Integrand size = 20, antiderivative size = 196 \[ \int \frac {x^4 \left (A+B x^3\right )}{\left (a+b x^3\right )^2} \, dx=-\frac {(2 A b-5 a B) x^2}{6 a b^2}+\frac {(A b-a B) x^5}{3 a b \left (a+b x^3\right )}-\frac {(2 A b-5 a B) \arctan \left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right )}{3 \sqrt {3} \sqrt [3]{a} b^{8/3}}-\frac {(2 A b-5 a B) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{9 \sqrt [3]{a} b^{8/3}}+\frac {(2 A b-5 a B) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{18 \sqrt [3]{a} b^{8/3}} \]
-1/6*(2*A*b-5*B*a)*x^2/a/b^2+1/3*(A*b-B*a)*x^5/a/b/(b*x^3+a)-1/9*(2*A*b-5* B*a)*ln(a^(1/3)+b^(1/3)*x)/a^(1/3)/b^(8/3)+1/18*(2*A*b-5*B*a)*ln(a^(2/3)-a ^(1/3)*b^(1/3)*x+b^(2/3)*x^2)/a^(1/3)/b^(8/3)-1/9*(2*A*b-5*B*a)*arctan(1/3 *(a^(1/3)-2*b^(1/3)*x)/a^(1/3)*3^(1/2))/a^(1/3)/b^(8/3)*3^(1/2)
Time = 0.12 (sec) , antiderivative size = 165, normalized size of antiderivative = 0.84 \[ \int \frac {x^4 \left (A+B x^3\right )}{\left (a+b x^3\right )^2} \, dx=\frac {9 b^{2/3} B x^2-\frac {6 b^{2/3} (A b-a B) x^2}{a+b x^3}+\frac {2 \sqrt {3} (-2 A b+5 a B) \arctan \left (\frac {1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}}{\sqrt {3}}\right )}{\sqrt [3]{a}}+\frac {2 (-2 A b+5 a B) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\sqrt [3]{a}}+\frac {(2 A b-5 a B) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{\sqrt [3]{a}}}{18 b^{8/3}} \]
(9*b^(2/3)*B*x^2 - (6*b^(2/3)*(A*b - a*B)*x^2)/(a + b*x^3) + (2*Sqrt[3]*(- 2*A*b + 5*a*B)*ArcTan[(1 - (2*b^(1/3)*x)/a^(1/3))/Sqrt[3]])/a^(1/3) + (2*( -2*A*b + 5*a*B)*Log[a^(1/3) + b^(1/3)*x])/a^(1/3) + ((2*A*b - 5*a*B)*Log[a ^(2/3) - a^(1/3)*b^(1/3)*x + b^(2/3)*x^2])/a^(1/3))/(18*b^(8/3))
Time = 0.35 (sec) , antiderivative size = 183, normalized size of antiderivative = 0.93, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {957, 843, 821, 16, 1142, 25, 27, 1082, 217, 1103}
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.
| \(\displaystyle \int \frac {x^4 \left (A+B x^3\right )}{\left (a+b x^3\right )^2} \, dx\) |
| \(\Big \downarrow \) 957 |
| \(\displaystyle \frac {x^5 (A b-a B)}{3 a b \left (a+b x^3\right )}-\frac {(2 A b-5 a B) \int \frac {x^4}{b x^3+a}dx}{3 a b}\) |
| \(\Big \downarrow \) 843 |
| \(\displaystyle \frac {x^5 (A b-a B)}{3 a b \left (a+b x^3\right )}-\frac {(2 A b-5 a B) \left (\frac {x^2}{2 b}-\frac {a \int \frac {x}{b x^3+a}dx}{b}\right )}{3 a b}\) |
| \(\Big \downarrow \) 821 |
| \(\displaystyle \frac {x^5 (A b-a B)}{3 a b \left (a+b x^3\right )}-\frac {(2 A b-5 a B) \left (\frac {x^2}{2 b}-\frac {a \left (\frac {\int \frac {\sqrt [3]{b} x+\sqrt [3]{a}}{b^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3}}dx}{3 \sqrt [3]{a} \sqrt [3]{b}}-\frac {\int \frac {1}{\sqrt [3]{b} x+\sqrt [3]{a}}dx}{3 \sqrt [3]{a} \sqrt [3]{b}}\right )}{b}\right )}{3 a b}\) |
| \(\Big \downarrow \) 16 |
| \(\displaystyle \frac {x^5 (A b-a B)}{3 a b \left (a+b x^3\right )}-\frac {(2 A b-5 a B) \left (\frac {x^2}{2 b}-\frac {a \left (\frac {\int \frac {\sqrt [3]{b} x+\sqrt [3]{a}}{b^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3}}dx}{3 \sqrt [3]{a} \sqrt [3]{b}}-\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 \sqrt [3]{a} b^{2/3}}\right )}{b}\right )}{3 a b}\) |
