\(\int \frac {1}{\sqrt [4]{a+b x^2} (c+d x^2)^{13/4}} \, dx\) [515]

Optimal result

Integrand size = 23, antiderivative size = 221 \[ \int \frac {1}{\sqrt [4]{a+b x^2} \left (c+d x^2\right )^{13/4}} \, dx=-\frac {2 d x \left (a+b x^2\right )^{3/4}}{9 c (b c-a d) \left (c+d x^2\right )^{9/4}}-\frac {2 b (3 b c-a d) x \left (a+b x^2\right )^{3/4}}{9 a c (b c-a d)^2 \left (c+d x^2\right )^{5/4}}+\frac {\left (15 b^2 c^2-18 a b c d+7 a^2 d^2\right ) x \left (a+b x^2\right )^{3/4} \operatorname {Hypergeometric2F1}\left (-\frac {3}{4},\frac {1}{2},\frac {3}{2},-\frac {(b c-a d) x^2}{a \left (c+d x^2\right )}\right )}{9 a c^2 (b c-a d)^2 \left (\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}\right )^{3/4} \left (c+d x^2\right )^{5/4}} \] Output:

-2/9*d*x*(b*x^2+a)^(3/4)/c/(-a*d+b*c)/(d*x^2+c)^(9/4)-2/9*b*(-a*d+3*b*c)*x 
*(b*x^2+a)^(3/4)/a/c/(-a*d+b*c)^2/(d*x^2+c)^(5/4)+1/9*(7*a^2*d^2-18*a*b*c* 
d+15*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/c^2/(-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 = 5.80 (sec) , antiderivative size = 218, normalized size of antiderivative = 0.99 \[ \int \frac {1}{\sqrt [4]{a+b x^2} \left (c+d x^2\right )^{13/4}} \, dx=\frac {x \left (7 c \left (a+b x^2\right ) \left (15 c^2+20 c d x^2+8 d^2 x^4\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{4},1,\frac {7}{2},\frac {(b c-a d) x^2}{c \left (a+b x^2\right )}\right )+2 (b c-a d) x^2 \left (4 c^2+7 c d x^2+3 d^2 x^4\right ) \operatorname {Hypergeometric2F1}\left (\frac {5}{4},2,\frac {9}{2},\frac {(b c-a d) x^2}{c \left (a+b x^2\right )}\right )+2 (b c-a d) x^2 \left (c+d x^2\right )^2 \, _3F_2\left (\frac {5}{4},2,2;1,\frac {9}{2};\frac {(b c-a d) x^2}{c \left (a+b x^2\right )}\right )\right )}{105 c^4 \left (a+b x^2\right )^{5/4} \left (c+d x^2\right )^{9/4}} \] Input:

Integrate[1/((a + b*x^2)^(1/4)*(c + d*x^2)^(13/4)),x]
 

Output:

(x*(7*c*(a + b*x^2)*(15*c^2 + 20*c*d*x^2 + 8*d^2*x^4)*Hypergeometric2F1[1/ 
4, 1, 7/2, ((b*c - a*d)*x^2)/(c*(a + b*x^2))] + 2*(b*c - a*d)*x^2*(4*c^2 + 
 7*c*d*x^2 + 3*d^2*x^4)*Hypergeometric2F1[5/4, 2, 9/2, ((b*c - a*d)*x^2)/( 
c*(a + b*x^2))] + 2*(b*c - a*d)*x^2*(c + d*x^2)^2*HypergeometricPFQ[{5/4, 
2, 2}, {1, 9/2}, ((b*c - a*d)*x^2)/(c*(a + b*x^2))]))/(105*c^4*(a + b*x^2) 
^(5/4)*(c + d*x^2)^(9/4))
 

Rubi [A] (warning: unable to verify)

Time = 2.20 (sec) , antiderivative size = 230, normalized size of antiderivative = 1.04, 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.

\(\displaystyle \int \frac {1}{\sqrt [4]{a+b x^2} \left (c+d x^2\right )^{13/4}} \, dx\)

\(\Big \downarrow \) 334

\(\displaystyle \frac {\sqrt [4]{\frac {b x^2}{a}+1} \int \frac {1}{\sqrt [4]{\frac {b x^2}{a}+1} \left (d x^2+c\right )^{13/4}}dx}{\sqrt [4]{a+b x^2}}\)

\(\Big \downarrow \) 334

\(\displaystyle \frac {\sqrt [4]{\frac {b x^2}{a}+1} \sqrt [4]{\frac {d x^2}{c}+1} \int \frac {1}{\sqrt [4]{\frac {b x^2}{a}+1} \left (\frac {d x^2}{c}+1\right )^{13/4}}dx}{c^3 \sqrt [4]{a+b x^2} \sqrt [4]{c+d x^2}}\)

\(\Big \downarrow \) 333

\(\displaystyle \frac {x \left (2 x^2 \left (c+d x^2\right )^2 (b c-a d) \, _3F_2\left (\frac {5}{4},2,2;1,\frac {9}{2};\frac {(b c-a d) x^2}{c \left (b x^2+a\right )}\right )+2 x^2 \left (4 c^2+7 c d x^2+3 d^2 x^4\right ) (b c-a d) \operatorname {Hypergeometric2F1}\left (\frac {5}{4},2,\frac {9}{2},\frac {(b c-a d) x^2}{c \left (b x^2+a\right )}\right )+7 c \left (a+b x^2\right ) \left (15 c^2+20 c d x^2+8 d^2 x^4\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{4},1,\frac {7}{2},\frac {(b c-a d) x^2}{c \left (b x^2+a\right )}\right )\right )}{105 c^6 \left (a+b x^2\right )^{5/4} \sqrt [4]{c+d x^2} \left (\frac {d x^2}{c}+1\right )^2}\)

