Integrand size = 24, antiderivative size = 220 \[ \int \frac {x^8 \sqrt [3]{a+b x^3}}{c+d x^3} \, dx=\frac {c^2 \sqrt [3]{a+b x^3}}{d^3}-\frac {(b c+a d) \left (a+b x^3\right )^{4/3}}{4 b^2 d^2}+\frac {\left (a+b x^3\right )^{7/3}}{7 b^2 d}+\frac {c^2 \sqrt [3]{b c-a d} \arctan \left (\frac {1-\frac {2 \sqrt [3]{d} \sqrt [3]{a+b x^3}}{\sqrt [3]{b c-a d}}}{\sqrt {3}}\right )}{\sqrt {3} d^{10/3}}+\frac {c^2 \sqrt [3]{b c-a d} \log \left (c+d x^3\right )}{6 d^{10/3}}-\frac {c^2 \sqrt [3]{b c-a d} \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{2 d^{10/3}} \]
c^2*(b*x^3+a)^(1/3)/d^3-1/4*(a*d+b*c)*(b*x^3+a)^(4/3)/b^2/d^2+1/7*(b*x^3+a )^(7/3)/b^2/d+1/6*c^2*(-a*d+b*c)^(1/3)*ln(d*x^3+c)/d^(10/3)-1/2*c^2*(-a*d+ b*c)^(1/3)*ln((-a*d+b*c)^(1/3)+d^(1/3)*(b*x^3+a)^(1/3))/d^(10/3)+1/3*c^2*( -a*d+b*c)^(1/3)*arctan(1/3*(1-2*d^(1/3)*(b*x^3+a)^(1/3)/(-a*d+b*c)^(1/3))* 3^(1/2))/d^(10/3)*3^(1/2)
Time = 0.63 (sec) , antiderivative size = 265, normalized size of antiderivative = 1.20 \[ \int \frac {x^8 \sqrt [3]{a+b x^3}}{c+d x^3} \, dx=\frac {\frac {3 \sqrt [3]{d} \sqrt [3]{a+b x^3} \left (-3 a^2 d^2+a b d \left (-7 c+d x^3\right )+b^2 \left (28 c^2-7 c d x^3+4 d^2 x^6\right )\right )}{b^2}+28 \sqrt {3} c^2 \sqrt [3]{b c-a d} \arctan \left (\frac {1-\frac {2 \sqrt [3]{d} \sqrt [3]{a+b x^3}}{\sqrt [3]{b c-a d}}}{\sqrt {3}}\right )-28 c^2 \sqrt [3]{b c-a d} \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )+14 c^2 \sqrt [3]{b c-a d} \log \left ((b c-a d)^{2/3}-\sqrt [3]{d} \sqrt [3]{b c-a d} \sqrt [3]{a+b x^3}+d^{2/3} \left (a+b x^3\right )^{2/3}\right )}{84 d^{10/3}} \]
((3*d^(1/3)*(a + b*x^3)^(1/3)*(-3*a^2*d^2 + a*b*d*(-7*c + d*x^3) + b^2*(28 *c^2 - 7*c*d*x^3 + 4*d^2*x^6)))/b^2 + 28*Sqrt[3]*c^2*(b*c - a*d)^(1/3)*Arc Tan[(1 - (2*d^(1/3)*(a + b*x^3)^(1/3))/(b*c - a*d)^(1/3))/Sqrt[3]] - 28*c^ 2*(b*c - a*d)^(1/3)*Log[(b*c - a*d)^(1/3) + d^(1/3)*(a + b*x^3)^(1/3)] + 1 4*c^2*(b*c - a*d)^(1/3)*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)])/(84*d^(10/3))
Time = 0.40 (sec) , antiderivative size = 225, 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.