| \(\Big \downarrow \) 1142 |
| \(\displaystyle \frac {x^5 (A b-a B)}{3 a b \left (a+b x^3\right )}-\frac {(2 A b-5 a B) \left (\frac {x^2}{2 b}-\frac {a \left (\frac {\frac {3}{2} \sqrt [3]{a} \int \frac {1}{b^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3}}dx+\frac {\int -\frac {\sqrt [3]{b} \left (\sqrt [3]{a}-2 \sqrt [3]{b} x\right )}{b^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3}}dx}{2 \sqrt [3]{b}}}{3 \sqrt [3]{a} \sqrt [3]{b}}-\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 \sqrt [3]{a} b^{2/3}}\right )}{b}\right )}{3 a b}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {x^5 (A b-a B)}{3 a b \left (a+b x^3\right )}-\frac {(2 A b-5 a B) \left (\frac {x^2}{2 b}-\frac {a \left (\frac {\frac {3}{2} \sqrt [3]{a} \int \frac {1}{b^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3}}dx-\frac {\int \frac {\sqrt [3]{b} \left (\sqrt [3]{a}-2 \sqrt [3]{b} x\right )}{b^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3}}dx}{2 \sqrt [3]{b}}}{3 \sqrt [3]{a} \sqrt [3]{b}}-\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 \sqrt [3]{a} b^{2/3}}\right )}{b}\right )}{3 a b}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {x^5 (A b-a B)}{3 a b \left (a+b x^3\right )}-\frac {(2 A b-5 a B) \left (\frac {x^2}{2 b}-\frac {a \left (\frac {\frac {3}{2} \sqrt [3]{a} \int \frac {1}{b^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3}}dx-\frac {1}{2} \int \frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{b^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3}}dx}{3 \sqrt [3]{a} \sqrt [3]{b}}-\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 \sqrt [3]{a} b^{2/3}}\right )}{b}\right )}{3 a b}\) |
| \(\Big \downarrow \) 1082 |
| \(\displaystyle \frac {x^5 (A b-a B)}{3 a b \left (a+b x^3\right )}-\frac {(2 A b-5 a B) \left (\frac {x^2}{2 b}-\frac {a \left (\frac {\frac {3 \int \frac {1}{-\left (1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}\right )^2-3}d\left (1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}\right )}{\sqrt [3]{b}}-\frac {1}{2} \int \frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{b^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3}}dx}{3 \sqrt [3]{a} \sqrt [3]{b}}-\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 \sqrt [3]{a} b^{2/3}}\right )}{b}\right )}{3 a b}\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle \frac {x^5 (A b-a B)}{3 a b \left (a+b x^3\right )}-\frac {(2 A b-5 a B) \left (\frac {x^2}{2 b}-\frac {a \left (\frac {-\frac {1}{2} \int \frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{b^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3}}dx-\frac {\sqrt {3} \arctan \left (\frac {1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}}{\sqrt {3}}\right )}{\sqrt [3]{b}}}{3 \sqrt [3]{a} \sqrt [3]{b}}-\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 \sqrt [3]{a} b^{2/3}}\right )}{b}\right )}{3 a b}\) |
