output:
(x*(1 + (b*x^2)/a)^(1/4)*Hypergeometric2F1[1/2, 5/4, 3/2, ((-(b*c) + a*d)* x^2)/(a*(c + d*x^2))])/(a*(a + b*x^2)^(1/4)*(c + d*x^2)^(1/4)*(1 + (d*x^2) /c)^(1/4))
Time = 0.18 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.01, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.043, Rules used = {294}
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.
\(\displaystyle \int \frac {1}{\left (a+b x^2\right )^{5/4} \sqrt [4]{c+d x^2}} \, dx\) |
\(\Big \downarrow \) 294 |
\(\displaystyle \frac {x \left (c+d x^2\right )^{3/4} \left (\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}\right )^{5/4} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5}{4},\frac {3}{2},-\frac {(b c-a d) x^2}{a \left (d x^2+c\right )}\right )}{c \left (a+b x^2\right )^{5/4}}\) |
Input:
Int[1/((a + b*x^2)^(5/4)*(c + d*x^2)^(1/4)),x]
Output:
(x*((c*(a + b*x^2))/(a*(c + d*x^2)))^(5/4)*(c + d*x^2)^(3/4)*Hypergeometri c2F1[1/2, 5/4, 3/2, -(((b*c - a*d)*x^2)/(a*(c + d*x^2)))])/(c*(a + b*x^2)^ (5/4))
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim p[x*((a + b*x^2)^p/(c*(c*((a + b*x^2)/(a*(c + d*x^2))))^p*(c + d*x^2)^(1/2 + p)))*Hypergeometric2F1[1/2, -p, 3/2, (-(b*c - a*d))*(x^2/(a*(c + d*x^2))) ], x] /; FreeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && EqQ[2*(p + q + 1) + 1, 0]
\[\int \frac {1}{\left (b \,x^{2}+a \right )^{\frac {5}{4}} \left (x^{2} d +c \right )^{\frac {1}{4}}}d x\]
Input:
int(1/(b*x^2+a)^(5/4)/(d*x^2+c)^(1/4),x)
Output:
int(1/(b*x^2+a)^(5/4)/(d*x^2+c)^(1/4),x)
\[ \int \frac {1}{\left (a+b x^2\right )^{5/4} \sqrt [4]{c+d x^2}} \, dx=\int { \frac {1}{{\left (b x^{2} + a\right )}^{\frac {5}{4}} {\left (d x^{2} + c\right )}^{\frac {1}{4}}} \,d x } \] Input:
integrate(1/(b*x^2+a)^(5/4)/(d*x^2+c)^(1/4),x, algorithm="fricas")
Output:
integral((b*x^2 + a)^(3/4)*(d*x^2 + c)^(3/4)/(b^2*d*x^6 + (b^2*c + 2*a*b*d )*x^4 + a^2*c + (2*a*b*c + a^2*d)*x^2), x)
\[ \int \frac {1}{\left (a+b x^2\right )^{5/4} \sqrt [4]{c+d x^2}} \, dx=\int \frac {1}{\left (a + b x^{2}\right )^{\frac {5}{4}} \sqrt [4]{c + d x^{2}}}\, dx \] Input:
integrate(1/(b*x**2+a)**(5/4)/(d*x**2+c)**(1/4),x)
Output:
Integral(1/((a + b*x**2)**(5/4)*(c + d*x**2)**(1/4)), x)
\[ \int \frac {1}{\left (a+b x^2\right )^{5/4} \sqrt [4]{c+d x^2}} \, dx=\int { \frac {1}{{\left (b x^{2} + a\right )}^{\frac {5}{4}} {\left (d x^{2} + c\right )}^{\frac {1}{4}}} \,d x } \] Input:
integrate(1/(b*x^2+a)^(5/4)/(d*x^2+c)^(1/4),x, algorithm="maxima")
Output:
integrate(1/((b*x^2 + a)^(5/4)*(d*x^2 + c)^(1/4)), x)
\[ \int \frac {1}{\left (a+b x^2\right )^{5/4} \sqrt [4]{c+d x^2}} \, dx=\int { \frac {1}{{\left (b x^{2} + a\right )}^{\frac {5}{4}} {\left (d x^{2} + c\right )}^{\frac {1}{4}}} \,d x } \] Input:
integrate(1/(b*x^2+a)^(5/4)/(d*x^2+c)^(1/4),x, algorithm="giac")
Output:
integrate(1/((b*x^2 + a)^(5/4)*(d*x^2 + c)^(1/4)), x)
Timed out. \[ \int \frac {1}{\left (a+b x^2\right )^{5/4} \sqrt [4]{c+d x^2}} \, dx=\int \frac {1}{{\left (b\,x^2+a\right )}^{5/4}\,{\left (d\,x^2+c\right )}^{1/4}} \,d x \] Input:
int(1/((a + b*x^2)^(5/4)*(c + d*x^2)^(1/4)),x)
Output:
int(1/((a + b*x^2)^(5/4)*(c + d*x^2)^(1/4)), x)
\[ \int \frac {1}{\left (a+b x^2\right )^{5/4} \sqrt [4]{c+d x^2}} \, dx=\int \frac {1}{\left (d \,x^{2}+c \right )^{\frac {1}{4}} \left (b \,x^{2}+a \right )^{\frac {1}{4}} a +\left (d \,x^{2}+c \right )^{\frac {1}{4}} \left (b \,x^{2}+a \right )^{\frac {1}{4}} b \,x^{2}}d x \] Input:
int(1/(b*x^2+a)^(5/4)/(d*x^2+c)^(1/4),x)
Output:
int(1/((c + d*x**2)**(1/4)*(a + b*x**2)**(1/4)*a + (c + d*x**2)**(1/4)*(a + b*x**2)**(1/4)*b*x**2),x)