Integrand size = 28, antiderivative size = 174 \[ \int \frac {x^8 \sqrt [3]{a+b x^3}}{a d-b d x^3} \, dx=-\frac {a^2 \sqrt [3]{a+b x^3}}{b^3 d}-\frac {\left (a+b x^3\right )^{7/3}}{7 b^3 d}+\frac {\sqrt [3]{2} a^{7/3} \arctan \left (\frac {\sqrt [3]{a}+2^{2/3} \sqrt [3]{a+b x^3}}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt {3} b^3 d}+\frac {a^{7/3} \log \left (a-b x^3\right )}{3\ 2^{2/3} b^3 d}-\frac {a^{7/3} \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+b x^3}\right )}{2^{2/3} b^3 d} \] Output:
-a^2*(b*x^3+a)^(1/3)/b^3/d-1/7*(b*x^3+a)^(7/3)/b^3/d+1/3*2^(1/3)*a^(7/3)*a rctan(1/3*(a^(1/3)+2^(2/3)*(b*x^3+a)^(1/3))*3^(1/2)/a^(1/3))*3^(1/2)/b^3/d +1/6*a^(7/3)*ln(-b*x^3+a)*2^(1/3)/b^3/d-1/2*a^(7/3)*ln(2^(1/3)*a^(1/3)-(b* x^3+a)^(1/3))*2^(1/3)/b^3/d
Time = 0.19 (sec) , antiderivative size = 210, normalized size of antiderivative = 1.21 \[ \int \frac {x^8 \sqrt [3]{a+b x^3}}{a d-b d x^3} \, dx=-\frac {48 a^2 \sqrt [3]{a+b x^3}+12 a b x^3 \sqrt [3]{a+b x^3}+6 b^2 x^6 \sqrt [3]{a+b x^3}-14 \sqrt [3]{2} \sqrt {3} a^{7/3} \arctan \left (\frac {1+\frac {2^{2/3} \sqrt [3]{a+b x^3}}{\sqrt [3]{a}}}{\sqrt {3}}\right )+14 \sqrt [3]{2} a^{7/3} \log \left (-2 \sqrt [3]{a}+2^{2/3} \sqrt [3]{a+b x^3}\right )-7 \sqrt [3]{2} a^{7/3} \log \left (2 a^{2/3}+2^{2/3} \sqrt [3]{a} \sqrt [3]{a+b x^3}+\sqrt [3]{2} \left (a+b x^3\right )^{2/3}\right )}{42 b^3 d} \] Input:
Integrate[(x^8*(a + b*x^3)^(1/3))/(a*d - b*d*x^3),x]
Output:
-1/42*(48*a^2*(a + b*x^3)^(1/3) + 12*a*b*x^3*(a + b*x^3)^(1/3) + 6*b^2*x^6 *(a + b*x^3)^(1/3) - 14*2^(1/3)*Sqrt[3]*a^(7/3)*ArcTan[(1 + (2^(2/3)*(a + b*x^3)^(1/3))/a^(1/3))/Sqrt[3]] + 14*2^(1/3)*a^(7/3)*Log[-2*a^(1/3) + 2^(2 /3)*(a + b*x^3)^(1/3)] - 7*2^(1/3)*a^(7/3)*Log[2*a^(2/3) + 2^(2/3)*a^(1/3) *(a + b*x^3)^(1/3) + 2^(1/3)*(a + b*x^3)^(2/3)])/(b^3*d)
Time = 0.55 (sec) , antiderivative size = 163, normalized size of antiderivative = 0.94, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {948, 27, 99, 2009}
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.