Input:

Int[1/((a + b*x^2)^(1/4)*(c + d*x^2)^(13/4)),x]
 

Output:

(x*(7*c*(a + b*x^2)*(15*c^2 + 20*c*d*x^2 + 8*d^2*x^4)*Hypergeometric2F1[1/ 
4, 1, 7/2, ((b*c - a*d)*x^2)/(c*(a + b*x^2))] + 2*(b*c - a*d)*x^2*(4*c^2 + 
 7*c*d*x^2 + 3*d^2*x^4)*Hypergeometric2F1[5/4, 2, 9/2, ((b*c - a*d)*x^2)/( 
c*(a + b*x^2))] + 2*(b*c - a*d)*x^2*(c + d*x^2)^2*HypergeometricPFQ[{5/4, 
2, 2}, {1, 9/2}, ((b*c - a*d)*x^2)/(c*(a + b*x^2))]))/(105*c^6*(a + b*x^2) 
^(5/4)*(c + d*x^2)^(1/4)*(1 + (d*x^2)/c)^2)
 

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]

\[\int \frac {1}{\left (b \,x^{2}+a \right )^{\frac {1}{4}} \left (x^{2} d +c \right )^{\frac {13}{4}}}d x\]

Input:

int(1/(b*x^2+a)^(1/4)/(d*x^2+c)^(13/4),x)
 

Output:

int(1/(b*x^2+a)^(1/4)/(d*x^2+c)^(13/4),x)
 

Fricas [F]

\[ \int \frac {1}{\sqrt [4]{a+b x^2} \left (c+d x^2\right )^{13/4}} \, dx=\int { \frac {1}{{\left (b x^{2} + a\right )}^{\frac {1}{4}} {\left (d x^{2} + c\right )}^{\frac {13}{4}}} \,d x } \] Input:

integrate(1/(b*x^2+a)^(1/4)/(d*x^2+c)^(13/4),x, algorithm="fricas")
 

Output:

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

Sympy [F]

\[ \int \frac {1}{\sqrt [4]{a+b x^2} \left (c+d x^2\right )^{13/4}} \, dx=\int \frac {1}{\sqrt [4]{a + b x^{2}} \left (c + d x^{2}\right )^{\frac {13}{4}}}\, dx \] Input:

integrate(1/(b*x**2+a)**(1/4)/(d*x**2+c)**(13/4),x)
 

Output:

Integral(1/((a + b*x**2)**(1/4)*(c + d*x**2)**(13/4)), x)
 

Maxima [F]

\[ \int \frac {1}{\sqrt [4]{a+b x^2} \left (c+d x^2\right )^{13/4}} \, dx=\int { \frac {1}{{\left (b x^{2} + a\right )}^{\frac {1}{4}} {\left (d x^{2} + c\right )}^{\frac {13}{4}}} \,d x } \] Input:

integrate(1/(b*x^2+a)^(1/4)/(d*x^2+c)^(13/4),x, algorithm="maxima")
 

Output:

integrate(1/((b*x^2 + a)^(1/4)*(d*x^2 + c)^(13/4)), x)
 

Giac [F]

\[ \int \frac {1}{\sqrt [4]{a+b x^2} \left (c+d x^2\right )^{13/4}} \, dx=\int { \frac {1}{{\left (b x^{2} + a\right )}^{\frac {1}{4}} {\left (d x^{2} + c\right )}^{\frac {13}{4}}} \,d x } \] Input:

integrate(1/(b*x^2+a)^(1/4)/(d*x^2+c)^(13/4),x, algorithm="giac")
 

Output:

integrate(1/((b*x^2 + a)^(1/4)*(d*x^2 + c)^(13/4)), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\sqrt [4]{a+b x^2} \left (c+d x^2\right )^{13/4}} \, dx=\int \frac {1}{{\left (b\,x^2+a\right )}^{1/4}\,{\left (d\,x^2+c\right )}^{13/4}} \,d x \] Input:

int(1/((a + b*x^2)^(1/4)*(c + d*x^2)^(13/4)),x)
 

Output:

int(1/((a + b*x^2)^(1/4)*(c + d*x^2)^(13/4)), x)
 

Reduce [F]

\[ \int \frac {1}{\sqrt [4]{a+b x^2} \left (c+d x^2\right )^{13/4}} \, dx=\int \frac {1}{\left (d \,x^{2}+c \right )^{\frac {1}{4}} \left (b \,x^{2}+a \right )^{\frac {1}{4}} c^{3}+3 \left (d \,x^{2}+c \right )^{\frac {1}{4}} \left (b \,x^{2}+a \right )^{\frac {1}{4}} c^{2} d \,x^{2}+3 \left (d \,x^{2}+c \right )^{\frac {1}{4}} \left (b \,x^{2}+a \right )^{\frac {1}{4}} c \,d^{2} x^{4}+\left (d \,x^{2}+c \right )^{\frac {1}{4}} \left (b \,x^{2}+a \right )^{\frac {1}{4}} d^{3} x^{6}}d x \] Input:

int(1/(b*x^2+a)^(1/4)/(d*x^2+c)^(13/4),x)
 

Output:

int(1/((c + d*x**2)**(1/4)*(a + b*x**2)**(1/4)*c**3 + 3*(c + d*x**2)**(1/4 
)*(a + b*x**2)**(1/4)*c**2*d*x**2 + 3*(c + d*x**2)**(1/4)*(a + b*x**2)**(1 
/4)*c*d**2*x**4 + (c + d*x**2)**(1/4)*(a + b*x**2)**(1/4)*d**3*x**6),x)