Integrand size = 15, antiderivative size = 134 \[ \int \frac {1}{\left (a x^3+b x^6\right )^{8/3}} \, dx=-\frac {1}{7 a x^2 \left (a x^3+b x^6\right )^{5/3}}+\frac {3 b x}{7 a^2 \left (a x^3+b x^6\right )^{5/3}}-\frac {27 b^2 x^4}{7 a^3 \left (a x^3+b x^6\right )^{5/3}}-\frac {81 b^3 x^7}{7 a^4 \left (a x^3+b x^6\right )^{5/3}}-\frac {243 b^4 x^{10}}{35 a^5 \left (a x^3+b x^6\right )^{5/3}} \] Output:
-1/7/a/x^2/(b*x^6+a*x^3)^(5/3)+3/7*b*x/a^2/(b*x^6+a*x^3)^(5/3)-27/7*b^2*x^ 4/a^3/(b*x^6+a*x^3)^(5/3)-81/7*b^3*x^7/a^4/(b*x^6+a*x^3)^(5/3)-243/35*b^4* x^10/a^5/(b*x^6+a*x^3)^(5/3)
Time = 1.10 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.51 \[ \int \frac {1}{\left (a x^3+b x^6\right )^{8/3}} \, dx=\frac {-5 a^4+15 a^3 b x^3-135 a^2 b^2 x^6-405 a b^3 x^9-243 b^4 x^{12}}{35 a^5 x^2 \left (x^3 \left (a+b x^3\right )\right )^{5/3}} \] Input:
Integrate[(a*x^3 + b*x^6)^(-8/3),x]
Output:
(-5*a^4 + 15*a^3*b*x^3 - 135*a^2*b^2*x^6 - 405*a*b^3*x^9 - 243*b^4*x^12)/( 35*a^5*x^2*(x^3*(a + b*x^3))^(5/3))
Time = 0.32 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.13, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {1907, 1921, 1922, 1922, 1906}
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.
\(\displaystyle \int \frac {1}{\left (a x^3+b x^6\right )^{8/3}} \, dx\) |
\(\Big \downarrow \) 1907 |
\(\displaystyle \frac {12 \int \frac {1}{x^3 \left (b x^6+a x^3\right )^{5/3}}dx}{5 a}+\frac {1}{5 a x^2 \left (a x^3+b x^6\right )^{5/3}}\) |
\(\Big \downarrow \) 1921 |
\(\displaystyle \frac {12 \left (\frac {9 \int \frac {1}{x^6 \left (b x^6+a x^3\right )^{2/3}}dx}{2 a}+\frac {1}{2 a x^5 \left (a x^3+b x^6\right )^{2/3}}\right )}{5 a}+\frac {1}{5 a x^2 \left (a x^3+b x^6\right )^{5/3}}\) |
\(\Big \downarrow \) 1922 |
\(\displaystyle \frac {12 \left (\frac {9 \left (-\frac {6 b \int \frac {1}{x^3 \left (b x^6+a x^3\right )^{2/3}}dx}{7 a}-\frac {\sqrt [3]{a x^3+b x^6}}{7 a x^8}\right )}{2 a}+\frac {1}{2 a x^5 \left (a x^3+b x^6\right )^{2/3}}\right )}{5 a}+\frac {1}{5 a x^2 \left (a x^3+b x^6\right )^{5/3}}\) |
\(\Big \downarrow \) 1922 |
\(\displaystyle \frac {12 \left (\frac {9 \left (-\frac {6 b \left (-\frac {3 b \int \frac {1}{\left (b x^6+a x^3\right )^{2/3}}dx}{4 a}-\frac {\sqrt [3]{a x^3+b x^6}}{4 a x^5}\right )}{7 a}-\frac {\sqrt [3]{a x^3+b x^6}}{7 a x^8}\right )}{2 a}+\frac {1}{2 a x^5 \left (a x^3+b x^6\right )^{2/3}}\right )}{5 a}+\frac {1}{5 a x^2 \left (a x^3+b x^6\right )^{5/3}}\) |