| \(\displaystyle \int \frac {x^8 \sqrt [3]{a+b x^3}}{a d-b d x^3} \, dx\) |
| \(\Big \downarrow \) 948 |
| \(\displaystyle \frac {1}{3} \int \frac {x^6 \sqrt [3]{b x^3+a}}{d \left (a-b x^3\right )}dx^3\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {x^6 \sqrt [3]{b x^3+a}}{a-b x^3}dx^3}{3 d}\) |
| \(\Big \downarrow \) 99 |
| \(\displaystyle \frac {\int \left (\frac {a^2 \sqrt [3]{b x^3+a}}{b^2 \left (a-b x^3\right )}-\frac {\left (b x^3+a\right )^{4/3}}{b^2}\right )dx^3}{3 d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\frac {\sqrt [3]{2} \sqrt {3} a^{7/3} \arctan \left (\frac {2^{2/3} \sqrt [3]{a+b x^3}+\sqrt [3]{a}}{\sqrt {3} \sqrt [3]{a}}\right )}{b^3}+\frac {a^{7/3} \log \left (a-b x^3\right )}{2^{2/3} b^3}-\frac {3 a^{7/3} \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+b x^3}\right )}{2^{2/3} b^3}-\frac {3 a^2 \sqrt [3]{a+b x^3}}{b^3}-\frac {3 \left (a+b x^3\right )^{7/3}}{7 b^3}}{3 d}\) |
Input:
Int[(x^8*(a + b*x^3)^(1/3))/(a*d - b*d*x^3),x]
Output:
((-3*a^2*(a + b*x^3)^(1/3))/b^3 - (3*(a + b*x^3)^(7/3))/(7*b^3) + (2^(1/3) *Sqrt[3]*a^(7/3)*ArcTan[(a^(1/3) + 2^(2/3)*(a + b*x^3)^(1/3))/(Sqrt[3]*a^( 1/3))])/b^3 + (a^(7/3)*Log[a - b*x^3])/(2^(2/3)*b^3) - (3*a^(7/3)*Log[2^(1 /3)*a^(1/3) - (a + b*x^3)^(1/3)])/(2^(2/3)*b^3))/(3*d)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] && (IntegerQ[p] | | (GtQ[m, 0] && GeQ[n, -1]))
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_. ), x_Symbol] :> Simp[1/n Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^ p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] && NeQ [b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]
Time = 2.02 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.78
| method | result | size |
| pseudoelliptic | \(\frac {\left (-6 b^{2} x^{6}-12 a b \,x^{3}-48 a^{2}\right ) \left (b \,x^{3}+a \right )^{\frac {1}{3}}+7 a^{\frac {7}{3}} \left (2 \arctan \left (\frac {\left (a^{\frac {1}{3}}+2^{\frac {2}{3}} \left (b \,x^{3}+a \right )^{\frac {1}{3}}\right ) \sqrt {3}}{3 a^{\frac {1}{3}}}\right ) \sqrt {3}+\ln \left (\left (b \,x^{3}+a \right )^{\frac {2}{3}}+2^{\frac {1}{3}} a^{\frac {1}{3}} \left (b \,x^{3}+a \right )^{\frac {1}{3}}+2^{\frac {2}{3}} a^{\frac {2}{3}}\right )-2 \ln \left (\left (b \,x^{3}+a \right )^{\frac {1}{3}}-2^{\frac {1}{3}} a^{\frac {1}{3}}\right )\right ) 2^{\frac {1}{3}}}{42 b^{3} d}\) | \(136\) |
