\(\int \frac {1}{\sqrt [4]{a+b x^2} (c+d x^2)^{17/4}} \, dx\) [516]

Optimal result

Integrand size = 23, antiderivative size = 301 \[ \int \frac {1}{\sqrt [4]{a+b x^2} \left (c+d x^2\right )^{17/4}} \, dx=-\frac {2 d x \left (a+b x^2\right )^{3/4}}{13 c (b c-a d) \left (c+d x^2\right )^{13/4}}-\frac {2 b (13 b c-3 a d) (3 b c-a d) x \left (a+b x^2\right )^{3/4}}{117 a c^2 (b c-a d)^3 \left (c+d x^2\right )^{5/4}}-\frac {2 d x \left (a+b x^2\right )^{3/4} \left (13 b c-11 a d-8 b d x^2\right )}{117 c^2 (b c-a d)^2 \left (c+d x^2\right )^{9/4}}+\frac {\left (195 b^3 c^3-351 a b^2 c^2 d+273 a^2 b c d^2-77 a^3 d^3\right ) x \left (a+b x^2\right )^{3/4} \operatorname {Hypergeometric2F1}\left (-\frac {3}{4},\frac {1}{2},\frac {3}{2},-\frac {(b c-a d) x^2}{a \left (c+d x^2\right )}\right )}{117 a c^3 (b c-a d)^3 \left (\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}\right )^{3/4} \left (c+d x^2\right )^{5/4}} \] Output:

-2/13*d*x*(b*x^2+a)^(3/4)/c/(-a*d+b*c)/(d*x^2+c)^(13/4)-2/117*b*(-3*a*d+13 
*b*c)*(-a*d+3*b*c)*x*(b*x^2+a)^(3/4)/a/c^2/(-a*d+b*c)^3/(d*x^2+c)^(5/4)-2/ 
117*d*x*(b*x^2+a)^(3/4)*(-8*b*d*x^2-11*a*d+13*b*c)/c^2/(-a*d+b*c)^2/(d*x^2 
+c)^(9/4)+1/117*(-77*a^3*d^3+273*a^2*b*c*d^2-351*a*b^2*c^2*d+195*b^3*c^3)* 
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^3/(-a*d+b*c)^3/(c*(b*x^2+a)/a/(d*x^2+c))^(3/4)/(d*x^2+c)^(5/4)
                                                                                    
                                                                                    
 

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(832\) vs. \(2(301)=602\).

Time = 7.65 (sec) , antiderivative size = 832, normalized size of antiderivative = 2.76 \[ \int \frac {1}{\sqrt [4]{a+b x^2} \left (c+d x^2\right )^{17/4}} \, dx =\text {Too large to display} \] Input:

Integrate[1/((a + b*x^2)^(1/4)*(c + d*x^2)^(17/4)),x]
 

Output:

(945*a*c^4*x*Hypergeometric2F1[1/4, 1, 9/2, ((b*c - a*d)*x^2)/(c*(a + b*x^ 
2))] + 945*b*c^4*x^3*Hypergeometric2F1[1/4, 1, 9/2, ((b*c - a*d)*x^2)/(c*( 
a + b*x^2))] + 1890*a*c^3*d*x^3*Hypergeometric2F1[1/4, 1, 9/2, ((b*c - a*d 
)*x^2)/(c*(a + b*x^2))] + 1890*b*c^3*d*x^5*Hypergeometric2F1[1/4, 1, 9/2, 
((b*c - a*d)*x^2)/(c*(a + b*x^2))] + 1512*a*c^2*d^2*x^5*Hypergeometric2F1[ 
1/4, 1, 9/2, ((b*c - a*d)*x^2)/(c*(a + b*x^2))] + 1512*b*c^2*d^2*x^7*Hyper 
geometric2F1[1/4, 1, 9/2, ((b*c - a*d)*x^2)/(c*(a + b*x^2))] + 432*a*c*d^3 
*x^7*Hypergeometric2F1[1/4, 1, 9/2, ((b*c - a*d)*x^2)/(c*(a + b*x^2))] + 4 
32*b*c*d^3*x^9*Hypergeometric2F1[1/4, 1, 9/2, ((b*c - a*d)*x^2)/(c*(a + b* 
x^2))] + 71*b*c^4*x^3*Hypergeometric2F1[5/4, 2, 11/2, ((b*c - a*d)*x^2)/(c 
*(a + b*x^2))] - 71*a*c^3*d*x^3*Hypergeometric2F1[5/4, 2, 11/2, ((b*c - a* 
d)*x^2)/(c*(a + b*x^2))] + 177*b*c^3*d*x^5*Hypergeometric2F1[5/4, 2, 11/2, 
 ((b*c - a*d)*x^2)/(c*(a + b*x^2))] - 177*a*c^2*d^2*x^5*Hypergeometric2F1[ 
5/4, 2, 11/2, ((b*c - a*d)*x^2)/(c*(a + b*x^2))] + 150*b*c^2*d^2*x^7*Hyper 
geometric2F1[5/4, 2, 11/2, ((b*c - a*d)*x^2)/(c*(a + b*x^2))] - 150*a*c*d^ 
3*x^7*Hypergeometric2F1[5/4, 2, 11/2, ((b*c - a*d)*x^2)/(c*(a + b*x^2))] + 
 44*b*c*d^3*x^9*Hypergeometric2F1[5/4, 2, 11/2, ((b*c - a*d)*x^2)/(c*(a + 
b*x^2))] - 44*a*d^4*x^9*Hypergeometric2F1[5/4, 2, 11/2, ((b*c - a*d)*x^2)/ 
(c*(a + b*x^2))] + 6*(b*c - a*d)*x^3*(c + d*x^2)^2*(5*c + 4*d*x^2)*Hyperge 
ometricPFQ[{5/4, 2, 2}, {1, 11/2}, ((b*c - a*d)*x^2)/(c*(a + b*x^2))] +...
 

