Integrand size = 27, antiderivative size = 90 \[ \int \frac {x^{11}}{\left (8 c-d x^3\right ) \sqrt {c+d x^3}} \, dx=-\frac {38 c^2 \sqrt {c+d x^3}}{d^4}-\frac {4 c \left (c+d x^3\right )^{3/2}}{3 d^4}-\frac {2 \left (c+d x^3\right )^{5/2}}{15 d^4}+\frac {1024 c^{5/2} \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{9 d^4} \]
-4/3*c*(d*x^3+c)^(3/2)/d^4-2/15*(d*x^3+c)^(5/2)/d^4+1024/9*c^(5/2)*arctanh (1/3*(d*x^3+c)^(1/2)/c^(1/2))/d^4-38*c^2*(d*x^3+c)^(1/2)/d^4
Time = 0.11 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.77 \[ \int \frac {x^{11}}{\left (8 c-d x^3\right ) \sqrt {c+d x^3}} \, dx=\frac {-6 \sqrt {c+d x^3} \left (296 c^2+12 c d x^3+d^2 x^6\right )+5120 c^{5/2} \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{45 d^4} \]
(-6*Sqrt[c + d*x^3]*(296*c^2 + 12*c*d*x^3 + d^2*x^6) + 5120*c^(5/2)*ArcTan h[Sqrt[c + d*x^3]/(3*Sqrt[c])])/(45*d^4)
Time = 0.24 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.02, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, 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.
\(\displaystyle \int \frac {x^{11}}{\left (8 c-d x^3\right ) \sqrt {c+d x^3}} \, dx\) |
\(\Big \downarrow \) 948 |
\(\displaystyle \frac {1}{3} \int \frac {x^9}{\left (8 c-d x^3\right ) \sqrt {d x^3+c}}dx^3\) |
\(\Big \downarrow \) 99 |
\(\displaystyle \frac {1}{3} \int \left (\frac {512 c^3}{d^3 \left (8 c-d x^3\right ) \sqrt {d x^3+c}}-\frac {57 c^2}{d^3 \sqrt {d x^3+c}}-\frac {6 \sqrt {d x^3+c} c}{d^3}-\frac {\left (d x^3+c\right )^{3/2}}{d^3}\right )dx^3\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{3} \left (\frac {1024 c^{5/2} \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{3 d^4}-\frac {114 c^2 \sqrt {c+d x^3}}{d^4}-\frac {4 c \left (c+d x^3\right )^{3/2}}{d^4}-\frac {2 \left (c+d x^3\right )^{5/2}}{5 d^4}\right )\) |
((-114*c^2*Sqrt[c + d*x^3])/d^4 - (4*c*(c + d*x^3)^(3/2))/d^4 - (2*(c + d* x^3)^(5/2))/(5*d^4) + (1024*c^(5/2)*ArcTanh[Sqrt[c + d*x^3]/(3*Sqrt[c])])/ (3*d^4))/3
3.4.8.3.1 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 = 4.50 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.64
method | result | size |
pseudoelliptic | \(\frac {5120 c^{\frac {5}{2}} \operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{3 \sqrt {c}}\right )-6 \sqrt {d \,x^{3}+c}\, \left (d^{2} x^{6}+12 c d \,x^{3}+296 c^{2}\right )}{45 d^{4}}\) | \(58\) |
risch | \(-\frac {2 \left (d^{2} x^{6}+12 c d \,x^{3}+296 c^{2}\right ) \sqrt {d \,x^{3}+c}}{15 d^{4}}+\frac {1024 c^{\frac {5}{2}} \operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{3 \sqrt {c}}\right )}{9 d^{4}}\) | \(59\) |
default | \(-\frac {\frac {2 x^{6} \sqrt {d \,x^{3}+c}}{15 d}-\frac {8 c \,x^{3} \sqrt {d \,x^{3}+c}}{45 d^{2}}+\frac {16 c^{2} \sqrt {d \,x^{3}+c}}{45 d^{3}}}{d}-\frac {8 c \left (\frac {2 x^{3} \sqrt {d \,x^{3}+c}}{9 d}-\frac {4 c \sqrt {d \,x^{3}+c}}{9 d^{2}}\right )}{d^{2}}-\frac {128 c^{2} \sqrt {d \,x^{3}+c}}{3 d^{4}}+\frac {1024 c^{\frac {5}{2}} \operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{3 \sqrt {c}}\right )}{9 d^{4}}\) | \(139\) |
elliptic | \(-\frac {2 x^{6} \sqrt {d \,x^{3}+c}}{15 d^{2}}-\frac {8 c \,x^{3} \sqrt {d \,x^{3}+c}}{5 d^{3}}-\frac {592 c^{2} \sqrt {d \,x^{3}+c}}{15 d^{4}}-\frac {512 i c^{2} \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 )}{27 d^{6}}\) | \(466\) |
