Integrand size = 27, antiderivative size = 64 \[ \int \frac {x^5}{\left (8 c-d x^3\right )^2 \sqrt {c+d x^3}} \, dx=\frac {8 \sqrt {c+d x^3}}{27 d^2 \left (8 c-d x^3\right )}-\frac {10 \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{81 \sqrt {c} d^2} \]
-10/81*arctanh(1/3*(d*x^3+c)^(1/2)/c^(1/2))/d^2/c^(1/2)+8/27*(d*x^3+c)^(1/ 2)/d^2/(-d*x^3+8*c)
Time = 0.12 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.98 \[ \int \frac {x^5}{\left (8 c-d x^3\right )^2 \sqrt {c+d x^3}} \, dx=-\frac {8 \sqrt {c+d x^3}}{27 d^2 \left (-8 c+d x^3\right )}-\frac {10 \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{81 \sqrt {c} d^2} \]
(-8*Sqrt[c + d*x^3])/(27*d^2*(-8*c + d*x^3)) - (10*ArcTanh[Sqrt[c + d*x^3] /(3*Sqrt[c])])/(81*Sqrt[c]*d^2)
Time = 0.19 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.06, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {948, 87, 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.
\(\displaystyle \int \frac {x^5}{\left (8 c-d x^3\right )^2 \sqrt {c+d x^3}} \, dx\) |
\(\Big \downarrow \) 948 |
\(\displaystyle \frac {1}{3} \int \frac {x^3}{\left (8 c-d x^3\right )^2 \sqrt {d x^3+c}}dx^3\) |
\(\Big \downarrow \) 87 |
\(\displaystyle \frac {1}{3} \left (\frac {8 \sqrt {c+d x^3}}{9 d^2 \left (8 c-d x^3\right )}-\frac {5 \int \frac {1}{\left (8 c-d x^3\right ) \sqrt {d x^3+c}}dx^3}{9 d}\right )\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {1}{3} \left (\frac {8 \sqrt {c+d x^3}}{9 d^2 \left (8 c-d x^3\right )}-\frac {10 \int \frac {1}{9 c-x^6}d\sqrt {d x^3+c}}{9 d^2}\right )\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {1}{3} \left (\frac {8 \sqrt {c+d x^3}}{9 d^2 \left (8 c-d x^3\right )}-\frac {10 \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{27 \sqrt {c} d^2}\right )\) |
((8*Sqrt[c + d*x^3])/(9*d^2*(8*c - d*x^3)) - (10*ArcTanh[Sqrt[c + d*x^3]/( 3*Sqrt[c])])/(27*Sqrt[c]*d^2))/3
3.5.26.3.1 Defintions of rubi rules used
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_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p _.), x_] :> Simp[(-(b*e - a*f))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p + 1)*(c*f - d*e))), x] - Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)) Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || Intege rQ[p] || !(IntegerQ[n] || !(EqQ[e, 0] || !(EqQ[c, 0] || LtQ[p, n]))))
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 = 4.35 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.78
method | result | size |
pseudoelliptic | \(\frac {\frac {8 \sqrt {d \,x^{3}+c}}{27 \left (-d \,x^{3}+8 c \right )}-\frac {10 \,\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{3 \sqrt {c}}\right )}{81 \sqrt {c}}}{d^{2}}\) | \(50\) |
