\(\int \frac {1}{\sqrt [4]{a+b x^2} (c+d x^2)^{9/4}} \, dx\) [514]

Optimal result

Integrand size = 23, antiderivative size = 151 \[ \int \frac {1}{\sqrt [4]{a+b x^2} \left (c+d x^2\right )^{9/4}} \, dx=-\frac {2 d x \left (a+b x^2\right )^{3/4}}{5 c (b c-a d) \left (c+d x^2\right )^{5/4}}+\frac {(5 b c-3 a d) x \sqrt [4]{\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {3}{2},-\frac {(b c-a d) x^2}{a \left (c+d x^2\right )}\right )}{5 c^2 (b c-a d) \sqrt [4]{a+b x^2} \sqrt [4]{c+d x^2}} \] Output:

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

Mathematica [A] (warning: unable to verify)

Time = 4.49 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.90 \[ \int \frac {1}{\sqrt [4]{a+b x^2} \left (c+d x^2\right )^{9/4}} \, dx=\frac {x \left (5 c \left (3 c+2 d x^2\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{4},1,\frac {5}{2},\frac {(b c-a d) x^2}{c \left (a+b x^2\right )}\right )+\frac {(b c-a d) x^2 \left (c+d x^2\right ) \operatorname {Hypergeometric2F1}\left (\frac {5}{4},2,\frac {7}{2},\frac {(b c-a d) x^2}{c \left (a+b x^2\right )}\right )}{a+b x^2}\right )}{15 c^3 \sqrt [4]{a+b x^2} \left (c+d x^2\right )^{5/4}} \] Input:

Integrate[1/((a + b*x^2)^(1/4)*(c + d*x^2)^(9/4)),x]
 

Output:

(x*(5*c*(3*c + 2*d*x^2)*Hypergeometric2F1[1/4, 1, 5/2, ((b*c - a*d)*x^2)/( 
c*(a + b*x^2))] + ((b*c - a*d)*x^2*(c + d*x^2)*Hypergeometric2F1[5/4, 2, 7 
/2, ((b*c - a*d)*x^2)/(c*(a + b*x^2))])/(a + b*x^2)))/(15*c^3*(a + b*x^2)^ 
(1/4)*(c + d*x^2)^(5/4))
 

Rubi [A] (verified)

Time = 0.25 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {296, 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.

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

\(\Big \downarrow \) 296

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

\(\Big \downarrow \) 294

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

Input:

Int[1/((a + b*x^2)^(1/4)*(c + d*x^2)^(9/4)),x]
 

Output:

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

Defintions of rubi rules used

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]
 

rule 296

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

Maple [F]

\[\int \frac {1}{\left (b \,x^{2}+a \right )^{\frac {1}{4}} \left (x^{2} d +c \right )^{\frac {9}{4}}}d x\]

Input:

int(1/(b*x^2+a)^(1/4)/(d*x^2+c)^(9/4),x)
 

Output:

int(1/(b*x^2+a)^(1/4)/(d*x^2+c)^(9/4),x)
 

Fricas [F]

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

integrate(1/(b*x^2+a)^(1/4)/(d*x^2+c)^(9/4),x, algorithm="fricas")
 

Output:

integral((b*x^2 + a)^(3/4)*(d*x^2 + c)^(3/4)/(b*d^3*x^8 + (3*b*c*d^2 + a*d 
^3)*x^6 + 3*(b*c^2*d + a*c*d^2)*x^4 + a*c^3 + (b*c^3 + 3*a*c^2*d)*x^2), x)
 

Sympy [F]

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

integrate(1/(b*x**2+a)**(1/4)/(d*x**2+c)**(9/4),x)
 

Output:

Integral(1/((a + b*x**2)**(1/4)*(c + d*x**2)**(9/4)), x)
 

Maxima [F]

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

integrate(1/(b*x^2+a)^(1/4)/(d*x^2+c)^(9/4),x, algorithm="maxima")
 

Output:

integrate(1/((b*x^2 + a)^(1/4)*(d*x^2 + c)^(9/4)), x)
 

Giac [F]

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

integrate(1/(b*x^2+a)^(1/4)/(d*x^2+c)^(9/4),x, algorithm="giac")
 

Output:

integrate(1/((b*x^2 + a)^(1/4)*(d*x^2 + c)^(9/4)), x)
 

Mupad [F(-1)]

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

int(1/((a + b*x^2)^(1/4)*(c + d*x^2)^(9/4)),x)
 

Output:

int(1/((a + b*x^2)^(1/4)*(c + d*x^2)^(9/4)), x)
 

Reduce [F]

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

int(1/(b*x^2+a)^(1/4)/(d*x^2+c)^(9/4),x)
 

Output:

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