| \(\Big \downarrow \) 1103 |
| \(\displaystyle \frac {x^5 (A b-a B)}{3 a b \left (a+b x^3\right )}-\frac {(2 A b-5 a B) \left (\frac {x^2}{2 b}-\frac {a \left (\frac {\frac {\log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{2 \sqrt [3]{b}}-\frac {\sqrt {3} \arctan \left (\frac {1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}}{\sqrt {3}}\right )}{\sqrt [3]{b}}}{3 \sqrt [3]{a} \sqrt [3]{b}}-\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 \sqrt [3]{a} b^{2/3}}\right )}{b}\right )}{3 a b}\) |
((A*b - a*B)*x^5)/(3*a*b*(a + b*x^3)) - ((2*A*b - 5*a*B)*(x^2/(2*b) - (a*( -1/3*Log[a^(1/3) + b^(1/3)*x]/(a^(1/3)*b^(2/3)) + (-((Sqrt[3]*ArcTan[(1 - (2*b^(1/3)*x)/a^(1/3))/Sqrt[3]])/b^(1/3)) + Log[a^(2/3) - a^(1/3)*b^(1/3)* x + b^(2/3)*x^2]/(2*b^(1/3)))/(3*a^(1/3)*b^(1/3))))/b))/(3*a*b)
3.1.76.3.1 Defintions of rubi rules used
Int[(c_.)/((a_.) + (b_.)*(x_)), x_Symbol] :> Simp[c*(Log[RemoveContent[a + b*x, x]]/b), x] /; FreeQ[{a, b, c}, x]
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( -1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & & (LtQ[a, 0] || LtQ[b, 0])
Int[(x_)/((a_) + (b_.)*(x_)^3), x_Symbol] :> Simp[-(3*Rt[a, 3]*Rt[b, 3])^(- 1) Int[1/(Rt[a, 3] + Rt[b, 3]*x), x], x] + Simp[1/(3*Rt[a, 3]*Rt[b, 3]) Int[(Rt[a, 3] + Rt[b, 3]*x)/(Rt[a, 3]^2 - Rt[a, 3]*Rt[b, 3]*x + Rt[b, 3]^2 *x^2), x], x] /; FreeQ[{a, b}, x]
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^(n - 1)*(c*x)^(m - n + 1)*((a + b*x^n)^(p + 1)/(b*(m + n*p + 1))), x] - Simp[ a*c^n*((m - n + 1)/(b*(m + n*p + 1))) Int[(c*x)^(m - n)*(a + b*x^n)^p, x] , x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n* p + 1, 0] && IntBinomialQ[a, b, c, n, m, p, x]
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n _)), x_Symbol] :> Simp[(-(b*c - a*d))*(e*x)^(m + 1)*((a + b*x^n)^(p + 1)/(a *b*e*n*(p + 1))), x] - Simp[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*b*n* (p + 1)) Int[(e*x)^m*(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && (( !IntegerQ[p + 1/2] && N eQ[p, -5/4]) || !RationalQ[m] || (IGtQ[n, 0] && ILtQ[p + 1/2, 0] && LeQ[-1 , m, (-n)*(p + 1)]))
Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*S implify[a*(c/b^2)]}, Simp[-2/b Subst[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b )], x] /; RationalQ[q] && (EqQ[q^2, 1] || !RationalQ[b^2 - 4*a*c])] /; Fre eQ[{a, b, c}, x]
Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S imp[d*(Log[RemoveContent[a + b*x + c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]
Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S imp[(2*c*d - b*e)/(2*c) Int[1/(a + b*x + c*x^2), x], x] + Simp[e/(2*c) Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 4.21 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.36
| method | result | size |
| risch | \(\frac {B \,x^{2}}{2 b^{2}}+\frac {\left (-\frac {A b}{3}+\frac {B a}{3}\right ) x^{2}}{b^{2} \left (b \,x^{3}+a \right )}+\frac {\munderset {\textit {\_R} =\operatorname {RootOf}\left (b \,\textit {\_Z}^{3}+a \right )}{\sum }\frac {\left (2 A b -5 B a \right ) \ln \left (x -\textit {\_R} \right )}{\textit {\_R}}}{9 b^{3}}\) | \(71\) |