\(\Big \downarrow \) 1906 |
\(\displaystyle \frac {12 \left (\frac {9 \left (-\frac {6 b \left (\frac {3 b \sqrt [3]{a x^3+b x^6}}{4 a^2 x^2}-\frac {\sqrt [3]{a x^3+b x^6}}{4 a x^5}\right )}{7 a}-\frac {\sqrt [3]{a x^3+b x^6}}{7 a x^8}\right )}{2 a}+\frac {1}{2 a x^5 \left (a x^3+b x^6\right )^{2/3}}\right )}{5 a}+\frac {1}{5 a x^2 \left (a x^3+b x^6\right )^{5/3}}\) |
Input:
Int[(a*x^3 + b*x^6)^(-8/3),x]
Output:
1/(5*a*x^2*(a*x^3 + b*x^6)^(5/3)) + (12*(1/(2*a*x^5*(a*x^3 + b*x^6)^(2/3)) + (9*(-1/7*(a*x^3 + b*x^6)^(1/3)/(a*x^8) - (6*b*(-1/4*(a*x^3 + b*x^6)^(1/ 3)/(a*x^5) + (3*b*(a*x^3 + b*x^6)^(1/3))/(4*a^2*x^2)))/(7*a)))/(2*a)))/(5* a)
Int[((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(a*x^j + b*x^n)^(p + 1)/(b*(n - j)*(p + 1)*x^(n - 1)), x] /; FreeQ[{a, b, j, n, p}, x] && !IntegerQ[p] && NeQ[n, j] && EqQ[j*p - n + j + 1, 0]
Int[((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[-(a*x^j + b*x^n)^(p + 1)/(a*(n - j)*(p + 1)*x^(j - 1)), x] + Simp[(n*p + n - j + 1)/ (a*(n - j)*(p + 1)) Int[(a*x^j + b*x^n)^(p + 1)/x^j, x], x] /; FreeQ[{a, b, j, n}, x] && !IntegerQ[p] && NeQ[n, j] && ILtQ[Simplify[(n*p + n - j + 1)/(n - j)], 0] && LtQ[p, -1]
Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol ] :> Simp[(-c^(j - 1))*(c*x)^(m - j + 1)*((a*x^j + b*x^n)^(p + 1)/(a*(n - j )*(p + 1))), x] + Simp[c^j*((m + n*p + n - j + 1)/(a*(n - j)*(p + 1))) In t[(c*x)^(m - j)*(a*x^j + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, c, j, m, n} , x] && !IntegerQ[p] && NeQ[n, j] && ILtQ[Simplify[(m + n*p + n - j + 1)/( n - j)], 0] && LtQ[p, -1] && (IntegerQ[j] || GtQ[c, 0])
Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol ] :> Simp[c^(j - 1)*(c*x)^(m - j + 1)*((a*x^j + b*x^n)^(p + 1)/(a*(m + j*p + 1))), x] - Simp[b*((m + n*p + n - j + 1)/(a*c^(n - j)*(m + j*p + 1))) I nt[(c*x)^(m + n - j)*(a*x^j + b*x^n)^p, x], x] /; FreeQ[{a, b, c, j, m, n, p}, x] && !IntegerQ[p] && NeQ[n, j] && ILtQ[Simplify[(m + n*p + n - j + 1) /(n - j)], 0] && NeQ[m + j*p + 1, 0] && (IntegersQ[j, n] || GtQ[c, 0])
Time = 0.24 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.52
method | result | size |
