Integrand size = 24, antiderivative size = 94 \[ \int \frac {x^2 \left (c+d x^3\right )^{3/2}}{\left (a+b x^3\right )^2} \, dx=\frac {d \sqrt {c+d x^3}}{b^2}-\frac {\left (c+d x^3\right )^{3/2}}{3 b \left (a+b x^3\right )}-\frac {d \sqrt {b c-a d} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x^3}}{\sqrt {b c-a d}}\right )}{b^{5/2}} \]
-1/3*(d*x^3+c)^(3/2)/b/(b*x^3+a)-d*arctanh(b^(1/2)*(d*x^3+c)^(1/2)/(-a*d+b *c)^(1/2))*(-a*d+b*c)^(1/2)/b^(5/2)+d*(d*x^3+c)^(1/2)/b^2
Time = 0.31 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.00 \[ \int \frac {x^2 \left (c+d x^3\right )^{3/2}}{\left (a+b x^3\right )^2} \, dx=\frac {\sqrt {c+d x^3} \left (-b c+3 a d+2 b d x^3\right )}{3 b^2 \left (a+b x^3\right )}-\frac {d \sqrt {-b c+a d} \arctan \left (\frac {\sqrt {b} \sqrt {c+d x^3}}{\sqrt {-b c+a d}}\right )}{b^{5/2}} \]
(Sqrt[c + d*x^3]*(-(b*c) + 3*a*d + 2*b*d*x^3))/(3*b^2*(a + b*x^3)) - (d*Sq rt[-(b*c) + a*d]*ArcTan[(Sqrt[b]*Sqrt[c + d*x^3])/Sqrt[-(b*c) + a*d]])/b^( 5/2)
Time = 0.23 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.11, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {946, 51, 60, 73, 221}
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.
| \(\displaystyle \int \frac {x^2 \left (c+d x^3\right )^{3/2}}{\left (a+b x^3\right )^2} \, dx\) |
| \(\Big \downarrow \) 946 |
| \(\displaystyle \frac {1}{3} \int \frac {\left (d x^3+c\right )^{3/2}}{\left (b x^3+a\right )^2}dx^3\) |
| \(\Big \downarrow \) 51 |
| \(\displaystyle \frac {1}{3} \left (\frac {3 d \int \frac {\sqrt {d x^3+c}}{b x^3+a}dx^3}{2 b}-\frac {\left (c+d x^3\right )^{3/2}}{b \left (a+b x^3\right )}\right )\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {1}{3} \left (\frac {3 d \left (\frac {(b c-a d) \int \frac {1}{\left (b x^3+a\right ) \sqrt {d x^3+c}}dx^3}{b}+\frac {2 \sqrt {c+d x^3}}{b}\right )}{2 b}-\frac {\left (c+d x^3\right )^{3/2}}{b \left (a+b x^3\right )}\right )\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {1}{3} \left (\frac {3 d \left (\frac {2 (b c-a d) \int \frac {1}{\frac {b x^6}{d}+a-\frac {b c}{d}}d\sqrt {d x^3+c}}{b d}+\frac {2 \sqrt {c+d x^3}}{b}\right )}{2 b}-\frac {\left (c+d x^3\right )^{3/2}}{b \left (a+b x^3\right )}\right )\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {1}{3} \left (\frac {3 d \left (\frac {2 \sqrt {c+d x^3}}{b}-\frac {2 \sqrt {b c-a d} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x^3}}{\sqrt {b c-a d}}\right )}{b^{3/2}}\right )}{2 b}-\frac {\left (c+d x^3\right )^{3/2}}{b \left (a+b x^3\right )}\right )\) |
(-((c + d*x^3)^(3/2)/(b*(a + b*x^3))) + (3*d*((2*Sqrt[c + d*x^3])/b - (2*S qrt[b*c - a*d]*ArcTanh[(Sqrt[b]*Sqrt[c + d*x^3])/Sqrt[b*c - a*d]])/b^(3/2) ))/(2*b))/3
