Integrand size = 24, antiderivative size = 264 \[ \int \frac {x^{11} \sqrt [3]{a+b x^3}}{c+d x^3} \, dx=-\frac {c^3 \sqrt [3]{a+b x^3}}{d^4}+\frac {\left (b^2 c^2+a b c d+a^2 d^2\right ) \left (a+b x^3\right )^{4/3}}{4 b^3 d^3}-\frac {(b c+2 a d) \left (a+b x^3\right )^{7/3}}{7 b^3 d^2}+\frac {\left (a+b x^3\right )^{10/3}}{10 b^3 d}-\frac {c^3 \sqrt [3]{b c-a d} \arctan \left (\frac {1-\frac {2 \sqrt [3]{d} \sqrt [3]{a+b x^3}}{\sqrt [3]{b c-a d}}}{\sqrt {3}}\right )}{\sqrt {3} d^{13/3}}-\frac {c^3 \sqrt [3]{b c-a d} \log \left (c+d x^3\right )}{6 d^{13/3}}+\frac {c^3 \sqrt [3]{b c-a d} \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{2 d^{13/3}} \]
-c^3*(b*x^3+a)^(1/3)/d^4+1/4*(a^2*d^2+a*b*c*d+b^2*c^2)*(b*x^3+a)^(4/3)/b^3 /d^3-1/7*(2*a*d+b*c)*(b*x^3+a)^(7/3)/b^3/d^2+1/10*(b*x^3+a)^(10/3)/b^3/d-1 /6*c^3*(-a*d+b*c)^(1/3)*ln(d*x^3+c)/d^(13/3)+1/2*c^3*(-a*d+b*c)^(1/3)*ln(( -a*d+b*c)^(1/3)+d^(1/3)*(b*x^3+a)^(1/3))/d^(13/3)-1/3*c^3*(-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^(13/ 3)*3^(1/2)
Time = 0.87 (sec) , antiderivative size = 308, normalized size of antiderivative = 1.17 \[ \int \frac {x^{11} \sqrt [3]{a+b x^3}}{c+d x^3} \, dx=\frac {\frac {3 \sqrt [3]{d} \sqrt [3]{a+b x^3} \left (9 a^3 d^3-3 a^2 b d^2 \left (-5 c+d x^3\right )+a b^2 d \left (35 c^2-5 c d x^3+2 d^2 x^6\right )+b^3 \left (-140 c^3+35 c^2 d x^3-20 c d^2 x^6+14 d^3 x^9\right )\right )}{b^3}-140 \sqrt {3} c^3 \sqrt [3]{b c-a d} \arctan \left (\frac {1-\frac {2 \sqrt [3]{d} \sqrt [3]{a+b x^3}}{\sqrt [3]{b c-a d}}}{\sqrt {3}}\right )+140 c^3 \sqrt [3]{b c-a d} \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )-70 c^3 \sqrt [3]{b c-a d} \log \left ((b c-a d)^{2/3}-\sqrt [3]{d} \sqrt [3]{b c-a d} \sqrt [3]{a+b x^3}+d^{2/3} \left (a+b x^3\right )^{2/3}\right )}{420 d^{13/3}} \]
((3*d^(1/3)*(a + b*x^3)^(1/3)*(9*a^3*d^3 - 3*a^2*b*d^2*(-5*c + d*x^3) + a* b^2*d*(35*c^2 - 5*c*d*x^3 + 2*d^2*x^6) + b^3*(-140*c^3 + 35*c^2*d*x^3 - 20 *c*d^2*x^6 + 14*d^3*x^9)))/b^3 - 140*Sqrt[3]*c^3*(b*c - a*d)^(1/3)*ArcTan[ (1 - (2*d^(1/3)*(a + b*x^3)^(1/3))/(b*c - a*d)^(1/3))/Sqrt[3]] + 140*c^3*( b*c - a*d)^(1/3)*Log[(b*c - a*d)^(1/3) + d^(1/3)*(a + b*x^3)^(1/3)] - 70*c ^3*(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)])/(420*d^(13/3))
Time = 0.49 (sec) , antiderivative size = 268, 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.