| default | \(\frac {B \,x^{2}}{2 b^{2}}+\frac {\frac {\left (-\frac {A b}{3}+\frac {B a}{3}\right ) x^{2}}{b \,x^{3}+a}+\left (-\frac {5 B a}{3}+\frac {2 A b}{3}\right ) \left (-\frac {\ln \left (x +\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {1}{3}}}+\frac {\ln \left (x^{2}-\left (\frac {a}{b}\right )^{\frac {1}{3}} x +\left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 b \left (\frac {a}{b}\right )^{\frac {1}{3}}}+\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {a}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{b^{2}}\) | \(138\) |
1/2*B*x^2/b^2+(-1/3*A*b+1/3*B*a)*x^2/b^2/(b*x^3+a)+1/9/b^3*sum((2*A*b-5*B* a)/_R*ln(x-_R),_R=RootOf(_Z^3*b+a))
Time = 0.28 (sec) , antiderivative size = 578, normalized size of antiderivative = 2.95 \[ \int \frac {x^4 \left (A+B x^3\right )}{\left (a+b x^3\right )^2} \, dx=\left [\frac {9 \, B a b^{3} x^{5} + 3 \, {\left (5 \, B a^{2} b^{2} - 2 \, A a b^{3}\right )} x^{2} - 3 \, \sqrt {\frac {1}{3}} {\left (5 \, B a^{3} b - 2 \, A a^{2} b^{2} + {\left (5 \, B a^{2} b^{2} - 2 \, A a b^{3}\right )} x^{3}\right )} \sqrt {\frac {\left (-a b^{2}\right )^{\frac {1}{3}}}{a}} \log \left (\frac {2 \, b^{2} x^{3} - a b + 3 \, \sqrt {\frac {1}{3}} {\left (a b x + 2 \, \left (-a b^{2}\right )^{\frac {2}{3}} x^{2} + \left (-a b^{2}\right )^{\frac {1}{3}} a\right )} \sqrt {\frac {\left (-a b^{2}\right )^{\frac {1}{3}}}{a}} - 3 \, \left (-a b^{2}\right )^{\frac {2}{3}} x}{b x^{3} + a}\right ) - {\left ({\left (5 \, B a b - 2 \, A b^{2}\right )} x^{3} + 5 \, B a^{2} - 2 \, A a b\right )} \left (-a b^{2}\right )^{\frac {2}{3}} \log \left (b^{2} x^{2} + \left (-a b^{2}\right )^{\frac {1}{3}} b x + \left (-a b^{2}\right )^{\frac {2}{3}}\right ) + 2 \, {\left ({\left (5 \, B a b - 2 \, A b^{2}\right )} x^{3} + 5 \, B a^{2} - 2 \, A a b\right )} \left (-a b^{2}\right )^{\frac {2}{3}} \log \left (b x - \left (-a b^{2}\right )^{\frac {1}{3}}\right )}{18 \, {\left (a b^{5} x^{3} + a^{2} b^{4}\right )}}, \frac {9 \, B a b^{3} x^{5} + 3 \, {\left (5 \, B a^{2} b^{2} - 2 \, A a b^{3}\right )} x^{2} - 6 \, \sqrt {\frac {1}{3}} {\left (5 \, B a^{3} b - 2 \, A a^{2} b^{2} + {\left (5 \, B a^{2} b^{2} - 2 \, A a b^{3}\right )} x^{3}\right )} \sqrt {-\frac {\left (-a b^{2}\right )^{\frac {1}{3}}}{a}} \arctan \left (\frac {\sqrt {\frac {1}{3}} {\left (2 \, b x + \left (-a b^{2}\right )^{\frac {1}{3}}\right )} \sqrt {-\frac {\left (-a b^{2}\right )^{\frac {1}{3}}}{a}}}{b}\right ) - {\left ({\left (5 \, B a b - 2 \, A b^{2}\right )} x^{3} + 5 \, B a^{2} - 2 \, A a b\right )} \left (-a b^{2}\right )^{\frac {2}{3}} \log \left (b^{2} x^{2} + \left (-a b^{2}\right )^{\frac {1}{3}} b x + \left (-a b^{2}\right )^{\frac {2}{3}}\right ) + 2 \, {\left ({\left (5 \, B a b - 2 \, A b^{2}\right )} x^{3} + 5 \, B a^{2} - 2 \, A a b\right )} \left (-a b^{2}\right )^{\frac {2}{3}} \log \left (b x - \left (-a b^{2}\right )^{\frac {1}{3}}\right )}{18 \, {\left (a b^{5} x^{3} + a^{2} b^{4}\right )}}\right ] \]
