\(\int \frac {(a+b x^3)^{5/4}}{(c+d x^3)^{31/12}} \, dx\) [174]

Optimal result

Integrand size = 23, antiderivative size = 87 \[ \int \frac {\left (a+b x^3\right )^{5/4}}{\left (c+d x^3\right )^{31/12}} \, dx=\frac {x \left (a+b x^3\right )^{5/4} \operatorname {Hypergeometric2F1}\left (-\frac {5}{4},\frac {1}{3},\frac {4}{3},-\frac {(b c-a d) x^3}{a \left (c+d x^3\right )}\right )}{c \left (\frac {c \left (a+b x^3\right )}{a \left (c+d x^3\right )}\right )^{5/4} \left (c+d x^3\right )^{19/12}} \] Output:

x*(b*x^3+a)^(5/4)*hypergeom([-5/4, 1/3],[4/3],-(-a*d+b*c)*x^3/a/(d*x^3+c)) 
/c/(c*(b*x^3+a)/a/(d*x^3+c))^(5/4)/(d*x^3+c)^(19/12)
 

Mathematica [A] (warning: unable to verify)

Time = 5.56 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.03 \[ \int \frac {\left (a+b x^3\right )^{5/4}}{\left (c+d x^3\right )^{31/12}} \, dx=\frac {a x \sqrt [4]{a+b x^3} \sqrt [4]{1+\frac {d x^3}{c}} \operatorname {Hypergeometric2F1}\left (-\frac {5}{4},\frac {1}{3},\frac {4}{3},\frac {(-b c+a d) x^3}{a \left (c+d x^3\right )}\right )}{c^2 \sqrt [4]{1+\frac {b x^3}{a}} \left (c+d x^3\right )^{7/12}} \] Input:

Integrate[(a + b*x^3)^(5/4)/(c + d*x^3)^(31/12),x]
 

Output:

(a*x*(a + b*x^3)^(1/4)*(1 + (d*x^3)/c)^(1/4)*Hypergeometric2F1[-5/4, 1/3, 
4/3, ((-(b*c) + a*d)*x^3)/(a*(c + d*x^3))])/(c^2*(1 + (b*x^3)/a)^(1/4)*(c 
+ d*x^3)^(7/12))
 

Rubi [A] (verified)

Time = 0.41 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.85, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {903, 903, 905}

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.

\(\displaystyle \int \frac {\left (a+b x^3\right )^{5/4}}{\left (c+d x^3\right )^{31/12}} \, dx\)

\(\Big \downarrow \) 903

\(\displaystyle \frac {15 a \int \frac {\sqrt [4]{b x^3+a}}{\left (d x^3+c\right )^{19/12}}dx}{19 c}+\frac {4 x \left (a+b x^3\right )^{5/4}}{19 c \left (c+d x^3\right )^{19/12}}\)

\(\Big \downarrow \) 903

\(\displaystyle \frac {15 a \left (\frac {3 a \int \frac {1}{\left (b x^3+a\right )^{3/4} \left (d x^3+c\right )^{7/12}}dx}{7 c}+\frac {4 x \sqrt [4]{a+b x^3}}{7 c \left (c+d x^3\right )^{7/12}}\right )}{19 c}+\frac {4 x \left (a+b x^3\right )^{5/4}}{19 c \left (c+d x^3\right )^{19/12}}\)

\(\Big \downarrow \) 905

\(\displaystyle \frac {15 a \left (\frac {3 a x \left (c+d x^3\right )^{5/12} \left (\frac {c \left (a+b x^3\right )}{a \left (c+d x^3\right )}\right )^{3/4} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {3}{4},\frac {4}{3},-\frac {(b c-a d) x^3}{a \left (d x^3+c\right )}\right )}{7 c^2 \left (a+b x^3\right )^{3/4}}+\frac {4 x \sqrt [4]{a+b x^3}}{7 c \left (c+d x^3\right )^{7/12}}\right )}{19 c}+\frac {4 x \left (a+b x^3\right )^{5/4}}{19 c \left (c+d x^3\right )^{19/12}}\)

Input:

Int[(a + b*x^3)^(5/4)/(c + d*x^3)^(31/12),x]
 

Output:

(4*x*(a + b*x^3)^(5/4))/(19*c*(c + d*x^3)^(19/12)) + (15*a*((4*x*(a + b*x^ 
3)^(1/4))/(7*c*(c + d*x^3)^(7/12)) + (3*a*x*((c*(a + b*x^3))/(a*(c + d*x^3 
)))^(3/4)*(c + d*x^3)^(5/12)*Hypergeometric2F1[1/3, 3/4, 4/3, -(((b*c - a* 
d)*x^3)/(a*(c + d*x^3)))])/(7*c^2*(a + b*x^3)^(3/4))))/(19*c)
 

Defintions of rubi rules used

rule 903

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

rule 905

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] 
:> Simp[x*((a + b*x^n)^p/(c*(c*((a + b*x^n)/(a*(c + d*x^n))))^p*(c + d*x^n) 
^(1/n + p)))*Hypergeometric2F1[1/n, -p, 1 + 1/n, (-(b*c - a*d))*(x^n/(a*(c 
+ d*x^n)))], x] /; FreeQ[{a, b, c, d, n, p, q}, x] && NeQ[b*c - a*d, 0] && 
EqQ[n*(p + q + 1) + 1, 0]
 

Maple [F]

\[\int \frac {\left (b \,x^{3}+a \right )^{\frac {5}{4}}}{\left (d \,x^{3}+c \right )^{\frac {31}{12}}}d x\]

Input:

int((b*x^3+a)^(5/4)/(d*x^3+c)^(31/12),x)
 

Output:

int((b*x^3+a)^(5/4)/(d*x^3+c)^(31/12),x)
 

