\(\int \frac {\sqrt [4]{a+b x^2}}{(c+d x^2)^{15/4}} \, dx\) [500]

Optimal result

Integrand size = 23, antiderivative size = 221 \[ \int \frac {\sqrt [4]{a+b x^2}}{\left (c+d x^2\right )^{15/4}} \, dx=-\frac {2 d x \left (a+b x^2\right )^{5/4}}{11 c (b c-a d) \left (c+d x^2\right )^{11/4}}-\frac {2 b (11 b c-5 a d) x \left (a+b x^2\right )^{5/4}}{55 a c (b c-a d)^2 \left (c+d x^2\right )^{7/4}}+\frac {\left (77 b^2 c^2-110 a b c d+45 a^2 d^2\right ) x \left (a+b x^2\right )^{5/4} \operatorname {Hypergeometric2F1}\left (-\frac {5}{4},\frac {1}{2},\frac {3}{2},-\frac {(b c-a d) x^2}{a \left (c+d x^2\right )}\right )}{55 a c^2 (b c-a d)^2 \left (\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}\right )^{5/4} \left (c+d x^2\right )^{7/4}} \] Output:

-2/11*d*x*(b*x^2+a)^(5/4)/c/(-a*d+b*c)/(d*x^2+c)^(11/4)-2/55*b*(-5*a*d+11* 
b*c)*x*(b*x^2+a)^(5/4)/a/c/(-a*d+b*c)^2/(d*x^2+c)^(7/4)+1/55*(45*a^2*d^2-1 
10*a*b*c*d+77*b^2*c^2)*x*(b*x^2+a)^(5/4)*hypergeom([-5/4, 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))^(5/4) 
/(d*x^2+c)^(7/4)
 

Mathematica [A] (warning: unable to verify)

Time = 5.08 (sec) , antiderivative size = 236, normalized size of antiderivative = 1.07 \[ \int \frac {\sqrt [4]{a+b x^2}}{\left (c+d x^2\right )^{15/4}} \, dx=\frac {x \left (7 c \left (a+b x^2\right ) \left (15 c^2+20 c d x^2+8 d^2 x^4\right ) \operatorname {Gamma}\left (-\frac {1}{4}\right ) \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},1,\frac {7}{2},\frac {(b c-a d) x^2}{c \left (a+b x^2\right )}\right )+8 (b c-a d) x^2 \left (4 c^2+7 c d x^2+3 d^2 x^4\right ) \operatorname {Gamma}\left (\frac {3}{4}\right ) \operatorname {Hypergeometric2F1}\left (\frac {3}{4},2,\frac {9}{2},\frac {(b c-a d) x^2}{c \left (a+b x^2\right )}\right )+8 (b c-a d) x^2 \left (c+d x^2\right )^2 \operatorname {Gamma}\left (\frac {3}{4}\right ) \, _3F_2\left (\frac {3}{4},2,2;1,\frac {9}{2};\frac {(b c-a d) x^2}{c \left (a+b x^2\right )}\right )\right )}{105 c^4 \left (a+b x^2\right )^{3/4} \left (c+d x^2\right )^{11/4} \operatorname {Gamma}\left (-\frac {1}{4}\right )} \] Input:

Integrate[(a + b*x^2)^(1/4)/(c + d*x^2)^(15/4),x]
 

Output:

(x*(7*c*(a + b*x^2)*(15*c^2 + 20*c*d*x^2 + 8*d^2*x^4)*Gamma[-1/4]*Hypergeo 
metric2F1[-1/4, 1, 7/2, ((b*c - a*d)*x^2)/(c*(a + b*x^2))] + 8*(b*c - a*d) 
*x^2*(4*c^2 + 7*c*d*x^2 + 3*d^2*x^4)*Gamma[3/4]*Hypergeometric2F1[3/4, 2, 
9/2, ((b*c - a*d)*x^2)/(c*(a + b*x^2))] + 8*(b*c - a*d)*x^2*(c + d*x^2)^2* 
Gamma[3/4]*HypergeometricPFQ[{3/4, 2, 2}, {1, 9/2}, ((b*c - a*d)*x^2)/(c*( 
a + b*x^2))]))/(105*c^4*(a + b*x^2)^(3/4)*(c + d*x^2)^(11/4)*Gamma[-1/4])
 

Rubi [A] (warning: unable to verify)

Time = 2.18 (sec) , antiderivative size = 248, normalized size of antiderivative = 1.12, 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.

\(\displaystyle \int \frac {\sqrt [4]{a+b x^2}}{\left (c+d x^2\right )^{15/4}} \, dx\)

\(\Big \downarrow \) 334

\(\displaystyle \frac {\sqrt [4]{a+b x^2} \int \frac {\sqrt [4]{\frac {b x^2}{a}+1}}{\left (d x^2+c\right )^{15/4}}dx}{\sqrt [4]{\frac {b x^2}{a}+1}}\)

\(\Big \downarrow \) 334

\(\displaystyle \frac {\sqrt [4]{a+b x^2} \left (\frac {d x^2}{c}+1\right )^{3/4} \int \frac {\sqrt [4]{\frac {b x^2}{a}+1}}{\left (\frac {d x^2}{c}+1\right )^{15/4}}dx}{c^3 \sqrt [4]{\frac {b x^2}{a}+1} \left (c+d x^2\right )^{3/4}}\)

