Integrand size = 27, antiderivative size = 67 \[ \int \frac {x^2 \left (c+d x^3\right )^{3/2}}{8 c-d x^3} \, dx=-\frac {6 c \sqrt {c+d x^3}}{d}-\frac {2 \left (c+d x^3\right )^{3/2}}{9 d}+\frac {18 c^{3/2} \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{d} \]
-2/9*(d*x^3+c)^(3/2)/d+18*c^(3/2)*arctanh(1/3*(d*x^3+c)^(1/2)/c^(1/2))/d-6 *c*(d*x^3+c)^(1/2)/d
Time = 0.09 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.88 \[ \int \frac {x^2 \left (c+d x^3\right )^{3/2}}{8 c-d x^3} \, dx=-\frac {2 \sqrt {c+d x^3} \left (28 c+d x^3\right )}{9 d}+\frac {18 c^{3/2} \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{d} \]
(-2*Sqrt[c + d*x^3]*(28*c + d*x^3))/(9*d) + (18*c^(3/2)*ArcTanh[Sqrt[c + d *x^3]/(3*Sqrt[c])])/d
Time = 0.21 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.10, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {946, 60, 60, 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.
\(\displaystyle \int \frac {x^2 \left (c+d x^3\right )^{3/2}}{8 c-d x^3} \, dx\) |
\(\Big \downarrow \) 946 |
\(\displaystyle \frac {1}{3} \int \frac {\left (d x^3+c\right )^{3/2}}{8 c-d x^3}dx^3\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {1}{3} \left (9 c \int \frac {\sqrt {d x^3+c}}{8 c-d x^3}dx^3-\frac {2 \left (c+d x^3\right )^{3/2}}{3 d}\right )\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {1}{3} \left (9 c \left (9 c \int \frac {1}{\left (8 c-d x^3\right ) \sqrt {d x^3+c}}dx^3-\frac {2 \sqrt {c+d x^3}}{d}\right )-\frac {2 \left (c+d x^3\right )^{3/2}}{3 d}\right )\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {1}{3} \left (9 c \left (\frac {18 c \int \frac {1}{9 c-x^6}d\sqrt {d x^3+c}}{d}-\frac {2 \sqrt {c+d x^3}}{d}\right )-\frac {2 \left (c+d x^3\right )^{3/2}}{3 d}\right )\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {1}{3} \left (9 c \left (\frac {6 \sqrt {c} \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{d}-\frac {2 \sqrt {c+d x^3}}{d}\right )-\frac {2 \left (c+d x^3\right )^{3/2}}{3 d}\right )\) |
((-2*(c + d*x^3)^(3/2))/(3*d) + 9*c*((-2*Sqrt[c + d*x^3])/d + (6*Sqrt[c]*A rcTanh[Sqrt[c + d*x^3]/(3*Sqrt[c])])/d))/3
3.3.98.3.1 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/(b*(m + n + 1))), x] + Simp[n*((b*c - a*d)/( b*(m + n + 1))) Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] && GtQ[n, 0] && NeQ[m + n + 1, 0] && !(IGtQ[m, 0] && ( !Integer Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) && !ILtQ[m + n + 2, 0] && IntLinear Q[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_)^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 = 4.16 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.70
method | result | size |
default | \(\frac {18 c^{\frac {3}{2}} \operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{3 \sqrt {c}}\right )-\frac {2 \left (d \,x^{3}+28 c \right ) \sqrt {d \,x^{3}+c}}{9}}{d}\) | \(47\) |
pseudoelliptic | \(\frac {18 c^{\frac {3}{2}} \operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{3 \sqrt {c}}\right )-\frac {2 \left (d \,x^{3}+28 c \right ) \sqrt {d \,x^{3}+c}}{9}}{d}\) | \(47\) |
