\(\int \frac {1}{(a+b x^3)^{10/3}} \, dx\) [29]

Optimal result

Integrand size = 11, antiderivative size = 58 \[ \int \frac {1}{\left (a+b x^3\right )^{10/3}} \, dx=\frac {x}{7 a \left (a+b x^3\right )^{7/3}}+\frac {3 x}{14 a^2 \left (a+b x^3\right )^{4/3}}+\frac {9 x}{14 a^3 \sqrt [3]{a+b x^3}} \] Output:

1/7*x/a/(b*x^3+a)^(7/3)+3/14*x/a^2/(b*x^3+a)^(4/3)+9/14*x/a^3/(b*x^3+a)^(1 
/3)
 

Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 5 vs. order 2 in optimal.

Time = 55.03 (sec) , antiderivative size = 11702, normalized size of antiderivative = 201.76 \[ \int \frac {1}{\left (a+b x^3\right )^{10/3}} \, dx=\text {Result too large to show} \] Input:

Integrate[(a + b*x^3)^(-10/3),x]
 

Output:

Result too large to show
 

Rubi [A] (verified)

Time = 0.17 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.14, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {749, 749, 746}

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.

\(\displaystyle \int \frac {1}{\left (a+b x^3\right )^{10/3}} \, dx\)

\(\Big \downarrow \) 749

\(\displaystyle \frac {6 \int \frac {1}{\left (b x^3+a\right )^{7/3}}dx}{7 a}+\frac {x}{7 a \left (a+b x^3\right )^{7/3}}\)

\(\Big \downarrow \) 749

\(\displaystyle \frac {6 \left (\frac {3 \int \frac {1}{\left (b x^3+a\right )^{4/3}}dx}{4 a}+\frac {x}{4 a \left (a+b x^3\right )^{4/3}}\right )}{7 a}+\frac {x}{7 a \left (a+b x^3\right )^{7/3}}\)

\(\Big \downarrow \) 746

\(\displaystyle \frac {6 \left (\frac {3 x}{4 a^2 \sqrt [3]{a+b x^3}}+\frac {x}{4 a \left (a+b x^3\right )^{4/3}}\right )}{7 a}+\frac {x}{7 a \left (a+b x^3\right )^{7/3}}\)

Input:

Int[(a + b*x^3)^(-10/3),x]
 

Output:

x/(7*a*(a + b*x^3)^(7/3)) + (6*(x/(4*a*(a + b*x^3)^(4/3)) + (3*x)/(4*a^2*( 
a + b*x^3)^(1/3))))/(7*a)
 

Defintions of rubi rules used

rule 746

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x*((a + b*x^n)^(p + 1) 
/a), x] /; FreeQ[{a, b, n, p}, x] && EqQ[1/n + p + 1, 0]
 

rule 749

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^n)^(p + 
 1)/(a*n*(p + 1))), x] + Simp[(n*(p + 1) + 1)/(a*n*(p + 1))   Int[(a + b*x^ 
n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (Inte 
gerQ[2*p] || Denominator[p + 1/n] < Denominator[p])
 

Maple [A] (verified)

Time = 0.48 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.64

method	result	size
gosper	\(\frac {x \left (9 b^{2} x^{6}+21 a b \,x^{3}+14 a^{2}\right )}{14 \left (b \,x^{3}+a \right )^{\frac {7}{3}} a^{3}}\)	\(37\)
trager	\(\frac {x \left (9 b^{2} x^{6}+21 a b \,x^{3}+14 a^{2}\right )}{14 \left (b \,x^{3}+a \right )^{\frac {7}{3}} a^{3}}\)	\(37\)
pseudoelliptic	\(\frac {x \left (9 b^{2} x^{6}+21 a b \,x^{3}+14 a^{2}\right )}{14 \left (b \,x^{3}+a \right )^{\frac {7}{3}} a^{3}}\)	\(37\)
orering	\(\frac {x \left (9 b^{2} x^{6}+21 a b \,x^{3}+14 a^{2}\right )}{14 \left (b \,x^{3}+a \right )^{\frac {7}{3}} a^{3}}\)	\(37\)

Input:

int(1/(b*x^3+a)^(10/3),x,method=_RETURNVERBOSE)
 

Output:

1/14*x*(9*b^2*x^6+21*a*b*x^3+14*a^2)/(b*x^3+a)^(7/3)/a^3
 

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.19 \[ \int \frac {1}{\left (a+b x^3\right )^{10/3}} \, dx=\frac {{\left (9 \, b^{2} x^{7} + 21 \, a b x^{4} + 14 \, a^{2} x\right )} {\left (b x^{3} + a\right )}^{\frac {2}{3}}}{14 \, {\left (a^{3} b^{3} x^{9} + 3 \, a^{4} b^{2} x^{6} + 3 \, a^{5} b x^{3} + a^{6}\right )}} \] Input:

integrate(1/(b*x^3+a)^(10/3),x, algorithm="fricas")
 

Output:

1/14*(9*b^2*x^7 + 21*a*b*x^4 + 14*a^2*x)*(b*x^3 + a)^(2/3)/(a^3*b^3*x^9 + 
3*a^4*b^2*x^6 + 3*a^5*b*x^3 + a^6)
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 515 vs. \(2 (51) = 102\).

Time = 0.99 (sec) , antiderivative size = 515, normalized size of antiderivative = 8.88 \[ \int \frac {1}{\left (a+b x^3\right )^{10/3}} \, dx=\frac {28 a^{5} x \Gamma \left (\frac {1}{3}\right )}{27 a^{\frac {25}{3}} \sqrt [3]{1 + \frac {b x^{3}}{a}} \Gamma \left (\frac {10}{3}\right ) + 81 a^{\frac {22}{3}} b x^{3} \sqrt [3]{1 + \frac {b x^{3}}{a}} \Gamma \left (\frac {10}{3}\right ) + 81 a^{\frac {19}{3}} b^{2} x^{6} \sqrt [3]{1 + \frac {b x^{3}}{a}} \Gamma \left (\frac {10}{3}\right ) + 27 a^{\frac {16}{3}} b^{3} x^{9} \sqrt [3]{1 + \frac {b x^{3}}{a}} \Gamma \left (\frac {10}{3}\right )} + \frac {70 a^{4} b x^{4} \Gamma \left (\frac {1}{3}\right )}{27 a^{\frac {25}{3}} \sqrt [3]{1 + \frac {b x^{3}}{a}} \Gamma \left (\frac {10}{3}\right ) + 81 a^{\frac {22}{3}} b x^{3} \sqrt [3]{1 + \frac {b x^{3}}{a}} \Gamma \left (\frac {10}{3}\right ) + 81 a^{\frac {19}{3}} b^{2} x^{6} \sqrt [3]{1 + \frac {b x^{3}}{a}} \Gamma \left (\frac {10}{3}\right ) + 27 a^{\frac {16}{3}} b^{3} x^{9} \sqrt [3]{1 + \frac {b x^{3}}{a}} \Gamma \left (\frac {10}{3}\right )} + \frac {60 a^{3} b^{2} x^{7} \Gamma \left (\frac {1}{3}\right )}{27 a^{\frac {25}{3}} \sqrt [3]{1 + \frac {b x^{3}}{a}} \Gamma \left (\frac {10}{3}\right ) + 81 a^{\frac {22}{3}} b x^{3} \sqrt [3]{1 + \frac {b x^{3}}{a}} \Gamma \left (\frac {10}{3}\right ) + 81 a^{\frac {19}{3}} b^{2} x^{6} \sqrt [3]{1 + \frac {b x^{3}}{a}} \Gamma \left (\frac {10}{3}\right ) + 27 a^{\frac {16}{3}} b^{3} x^{9} \sqrt [3]{1 + \frac {b x^{3}}{a}} \Gamma \left (\frac {10}{3}\right )} + \frac {18 a^{2} b^{3} x^{10} \Gamma \left (\frac {1}{3}\right )}{27 a^{\frac {25}{3}} \sqrt [3]{1 + \frac {b x^{3}}{a}} \Gamma \left (\frac {10}{3}\right ) + 81 a^{\frac {22}{3}} b x^{3} \sqrt [3]{1 + \frac {b x^{3}}{a}} \Gamma \left (\frac {10}{3}\right ) + 81 a^{\frac {19}{3}} b^{2} x^{6} \sqrt [3]{1 + \frac {b x^{3}}{a}} \Gamma \left (\frac {10}{3}\right ) + 27 a^{\frac {16}{3}} b^{3} x^{9} \sqrt [3]{1 + \frac {b x^{3}}{a}} \Gamma \left (\frac {10}{3}\right )} \] Input:

integrate(1/(b*x**3+a)**(10/3),x)
 

Output:

28*a**5*x*gamma(1/3)/(27*a**(25/3)*(1 + b*x**3/a)**(1/3)*gamma(10/3) + 81* 
a**(22/3)*b*x**3*(1 + b*x**3/a)**(1/3)*gamma(10/3) + 81*a**(19/3)*b**2*x** 
6*(1 + b*x**3/a)**(1/3)*gamma(10/3) + 27*a**(16/3)*b**3*x**9*(1 + b*x**3/a 
)**(1/3)*gamma(10/3)) + 70*a**4*b*x**4*gamma(1/3)/(27*a**(25/3)*(1 + b*x** 
3/a)**(1/3)*gamma(10/3) + 81*a**(22/3)*b*x**3*(1 + b*x**3/a)**(1/3)*gamma( 
10/3) + 81*a**(19/3)*b**2*x**6*(1 + b*x**3/a)**(1/3)*gamma(10/3) + 27*a**( 
16/3)*b**3*x**9*(1 + b*x**3/a)**(1/3)*gamma(10/3)) + 60*a**3*b**2*x**7*gam 
ma(1/3)/(27*a**(25/3)*(1 + b*x**3/a)**(1/3)*gamma(10/3) + 81*a**(22/3)*b*x 
**3*(1 + b*x**3/a)**(1/3)*gamma(10/3) + 81*a**(19/3)*b**2*x**6*(1 + b*x**3 
/a)**(1/3)*gamma(10/3) + 27*a**(16/3)*b**3*x**9*(1 + b*x**3/a)**(1/3)*gamm 
a(10/3)) + 18*a**2*b**3*x**10*gamma(1/3)/(27*a**(25/3)*(1 + b*x**3/a)**(1/ 
3)*gamma(10/3) + 81*a**(22/3)*b*x**3*(1 + b*x**3/a)**(1/3)*gamma(10/3) + 8 
1*a**(19/3)*b**2*x**6*(1 + b*x**3/a)**(1/3)*gamma(10/3) + 27*a**(16/3)*b** 
3*x**9*(1 + b*x**3/a)**(1/3)*gamma(10/3))
 

Maxima [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.86 \[ \int \frac {1}{\left (a+b x^3\right )^{10/3}} \, dx=\frac {{\left (2 \, b^{2} - \frac {7 \, {\left (b x^{3} + a\right )} b}{x^{3}} + \frac {14 \, {\left (b x^{3} + a\right )}^{2}}{x^{6}}\right )} x^{7}}{14 \, {\left (b x^{3} + a\right )}^{\frac {7}{3}} a^{3}} \] Input:

integrate(1/(b*x^3+a)^(10/3),x, algorithm="maxima")
 

Output:

1/14*(2*b^2 - 7*(b*x^3 + a)*b/x^3 + 14*(b*x^3 + a)^2/x^6)*x^7/((b*x^3 + a) 
^(7/3)*a^3)
 

Giac [F]

\[ \int \frac {1}{\left (a+b x^3\right )^{10/3}} \, dx=\int { \frac {1}{{\left (b x^{3} + a\right )}^{\frac {10}{3}}} \,d x } \] Input:

integrate(1/(b*x^3+a)^(10/3),x, algorithm="giac")
 

Output:

integrate((b*x^3 + a)^(-10/3), x)
 

Mupad [B] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.76 \[ \int \frac {1}{\left (a+b x^3\right )^{10/3}} \, dx=\frac {9\,x\,{\left (b\,x^3+a\right )}^2+2\,a^2\,x+3\,a\,x\,\left (b\,x^3+a\right )}{14\,a^3\,{\left (b\,x^3+a\right )}^{7/3}} \] Input:

int(1/(a + b*x^3)^(10/3),x)
 

Output:

(9*x*(a + b*x^3)^2 + 2*a^2*x + 3*a*x*(a + b*x^3))/(14*a^3*(a + b*x^3)^(7/3 
))
 

Reduce [F]

\[ \int \frac {1}{\left (a+b x^3\right )^{10/3}} \, dx=\int \frac {1}{\left (b \,x^{3}+a \right )^{\frac {1}{3}} a^{3}+3 \left (b \,x^{3}+a \right )^{\frac {1}{3}} a^{2} b \,x^{3}+3 \left (b \,x^{3}+a \right )^{\frac {1}{3}} a \,b^{2} x^{6}+\left (b \,x^{3}+a \right )^{\frac {1}{3}} b^{3} x^{9}}d x \] Input:

int(1/(b*x^3+a)^(10/3),x)
 

Output:

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