\(\Big \downarrow \) 333

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

Input:

Int[(a + b*x^2)^(1/4)/(c + d*x^2)^(15/4),x]
 

Output:

(x*(7*c*(a + b*x^2)*(15*c^2 + 20*c*d*x^2 + 8*d^2*x^4)*Gamma[-1/4]*Hypergeo 
metric2F1[-1/4, 1, 7/2, ((b*c - a*d)*x^2)/(c*(a + b*x^2))] + 8*(b*c - a*d) 
*x^2*(4*c^2 + 7*c*d*x^2 + 3*d^2*x^4)*Gamma[3/4]*Hypergeometric2F1[3/4, 2, 
9/2, ((b*c - a*d)*x^2)/(c*(a + b*x^2))] + 8*(b*c - a*d)*x^2*(c + d*x^2)^2* 
Gamma[3/4]*HypergeometricPFQ[{3/4, 2, 2}, {1, 9/2}, ((b*c - a*d)*x^2)/(c*( 
a + b*x^2))]))/(105*c^6*(a + b*x^2)^(3/4)*(c + d*x^2)^(3/4)*(1 + (d*x^2)/c 
)^2*Gamma[-1/4])
 

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]

\[\int \frac {\left (b \,x^{2}+a \right )^{\frac {1}{4}}}{\left (x^{2} d +c \right )^{\frac {15}{4}}}d x\]

Input:

int((b*x^2+a)^(1/4)/(d*x^2+c)^(15/4),x)
 

Output:

int((b*x^2+a)^(1/4)/(d*x^2+c)^(15/4),x)
 

Fricas [F]

\[ \int \frac {\sqrt [4]{a+b x^2}}{\left (c+d x^2\right )^{15/4}} \, dx=\int { \frac {{\left (b x^{2} + a\right )}^{\frac {1}{4}}}{{\left (d x^{2} + c\right )}^{\frac {15}{4}}} \,d x } \] Input:

integrate((b*x^2+a)^(1/4)/(d*x^2+c)^(15/4),x, algorithm="fricas")
 

Output:

integral((b*x^2 + a)^(1/4)*(d*x^2 + c)^(1/4)/(d^4*x^8 + 4*c*d^3*x^6 + 6*c^ 
2*d^2*x^4 + 4*c^3*d*x^2 + c^4), x)
 

Sympy [F]

\[ \int \frac {\sqrt [4]{a+b x^2}}{\left (c+d x^2\right )^{15/4}} \, dx=\int \frac {\sqrt [4]{a + b x^{2}}}{\left (c + d x^{2}\right )^{\frac {15}{4}}}\, dx \] Input:

integrate((b*x**2+a)**(1/4)/(d*x**2+c)**(15/4),x)
 

Output:

Integral((a + b*x**2)**(1/4)/(c + d*x**2)**(15/4), x)
 

Maxima [F]

\[ \int \frac {\sqrt [4]{a+b x^2}}{\left (c+d x^2\right )^{15/4}} \, dx=\int { \frac {{\left (b x^{2} + a\right )}^{\frac {1}{4}}}{{\left (d x^{2} + c\right )}^{\frac {15}{4}}} \,d x } \] Input:

integrate((b*x^2+a)^(1/4)/(d*x^2+c)^(15/4),x, algorithm="maxima")
 

Output:

integrate((b*x^2 + a)^(1/4)/(d*x^2 + c)^(15/4), x)
 

Giac [F]

\[ \int \frac {\sqrt [4]{a+b x^2}}{\left (c+d x^2\right )^{15/4}} \, dx=\int { \frac {{\left (b x^{2} + a\right )}^{\frac {1}{4}}}{{\left (d x^{2} + c\right )}^{\frac {15}{4}}} \,d x } \] Input:

integrate((b*x^2+a)^(1/4)/(d*x^2+c)^(15/4),x, algorithm="giac")
 

Output:

integrate((b*x^2 + a)^(1/4)/(d*x^2 + c)^(15/4), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt [4]{a+b x^2}}{\left (c+d x^2\right )^{15/4}} \, dx=\int \frac {{\left (b\,x^2+a\right )}^{1/4}}{{\left (d\,x^2+c\right )}^{15/4}} \,d x \] Input:

int((a + b*x^2)^(1/4)/(c + d*x^2)^(15/4),x)
 

Output:

int((a + b*x^2)^(1/4)/(c + d*x^2)^(15/4), x)
 

Reduce [F]

\[ \int \frac {\sqrt [4]{a+b x^2}}{\left (c+d x^2\right )^{15/4}} \, dx=\int \frac {\left (b \,x^{2}+a \right )^{\frac {1}{4}}}{\left (d \,x^{2}+c \right )^{\frac {3}{4}} c^{3}+3 \left (d \,x^{2}+c \right )^{\frac {3}{4}} c^{2} d \,x^{2}+3 \left (d \,x^{2}+c \right )^{\frac {3}{4}} c \,d^{2} x^{4}+\left (d \,x^{2}+c \right )^{\frac {3}{4}} d^{3} x^{6}}d x \] Input:

int((b*x^2+a)^(1/4)/(d*x^2+c)^(15/4),x)
 

Output:

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