\(\int \frac {1}{(a+b x^2)^{5/4} (c+d x^2)^{9/4}} \, dx\) [531]

Optimal result

Integrand size = 23, antiderivative size = 215 \[ \int \frac {1}{\left (a+b x^2\right )^{5/4} \left (c+d x^2\right )^{9/4}} \, dx=\frac {2 b x}{a (b c-a d) \sqrt [4]{a+b x^2} \left (c+d x^2\right )^{5/4}}+\frac {2 b (b c+5 a d) x \left (a+b x^2\right )^{3/4}}{3 a^2 (b c-a d)^2 \left (c+d x^2\right )^{5/4}}-\frac {\left (5 b^2 c^2+10 a b c d-3 a^2 d^2\right ) x \left (a+b x^2\right )^{3/4} \operatorname {Hypergeometric2F1}\left (-\frac {3}{4},\frac {1}{2},\frac {3}{2},-\frac {(b c-a d) x^2}{a \left (c+d x^2\right )}\right )}{3 a^2 c (b c-a d)^2 \left (\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}\right )^{3/4} \left (c+d x^2\right )^{5/4}} \] Output:

2*b*x/a/(-a*d+b*c)/(b*x^2+a)^(1/4)/(d*x^2+c)^(5/4)+2/3*b*(5*a*d+b*c)*x*(b* 
x^2+a)^(3/4)/a^2/(-a*d+b*c)^2/(d*x^2+c)^(5/4)-1/3*(-3*a^2*d^2+10*a*b*c*d+5 
*b^2*c^2)*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))/a^2/c/(-a*d+b*c)^2/(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 = 8.23 (sec) , antiderivative size = 228, normalized size of antiderivative = 1.06 \[ \int \frac {1}{\left (a+b x^2\right )^{5/4} \left (c+d x^2\right )^{9/4}} \, dx=\frac {x \operatorname {Gamma}\left (\frac {1}{4}\right ) \left (7 c \left (a+b x^2\right ) \left (15 c^2+20 c d x^2+8 d^2 x^4\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {5}{4},\frac {7}{2},\frac {(b c-a d) x^2}{c \left (a+b x^2\right )}\right )+10 (b c-a d) x^2 \left (4 c^2+7 c d x^2+3 d^2 x^4\right ) \operatorname {Hypergeometric2F1}\left (2,\frac {9}{4},\frac {9}{2},\frac {(b c-a d) x^2}{c \left (a+b x^2\right )}\right )+10 (b c-a d) x^2 \left (c+d x^2\right )^2 \, _3F_2\left (2,2,\frac {9}{4};1,\frac {9}{2};\frac {(b c-a d) x^2}{c \left (a+b x^2\right )}\right )\right )}{420 c^4 \left (a+b x^2\right )^{9/4} \left (c+d x^2\right )^{5/4} \operatorname {Gamma}\left (\frac {5}{4}\right )} \] Input:

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

Output:

(x*Gamma[1/4]*(7*c*(a + b*x^2)*(15*c^2 + 20*c*d*x^2 + 8*d^2*x^4)*Hypergeom 
etric2F1[1, 5/4, 7/2, ((b*c - a*d)*x^2)/(c*(a + b*x^2))] + 10*(b*c - a*d)* 
x^2*(4*c^2 + 7*c*d*x^2 + 3*d^2*x^4)*Hypergeometric2F1[2, 9/4, 9/2, ((b*c - 
 a*d)*x^2)/(c*(a + b*x^2))] + 10*(b*c - a*d)*x^2*(c + d*x^2)^2*Hypergeomet 
ricPFQ[{2, 2, 9/4}, {1, 9/2}, ((b*c - a*d)*x^2)/(c*(a + b*x^2))]))/(420*c^ 
4*(a + b*x^2)^(9/4)*(c + d*x^2)^(5/4)*Gamma[5/4])
 

Rubi [A] (warning: unable to verify)

Time = 2.93 (sec) , antiderivative size = 255, normalized size of antiderivative = 1.19, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {334, 334, 333}

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.

