\(\int \frac {(a+b x^3)^{3/4}}{(c+d x^3)^{25/12}} \, dx\) [175]

Optimal result

Integrand size = 23, antiderivative size = 87 \[ \int \frac {\left (a+b x^3\right )^{3/4}}{\left (c+d x^3\right )^{25/12}} \, dx=\frac {x \left (a+b x^3\right )^{3/4} \operatorname {Hypergeometric2F1}\left (-\frac {3}{4},\frac {1}{3},\frac {4}{3},-\frac {(b c-a d) x^3}{a \left (c+d x^3\right )}\right )}{c \left (\frac {c \left (a+b x^3\right )}{a \left (c+d x^3\right )}\right )^{3/4} \left (c+d x^3\right )^{13/12}} \] Output:

x*(b*x^3+a)^(3/4)*hypergeom([-3/4, 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))^(3/4)/(d*x^3+c)^(13/12)
 

Mathematica [A] (warning: unable to verify)

Time = 5.82 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.02 \[ \int \frac {\left (a+b x^3\right )^{3/4}}{\left (c+d x^3\right )^{25/12}} \, dx=\frac {x \left (a+b x^3\right )^{3/4} \operatorname {Hypergeometric2F1}\left (-\frac {3}{4},\frac {1}{3},\frac {4}{3},\frac {(-b c+a d) x^3}{a \left (c+d x^3\right )}\right )}{c^2 \left (1+\frac {b x^3}{a}\right )^{3/4} \sqrt [12]{c+d x^3} \sqrt [4]{1+\frac {d x^3}{c}}} \] Input:

Integrate[(a + b*x^3)^(3/4)/(c + d*x^3)^(25/12),x]
 

Output:

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

Rubi [A] (verified)

Time = 0.35 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.40, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {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.

\(\displaystyle \int \frac {\left (a+b x^3\right )^{3/4}}{\left (c+d x^3\right )^{25/12}} \, dx\)

\(\Big \downarrow \) 903

\(\displaystyle \frac {9 a \int \frac {1}{\sqrt [4]{b x^3+a} \left (d x^3+c\right )^{13/12}}dx}{13 c}+\frac {4 x \left (a+b x^3\right )^{3/4}}{13 c \left (c+d x^3\right )^{13/12}}\)

\(\Big \downarrow \) 905

\(\displaystyle \frac {9 a x \sqrt [4]{\frac {c \left (a+b x^3\right )}{a \left (c+d x^3\right )}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{3},\frac {4}{3},-\frac {(b c-a d) x^3}{a \left (d x^3+c\right )}\right )}{13 c^2 \sqrt [4]{a+b x^3} \sqrt [12]{c+d x^3}}+\frac {4 x \left (a+b x^3\right )^{3/4}}{13 c \left (c+d x^3\right )^{13/12}}\)

Input:

Int[(a + b*x^3)^(3/4)/(c + d*x^3)^(25/12),x]
 

Output:

(4*x*(a + b*x^3)^(3/4))/(13*c*(c + d*x^3)^(13/12)) + (9*a*x*((c*(a + b*x^3 
))/(a*(c + d*x^3)))^(1/4)*Hypergeometric2F1[1/4, 1/3, 4/3, -(((b*c - a*d)* 
x^3)/(a*(c + d*x^3)))])/(13*c^2*(a + b*x^3)^(1/4)*(c + d*x^3)^(1/12))
 

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]

\[\int \frac {\left (b \,x^{3}+a \right )^{\frac {3}{4}}}{\left (d \,x^{3}+c \right )^{\frac {25}{12}}}d x\]

Input:

int((b*x^3+a)^(3/4)/(d*x^3+c)^(25/12),x)
 

Output:

int((b*x^3+a)^(3/4)/(d*x^3+c)^(25/12),x)
 

Fricas [F]

\[ \int \frac {\left (a+b x^3\right )^{3/4}}{\left (c+d x^3\right )^{25/12}} \, dx=\int { \frac {{\left (b x^{3} + a\right )}^{\frac {3}{4}}}{{\left (d x^{3} + c\right )}^{\frac {25}{12}}} \,d x } \] Input:

integrate((b*x^3+a)^(3/4)/(d*x^3+c)^(25/12),x, algorithm="fricas")
 

Output:

integral((b*x^3 + a)^(3/4)*(d*x^3 + c)^(11/12)/(d^3*x^9 + 3*c*d^2*x^6 + 3* 
c^2*d*x^3 + c^3), x)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\left (a+b x^3\right )^{3/4}}{\left (c+d x^3\right )^{25/12}} \, dx=\text {Timed out} \] Input:

integrate((b*x**3+a)**(3/4)/(d*x**3+c)**(25/12),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {\left (a+b x^3\right )^{3/4}}{\left (c+d x^3\right )^{25/12}} \, dx=\int { \frac {{\left (b x^{3} + a\right )}^{\frac {3}{4}}}{{\left (d x^{3} + c\right )}^{\frac {25}{12}}} \,d x } \] Input:

integrate((b*x^3+a)^(3/4)/(d*x^3+c)^(25/12),x, algorithm="maxima")
 

