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

Optimal result

Integrand size = 23, antiderivative size = 86 \[ \int \frac {\sqrt [4]{c+\frac {d}{x^4}}}{\left (a+\frac {b}{x^4}\right )^{15/4}} \, dx=\frac {\left (1+\frac {b}{a x^4}\right )^{3/4} \sqrt [4]{c+\frac {d}{x^4}} x \operatorname {AppellF1}\left (-\frac {1}{4},\frac {15}{4},-\frac {1}{4},\frac {3}{4},-\frac {b}{a x^4},-\frac {d}{c x^4}\right )}{a^3 \left (a+\frac {b}{x^4}\right )^{3/4} \sqrt [4]{1+\frac {d}{c x^4}}} \] Output:

(1+b/a/x^4)^(3/4)*(c+d/x^4)^(1/4)*x*AppellF1(-1/4,15/4,-1/4,3/4,-b/a/x^4,- 
d/c/x^4)/a^3/(a+b/x^4)^(3/4)/(1+d/c/x^4)^(1/4)
                                                                                    
                                                                                    
 

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(342\) vs. \(2(86)=172\).

Time = 5.41 (sec) , antiderivative size = 342, normalized size of antiderivative = 3.98 \[ \int \frac {\sqrt [4]{c+\frac {d}{x^4}}}{\left (a+\frac {b}{x^4}\right )^{15/4}} \, dx=\frac {\sqrt [4]{c+\frac {d}{x^4}} x \left (-7 \left (d+c x^4\right ) \left (96 b^4 c^2+131 a^4 d^2 x^8+a^3 b d x^4 \left (187 d-290 c x^4\right )+a b^3 c \left (-179 d+228 c x^4\right )+a^2 b^2 \left (77 d^2-427 c d x^4+153 c^2 x^8\right )\right )+7 d \left (96 b^2 c^2-179 a b c d+77 a^2 d^2\right ) \left (b+a x^4\right )^2 \left (1+\frac {a x^4}{b}\right )^{3/4} \left (1+\frac {c x^4}{d}\right )^{3/4} \operatorname {AppellF1}\left (\frac {3}{4},\frac {3}{4},\frac {3}{4},\frac {7}{4},-\frac {a x^4}{b},-\frac {c x^4}{d}\right )+2 c \left (192 b^2 c^2-376 a b c d+181 a^2 d^2\right ) x^4 \left (b+a x^4\right )^2 \left (1+\frac {a x^4}{b}\right )^{3/4} \left (1+\frac {c x^4}{d}\right )^{3/4} \operatorname {AppellF1}\left (\frac {7}{4},\frac {3}{4},\frac {3}{4},\frac {11}{4},-\frac {a x^4}{b},-\frac {c x^4}{d}\right )\right )}{1617 a^3 (b c-a d)^2 \left (a+\frac {b}{x^4}\right )^{3/4} \left (b+a x^4\right )^2 \left (d+c x^4\right )} \] Input:

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

Output:

((c + d/x^4)^(1/4)*x*(-7*(d + c*x^4)*(96*b^4*c^2 + 131*a^4*d^2*x^8 + a^3*b 
*d*x^4*(187*d - 290*c*x^4) + a*b^3*c*(-179*d + 228*c*x^4) + a^2*b^2*(77*d^ 
2 - 427*c*d*x^4 + 153*c^2*x^8)) + 7*d*(96*b^2*c^2 - 179*a*b*c*d + 77*a^2*d 
^2)*(b + a*x^4)^2*(1 + (a*x^4)/b)^(3/4)*(1 + (c*x^4)/d)^(3/4)*AppellF1[3/4 
, 3/4, 3/4, 7/4, -((a*x^4)/b), -((c*x^4)/d)] + 2*c*(192*b^2*c^2 - 376*a*b* 
c*d + 181*a^2*d^2)*x^4*(b + a*x^4)^2*(1 + (a*x^4)/b)^(3/4)*(1 + (c*x^4)/d) 
^(3/4)*AppellF1[7/4, 3/4, 3/4, 11/4, -((a*x^4)/b), -((c*x^4)/d)]))/(1617*a 
^3*(b*c - a*d)^2*(a + b/x^4)^(3/4)*(b + a*x^4)^2*(d + c*x^4))
 

Rubi [A] (verified)

Time = 0.46 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {899, 1013, 1013, 1012}

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.

