Integrand size = 27, antiderivative size = 85 \[ \int \frac {x^5}{\left (8 c-d x^3\right )^2 \left (c+d x^3\right )^{3/2}} \, dx=-\frac {2}{81 c d^2 \sqrt {c+d x^3}}+\frac {8}{27 d^2 \left (8 c-d x^3\right ) \sqrt {c+d x^3}}+\frac {2 \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{243 c^{3/2} d^2} \]
2/243*arctanh(1/3*(d*x^3+c)^(1/2)/c^(1/2))/c^(3/2)/d^2-2/81/c/d^2/(d*x^3+c )^(1/2)+8/27/d^2/(-d*x^3+8*c)/(d*x^3+c)^(1/2)
Time = 0.15 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.86 \[ \int \frac {x^5}{\left (8 c-d x^3\right )^2 \left (c+d x^3\right )^{3/2}} \, dx=\frac {2 \left (\frac {3 \sqrt {c} \left (4 c+d x^3\right )}{\left (8 c-d x^3\right ) \sqrt {c+d x^3}}+\text {arctanh}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )\right )}{243 c^{3/2} d^2} \]
(2*((3*Sqrt[c]*(4*c + d*x^3))/((8*c - d*x^3)*Sqrt[c + d*x^3]) + ArcTanh[Sq rt[c + d*x^3]/(3*Sqrt[c])]))/(243*c^(3/2)*d^2)
Time = 0.20 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.14, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {948, 87, 61, 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.
\(\displaystyle \int \frac {x^5}{\left (8 c-d x^3\right )^2 \left (c+d x^3\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 948 |
\(\displaystyle \frac {1}{3} \int \frac {x^3}{\left (8 c-d x^3\right )^2 \left (d x^3+c\right )^{3/2}}dx^3\) |
\(\Big \downarrow \) 87 |
\(\displaystyle \frac {1}{3} \left (\frac {\int \frac {1}{\left (8 c-d x^3\right ) \left (d x^3+c\right )^{3/2}}dx^3}{3 d}+\frac {8}{9 d^2 \left (8 c-d x^3\right ) \sqrt {c+d x^3}}\right )\) |
\(\Big \downarrow \) 61 |
\(\displaystyle \frac {1}{3} \left (\frac {\frac {\int \frac {1}{\left (8 c-d x^3\right ) \sqrt {d x^3+c}}dx^3}{9 c}-\frac {2}{9 c d \sqrt {c+d x^3}}}{3 d}+\frac {8}{9 d^2 \left (8 c-d x^3\right ) \sqrt {c+d x^3}}\right )\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {1}{3} \left (\frac {\frac {2 \int \frac {1}{9 c-x^6}d\sqrt {d x^3+c}}{9 c d}-\frac {2}{9 c d \sqrt {c+d x^3}}}{3 d}+\frac {8}{9 d^2 \left (8 c-d x^3\right ) \sqrt {c+d x^3}}\right )\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {1}{3} \left (\frac {\frac {2 \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{27 c^{3/2} d}-\frac {2}{9 c d \sqrt {c+d x^3}}}{3 d}+\frac {8}{9 d^2 \left (8 c-d x^3\right ) \sqrt {c+d x^3}}\right )\) |
(8/(9*d^2*(8*c - d*x^3)*Sqrt[c + d*x^3]) + (-2/(9*c*d*Sqrt[c + d*x^3]) + ( 2*ArcTanh[Sqrt[c + d*x^3]/(3*Sqrt[c])])/(27*c^(3/2)*d))/(3*d))/3
3.5.44.3.1 Defintions of rubi rules used
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(( m + n + 2)/((b*c - a*d)*(m + 1))) Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && LtQ[m, -1] && !(LtQ[n, -1] && (EqQ[a, 0 ] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d , m, n, 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_))*((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 = 4.38 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.72
method | result | size |
pseudoelliptic | \(\frac {\frac {8 \sqrt {d \,x^{3}+c}}{243 \left (-d \,x^{3}+8 c \right )}+\frac {2 \,\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{3 \sqrt {c}}\right )}{243 \sqrt {c}}+\frac {2}{243 \sqrt {d \,x^{3}+c}}}{c \,d^{2}}\) | \(61\) |
default | \(\frac {-\frac {2 \,\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{3 \sqrt {c}}\right ) \sqrt {d \,x^{3}+c}}{81}+\frac {2 \sqrt {c}}{27}}{d^{2} \sqrt {d \,x^{3}+c}\, c^{\frac {3}{2}}}+\frac {-\frac {16}{243 \sqrt {d \,x^{3}+c}}+\frac {8 \sqrt {d \,x^{3}+c}}{243 \left (-d \,x^{3}+8 c \right )}+\frac {8 \,\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{3 \sqrt {c}}\right )}{243 \sqrt {c}}}{c \,d^{2}}\) | \(110\) |
