Integrand size = 27, antiderivative size = 77 \[ \int \frac {x^2 \left (c+d x^3\right )^{3/2}}{\left (8 c-d x^3\right )^2} \, dx=\frac {\sqrt {c+d x^3}}{d}+\frac {\left (c+d x^3\right )^{3/2}}{3 d \left (8 c-d x^3\right )}-\frac {3 \sqrt {c} \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{d} \] Output:
(d*x^3+c)^(1/2)/d+1/3*(d*x^3+c)^(3/2)/d/(-d*x^3+8*c)-3*c^(1/2)*arctanh(1/3 *(d*x^3+c)^(1/2)/c^(1/2))/d
Time = 0.08 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.94 \[ \int \frac {x^2 \left (c+d x^3\right )^{3/2}}{\left (8 c-d x^3\right )^2} \, dx=\frac {\left (25 c-2 d x^3\right ) \sqrt {c+d x^3}}{3 d \left (8 c-d x^3\right )}-\frac {3 \sqrt {c} \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{d} \] Input:
Integrate[(x^2*(c + d*x^3)^(3/2))/(8*c - d*x^3)^2,x]
Output:
((25*c - 2*d*x^3)*Sqrt[c + d*x^3])/(3*d*(8*c - d*x^3)) - (3*Sqrt[c]*ArcTan h[Sqrt[c + d*x^3]/(3*Sqrt[c])])/d
Time = 0.34 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.09, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {946, 51, 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.
| \(\displaystyle \int \frac {x^2 \left (c+d x^3\right )^{3/2}}{\left (8 c-d x^3\right )^2} \, dx\) |
| \(\Big \downarrow \) 946 |
| \(\displaystyle \frac {1}{3} \int \frac {\left (d x^3+c\right )^{3/2}}{\left (8 c-d x^3\right )^2}dx^3\) |
| \(\Big \downarrow \) 51 |
| \(\displaystyle \frac {1}{3} \left (\frac {\left (c+d x^3\right )^{3/2}}{d \left (8 c-d x^3\right )}-\frac {3}{2} \int \frac {\sqrt {d x^3+c}}{8 c-d x^3}dx^3\right )\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {1}{3} \left (\frac {\left (c+d x^3\right )^{3/2}}{d \left (8 c-d x^3\right )}-\frac {3}{2} \left (9 c \int \frac {1}{\left (8 c-d x^3\right ) \sqrt {d x^3+c}}dx^3-\frac {2 \sqrt {c+d x^3}}{d}\right )\right )\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {1}{3} \left (\frac {\left (c+d x^3\right )^{3/2}}{d \left (8 c-d x^3\right )}-\frac {3}{2} \left (\frac {18 c \int \frac {1}{9 c-x^6}d\sqrt {d x^3+c}}{d}-\frac {2 \sqrt {c+d x^3}}{d}\right )\right )\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {1}{3} \left (\frac {\left (c+d x^3\right )^{3/2}}{d \left (8 c-d x^3\right )}-\frac {3}{2} \left (\frac {6 \sqrt {c} \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{d}-\frac {2 \sqrt {c+d x^3}}{d}\right )\right )\) |
Input:
Int[(x^2*(c + d*x^3)^(3/2))/(8*c - d*x^3)^2,x]
Output:
((c + d*x^3)^(3/2)/(d*(8*c - d*x^3)) - (3*((-2*Sqrt[c + d*x^3])/d + (6*Sqr t[c]*ArcTanh[Sqrt[c + d*x^3]/(3*Sqrt[c])])/d))/2)/3
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + 1))), x] - Simp[d*(n/(b*(m + 1))) Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d, n}, x ] && ILtQ[m, -1] && FractionQ[n] && GtQ[n, 0]