| \(\displaystyle \int \frac {x^{11} \sqrt [3]{a+b x^3}}{c+d x^3} \, dx\) |
| \(\Big \downarrow \) 948 |
| \(\displaystyle \frac {1}{3} \int \frac {x^9 \sqrt [3]{b x^3+a}}{d x^3+c}dx^3\) |
| \(\Big \downarrow \) 99 |
| \(\displaystyle \frac {1}{3} \int \left (-\frac {\sqrt [3]{b x^3+a} c^3}{d^3 \left (d x^3+c\right )}+\frac {\left (b x^3+a\right )^{7/3}}{b^2 d}+\frac {(-b c-2 a d) \left (b x^3+a\right )^{4/3}}{b^2 d^2}+\frac {\left (b^2 c^2+a b d c+a^2 d^2\right ) \sqrt [3]{b x^3+a}}{b^2 d^3}\right )dx^3\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {1}{3} \left (\frac {3 \left (a+b x^3\right )^{4/3} \left (a^2 d^2+a b c d+b^2 c^2\right )}{4 b^3 d^3}-\frac {\sqrt {3} c^3 \sqrt [3]{b c-a d} \arctan \left (\frac {1-\frac {2 \sqrt [3]{d} \sqrt [3]{a+b x^3}}{\sqrt [3]{b c-a d}}}{\sqrt {3}}\right )}{d^{13/3}}-\frac {3 \left (a+b x^3\right )^{7/3} (2 a d+b c)}{7 b^3 d^2}+\frac {3 \left (a+b x^3\right )^{10/3}}{10 b^3 d}-\frac {c^3 \sqrt [3]{b c-a d} \log \left (c+d x^3\right )}{2 d^{13/3}}+\frac {3 c^3 \sqrt [3]{b c-a d} \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{2 d^{13/3}}-\frac {3 c^3 \sqrt [3]{a+b x^3}}{d^4}\right )\) |
((-3*c^3*(a + b*x^3)^(1/3))/d^4 + (3*(b^2*c^2 + a*b*c*d + a^2*d^2)*(a + b* x^3)^(4/3))/(4*b^3*d^3) - (3*(b*c + 2*a*d)*(a + b*x^3)^(7/3))/(7*b^3*d^2) + (3*(a + b*x^3)^(10/3))/(10*b^3*d) - (Sqrt[3]*c^3*(b*c - a*d)^(1/3)*ArcTa n[(1 - (2*d^(1/3)*(a + b*x^3)^(1/3))/(b*c - a*d)^(1/3))/Sqrt[3]])/d^(13/3) - (c^3*(b*c - a*d)^(1/3)*Log[c + d*x^3])/(2*d^(13/3)) + (3*c^3*(b*c - a*d )^(1/3)*Log[(b*c - a*d)^(1/3) + d^(1/3)*(a + b*x^3)^(1/3)])/(2*d^(13/3)))/ 3
3.7.58.3.1 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 = 6.04 (sec) , antiderivative size = 283, normalized size of antiderivative = 1.07
| method | result | size |
| pseudoelliptic | \(\frac {\frac {27 \left (\frac {a d -b c}{d}\right )^{\frac {2}{3}} \left (b \,x^{3}+a \right )^{\frac {1}{3}} d \left (\frac {\left (14 d^{3} x^{9}-20 c \,d^{2} x^{6}+35 c^{2} d \,x^{3}-140 c^{3}\right ) b^{3}}{9}+\frac {35 \left (\frac {2}{35} d^{2} x^{6}-\frac {1}{7} c d \,x^{3}+c^{2}\right ) d a \,b^{2}}{9}+\frac {5 \left (-\frac {d \,x^{3}}{5}+c \right ) d^{2} a^{2} b}{3}+a^{3} d^{3}\right )}{70}+b^{3} c^{3} \left (a d -b c \right ) \left (2 \arctan \left (\frac {\sqrt {3}\, \left (2 \left (b \,x^{3}+a \right )^{\frac {1}{3}}+\left (\frac {a d -b c}{d}\right )^{\frac {1}{3}}\right )}{3 \left (\frac {a d -b c}{d}\right )^{\frac {1}{3}}}\right ) \sqrt {3}+\ln \left (\left (b \,x^{3}+a \right )^{\frac {2}{3}}+\left (\frac {a d -b c}{d}\right )^{\frac {1}{3}} \left (b \,x^{3}+a \right )^{\frac {1}{3}}+\left (\frac {a d -b c}{d}\right )^{\frac {2}{3}}\right )-2 \ln \left (\left (b \,x^{3}+a \right )^{\frac {1}{3}}-\left (\frac {a d -b c}{d}\right )^{\frac {1}{3}}\right )\right )}{6 \left (\frac {a d -b c}{d}\right )^{\frac {2}{3}} b^{3} d^{5}}\) | \(283\) |
1/6*(27/70*(1/d*(a*d-b*c))^(2/3)*(b*x^3+a)^(1/3)*d*(1/9*(14*d^3*x^9-20*c*d ^2*x^6+35*c^2*d*x^3-140*c^3)*b^3+35/9*(2/35*d^2*x^6-1/7*c*d*x^3+c^2)*d*a*b ^2+5/3*(-1/5*d*x^3+c)*d^2*a^2*b+a^3*d^3)+b^3*c^3*(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))))/(1/d*(a*d-b*c))^ (2/3)/b^3/d^5
Time = 0.33 (sec) , antiderivative size = 325, normalized size of antiderivative = 1.23 \[ \int \frac {x^{11} \sqrt [3]{a+b x^3}}{c+d x^3} \, dx=-\frac {140 \, \sqrt {3} b^{3} c^{3} \left (\frac {b c - a d}{d}\right )^{\frac {1}{3}} \arctan \left (-\frac {2 \, \sqrt {3} {\left (b x^{3} + a\right )}^{\frac {1}{3}} d \left (\frac {b c - a d}{d}\right )^{\frac {2}{3}} - \sqrt {3} {\left (b c - a d\right )}}{3 \, {\left (b c - a d\right )}}\right ) + 70 \, b^{3} c^{3} \left (\frac {b c - a d}{d}\right )^{\frac {1}{3}} \log \left ({\left (b x^{3} + a\right )}^{\frac {2}{3}} - {\left (b x^{3} + a\right )}^{\frac {1}{3}} \left (\frac {b c - a d}{d}\right )^{\frac {1}{3}} + \left (\frac {b c - a d}{d}\right )^{\frac {2}{3}}\right ) - 140 \, b^{3} c^{3} \left (\frac {b c - a d}{d}\right )^{\frac {1}{3}} \log \left ({\left (b x^{3} + a\right )}^{\frac {1}{3}} + \left (\frac {b c - a d}{d}\right )^{\frac {1}{3}}\right ) - 3 \, {\left (14 \, b^{3} d^{3} x^{9} - 2 \, {\left (10 \, b^{3} c d^{2} - a b^{2} d^{3}\right )} x^{6} - 140 \, b^{3} c^{3} + 35 \, a b^{2} c^{2} d + 15 \, a^{2} b c d^{2} + 9 \, a^{3} d^{3} + {\left (35 \, b^{3} c^{2} d - 5 \, a b^{2} c d^{2} - 3 \, a^{2} b d^{3}\right )} x^{3}\right )} {\left (b x^{3} + a\right )}^{\frac {1}{3}}}{420 \, b^{3} d^{4}} \]
-1/420*(140*sqrt(3)*b^3*c^3*((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 )) + 70*b^3*c^3*((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)) - 140*b^3*c^3*((b*c - a*d)/d)^(1/3)*log((b*x^3 + a)^(1/3) + ((b*c - a*d)/d)^(1/3)) - 3*(14*b^3* d^3*x^9 - 2*(10*b^3*c*d^2 - a*b^2*d^3)*x^6 - 140*b^3*c^3 + 35*a*b^2*c^2*d + 15*a^2*b*c*d^2 + 9*a^3*d^3 + (35*b^3*c^2*d - 5*a*b^2*c*d^2 - 3*a^2*b*d^3 )*x^3)*(b*x^3 + a)^(1/3))/(b^3*d^4)
