Integrand size = 24, antiderivative size = 203 \[ \int \frac {x^8}{\sqrt [3]{a+b x^3} \left (c+d x^3\right )} \, dx=-\frac {(b c+a d) \left (a+b x^3\right )^{2/3}}{2 b^2 d^2}+\frac {\left (a+b x^3\right )^{5/3}}{5 b^2 d}-\frac {c^2 \arctan \left (\frac {1-\frac {2 \sqrt [3]{d} \sqrt [3]{a+b x^3}}{\sqrt [3]{b c-a d}}}{\sqrt {3}}\right )}{\sqrt {3} d^{8/3} \sqrt [3]{b c-a d}}+\frac {c^2 \log \left (c+d x^3\right )}{6 d^{8/3} \sqrt [3]{b c-a d}}-\frac {c^2 \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{2 d^{8/3} \sqrt [3]{b c-a d}} \]
-1/2*(a*d+b*c)*(b*x^3+a)^(2/3)/b^2/d^2+1/5*(b*x^3+a)^(5/3)/b^2/d+1/6*c^2*l n(d*x^3+c)/d^(8/3)/(-a*d+b*c)^(1/3)-1/2*c^2*ln((-a*d+b*c)^(1/3)+d^(1/3)*(b *x^3+a)^(1/3))/d^(8/3)/(-a*d+b*c)^(1/3)-1/3*c^2*arctan(1/3*(1-2*d^(1/3)*(b *x^3+a)^(1/3)/(-a*d+b*c)^(1/3))*3^(1/2))/d^(8/3)/(-a*d+b*c)^(1/3)*3^(1/2)
Time = 0.52 (sec) , antiderivative size = 231, normalized size of antiderivative = 1.14 \[ \int \frac {x^8}{\sqrt [3]{a+b x^3} \left (c+d x^3\right )} \, dx=\frac {-3 d^{2/3} \sqrt [3]{b c-a d} \left (a+b x^3\right )^{2/3} \left (5 b c+3 a d-2 b d x^3\right )-10 \sqrt {3} b^2 c^2 \arctan \left (\frac {1-\frac {2 \sqrt [3]{d} \sqrt [3]{a+b x^3}}{\sqrt [3]{b c-a d}}}{\sqrt {3}}\right )-10 b^2 c^2 \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )+5 b^2 c^2 \log \left ((b c-a d)^{2/3}-\sqrt [3]{d} \sqrt [3]{b c-a d} \sqrt [3]{a+b x^3}+d^{2/3} \left (a+b x^3\right )^{2/3}\right )}{30 b^2 d^{8/3} \sqrt [3]{b c-a d}} \]
(-3*d^(2/3)*(b*c - a*d)^(1/3)*(a + b*x^3)^(2/3)*(5*b*c + 3*a*d - 2*b*d*x^3 ) - 10*Sqrt[3]*b^2*c^2*ArcTan[(1 - (2*d^(1/3)*(a + b*x^3)^(1/3))/(b*c - a* d)^(1/3))/Sqrt[3]] - 10*b^2*c^2*Log[(b*c - a*d)^(1/3) + d^(1/3)*(a + b*x^3 )^(1/3)] + 5*b^2*c^2*Log[(b*c - a*d)^(2/3) - d^(1/3)*(b*c - a*d)^(1/3)*(a + b*x^3)^(1/3) + d^(2/3)*(a + b*x^3)^(2/3)])/(30*b^2*d^(8/3)*(b*c - a*d)^( 1/3))
Time = 0.37 (sec) , antiderivative size = 207, normalized size of antiderivative = 1.02, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {948, 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.
