\(\int \frac {\sqrt [4]{c+d x^4}}{(a+b x^4)^{15/4}} \, dx\) [9]

Optimal result

Integrand size = 23, antiderivative size = 86 \[ \int \frac {\sqrt [4]{c+d x^4}}{\left (a+b x^4\right )^{15/4}} \, dx=\frac {x \left (1+\frac {b x^4}{a}\right )^{3/4} \sqrt [4]{c+d x^4} \operatorname {AppellF1}\left (\frac {1}{4},\frac {15}{4},-\frac {1}{4},\frac {5}{4},-\frac {b x^4}{a},-\frac {d x^4}{c}\right )}{a^3 \left (a+b x^4\right )^{3/4} \sqrt [4]{1+\frac {d x^4}{c}}} \] Output:

x*(1+b*x^4/a)^(3/4)*(d*x^4+c)^(1/4)*AppellF1(1/4,15/4,-1/4,5/4,-b*x^4/a,-d 
*x^4/c)/a^3/(b*x^4+a)^(3/4)/(1+d*x^4/c)^(1/4)
 

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(401\) vs. \(2(86)=172\).

Time = 4.49 (sec) , antiderivative size = 401, normalized size of antiderivative = 4.66 \[ \int \frac {\sqrt [4]{c+d x^4}}{\left (a+b x^4\right )^{15/4}} \, dx=\frac {x \left (\frac {5 \left (c+d x^4\right ) \left (31 a^4 d^2+20 b^4 c^2 x^8+a b^3 c x^4 \left (50 c-37 d x^4\right )+a^3 b d \left (-70 c+39 d x^4\right )+a^2 b^2 \left (37 c^2-93 c d x^4+15 d^2 x^8\right )\right )}{\left (a+b x^4\right )^2}+d \left (20 b^2 c^2-37 a b c d+15 a^2 d^2\right ) x^4 \left (1+\frac {b x^4}{a}\right )^{3/4} \left (1+\frac {d x^4}{c}\right )^{3/4} \operatorname {AppellF1}\left (\frac {5}{4},\frac {3}{4},\frac {3}{4},\frac {9}{4},-\frac {b x^4}{a},-\frac {d x^4}{c}\right )+\frac {50 a c^2 \left (20 b^2 c^2-42 a b c d+23 a^2 d^2\right ) \operatorname {AppellF1}\left (\frac {1}{4},\frac {3}{4},\frac {3}{4},\frac {5}{4},-\frac {b x^4}{a},-\frac {d x^4}{c}\right )}{5 a c \operatorname {AppellF1}\left (\frac {1}{4},\frac {3}{4},\frac {3}{4},\frac {5}{4},-\frac {b x^4}{a},-\frac {d x^4}{c}\right )-3 x^4 \left (a d \operatorname {AppellF1}\left (\frac {5}{4},\frac {3}{4},\frac {7}{4},\frac {9}{4},-\frac {b x^4}{a},-\frac {d x^4}{c}\right )+b c \operatorname {AppellF1}\left (\frac {5}{4},\frac {7}{4},\frac {3}{4},\frac {9}{4},-\frac {b x^4}{a},-\frac {d x^4}{c}\right )\right )}\right )}{385 a^3 (b c-a d)^2 \left (a+b x^4\right )^{3/4} \left (c+d x^4\right )^{3/4}} \] Input:

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

Output:

(x*((5*(c + d*x^4)*(31*a^4*d^2 + 20*b^4*c^2*x^8 + a*b^3*c*x^4*(50*c - 37*d 
*x^4) + a^3*b*d*(-70*c + 39*d*x^4) + a^2*b^2*(37*c^2 - 93*c*d*x^4 + 15*d^2 
*x^8)))/(a + b*x^4)^2 + d*(20*b^2*c^2 - 37*a*b*c*d + 15*a^2*d^2)*x^4*(1 + 
(b*x^4)/a)^(3/4)*(1 + (d*x^4)/c)^(3/4)*AppellF1[5/4, 3/4, 3/4, 9/4, -((b*x 
^4)/a), -((d*x^4)/c)] + (50*a*c^2*(20*b^2*c^2 - 42*a*b*c*d + 23*a^2*d^2)*A 
ppellF1[1/4, 3/4, 3/4, 5/4, -((b*x^4)/a), -((d*x^4)/c)])/(5*a*c*AppellF1[1 
/4, 3/4, 3/4, 5/4, -((b*x^4)/a), -((d*x^4)/c)] - 3*x^4*(a*d*AppellF1[5/4, 
3/4, 7/4, 9/4, -((b*x^4)/a), -((d*x^4)/c)] + b*c*AppellF1[5/4, 7/4, 3/4, 9 
/4, -((b*x^4)/a), -((d*x^4)/c)]))))/(385*a^3*(b*c - a*d)^2*(a + b*x^4)^(3/ 
4)*(c + d*x^4)^(3/4))
 

Rubi [A] (verified)

Time = 0.37 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {937, 937, 936}

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.

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

\(\Big \downarrow \) 937

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

\(\Big \downarrow \) 937

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

\(\Big \downarrow \) 936

\(\displaystyle \frac {x \left (\frac {b x^4}{a}+1\right )^{3/4} \sqrt [4]{c+d x^4} \operatorname {AppellF1}\left (\frac {1}{4},\frac {15}{4},-\frac {1}{4},\frac {5}{4},-\frac {b x^4}{a},-\frac {d x^4}{c}\right )}{a^3 \left (a+b x^4\right )^{3/4} \sqrt [4]{\frac {d x^4}{c}+1}}\)

Input:

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

Output:

(x*(1 + (b*x^4)/a)^(3/4)*(c + d*x^4)^(1/4)*AppellF1[1/4, 15/4, -1/4, 5/4, 
-((b*x^4)/a), -((d*x^4)/c)])/(a^3*(a + b*x^4)^(3/4)*(1 + (d*x^4)/c)^(1/4))
 

Defintions of rubi rules used

rule 936

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

rule 937

Int[((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[(1 + b*(x^n/a))^p*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, n, p, q 
}, x] && NeQ[b*c - a*d, 0] && NeQ[n, -1] &&  !(IntegerQ[p] || GtQ[a, 0])
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

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

Output:

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