\(\int \frac {(a-b x^3)^2}{(a+b x^3)^{16/3}} \, dx\) [86]

Optimal result

Integrand size = 22, antiderivative size = 91 \[ \int \frac {\left (a-b x^3\right )^2}{\left (a+b x^3\right )^{16/3}} \, dx=\frac {4 a x}{13 \left (a+b x^3\right )^{13/3}}-\frac {2 x}{65 \left (a+b x^3\right )^{10/3}}+\frac {47 x}{455 a \left (a+b x^3\right )^{7/3}}+\frac {141 x}{910 a^2 \left (a+b x^3\right )^{4/3}}+\frac {423 x}{910 a^3 \sqrt [3]{a+b x^3}} \] Output:

4/13*a*x/(b*x^3+a)^(13/3)-2/65*x/(b*x^3+a)^(10/3)+47/455*x/a/(b*x^3+a)^(7/ 
3)+141/910*x/a^2/(b*x^3+a)^(4/3)+423/910*x/a^3/(b*x^3+a)^(1/3)
 

Mathematica [A] (verified)

Time = 0.62 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.68 \[ \int \frac {\left (a-b x^3\right )^2}{\left (a+b x^3\right )^{16/3}} \, dx=\frac {910 a^4 x+2275 a^3 b x^4+3055 a^2 b^2 x^7+1833 a b^3 x^{10}+423 b^4 x^{13}}{910 a^3 \left (a+b x^3\right )^{13/3}} \] Input:

Integrate[(a - b*x^3)^2/(a + b*x^3)^(16/3),x]
 

Output:

(910*a^4*x + 2275*a^3*b*x^4 + 3055*a^2*b^2*x^7 + 1833*a*b^3*x^10 + 423*b^4 
*x^13)/(910*a^3*(a + b*x^3)^(13/3))
 

Rubi [A] (verified)

Time = 0.40 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.27, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {930, 27, 910, 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.

\(\displaystyle \int \frac {\left (a-b x^3\right )^2}{\left (a+b x^3\right )^{16/3}} \, dx\)

\(\Big \downarrow \) 930

\(\displaystyle \frac {\int \frac {a b \left (11 a-5 b x^3\right )}{\left (b x^3+a\right )^{13/3}}dx}{13 a b}+\frac {2 x \left (a-b x^3\right )}{13 \left (a+b x^3\right )^{13/3}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{13} \int \frac {11 a-5 b x^3}{\left (b x^3+a\right )^{13/3}}dx+\frac {2 x \left (a-b x^3\right )}{13 \left (a+b x^3\right )^{13/3}}\)

\(\Big \downarrow \) 910

\(\displaystyle \frac {1}{13} \left (\frac {47}{5} \int \frac {1}{\left (b x^3+a\right )^{10/3}}dx+\frac {8 x}{5 \left (a+b x^3\right )^{10/3}}\right )+\frac {2 x \left (a-b x^3\right )}{13 \left (a+b x^3\right )^{13/3}}\)

\(\Big \downarrow \) 749

\(\displaystyle \frac {1}{13} \left (\frac {47}{5} \left (\frac {6 \int \frac {1}{\left (b x^3+a\right )^{7/3}}dx}{7 a}+\frac {x}{7 a \left (a+b x^3\right )^{7/3}}\right )+\frac {8 x}{5 \left (a+b x^3\right )^{10/3}}\right )+\frac {2 x \left (a-b x^3\right )}{13 \left (a+b x^3\right )^{13/3}}\)

\(\Big \downarrow \) 749

\(\displaystyle \frac {1}{13} \left (\frac {47}{5} \left (\frac {6 \left (\frac {3 \int \frac {1}{\left (b x^3+a\right )^{4/3}}dx}{4 a}+\frac {x}{4 a \left (a+b x^3\right )^{4/3}}\right )}{7 a}+\frac {x}{7 a \left (a+b x^3\right )^{7/3}}\right )+\frac {8 x}{5 \left (a+b x^3\right )^{10/3}}\right )+\frac {2 x \left (a-b x^3\right )}{13 \left (a+b x^3\right )^{13/3}}\)