risch | \(-\frac {2 \left (d \,x^{3}+28 c \right ) \sqrt {d \,x^{3}+c}}{9 d}+\frac {18 c^{\frac {3}{2}} \operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{3 \sqrt {c}}\right )}{d}\) | \(48\) |
elliptic | \(-\frac {2 x^{3} \sqrt {d \,x^{3}+c}}{9}-\frac {56 c \sqrt {d \,x^{3}+c}}{9 d}-\frac {3 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 )}{d^{3}}\) | \(441\) |
Time = 0.28 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.81 \[ \int \frac {x^2 \left (c+d x^3\right )^{3/2}}{8 c-d x^3} \, dx=\left [\frac {81 \, c^{\frac {3}{2}} \log \left (\frac {d x^{3} + 6 \, \sqrt {d x^{3} + c} \sqrt {c} + 10 \, c}{d x^{3} - 8 \, c}\right ) - 2 \, {\left (d x^{3} + 28 \, c\right )} \sqrt {d x^{3} + c}}{9 \, d}, -\frac {2 \, {\left (81 \, \sqrt {-c} c \arctan \left (\frac {\sqrt {d x^{3} + c} \sqrt {-c}}{3 \, c}\right ) + {\left (d x^{3} + 28 \, c\right )} \sqrt {d x^{3} + c}\right )}}{9 \, d}\right ] \]
[1/9*(81*c^(3/2)*log((d*x^3 + 6*sqrt(d*x^3 + c)*sqrt(c) + 10*c)/(d*x^3 - 8 *c)) - 2*(d*x^3 + 28*c)*sqrt(d*x^3 + c))/d, -2/9*(81*sqrt(-c)*c*arctan(1/3 *sqrt(d*x^3 + c)*sqrt(-c)/c) + (d*x^3 + 28*c)*sqrt(d*x^3 + c))/d]
Time = 6.34 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.06 \[ \int \frac {x^2 \left (c+d x^3\right )^{3/2}}{8 c-d x^3} \, dx=\begin {cases} \frac {2 \left (- \frac {9 c^{2} \operatorname {atan}{\left (\frac {\sqrt {c + d x^{3}}}{3 \sqrt {- c}} \right )}}{\sqrt {- c}} - 3 c \sqrt {c + d x^{3}} - \frac {\left (c + d x^{3}\right )^{\frac {3}{2}}}{9}\right )}{d} & \text {for}\: d \neq 0 \\\frac {\sqrt {c} x^{3}}{24} & \text {otherwise} \end {cases} \]
Piecewise((2*(-9*c**2*atan(sqrt(c + d*x**3)/(3*sqrt(-c)))/sqrt(-c) - 3*c*s qrt(c + d*x**3) - (c + d*x**3)**(3/2)/9)/d, Ne(d, 0)), (sqrt(c)*x**3/24, T rue))
Time = 0.27 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.01 \[ \int \frac {x^2 \left (c+d x^3\right )^{3/2}}{8 c-d x^3} \, dx=-\frac {81 \, c^{\frac {3}{2}} \log \left (\frac {\sqrt {d x^{3} + c} - 3 \, \sqrt {c}}{\sqrt {d x^{3} + c} + 3 \, \sqrt {c}}\right ) + 2 \, {\left (d x^{3} + c\right )}^{\frac {3}{2}} + 54 \, \sqrt {d x^{3} + c} c}{9 \, d} \]
-1/9*(81*c^(3/2)*log((sqrt(d*x^3 + c) - 3*sqrt(c))/(sqrt(d*x^3 + c) + 3*sq rt(c))) + 2*(d*x^3 + c)^(3/2) + 54*sqrt(d*x^3 + c)*c)/d
Time = 0.27 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.97 \[ \int \frac {x^2 \left (c+d x^3\right )^{3/2}}{8 c-d x^3} \, dx=-\frac {18 \, c^{2} \arctan \left (\frac {\sqrt {d x^{3} + c}}{3 \, \sqrt {-c}}\right )}{\sqrt {-c} d} - \frac {2 \, {\left ({\left (d x^{3} + c\right )}^{\frac {3}{2}} d^{2} + 27 \, \sqrt {d x^{3} + c} c d^{2}\right )}}{9 \, d^{3}} \]
-18*c^2*arctan(1/3*sqrt(d*x^3 + c)/sqrt(-c))/(sqrt(-c)*d) - 2/9*((d*x^3 + c)^(3/2)*d^2 + 27*sqrt(d*x^3 + c)*c*d^2)/d^3
Time = 7.43 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.12 \[ \int \frac {x^2 \left (c+d x^3\right )^{3/2}}{8 c-d x^3} \, dx=\frac {9\,c^{3/2}\,\ln \left (\frac {10\,c+d\,x^3+6\,\sqrt {c}\,\sqrt {d\,x^3+c}}{8\,c-d\,x^3}\right )}{d}-\frac {56\,c\,\sqrt {d\,x^3+c}}{9\,d}-\frac {2\,x^3\,\sqrt {d\,x^3+c}}{9} \]