Integrand size = 27, antiderivative size = 95 \[ \int \frac {x^{11}}{\left (8 c-d x^3\right )^2 \sqrt {c+d x^3}} \, dx=\frac {8 x^6 \sqrt {c+d x^3}}{27 d^2 \left (8 c-d x^3\right )}+\frac {2 \sqrt {c+d x^3} \left (170 c+7 d x^3\right )}{27 d^4}-\frac {2944 c^{3/2} \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{81 d^4} \]
-2944/81*c^(3/2)*arctanh(1/3*(d*x^3+c)^(1/2)/c^(1/2))/d^4+8/27*x^6*(d*x^3+ c)^(1/2)/d^2/(-d*x^3+8*c)+2/27*(7*d*x^3+170*c)*(d*x^3+c)^(1/2)/d^4
Time = 0.19 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.85 \[ \int \frac {x^{11}}{\left (8 c-d x^3\right )^2 \sqrt {c+d x^3}} \, dx=\frac {2 \left (\frac {3 \sqrt {c+d x^3} \left (-1360 c^2+114 c d x^3+3 d^2 x^6\right )}{-8 c+d x^3}-1472 c^{3/2} \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )\right )}{81 d^4} \]
(2*((3*Sqrt[c + d*x^3]*(-1360*c^2 + 114*c*d*x^3 + 3*d^2*x^6))/(-8*c + d*x^ 3) - 1472*c^(3/2)*ArcTanh[Sqrt[c + d*x^3]/(3*Sqrt[c])]))/(81*d^4)
Time = 0.23 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.11, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {948, 109, 27, 164, 73, 219}
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.
| \(\displaystyle \int \frac {x^{11}}{\left (8 c-d x^3\right )^2 \sqrt {c+d x^3}} \, dx\) |
| \(\Big \downarrow \) 948 |
| \(\displaystyle \frac {1}{3} \int \frac {x^9}{\left (8 c-d x^3\right )^2 \sqrt {d x^3+c}}dx^3\) |
| \(\Big \downarrow \) 109 |
| \(\displaystyle \frac {1}{3} \left (\frac {8 x^6 \sqrt {c+d x^3}}{9 d^2 \left (8 c-d x^3\right )}-\frac {\int \frac {c x^3 \left (21 d x^3+16 c\right )}{\left (8 c-d x^3\right ) \sqrt {d x^3+c}}dx^3}{9 c d^2}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{3} \left (\frac {8 x^6 \sqrt {c+d x^3}}{9 d^2 \left (8 c-d x^3\right )}-\frac {\int \frac {x^3 \left (21 d x^3+16 c\right )}{\left (8 c-d x^3\right ) \sqrt {d x^3+c}}dx^3}{9 d^2}\right )\) |
| \(\Big \downarrow \) 164 |
| \(\displaystyle \frac {1}{3} \left (\frac {8 x^6 \sqrt {c+d x^3}}{9 d^2 \left (8 c-d x^3\right )}-\frac {\frac {1472 c^2 \int \frac {1}{\left (8 c-d x^3\right ) \sqrt {d x^3+c}}dx^3}{d}-\frac {2 \sqrt {c+d x^3} \left (170 c+7 d x^3\right )}{d^2}}{9 d^2}\right )\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {1}{3} \left (\frac {8 x^6 \sqrt {c+d x^3}}{9 d^2 \left (8 c-d x^3\right )}-\frac {\frac {2944 c^2 \int \frac {1}{9 c-x^6}d\sqrt {d x^3+c}}{d^2}-\frac {2 \sqrt {c+d x^3} \left (170 c+7 d x^3\right )}{d^2}}{9 d^2}\right )\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {1}{3} \left (\frac {8 x^6 \sqrt {c+d x^3}}{9 d^2 \left (8 c-d x^3\right )}-\frac {\frac {2944 c^{3/2} \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{3 d^2}-\frac {2 \sqrt {c+d x^3} \left (170 c+7 d x^3\right )}{d^2}}{9 d^2}\right )\) |