Rubi [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(844\) vs. \(2(301)=602\).

Time = 4.13 (sec) , antiderivative size = 844, normalized size of antiderivative = 2.80, 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.

\(\displaystyle \int \frac {1}{\sqrt [4]{a+b x^2} \left (c+d x^2\right )^{17/4}} \, dx\)

\(\Big \downarrow \) 334

\(\displaystyle \frac {\sqrt [4]{\frac {b x^2}{a}+1} \int \frac {1}{\sqrt [4]{\frac {b x^2}{a}+1} \left (d x^2+c\right )^{17/4}}dx}{\sqrt [4]{a+b x^2}}\)

\(\Big \downarrow \) 334

\(\displaystyle \frac {\sqrt [4]{\frac {b x^2}{a}+1} \sqrt [4]{\frac {d x^2}{c}+1} \int \frac {1}{\sqrt [4]{\frac {b x^2}{a}+1} \left (\frac {d x^2}{c}+1\right )^{17/4}}dx}{c^4 \sqrt [4]{a+b x^2} \sqrt [4]{c+d x^2}}\)

\(\Big \downarrow \) 333

\(\displaystyle \frac {x \left (432 b c d^3 \operatorname {Hypergeometric2F1}\left (\frac {1}{4},1,\frac {9}{2},\frac {(b c-a d) x^2}{c \left (b x^2+a\right )}\right ) x^8-44 a d^4 \operatorname {Hypergeometric2F1}\left (\frac {5}{4},2,\frac {11}{2},\frac {(b c-a d) x^2}{c \left (b x^2+a\right )}\right ) x^8+44 b c d^3 \operatorname {Hypergeometric2F1}\left (\frac {5}{4},2,\frac {11}{2},\frac {(b c-a d) x^2}{c \left (b x^2+a\right )}\right ) x^8+432 a c d^3 \operatorname {Hypergeometric2F1}\left (\frac {1}{4},1,\frac {9}{2},\frac {(b c-a d) x^2}{c \left (b x^2+a\right )}\right ) x^6+1512 b c^2 d^2 \operatorname {Hypergeometric2F1}\left (\frac {1}{4},1,\frac {9}{2},\frac {(b c-a d) x^2}{c \left (b x^2+a\right )}\right ) x^6-150 a c d^3 \operatorname {Hypergeometric2F1}\left (\frac {5}{4},2,\frac {11}{2},\frac {(b c-a d) x^2}{c \left (b x^2+a\right )}\right ) x^6+150 b c^2 d^2 \operatorname {Hypergeometric2F1}\left (\frac {5}{4},2,\frac {11}{2},\frac {(b c-a d) x^2}{c \left (b x^2+a\right )}\right ) x^6+1512 a c^2 d^2 \operatorname {Hypergeometric2F1}\left (\frac {1}{4},1,\frac {9}{2},\frac {(b c-a d) x^2}{c \left (b x^2+a\right )}\right ) x^4+1890 b c^3 d \operatorname {Hypergeometric2F1}\left (\frac {1}{4},1,\frac {9}{2},\frac {(b c-a d) x^2}{c \left (b x^2+a\right )}\right ) x^4-177 a c^2 d^2 \operatorname {Hypergeometric2F1}\left (\frac {5}{4},2,\frac {11}{2},\frac {(b c-a d) x^2}{c \left (b x^2+a\right )}\right ) x^4+177 b c^3 d \operatorname {Hypergeometric2F1}\left (\frac {5}{4},2,\frac {11}{2},\frac {(b c-a d) x^2}{c \left (b x^2+a\right )}\right ) x^4+945 b c^4 \operatorname {Hypergeometric2F1}\left (\frac {1}{4},1,\frac {9}{2},\frac {(b c-a d) x^2}{c \left (b x^2+a\right )}\right ) x^2+1890 a c^3 d \operatorname {Hypergeometric2F1}\left (\frac {1}{4},1,\frac {9}{2},\frac {(b c-a d) x^2}{c \left (b x^2+a\right )}\right ) x^2+71 b c^4 \operatorname {Hypergeometric2F1}\left (\frac {5}{4},2,\frac {11}{2},\frac {(b c-a d) x^2}{c \left (b x^2+a\right )}\right ) x^2-71 a c^3 d \operatorname {Hypergeometric2F1}\left (\frac {5}{4},2,\frac {11}{2},\frac {(b c-a d) x^2}{c \left (b x^2+a\right )}\right ) x^2+6 (b c-a d) \left (d x^2+c\right )^2 \left (4 d x^2+5 c\right ) \, _3F_2\left (\frac {5}{4},2,2;1,\frac {11}{2};\frac {(b c-a d) x^2}{c \left (b x^2+a\right )}\right ) x^2+4 (b c-a d) \left (d x^2+c\right )^3 \, _4F_3\left (\frac {5}{4},2,2,2;1,1,\frac {11}{2};\frac {(b c-a d) x^2}{c \left (b x^2+a\right )}\right ) x^2+945 a c^4 \operatorname {Hypergeometric2F1}\left (\frac {1}{4},1,\frac {9}{2},\frac {(b c-a d) x^2}{c \left (b x^2+a\right )}\right )\right )}{945 c^8 \left (b x^2+a\right )^{5/4} \sqrt [4]{d x^2+c} \left (\frac {d x^2}{c}+1\right )^3}\)

