\(\int \frac {(c+d x^2)^{3/4}}{(a+b x^2)^{9/4}} \, dx\) [543]

Optimal result

Integrand size = 23, antiderivative size = 86 \[ \int \frac {\left (c+d x^2\right )^{3/4}}{\left (a+b x^2\right )^{9/4}} \, dx=\frac {x \left (c+d x^2\right )^{3/4} \operatorname {Hypergeometric2F1}\left (-\frac {3}{4},\frac {1}{2},\frac {3}{2},\frac {(b c-a d) x^2}{c \left (a+b x^2\right )}\right )}{a \left (a+b x^2\right )^{5/4} \left (\frac {a \left (c+d x^2\right )}{c \left (a+b x^2\right )}\right )^{3/4}} \] Output:

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

Mathematica [A] (warning: unable to verify)

Time = 5.11 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.03 \[ \int \frac {\left (c+d x^2\right )^{3/4}}{\left (a+b x^2\right )^{9/4}} \, dx=\frac {x \sqrt [4]{1+\frac {b x^2}{a}} \left (c+d x^2\right )^{3/4} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {9}{4},\frac {3}{2},\frac {(-b c+a d) x^2}{a \left (c+d x^2\right )}\right )}{a^2 \sqrt [4]{a+b x^2} \left (1+\frac {d x^2}{c}\right )^{5/4}} \] Input:

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

Output:

(x*(1 + (b*x^2)/a)^(1/4)*(c + d*x^2)^(3/4)*Hypergeometric2F1[1/2, 9/4, 3/2 
, ((-(b*c) + a*d)*x^2)/(a*(c + d*x^2))])/(a^2*(a + b*x^2)^(1/4)*(1 + (d*x^ 
2)/c)^(5/4))
 

Rubi [A] (verified)

Time = 0.20 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.41, 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.

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

\(\Big \downarrow \) 292

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

\(\Big \downarrow \) 294

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

Input:

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

Output:

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

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]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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