\(\int \frac {(c+d x^3)^{11/12}}{(a+b x^3)^{9/4}} \, dx\) [181]

Optimal result

Integrand size = 23, antiderivative size = 86 \[ \int \frac {\left (c+d x^3\right )^{11/12}}{\left (a+b x^3\right )^{9/4}} \, dx=\frac {x \left (c+d x^3\right )^{11/12} \operatorname {Hypergeometric2F1}\left (-\frac {11}{12},\frac {1}{3},\frac {4}{3},\frac {(b c-a d) x^3}{c \left (a+b x^3\right )}\right )}{a \left (a+b x^3\right )^{5/4} \left (\frac {a \left (c+d x^3\right )}{c \left (a+b x^3\right )}\right )^{11/12}} \] Output:

x*(d*x^3+c)^(11/12)*hypergeom([-11/12, 1/3],[4/3],(-a*d+b*c)*x^3/c/(b*x^3+ 
a))/a/(b*x^3+a)^(5/4)/(a*(d*x^3+c)/c/(b*x^3+a))^(11/12)
 

Mathematica [A] (warning: unable to verify)

Time = 5.92 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.03 \[ \int \frac {\left (c+d x^3\right )^{11/12}}{\left (a+b x^3\right )^{9/4}} \, dx=\frac {x \sqrt [4]{1+\frac {b x^3}{a}} \left (c+d x^3\right )^{11/12} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {9}{4},\frac {4}{3},\frac {(-b c+a d) x^3}{a \left (c+d x^3\right )}\right )}{a^2 \sqrt [4]{a+b x^3} \left (1+\frac {d x^3}{c}\right )^{5/4}} \] Input:

Integrate[(c + d*x^3)^(11/12)/(a + b*x^3)^(9/4),x]
 

Output:

(x*(1 + (b*x^3)/a)^(1/4)*(c + d*x^3)^(11/12)*Hypergeometric2F1[1/3, 9/4, 4 
/3, ((-(b*c) + a*d)*x^3)/(a*(c + d*x^3))])/(a^2*(a + b*x^3)^(1/4)*(1 + (d* 
x^3)/c)^(5/4))
 

Rubi [A] (verified)

Time = 0.36 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.41, 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.

\(\displaystyle \int \frac {\left (c+d x^3\right )^{11/12}}{\left (a+b x^3\right )^{9/4}} \, dx\)

\(\Big \downarrow \) 903

\(\displaystyle \frac {11 c \int \frac {1}{\left (b x^3+a\right )^{5/4} \sqrt [12]{d x^3+c}}dx}{15 a}+\frac {4 x \left (c+d x^3\right )^{11/12}}{15 a \left (a+b x^3\right )^{5/4}}\)

\(\Big \downarrow \) 905

\(\displaystyle \frac {11 x \left (c+d x^3\right )^{11/12} \left (\frac {c \left (a+b x^3\right )}{a \left (c+d x^3\right )}\right )^{5/4} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {5}{4},\frac {4}{3},-\frac {(b c-a d) x^3}{a \left (d x^3+c\right )}\right )}{15 a \left (a+b x^3\right )^{5/4}}+\frac {4 x \left (c+d x^3\right )^{11/12}}{15 a \left (a+b x^3\right )^{5/4}}\)

Input:

Int[(c + d*x^3)^(11/12)/(a + b*x^3)^(9/4),x]
 

Output:

(4*x*(c + d*x^3)^(11/12))/(15*a*(a + b*x^3)^(5/4)) + (11*x*((c*(a + b*x^3) 
)/(a*(c + d*x^3)))^(5/4)*(c + d*x^3)^(11/12)*Hypergeometric2F1[1/3, 5/4, 4 
/3, -(((b*c - a*d)*x^3)/(a*(c + d*x^3)))])/(15*a*(a + b*x^3)^(5/4))
 

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]

\[\int \frac {\left (d \,x^{3}+c \right )^{\frac {11}{12}}}{\left (b \,x^{3}+a \right )^{\frac {9}{4}}}d x\]

Input:

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

Output:

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

Fricas [F]

\[ \int \frac {\left (c+d x^3\right )^{11/12}}{\left (a+b x^3\right )^{9/4}} \, dx=\int { \frac {{\left (d x^{3} + c\right )}^{\frac {11}{12}}}{{\left (b x^{3} + a\right )}^{\frac {9}{4}}} \,d x } \] Input:

integrate((d*x^3+c)^(11/12)/(b*x^3+a)^(9/4),x, algorithm="fricas")
 

Output:

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

Sympy [F(-1)]

Timed out. \[ \int \frac {\left (c+d x^3\right )^{11/12}}{\left (a+b x^3\right )^{9/4}} \, dx=\text {Timed out} \] Input:

integrate((d*x**3+c)**(11/12)/(b*x**3+a)**(9/4),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {\left (c+d x^3\right )^{11/12}}{\left (a+b x^3\right )^{9/4}} \, dx=\int { \frac {{\left (d x^{3} + c\right )}^{\frac {11}{12}}}{{\left (b x^{3} + a\right )}^{\frac {9}{4}}} \,d x } \] Input:

integrate((d*x^3+c)^(11/12)/(b*x^3+a)^(9/4),x, algorithm="maxima")
 

Output:

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

Giac [F]

\[ \int \frac {\left (c+d x^3\right )^{11/12}}{\left (a+b x^3\right )^{9/4}} \, dx=\int { \frac {{\left (d x^{3} + c\right )}^{\frac {11}{12}}}{{\left (b x^{3} + a\right )}^{\frac {9}{4}}} \,d x } \] Input:

integrate((d*x^3+c)^(11/12)/(b*x^3+a)^(9/4),x, algorithm="giac")
 

Output:

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

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (c+d x^3\right )^{11/12}}{\left (a+b x^3\right )^{9/4}} \, dx=\int \frac {{\left (d\,x^3+c\right )}^{11/12}}{{\left (b\,x^3+a\right )}^{9/4}} \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int \frac {\left (c+d x^3\right )^{11/12}}{\left (a+b x^3\right )^{9/4}} \, dx=\int \frac {\left (d \,x^{3}+c \right )^{\frac {11}{12}}}{\left (b \,x^{3}+a \right )^{\frac {1}{4}} a^{2}+2 \left (b \,x^{3}+a \right )^{\frac {1}{4}} a b \,x^{3}+\left (b \,x^{3}+a \right )^{\frac {1}{4}} b^{2} x^{6}}d x \] Input:

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

Output:

int((c + d*x**3)**(11/12)/((a + b*x**3)**(1/4)*a**2 + 2*(a + b*x**3)**(1/4 
)*a*b*x**3 + (a + b*x**3)**(1/4)*b**2*x**6),x)