Input:
int(x^8*(b*x^3+a)^(1/3)/(-b*d*x^3+a*d),x,method=_RETURNVERBOSE)
Output:
1/42*((-6*b^2*x^6-12*a*b*x^3-48*a^2)*(b*x^3+a)^(1/3)+7*a^(7/3)*(2*arctan(1 /3*(a^(1/3)+2^(2/3)*(b*x^3+a)^(1/3))*3^(1/2)/a^(1/3))*3^(1/2)+ln((b*x^3+a) ^(2/3)+2^(1/3)*a^(1/3)*(b*x^3+a)^(1/3)+2^(2/3)*a^(2/3))-2*ln((b*x^3+a)^(1/ 3)-2^(1/3)*a^(1/3)))*2^(1/3))/b^3/d
Time = 0.09 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.00 \[ \int \frac {x^8 \sqrt [3]{a+b x^3}}{a d-b d x^3} \, dx=-\frac {14 \, \sqrt {3} 2^{\frac {1}{3}} \left (-a\right )^{\frac {1}{3}} a^{2} \arctan \left (\frac {\sqrt {3} 2^{\frac {2}{3}} {\left (b x^{3} + a\right )}^{\frac {1}{3}} \left (-a\right )^{\frac {2}{3}} + \sqrt {3} a}{3 \, a}\right ) + 7 \cdot 2^{\frac {1}{3}} \left (-a\right )^{\frac {1}{3}} a^{2} \log \left (2^{\frac {2}{3}} \left (-a\right )^{\frac {2}{3}} - 2^{\frac {1}{3}} {\left (b x^{3} + a\right )}^{\frac {1}{3}} \left (-a\right )^{\frac {1}{3}} + {\left (b x^{3} + a\right )}^{\frac {2}{3}}\right ) - 14 \cdot 2^{\frac {1}{3}} \left (-a\right )^{\frac {1}{3}} a^{2} \log \left (2^{\frac {1}{3}} \left (-a\right )^{\frac {1}{3}} + {\left (b x^{3} + a\right )}^{\frac {1}{3}}\right ) + 6 \, {\left (b^{2} x^{6} + 2 \, a b x^{3} + 8 \, a^{2}\right )} {\left (b x^{3} + a\right )}^{\frac {1}{3}}}{42 \, b^{3} d} \] Input:
integrate(x^8*(b*x^3+a)^(1/3)/(-b*d*x^3+a*d),x, algorithm="fricas")
Output:
-1/42*(14*sqrt(3)*2^(1/3)*(-a)^(1/3)*a^2*arctan(1/3*(sqrt(3)*2^(2/3)*(b*x^ 3 + a)^(1/3)*(-a)^(2/3) + sqrt(3)*a)/a) + 7*2^(1/3)*(-a)^(1/3)*a^2*log(2^( 2/3)*(-a)^(2/3) - 2^(1/3)*(b*x^3 + a)^(1/3)*(-a)^(1/3) + (b*x^3 + a)^(2/3) ) - 14*2^(1/3)*(-a)^(1/3)*a^2*log(2^(1/3)*(-a)^(1/3) + (b*x^3 + a)^(1/3)) + 6*(b^2*x^6 + 2*a*b*x^3 + 8*a^2)*(b*x^3 + a)^(1/3))/(b^3*d)
\[ \int \frac {x^8 \sqrt [3]{a+b x^3}}{a d-b d x^3} \, dx=- \frac {\int \frac {x^{8} \sqrt [3]{a + b x^{3}}}{- a + b x^{3}}\, dx}{d} \] Input:
integrate(x**8*(b*x**3+a)**(1/3)/(-b*d*x**3+a*d),x)
Output:
-Integral(x**8*(a + b*x**3)**(1/3)/(-a + b*x**3), x)/d
