\(\int \frac {\sqrt [4]{c+d x^3}}{(a+b x^3)^{15/4}} \, dx\) [8]

Optimal result

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

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

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(402\) vs. \(2(86)=172\).

Time = 4.40 (sec) , antiderivative size = 402, normalized size of antiderivative = 4.67 \[ \int \frac {\sqrt [4]{c+d x^3}}{\left (a+b x^3\right )^{15/4}} \, dx=\frac {x \left (\frac {8 \left (c+d x^3\right ) \left (787 a^4 d^2+493 b^4 c^2 x^6+a b^3 c x^3 \left (1247 c-911 d x^3\right )+2 a^3 b d \left (-892 c+481 d x^3\right )+a^2 b^2 \left (943 c^2-2317 c d x^3+364 d^2 x^6\right )\right )}{\left (a+b x^3\right )^2}+d \left (493 b^2 c^2-911 a b c d+364 a^2 d^2\right ) x^3 \left (1+\frac {b x^3}{a}\right )^{3/4} \left (1+\frac {d x^3}{c}\right )^{3/4} \operatorname {AppellF1}\left (\frac {4}{3},\frac {3}{4},\frac {3}{4},\frac {7}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )+\frac {32 a c^2 \left (2465 b^2 c^2-5338 a b c d+3089 a^2 d^2\right ) \operatorname {AppellF1}\left (\frac {1}{3},\frac {3}{4},\frac {3}{4},\frac {4}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{16 a c \operatorname {AppellF1}\left (\frac {1}{3},\frac {3}{4},\frac {3}{4},\frac {4}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )-9 x^3 \left (a d \operatorname {AppellF1}\left (\frac {4}{3},\frac {3}{4},\frac {7}{4},\frac {7}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )+b c \operatorname {AppellF1}\left (\frac {4}{3},\frac {7}{4},\frac {3}{4},\frac {7}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )\right )}\right )}{12474 a^3 (b c-a d)^2 \left (a+b x^3\right )^{3/4} \left (c+d x^3\right )^{3/4}} \] Input:

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

Output:

(x*((8*(c + d*x^3)*(787*a^4*d^2 + 493*b^4*c^2*x^6 + a*b^3*c*x^3*(1247*c - 
911*d*x^3) + 2*a^3*b*d*(-892*c + 481*d*x^3) + a^2*b^2*(943*c^2 - 2317*c*d* 
x^3 + 364*d^2*x^6)))/(a + b*x^3)^2 + d*(493*b^2*c^2 - 911*a*b*c*d + 364*a^ 
2*d^2)*x^3*(1 + (b*x^3)/a)^(3/4)*(1 + (d*x^3)/c)^(3/4)*AppellF1[4/3, 3/4, 
3/4, 7/3, -((b*x^3)/a), -((d*x^3)/c)] + (32*a*c^2*(2465*b^2*c^2 - 5338*a*b 
*c*d + 3089*a^2*d^2)*AppellF1[1/3, 3/4, 3/4, 4/3, -((b*x^3)/a), -((d*x^3)/ 
c)])/(16*a*c*AppellF1[1/3, 3/4, 3/4, 4/3, -((b*x^3)/a), -((d*x^3)/c)] - 9* 
x^3*(a*d*AppellF1[4/3, 3/4, 7/4, 7/3, -((b*x^3)/a), -((d*x^3)/c)] + b*c*Ap 
pellF1[4/3, 7/4, 3/4, 7/3, -((b*x^3)/a), -((d*x^3)/c)]))))/(12474*a^3*(b*c 
 - a*d)^2*(a + b*x^3)^(3/4)*(c + d*x^3)^(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.

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

\(\Big \downarrow \) 937

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

\(\Big \downarrow \) 937

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

\(\Big \downarrow \) 936

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

Input:

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

Output:

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

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

Input:

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

Output:

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

Fricas [F]

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

integrate((d*x^3+c)^(1/4)/(b*x^3+a)^(15/4),x, algorithm="fricas")
 

Output:

integral((b*x^3 + a)^(1/4)*(d*x^3 + c)^(1/4)/(b^4*x^12 + 4*a*b^3*x^9 + 6*a 
^2*b^2*x^6 + 4*a^3*b*x^3 + a^4), x)
 

Sympy [F]

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

integrate((d*x**3+c)**(1/4)/(b*x**3+a)**(15/4),x)
 

Output:

Integral((c + d*x**3)**(1/4)/(a + b*x**3)**(15/4), x)
 

Maxima [F]

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

integrate((d*x^3+c)^(1/4)/(b*x^3+a)^(15/4),x, algorithm="maxima")
 

Output:

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

Giac [F]

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

integrate((d*x^3+c)^(1/4)/(b*x^3+a)^(15/4),x, algorithm="giac")
 

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {\sqrt [4]{c+d x^3}}{\left (a+b x^3\right )^{15/4}} \, dx=\int \frac {\left (d \,x^{3}+c \right )^{\frac {1}{4}}}{\left (b \,x^{3}+a \right )^{\frac {3}{4}} a^{3}+3 \left (b \,x^{3}+a \right )^{\frac {3}{4}} a^{2} b \,x^{3}+3 \left (b \,x^{3}+a \right )^{\frac {3}{4}} a \,b^{2} x^{6}+\left (b \,x^{3}+a \right )^{\frac {3}{4}} b^{3} x^{9}}d x \] Input:

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

Output:

int((c + d*x**3)**(1/4)/((a + b*x**3)**(3/4)*a**3 + 3*(a + b*x**3)**(3/4)* 
a**2*b*x**3 + 3*(a + b*x**3)**(3/4)*a*b**2*x**6 + (a + b*x**3)**(3/4)*b**3 
*x**9),x)