Integrand size = 24, antiderivative size = 96 \[ \int \frac {x^2 \left (c+d x^3\right )^{3/2}}{a+b x^3} \, dx=\frac {2 (b c-a d) \sqrt {c+d x^3}}{3 b^2}+\frac {2 \left (c+d x^3\right )^{3/2}}{9 b}-\frac {2 (b c-a d)^{3/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x^3}}{\sqrt {b c-a d}}\right )}{3 b^{5/2}} \]
2/9*(d*x^3+c)^(3/2)/b-2/3*(-a*d+b*c)^(3/2)*arctanh(b^(1/2)*(d*x^3+c)^(1/2) /(-a*d+b*c)^(1/2))/b^(5/2)+2/3*(-a*d+b*c)*(d*x^3+c)^(1/2)/b^2
Time = 0.27 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.89 \[ \int \frac {x^2 \left (c+d x^3\right )^{3/2}}{a+b x^3} \, dx=\frac {2 \sqrt {c+d x^3} \left (4 b c-3 a d+b d x^3\right )}{9 b^2}+\frac {2 (-b c+a d)^{3/2} \arctan \left (\frac {\sqrt {b} \sqrt {c+d x^3}}{\sqrt {-b c+a d}}\right )}{3 b^{5/2}} \]
(2*Sqrt[c + d*x^3]*(4*b*c - 3*a*d + b*d*x^3))/(9*b^2) + (2*(-(b*c) + a*d)^ (3/2)*ArcTan[(Sqrt[b]*Sqrt[c + d*x^3])/Sqrt[-(b*c) + a*d]])/(3*b^(5/2))
Time = 0.23 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.05, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {946, 60, 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.
\(\displaystyle \int \frac {x^2 \left (c+d x^3\right )^{3/2}}{a+b x^3} \, dx\) |
\(\Big \downarrow \) 946 |
\(\displaystyle \frac {1}{3} \int \frac {\left (d x^3+c\right )^{3/2}}{b x^3+a}dx^3\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {1}{3} \left (\frac {(b c-a d) \int \frac {\sqrt {d x^3+c}}{b x^3+a}dx^3}{b}+\frac {2 \left (c+d x^3\right )^{3/2}}{3 b}\right )\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {1}{3} \left (\frac {(b c-a d) \left (\frac {(b c-a d) \int \frac {1}{\left (b x^3+a\right ) \sqrt {d x^3+c}}dx^3}{b}+\frac {2 \sqrt {c+d x^3}}{b}\right )}{b}+\frac {2 \left (c+d x^3\right )^{3/2}}{3 b}\right )\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {1}{3} \left (\frac {(b c-a d) \left (\frac {2 (b c-a d) \int \frac {1}{\frac {b x^6}{d}+a-\frac {b c}{d}}d\sqrt {d x^3+c}}{b d}+\frac {2 \sqrt {c+d x^3}}{b}\right )}{b}+\frac {2 \left (c+d x^3\right )^{3/2}}{3 b}\right )\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {1}{3} \left (\frac {(b c-a d) \left (\frac {2 \sqrt {c+d x^3}}{b}-\frac {2 \sqrt {b c-a d} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x^3}}{\sqrt {b c-a d}}\right )}{b^{3/2}}\right )}{b}+\frac {2 \left (c+d x^3\right )^{3/2}}{3 b}\right )\) |
((2*(c + d*x^3)^(3/2))/(3*b) + ((b*c - a*d)*((2*Sqrt[c + d*x^3])/b - (2*Sq rt[b*c - a*d]*ArcTanh[(Sqrt[b]*Sqrt[c + d*x^3])/Sqrt[b*c - a*d]])/b^(3/2)) )/b)/3
3.4.70.3.1 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[(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 = 4.42 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.98