\[ \int \frac {x^{11} \sqrt [3]{a+b x^3}}{c+d x^3} \, dx=\int \frac {x^{11} \sqrt [3]{a + b x^{3}}}{c + d x^{3}}\, dx \]
Exception generated. \[ \int \frac {x^{11} \sqrt [3]{a+b x^3}}{c+d x^3} \, dx=\text {Exception raised: ValueError} \]
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 = 0.33 (sec) , antiderivative size = 379, normalized size of antiderivative = 1.44 \[ \int \frac {x^{11} \sqrt [3]{a+b x^3}}{c+d x^3} \, dx=-\frac {{\left (b^{34} c^{4} d^{6} - a b^{33} c^{3} d^{7}\right )} \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}} \log \left ({\left | {\left (b x^{3} + a\right )}^{\frac {1}{3}} - \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}} \right |}\right )}{3 \, {\left (b^{34} c d^{10} - a b^{33} d^{11}\right )}} + \frac {\sqrt {3} {\left (-b c d^{2} + a d^{3}\right )}^{\frac {1}{3}} c^{3} \arctan \left (\frac {\sqrt {3} {\left (2 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}} + \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}}\right )}}{3 \, \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}}}\right )}{3 \, d^{5}} + \frac {{\left (-b c d^{2} + a d^{3}\right )}^{\frac {1}{3}} c^{3} \log \left ({\left (b x^{3} + a\right )}^{\frac {2}{3}} + {\left (b x^{3} + a\right )}^{\frac {1}{3}} \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}} + \left (-\frac {b c - a d}{d}\right )^{\frac {2}{3}}\right )}{6 \, d^{5}} - \frac {140 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}} b^{30} c^{3} d^{6} - 35 \, {\left (b x^{3} + a\right )}^{\frac {4}{3}} b^{29} c^{2} d^{7} + 20 \, {\left (b x^{3} + a\right )}^{\frac {7}{3}} b^{28} c d^{8} - 35 \, {\left (b x^{3} + a\right )}^{\frac {4}{3}} a b^{28} c d^{8} - 14 \, {\left (b x^{3} + a\right )}^{\frac {10}{3}} b^{27} d^{9} + 40 \, {\left (b x^{3} + a\right )}^{\frac {7}{3}} a b^{27} d^{9} - 35 \, {\left (b x^{3} + a\right )}^{\frac {4}{3}} a^{2} b^{27} d^{9}}{140 \, b^{30} d^{10}} \]
-1/3*(b^34*c^4*d^6 - a*b^33*c^3*d^7)*(-(b*c - a*d)/d)^(1/3)*log(abs((b*x^3 + a)^(1/3) - (-(b*c - a*d)/d)^(1/3)))/(b^34*c*d^10 - a*b^33*d^11) + 1/3*s qrt(3)*(-b*c*d^2 + a*d^3)^(1/3)*c^3*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^5 + 1/6*(-b*c*d^2 + a*d^3)^(1/3)*c^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))/d^5 - 1/140*(140*(b*x^3 + a)^(1/3)*b^30* c^3*d^6 - 35*(b*x^3 + a)^(4/3)*b^29*c^2*d^7 + 20*(b*x^3 + a)^(7/3)*b^28*c* d^8 - 35*(b*x^3 + a)^(4/3)*a*b^28*c*d^8 - 14*(b*x^3 + a)^(10/3)*b^27*d^9 + 40*(b*x^3 + a)^(7/3)*a*b^27*d^9 - 35*(b*x^3 + a)^(4/3)*a^2*b^27*d^9)/(b^3 0*d^10)