| \(\displaystyle \int \frac {x^8 \sqrt [3]{a+b x^3}}{c+d x^3} \, dx\) |
| \(\Big \downarrow \) 948 |
| \(\displaystyle \frac {1}{3} \int \frac {x^6 \sqrt [3]{b x^3+a}}{d x^3+c}dx^3\) |
| \(\Big \downarrow \) 99 |
| \(\displaystyle \frac {1}{3} \int \left (\frac {\sqrt [3]{b x^3+a} c^2}{d^2 \left (d x^3+c\right )}+\frac {\left (b x^3+a\right )^{4/3}}{b d}+\frac {(-b c-a d) \sqrt [3]{b x^3+a}}{b d^2}\right )dx^3\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {1}{3} \left (\frac {\sqrt {3} c^2 \sqrt [3]{b c-a d} \arctan \left (\frac {1-\frac {2 \sqrt [3]{d} \sqrt [3]{a+b x^3}}{\sqrt [3]{b c-a d}}}{\sqrt {3}}\right )}{d^{10/3}}-\frac {3 \left (a+b x^3\right )^{4/3} (a d+b c)}{4 b^2 d^2}+\frac {3 \left (a+b x^3\right )^{7/3}}{7 b^2 d}+\frac {c^2 \sqrt [3]{b c-a d} \log \left (c+d x^3\right )}{2 d^{10/3}}-\frac {3 c^2 \sqrt [3]{b c-a d} \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{2 d^{10/3}}+\frac {3 c^2 \sqrt [3]{a+b x^3}}{d^3}\right )\) |
((3*c^2*(a + b*x^3)^(1/3))/d^3 - (3*(b*c + a*d)*(a + b*x^3)^(4/3))/(4*b^2* d^2) + (3*(a + b*x^3)^(7/3))/(7*b^2*d) + (Sqrt[3]*c^2*(b*c - a*d)^(1/3)*Ar cTan[(1 - (2*d^(1/3)*(a + b*x^3)^(1/3))/(b*c - a*d)^(1/3))/Sqrt[3]])/d^(10 /3) + (c^2*(b*c - a*d)^(1/3)*Log[c + d*x^3])/(2*d^(10/3)) - (3*c^2*(b*c - a*d)^(1/3)*Log[(b*c - a*d)^(1/3) + d^(1/3)*(a + b*x^3)^(1/3)])/(2*d^(10/3) ))/3
3.7.59.3.1 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 = 4.73 (sec) , antiderivative size = 235, normalized size of antiderivative = 1.07
| method | result | size |
| pseudoelliptic | \(-\frac {\frac {9 \left (\frac {a d -b c}{d}\right )^{\frac {2}{3}} d \left (\left (-\frac {4 b \,x^{3}}{3}+a \right ) \left (b \,x^{3}+a \right ) d^{2}+\frac {7 \left (b \,x^{3}+a \right ) b c d}{3}-\frac {28 b^{2} c^{2}}{3}\right ) \left (b \,x^{3}+a \right )^{\frac {1}{3}}}{14}+b^{2} c^{2} \left (a d -b c \right ) \left (2 \arctan \left (\frac {\sqrt {3}\, \left (2 \left (b \,x^{3}+a \right )^{\frac {1}{3}}+\left (\frac {a d -b c}{d}\right )^{\frac {1}{3}}\right )}{3 \left (\frac {a d -b c}{d}\right )^{\frac {1}{3}}}\right ) \sqrt {3}+\ln \left (\left (b \,x^{3}+a \right )^{\frac {2}{3}}+\left (\frac {a d -b c}{d}\right )^{\frac {1}{3}} \left (b \,x^{3}+a \right )^{\frac {1}{3}}+\left (\frac {a d -b c}{d}\right )^{\frac {2}{3}}\right )-2 \ln \left (\left (b \,x^{3}+a \right )^{\frac {1}{3}}-\left (\frac {a d -b c}{d}\right )^{\frac {1}{3}}\right )\right )}{6 \left (\frac {a d -b c}{d}\right )^{\frac {2}{3}} b^{2} d^{4}}\) | \(235\) |
-1/6/(1/d*(a*d-b*c))^(2/3)*(9/14*(1/d*(a*d-b*c))^(2/3)*d*((-4/3*b*x^3+a)*( b*x^3+a)*d^2+7/3*(b*x^3+a)*b*c*d-28/3*b^2*c^2)*(b*x^3+a)^(1/3)+b^2*c^2*(a* d-b*c)*(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))))/b^2/d^4
