\(\int \frac {\sqrt [4]{a+b x^2}}{(c+d x^2)^{7/4}} \, dx\) [498]

Optimal result

Integrand size = 23, antiderivative size = 87 \[ \int \frac {\sqrt [4]{a+b x^2}}{\left (c+d x^2\right )^{7/4}} \, dx=\frac {x \sqrt [4]{a+b x^2} \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},\frac {1}{2},\frac {3}{2},-\frac {(b c-a d) x^2}{a \left (c+d x^2\right )}\right )}{c \sqrt [4]{\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}} \left (c+d x^2\right )^{3/4}} \] Output:

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

Mathematica [A] (warning: unable to verify)

Time = 3.09 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.02 \[ \int \frac {\sqrt [4]{a+b x^2}}{\left (c+d x^2\right )^{7/4}} \, dx=\frac {x \sqrt [4]{a+b x^2} \sqrt [4]{1+\frac {d x^2}{c}} \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},\frac {1}{2},\frac {3}{2},\frac {(-b c+a d) x^2}{a \left (c+d x^2\right )}\right )}{c \sqrt [4]{1+\frac {b x^2}{a}} \left (c+d x^2\right )^{3/4}} \] Input:

Integrate[(a + b*x^2)^(1/4)/(c + d*x^2)^(7/4),x]
 

Output:

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

Rubi [A] (verified)

Time = 0.21 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.40, 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.

\(\displaystyle \int \frac {\sqrt [4]{a+b x^2}}{\left (c+d x^2\right )^{7/4}} \, dx\)

\(\Big \downarrow \) 292

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

\(\Big \downarrow \) 294

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

Input:

Int[(a + b*x^2)^(1/4)/(c + d*x^2)^(7/4),x]
 

Output:

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

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]

\[\int \frac {\left (b \,x^{2}+a \right )^{\frac {1}{4}}}{\left (x^{2} d +c \right )^{\frac {7}{4}}}d x\]

Input:

int((b*x^2+a)^(1/4)/(d*x^2+c)^(7/4),x)
 

Output:

int((b*x^2+a)^(1/4)/(d*x^2+c)^(7/4),x)
 

Fricas [F]

\[ \int \frac {\sqrt [4]{a+b x^2}}{\left (c+d x^2\right )^{7/4}} \, dx=\int { \frac {{\left (b x^{2} + a\right )}^{\frac {1}{4}}}{{\left (d x^{2} + c\right )}^{\frac {7}{4}}} \,d x } \] Input:

integrate((b*x^2+a)^(1/4)/(d*x^2+c)^(7/4),x, algorithm="fricas")
 

Output:

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

Sympy [F]

\[ \int \frac {\sqrt [4]{a+b x^2}}{\left (c+d x^2\right )^{7/4}} \, dx=\int \frac {\sqrt [4]{a + b x^{2}}}{\left (c + d x^{2}\right )^{\frac {7}{4}}}\, dx \] Input:

integrate((b*x**2+a)**(1/4)/(d*x**2+c)**(7/4),x)
 

Output:

Integral((a + b*x**2)**(1/4)/(c + d*x**2)**(7/4), x)
 

Maxima [F]

\[ \int \frac {\sqrt [4]{a+b x^2}}{\left (c+d x^2\right )^{7/4}} \, dx=\int { \frac {{\left (b x^{2} + a\right )}^{\frac {1}{4}}}{{\left (d x^{2} + c\right )}^{\frac {7}{4}}} \,d x } \] Input:

integrate((b*x^2+a)^(1/4)/(d*x^2+c)^(7/4),x, algorithm="maxima")
 

Output:

integrate((b*x^2 + a)^(1/4)/(d*x^2 + c)^(7/4), x)
 

Giac [F]

\[ \int \frac {\sqrt [4]{a+b x^2}}{\left (c+d x^2\right )^{7/4}} \, dx=\int { \frac {{\left (b x^{2} + a\right )}^{\frac {1}{4}}}{{\left (d x^{2} + c\right )}^{\frac {7}{4}}} \,d x } \] Input:

integrate((b*x^2+a)^(1/4)/(d*x^2+c)^(7/4),x, algorithm="giac")
 

Output:

integrate((b*x^2 + a)^(1/4)/(d*x^2 + c)^(7/4), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt [4]{a+b x^2}}{\left (c+d x^2\right )^{7/4}} \, dx=\int \frac {{\left (b\,x^2+a\right )}^{1/4}}{{\left (d\,x^2+c\right )}^{7/4}} \,d x \] Input:

int((a + b*x^2)^(1/4)/(c + d*x^2)^(7/4),x)
 

Output:

int((a + b*x^2)^(1/4)/(c + d*x^2)^(7/4), x)
 

Reduce [F]

\[ \int \frac {\sqrt [4]{a+b x^2}}{\left (c+d x^2\right )^{7/4}} \, dx=\int \frac {\left (b \,x^{2}+a \right )^{\frac {1}{4}}}{\left (d \,x^{2}+c \right )^{\frac {3}{4}} c +\left (d \,x^{2}+c \right )^{\frac {3}{4}} d \,x^{2}}d x \] Input:

int((b*x^2+a)^(1/4)/(d*x^2+c)^(7/4),x)
 

Output:

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