((8*x^6*Sqrt[c + d*x^3])/(9*d^2*(8*c - d*x^3)) - ((-2*Sqrt[c + d*x^3]*(170 *c + 7*d*x^3))/d^2 + (2944*c^(3/2)*ArcTanh[Sqrt[c + d*x^3]/(3*Sqrt[c])])/( 3*d^2))/(9*d^2))/3
3.5.24.3.1 Defintions of rubi rules used
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_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Simp[(b*c - a*d)*(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*((e + f *x)^(p + 1)/(b*(b*e - a*f)*(m + 1))), x] + Simp[1/(b*(b*e - a*f)*(m + 1)) Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 2)*(e + f*x)^p*Simp[a*d*(d*e*(n - 1) + c*f*(p + 1)) + b*c*(d*e*(m - n + 2) - c*f*(m + p + 2)) + d*(a*d*f*(n + p) + b*(d*e*(m + 1) - c*f*(m + n + p + 1)))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && LtQ[m, -1] && GtQ[n, 1] && (IntegersQ[2*m, 2*n, 2*p] || IntegersQ[m, n + p] || IntegersQ[p, m + n])
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_) + (f_.)*(x_ ))*((g_.) + (h_.)*(x_)), x_] :> Simp[(-(a*d*f*h*(n + 2) + b*c*f*h*(m + 2) - b*d*(f*g + e*h)*(m + n + 3) - b*d*f*h*(m + n + 2)*x))*(a + b*x)^(m + 1)*(( c + d*x)^(n + 1)/(b^2*d^2*(m + n + 2)*(m + n + 3))), x] + Simp[(a^2*d^2*f*h *(n + 1)*(n + 2) + a*b*d*(n + 1)*(2*c*f*h*(m + 1) - d*(f*g + e*h)*(m + n + 3)) + b^2*(c^2*f*h*(m + 1)*(m + 2) - c*d*(f*g + e*h)*(m + 1)*(m + n + 3) + d^2*e*g*(m + n + 2)*(m + n + 3)))/(b^2*d^2*(m + n + 2)*(m + n + 3)) Int[( a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, m, n}, x] && NeQ[m + n + 2, 0] && NeQ[m + n + 3, 0]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
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 = 4.58 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.83
| method | result | size |
| pseudoelliptic | \(\frac {\frac {2 \left (d \,x^{3}+c \right )^{\frac {3}{2}}}{9}+10 c \sqrt {d \,x^{3}+c}+\frac {128 c^{2} \left (\frac {4 \sqrt {d \,x^{3}+c}}{-d \,x^{3}+8 c}-\frac {23 \,\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{3 \sqrt {c}}\right )}{3 \sqrt {c}}\right )}{27}}{d^{4}}\) | \(79\) |
| risch | \(\frac {2 \left (d \,x^{3}+46 c \right ) \sqrt {d \,x^{3}+c}}{9 d^{4}}+\frac {64 c^{2} \left (-\frac {2 \,\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{3 \sqrt {c}}\right )}{3 d \sqrt {c}}+\frac {8 c \left (-\frac {\sqrt {d \,x^{3}+c}}{c \left (d \,x^{3}-8 c \right )}+\frac {\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{3 \sqrt {c}}\right )}{3 c^{\frac {3}{2}}}\right )}{27 d}\right )}{d^{3}}\) | \(109\) |
| default | \(\frac {d \left (\frac {2 x^{3} \sqrt {d \,x^{3}+c}}{9 d}-\frac {4 c \sqrt {d \,x^{3}+c}}{9 d^{2}}\right )+\frac {32 c \sqrt {d \,x^{3}+c}}{3 d}}{d^{3}}-\frac {128 c^{\frac {3}{2}} \operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{3 \sqrt {c}}\right )}{3 d^{4}}+\frac {512 c^{3} \left (\frac {\sqrt {d \,x^{3}+c}}{c \left (-d \,x^{3}+8 c \right )}+\frac {\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{3 \sqrt {c}}\right )}{3 c^{\frac {3}{2}}}\right )}{27 d^{4}}\) | \(134\) |
