Integrand size = 27, antiderivative size = 71 \[ \int \frac {x^8}{\left (8 c-d x^3\right ) \sqrt {c+d x^3}} \, dx=-\frac {14 c \sqrt {c+d x^3}}{3 d^3}-\frac {2 \left (c+d x^3\right )^{3/2}}{9 d^3}+\frac {128 c^{3/2} \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{9 d^3} \]
-2/9*(d*x^3+c)^(3/2)/d^3+128/9*c^(3/2)*arctanh(1/3*(d*x^3+c)^(1/2)/c^(1/2) )/d^3-14/3*c*(d*x^3+c)^(1/2)/d^3
Time = 0.09 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.82 \[ \int \frac {x^8}{\left (8 c-d x^3\right ) \sqrt {c+d x^3}} \, dx=\frac {-2 \sqrt {c+d x^3} \left (22 c+d x^3\right )+128 c^{3/2} \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{9 d^3} \]
(-2*Sqrt[c + d*x^3]*(22*c + d*x^3) + 128*c^(3/2)*ArcTanh[Sqrt[c + d*x^3]/( 3*Sqrt[c])])/(9*d^3)
Time = 0.23 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.03, 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.
\(\displaystyle \int \frac {x^8}{\left (8 c-d x^3\right ) \sqrt {c+d x^3}} \, dx\) |
\(\Big \downarrow \) 948 |
\(\displaystyle \frac {1}{3} \int \frac {x^6}{\left (8 c-d x^3\right ) \sqrt {d x^3+c}}dx^3\) |
\(\Big \downarrow \) 99 |
\(\displaystyle \frac {1}{3} \int \left (\frac {64 c^2}{d^2 \left (8 c-d x^3\right ) \sqrt {d x^3+c}}-\frac {7 c}{d^2 \sqrt {d x^3+c}}-\frac {\sqrt {d x^3+c}}{d^2}\right )dx^3\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{3} \left (\frac {128 c^{3/2} \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{3 d^3}-\frac {14 c \sqrt {c+d x^3}}{d^3}-\frac {2 \left (c+d x^3\right )^{3/2}}{3 d^3}\right )\) |
((-14*c*Sqrt[c + d*x^3])/d^3 - (2*(c + d*x^3)^(3/2))/(3*d^3) + (128*c^(3/2 )*ArcTanh[Sqrt[c + d*x^3]/(3*Sqrt[c])])/(3*d^3))/3
3.4.9.3.1 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 = 4.45 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.66
method | result | size |
pseudoelliptic | \(\frac {\frac {128 c^{\frac {3}{2}} \operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{3 \sqrt {c}}\right )}{9}-\frac {2 \left (d \,x^{3}+22 c \right ) \sqrt {d \,x^{3}+c}}{9}}{d^{3}}\) | \(47\) |
risch | \(-\frac {2 \left (d \,x^{3}+22 c \right ) \sqrt {d \,x^{3}+c}}{9 d^{3}}+\frac {128 c^{\frac {3}{2}} \operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{3 \sqrt {c}}\right )}{9 d^{3}}\) | \(48\) |
default | \(-\frac {\frac {2 x^{3} \sqrt {d \,x^{3}+c}}{9 d}-\frac {4 c \sqrt {d \,x^{3}+c}}{9 d^{2}}}{d}-\frac {16 c \sqrt {d \,x^{3}+c}}{3 d^{3}}+\frac {128 c^{\frac {3}{2}} \operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{3 \sqrt {c}}\right )}{9 d^{3}}\) | \(78\) |
elliptic | \(-\frac {2 x^{3} \sqrt {d \,x^{3}+c}}{9 d^{2}}-\frac {44 c \sqrt {d \,x^{3}+c}}{9 d^{3}}-\frac {64 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 )}{27 d^{5}}\) | \(444\) |