| \(\displaystyle \int \frac {x^8}{\sqrt [3]{a+b x^3} \left (c+d x^3\right )} \, dx\) |
| \(\Big \downarrow \) 948 |
| \(\displaystyle \frac {1}{3} \int \frac {x^6}{\sqrt [3]{b x^3+a} \left (d x^3+c\right )}dx^3\) |
| \(\Big \downarrow \) 99 |
| \(\displaystyle \frac {1}{3} \int \left (\frac {c^2}{d^2 \sqrt [3]{b x^3+a} \left (d x^3+c\right )}+\frac {\left (b x^3+a\right )^{2/3}}{b d}+\frac {-b c-a d}{b d^2 \sqrt [3]{b x^3+a}}\right )dx^3\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {1}{3} \left (-\frac {\sqrt {3} c^2 \arctan \left (\frac {1-\frac {2 \sqrt [3]{d} \sqrt [3]{a+b x^3}}{\sqrt [3]{b c-a d}}}{\sqrt {3}}\right )}{d^{8/3} \sqrt [3]{b c-a d}}-\frac {3 \left (a+b x^3\right )^{2/3} (a d+b c)}{2 b^2 d^2}+\frac {3 \left (a+b x^3\right )^{5/3}}{5 b^2 d}+\frac {c^2 \log \left (c+d x^3\right )}{2 d^{8/3} \sqrt [3]{b c-a d}}-\frac {3 c^2 \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{2 d^{8/3} \sqrt [3]{b c-a d}}\right )\) |
((-3*(b*c + a*d)*(a + b*x^3)^(2/3))/(2*b^2*d^2) + (3*(a + b*x^3)^(5/3))/(5 *b^2*d) - (Sqrt[3]*c^2*ArcTan[(1 - (2*d^(1/3)*(a + b*x^3)^(1/3))/(b*c - a* d)^(1/3))/Sqrt[3]])/(d^(8/3)*(b*c - a*d)^(1/3)) + (c^2*Log[c + d*x^3])/(2* d^(8/3)*(b*c - a*d)^(1/3)) - (3*c^2*Log[(b*c - a*d)^(1/3) + d^(1/3)*(a + b *x^3)^(1/3)])/(2*d^(8/3)*(b*c - a*d)^(1/3)))/3
3.8.16.3.1 Defintions of rubi rules used
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 = 5.06 (sec) , antiderivative size = 204, normalized size of antiderivative = 1.00
| method | result | size |
| pseudoelliptic | \(-\frac {\frac {9 \left (\frac {a d -b c}{d}\right )^{\frac {1}{3}} d \left (\frac {\left (-2 d \,x^{3}+5 c \right ) b}{3}+a d \right ) \left (b \,x^{3}+a \right )^{\frac {2}{3}}}{5}+b^{2} c^{2} \left (-2 \arctan \left (\frac {\sqrt {3}\, \left (2 \left (b \,x^{3}+a \right )^{\frac {1}{3}}+\left (\frac {a d -b c}{d}\right )^{\frac {1}{3}}\right )}{3 \left (\frac {a d -b c}{d}\right )^{\frac {1}{3}}}\right ) \sqrt {3}+\ln \left (\left (b \,x^{3}+a \right )^{\frac {2}{3}}+\left (\frac {a d -b c}{d}\right )^{\frac {1}{3}} \left (b \,x^{3}+a \right )^{\frac {1}{3}}+\left (\frac {a d -b c}{d}\right )^{\frac {2}{3}}\right )-2 \ln \left (\left (b \,x^{3}+a \right )^{\frac {1}{3}}-\left (\frac {a d -b c}{d}\right )^{\frac {1}{3}}\right )\right )}{6 \left (\frac {a d -b c}{d}\right )^{\frac {1}{3}} b^{2} d^{3}}\) | \(204\) |
-1/6*(9/5*(1/d*(a*d-b*c))^(1/3)*d*(1/3*(-2*d*x^3+5*c)*b+a*d)*(b*x^3+a)^(2/ 3)+b^2*c^2*(-2*arctan(1/3*3^(1/2)*(2*(b*x^3+a)^(1/3)+(1/d*(a*d-b*c))^(1/3) )/(1/d*(a*d-b*c))^(1/3))*3^(1/2)+ln((b*x^3+a)^(2/3)+(1/d*(a*d-b*c))^(1/3)* (b*x^3+a)^(1/3)+(1/d*(a*d-b*c))^(2/3))-2*ln((b*x^3+a)^(1/3)-(1/d*(a*d-b*c) )^(1/3))))/(1/d*(a*d-b*c))^(1/3)/b^2/d^3
Leaf count of result is larger than twice the leaf count of optimal. 335 vs. \(2 (164) = 328\).