elliptic | \(\frac {8 \sqrt {d \,x^{3}+c}}{243 d^{2} c \left (-d \,x^{3}+8 c \right )}+\frac {2}{243 d^{2} c \sqrt {\left (x^{3}+\frac {c}{d}\right ) d}}-\frac {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 )}{729 d^{4} c^{2}}\) | \(464\) |
2/243/c*(4*(d*x^3+c)^(1/2)/(-d*x^3+8*c)+arctanh(1/3*(d*x^3+c)^(1/2)/c^(1/2 ))/c^(1/2)+1/(d*x^3+c)^(1/2))/d^2
Time = 0.35 (sec) , antiderivative size = 223, normalized size of antiderivative = 2.62 \[ \int \frac {x^5}{\left (8 c-d x^3\right )^2 \left (c+d x^3\right )^{3/2}} \, dx=\left [\frac {{\left (d^{2} x^{6} - 7 \, c d x^{3} - 8 \, c^{2}\right )} \sqrt {c} \log \left (\frac {d x^{3} + 6 \, \sqrt {d x^{3} + c} \sqrt {c} + 10 \, c}{d x^{3} - 8 \, c}\right ) - 6 \, {\left (c d x^{3} + 4 \, c^{2}\right )} \sqrt {d x^{3} + c}}{243 \, {\left (c^{2} d^{4} x^{6} - 7 \, c^{3} d^{3} x^{3} - 8 \, c^{4} d^{2}\right )}}, -\frac {2 \, {\left ({\left (d^{2} x^{6} - 7 \, c d x^{3} - 8 \, c^{2}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {d x^{3} + c} \sqrt {-c}}{3 \, c}\right ) + 3 \, {\left (c d x^{3} + 4 \, c^{2}\right )} \sqrt {d x^{3} + c}\right )}}{243 \, {\left (c^{2} d^{4} x^{6} - 7 \, c^{3} d^{3} x^{3} - 8 \, c^{4} d^{2}\right )}}\right ] \]
[1/243*((d^2*x^6 - 7*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)) - 6*(c*d*x^3 + 4*c^2)*sqrt(d*x^3 + c))/( c^2*d^4*x^6 - 7*c^3*d^3*x^3 - 8*c^4*d^2), -2/243*((d^2*x^6 - 7*c*d*x^3 - 8 *c^2)*sqrt(-c)*arctan(1/3*sqrt(d*x^3 + c)*sqrt(-c)/c) + 3*(c*d*x^3 + 4*c^2 )*sqrt(d*x^3 + c))/(c^2*d^4*x^6 - 7*c^3*d^3*x^3 - 8*c^4*d^2)]
\[ \int \frac {x^5}{\left (8 c-d x^3\right )^2 \left (c+d x^3\right )^{3/2}} \, dx=\int \frac {x^{5}}{\left (- 8 c + d x^{3}\right )^{2} \left (c + d x^{3}\right )^{\frac {3}{2}}}\, dx \]
Time = 0.29 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.98 \[ \int \frac {x^5}{\left (8 c-d x^3\right )^2 \left (c+d x^3\right )^{3/2}} \, dx=-\frac {\frac {6 \, {\left (d x^{3} + 4 \, c\right )}}{{\left (d x^{3} + c\right )}^{\frac {3}{2}} c - 9 \, \sqrt {d x^{3} + c} c^{2}} + \frac {\log \left (\frac {\sqrt {d x^{3} + c} - 3 \, \sqrt {c}}{\sqrt {d x^{3} + c} + 3 \, \sqrt {c}}\right )}{c^{\frac {3}{2}}}}{243 \, d^{2}} \]
-1/243*(6*(d*x^3 + 4*c)/((d*x^3 + c)^(3/2)*c - 9*sqrt(d*x^3 + c)*c^2) + lo g((sqrt(d*x^3 + c) - 3*sqrt(c))/(sqrt(d*x^3 + c) + 3*sqrt(c)))/c^(3/2))/d^ 2
Time = 0.30 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.89 \[ \int \frac {x^5}{\left (8 c-d x^3\right )^2 \left (c+d x^3\right )^{3/2}} \, dx=-\frac {2 \, {\left (\frac {\arctan \left (\frac {\sqrt {d x^{3} + c}}{3 \, \sqrt {-c}}\right )}{\sqrt {-c} c d} + \frac {3 \, {\left (d x^{3} + 4 \, c\right )}}{{\left ({\left (d x^{3} + c\right )}^{\frac {3}{2}} - 9 \, \sqrt {d x^{3} + c} c\right )} c d}\right )}}{243 \, d} \]
-2/243*(arctan(1/3*sqrt(d*x^3 + c)/sqrt(-c))/(sqrt(-c)*c*d) + 3*(d*x^3 + 4 *c)/(((d*x^3 + c)^(3/2) - 9*sqrt(d*x^3 + c)*c)*c*d))/d
Time = 8.28 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.13 \[ \int \frac {x^5}{\left (8 c-d x^3\right )^2 \left (c+d x^3\right )^{3/2}} \, dx=\frac {\left (\frac {8}{81\,d^2}+\frac {2\,x^3}{81\,c\,d}\right )\,\sqrt {d\,x^3+c}}{8\,c^2+7\,c\,d\,x^3-d^2\,x^6}+\frac {\ln \left (\frac {10\,c+d\,x^3+6\,\sqrt {c}\,\sqrt {d\,x^3+c}}{8\,c-d\,x^3}\right )}{243\,c^{3/2}\,d^2} \]