Output:

integrate((b*x^3 + a)^(3/4)/(d*x^3 + c)^(25/12), x)
 

Giac [F]

\[ \int \frac {\left (a+b x^3\right )^{3/4}}{\left (c+d x^3\right )^{25/12}} \, dx=\int { \frac {{\left (b x^{3} + a\right )}^{\frac {3}{4}}}{{\left (d x^{3} + c\right )}^{\frac {25}{12}}} \,d x } \] Input:

integrate((b*x^3+a)^(3/4)/(d*x^3+c)^(25/12),x, algorithm="giac")
 

Output:

integrate((b*x^3 + a)^(3/4)/(d*x^3 + c)^(25/12), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (a+b x^3\right )^{3/4}}{\left (c+d x^3\right )^{25/12}} \, dx=\int \frac {{\left (b\,x^3+a\right )}^{3/4}}{{\left (d\,x^3+c\right )}^{25/12}} \,d x \] Input:

int((a + b*x^3)^(3/4)/(c + d*x^3)^(25/12),x)
 

Output:

int((a + b*x^3)^(3/4)/(c + d*x^3)^(25/12), x)
 

Reduce [F]

\[ \int \frac {\left (a+b x^3\right )^{3/4}}{\left (c+d x^3\right )^{25/12}} \, dx=\text {too large to display} \] Input:

int((b*x^3+a)^(3/4)/(d*x^3+c)^(25/12),x)
 

Output:

( - 4*(c + d*x**3)**(3/4)*(a + b*x**3)**(3/4)*b*x - 136*(c + d*x**3)**(5/6 
)*int(((c + d*x**3)**(2/3)*(a + b*x**3)**(3/4)*x**6)/(17*(c + d*x**3)**(3/ 
4)*a**2*c**2*d + 34*(c + d*x**3)**(3/4)*a**2*c*d**2*x**3 + 17*(c + d*x**3) 
**(3/4)*a**2*d**3*x**6 - 5*(c + d*x**3)**(3/4)*a*b*c**3 + 7*(c + d*x**3)** 
(3/4)*a*b*c**2*d*x**3 + 29*(c + d*x**3)**(3/4)*a*b*c*d**2*x**6 + 17*(c + d 
*x**3)**(3/4)*a*b*d**3*x**9 - 5*(c + d*x**3)**(3/4)*b**2*c**3*x**3 - 10*(c 
 + d*x**3)**(3/4)*b**2*c**2*d*x**6 - 5*(c + d*x**3)**(3/4)*b**2*c*d**2*x** 
9),x)*a*b**2*c*d**2 - 136*(c + d*x**3)**(5/6)*int(((c + d*x**3)**(2/3)*(a 
+ b*x**3)**(3/4)*x**6)/(17*(c + d*x**3)**(3/4)*a**2*c**2*d + 34*(c + d*x** 
3)**(3/4)*a**2*c*d**2*x**3 + 17*(c + d*x**3)**(3/4)*a**2*d**3*x**6 - 5*(c 
+ d*x**3)**(3/4)*a*b*c**3 + 7*(c + d*x**3)**(3/4)*a*b*c**2*d*x**3 + 29*(c 
+ d*x**3)**(3/4)*a*b*c*d**2*x**6 + 17*(c + d*x**3)**(3/4)*a*b*d**3*x**9 - 
5*(c + d*x**3)**(3/4)*b**2*c**3*x**3 - 10*(c + d*x**3)**(3/4)*b**2*c**2*d* 
x**6 - 5*(c + d*x**3)**(3/4)*b**2*c*d**2*x**9),x)*a*b**2*d**3*x**3 + 40*(c 
 + d*x**3)**(5/6)*int(((c + d*x**3)**(2/3)*(a + b*x**3)**(3/4)*x**6)/(17*( 
c + d*x**3)**(3/4)*a**2*c**2*d + 34*(c + d*x**3)**(3/4)*a**2*c*d**2*x**3 + 
 17*(c + d*x**3)**(3/4)*a**2*d**3*x**6 - 5*(c + d*x**3)**(3/4)*a*b*c**3 + 
7*(c + d*x**3)**(3/4)*a*b*c**2*d*x**3 + 29*(c + d*x**3)**(3/4)*a*b*c*d**2* 
x**6 + 17*(c + d*x**3)**(3/4)*a*b*d**3*x**9 - 5*(c + d*x**3)**(3/4)*b**2*c 
**3*x**3 - 10*(c + d*x**3)**(3/4)*b**2*c**2*d*x**6 - 5*(c + d*x**3)**(3...