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

\(\Big \downarrow \) 899

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

\(\Big \downarrow \) 1013

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

\(\Big \downarrow \) 1013

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

\(\Big \downarrow \) 1012

\(\displaystyle \frac {x \left (\frac {b}{a x^4}+1\right )^{3/4} \sqrt [4]{c+\frac {d}{x^4}} \operatorname {AppellF1}\left (-\frac {1}{4},\frac {15}{4},-\frac {1}{4},\frac {3}{4},-\frac {b}{a x^4},-\frac {d}{c x^4}\right )}{a^3 \left (a+\frac {b}{x^4}\right )^{3/4} \sqrt [4]{\frac {d}{c x^4}+1}}\)

Input:

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

Output:

((1 + b/(a*x^4))^(3/4)*(c + d/x^4)^(1/4)*x*AppellF1[-1/4, 15/4, -1/4, 3/4, 
 -(b/(a*x^4)), -(d/(c*x^4))])/(a^3*(a + b/x^4)^(3/4)*(1 + d/(c*x^4))^(1/4) 
)
 

Defintions of rubi rules used

rule 899

Int[((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol 
] :> -Subst[Int[(a + b/x^n)^p*((c + d/x^n)^q/x^2), x], x, 1/x] /; FreeQ[{a, 
 b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && ILtQ[n, 0]
 

rule 1012

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_ 
))^(q_), x_Symbol] :> Simp[a^p*c^q*((e*x)^(m + 1)/(e*(m + 1)))*AppellF1[(m 
+ 1)/n, -p, -q, 1 + (m + 1)/n, (-b)*(x^n/a), (-d)*(x^n/c)], x] /; FreeQ[{a, 
 b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, n 
 - 1] && (IntegerQ[p] || GtQ[a, 0]) && (IntegerQ[q] || GtQ[c, 0])
 

rule 1013

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_ 
))^(q_), x_Symbol] :> Simp[a^IntPart[p]*((a + b*x^n)^FracPart[p]/(1 + b*(x^ 
n/a))^FracPart[p])   Int[(e*x)^m*(1 + b*(x^n/a))^p*(c + d*x^n)^q, x], x] /; 
 FreeQ[{a, b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] & 
& NeQ[m, n - 1] &&  !(IntegerQ[p] || GtQ[a, 0])
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

integral(x^16*((a*x^4 + b)/x^4)^(1/4)*((c*x^4 + d)/x^4)^(1/4)/(a^4*x^16 + 
4*a^3*b*x^12 + 6*a^2*b^2*x^8 + 4*a*b^3*x^4 + b^4), x)
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {\sqrt [4]{c+\frac {d}{x^4}}}{\left (a+\frac {b}{x^4}\right )^{15/4}} \, dx=\int \frac {\left (c \,x^{4}+d \right )^{\frac {1}{4}} x^{14}}{\left (a \,x^{4}+b \right )^{\frac {3}{4}} a^{3} x^{12}+3 \left (a \,x^{4}+b \right )^{\frac {3}{4}} a^{2} b \,x^{8}+3 \left (a \,x^{4}+b \right )^{\frac {3}{4}} a \,b^{2} x^{4}+\left (a \,x^{4}+b \right )^{\frac {3}{4}} b^{3}}d x \] Input:

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

Output:

int(((c*x**4 + d)**(1/4)*x**14)/((a*x**4 + b)**(3/4)*a**3*x**12 + 3*(a*x** 
4 + b)**(3/4)*a**2*b*x**8 + 3*(a*x**4 + b)**(3/4)*a*b**2*x**4 + (a*x**4 + 
b)**(3/4)*b**3),x)