\(\Big \downarrow \) 746

\(\displaystyle \frac {1}{13} \left (\frac {47}{5} \left (\frac {6 \left (\frac {3 x}{4 a^2 \sqrt [3]{a+b x^3}}+\frac {x}{4 a \left (a+b x^3\right )^{4/3}}\right )}{7 a}+\frac {x}{7 a \left (a+b x^3\right )^{7/3}}\right )+\frac {8 x}{5 \left (a+b x^3\right )^{10/3}}\right )+\frac {2 x \left (a-b x^3\right )}{13 \left (a+b x^3\right )^{13/3}}\)

Input:

Int[(a - b*x^3)^2/(a + b*x^3)^(16/3),x]
 

Output:

(2*x*(a - b*x^3))/(13*(a + b*x^3)^(13/3)) + ((8*x)/(5*(a + b*x^3)^(10/3)) 
+ (47*(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)))/5)/13
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

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])
 

rule 910

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

rule 930

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

Maple [A] (verified)

Time = 0.96 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.65

method	result	size
gosper	\(\frac {x \left (423 b^{4} x^{12}+1833 a \,b^{3} x^{9}+3055 a^{2} b^{2} x^{6}+2275 a^{3} b \,x^{3}+910 a^{4}\right )}{910 \left (b \,x^{3}+a \right )^{\frac {13}{3}} a^{3}}\)	\(59\)
trager	\(\frac {x \left (423 b^{4} x^{12}+1833 a \,b^{3} x^{9}+3055 a^{2} b^{2} x^{6}+2275 a^{3} b \,x^{3}+910 a^{4}\right )}{910 \left (b \,x^{3}+a \right )^{\frac {13}{3}} a^{3}}\)	\(59\)
pseudoelliptic	\(\frac {x \left (423 b^{4} x^{12}+1833 a \,b^{3} x^{9}+3055 a^{2} b^{2} x^{6}+2275 a^{3} b \,x^{3}+910 a^{4}\right )}{910 \left (b \,x^{3}+a \right )^{\frac {13}{3}} a^{3}}\)	\(59\)
orering	\(\frac {x \left (423 b^{4} x^{12}+1833 a \,b^{3} x^{9}+3055 a^{2} b^{2} x^{6}+2275 a^{3} b \,x^{3}+910 a^{4}\right )}{910 \left (b \,x^{3}+a \right )^{\frac {13}{3}} a^{3}}\)	\(59\)

Input:

int((-b*x^3+a)^2/(b*x^3+a)^(16/3),x,method=_RETURNVERBOSE)
 

Output:

1/910*x*(423*b^4*x^12+1833*a*b^3*x^9+3055*a^2*b^2*x^6+2275*a^3*b*x^3+910*a 
^4)/(b*x^3+a)^(13/3)/a^3
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.24 \[ \int \frac {\left (a-b x^3\right )^2}{\left (a+b x^3\right )^{16/3}} \, dx=\frac {{\left (423 \, b^{4} x^{13} + 1833 \, a b^{3} x^{10} + 3055 \, a^{2} b^{2} x^{7} + 2275 \, a^{3} b x^{4} + 910 \, a^{4} x\right )} {\left (b x^{3} + a\right )}^{\frac {2}{3}}}{910 \, {\left (a^{3} b^{5} x^{15} + 5 \, a^{4} b^{4} x^{12} + 10 \, a^{5} b^{3} x^{9} + 10 \, a^{6} b^{2} x^{6} + 5 \, a^{7} b x^{3} + a^{8}\right )}} \] Input:

integrate((-b*x^3+a)^2/(b*x^3+a)^(16/3),x, algorithm="fricas")
 

Output:

1/910*(423*b^4*x^13 + 1833*a*b^3*x^10 + 3055*a^2*b^2*x^7 + 2275*a^3*b*x^4 
+ 910*a^4*x)*(b*x^3 + a)^(2/3)/(a^3*b^5*x^15 + 5*a^4*b^4*x^12 + 10*a^5*b^3 
*x^9 + 10*a^6*b^2*x^6 + 5*a^7*b*x^3 + a^8)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\left (a-b x^3\right )^2}{\left (a+b x^3\right )^{16/3}} \, dx=\text {Timed out} \] Input:

integrate((-b*x**3+a)**2/(b*x**3+a)**(16/3),x)
 

Output:

Timed out
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 206 vs. \(2 (71) = 142\).