Time = 0.32 (sec) , antiderivative size = 768, normalized size of antiderivative = 3.78 \[ \int \frac {x^8}{\sqrt [3]{a+b x^3} \left (c+d x^3\right )} \, dx=\left [\frac {5 \, {\left (-b c d^{2} + a d^{3}\right )}^{\frac {2}{3}} b^{2} c^{2} \log \left ({\left (b x^{3} + a\right )}^{\frac {2}{3}} d^{2} + {\left (-b c d^{2} + a d^{3}\right )}^{\frac {1}{3}} {\left (b x^{3} + a\right )}^{\frac {1}{3}} d + {\left (-b c d^{2} + a d^{3}\right )}^{\frac {2}{3}}\right ) - 10 \, {\left (-b c d^{2} + a d^{3}\right )}^{\frac {2}{3}} b^{2} c^{2} \log \left ({\left (b x^{3} + a\right )}^{\frac {1}{3}} d - {\left (-b c d^{2} + a d^{3}\right )}^{\frac {1}{3}}\right ) + 15 \, \sqrt {\frac {1}{3}} {\left (b^{3} c^{3} d - a b^{2} c^{2} d^{2}\right )} \sqrt {\frac {{\left (-b c d^{2} + a d^{3}\right )}^{\frac {1}{3}}}{b c - a d}} \log \left (\frac {2 \, b d^{2} x^{3} - b c d + 3 \, a d^{2} + 3 \, \sqrt {\frac {1}{3}} {\left (2 \, {\left (-b c d^{2} + a d^{3}\right )}^{\frac {2}{3}} {\left (b x^{3} + a\right )}^{\frac {2}{3}} + {\left (b x^{3} + a\right )}^{\frac {1}{3}} {\left (b c d - a d^{2}\right )} + {\left (-b c d^{2} + a d^{3}\right )}^{\frac {1}{3}} {\left (b c - a d\right )}\right )} \sqrt {\frac {{\left (-b c d^{2} + a d^{3}\right )}^{\frac {1}{3}}}{b c - a d}} - 3 \, {\left (-b c d^{2} + a d^{3}\right )}^{\frac {2}{3}} {\left (b x^{3} + a\right )}^{\frac {1}{3}}}{d x^{3} + c}\right ) - 3 \, {\left (5 \, b^{2} c^{2} d^{2} - 2 \, a b c d^{3} - 3 \, a^{2} d^{4} - 2 \, {\left (b^{2} c d^{3} - a b d^{4}\right )} x^{3}\right )} {\left (b x^{3} + a\right )}^{\frac {2}{3}}}{30 \, {\left (b^{3} c d^{4} - a b^{2} d^{5}\right )}}, \frac {5 \, {\left (-b c d^{2} + a d^{3}\right )}^{\frac {2}{3}} b^{2} c^{2} \log \left ({\left (b x^{3} + a\right )}^{\frac {2}{3}} d^{2} + {\left (-b c d^{2} + a d^{3}\right )}^{\frac {1}{3}} {\left (b x^{3} + a\right )}^{\frac {1}{3}} d + {\left (-b c d^{2} + a d^{3}\right )}^{\frac {2}{3}}\right ) - 10 \, {\left (-b c d^{2} + a d^{3}\right )}^{\frac {2}{3}} b^{2} c^{2} \log \left ({\left (b x^{3} + a\right )}^{\frac {1}{3}} d - {\left (-b c d^{2} + a d^{3}\right )}^{\frac {1}{3}}\right ) + 30 \, \sqrt {\frac {1}{3}} {\left (b^{3} c^{3} d - a b^{2} c^{2} d^{2}\right )} \sqrt {-\frac {{\left (-b c d^{2} + a d^{3}\right )}^{\frac {1}{3}}}{b c - a d}} \arctan \left (\frac {\sqrt {\frac {1}{3}} {\left (2 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}} d + {\left (-b c d^{2} + a d^{3}\right )}^{\frac {1}{3}}\right )} \sqrt {-\frac {{\left (-b c d^{2} + a d^{3}\right )}^{\frac {1}{3}}}{b c - a d}}}{d}\right ) - 3 \, {\left (5 \, b^{2} c^{2} d^{2} - 2 \, a b c d^{3} - 3 \, a^{2} d^{4} - 2 \, {\left (b^{2} c d^{3} - a b d^{4}\right )} x^{3}\right )} {\left (b x^{3} + a\right )}^{\frac {2}{3}}}{30 \, {\left (b^{3} c d^{4} - a b^{2} d^{5}\right )}}\right ] \]