\(\displaystyle \int \frac {1}{\left (a+b x^2\right )^{5/4} \left (c+d x^2\right )^{9/4}} \, dx\)

\(\Big \downarrow \) 334

\(\displaystyle \frac {\sqrt [4]{\frac {b x^2}{a}+1} \int \frac {1}{\left (\frac {b x^2}{a}+1\right )^{5/4} \left (d x^2+c\right )^{9/4}}dx}{a \sqrt [4]{a+b x^2}}\)

\(\Big \downarrow \) 334

\(\displaystyle \frac {\sqrt [4]{\frac {b x^2}{a}+1} \sqrt [4]{\frac {d x^2}{c}+1} \int \frac {1}{\left (\frac {b x^2}{a}+1\right )^{5/4} \left (\frac {d x^2}{c}+1\right )^{9/4}}dx}{a c^2 \sqrt [4]{a+b x^2} \sqrt [4]{c+d x^2}}\)

\(\Big \downarrow \) 333

\(\displaystyle \frac {x \operatorname {Gamma}\left (\frac {1}{4}\right ) \left (10 x^2 \left (c+d x^2\right )^2 (b c-a d) \, _3F_2\left (2,2,\frac {9}{4};1,\frac {9}{2};\frac {(b c-a d) x^2}{c \left (b x^2+a\right )}\right )+10 x^2 \left (4 c^2+7 c d x^2+3 d^2 x^4\right ) (b c-a d) \operatorname {Hypergeometric2F1}\left (2,\frac {9}{4},\frac {9}{2},\frac {(b c-a d) x^2}{c \left (b x^2+a\right )}\right )+7 c \left (a+b x^2\right ) \left (15 c^2+20 c d x^2+8 d^2 x^4\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {5}{4},\frac {7}{2},\frac {(b c-a d) x^2}{c \left (b x^2+a\right )}\right )\right )}{420 a c^5 \operatorname {Gamma}\left (\frac {5}{4}\right ) \left (a+b x^2\right )^{5/4} \left (\frac {b x^2}{a}+1\right ) \sqrt [4]{c+d x^2} \left (\frac {d x^2}{c}+1\right )}\)

Input:

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

Output:

(x*Gamma[1/4]*(7*c*(a + b*x^2)*(15*c^2 + 20*c*d*x^2 + 8*d^2*x^4)*Hypergeom 
etric2F1[1, 5/4, 7/2, ((b*c - a*d)*x^2)/(c*(a + b*x^2))] + 10*(b*c - a*d)* 
x^2*(4*c^2 + 7*c*d*x^2 + 3*d^2*x^4)*Hypergeometric2F1[2, 9/4, 9/2, ((b*c - 
 a*d)*x^2)/(c*(a + b*x^2))] + 10*(b*c - a*d)*x^2*(c + d*x^2)^2*Hypergeomet 
ricPFQ[{2, 2, 9/4}, {1, 9/2}, ((b*c - a*d)*x^2)/(c*(a + b*x^2))]))/(420*a* 
c^5*(a + b*x^2)^(5/4)*(1 + (b*x^2)/a)*(c + d*x^2)^(1/4)*(1 + (d*x^2)/c)*Ga 
mma[5/4])
 

Defintions of rubi rules used

rule 333

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim 
p[a^p*c^q*x*AppellF1[1/2, -p, -q, 3/2, (-b)*(x^2/a), (-d)*(x^2/c)], x] /; F 
reeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && (IntegerQ[p] || GtQ[a, 
0]) && (IntegerQ[q] || GtQ[c, 0])
 

rule 334

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

Maple [F]

\[\int \frac {1}{\left (b \,x^{2}+a \right )^{\frac {5}{4}} \left (x^{2} d +c \right )^{\frac {9}{4}}}d x\]

Input:

int(1/(b*x^2+a)^(5/4)/(d*x^2+c)^(9/4),x)
 

Output:

int(1/(b*x^2+a)^(5/4)/(d*x^2+c)^(9/4),x)
 

Fricas [F]

\[ \int \frac {1}{\left (a+b x^2\right )^{5/4} \left (c+d x^2\right )^{9/4}} \, dx=\int { \frac {1}{{\left (b x^{2} + a\right )}^{\frac {5}{4}} {\left (d x^{2} + c\right )}^{\frac {9}{4}}} \,d x } \] Input:

integrate(1/(b*x^2+a)^(5/4)/(d*x^2+c)^(9/4),x, algorithm="fricas")
 

Output:

integral((b*x^2 + a)^(3/4)*(d*x^2 + c)^(3/4)/(b^2*d^3*x^10 + (3*b^2*c*d^2 
+ 2*a*b*d^3)*x^8 + (3*b^2*c^2*d + 6*a*b*c*d^2 + a^2*d^3)*x^6 + a^2*c^3 + ( 
b^2*c^3 + 6*a*b*c^2*d + 3*a^2*c*d^2)*x^4 + (2*a*b*c^3 + 3*a^2*c^2*d)*x^2), 
 x)
 