default | \(-\frac {2 \,\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{3 \sqrt {c}}\right )}{9 d^{2} \sqrt {c}}+\frac {8 c \left (\frac {\sqrt {d \,x^{3}+c}}{c \left (-d \,x^{3}+8 c \right )}+\frac {\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{3 \sqrt {c}}\right )}{3 c^{\frac {3}{2}}}\right )}{27 d^{2}}\) | \(77\) |
elliptic | \(\frac {8 \sqrt {d \,x^{3}+c}}{27 d^{2} \left (-d \,x^{3}+8 c \right )}+\frac {5 i \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^{4} c}\) | \(440\) |
Time = 0.30 (sec) , antiderivative size = 155, normalized size of antiderivative = 2.42 \[ \int \frac {x^5}{\left (8 c-d x^3\right )^2 \sqrt {c+d x^3}} \, dx=\left [\frac {5 \, {\left (d x^{3} - 8 \, c\right )} \sqrt {c} \log \left (\frac {d x^{3} - 6 \, \sqrt {d x^{3} + c} \sqrt {c} + 10 \, c}{d x^{3} - 8 \, c}\right ) - 24 \, \sqrt {d x^{3} + c} c}{81 \, {\left (c d^{3} x^{3} - 8 \, c^{2} d^{2}\right )}}, \frac {2 \, {\left (5 \, {\left (d x^{3} - 8 \, c\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {d x^{3} + c} \sqrt {-c}}{3 \, c}\right ) - 12 \, \sqrt {d x^{3} + c} c\right )}}{81 \, {\left (c d^{3} x^{3} - 8 \, c^{2} d^{2}\right )}}\right ] \]
[1/81*(5*(d*x^3 - 8*c)*sqrt(c)*log((d*x^3 - 6*sqrt(d*x^3 + c)*sqrt(c) + 10 *c)/(d*x^3 - 8*c)) - 24*sqrt(d*x^3 + c)*c)/(c*d^3*x^3 - 8*c^2*d^2), 2/81*( 5*(d*x^3 - 8*c)*sqrt(-c)*arctan(1/3*sqrt(d*x^3 + c)*sqrt(-c)/c) - 12*sqrt( d*x^3 + c)*c)/(c*d^3*x^3 - 8*c^2*d^2)]
\[ \int \frac {x^5}{\left (8 c-d x^3\right )^2 \sqrt {c+d x^3}} \, dx=\int \frac {x^{5}}{\left (- 8 c + d x^{3}\right )^{2} \sqrt {c + d x^{3}}}\, dx \]
Time = 0.29 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.05 \[ \int \frac {x^5}{\left (8 c-d x^3\right )^2 \sqrt {c+d x^3}} \, dx=\frac {\frac {5 \, \log \left (\frac {\sqrt {d x^{3} + c} - 3 \, \sqrt {c}}{\sqrt {d x^{3} + c} + 3 \, \sqrt {c}}\right )}{\sqrt {c}} - \frac {24 \, \sqrt {d x^{3} + c}}{d x^{3} - 8 \, c}}{81 \, d^{2}} \]
1/81*(5*log((sqrt(d*x^3 + c) - 3*sqrt(c))/(sqrt(d*x^3 + c) + 3*sqrt(c)))/s qrt(c) - 24*sqrt(d*x^3 + c)/(d*x^3 - 8*c))/d^2
Time = 0.28 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.91 \[ \int \frac {x^5}{\left (8 c-d x^3\right )^2 \sqrt {c+d x^3}} \, dx=\frac {2 \, {\left (\frac {5 \, \arctan \left (\frac {\sqrt {d x^{3} + c}}{3 \, \sqrt {-c}}\right )}{\sqrt {-c} d} - \frac {12 \, \sqrt {d x^{3} + c}}{{\left (d x^{3} - 8 \, c\right )} d}\right )}}{81 \, d} \]
2/81*(5*arctan(1/3*sqrt(d*x^3 + c)/sqrt(-c))/(sqrt(-c)*d) - 12*sqrt(d*x^3 + c)/((d*x^3 - 8*c)*d))/d
Time = 7.96 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.12 \[ \int \frac {x^5}{\left (8 c-d x^3\right )^2 \sqrt {c+d x^3}} \, dx=\frac {5\,\ln \left (\frac {10\,c+d\,x^3-6\,\sqrt {c}\,\sqrt {d\,x^3+c}}{8\,c-d\,x^3}\right )}{81\,\sqrt {c}\,d^2}+\frac {8\,\sqrt {d\,x^3+c}}{27\,d^2\,\left (8\,c-d\,x^3\right )} \]