| elliptic | \(\frac {512 c^{2} \sqrt {d \,x^{3}+c}}{27 d^{4} \left (-d \,x^{3}+8 c \right )}+\frac {2 x^{3} \sqrt {d \,x^{3}+c}}{9 d^{3}}+\frac {92 c \sqrt {d \,x^{3}+c}}{9 d^{4}}+\frac {1472 i c \sqrt {2}\, \left (\munderset {\underline {\hspace {1.25 ex}}\alpha =\operatorname {RootOf}\left (d \,\textit {\_Z}^{3}-8 c \right )}{\sum }\frac {\left (-c \,d^{2}\right )^{\frac {1}{3}} \sqrt {2}\, \sqrt {\frac {i d \left (2 x +\frac {-i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {1}{3}}+\left (-c \,d^{2}\right )^{\frac {1}{3}}}{d}\right )}{\left (-c \,d^{2}\right )^{\frac {1}{3}}}}\, \sqrt {\frac {d \left (x -\frac {\left (-c \,d^{2}\right )^{\frac {1}{3}}}{d}\right )}{-3 \left (-c \,d^{2}\right )^{\frac {1}{3}}+i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {1}{3}}}}\, \sqrt {-\frac {i d \left (2 x +\frac {i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {1}{3}}+\left (-c \,d^{2}\right )^{\frac {1}{3}}}{d}\right )}{2 \left (-c \,d^{2}\right )^{\frac {1}{3}}}}\, \left (i \left (-c \,d^{2}\right )^{\frac {1}{3}} \underline {\hspace {1.25 ex}}\alpha \sqrt {3}\, d -i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {2}{3}}+2 \underline {\hspace {1.25 ex}}\alpha ^{2} d^{2}-\left (-c \,d^{2}\right )^{\frac {1}{3}} \underline {\hspace {1.25 ex}}\alpha d -\left (-c \,d^{2}\right )^{\frac {2}{3}}\right ) \Pi \left (\frac {\sqrt {3}\, \sqrt {\frac {i \left (x +\frac {\left (-c \,d^{2}\right )^{\frac {1}{3}}}{2 d}-\frac {i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {1}{3}}}{2 d}\right ) \sqrt {3}\, d}{\left (-c \,d^{2}\right )^{\frac {1}{3}}}}}{3}, -\frac {2 i \left (-c \,d^{2}\right )^{\frac {1}{3}} \sqrt {3}\, \underline {\hspace {1.25 ex}}\alpha ^{2} d -i \left (-c \,d^{2}\right )^{\frac {2}{3}} \sqrt {3}\, \underline {\hspace {1.25 ex}}\alpha +i \sqrt {3}\, c d -3 \left (-c \,d^{2}\right )^{\frac {2}{3}} \underline {\hspace {1.25 ex}}\alpha -3 c d}{18 d c}, \sqrt {\frac {i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {1}{3}}}{d \left (-\frac {3 \left (-c \,d^{2}\right )^{\frac {1}{3}}}{2 d}+\frac {i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {1}{3}}}{2 d}\right )}}\right )}{2 \sqrt {d \,x^{3}+c}}\right )}{243 d^{6}}\) | \(473\) |
2*(1/9*(d*x^3+c)^(3/2)+5*c*(d*x^3+c)^(1/2)+64/27*c^2*(4*(d*x^3+c)^(1/2)/(- d*x^3+8*c)-23/3*arctanh(1/3*(d*x^3+c)^(1/2)/c^(1/2))/c^(1/2)))/d^4
