\(\int \frac {(a+b x^2)^{3/4}}{(c+d x^2)^{9/4}} \, dx\) [505]

Optimal result

Integrand size = 23, antiderivative size = 87 \[ \int \frac {\left (a+b x^2\right )^{3/4}}{\left (c+d x^2\right )^{9/4}} \, dx=\frac {x \left (a+b x^2\right )^{3/4} \operatorname {Hypergeometric2F1}\left (-\frac {3}{4},\frac {1}{2},\frac {3}{2},-\frac {(b c-a d) x^2}{a \left (c+d x^2\right )}\right )}{c \left (\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}\right )^{3/4} \left (c+d x^2\right )^{5/4}} \] Output:

x*(b*x^2+a)^(3/4)*hypergeom([-3/4, 1/2],[3/2],-(-a*d+b*c)*x^2/a/(d*x^2+c)) 
/c/(c*(b*x^2+a)/a/(d*x^2+c))^(3/4)/(d*x^2+c)^(5/4)
 

Mathematica [A] (warning: unable to verify)

Time = 5.12 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.02 \[ \int \frac {\left (a+b x^2\right )^{3/4}}{\left (c+d x^2\right )^{9/4}} \, dx=\frac {x \left (a+b x^2\right )^{3/4} \operatorname {Hypergeometric2F1}\left (-\frac {3}{4},\frac {1}{2},\frac {3}{2},\frac {(-b c+a d) x^2}{a \left (c+d x^2\right )}\right )}{c^2 \left (1+\frac {b x^2}{a}\right )^{3/4} \sqrt [4]{c+d x^2} \sqrt [4]{1+\frac {d x^2}{c}}} \] Input:

Integrate[(a + b*x^2)^(3/4)/(c + d*x^2)^(9/4),x]
 

Output:

(x*(a + b*x^2)^(3/4)*Hypergeometric2F1[-3/4, 1/2, 3/2, ((-(b*c) + a*d)*x^2 
)/(a*(c + d*x^2))])/(c^2*(1 + (b*x^2)/a)^(3/4)*(c + d*x^2)^(1/4)*(1 + (d*x 
^2)/c)^(1/4))
 

Rubi [A] (verified)

Time = 0.21 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.40, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {292, 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.

\(\displaystyle \int \frac {\left (a+b x^2\right )^{3/4}}{\left (c+d x^2\right )^{9/4}} \, dx\)

\(\Big \downarrow \) 292

\(\displaystyle \frac {3 a \int \frac {1}{\sqrt [4]{b x^2+a} \left (d x^2+c\right )^{5/4}}dx}{5 c}+\frac {2 x \left (a+b x^2\right )^{3/4}}{5 c \left (c+d x^2\right )^{5/4}}\)

\(\Big \downarrow \) 294

\(\displaystyle \frac {3 a x \sqrt [4]{\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {3}{2},-\frac {(b c-a d) x^2}{a \left (d x^2+c\right )}\right )}{5 c^2 \sqrt [4]{a+b x^2} \sqrt [4]{c+d x^2}}+\frac {2 x \left (a+b x^2\right )^{3/4}}{5 c \left (c+d x^2\right )^{5/4}}\)

Input:

Int[(a + b*x^2)^(3/4)/(c + d*x^2)^(9/4),x]
 

Output:

(2*x*(a + b*x^2)^(3/4))/(5*c*(c + d*x^2)^(5/4)) + (3*a*x*((c*(a + b*x^2))/ 
(a*(c + d*x^2)))^(1/4)*Hypergeometric2F1[1/4, 1/2, 3/2, -(((b*c - a*d)*x^2 
)/(a*(c + d*x^2)))])/(5*c^2*(a + b*x^2)^(1/4)*(c + d*x^2)^(1/4))
 

Defintions of rubi rules used

rule 292

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_.), x_Symbol] :> Si 
mp[(-x)*(a + b*x^2)^(p + 1)*((c + d*x^2)^q/(2*a*(p + 1))), x] - Simp[c*(q/( 
a*(p + 1)))   Int[(a + b*x^2)^(p + 1)*(c + d*x^2)^(q - 1), x], x] /; FreeQ[ 
{a, b, c, d, p}, x] && NeQ[b*c - a*d, 0] && EqQ[2*(p + q + 1) + 1, 0] && Gt 
Q[q, 0] && NeQ[p, -1]
 

rule 294

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]
 