Time = 0.30 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.70 \[ \int \frac {x^8}{\left (8 c-d x^3\right ) \sqrt {c+d x^3}} \, dx=\left [\frac {2 \, {\left (32 \, c^{\frac {3}{2}} \log \left (\frac {d x^{3} + 6 \, \sqrt {d x^{3} + c} \sqrt {c} + 10 \, c}{d x^{3} - 8 \, c}\right ) - {\left (d x^{3} + 22 \, c\right )} \sqrt {d x^{3} + c}\right )}}{9 \, d^{3}}, -\frac {2 \, {\left (64 \, \sqrt {-c} c \arctan \left (\frac {\sqrt {d x^{3} + c} \sqrt {-c}}{3 \, c}\right ) + {\left (d x^{3} + 22 \, c\right )} \sqrt {d x^{3} + c}\right )}}{9 \, d^{3}}\right ] \]
[2/9*(32*c^(3/2)*log((d*x^3 + 6*sqrt(d*x^3 + c)*sqrt(c) + 10*c)/(d*x^3 - 8 *c)) - (d*x^3 + 22*c)*sqrt(d*x^3 + c))/d^3, -2/9*(64*sqrt(-c)*c*arctan(1/3 *sqrt(d*x^3 + c)*sqrt(-c)/c) + (d*x^3 + 22*c)*sqrt(d*x^3 + c))/d^3]
Time = 8.64 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.07 \[ \int \frac {x^8}{\left (8 c-d x^3\right ) \sqrt {c+d x^3}} \, dx=\begin {cases} \frac {2 \left (- \frac {64 c^{2} \operatorname {atan}{\left (\frac {\sqrt {c + d x^{3}}}{3 \sqrt {- c}} \right )}}{9 \sqrt {- c}} - \frac {7 c \sqrt {c + d x^{3}}}{3} - \frac {\left (c + d x^{3}\right )^{\frac {3}{2}}}{9}\right )}{d^{3}} & \text {for}\: d \neq 0 \\\frac {x^{9}}{72 c^{\frac {3}{2}}} & \text {otherwise} \end {cases} \]
Piecewise((2*(-64*c**2*atan(sqrt(c + d*x**3)/(3*sqrt(-c)))/(9*sqrt(-c)) - 7*c*sqrt(c + d*x**3)/3 - (c + d*x**3)**(3/2)/9)/d**3, Ne(d, 0)), (x**9/(72 *c**(3/2)), True))
Time = 0.27 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.93 \[ \int \frac {x^8}{\left (8 c-d x^3\right ) \sqrt {c+d x^3}} \, dx=-\frac {2 \, {\left (32 \, c^{\frac {3}{2}} \log \left (\frac {\sqrt {d x^{3} + c} - 3 \, \sqrt {c}}{\sqrt {d x^{3} + c} + 3 \, \sqrt {c}}\right ) + {\left (d x^{3} + c\right )}^{\frac {3}{2}} + 21 \, \sqrt {d x^{3} + c} c\right )}}{9 \, d^{3}} \]
-2/9*(32*c^(3/2)*log((sqrt(d*x^3 + c) - 3*sqrt(c))/(sqrt(d*x^3 + c) + 3*sq rt(c))) + (d*x^3 + c)^(3/2) + 21*sqrt(d*x^3 + c)*c)/d^3
Time = 0.28 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.92 \[ \int \frac {x^8}{\left (8 c-d x^3\right ) \sqrt {c+d x^3}} \, dx=-\frac {128 \, c^{2} \arctan \left (\frac {\sqrt {d x^{3} + c}}{3 \, \sqrt {-c}}\right )}{9 \, \sqrt {-c} d^{3}} - \frac {2 \, {\left ({\left (d x^{3} + c\right )}^{\frac {3}{2}} d^{6} + 21 \, \sqrt {d x^{3} + c} c d^{6}\right )}}{9 \, d^{9}} \]
-128/9*c^2*arctan(1/3*sqrt(d*x^3 + c)/sqrt(-c))/(sqrt(-c)*d^3) - 2/9*((d*x ^3 + c)^(3/2)*d^6 + 21*sqrt(d*x^3 + c)*c*d^6)/d^9
Time = 7.44 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.10 \[ \int \frac {x^8}{\left (8 c-d x^3\right ) \sqrt {c+d x^3}} \, dx=\frac {64\,c^{3/2}\,\ln \left (\frac {10\,c+d\,x^3+6\,\sqrt {c}\,\sqrt {d\,x^3+c}}{8\,c-d\,x^3}\right )}{9\,d^3}-\frac {44\,c\,\sqrt {d\,x^3+c}}{9\,d^3}-\frac {2\,x^3\,\sqrt {d\,x^3+c}}{9\,d^2} \]