Time = 9.20 (sec) , antiderivative size = 442, normalized size of antiderivative = 1.67 \[ \int \frac {x^{11} \sqrt [3]{a+b x^3}}{c+d x^3} \, dx=\left (\frac {3\,a^2}{4\,b^3\,d}+\frac {\left (\frac {3\,a}{b^3\,d}+\frac {b^4\,c-a\,b^3\,d}{b^6\,d^2}\right )\,\left (b^4\,c-a\,b^3\,d\right )}{4\,b^3\,d}\right )\,{\left (b\,x^3+a\right )}^{4/3}-\left (\frac {3\,a}{7\,b^3\,d}+\frac {b^4\,c-a\,b^3\,d}{7\,b^6\,d^2}\right )\,{\left (b\,x^3+a\right )}^{7/3}-{\left (b\,x^3+a\right )}^{1/3}\,\left (\frac {a^3}{b^3\,d}+\frac {\left (\frac {3\,a^2}{b^3\,d}+\frac {\left (\frac {3\,a}{b^3\,d}+\frac {b^4\,c-a\,b^3\,d}{b^6\,d^2}\right )\,\left (b^4\,c-a\,b^3\,d\right )}{b^3\,d}\right )\,\left (b^4\,c-a\,b^3\,d\right )}{b^3\,d}\right )+\frac {{\left (b\,x^3+a\right )}^{10/3}}{10\,b^3\,d}-\frac {c^3\,\ln \left ({\left (a\,d-b\,c\right )}^{1/3}-d^{1/3}\,{\left (b\,x^3+a\right )}^{1/3}\right )\,{\left (a\,d-b\,c\right )}^{1/3}}{3\,d^{13/3}}-\frac {c^3\,\ln \left (\frac {3\,{\left (b\,x^3+a\right )}^{1/3}\,\left (b\,c^4-a\,c^3\,d\right )}{d^2}+\frac {3\,c^3\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,{\left (a\,d-b\,c\right )}^{4/3}}{d^{7/3}}\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,{\left (a\,d-b\,c\right )}^{1/3}}{3\,d^{13/3}}+\frac {c^3\,\ln \left (\frac {3\,{\left (b\,x^3+a\right )}^{1/3}\,\left (b\,c^4-a\,c^3\,d\right )}{d^2}-\frac {9\,c^3\,\left (\frac {1}{6}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right )\,{\left (a\,d-b\,c\right )}^{4/3}}{d^{7/3}}\right )\,\left (\frac {1}{6}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right )\,{\left (a\,d-b\,c\right )}^{1/3}}{d^{13/3}} \]
((3*a^2)/(4*b^3*d) + (((3*a)/(b^3*d) + (b^4*c - a*b^3*d)/(b^6*d^2))*(b^4*c - a*b^3*d))/(4*b^3*d))*(a + b*x^3)^(4/3) - ((3*a)/(7*b^3*d) + (b^4*c - a* b^3*d)/(7*b^6*d^2))*(a + b*x^3)^(7/3) - (a + b*x^3)^(1/3)*(a^3/(b^3*d) + ( ((3*a^2)/(b^3*d) + (((3*a)/(b^3*d) + (b^4*c - a*b^3*d)/(b^6*d^2))*(b^4*c - a*b^3*d))/(b^3*d))*(b^4*c - a*b^3*d))/(b^3*d)) + (a + b*x^3)^(10/3)/(10*b ^3*d) - (c^3*log((a*d - b*c)^(1/3) - d^(1/3)*(a + b*x^3)^(1/3))*(a*d - b*c )^(1/3))/(3*d^(13/3)) - (c^3*log((3*(a + b*x^3)^(1/3)*(b*c^4 - a*c^3*d))/d ^2 + (3*c^3*((3^(1/2)*1i)/2 - 1/2)*(a*d - b*c)^(4/3))/d^(7/3))*((3^(1/2)*1 i)/2 - 1/2)*(a*d - b*c)^(1/3))/(3*d^(13/3)) + (c^3*log((3*(a + b*x^3)^(1/3 )*(b*c^4 - a*c^3*d))/d^2 - (9*c^3*((3^(1/2)*1i)/6 + 1/6)*(a*d - b*c)^(4/3) )/d^(7/3))*((3^(1/2)*1i)/6 + 1/6)*(a*d - b*c)^(1/3))/d^(13/3)