\(\int \frac {\sqrt [4]{c+d x^2}}{(a+b x^2)^{15/4}} \, dx\) [7]

Optimal result

Integrand size = 23, antiderivative size = 214 \[ \int \frac {\sqrt [4]{c+d x^2}}{\left (a+b x^2\right )^{15/4}} \, dx=\frac {2 b x \left (c+d x^2\right )^{5/4}}{11 a (b c-a d) \left (a+b x^2\right )^{11/4}}+\frac {6 b (3 b c-5 a d) x \left (c+d x^2\right )^{5/4}}{77 a^2 (b c-a d)^2 \left (a+b x^2\right )^{7/4}}+\frac {\left (45 b^2 c^2-110 a b c d+77 a^2 d^2\right ) x \sqrt [4]{c+d x^2} \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},\frac {1}{2},\frac {3}{2},\frac {(b c-a d) x^2}{c \left (a+b x^2\right )}\right )}{77 a^3 (b c-a d)^2 \left (a+b x^2\right )^{3/4} \sqrt [4]{\frac {a \left (c+d x^2\right )}{c \left (a+b x^2\right )}}} \] Output:

2/11*b*x*(d*x^2+c)^(5/4)/a/(-a*d+b*c)/(b*x^2+a)^(11/4)+6/77*b*(-5*a*d+3*b* 
c)*x*(d*x^2+c)^(5/4)/a^2/(-a*d+b*c)^2/(b*x^2+a)^(7/4)+1/77*(77*a^2*d^2-110 
*a*b*c*d+45*b^2*c^2)*x*(d*x^2+c)^(1/4)*hypergeom([-1/4, 1/2],[3/2],(-a*d+b 
*c)*x^2/c/(b*x^2+a))/a^3/(-a*d+b*c)^2/(b*x^2+a)^(3/4)/(a*(d*x^2+c)/c/(b*x^ 
2+a))^(1/4)
 

Mathematica [A] (verified)

Time = 8.59 (sec) , antiderivative size = 216, normalized size of antiderivative = 1.01 \[ \int \frac {\sqrt [4]{c+d x^2}}{\left (a+b x^2\right )^{15/4}} \, dx=\frac {x \sqrt [4]{c+d x^2} \left (2 c \left (21 a^2 (b c-a d)^2+3 a (-b c+a d) (-9 b c+8 a d) \left (a+b x^2\right )+\left (45 b^2 c^2-83 a b c d+32 a^2 d^2\right ) \left (a+b x^2\right )^2\right )+c \left (45 b^2 c^2-110 a b c d+77 a^2 d^2\right ) \left (a+b x^2\right )^2 \left (\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}\right )^{3/4} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {3}{2},\frac {(-b c+a d) x^2}{a \left (c+d x^2\right )}\right )\right )}{231 a^3 c (b c-a d)^2 \left (a+b x^2\right )^{11/4}} \] Input:

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

Output:

(x*(c + d*x^2)^(1/4)*(2*c*(21*a^2*(b*c - a*d)^2 + 3*a*(-(b*c) + a*d)*(-9*b 
*c + 8*a*d)*(a + b*x^2) + (45*b^2*c^2 - 83*a*b*c*d + 32*a^2*d^2)*(a + b*x^ 
2)^2) + c*(45*b^2*c^2 - 110*a*b*c*d + 77*a^2*d^2)*(a + b*x^2)^2*((c*(a + b 
*x^2))/(a*(c + d*x^2)))^(3/4)*Hypergeometric2F1[1/2, 3/4, 3/2, ((-(b*c) + 
a*d)*x^2)/(a*(c + d*x^2))]))/(231*a^3*c*(b*c - a*d)^2*(a + b*x^2)^(11/4))
 

Rubi [A] (warning: unable to verify)

Time = 6.76 (sec) , antiderivative size = 253, normalized size of antiderivative = 1.18, 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.

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

\(\Big \downarrow \) 334

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

\(\Big \downarrow \) 334

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

\(\Big \downarrow \) 333

\(\displaystyle \frac {11 x \operatorname {Gamma}\left (\frac {3}{4}\right ) \sqrt [4]{c+d x^2} \left (\frac {d x^2}{c}+1\right ) \left (30 x^2 \left (c+d x^2\right )^2 (b c-a d) \, _3F_2\left (2,2,\frac {19}{4};1,\frac {9}{2};\frac {(b c-a d) x^2}{c \left (b x^2+a\right )}\right )+30 x^2 \left (4 c^2+7 c d x^2+3 d^2 x^4\right ) (b c-a d) \operatorname {Hypergeometric2F1}\left (2,\frac {19}{4},\frac {9}{2},\frac {(b c-a d) x^2}{c \left (b x^2+a\right )}\right )+7 c \left (a+b x^2\right ) \left (15 c^2+20 c d x^2+8 d^2 x^4\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {15}{4},\frac {7}{2},\frac {(b c-a d) x^2}{c \left (b x^2+a\right )}\right )\right )}{320 a^3 c^3 \operatorname {Gamma}\left (\frac {15}{4}\right ) \left (a+b x^2\right )^{7/4} \left (\frac {b x^2}{a}+1\right )^3}\)

Input:

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

Output:

(11*x*(c + d*x^2)^(1/4)*(1 + (d*x^2)/c)*Gamma[3/4]*(7*c*(a + b*x^2)*(15*c^ 
2 + 20*c*d*x^2 + 8*d^2*x^4)*Hypergeometric2F1[1, 15/4, 7/2, ((b*c - a*d)*x 
^2)/(c*(a + b*x^2))] + 30*(b*c - a*d)*x^2*(4*c^2 + 7*c*d*x^2 + 3*d^2*x^4)* 
Hypergeometric2F1[2, 19/4, 9/2, ((b*c - a*d)*x^2)/(c*(a + b*x^2))] + 30*(b 
*c - a*d)*x^2*(c + d*x^2)^2*HypergeometricPFQ[{2, 2, 19/4}, {1, 9/2}, ((b* 
c - a*d)*x^2)/(c*(a + b*x^2))]))/(320*a^3*c^3*(a + b*x^2)^(7/4)*(1 + (b*x^ 
2)/a)^3*Gamma[15/4])
 

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]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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