[1/30*(5*(-b*c*d^2 + a*d^3)^(2/3)*b^2*c^2*log((b*x^3 + a)^(2/3)*d^2 + (-b* c*d^2 + a*d^3)^(1/3)*(b*x^3 + a)^(1/3)*d + (-b*c*d^2 + a*d^3)^(2/3)) - 10* (-b*c*d^2 + a*d^3)^(2/3)*b^2*c^2*log((b*x^3 + a)^(1/3)*d - (-b*c*d^2 + a*d ^3)^(1/3)) + 15*sqrt(1/3)*(b^3*c^3*d - a*b^2*c^2*d^2)*sqrt((-b*c*d^2 + a*d ^3)^(1/3)/(b*c - a*d))*log((2*b*d^2*x^3 - b*c*d + 3*a*d^2 + 3*sqrt(1/3)*(2 *(-b*c*d^2 + a*d^3)^(2/3)*(b*x^3 + a)^(2/3) + (b*x^3 + a)^(1/3)*(b*c*d - a *d^2) + (-b*c*d^2 + a*d^3)^(1/3)*(b*c - a*d))*sqrt((-b*c*d^2 + a*d^3)^(1/3 )/(b*c - a*d)) - 3*(-b*c*d^2 + a*d^3)^(2/3)*(b*x^3 + a)^(1/3))/(d*x^3 + c) ) - 3*(5*b^2*c^2*d^2 - 2*a*b*c*d^3 - 3*a^2*d^4 - 2*(b^2*c*d^3 - a*b*d^4)*x ^3)*(b*x^3 + a)^(2/3))/(b^3*c*d^4 - a*b^2*d^5), 1/30*(5*(-b*c*d^2 + a*d^3) ^(2/3)*b^2*c^2*log((b*x^3 + a)^(2/3)*d^2 + (-b*c*d^2 + a*d^3)^(1/3)*(b*x^3 + a)^(1/3)*d + (-b*c*d^2 + a*d^3)^(2/3)) - 10*(-b*c*d^2 + a*d^3)^(2/3)*b^ 2*c^2*log((b*x^3 + a)^(1/3)*d - (-b*c*d^2 + a*d^3)^(1/3)) + 30*sqrt(1/3)*( b^3*c^3*d - a*b^2*c^2*d^2)*sqrt(-(-b*c*d^2 + a*d^3)^(1/3)/(b*c - a*d))*arc tan(sqrt(1/3)*(2*(b*x^3 + a)^(1/3)*d + (-b*c*d^2 + a*d^3)^(1/3))*sqrt(-(-b *c*d^2 + a*d^3)^(1/3)/(b*c - a*d))/d) - 3*(5*b^2*c^2*d^2 - 2*a*b*c*d^3 - 3 *a^2*d^4 - 2*(b^2*c*d^3 - a*b*d^4)*x^3)*(b*x^3 + a)^(2/3))/(b^3*c*d^4 - a* b^2*d^5)]
\[ \int \frac {x^8}{\sqrt [3]{a+b x^3} \left (c+d x^3\right )} \, dx=\int \frac {x^{8}}{\sqrt [3]{a + b x^{3}} \left (c + d x^{3}\right )}\, dx \]
Exception generated. \[ \int \frac {x^8}{\sqrt [3]{a+b x^3} \left (c+d x^3\right )} \, dx=\text {Exception raised: ValueError} \]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(a*d-b*c>0)', see `assume?` for m ore detail
Time = 0.30 (sec) , antiderivative size = 313, normalized size of antiderivative = 1.54 \[ \int \frac {x^8}{\sqrt [3]{a+b x^3} \left (c+d x^3\right )} \, dx=-\frac {b^{12} c^{2} d^{3} \left (-\frac {b c - a d}{d}\right )^{\frac {2}{3}} \log \left ({\left | {\left (b x^{3} + a\right )}^{\frac {1}{3}} - \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}} \right |}\right )}{3 \, {\left (b^{13} c d^{5} - a b^{12} d^{6}\right )}} - \frac {{\left (-b c d^{2} + a d^{3}\right )}^{\frac {2}{3}} c^{2} \arctan \left (\frac {\sqrt {3} {\left (2 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}} + \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}}\right )}}{3 \, \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}}}\right )}{\sqrt {3} b c d^{4} - \sqrt {3} a d^{5}} + \frac {{\left (-b c d^{2} + a d^{3}\right )}^{\frac {2}{3}} c^{2} \log \left ({\left (b x^{3} + a\right )}^{\frac {2}{3}} + {\left (b x^{3} + a\right )}^{\frac {1}{3}} \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}} + \left (-\frac {b c - a d}{d}\right )^{\frac {2}{3}}\right )}{6 \, {\left (b c d^{4} - a d^{5}\right )}} - \frac {5 \, {\left (b x^{3} + a\right )}^{\frac {2}{3}} b^{9} c d^{3} - 2 \, {\left (b x^{3} + a\right )}^{\frac {5}{3}} b^{8} d^{4} + 5 \, {\left (b x^{3} + a\right )}^{\frac {2}{3}} a b^{8} d^{4}}{10 \, b^{10} d^{5}} \]
-1/3*b^12*c^2*d^3*(-(b*c - a*d)/d)^(2/3)*log(abs((b*x^3 + a)^(1/3) - (-(b* c - a*d)/d)^(1/3)))/(b^13*c*d^5 - a*b^12*d^6) - (-b*c*d^2 + a*d^3)^(2/3)*c ^2*arctan(1/3*sqrt(3)*(2*(b*x^3 + a)^(1/3) + (-(b*c - a*d)/d)^(1/3))/(-(b* c - a*d)/d)^(1/3))/(sqrt(3)*b*c*d^4 - sqrt(3)*a*d^5) + 1/6*(-b*c*d^2 + a*d ^3)^(2/3)*c^2*log((b*x^3 + a)^(2/3) + (b*x^3 + a)^(1/3)*(-(b*c - a*d)/d)^( 1/3) + (-(b*c - a*d)/d)^(2/3))/(b*c*d^4 - a*d^5) - 1/10*(5*(b*x^3 + a)^(2/ 3)*b^9*c*d^3 - 2*(b*x^3 + a)^(5/3)*b^8*d^4 + 5*(b*x^3 + a)^(2/3)*a*b^8*d^4 )/(b^10*d^5)
Time = 9.03 (sec) , antiderivative size = 267, normalized size of antiderivative = 1.32 \[ \int \frac {x^8}{\sqrt [3]{a+b x^3} \left (c+d x^3\right )} \, dx=\frac {{\left (b\,x^3+a\right )}^{5/3}}{5\,b^2\,d}-\left (\frac {a}{b^2\,d}+\frac {b^3\,c-a\,b^2\,d}{2\,b^4\,d^2}\right )\,{\left (b\,x^3+a\right )}^{2/3}+\frac {c^2\,\ln \left (\frac {c^4\,{\left (b\,x^3+a\right )}^{1/3}}{d^3}+\frac {b\,c^5-a\,c^4\,d}{d^{10/3}\,{\left (a\,d-b\,c\right )}^{2/3}}\right )}{3\,d^{8/3}\,{\left (a\,d-b\,c\right )}^{1/3}}-\frac {\ln \left (\frac {c^4\,{\left (b\,x^3+a\right )}^{1/3}}{d^3}-\frac {c^4\,{\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}^2\,{\left (a\,d-b\,c\right )}^{1/3}}{4\,d^{10/3}}\right )\,\left (c^2+\sqrt {3}\,c^2\,1{}\mathrm {i}\right )}{6\,d^{8/3}\,{\left (a\,d-b\,c\right )}^{1/3}}+\frac {c^2\,\ln \left (\frac {c^4\,{\left (b\,x^3+a\right )}^{1/3}}{d^3}-\frac {c^4\,{\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}^2\,{\left (a\,d-b\,c\right )}^{1/3}}{4\,d^{10/3}}\right )\,\left (-\frac {1}{6}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right )}{d^{8/3}\,{\left (a\,d-b\,c\right )}^{1/3}} \]
(a + b*x^3)^(5/3)/(5*b^2*d) - (a/(b^2*d) + (b^3*c - a*b^2*d)/(2*b^4*d^2))* (a + b*x^3)^(2/3) + (c^2*log((c^4*(a + b*x^3)^(1/3))/d^3 + (b*c^5 - a*c^4* d)/(d^(10/3)*(a*d - b*c)^(2/3))))/(3*d^(8/3)*(a*d - b*c)^(1/3)) - (log((c^ 4*(a + b*x^3)^(1/3))/d^3 - (c^4*(3^(1/2)*1i + 1)^2*(a*d - b*c)^(1/3))/(4*d ^(10/3)))*(3^(1/2)*c^2*1i + c^2))/(6*d^(8/3)*(a*d - b*c)^(1/3)) + (c^2*log ((c^4*(a + b*x^3)^(1/3))/d^3 - (c^4*(3^(1/2)*1i - 1)^2*(a*d - b*c)^(1/3))/ (4*d^(10/3)))*((3^(1/2)*1i)/6 - 1/6))/(d^(8/3)*(a*d - b*c)^(1/3))