3.5.72.3.1 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 + 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[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
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 = 4.26 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.15
| method | result | size |
| default | \(\frac {-d \left (b \,x^{3}+a \right ) \left (a d -b c \right ) \arctan \left (\frac {b \sqrt {d \,x^{3}+c}}{\sqrt {\left (a d -b c \right ) b}}\right )+\left (\frac {\left (2 d \,x^{3}-c \right ) b}{3}+a d \right ) \sqrt {d \,x^{3}+c}\, \sqrt {\left (a d -b c \right ) b}}{\sqrt {\left (a d -b c \right ) b}\, b^{2} \left (b \,x^{3}+a \right )}\) | \(108\) |
| pseudoelliptic | \(\frac {-d \left (b \,x^{3}+a \right ) \left (a d -b c \right ) \arctan \left (\frac {b \sqrt {d \,x^{3}+c}}{\sqrt {\left (a d -b c \right ) b}}\right )+\left (\frac {\left (2 d \,x^{3}-c \right ) b}{3}+a d \right ) \sqrt {d \,x^{3}+c}\, \sqrt {\left (a d -b c \right ) b}}{\sqrt {\left (a d -b c \right ) b}\, b^{2} \left (b \,x^{3}+a \right )}\) | \(108\) |
| risch | \(\frac {2 d \sqrt {d \,x^{3}+c}}{3 b^{2}}-\frac {\frac {4 \left (a d -b c \right ) d \arctan \left (\frac {b \sqrt {d \,x^{3}+c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{3 \sqrt {\left (a d -b c \right ) b}}+\frac {\left (-a^{2} d^{2}+2 a b c d -b^{2} c^{2}\right ) \left (d \arctan \left (\frac {b \sqrt {d \,x^{3}+c}}{\sqrt {\left (a d -b c \right ) b}}\right ) \left (b \,x^{3}+a \right )+\sqrt {d \,x^{3}+c}\, \sqrt {\left (a d -b c \right ) b}\right )}{3 \sqrt {\left (a d -b c \right ) b}\, \left (a d -b c \right ) \left (b \,x^{3}+a \right )}}{b^{2}}\) | \(182\) |
| elliptic | \(\frac {\left (a d -b c \right ) \sqrt {d \,x^{3}+c}}{3 b^{2} \left (b \,x^{3}+a \right )}+\frac {2 d \sqrt {d \,x^{3}+c}}{3 b^{2}}+\frac {i \sqrt {2}\, \left (\munderset {\underline {\hspace {1.25 ex}}\alpha =\operatorname {RootOf}\left (b \,\textit {\_Z}^{3}+a \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 {b \left (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 \right )}{2 d \left (a d -b c \right )}, \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 \,b^{2}}\) | \(466\) |
1/((a*d-b*c)*b)^(1/2)*(-d*(b*x^3+a)*(a*d-b*c)*arctan(b*(d*x^3+c)^(1/2)/((a *d-b*c)*b)^(1/2))+(1/3*(2*d*x^3-c)*b+a*d)*(d*x^3+c)^(1/2)*((a*d-b*c)*b)^(1 /2))/b^2/(b*x^3+a)
Time = 0.32 (sec) , antiderivative size = 234, normalized size of antiderivative = 2.49 \[ \int \frac {x^2 \left (c+d x^3\right )^{3/2}}{\left (a+b x^3\right )^2} \, dx=\left [\frac {3 \, {\left (b d x^{3} + a d\right )} \sqrt {\frac {b c - a d}{b}} \log \left (\frac {b d x^{3} + 2 \, b c - a d - 2 \, \sqrt {d x^{3} + c} b \sqrt {\frac {b c - a d}{b}}}{b x^{3} + a}\right ) + 2 \, {\left (2 \, b d x^{3} - b c + 3 \, a d\right )} \sqrt {d x^{3} + c}}{6 \, {\left (b^{3} x^{3} + a b^{2}\right )}}, -\frac {3 \, {\left (b d x^{3} + a d\right )} \sqrt {-\frac {b c - a d}{b}} \arctan \left (-\frac {\sqrt {d x^{3} + c} b \sqrt {-\frac {b c - a d}{b}}}{b c - a d}\right ) - {\left (2 \, b d x^{3} - b c + 3 \, a d\right )} \sqrt {d x^{3} + c}}{3 \, {\left (b^{3} x^{3} + a b^{2}\right )}}\right ] \]