Time = 0.11 (sec) , antiderivative size = 155, normalized size of antiderivative = 0.89 \[ \int \frac {x^8 \sqrt [3]{a+b x^3}}{a d-b d x^3} \, dx=\frac {\frac {14 \, \sqrt {3} 2^{\frac {1}{3}} a^{\frac {7}{3}} \arctan \left (\frac {\sqrt {3} 2^{\frac {2}{3}} {\left (2^{\frac {1}{3}} a^{\frac {1}{3}} + 2 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}}\right )}}{6 \, a^{\frac {1}{3}}}\right )}{d} + \frac {7 \cdot 2^{\frac {1}{3}} a^{\frac {7}{3}} \log \left (2^{\frac {2}{3}} a^{\frac {2}{3}} + 2^{\frac {1}{3}} {\left (b x^{3} + a\right )}^{\frac {1}{3}} a^{\frac {1}{3}} + {\left (b x^{3} + a\right )}^{\frac {2}{3}}\right )}{d} - \frac {14 \cdot 2^{\frac {1}{3}} a^{\frac {7}{3}} \log \left (-2^{\frac {1}{3}} a^{\frac {1}{3}} + {\left (b x^{3} + a\right )}^{\frac {1}{3}}\right )}{d} - \frac {6 \, {\left ({\left (b x^{3} + a\right )}^{\frac {7}{3}} + 7 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}} a^{2}\right )}}{d}}{42 \, b^{3}} \] Input:
integrate(x^8*(b*x^3+a)^(1/3)/(-b*d*x^3+a*d),x, algorithm="maxima")
Output:
1/42*(14*sqrt(3)*2^(1/3)*a^(7/3)*arctan(1/6*sqrt(3)*2^(2/3)*(2^(1/3)*a^(1/ 3) + 2*(b*x^3 + a)^(1/3))/a^(1/3))/d + 7*2^(1/3)*a^(7/3)*log(2^(2/3)*a^(2/ 3) + 2^(1/3)*(b*x^3 + a)^(1/3)*a^(1/3) + (b*x^3 + a)^(2/3))/d - 14*2^(1/3) *a^(7/3)*log(-2^(1/3)*a^(1/3) + (b*x^3 + a)^(1/3))/d - 6*((b*x^3 + a)^(7/3 ) + 7*(b*x^3 + a)^(1/3)*a^2)/d)/b^3
Time = 0.58 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.01 \[ \int \frac {x^8 \sqrt [3]{a+b x^3}}{a d-b d x^3} \, dx=\frac {\sqrt {3} 2^{\frac {1}{3}} a^{\frac {7}{3}} \arctan \left (\frac {\sqrt {3} 2^{\frac {2}{3}} {\left (2^{\frac {1}{3}} a^{\frac {1}{3}} + 2 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}}\right )}}{6 \, a^{\frac {1}{3}}}\right )}{3 \, b^{3} d} + \frac {2^{\frac {1}{3}} a^{\frac {7}{3}} \log \left (2^{\frac {2}{3}} a^{\frac {2}{3}} + 2^{\frac {1}{3}} {\left (b x^{3} + a\right )}^{\frac {1}{3}} a^{\frac {1}{3}} + {\left (b x^{3} + a\right )}^{\frac {2}{3}}\right )}{6 \, b^{3} d} - \frac {2^{\frac {1}{3}} a^{\frac {7}{3}} \log \left ({\left | -2^{\frac {1}{3}} a^{\frac {1}{3}} + {\left (b x^{3} + a\right )}^{\frac {1}{3}} \right |}\right )}{3 \, b^{3} d} - \frac {{\left (b x^{3} + a\right )}^{\frac {7}{3}} b^{18} d^{6} + 7 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}} a^{2} b^{18} d^{6}}{7 \, b^{21} d^{7}} \] Input:
integrate(x^8*(b*x^3+a)^(1/3)/(-b*d*x^3+a*d),x, algorithm="giac")
Output:
1/3*sqrt(3)*2^(1/3)*a^(7/3)*arctan(1/6*sqrt(3)*2^(2/3)*(2^(1/3)*a^(1/3) + 2*(b*x^3 + a)^(1/3))/a^(1/3))/(b^3*d) + 1/6*2^(1/3)*a^(7/3)*log(2^(2/3)*a^ (2/3) + 2^(1/3)*(b*x^3 + a)^(1/3)*a^(1/3) + (b*x^3 + a)^(2/3))/(b^3*d) - 1 /3*2^(1/3)*a^(7/3)*log(abs(-2^(1/3)*a^(1/3) + (b*x^3 + a)^(1/3)))/(b^3*d) - 1/7*((b*x^3 + a)^(7/3)*b^18*d^6 + 7*(b*x^3 + a)^(1/3)*a^2*b^18*d^6)/(b^2 1*d^7)