Maple [F]

\[\int \frac {\left (b \,x^{2}+a \right )^{\frac {3}{4}}}{\left (x^{2} d +c \right )^{\frac {9}{4}}}d x\]

Input:

int((b*x^2+a)^(3/4)/(d*x^2+c)^(9/4),x)
 

Output:

int((b*x^2+a)^(3/4)/(d*x^2+c)^(9/4),x)
 

Fricas [F]

\[ \int \frac {\left (a+b x^2\right )^{3/4}}{\left (c+d x^2\right )^{9/4}} \, dx=\int { \frac {{\left (b x^{2} + a\right )}^{\frac {3}{4}}}{{\left (d x^{2} + c\right )}^{\frac {9}{4}}} \,d x } \] Input:

integrate((b*x^2+a)^(3/4)/(d*x^2+c)^(9/4),x, algorithm="fricas")
 

Output:

integral((b*x^2 + a)^(3/4)*(d*x^2 + c)^(3/4)/(d^3*x^6 + 3*c*d^2*x^4 + 3*c^ 
2*d*x^2 + c^3), x)
 

Sympy [F]

\[ \int \frac {\left (a+b x^2\right )^{3/4}}{\left (c+d x^2\right )^{9/4}} \, dx=\int \frac {\left (a + b x^{2}\right )^{\frac {3}{4}}}{\left (c + d x^{2}\right )^{\frac {9}{4}}}\, dx \] Input:

integrate((b*x**2+a)**(3/4)/(d*x**2+c)**(9/4),x)
 

Output:

Integral((a + b*x**2)**(3/4)/(c + d*x**2)**(9/4), x)
 

Maxima [F]

\[ \int \frac {\left (a+b x^2\right )^{3/4}}{\left (c+d x^2\right )^{9/4}} \, dx=\int { \frac {{\left (b x^{2} + a\right )}^{\frac {3}{4}}}{{\left (d x^{2} + c\right )}^{\frac {9}{4}}} \,d x } \] Input:

integrate((b*x^2+a)^(3/4)/(d*x^2+c)^(9/4),x, algorithm="maxima")
 

Output:

integrate((b*x^2 + a)^(3/4)/(d*x^2 + c)^(9/4), x)
 

Giac [F]

\[ \int \frac {\left (a+b x^2\right )^{3/4}}{\left (c+d x^2\right )^{9/4}} \, dx=\int { \frac {{\left (b x^{2} + a\right )}^{\frac {3}{4}}}{{\left (d x^{2} + c\right )}^{\frac {9}{4}}} \,d x } \] Input:

integrate((b*x^2+a)^(3/4)/(d*x^2+c)^(9/4),x, algorithm="giac")
 

Output:

integrate((b*x^2 + a)^(3/4)/(d*x^2 + c)^(9/4), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (a+b x^2\right )^{3/4}}{\left (c+d x^2\right )^{9/4}} \, dx=\int \frac {{\left (b\,x^2+a\right )}^{3/4}}{{\left (d\,x^2+c\right )}^{9/4}} \,d x \] Input:

int((a + b*x^2)^(3/4)/(c + d*x^2)^(9/4),x)
 

Output:

int((a + b*x^2)^(3/4)/(c + d*x^2)^(9/4), x)
 

Reduce [F]

\[ \int \frac {\left (a+b x^2\right )^{3/4}}{\left (c+d x^2\right )^{9/4}} \, dx=\int \frac {\left (b \,x^{2}+a \right )^{\frac {3}{4}}}{\left (d \,x^{2}+c \right )^{\frac {1}{4}} c^{2}+2 \left (d \,x^{2}+c \right )^{\frac {1}{4}} c d \,x^{2}+\left (d \,x^{2}+c \right )^{\frac {1}{4}} d^{2} x^{4}}d x \] Input:

int((b*x^2+a)^(3/4)/(d*x^2+c)^(9/4),x)
 

Output:

int((a + b*x**2)**(3/4)/((c + d*x**2)**(1/4)*c**2 + 2*(c + d*x**2)**(1/4)* 
c*d*x**2 + (c + d*x**2)**(1/4)*d**2*x**4),x)