Input:

Int[1/((a + b*x^2)^(1/4)*(c + d*x^2)^(17/4)),x]
 

Output:

(x*(945*a*c^4*Hypergeometric2F1[1/4, 1, 9/2, ((b*c - a*d)*x^2)/(c*(a + b*x 
^2))] + 945*b*c^4*x^2*Hypergeometric2F1[1/4, 1, 9/2, ((b*c - a*d)*x^2)/(c* 
(a + b*x^2))] + 1890*a*c^3*d*x^2*Hypergeometric2F1[1/4, 1, 9/2, ((b*c - a* 
d)*x^2)/(c*(a + b*x^2))] + 1890*b*c^3*d*x^4*Hypergeometric2F1[1/4, 1, 9/2, 
 ((b*c - a*d)*x^2)/(c*(a + b*x^2))] + 1512*a*c^2*d^2*x^4*Hypergeometric2F1 
[1/4, 1, 9/2, ((b*c - a*d)*x^2)/(c*(a + b*x^2))] + 1512*b*c^2*d^2*x^6*Hype 
rgeometric2F1[1/4, 1, 9/2, ((b*c - a*d)*x^2)/(c*(a + b*x^2))] + 432*a*c*d^ 
3*x^6*Hypergeometric2F1[1/4, 1, 9/2, ((b*c - a*d)*x^2)/(c*(a + b*x^2))] + 
432*b*c*d^3*x^8*Hypergeometric2F1[1/4, 1, 9/2, ((b*c - a*d)*x^2)/(c*(a + b 
*x^2))] + 71*b*c^4*x^2*Hypergeometric2F1[5/4, 2, 11/2, ((b*c - a*d)*x^2)/( 
c*(a + b*x^2))] - 71*a*c^3*d*x^2*Hypergeometric2F1[5/4, 2, 11/2, ((b*c - a 
*d)*x^2)/(c*(a + b*x^2))] + 177*b*c^3*d*x^4*Hypergeometric2F1[5/4, 2, 11/2 
, ((b*c - a*d)*x^2)/(c*(a + b*x^2))] - 177*a*c^2*d^2*x^4*Hypergeometric2F1 
[5/4, 2, 11/2, ((b*c - a*d)*x^2)/(c*(a + b*x^2))] + 150*b*c^2*d^2*x^6*Hype 
rgeometric2F1[5/4, 2, 11/2, ((b*c - a*d)*x^2)/(c*(a + b*x^2))] - 150*a*c*d 
^3*x^6*Hypergeometric2F1[5/4, 2, 11/2, ((b*c - a*d)*x^2)/(c*(a + b*x^2))] 
+ 44*b*c*d^3*x^8*Hypergeometric2F1[5/4, 2, 11/2, ((b*c - a*d)*x^2)/(c*(a + 
 b*x^2))] - 44*a*d^4*x^8*Hypergeometric2F1[5/4, 2, 11/2, ((b*c - a*d)*x^2) 
/(c*(a + b*x^2))] + 6*(b*c - a*d)*x^2*(c + d*x^2)^2*(5*c + 4*d*x^2)*Hyperg 
eometricPFQ[{5/4, 2, 2}, {1, 11/2}, ((b*c - a*d)*x^2)/(c*(a + b*x^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]

\[\int \frac {1}{\left (b \,x^{2}+a \right )^{\frac {1}{4}} \left (x^{2} d +c \right )^{\frac {17}{4}}}d x\]