gosper | \(-\frac {x \left (b \,x^{3}+a \right ) \left (243 b^{4} x^{12}+405 a \,b^{3} x^{9}+135 b^{2} x^{6} a^{2}-15 b \,x^{3} a^{3}+5 a^{4}\right )}{35 a^{5} \left (b \,x^{6}+a \,x^{3}\right )^{\frac {8}{3}}}\) | \(70\) |
orering | \(-\frac {x \left (b \,x^{3}+a \right ) \left (243 b^{4} x^{12}+405 a \,b^{3} x^{9}+135 b^{2} x^{6} a^{2}-15 b \,x^{3} a^{3}+5 a^{4}\right )}{35 a^{5} \left (b \,x^{6}+a \,x^{3}\right )^{\frac {8}{3}}}\) | \(70\) |
pseudoelliptic | \(-\frac {\frac {243}{5} b^{4} x^{12}+81 a \,b^{3} x^{9}+27 b^{2} x^{6} a^{2}-3 b \,x^{3} a^{3}+a^{4}}{7 \left (x^{3} \left (b \,x^{3}+a \right )\right )^{\frac {2}{3}} x^{5} \left (b \,x^{3}+a \right ) a^{5}}\) | \(72\) |
trager | \(-\frac {\left (243 b^{4} x^{12}+405 a \,b^{3} x^{9}+135 b^{2} x^{6} a^{2}-15 b \,x^{3} a^{3}+5 a^{4}\right ) \left (b \,x^{6}+a \,x^{3}\right )^{\frac {1}{3}}}{35 x^{8} a^{5} \left (b \,x^{3}+a \right )^{2}}\) | \(74\) |
risch | \(-\frac {\left (b \,x^{3}+a \right ) \left (36 b^{2} x^{6}-5 a \,x^{3} b +a^{2}\right )}{7 a^{5} x^{5} \left (x^{3} \left (b \,x^{3}+a \right )\right )^{\frac {2}{3}}}-\frac {b^{3} x^{4} \left (9 b \,x^{3}+10 a \right )}{5 \left (b \,x^{3}+a \right ) a^{5} \left (x^{3} \left (b \,x^{3}+a \right )\right )^{\frac {2}{3}}}\) | \(92\) |
Input:
int(1/(b*x^6+a*x^3)^(8/3),x,method=_RETURNVERBOSE)
Output:
-1/35*x*(b*x^3+a)*(243*b^4*x^12+405*a*b^3*x^9+135*a^2*b^2*x^6-15*a^3*b*x^3 +5*a^4)/a^5/(b*x^6+a*x^3)^(8/3)
Time = 0.08 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.65 \[ \int \frac {1}{\left (a x^3+b x^6\right )^{8/3}} \, dx=-\frac {{\left (243 \, b^{4} x^{12} + 405 \, a b^{3} x^{9} + 135 \, a^{2} b^{2} x^{6} - 15 \, a^{3} b x^{3} + 5 \, a^{4}\right )} {\left (b x^{6} + a x^{3}\right )}^{\frac {1}{3}}}{35 \, {\left (a^{5} b^{2} x^{14} + 2 \, a^{6} b x^{11} + a^{7} x^{8}\right )}} \] Input:
integrate(1/(b*x^6+a*x^3)^(8/3),x, algorithm="fricas")
Output:
-1/35*(243*b^4*x^12 + 405*a*b^3*x^9 + 135*a^2*b^2*x^6 - 15*a^3*b*x^3 + 5*a ^4)*(b*x^6 + a*x^3)^(1/3)/(a^5*b^2*x^14 + 2*a^6*b*x^11 + a^7*x^8)
\[ \int \frac {1}{\left (a x^3+b x^6\right )^{8/3}} \, dx=\int \frac {1}{\left (a x^{3} + b x^{6}\right )^{\frac {8}{3}}}\, dx \] Input:
integrate(1/(b*x**6+a*x**3)**(8/3),x)
Output:
Integral((a*x**3 + b*x**6)**(-8/3), x)