[1/6*(3*(b*d*x^3 + a*d)*sqrt((b*c - a*d)/b)*log((b*d*x^3 + 2*b*c - a*d - 2 *sqrt(d*x^3 + c)*b*sqrt((b*c - a*d)/b))/(b*x^3 + a)) + 2*(2*b*d*x^3 - b*c + 3*a*d)*sqrt(d*x^3 + c))/(b^3*x^3 + a*b^2), -1/3*(3*(b*d*x^3 + a*d)*sqrt( -(b*c - a*d)/b)*arctan(-sqrt(d*x^3 + c)*b*sqrt(-(b*c - a*d)/b)/(b*c - a*d) ) - (2*b*d*x^3 - b*c + 3*a*d)*sqrt(d*x^3 + c))/(b^3*x^3 + a*b^2)]
\[ \int \frac {x^2 \left (c+d x^3\right )^{3/2}}{\left (a+b x^3\right )^2} \, dx=\int \frac {x^{2} \left (c + d x^{3}\right )^{\frac {3}{2}}}{\left (a + b x^{3}\right )^{2}}\, dx \]
Exception generated. \[ \int \frac {x^2 \left (c+d x^3\right )^{3/2}}{\left (a+b x^3\right )^2} \, dx=\text {Exception raised: ValueError} \]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(a*d-b*c>0)', see `assume?` for m ore detail
Time = 0.27 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.30 \[ \int \frac {x^2 \left (c+d x^3\right )^{3/2}}{\left (a+b x^3\right )^2} \, dx=\frac {2 \, \sqrt {d x^{3} + c} d}{3 \, b^{2}} + \frac {{\left (b c d - a d^{2}\right )} \arctan \left (\frac {\sqrt {d x^{3} + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{\sqrt {-b^{2} c + a b d} b^{2}} - \frac {\sqrt {d x^{3} + c} b c d - \sqrt {d x^{3} + c} a d^{2}}{3 \, {\left ({\left (d x^{3} + c\right )} b - b c + a d\right )} b^{2}} \]
2/3*sqrt(d*x^3 + c)*d/b^2 + (b*c*d - a*d^2)*arctan(sqrt(d*x^3 + c)*b/sqrt( -b^2*c + a*b*d))/(sqrt(-b^2*c + a*b*d)*b^2) - 1/3*(sqrt(d*x^3 + c)*b*c*d - sqrt(d*x^3 + c)*a*d^2)/(((d*x^3 + c)*b - b*c + a*d)*b^2)
Time = 11.37 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.81 \[ \int \frac {x^2 \left (c+d x^3\right )^{3/2}}{\left (a+b x^3\right )^2} \, dx=\frac {2\,d\,\sqrt {d\,x^3+c}}{3\,b^2}-\frac {\left (\frac {2\,b\,c^2}{3\,\left (2\,b^2\,c-2\,a\,b\,d\right )}+\frac {a\,\left (\frac {2\,a\,d^2}{3\,\left (2\,b^2\,c-2\,a\,b\,d\right )}-\frac {4\,b\,c\,d}{3\,\left (2\,b^2\,c-2\,a\,b\,d\right )}\right )}{b}\right )\,\sqrt {d\,x^3+c}}{b\,x^3+a}+\frac {d\,\ln \left (\frac {a\,d-2\,b\,c-b\,d\,x^3+\sqrt {b}\,\sqrt {d\,x^3+c}\,\sqrt {a\,d-b\,c}\,2{}\mathrm {i}}{b\,x^3+a}\right )\,\sqrt {a\,d-b\,c}\,1{}\mathrm {i}}{2\,b^{5/2}} \]
(2*d*(c + d*x^3)^(1/2))/(3*b^2) - (((2*b*c^2)/(3*(2*b^2*c - 2*a*b*d)) + (a *((2*a*d^2)/(3*(2*b^2*c - 2*a*b*d)) - (4*b*c*d)/(3*(2*b^2*c - 2*a*b*d))))/ b)*(c + d*x^3)^(1/2))/(a + b*x^3) + (d*log((a*d - 2*b*c + b^(1/2)*(c + d*x ^3)^(1/2)*(a*d - b*c)^(1/2)*2i - b*d*x^3)/(a + b*x^3))*(a*d - b*c)^(1/2)*1 i)/(2*b^(5/2))