Time = 3.45 (sec) , antiderivative size = 219, normalized size of antiderivative = 1.26 \[ \int \frac {x^8 \sqrt [3]{a+b x^3}}{a d-b d x^3} \, dx=\frac {2^{1/3}\,{\left (-a\right )}^{7/3}\,\ln \left (6\,a^3\,{\left (b\,x^3+a\right )}^{1/3}-6\,2^{1/3}\,{\left (-a\right )}^{10/3}\right )}{3\,b^3\,d}-\frac {a^2\,{\left (b\,x^3+a\right )}^{1/3}}{b^3\,d}-\frac {{\left (b\,x^3+a\right )}^{7/3}}{7\,b^3\,d}-\frac {2^{1/3}\,{\left (-a\right )}^{7/3}\,\ln \left (\frac {6\,a^3\,{\left (b\,x^3+a\right )}^{1/3}}{b^3\,d}+\frac {6\,2^{1/3}\,{\left (-a\right )}^{10/3}\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{b^3\,d}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{3\,b^3\,d}+\frac {2^{1/3}\,{\left (-a\right )}^{7/3}\,\ln \left (\frac {6\,a^3\,{\left (b\,x^3+a\right )}^{1/3}}{b^3\,d}-\frac {18\,2^{1/3}\,{\left (-a\right )}^{10/3}\,\left (-\frac {1}{6}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right )}{b^3\,d}\right )\,\left (-\frac {1}{6}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right )}{b^3\,d} \] Input:
int((x^8*(a + b*x^3)^(1/3))/(a*d - b*d*x^3),x)
Output:
(2^(1/3)*(-a)^(7/3)*log(6*a^3*(a + b*x^3)^(1/3) - 6*2^(1/3)*(-a)^(10/3)))/ (3*b^3*d) - (a^2*(a + b*x^3)^(1/3))/(b^3*d) - (a + b*x^3)^(7/3)/(7*b^3*d) - (2^(1/3)*(-a)^(7/3)*log((6*a^3*(a + b*x^3)^(1/3))/(b^3*d) + (6*2^(1/3)*( -a)^(10/3)*((3^(1/2)*1i)/2 + 1/2))/(b^3*d))*((3^(1/2)*1i)/2 + 1/2))/(3*b^3 *d) + (2^(1/3)*(-a)^(7/3)*log((6*a^3*(a + b*x^3)^(1/3))/(b^3*d) - (18*2^(1 /3)*(-a)^(10/3)*((3^(1/2)*1i)/6 - 1/6))/(b^3*d))*((3^(1/2)*1i)/6 - 1/6))/( b^3*d)
\[ \int \frac {x^8 \sqrt [3]{a+b x^3}}{a d-b d x^3} \, dx=\frac {6 \left (b \,x^{3}+a \right )^{\frac {1}{3}} a^{2}-2 \left (b \,x^{3}+a \right )^{\frac {1}{3}} a b \,x^{3}-\left (b \,x^{3}+a \right )^{\frac {1}{3}} b^{2} x^{6}+14 \left (\int \frac {\left (b \,x^{3}+a \right )^{\frac {1}{3}} x^{5}}{-b^{2} x^{6}+a^{2}}d x \right ) a^{2} b^{2}}{7 b^{3} d} \] Input:
int(x^8*(b*x^3+a)^(1/3)/(-b*d*x^3+a*d),x)
Output:
(6*(a + b*x**3)**(1/3)*a**2 - 2*(a + b*x**3)**(1/3)*a*b*x**3 - (a + b*x**3 )**(1/3)*b**2*x**6 + 14*int(((a + b*x**3)**(1/3)*x**5)/(a**2 - b**2*x**6), x)*a**2*b**2)/(7*b**3*d)