Time = 0.03 (sec) , antiderivative size = 206, normalized size of antiderivative = 2.26 \[ \int \frac {\left (a-b x^3\right )^2}{\left (a+b x^3\right )^{16/3}} \, dx=\frac {{\left (35 \, b^{2} - \frac {91 \, {\left (b x^{3} + a\right )} b}{x^{3}} + \frac {65 \, {\left (b x^{3} + a\right )}^{2}}{x^{6}}\right )} b^{2} x^{13}}{455 \, {\left (b x^{3} + a\right )}^{\frac {13}{3}} a^{3}} + \frac {{\left (140 \, b^{3} - \frac {546 \, {\left (b x^{3} + a\right )} b^{2}}{x^{3}} + \frac {780 \, {\left (b x^{3} + a\right )}^{2} b}{x^{6}} - \frac {455 \, {\left (b x^{3} + a\right )}^{3}}{x^{9}}\right )} b x^{13}}{910 \, {\left (b x^{3} + a\right )}^{\frac {13}{3}} a^{3}} + \frac {{\left (35 \, b^{4} - \frac {182 \, {\left (b x^{3} + a\right )} b^{3}}{x^{3}} + \frac {390 \, {\left (b x^{3} + a\right )}^{2} b^{2}}{x^{6}} - \frac {455 \, {\left (b x^{3} + a\right )}^{3} b}{x^{9}} + \frac {455 \, {\left (b x^{3} + a\right )}^{4}}{x^{12}}\right )} x^{13}}{455 \, {\left (b x^{3} + a\right )}^{\frac {13}{3}} a^{3}} \] Input:

integrate((-b*x^3+a)^2/(b*x^3+a)^(16/3),x, algorithm="maxima")
 

Output:

1/455*(35*b^2 - 91*(b*x^3 + a)*b/x^3 + 65*(b*x^3 + a)^2/x^6)*b^2*x^13/((b* 
x^3 + a)^(13/3)*a^3) + 1/910*(140*b^3 - 546*(b*x^3 + a)*b^2/x^3 + 780*(b*x 
^3 + a)^2*b/x^6 - 455*(b*x^3 + a)^3/x^9)*b*x^13/((b*x^3 + a)^(13/3)*a^3) + 
 1/455*(35*b^4 - 182*(b*x^3 + a)*b^3/x^3 + 390*(b*x^3 + a)^2*b^2/x^6 - 455 
*(b*x^3 + a)^3*b/x^9 + 455*(b*x^3 + a)^4/x^12)*x^13/((b*x^3 + a)^(13/3)*a^ 
3)
 

Giac [F]

\[ \int \frac {\left (a-b x^3\right )^2}{\left (a+b x^3\right )^{16/3}} \, dx=\int { \frac {{\left (b x^{3} - a\right )}^{2}}{{\left (b x^{3} + a\right )}^{\frac {16}{3}}} \,d x } \] Input:

integrate((-b*x^3+a)^2/(b*x^3+a)^(16/3),x, algorithm="giac")
 

Output:

integrate((b*x^3 - a)^2/(b*x^3 + a)^(16/3), x)
 

Mupad [B] (verification not implemented)

Time = 0.66 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.78 \[ \int \frac {\left (a-b x^3\right )^2}{\left (a+b x^3\right )^{16/3}} \, dx=\frac {423\,x}{910\,a^3\,{\left (b\,x^3+a\right )}^{1/3}}-\frac {2\,x}{65\,{\left (b\,x^3+a\right )}^{10/3}}+\frac {141\,x}{910\,a^2\,{\left (b\,x^3+a\right )}^{4/3}}+\frac {47\,x}{455\,a\,{\left (b\,x^3+a\right )}^{7/3}}+\frac {4\,a\,x}{13\,{\left (b\,x^3+a\right )}^{13/3}} \] Input:

int((a - b*x^3)^2/(a + b*x^3)^(16/3),x)
 

Output:

(423*x)/(910*a^3*(a + b*x^3)^(1/3)) - (2*x)/(65*(a + b*x^3)^(10/3)) + (141 
*x)/(910*a^2*(a + b*x^3)^(4/3)) + (47*x)/(455*a*(a + b*x^3)^(7/3)) + (4*a* 
x)/(13*(a + b*x^3)^(13/3))
 

Reduce [F]

\[ \int \frac {\left (a-b x^3\right )^2}{\left (a+b x^3\right )^{16/3}} \, dx=\left (\int \frac {x^{6}}{\left (b \,x^{3}+a \right )^{\frac {1}{3}} a^{5}+5 \left (b \,x^{3}+a \right )^{\frac {1}{3}} a^{4} b \,x^{3}+10 \left (b \,x^{3}+a \right )^{\frac {1}{3}} a^{3} b^{2} x^{6}+10 \left (b \,x^{3}+a \right )^{\frac {1}{3}} a^{2} b^{3} x^{9}+5 \left (b \,x^{3}+a \right )^{\frac {1}{3}} a \,b^{4} x^{12}+\left (b \,x^{3}+a \right )^{\frac {1}{3}} b^{5} x^{15}}d x \right ) b^{2}-2 \left (\int \frac {x^{3}}{\left (b \,x^{3}+a \right )^{\frac {1}{3}} a^{5}+5 \left (b \,x^{3}+a \right )^{\frac {1}{3}} a^{4} b \,x^{3}+10 \left (b \,x^{3}+a \right )^{\frac {1}{3}} a^{3} b^{2} x^{6}+10 \left (b \,x^{3}+a \right )^{\frac {1}{3}} a^{2} b^{3} x^{9}+5 \left (b \,x^{3}+a \right )^{\frac {1}{3}} a \,b^{4} x^{12}+\left (b \,x^{3}+a \right )^{\frac {1}{3}} b^{5} x^{15}}d x \right ) a b +\left (\int \frac {1}{\left (b \,x^{3}+a \right )^{\frac {1}{3}} a^{5}+5 \left (b \,x^{3}+a \right )^{\frac {1}{3}} a^{4} b \,x^{3}+10 \left (b \,x^{3}+a \right )^{\frac {1}{3}} a^{3} b^{2} x^{6}+10 \left (b \,x^{3}+a \right )^{\frac {1}{3}} a^{2} b^{3} x^{9}+5 \left (b \,x^{3}+a \right )^{\frac {1}{3}} a \,b^{4} x^{12}+\left (b \,x^{3}+a \right )^{\frac {1}{3}} b^{5} x^{15}}d x \right ) a^{2} \] Input:

int((-b*x^3+a)^2/(b*x^3+a)^(16/3),x)
 

Output:

int(x**6/((a + b*x**3)**(1/3)*a**5 + 5*(a + b*x**3)**(1/3)*a**4*b*x**3 + 1 
0*(a + b*x**3)**(1/3)*a**3*b**2*x**6 + 10*(a + b*x**3)**(1/3)*a**2*b**3*x* 
*9 + 5*(a + b*x**3)**(1/3)*a*b**4*x**12 + (a + b*x**3)**(1/3)*b**5*x**15), 
x)*b**2 - 2*int(x**3/((a + b*x**3)**(1/3)*a**5 + 5*(a + b*x**3)**(1/3)*a** 
4*b*x**3 + 10*(a + b*x**3)**(1/3)*a**3*b**2*x**6 + 10*(a + b*x**3)**(1/3)* 
a**2*b**3*x**9 + 5*(a + b*x**3)**(1/3)*a*b**4*x**12 + (a + b*x**3)**(1/3)* 
b**5*x**15),x)*a*b + int(1/((a + b*x**3)**(1/3)*a**5 + 5*(a + b*x**3)**(1/ 
3)*a**4*b*x**3 + 10*(a + b*x**3)**(1/3)*a**3*b**2*x**6 + 10*(a + b*x**3)** 
(1/3)*a**2*b**3*x**9 + 5*(a + b*x**3)**(1/3)*a*b**4*x**12 + (a + b*x**3)** 
(1/3)*b**5*x**15),x)*a**2