Sympy [F]

\[ \int \frac {1}{\left (a+b x^2\right )^{5/4} \left (c+d x^2\right )^{9/4}} \, dx=\int \frac {1}{\left (a + b x^{2}\right )^{\frac {5}{4}} \left (c + d x^{2}\right )^{\frac {9}{4}}}\, dx \] Input:

integrate(1/(b*x**2+a)**(5/4)/(d*x**2+c)**(9/4),x)
 

Output:

Integral(1/((a + b*x**2)**(5/4)*(c + d*x**2)**(9/4)), x)
 

Maxima [F]

\[ \int \frac {1}{\left (a+b x^2\right )^{5/4} \left (c+d x^2\right )^{9/4}} \, dx=\int { \frac {1}{{\left (b x^{2} + a\right )}^{\frac {5}{4}} {\left (d x^{2} + c\right )}^{\frac {9}{4}}} \,d x } \] Input:

integrate(1/(b*x^2+a)^(5/4)/(d*x^2+c)^(9/4),x, algorithm="maxima")
 

Output:

integrate(1/((b*x^2 + a)^(5/4)*(d*x^2 + c)^(9/4)), x)
 

Giac [F]

\[ \int \frac {1}{\left (a+b x^2\right )^{5/4} \left (c+d x^2\right )^{9/4}} \, dx=\int { \frac {1}{{\left (b x^{2} + a\right )}^{\frac {5}{4}} {\left (d x^{2} + c\right )}^{\frac {9}{4}}} \,d x } \] Input:

integrate(1/(b*x^2+a)^(5/4)/(d*x^2+c)^(9/4),x, algorithm="giac")
 

Output:

integrate(1/((b*x^2 + a)^(5/4)*(d*x^2 + c)^(9/4)), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\left (a+b x^2\right )^{5/4} \left (c+d x^2\right )^{9/4}} \, dx=\int \frac {1}{{\left (b\,x^2+a\right )}^{5/4}\,{\left (d\,x^2+c\right )}^{9/4}} \,d x \] Input:

int(1/((a + b*x^2)^(5/4)*(c + d*x^2)^(9/4)),x)
 

Output:

int(1/((a + b*x^2)^(5/4)*(c + d*x^2)^(9/4)), x)
 

Reduce [F]

\[ \int \frac {1}{\left (a+b x^2\right )^{5/4} \left (c+d x^2\right )^{9/4}} \, dx=\int \frac {1}{\left (d \,x^{2}+c \right )^{\frac {1}{4}} \left (b \,x^{2}+a \right )^{\frac {1}{4}} a \,c^{2}+2 \left (d \,x^{2}+c \right )^{\frac {1}{4}} \left (b \,x^{2}+a \right )^{\frac {1}{4}} a c d \,x^{2}+\left (d \,x^{2}+c \right )^{\frac {1}{4}} \left (b \,x^{2}+a \right )^{\frac {1}{4}} a \,d^{2} x^{4}+\left (d \,x^{2}+c \right )^{\frac {1}{4}} \left (b \,x^{2}+a \right )^{\frac {1}{4}} b \,c^{2} x^{2}+2 \left (d \,x^{2}+c \right )^{\frac {1}{4}} \left (b \,x^{2}+a \right )^{\frac {1}{4}} b c d \,x^{4}+\left (d \,x^{2}+c \right )^{\frac {1}{4}} \left (b \,x^{2}+a \right )^{\frac {1}{4}} b \,d^{2} x^{6}}d x \] Input:

int(1/(b*x^2+a)^(5/4)/(d*x^2+c)^(9/4),x)
 

Output:

int(1/((c + d*x**2)**(1/4)*(a + b*x**2)**(1/4)*a*c**2 + 2*(c + d*x**2)**(1 
/4)*(a + b*x**2)**(1/4)*a*c*d*x**2 + (c + d*x**2)**(1/4)*(a + b*x**2)**(1/ 
4)*a*d**2*x**4 + (c + d*x**2)**(1/4)*(a + b*x**2)**(1/4)*b*c**2*x**2 + 2*( 
c + d*x**2)**(1/4)*(a + b*x**2)**(1/4)*b*c*d*x**4 + (c + d*x**2)**(1/4)*(a 
 + b*x**2)**(1/4)*b*d**2*x**6),x)