method | result | size |
default | \(-\frac {2 \left (-\left (a d -b c \right )^{2} \arctan \left (\frac {b \sqrt {d \,x^{3}+c}}{\sqrt {\left (a d -b c \right ) b}}\right )+\left (\frac {\left (-d \,x^{3}-4 c \right ) b}{3}+a d \right ) \sqrt {d \,x^{3}+c}\, \sqrt {\left (a d -b c \right ) b}\right )}{3 \sqrt {\left (a d -b c \right ) b}\, b^{2}}\) | \(94\) |
risch | \(-\frac {2 \sqrt {d \,x^{3}+c}\, \left (-b d \,x^{3}+3 a d -4 b c \right )}{9 b^{2}}+\frac {2 \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) \arctan \left (\frac {b \sqrt {d \,x^{3}+c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{3 b^{2} \sqrt {\left (a d -b c \right ) b}}\) | \(94\) |
pseudoelliptic | \(-\frac {2 \left (-\left (a d -b c \right )^{2} \arctan \left (\frac {b \sqrt {d \,x^{3}+c}}{\sqrt {\left (a d -b c \right ) b}}\right )+\left (\frac {\left (-d \,x^{3}-4 c \right ) b}{3}+a d \right ) \sqrt {d \,x^{3}+c}\, \sqrt {\left (a d -b c \right ) b}\right )}{3 \sqrt {\left (a d -b c \right ) b}\, b^{2}}\) | \(94\) |
elliptic | \(\frac {2 d \,x^{3} \sqrt {d \,x^{3}+c}}{9 b}+\frac {2 \left (-\frac {d \left (a d -2 b c \right )}{b^{2}}-\frac {2 c d}{3 b}\right ) \sqrt {d \,x^{3}+c}}{3 d}+\frac {i \sqrt {2}\, \left (\munderset {\underline {\hspace {1.25 ex}}\alpha =\operatorname {RootOf}\left (b \,\textit {\_Z}^{3}+a \right )}{\sum }\frac {\left (-a^{2} d^{2}+2 a b c d -b^{2} c^{2}\right ) \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 \left (a d -b c \right ) \sqrt {d \,x^{3}+c}}\right )}{3 b^{2} d^{2}}\) | \(507\) |
-2/3*(-(a*d-b*c)^2*arctan(b*(d*x^3+c)^(1/2)/((a*d-b*c)*b)^(1/2))+(1/3*(-d* x^3-4*c)*b+a*d)*(d*x^3+c)^(1/2)*((a*d-b*c)*b)^(1/2))/((a*d-b*c)*b)^(1/2)/b ^2
Time = 0.28 (sec) , antiderivative size = 204, normalized size of antiderivative = 2.12 \[ \int \frac {x^2 \left (c+d x^3\right )^{3/2}}{a+b x^3} \, dx=\left [-\frac {3 \, {\left (b c - a d\right )} \sqrt {\frac {b c - a d}{b}} \log \left (\frac {b d x^{3} + 2 \, b c - a d + 2 \, \sqrt {d x^{3} + c} b \sqrt {\frac {b c - a d}{b}}}{b x^{3} + a}\right ) - 2 \, {\left (b d x^{3} + 4 \, b c - 3 \, a d\right )} \sqrt {d x^{3} + c}}{9 \, b^{2}}, -\frac {2 \, {\left (3 \, {\left (b c - a d\right )} \sqrt {-\frac {b c - a d}{b}} \arctan \left (-\frac {\sqrt {d x^{3} + c} b \sqrt {-\frac {b c - a d}{b}}}{b c - a d}\right ) - {\left (b d x^{3} + 4 \, b c - 3 \, a d\right )} \sqrt {d x^{3} + c}\right )}}{9 \, b^{2}}\right ] \]