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 = 1.11 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.82
| method | result | size |
| default | \(\frac {\frac {2 \sqrt {d \,x^{3}+c}}{3}-3 c \left (-\frac {\sqrt {d \,x^{3}+c}}{-d \,x^{3}+8 c}+\frac {\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{3 \sqrt {c}}\right )}{\sqrt {c}}\right )}{d}\) | \(63\) |
| pseudoelliptic | \(\frac {\frac {2 \sqrt {d \,x^{3}+c}}{3}+3 c \left (\frac {\sqrt {d \,x^{3}+c}}{-d \,x^{3}+8 c}-\frac {\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{3 \sqrt {c}}\right )}{\sqrt {c}}\right )}{d}\) | \(63\) |
| risch | \(\frac {2 \sqrt {d \,x^{3}+c}}{3 d}+9 c \left (-\frac {4 \,\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{3 \sqrt {c}}\right )}{9 \sqrt {c}\, d}+\frac {c \left (-\frac {\sqrt {d \,x^{3}+c}}{c \left (d \,x^{3}-8 c \right )}+\frac {\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{3 \sqrt {c}}\right )}{3 c^{\frac {3}{2}}}\right )}{3 d}\right )\) | \(95\) |
| elliptic | \(\frac {3 c \sqrt {d \,x^{3}+c}}{d \left (-d \,x^{3}+8 c \right )}+\frac {2 \sqrt {d \,x^{3}+c}}{3 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 ) \operatorname {EllipticPi}\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}} \underline {\hspace {1.25 ex}}\alpha ^{2} \sqrt {3}\, d -i \left (-c \,d^{2}\right )^{\frac {2}{3}} \underline {\hspace {1.25 ex}}\alpha \sqrt {3}+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 )}{2 d^{3}}\) | \(452\) |
Input:
int(x^2*(d*x^3+c)^(3/2)/(-d*x^3+8*c)^2,x,method=_RETURNVERBOSE)
Output:
(2/3*(d*x^3+c)^(1/2)-3*c*(-(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)))/d
Time = 0.09 (sec) , antiderivative size = 159, normalized size of antiderivative = 2.06 \[ \int \frac {x^2 \left (c+d x^3\right )^{3/2}}{\left (8 c-d x^3\right )^2} \, dx=\left [\frac {9 \, {\left (d x^{3} - 8 \, c\right )} \sqrt {c} \log \left (\frac {d x^{3} - 6 \, \sqrt {d x^{3} + c} \sqrt {c} + 10 \, c}{d x^{3} - 8 \, c}\right ) + 2 \, {\left (2 \, d x^{3} - 25 \, c\right )} \sqrt {d x^{3} + c}}{6 \, {\left (d^{2} x^{3} - 8 \, c d\right )}}, \frac {9 \, {\left (d x^{3} - 8 \, c\right )} \sqrt {-c} \arctan \left (\frac {3 \, \sqrt {-c}}{\sqrt {d x^{3} + c}}\right ) + {\left (2 \, d x^{3} - 25 \, c\right )} \sqrt {d x^{3} + c}}{3 \, {\left (d^{2} x^{3} - 8 \, c d\right )}}\right ] \] Input:
integrate(x^2*(d*x^3+c)^(3/2)/(-d*x^3+8*c)^2,x, algorithm="fricas")
Output:
[1/6*(9*(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)) + 2*(2*d*x^3 - 25*c)*sqrt(d*x^3 + c))/(d^2*x^3 - 8*c*d), 1/3*(9*(d*x^3 - 8*c)*sqrt(-c)*arctan(3*sqrt(-c)/sqrt(d*x^3 + c)) + (2*d*x ^3 - 25*c)*sqrt(d*x^3 + c))/(d^2*x^3 - 8*c*d)]
\[ \int \frac {x^2 \left (c+d x^3\right )^{3/2}}{\left (8 c-d x^3\right )^2} \, dx=\int \frac {x^{2} \left (c + d x^{3}\right )^{\frac {3}{2}}}{\left (- 8 c + d x^{3}\right )^{2}}\, dx \] Input:
integrate(x**2*(d*x**3+c)**(3/2)/(-d*x**3+8*c)**2,x)
Output:
Integral(x**2*(c + d*x**3)**(3/2)/(-8*c + d*x**3)**2, x)
Time = 0.11 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.03 \[ \int \frac {x^2 \left (c+d x^3\right )^{3/2}}{\left (8 c-d x^3\right )^2} \, dx=\frac {9 \, \sqrt {c} \log \left (\frac {\sqrt {d x^{3} + c} - 3 \, \sqrt {c}}{\sqrt {d x^{3} + c} + 3 \, \sqrt {c}}\right ) + 4 \, \sqrt {d x^{3} + c} - \frac {18 \, \sqrt {d x^{3} + c} c}{d x^{3} - 8 \, c}}{6 \, d} \] Input:
integrate(x^2*(d*x^3+c)^(3/2)/(-d*x^3+8*c)^2,x, algorithm="maxima")
Output:
1/6*(9*sqrt(c)*log((sqrt(d*x^3 + c) - 3*sqrt(c))/(sqrt(d*x^3 + c) + 3*sqrt (c))) + 4*sqrt(d*x^3 + c) - 18*sqrt(d*x^3 + c)*c/(d*x^3 - 8*c))/d
Time = 0.13 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.90 \[ \int \frac {x^2 \left (c+d x^3\right )^{3/2}}{\left (8 c-d x^3\right )^2} \, dx=\frac {3 \, c \arctan \left (\frac {\sqrt {d x^{3} + c}}{3 \, \sqrt {-c}}\right )}{\sqrt {-c} d} + \frac {2 \, \sqrt {d x^{3} + c}}{3 \, d} - \frac {3 \, \sqrt {d x^{3} + c} c}{{\left (d x^{3} - 8 \, c\right )} d} \] Input:
integrate(x^2*(d*x^3+c)^(3/2)/(-d*x^3+8*c)^2,x, algorithm="giac")
Output:
3*c*arctan(1/3*sqrt(d*x^3 + c)/sqrt(-c))/(sqrt(-c)*d) + 2/3*sqrt(d*x^3 + c )/d - 3*sqrt(d*x^3 + c)*c/((d*x^3 - 8*c)*d)
Time = 2.62 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.13 \[ \int \frac {x^2 \left (c+d x^3\right )^{3/2}}{\left (8 c-d x^3\right )^2} \, dx=\frac {2\,\sqrt {d\,x^3+c}}{3\,d}+\frac {3\,\sqrt {c}\,\ln \left (\frac {10\,c+d\,x^3-6\,\sqrt {c}\,\sqrt {d\,x^3+c}}{8\,c-d\,x^3}\right )}{2\,d}+\frac {3\,c\,\sqrt {d\,x^3+c}}{d\,\left (8\,c-d\,x^3\right )} \] Input:
int((x^2*(c + d*x^3)^(3/2))/(8*c - d*x^3)^2,x)
Output:
(2*(c + d*x^3)^(1/2))/(3*d) + (3*c^(1/2)*log((10*c + d*x^3 - 6*c^(1/2)*(c + d*x^3)^(1/2))/(8*c - d*x^3)))/(2*d) + (3*c*(c + d*x^3)^(1/2))/(d*(8*c - d*x^3))
\[ \int \frac {x^2 \left (c+d x^3\right )^{3/2}}{\left (8 c-d x^3\right )^2} \, dx=\frac {34 \sqrt {d \,x^{3}+c}\, c -20 \sqrt {d \,x^{3}+c}\, d \,x^{3}+5832 \left (\int \frac {\sqrt {d \,x^{3}+c}\, x^{5}}{d^{3} x^{9}-15 c \,d^{2} x^{6}+48 c^{2} d \,x^{3}+64 c^{3}}d x \right ) c^{2} d^{2}-729 \left (\int \frac {\sqrt {d \,x^{3}+c}\, x^{5}}{d^{3} x^{9}-15 c \,d^{2} x^{6}+48 c^{2} d \,x^{3}+64 c^{3}}d x \right ) c \,d^{3} x^{3}}{30 d \left (-d \,x^{3}+8 c \right )} \] Input:
int(x^2*(d*x^3+c)^(3/2)/(-d*x^3+8*c)^2,x)
Output:
(34*sqrt(c + d*x**3)*c - 20*sqrt(c + d*x**3)*d*x**3 + 5832*int((sqrt(c + d *x**3)*x**5)/(64*c**3 + 48*c**2*d*x**3 - 15*c*d**2*x**6 + d**3*x**9),x)*c* *2*d**2 - 729*int((sqrt(c + d*x**3)*x**5)/(64*c**3 + 48*c**2*d*x**3 - 15*c *d**2*x**6 + d**3*x**9),x)*c*d**3*x**3)/(30*d*(8*c - d*x**3))