\(\int \frac {(a+b x^2)^{3/4}}{(c+d x^2)^{13/4}} \, dx\) [506]

Optimal result

Integrand size = 23, antiderivative size = 151 \[ \int \frac {\left (a+b x^2\right )^{3/4}}{\left (c+d x^2\right )^{13/4}} \, dx=-\frac {2 d x \left (a+b x^2\right )^{7/4}}{9 c (b c-a d) \left (c+d x^2\right )^{9/4}}+\frac {(9 b c-7 a d) x \left (a+b x^2\right )^{3/4} \operatorname {Hypergeometric2F1}\left (-\frac {3}{4},\frac {1}{2},\frac {3}{2},-\frac {(b c-a d) x^2}{a \left (c+d x^2\right )}\right )}{9 c^2 (b c-a d) \left (\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}\right )^{3/4} \left (c+d x^2\right )^{5/4}} \] Output:

-2/9*d*x*(b*x^2+a)^(7/4)/c/(-a*d+b*c)/(d*x^2+c)^(9/4)+1/9*(-7*a*d+9*b*c)*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))/ 
c^2/(-a*d+b*c)/(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 = 6.12 (sec) , antiderivative size = 147, normalized size of antiderivative = 0.97 \[ \int \frac {\left (a+b x^2\right )^{3/4}}{\left (c+d x^2\right )^{13/4}} \, dx=\frac {x \left (a+b x^2\right )^{3/4} \left (5 c \left (3 c+2 d x^2\right ) \operatorname {Hypergeometric2F1}\left (-\frac {3}{4},1,\frac {5}{2},\frac {(b c-a d) x^2}{c \left (a+b x^2\right )}\right )+\frac {4 (b c-a d) x^2 \left (c+d x^2\right ) \operatorname {Gamma}\left (\frac {1}{4}\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{4},2,\frac {7}{2},\frac {(b c-a d) x^2}{c \left (a+b x^2\right )}\right )}{\left (a+b x^2\right ) \operatorname {Gamma}\left (-\frac {3}{4}\right )}\right )}{15 c^3 \left (c+d x^2\right )^{9/4}} \] Input:

Integrate[(a + b*x^2)^(3/4)/(c + d*x^2)^(13/4),x]
 

Output:

(x*(a + b*x^2)^(3/4)*(5*c*(3*c + 2*d*x^2)*Hypergeometric2F1[-3/4, 1, 5/2, 
((b*c - a*d)*x^2)/(c*(a + b*x^2))] + (4*(b*c - a*d)*x^2*(c + d*x^2)*Gamma[ 
1/4]*Hypergeometric2F1[1/4, 2, 7/2, ((b*c - a*d)*x^2)/(c*(a + b*x^2))])/(( 
a + b*x^2)*Gamma[-3/4])))/(15*c^3*(c + d*x^2)^(9/4))
 

Rubi [A] (verified)

Time = 0.27 (sec) , antiderivative size = 190, normalized size of antiderivative = 1.26, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {296, 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.

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

\(\Big \downarrow \) 296

\(\displaystyle \frac {(9 b c-7 a d) \int \frac {\left (b x^2+a\right )^{3/4}}{\left (d x^2+c\right )^{9/4}}dx}{9 c (b c-a d)}-\frac {2 d x \left (a+b x^2\right )^{7/4}}{9 c \left (c+d x^2\right )^{9/4} (b c-a d)}\)

\(\Big \downarrow \) 292

\(\displaystyle \frac {(9 b c-7 a d) \left (\frac {3 a \int \frac {1}{\sqrt [4]{b x^2+a} \left (d x^2+c\right )^{5/4}}dx}{5 c}+\frac {2 x \left (a+b x^2\right )^{3/4}}{5 c \left (c+d x^2\right )^{5/4}}\right )}{9 c (b c-a d)}-\frac {2 d x \left (a+b x^2\right )^{7/4}}{9 c \left (c+d x^2\right )^{9/4} (b c-a d)}\)

\(\Big \downarrow \) 294

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

Input:

Int[(a + b*x^2)^(3/4)/(c + d*x^2)^(13/4),x]
 

Output:

(-2*d*x*(a + b*x^2)^(7/4))/(9*c*(b*c - a*d)*(c + d*x^2)^(9/4)) + ((9*b*c - 
 7*a*d)*((2*x*(a + b*x^2)^(3/4))/(5*c*(c + d*x^2)^(5/4)) + (3*a*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*(a + b*x^2)^(1/4)*(c + d*x^2)^(1/4))) 
)/(9*c*(b*c - a*d))
 

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]
 

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]

\[\int \frac {\left (b \,x^{2}+a \right )^{\frac {3}{4}}}{\left (x^{2} d +c \right )^{\frac {13}{4}}}d x\]

Input:

int((b*x^2+a)^(3/4)/(d*x^2+c)^(13/4),x)
 

Output:

int((b*x^2+a)^(3/4)/(d*x^2+c)^(13/4),x)
 

Fricas [F]

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

integrate((b*x^2+a)^(3/4)/(d*x^2+c)^(13/4),x, algorithm="fricas")
 

Output:

integral((b*x^2 + a)^(3/4)*(d*x^2 + c)^(3/4)/(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]

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

integrate((b*x**2+a)**(3/4)/(d*x**2+c)**(13/4),x)
 

Output:

Integral((a + b*x**2)**(3/4)/(c + d*x**2)**(13/4), x)
 

Maxima [F]

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

integrate((b*x^2+a)^(3/4)/(d*x^2+c)^(13/4),x, algorithm="maxima")
 

Output:

integrate((b*x^2 + a)^(3/4)/(d*x^2 + c)^(13/4), x)
 

Giac [F]

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

integrate((b*x^2+a)^(3/4)/(d*x^2+c)^(13/4),x, algorithm="giac")
 

Output:

integrate((b*x^2 + a)^(3/4)/(d*x^2 + c)^(13/4), x)
 

Mupad [F(-1)]

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

int((a + b*x^2)^(3/4)/(c + d*x^2)^(13/4),x)
 

Output:

int((a + b*x^2)^(3/4)/(c + d*x^2)^(13/4), x)
 

Reduce [F]

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

int((b*x^2+a)^(3/4)/(d*x^2+c)^(13/4),x)
 

Output:

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