[-1/9*(3*(b*c - a*d)*sqrt((b*c - a*d)/b)*log((b*d*x^3 + 2*b*c - a*d + 2*sq rt(d*x^3 + c)*b*sqrt((b*c - a*d)/b))/(b*x^3 + a)) - 2*(b*d*x^3 + 4*b*c - 3 *a*d)*sqrt(d*x^3 + c))/b^2, -2/9*(3*(b*c - a*d)*sqrt(-(b*c - a*d)/b)*arcta n(-sqrt(d*x^3 + c)*b*sqrt(-(b*c - a*d)/b)/(b*c - a*d)) - (b*d*x^3 + 4*b*c - 3*a*d)*sqrt(d*x^3 + c))/b^2]
Time = 7.95 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.29 \[ \int \frac {x^2 \left (c+d x^3\right )^{3/2}}{a+b x^3} \, dx=\begin {cases} \frac {2 \left (\frac {d \left (c + d x^{3}\right )^{\frac {3}{2}}}{9 b} + \frac {\sqrt {c + d x^{3}} \left (- a d^{2} + b c d\right )}{3 b^{2}} + \frac {d \left (a d - b c\right )^{2} \operatorname {atan}{\left (\frac {\sqrt {c + d x^{3}}}{\sqrt {\frac {a d - b c}{b}}} \right )}}{3 b^{3} \sqrt {\frac {a d - b c}{b}}}\right )}{d} & \text {for}\: d \neq 0 \\c^{\frac {3}{2}} \left (\begin {cases} \frac {x^{3}}{3 a} & \text {for}\: b = 0 \\\frac {\log {\left (3 a + 3 b x^{3} \right )}}{3 b} & \text {otherwise} \end {cases}\right ) & \text {otherwise} \end {cases} \]
Piecewise((2*(d*(c + d*x**3)**(3/2)/(9*b) + sqrt(c + d*x**3)*(-a*d**2 + b* c*d)/(3*b**2) + d*(a*d - b*c)**2*atan(sqrt(c + d*x**3)/sqrt((a*d - b*c)/b) )/(3*b**3*sqrt((a*d - b*c)/b)))/d, Ne(d, 0)), (c**(3/2)*Piecewise((x**3/(3 *a), Eq(b, 0)), (log(3*a + 3*b*x**3)/(3*b), True)), True))
Exception generated. \[ \int \frac {x^2 \left (c+d x^3\right )^{3/2}}{a+b x^3} \, dx=\text {Exception raised: ValueError} \]
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 = 0.29 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.18 \[ \int \frac {x^2 \left (c+d x^3\right )^{3/2}}{a+b x^3} \, dx=\frac {2 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \arctan \left (\frac {\sqrt {d x^{3} + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{3 \, \sqrt {-b^{2} c + a b d} b^{2}} + \frac {2 \, {\left ({\left (d x^{3} + c\right )}^{\frac {3}{2}} b^{2} + 3 \, \sqrt {d x^{3} + c} b^{2} c - 3 \, \sqrt {d x^{3} + c} a b d\right )}}{9 \, b^{3}} \]
2/3*(b^2*c^2 - 2*a*b*c*d + a^2*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) + 2/9*((d*x^3 + c)^(3/2)*b^2 + 3*sqrt( d*x^3 + c)*b^2*c - 3*sqrt(d*x^3 + c)*a*b*d)/b^3
Time = 10.18 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.49 \[ \int \frac {x^2 \left (c+d x^3\right )^{3/2}}{a+b x^3} \, dx=\frac {2\,d\,x^3\,\sqrt {d\,x^3+c}}{9\,b}-\frac {\sqrt {d\,x^3+c}\,\left (\frac {2\,a\,d^2}{b^2}-\frac {8\,c\,d}{3\,b}\right )}{3\,d}+\frac {\ln \left (\frac {a^2\,d^2+2\,b^2\,c^2-a\,b\,d^2\,x^3+b^2\,c\,d\,x^3-3\,a\,b\,c\,d-\sqrt {b}\,\sqrt {d\,x^3+c}\,{\left (a\,d-b\,c\right )}^{3/2}\,2{}\mathrm {i}}{b\,x^3+a}\right )\,{\left (a\,d-b\,c\right )}^{3/2}\,1{}\mathrm {i}}{3\,b^{5/2}} \]