\(\int \frac {1}{\sqrt [4]{a+b x^3} (c+d x^3)^{13/12}} \, dx\) [177]

Optimal result

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

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

Mathematica [A] (warning: unable to verify)

Time = 3.57 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.99 \[ \int \frac {1}{\sqrt [4]{a+b x^3} \left (c+d x^3\right )^{13/12}} \, dx=\frac {x \sqrt [4]{1+\frac {b x^3}{a}} \left (1+\frac {d x^3}{c}\right )^{3/4} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{3},\frac {4}{3},\frac {(-b c+a d) x^3}{a \left (c+d x^3\right )}\right )}{\sqrt [4]{a+b x^3} \left (c+d x^3\right )^{13/12}} \] Input:

Integrate[1/((a + b*x^3)^(1/4)*(c + d*x^3)^(13/12)),x]
 

Output:

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

Rubi [A] (verified)

Time = 0.31 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.043, Rules used = {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.

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

\(\Big \downarrow \) 905

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

Input:

Int[1/((a + b*x^3)^(1/4)*(c + d*x^3)^(13/12)),x]
 

Output:

(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)))])/(c*(a + b*x^3)^(1/4)*(c + d*x^3)^ 
(1/12))
 

Defintions of rubi rules used

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]

\[\int \frac {1}{\left (b \,x^{3}+a \right )^{\frac {1}{4}} \left (d \,x^{3}+c \right )^{\frac {13}{12}}}d x\]

Input:

int(1/(b*x^3+a)^(1/4)/(d*x^3+c)^(13/12),x)
 

Output:

int(1/(b*x^3+a)^(1/4)/(d*x^3+c)^(13/12),x)
 

Fricas [F]

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

integrate(1/(b*x^3+a)^(1/4)/(d*x^3+c)^(13/12),x, algorithm="fricas")
 

Output:

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

Sympy [F]

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

integrate(1/(b*x**3+a)**(1/4)/(d*x**3+c)**(13/12),x)
 

Output:

Integral(1/((a + b*x**3)**(1/4)*(c + d*x**3)**(13/12)), x)
 

Maxima [F]

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

integrate(1/(b*x^3+a)^(1/4)/(d*x^3+c)^(13/12),x, algorithm="maxima")
 

Output:

integrate(1/((b*x^3 + a)^(1/4)*(d*x^3 + c)^(13/12)), x)
 

Giac [F]

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

integrate(1/(b*x^3+a)^(1/4)/(d*x^3+c)^(13/12),x, algorithm="giac")
 

Output:

integrate(1/((b*x^3 + a)^(1/4)*(d*x^3 + c)^(13/12)), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\sqrt [4]{a+b x^3} \left (c+d x^3\right )^{13/12}} \, dx=\int \frac {1}{{\left (b\,x^3+a\right )}^{1/4}\,{\left (d\,x^3+c\right )}^{13/12}} \,d x \] Input:

int(1/((a + b*x^3)^(1/4)*(c + d*x^3)^(13/12)),x)
 

Output:

int(1/((a + b*x^3)^(1/4)*(c + d*x^3)^(13/12)), x)
 

Reduce [F]

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

int(1/(b*x^3+a)^(1/4)/(d*x^3+c)^(13/12),x)
 

Output:

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