Time = 0.40 (sec) , antiderivative size = 282, normalized size of antiderivative = 1.28 \[ \int \frac {x^8 \sqrt [3]{a+b x^3}}{c+d x^3} \, dx=-\frac {28 \, \sqrt {3} b^{2} c^{2} \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}} \arctan \left (-\frac {2 \, \sqrt {3} {\left (b x^{3} + a\right )}^{\frac {1}{3}} d \left (-\frac {b c - a d}{d}\right )^{\frac {2}{3}} - \sqrt {3} {\left (b c - a d\right )}}{3 \, {\left (b c - a d\right )}}\right ) + 14 \, b^{2} c^{2} \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}} \log \left ({\left (b x^{3} + a\right )}^{\frac {2}{3}} + {\left (b x^{3} + a\right )}^{\frac {1}{3}} \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}} + \left (-\frac {b c - a d}{d}\right )^{\frac {2}{3}}\right ) - 28 \, b^{2} c^{2} \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}} \log \left ({\left (b x^{3} + a\right )}^{\frac {1}{3}} - \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}}\right ) - 3 \, {\left (4 \, b^{2} d^{2} x^{6} + 28 \, b^{2} c^{2} - 7 \, a b c d - 3 \, a^{2} d^{2} - {\left (7 \, b^{2} c d - a b d^{2}\right )} x^{3}\right )} {\left (b x^{3} + a\right )}^{\frac {1}{3}}}{84 \, b^{2} d^{3}} \]
-1/84*(28*sqrt(3)*b^2*c^2*(-(b*c - a*d)/d)^(1/3)*arctan(-1/3*(2*sqrt(3)*(b *x^3 + a)^(1/3)*d*(-(b*c - a*d)/d)^(2/3) - sqrt(3)*(b*c - a*d))/(b*c - a*d )) + 14*b^2*c^2*(-(b*c - a*d)/d)^(1/3)*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)) - 28*b^2*c^2*(-(b* c - a*d)/d)^(1/3)*log((b*x^3 + a)^(1/3) - (-(b*c - a*d)/d)^(1/3)) - 3*(4*b ^2*d^2*x^6 + 28*b^2*c^2 - 7*a*b*c*d - 3*a^2*d^2 - (7*b^2*c*d - a*b*d^2)*x^ 3)*(b*x^3 + a)^(1/3))/(b^2*d^3)
\[ \int \frac {x^8 \sqrt [3]{a+b x^3}}{c+d x^3} \, dx=\int \frac {x^{8} \sqrt [3]{a + b x^{3}}}{c + d x^{3}}\, dx \]
Exception generated. \[ \int \frac {x^8 \sqrt [3]{a+b x^3}}{c+d x^3} \, dx=\text {Exception raised: ValueError} \]
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 = 320, normalized size of antiderivative = 1.45 \[ \int \frac {x^8 \sqrt [3]{a+b x^3}}{c+d x^3} \, dx=\frac {{\left (b^{17} c^{3} d^{4} - a b^{16} c^{2} d^{5}\right )} \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}} \log \left ({\left | {\left (b x^{3} + a\right )}^{\frac {1}{3}} - \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}} \right |}\right )}{3 \, {\left (b^{17} c d^{7} - a b^{16} d^{8}\right )}} - \frac {\sqrt {3} {\left (-b c d^{2} + a d^{3}\right )}^{\frac {1}{3}} c^{2} \arctan \left (\frac {\sqrt {3} {\left (2 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}} + \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}}\right )}}{3 \, \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}}}\right )}{3 \, d^{4}} - \frac {{\left (-b c d^{2} + a d^{3}\right )}^{\frac {1}{3}} c^{2} \log \left ({\left (b x^{3} + a\right )}^{\frac {2}{3}} + {\left (b x^{3} + a\right )}^{\frac {1}{3}} \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}} + \left (-\frac {b c - a d}{d}\right )^{\frac {2}{3}}\right )}{6 \, d^{4}} + \frac {28 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}} b^{14} c^{2} d^{4} - 7 \, {\left (b x^{3} + a\right )}^{\frac {4}{3}} b^{13} c d^{5} + 4 \, {\left (b x^{3} + a\right )}^{\frac {7}{3}} b^{12} d^{6} - 7 \, {\left (b x^{3} + a\right )}^{\frac {4}{3}} a b^{12} d^{6}}{28 \, b^{14} d^{7}} \]