1/45*(5120*c^(5/2)*arctanh(1/3*(d*x^3+c)^(1/2)/c^(1/2))-6*(d*x^3+c)^(1/2)* (d^2*x^6+12*c*d*x^3+296*c^2))/d^4
Time = 0.31 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.62 \[ \int \frac {x^{11}}{\left (8 c-d x^3\right ) \sqrt {c+d x^3}} \, dx=\left [\frac {2 \, {\left (1280 \, c^{\frac {5}{2}} \log \left (\frac {d x^{3} + 6 \, \sqrt {d x^{3} + c} \sqrt {c} + 10 \, c}{d x^{3} - 8 \, c}\right ) - 3 \, {\left (d^{2} x^{6} + 12 \, c d x^{3} + 296 \, c^{2}\right )} \sqrt {d x^{3} + c}\right )}}{45 \, d^{4}}, -\frac {2 \, {\left (2560 \, \sqrt {-c} c^{2} \arctan \left (\frac {\sqrt {d x^{3} + c} \sqrt {-c}}{3 \, c}\right ) + 3 \, {\left (d^{2} x^{6} + 12 \, c d x^{3} + 296 \, c^{2}\right )} \sqrt {d x^{3} + c}\right )}}{45 \, d^{4}}\right ] \]
[2/45*(1280*c^(5/2)*log((d*x^3 + 6*sqrt(d*x^3 + c)*sqrt(c) + 10*c)/(d*x^3 - 8*c)) - 3*(d^2*x^6 + 12*c*d*x^3 + 296*c^2)*sqrt(d*x^3 + c))/d^4, -2/45*( 2560*sqrt(-c)*c^2*arctan(1/3*sqrt(d*x^3 + c)*sqrt(-c)/c) + 3*(d^2*x^6 + 12 *c*d*x^3 + 296*c^2)*sqrt(d*x^3 + c))/d^4]
Time = 16.60 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.02 \[ \int \frac {x^{11}}{\left (8 c-d x^3\right ) \sqrt {c+d x^3}} \, dx=\begin {cases} \frac {2 \left (- \frac {512 c^{3} \operatorname {atan}{\left (\frac {\sqrt {c + d x^{3}}}{3 \sqrt {- c}} \right )}}{9 \sqrt {- c}} - 19 c^{2} \sqrt {c + d x^{3}} - \frac {2 c \left (c + d x^{3}\right )^{\frac {3}{2}}}{3} - \frac {\left (c + d x^{3}\right )^{\frac {5}{2}}}{15}\right )}{d^{4}} & \text {for}\: d \neq 0 \\\frac {x^{12}}{96 c^{\frac {3}{2}}} & \text {otherwise} \end {cases} \]
Piecewise((2*(-512*c**3*atan(sqrt(c + d*x**3)/(3*sqrt(-c)))/(9*sqrt(-c)) - 19*c**2*sqrt(c + d*x**3) - 2*c*(c + d*x**3)**(3/2)/3 - (c + d*x**3)**(5/2 )/15)/d**4, Ne(d, 0)), (x**12/(96*c**(3/2)), True))
Time = 0.27 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.91 \[ \int \frac {x^{11}}{\left (8 c-d x^3\right ) \sqrt {c+d x^3}} \, dx=-\frac {2 \, {\left (1280 \, c^{\frac {5}{2}} \log \left (\frac {\sqrt {d x^{3} + c} - 3 \, \sqrt {c}}{\sqrt {d x^{3} + c} + 3 \, \sqrt {c}}\right ) + 3 \, {\left (d x^{3} + c\right )}^{\frac {5}{2}} + 30 \, {\left (d x^{3} + c\right )}^{\frac {3}{2}} c + 855 \, \sqrt {d x^{3} + c} c^{2}\right )}}{45 \, d^{4}} \]
-2/45*(1280*c^(5/2)*log((sqrt(d*x^3 + c) - 3*sqrt(c))/(sqrt(d*x^3 + c) + 3 *sqrt(c))) + 3*(d*x^3 + c)^(5/2) + 30*(d*x^3 + c)^(3/2)*c + 855*sqrt(d*x^3 + c)*c^2)/d^4
Time = 0.31 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.91 \[ \int \frac {x^{11}}{\left (8 c-d x^3\right ) \sqrt {c+d x^3}} \, dx=-\frac {1024 \, c^{3} \arctan \left (\frac {\sqrt {d x^{3} + c}}{3 \, \sqrt {-c}}\right )}{9 \, \sqrt {-c} d^{4}} - \frac {2 \, {\left ({\left (d x^{3} + c\right )}^{\frac {5}{2}} d^{16} + 10 \, {\left (d x^{3} + c\right )}^{\frac {3}{2}} c d^{16} + 285 \, \sqrt {d x^{3} + c} c^{2} d^{16}\right )}}{15 \, d^{20}} \]
-1024/9*c^3*arctan(1/3*sqrt(d*x^3 + c)/sqrt(-c))/(sqrt(-c)*d^4) - 2/15*((d *x^3 + c)^(5/2)*d^16 + 10*(d*x^3 + c)^(3/2)*c*d^16 + 285*sqrt(d*x^3 + c)*c ^2*d^16)/d^20
Time = 7.47 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.09 \[ \int \frac {x^{11}}{\left (8 c-d x^3\right ) \sqrt {c+d x^3}} \, dx=\frac {512\,c^{5/2}\,\ln \left (\frac {10\,c+d\,x^3+6\,\sqrt {c}\,\sqrt {d\,x^3+c}}{8\,c-d\,x^3}\right )}{9\,d^4}-\frac {592\,c^2\,\sqrt {d\,x^3+c}}{15\,d^4}-\frac {2\,x^6\,\sqrt {d\,x^3+c}}{15\,d^2}-\frac {8\,c\,x^3\,\sqrt {d\,x^3+c}}{5\,d^3} \]