Integrand size = 27, antiderivative size = 67 \[ \int \frac {x^2}{\left (8 c-d x^3\right )^2 \sqrt {c+d x^3}} \, dx=\frac {\sqrt {c+d x^3}}{27 c d \left (8 c-d x^3\right )}+\frac {\text {arctanh}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{81 c^{3/2} d} \]
Time = 0.09 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.96 \[ \int \frac {x^2}{\left (8 c-d x^3\right )^2 \sqrt {c+d x^3}} \, dx=\frac {\frac {3 \sqrt {c} \sqrt {c+d x^3}}{8 c-d x^3}+\text {arctanh}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{81 c^{3/2} d} \]
((3*Sqrt[c]*Sqrt[c + d*x^3])/(8*c - d*x^3) + ArcTanh[Sqrt[c + d*x^3]/(3*Sq rt[c])])/(81*c^(3/2)*d)
Time = 0.20 (sec) , antiderivative size = 71, 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 = {946, 52, 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.
\(\displaystyle \int \frac {x^2}{\left (8 c-d x^3\right )^2 \sqrt {c+d x^3}} \, dx\) |
\(\Big \downarrow \) 946 |
\(\displaystyle \frac {1}{3} \int \frac {1}{\left (8 c-d x^3\right )^2 \sqrt {d x^3+c}}dx^3\) |
\(\Big \downarrow \) 52 |
\(\displaystyle \frac {1}{3} \left (\frac {\int \frac {1}{\left (8 c-d x^3\right ) \sqrt {d x^3+c}}dx^3}{18 c}+\frac {\sqrt {c+d x^3}}{9 c d \left (8 c-d x^3\right )}\right )\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {1}{3} \left (\frac {\int \frac {1}{9 c-x^6}d\sqrt {d x^3+c}}{9 c d}+\frac {\sqrt {c+d x^3}}{9 c d \left (8 c-d x^3\right )}\right )\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {1}{3} \left (\frac {\text {arctanh}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{27 c^{3/2} d}+\frac {\sqrt {c+d x^3}}{9 c d \left (8 c-d x^3\right )}\right )\) |
(Sqrt[c + d*x^3]/(9*c*d*(8*c - d*x^3)) + ArcTanh[Sqrt[c + d*x^3]/(3*Sqrt[c ])]/(27*c^(3/2)*d))/3
3.5.27.3.1 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] && ILtQ[m, -1] && FractionQ[n] && LtQ[n, 0]
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_)^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[(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] && EqQ[m - n + 1, 0]
Time = 4.25 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.78
method | result | size |
default | \(\frac {\frac {\sqrt {d \,x^{3}+c}}{c \left (-d \,x^{3}+8 c \right )}+\frac {\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{3 \sqrt {c}}\right )}{3 c^{\frac {3}{2}}}}{27 d}\) | \(52\) |
pseudoelliptic | \(\frac {\frac {\sqrt {d \,x^{3}+c}}{c \left (-d \,x^{3}+8 c \right )}+\frac {\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{3 \sqrt {c}}\right )}{3 c^{\frac {3}{2}}}}{27 d}\) | \(52\) |
elliptic | \(\frac {\sqrt {d \,x^{3}+c}}{27 c d \left (-d \,x^{3}+8 c \right )}-\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 )}{486 d^{3} c^{2}}\) | \(443\) |
Time = 0.28 (sec) , antiderivative size = 153, normalized size of antiderivative = 2.28 \[ \int \frac {x^2}{\left (8 c-d x^3\right )^2 \sqrt {c+d x^3}} \, dx=\left [\frac {{\left (d x^{3} - 8 \, c\right )} \sqrt {c} \log \left (\frac {d x^{3} + 6 \, \sqrt {d x^{3} + c} \sqrt {c} + 10 \, c}{d x^{3} - 8 \, c}\right ) - 6 \, \sqrt {d x^{3} + c} c}{162 \, {\left (c^{2} d^{2} x^{3} - 8 \, c^{3} d\right )}}, -\frac {{\left (d x^{3} - 8 \, c\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {d x^{3} + c} \sqrt {-c}}{3 \, c}\right ) + 3 \, \sqrt {d x^{3} + c} c}{81 \, {\left (c^{2} d^{2} x^{3} - 8 \, c^{3} d\right )}}\right ] \]
[1/162*((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)) - 6*sqrt(d*x^3 + c)*c)/(c^2*d^2*x^3 - 8*c^3*d), -1/81*(( d*x^3 - 8*c)*sqrt(-c)*arctan(1/3*sqrt(d*x^3 + c)*sqrt(-c)/c) + 3*sqrt(d*x^ 3 + c)*c)/(c^2*d^2*x^3 - 8*c^3*d)]
\[ \int \frac {x^2}{\left (8 c-d x^3\right )^2 \sqrt {c+d x^3}} \, dx=\int \frac {x^{2}}{\left (- 8 c + d x^{3}\right )^{2} \sqrt {c + d x^{3}}}\, dx \]
Time = 0.28 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.07 \[ \int \frac {x^2}{\left (8 c-d x^3\right )^2 \sqrt {c+d x^3}} \, dx=-\frac {\frac {6 \, \sqrt {d x^{3} + c}}{{\left (d x^{3} + c\right )} c - 9 \, c^{2}} + \frac {\log \left (\frac {\sqrt {d x^{3} + c} - 3 \, \sqrt {c}}{\sqrt {d x^{3} + c} + 3 \, \sqrt {c}}\right )}{c^{\frac {3}{2}}}}{162 \, d} \]
-1/162*(6*sqrt(d*x^3 + c)/((d*x^3 + c)*c - 9*c^2) + log((sqrt(d*x^3 + c) - 3*sqrt(c))/(sqrt(d*x^3 + c) + 3*sqrt(c)))/c^(3/2))/d
Time = 0.28 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.88 \[ \int \frac {x^2}{\left (8 c-d x^3\right )^2 \sqrt {c+d x^3}} \, dx=-\frac {\arctan \left (\frac {\sqrt {d x^{3} + c}}{3 \, \sqrt {-c}}\right )}{81 \, \sqrt {-c} c d} - \frac {\sqrt {d x^{3} + c}}{27 \, {\left (d x^{3} - 8 \, c\right )} c d} \]
-1/81*arctan(1/3*sqrt(d*x^3 + c)/sqrt(-c))/(sqrt(-c)*c*d) - 1/27*sqrt(d*x^ 3 + c)/((d*x^3 - 8*c)*c*d)
Time = 7.92 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.12 \[ \int \frac {x^2}{\left (8 c-d x^3\right )^2 \sqrt {c+d x^3}} \, dx=\frac {\ln \left (\frac {10\,c+d\,x^3+6\,\sqrt {c}\,\sqrt {d\,x^3+c}}{8\,c-d\,x^3}\right )}{162\,c^{3/2}\,d}+\frac {\sqrt {d\,x^3+c}}{27\,c\,d\,\left (8\,c-d\,x^3\right )} \]