1/3*(b^17*c^3*d^4 - a*b^16*c^2*d^5)*(-(b*c - a*d)/d)^(1/3)*log(abs((b*x^3 + a)^(1/3) - (-(b*c - a*d)/d)^(1/3)))/(b^17*c*d^7 - a*b^16*d^8) - 1/3*sqrt (3)*(-b*c*d^2 + a*d^3)^(1/3)*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))/d^4 - 1/6*(-b*c*d^2 + a*d ^3)^(1/3)*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))/d^4 + 1/28*(28*(b*x^3 + a)^(1/3)*b^14*c^2*d ^4 - 7*(b*x^3 + a)^(4/3)*b^13*c*d^5 + 4*(b*x^3 + a)^(7/3)*b^12*d^6 - 7*(b* x^3 + a)^(4/3)*a*b^12*d^6)/(b^14*d^7)
Time = 8.89 (sec) , antiderivative size = 336, normalized size of antiderivative = 1.53 \[ \int \frac {x^8 \sqrt [3]{a+b x^3}}{c+d x^3} \, dx=\left (\frac {a^2}{b^2\,d}+\frac {\left (\frac {2\,a}{b^2\,d}+\frac {b^3\,c-a\,b^2\,d}{b^4\,d^2}\right )\,\left (b^3\,c-a\,b^2\,d\right )}{b^2\,d}\right )\,{\left (b\,x^3+a\right )}^{1/3}-\left (\frac {a}{2\,b^2\,d}+\frac {b^3\,c-a\,b^2\,d}{4\,b^4\,d^2}\right )\,{\left (b\,x^3+a\right )}^{4/3}+\frac {{\left (b\,x^3+a\right )}^{7/3}}{7\,b^2\,d}+\frac {c^2\,\ln \left ({\left (a\,d-b\,c\right )}^{1/3}-d^{1/3}\,{\left (b\,x^3+a\right )}^{1/3}\right )\,{\left (a\,d-b\,c\right )}^{1/3}}{3\,d^{10/3}}-\frac {c^2\,\ln \left (\frac {3\,{\left (b\,x^3+a\right )}^{1/3}\,\left (b\,c^3-a\,c^2\,d\right )}{d}-\frac {3\,c^2\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,{\left (a\,d-b\,c\right )}^{4/3}}{d^{4/3}}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,{\left (a\,d-b\,c\right )}^{1/3}}{3\,d^{10/3}}+\frac {c^2\,\ln \left (\frac {3\,{\left (b\,x^3+a\right )}^{1/3}\,\left (b\,c^3-a\,c^2\,d\right )}{d}+\frac {9\,c^2\,\left (-\frac {1}{6}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right )\,{\left (a\,d-b\,c\right )}^{4/3}}{d^{4/3}}\right )\,\left (-\frac {1}{6}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right )\,{\left (a\,d-b\,c\right )}^{1/3}}{d^{10/3}} \]
(a^2/(b^2*d) + (((2*a)/(b^2*d) + (b^3*c - a*b^2*d)/(b^4*d^2))*(b^3*c - a*b ^2*d))/(b^2*d))*(a + b*x^3)^(1/3) - (a/(2*b^2*d) + (b^3*c - a*b^2*d)/(4*b^ 4*d^2))*(a + b*x^3)^(4/3) + (a + b*x^3)^(7/3)/(7*b^2*d) + (c^2*log((a*d - b*c)^(1/3) - d^(1/3)*(a + b*x^3)^(1/3))*(a*d - b*c)^(1/3))/(3*d^(10/3)) - (c^2*log((3*(a + b*x^3)^(1/3)*(b*c^3 - a*c^2*d))/d - (3*c^2*((3^(1/2)*1i)/ 2 + 1/2)*(a*d - b*c)^(4/3))/d^(4/3))*((3^(1/2)*1i)/2 + 1/2)*(a*d - b*c)^(1 /3))/(3*d^(10/3)) + (c^2*log((3*(a + b*x^3)^(1/3)*(b*c^3 - a*c^2*d))/d + ( 9*c^2*((3^(1/2)*1i)/6 - 1/6)*(a*d - b*c)^(4/3))/d^(4/3))*((3^(1/2)*1i)/6 - 1/6)*(a*d - b*c)^(1/3))/d^(10/3)