Time = 0.04 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.54 \[ \int \frac {1}{\left (a x^3+b x^6\right )^{8/3}} \, dx=-\frac {243 \, b^{4} x^{12} + 405 \, a b^{3} x^{9} + 135 \, a^{2} b^{2} x^{6} - 15 \, a^{3} b x^{3} + 5 \, a^{4}}{35 \, {\left (a^{5} b x^{10} + a^{6} x^{7}\right )} {\left (b x^{3} + a\right )}^{\frac {2}{3}}} \] Input:
integrate(1/(b*x^6+a*x^3)^(8/3),x, algorithm="maxima")
Output:
-1/35*(243*b^4*x^12 + 405*a*b^3*x^9 + 135*a^2*b^2*x^6 - 15*a^3*b*x^3 + 5*a ^4)/((a^5*b*x^10 + a^6*x^7)*(b*x^3 + a)^(2/3))
Time = 0.13 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.66 \[ \int \frac {1}{\left (a x^3+b x^6\right )^{8/3}} \, dx=-\frac {\frac {7 \, {\left (10 \, {\left (b + \frac {a}{x^{3}}\right )} b^{3} - b^{4}\right )}}{a {\left (b + \frac {a}{x^{3}}\right )}^{\frac {5}{3}}} + \frac {5 \, {\left (a^{6} {\left (b + \frac {a}{x^{3}}\right )}^{\frac {7}{3}} - 7 \, a^{6} {\left (b + \frac {a}{x^{3}}\right )}^{\frac {4}{3}} b + 42 \, a^{6} {\left (b + \frac {a}{x^{3}}\right )}^{\frac {1}{3}} b^{2}\right )}}{a^{7}}}{35 \, a^{4}} \] Input:
integrate(1/(b*x^6+a*x^3)^(8/3),x, algorithm="giac")
Output:
-1/35*(7*(10*(b + a/x^3)*b^3 - b^4)/(a*(b + a/x^3)^(5/3)) + 5*(a^6*(b + a/ x^3)^(7/3) - 7*a^6*(b + a/x^3)^(4/3)*b + 42*a^6*(b + a/x^3)^(1/3)*b^2)/a^7 )/a^4
Time = 20.41 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.93 \[ \int \frac {1}{\left (a x^3+b x^6\right )^{8/3}} \, dx=\frac {5\,b\,{\left (b\,x^6+a\,x^3\right )}^{1/3}}{7\,a^4\,x^5}-\frac {{\left (b\,x^6+a\,x^3\right )}^{1/3}}{7\,a^3\,x^8}-\frac {{\left (b\,x^6+a\,x^3\right )}^{1/3}\,\left (\frac {187\,b^2}{35\,a^4}+\frac {243\,b^3\,x^3}{35\,a^5}\right )}{x^2\,\left (b\,x^3+a\right )}+\frac {b^2\,{\left (b\,x^6+a\,x^3\right )}^{1/3}}{5\,a^3\,x^2\,{\left (b\,x^3+a\right )}^2} \] Input:
int(1/(a*x^3 + b*x^6)^(8/3),x)
Output:
(5*b*(a*x^3 + b*x^6)^(1/3))/(7*a^4*x^5) - (a*x^3 + b*x^6)^(1/3)/(7*a^3*x^8 ) - ((a*x^3 + b*x^6)^(1/3)*((187*b^2)/(35*a^4) + (243*b^3*x^3)/(35*a^5)))/ (x^2*(a + b*x^3)) + (b^2*(a*x^3 + b*x^6)^(1/3))/(5*a^3*x^2*(a + b*x^3)^2)
\[ \int \frac {1}{\left (a x^3+b x^6\right )^{8/3}} \, dx=\int \frac {1}{\left (b \,x^{3}+a \right )^{\frac {2}{3}} a^{2} x^{8}+2 \left (b \,x^{3}+a \right )^{\frac {2}{3}} a b \,x^{11}+\left (b \,x^{3}+a \right )^{\frac {2}{3}} b^{2} x^{14}}d x \] Input:
int(1/(b*x^6+a*x^3)^(8/3),x)
Output:
int(1/((a + b*x**3)**(2/3)*a**2*x**8 + 2*(a + b*x**3)**(2/3)*a*b*x**11 + ( a + b*x**3)**(2/3)*b**2*x**14),x)