Fricas [F]

\[ \int \frac {\left (a+b x^3\right )^{5/4}}{\left (c+d x^3\right )^{31/12}} \, dx=\int { \frac {{\left (b x^{3} + a\right )}^{\frac {5}{4}}}{{\left (d x^{3} + c\right )}^{\frac {31}{12}}} \,d x } \] Input:

integrate((b*x^3+a)^(5/4)/(d*x^3+c)^(31/12),x, algorithm="fricas")
 

Output:

integral((b*x^3 + a)^(5/4)*(d*x^3 + c)^(5/12)/(d^3*x^9 + 3*c*d^2*x^6 + 3*c 
^2*d*x^3 + c^3), x)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\left (a+b x^3\right )^{5/4}}{\left (c+d x^3\right )^{31/12}} \, dx=\text {Timed out} \] Input:

integrate((b*x**3+a)**(5/4)/(d*x**3+c)**(31/12),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {\left (a+b x^3\right )^{5/4}}{\left (c+d x^3\right )^{31/12}} \, dx=\int { \frac {{\left (b x^{3} + a\right )}^{\frac {5}{4}}}{{\left (d x^{3} + c\right )}^{\frac {31}{12}}} \,d x } \] Input:

integrate((b*x^3+a)^(5/4)/(d*x^3+c)^(31/12),x, algorithm="maxima")
 

Output:

integrate((b*x^3 + a)^(5/4)/(d*x^3 + c)^(31/12), x)
 

Giac [F]

\[ \int \frac {\left (a+b x^3\right )^{5/4}}{\left (c+d x^3\right )^{31/12}} \, dx=\int { \frac {{\left (b x^{3} + a\right )}^{\frac {5}{4}}}{{\left (d x^{3} + c\right )}^{\frac {31}{12}}} \,d x } \] Input:

integrate((b*x^3+a)^(5/4)/(d*x^3+c)^(31/12),x, algorithm="giac")
 

Output:

integrate((b*x^3 + a)^(5/4)/(d*x^3 + c)^(31/12), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (a+b x^3\right )^{5/4}}{\left (c+d x^3\right )^{31/12}} \, dx=\int \frac {{\left (b\,x^3+a\right )}^{5/4}}{{\left (d\,x^3+c\right )}^{31/12}} \,d x \] Input:

int((a + b*x^3)^(5/4)/(c + d*x^3)^(31/12),x)
 

Output:

int((a + b*x^3)^(5/4)/(c + d*x^3)^(31/12), x)
 

Reduce [F]

\[ \int \frac {\left (a+b x^3\right )^{5/4}}{\left (c+d x^3\right )^{31/12}} \, dx=\text {too large to display} \] Input:

int((b*x^3+a)^(5/4)/(d*x^3+c)^(31/12),x)
 

Output:

( - 68*(c + d*x**3)**(1/4)*(a + b*x**3)**(1/4)*a**3*d**3*x - 164*(c + d*x* 
*3)**(1/4)*(a + b*x**3)**(1/4)*a**2*b*c*d**2*x - 136*(c + d*x**3)**(1/4)*( 
a + b*x**3)**(1/4)*a**2*b*d**3*x**4 - 20*(c + d*x**3)**(1/4)*(a + b*x**3)* 
*(1/4)*a*b**2*c**2*d*x - 4*(c + d*x**3)**(1/4)*(a + b*x**3)**(1/4)*a*b**2* 
c*d**2*x**4 - 28*(c + d*x**3)**(1/4)*(a + b*x**3)**(1/4)*b**3*c**3*x - 28* 
(c + d*x**3)**(1/4)*(a + b*x**3)**(1/4)*b**3*c**2*d*x**4 + 63869*(c + d*x* 
*3)**(5/6)*int((a + b*x**3)**(1/4)/(221*(c + d*x**3)**(7/12)*a**3*c**2*d** 
2 + 442*(c + d*x**3)**(7/12)*a**3*c*d**3*x**3 + 221*(c + d*x**3)**(7/12)*a 
**3*d**4*x**6 - 28*(c + d*x**3)**(7/12)*a**2*b*c**3*d + 165*(c + d*x**3)** 
(7/12)*a**2*b*c**2*d**2*x**3 + 414*(c + d*x**3)**(7/12)*a**2*b*c*d**3*x**6 
 + 221*(c + d*x**3)**(7/12)*a**2*b*d**4*x**9 - 49*(c + d*x**3)**(7/12)*a*b 
**2*c**4 - 126*(c + d*x**3)**(7/12)*a*b**2*c**3*d*x**3 - 105*(c + d*x**3)* 
*(7/12)*a*b**2*c**2*d**2*x**6 - 28*(c + d*x**3)**(7/12)*a*b**2*c*d**3*x**9 
 - 49*(c + d*x**3)**(7/12)*b**3*c**4*x**3 - 98*(c + d*x**3)**(7/12)*b**3*c 
**3*d*x**6 - 49*(c + d*x**3)**(7/12)*b**3*c**2*d**2*x**9),x)*a**6*c**2*d** 
5 + 63869*(c + d*x**3)**(5/6)*int((a + b*x**3)**(1/4)/(221*(c + d*x**3)**( 
7/12)*a**3*c**2*d**2 + 442*(c + d*x**3)**(7/12)*a**3*c*d**3*x**3 + 221*(c 
+ d*x**3)**(7/12)*a**3*d**4*x**6 - 28*(c + d*x**3)**(7/12)*a**2*b*c**3*d + 
 165*(c + d*x**3)**(7/12)*a**2*b*c**2*d**2*x**3 + 414*(c + d*x**3)**(7/12) 
*a**2*b*c*d**3*x**6 + 221*(c + d*x**3)**(7/12)*a**2*b*d**4*x**9 - 49*(c...