Time = 0.28 (sec) , antiderivative size = 195, normalized size of antiderivative = 2.05 \[ \int \frac {x^{11}}{\left (8 c-d x^3\right )^2 \sqrt {c+d x^3}} \, dx=\left [\frac {2 \, {\left (736 \, {\left (c d x^{3} - 8 \, c^{2}\right )} \sqrt {c} \log \left (\frac {d x^{3} - 6 \, \sqrt {d x^{3} + c} \sqrt {c} + 10 \, c}{d x^{3} - 8 \, c}\right ) + 3 \, {\left (3 \, d^{2} x^{6} + 114 \, c d x^{3} - 1360 \, c^{2}\right )} \sqrt {d x^{3} + c}\right )}}{81 \, {\left (d^{5} x^{3} - 8 \, c d^{4}\right )}}, \frac {2 \, {\left (1472 \, {\left (c d x^{3} - 8 \, c^{2}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {d x^{3} + c} \sqrt {-c}}{3 \, c}\right ) + 3 \, {\left (3 \, d^{2} x^{6} + 114 \, c d x^{3} - 1360 \, c^{2}\right )} \sqrt {d x^{3} + c}\right )}}{81 \, {\left (d^{5} x^{3} - 8 \, c d^{4}\right )}}\right ] \]
[2/81*(736*(c*d*x^3 - 8*c^2)*sqrt(c)*log((d*x^3 - 6*sqrt(d*x^3 + c)*sqrt(c ) + 10*c)/(d*x^3 - 8*c)) + 3*(3*d^2*x^6 + 114*c*d*x^3 - 1360*c^2)*sqrt(d*x ^3 + c))/(d^5*x^3 - 8*c*d^4), 2/81*(1472*(c*d*x^3 - 8*c^2)*sqrt(-c)*arctan (1/3*sqrt(d*x^3 + c)*sqrt(-c)/c) + 3*(3*d^2*x^6 + 114*c*d*x^3 - 1360*c^2)* sqrt(d*x^3 + c))/(d^5*x^3 - 8*c*d^4)]
\[ \int \frac {x^{11}}{\left (8 c-d x^3\right )^2 \sqrt {c+d x^3}} \, dx=\int \frac {x^{11}}{\left (- 8 c + d x^{3}\right )^{2} \sqrt {c + d x^{3}}}\, dx \]
Time = 0.30 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.98 \[ \int \frac {x^{11}}{\left (8 c-d x^3\right )^2 \sqrt {c+d x^3}} \, dx=\frac {2 \, {\left (736 \, c^{\frac {3}{2}} \log \left (\frac {\sqrt {d x^{3} + c} - 3 \, \sqrt {c}}{\sqrt {d x^{3} + c} + 3 \, \sqrt {c}}\right ) + 9 \, {\left (d x^{3} + c\right )}^{\frac {3}{2}} + 405 \, \sqrt {d x^{3} + c} c - \frac {768 \, \sqrt {d x^{3} + c} c^{2}}{d x^{3} - 8 \, c}\right )}}{81 \, d^{4}} \]
2/81*(736*c^(3/2)*log((sqrt(d*x^3 + c) - 3*sqrt(c))/(sqrt(d*x^3 + c) + 3*s qrt(c))) + 9*(d*x^3 + c)^(3/2) + 405*sqrt(d*x^3 + c)*c - 768*sqrt(d*x^3 + c)*c^2/(d*x^3 - 8*c))/d^4
Time = 0.27 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.98 \[ \int \frac {x^{11}}{\left (8 c-d x^3\right )^2 \sqrt {c+d x^3}} \, dx=\frac {2944 \, c^{2} \arctan \left (\frac {\sqrt {d x^{3} + c}}{3 \, \sqrt {-c}}\right )}{81 \, \sqrt {-c} d^{4}} - \frac {512 \, \sqrt {d x^{3} + c} c^{2}}{27 \, {\left (d x^{3} - 8 \, c\right )} d^{4}} + \frac {2 \, {\left ({\left (d x^{3} + c\right )}^{\frac {3}{2}} d^{8} + 45 \, \sqrt {d x^{3} + c} c d^{8}\right )}}{9 \, d^{12}} \]
2944/81*c^2*arctan(1/3*sqrt(d*x^3 + c)/sqrt(-c))/(sqrt(-c)*d^4) - 512/27*s qrt(d*x^3 + c)*c^2/((d*x^3 - 8*c)*d^4) + 2/9*((d*x^3 + c)^(3/2)*d^8 + 45*s qrt(d*x^3 + c)*c*d^8)/d^12
Time = 8.05 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.13 \[ \int \frac {x^{11}}{\left (8 c-d x^3\right )^2 \sqrt {c+d x^3}} \, dx=\frac {92\,c\,\sqrt {d\,x^3+c}}{9\,d^4}+\frac {1472\,c^{3/2}\,\ln \left (\frac {10\,c+d\,x^3-6\,\sqrt {c}\,\sqrt {d\,x^3+c}}{8\,c-d\,x^3}\right )}{81\,d^4}+\frac {2\,x^3\,\sqrt {d\,x^3+c}}{9\,d^3}+\frac {512\,c^2\,\sqrt {d\,x^3+c}}{27\,d^4\,\left (8\,c-d\,x^3\right )} \]