Input:

int(1/(b*x^2+a)^(1/4)/(d*x^2+c)^(17/4),x)
 

Output:

int(1/(b*x^2+a)^(1/4)/(d*x^2+c)^(17/4),x)
 

Fricas [F]

\[ \int \frac {1}{\sqrt [4]{a+b x^2} \left (c+d x^2\right )^{17/4}} \, dx=\int { \frac {1}{{\left (b x^{2} + a\right )}^{\frac {1}{4}} {\left (d x^{2} + c\right )}^{\frac {17}{4}}} \,d x } \] Input:

integrate(1/(b*x^2+a)^(1/4)/(d*x^2+c)^(17/4),x, algorithm="fricas")
 

Output:

integral((b*x^2 + a)^(3/4)*(d*x^2 + c)^(3/4)/(b*d^5*x^12 + (5*b*c*d^4 + a* 
d^5)*x^10 + 5*(2*b*c^2*d^3 + a*c*d^4)*x^8 + 10*(b*c^3*d^2 + a*c^2*d^3)*x^6 
 + a*c^5 + 5*(b*c^4*d + 2*a*c^3*d^2)*x^4 + (b*c^5 + 5*a*c^4*d)*x^2), x)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {1}{\sqrt [4]{a+b x^2} \left (c+d x^2\right )^{17/4}} \, dx=\text {Timed out} \] Input:

integrate(1/(b*x**2+a)**(1/4)/(d*x**2+c)**(17/4),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {1}{\sqrt [4]{a+b x^2} \left (c+d x^2\right )^{17/4}} \, dx=\int { \frac {1}{{\left (b x^{2} + a\right )}^{\frac {1}{4}} {\left (d x^{2} + c\right )}^{\frac {17}{4}}} \,d x } \] Input:

integrate(1/(b*x^2+a)^(1/4)/(d*x^2+c)^(17/4),x, algorithm="maxima")
 

Output:

integrate(1/((b*x^2 + a)^(1/4)*(d*x^2 + c)^(17/4)), x)
 

Giac [F]

\[ \int \frac {1}{\sqrt [4]{a+b x^2} \left (c+d x^2\right )^{17/4}} \, dx=\int { \frac {1}{{\left (b x^{2} + a\right )}^{\frac {1}{4}} {\left (d x^{2} + c\right )}^{\frac {17}{4}}} \,d x } \] Input:

integrate(1/(b*x^2+a)^(1/4)/(d*x^2+c)^(17/4),x, algorithm="giac")
 

Output:

integrate(1/((b*x^2 + a)^(1/4)*(d*x^2 + c)^(17/4)), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\sqrt [4]{a+b x^2} \left (c+d x^2\right )^{17/4}} \, dx=\int \frac {1}{{\left (b\,x^2+a\right )}^{1/4}\,{\left (d\,x^2+c\right )}^{17/4}} \,d x \] Input:

int(1/((a + b*x^2)^(1/4)*(c + d*x^2)^(17/4)),x)
 

Output:

int(1/((a + b*x^2)^(1/4)*(c + d*x^2)^(17/4)), x)
 

Reduce [F]

\[ \int \frac {1}{\sqrt [4]{a+b x^2} \left (c+d x^2\right )^{17/4}} \, dx=\int \frac {1}{\left (d \,x^{2}+c \right )^{\frac {1}{4}} \left (b \,x^{2}+a \right )^{\frac {1}{4}} c^{4}+4 \left (d \,x^{2}+c \right )^{\frac {1}{4}} \left (b \,x^{2}+a \right )^{\frac {1}{4}} c^{3} d \,x^{2}+6 \left (d \,x^{2}+c \right )^{\frac {1}{4}} \left (b \,x^{2}+a \right )^{\frac {1}{4}} c^{2} d^{2} x^{4}+4 \left (d \,x^{2}+c \right )^{\frac {1}{4}} \left (b \,x^{2}+a \right )^{\frac {1}{4}} c \,d^{3} x^{6}+\left (d \,x^{2}+c \right )^{\frac {1}{4}} \left (b \,x^{2}+a \right )^{\frac {1}{4}} d^{4} x^{8}}d x \] Input:

int(1/(b*x^2+a)^(1/4)/(d*x^2+c)^(17/4),x)
 

Output:

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