\(\int \frac {\sqrt [4]{c+\frac {d}{x}}}{(a+\frac {b}{x})^{15/4}} \, dx\) [5]

Optimal result

Integrand size = 23, antiderivative size = 94 \[ \int \frac {\sqrt [4]{c+\frac {d}{x}}}{\left (a+\frac {b}{x}\right )^{15/4}} \, dx=\frac {4 b \sqrt [4]{c+\frac {d}{x}} \operatorname {AppellF1}\left (-\frac {11}{4},2,-\frac {1}{4},-\frac {7}{4},1+\frac {b}{a x},-\frac {d \left (a+\frac {b}{x}\right )}{b c-a d}\right )}{11 a^2 \left (a+\frac {b}{x}\right )^{11/4} \sqrt [4]{\frac {b \left (c+\frac {d}{x}\right )}{b c-a d}}} \] Output:

4/11*b*(c+d/x)^(1/4)*AppellF1(-11/4,-1/4,2,-7/4,-d*(a+b/x)/(-a*d+b*c),1+b/ 
a/x)/a^2/(a+b/x)^(11/4)/(b*(c+d/x)/(-a*d+b*c))^(1/4)
                                                                                    
                                                                                    
 

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(1290\) vs. \(2(94)=188\).

Time = 22.69 (sec) , antiderivative size = 1290, normalized size of antiderivative = 13.72 \[ \int \frac {\sqrt [4]{c+\frac {d}{x}}}{\left (a+\frac {b}{x}\right )^{15/4}} \, dx =\text {Too large to display} \] Input:

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

Output:

((a + b/x)^(1/4)*(c + d/x)^(1/4)*x*((5*a*(1155*b^5*c^2 + 231*a^5*d^2*x^3 + 
 7*a^4*b*d*x^2*(263*d - 66*c*x) + 11*a*b^4*c*(-202*d + 255*c*x) + a^3*b^2* 
x*(2569*d^2 - 3830*c*d*x + 231*c^2*x^2) + a^2*b^3*(1043*d^2 - 5422*c*d*x + 
 1965*c^2*x^2)))/(b + a*x)^3 - (x*(236775*a^3*b^2*c^2*d*(b*c - a*d)*Appell 
F1[1/4, 3/4, 1, 5/4, (d*(b + a*x))/((-(b*c) + a*d)*x), 1 + b/(a*x)] + 5792 
5*a^5*d^3*(b*c - a*d)*AppellF1[1/4, 3/4, 1, 5/4, (d*(b + a*x))/((-(b*c) + 
a*d)*x), 1 + b/(a*x)] + 86625*a^2*b^3*c^3*(-(b*c) + a*d)*AppellF1[1/4, 3/4 
, 1, 5/4, (d*(b + a*x))/((-(b*c) + a*d)*x), 1 + b/(a*x)] + 209275*a^4*b*c* 
d^2*(-(b*c) + a*d)*AppellF1[1/4, 3/4, 1, 5/4, (d*(b + a*x))/((-(b*c) + a*d 
)*x), 1 + b/(a*x)] + (2310*b^3*c^2*d*(b + a*x)*((b*d + b*c*x)/(b*c*x - a*d 
*x))^(3/4)*AppellF1[5/4, 3/4, 1, 9/4, (d*(b + a*x))/((-(b*c) + a*d)*x), 1 
+ b/(a*x)]*(5*a*(b*c - a*d)*x*AppellF1[1/4, 3/4, 1, 5/4, (d*(b + a*x))/((- 
(b*c) + a*d)*x), 1 + b/(a*x)] + (b + a*x)*(4*(b*c - a*d)*AppellF1[5/4, 3/4 
, 2, 9/4, (d*(b + a*x))/((-(b*c) + a*d)*x), 1 + b/(a*x)] - 3*a*d*AppellF1[ 
5/4, 7/4, 1, 9/4, (d*(b + a*x))/((-(b*c) + a*d)*x), 1 + b/(a*x)])))/x^3 - 
(4444*a*b^2*c*d^2*(b + a*x)*((b*d + b*c*x)/(b*c*x - a*d*x))^(3/4)*AppellF1 
[5/4, 3/4, 1, 9/4, (d*(b + a*x))/((-(b*c) + a*d)*x), 1 + b/(a*x)]*(5*a*(b* 
c - a*d)*x*AppellF1[1/4, 3/4, 1, 5/4, (d*(b + a*x))/((-(b*c) + a*d)*x), 1 
+ b/(a*x)] + (b + a*x)*(4*(b*c - a*d)*AppellF1[5/4, 3/4, 2, 9/4, (d*(b + a 
*x))/((-(b*c) + a*d)*x), 1 + b/(a*x)] - 3*a*d*AppellF1[5/4, 7/4, 1, 9/4...
 

Rubi [A] (verified)

Time = 0.38 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.01, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {899, 154, 153}

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.

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

\(\Big \downarrow \) 899

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

\(\Big \downarrow \) 154

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

\(\Big \downarrow \) 153

\(\displaystyle \frac {4 b \sqrt [4]{c+\frac {d}{x}} \operatorname {AppellF1}\left (-\frac {11}{4},-\frac {1}{4},2,-\frac {7}{4},-\frac {d \left (a+\frac {b}{x}\right )}{b c-a d},\frac {a+\frac {b}{x}}{a}\right )}{11 a^2 \left (a+\frac {b}{x}\right )^{11/4} \sqrt [4]{\frac {b \left (c+\frac {d}{x}\right )}{b c-a d}}}\)

Input:

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

Output:

(4*b*(c + d/x)^(1/4)*AppellF1[-11/4, -1/4, 2, -7/4, -((d*(a + b/x))/(b*c - 
 a*d)), (a + b/x)/a])/(11*a^2*(a + b/x)^(11/4)*((b*(c + d/x))/(b*c - a*d)) 
^(1/4))
 

Defintions of rubi rules used

rule 153

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_)) 
^(p_), x_] :> Simp[(b*e - a*f)^p*((a + b*x)^(m + 1)/(b^(p + 1)*(m + 1)*Simp 
lify[b/(b*c - a*d)]^n))*AppellF1[m + 1, -n, -p, m + 2, (-d)*((a + b*x)/(b*c 
 - a*d)), (-f)*((a + b*x)/(b*e - a*f))], x] /; FreeQ[{a, b, c, d, e, f, m, 
n}, x] &&  !IntegerQ[m] &&  !IntegerQ[n] && IntegerQ[p] && GtQ[Simplify[b/( 
b*c - a*d)], 0] &&  !(GtQ[Simplify[d/(d*a - c*b)], 0] && SimplerQ[c + d*x, 
a + b*x])
 

rule 154

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_)) 
^(p_), x_] :> Simp[(c + d*x)^FracPart[n]/(Simplify[b/(b*c - a*d)]^IntPart[n 
]*(b*((c + d*x)/(b*c - a*d)))^FracPart[n])   Int[(a + b*x)^m*Simp[b*(c/(b*c 
 - a*d)) + b*d*(x/(b*c - a*d)), x]^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, 
 d, e, f, m, n}, x] &&  !IntegerQ[m] &&  !IntegerQ[n] && IntegerQ[p] &&  !G 
tQ[Simplify[b/(b*c - a*d)], 0] &&  !SimplerQ[c + d*x, a + b*x]
 

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]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F(-1)]

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

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

Output:

Timed out
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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