[1/18*(9*B*a*b^3*x^5 + 3*(5*B*a^2*b^2 - 2*A*a*b^3)*x^2 - 3*sqrt(1/3)*(5*B* a^3*b - 2*A*a^2*b^2 + (5*B*a^2*b^2 - 2*A*a*b^3)*x^3)*sqrt((-a*b^2)^(1/3)/a )*log((2*b^2*x^3 - a*b + 3*sqrt(1/3)*(a*b*x + 2*(-a*b^2)^(2/3)*x^2 + (-a*b ^2)^(1/3)*a)*sqrt((-a*b^2)^(1/3)/a) - 3*(-a*b^2)^(2/3)*x)/(b*x^3 + a)) - ( (5*B*a*b - 2*A*b^2)*x^3 + 5*B*a^2 - 2*A*a*b)*(-a*b^2)^(2/3)*log(b^2*x^2 + (-a*b^2)^(1/3)*b*x + (-a*b^2)^(2/3)) + 2*((5*B*a*b - 2*A*b^2)*x^3 + 5*B*a^ 2 - 2*A*a*b)*(-a*b^2)^(2/3)*log(b*x - (-a*b^2)^(1/3)))/(a*b^5*x^3 + a^2*b^ 4), 1/18*(9*B*a*b^3*x^5 + 3*(5*B*a^2*b^2 - 2*A*a*b^3)*x^2 - 6*sqrt(1/3)*(5 *B*a^3*b - 2*A*a^2*b^2 + (5*B*a^2*b^2 - 2*A*a*b^3)*x^3)*sqrt(-(-a*b^2)^(1/ 3)/a)*arctan(sqrt(1/3)*(2*b*x + (-a*b^2)^(1/3))*sqrt(-(-a*b^2)^(1/3)/a)/b) - ((5*B*a*b - 2*A*b^2)*x^3 + 5*B*a^2 - 2*A*a*b)*(-a*b^2)^(2/3)*log(b^2*x^ 2 + (-a*b^2)^(1/3)*b*x + (-a*b^2)^(2/3)) + 2*((5*B*a*b - 2*A*b^2)*x^3 + 5* B*a^2 - 2*A*a*b)*(-a*b^2)^(2/3)*log(b*x - (-a*b^2)^(1/3)))/(a*b^5*x^3 + a^ 2*b^4)]
Time = 0.55 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.64 \[ \int \frac {x^4 \left (A+B x^3\right )}{\left (a+b x^3\right )^2} \, dx=\frac {B x^{2}}{2 b^{2}} + \frac {x^{2} \left (- A b + B a\right )}{3 a b^{2} + 3 b^{3} x^{3}} + \operatorname {RootSum} {\left (729 t^{3} a b^{8} + 8 A^{3} b^{3} - 60 A^{2} B a b^{2} + 150 A B^{2} a^{2} b - 125 B^{3} a^{3}, \left ( t \mapsto t \log {\left (\frac {81 t^{2} a b^{5}}{4 A^{2} b^{2} - 20 A B a b + 25 B^{2} a^{2}} + x \right )} \right )\right )} \]
B*x**2/(2*b**2) + x**2*(-A*b + B*a)/(3*a*b**2 + 3*b**3*x**3) + RootSum(729 *_t**3*a*b**8 + 8*A**3*b**3 - 60*A**2*B*a*b**2 + 150*A*B**2*a**2*b - 125*B **3*a**3, Lambda(_t, _t*log(81*_t**2*a*b**5/(4*A**2*b**2 - 20*A*B*a*b + 25 *B**2*a**2) + x)))
Time = 0.29 (sec) , antiderivative size = 162, normalized size of antiderivative = 0.83 \[ \int \frac {x^4 \left (A+B x^3\right )}{\left (a+b x^3\right )^2} \, dx=\frac {{\left (B a - A b\right )} x^{2}}{3 \, {\left (b^{3} x^{3} + a b^{2}\right )}} + \frac {B x^{2}}{2 \, b^{2}} - \frac {\sqrt {3} {\left (5 \, B a - 2 \, A b\right )} \arctan \left (\frac {\sqrt {3} {\left (2 \, x - \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{9 \, b^{3} \left (\frac {a}{b}\right )^{\frac {1}{3}}} - \frac {{\left (5 \, B a - 2 \, A b\right )} \log \left (x^{2} - x \left (\frac {a}{b}\right )^{\frac {1}{3}} + \left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{18 \, b^{3} \left (\frac {a}{b}\right )^{\frac {1}{3}}} + \frac {{\left (5 \, B a - 2 \, A b\right )} \log \left (x + \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{9 \, b^{3} \left (\frac {a}{b}\right )^{\frac {1}{3}}} \]
1/3*(B*a - A*b)*x^2/(b^3*x^3 + a*b^2) + 1/2*B*x^2/b^2 - 1/9*sqrt(3)*(5*B*a - 2*A*b)*arctan(1/3*sqrt(3)*(2*x - (a/b)^(1/3))/(a/b)^(1/3))/(b^3*(a/b)^( 1/3)) - 1/18*(5*B*a - 2*A*b)*log(x^2 - x*(a/b)^(1/3) + (a/b)^(2/3))/(b^3*( a/b)^(1/3)) + 1/9*(5*B*a - 2*A*b)*log(x + (a/b)^(1/3))/(b^3*(a/b)^(1/3))
Time = 0.31 (sec) , antiderivative size = 189, normalized size of antiderivative = 0.96 \[ \int \frac {x^4 \left (A+B x^3\right )}{\left (a+b x^3\right )^2} \, dx=\frac {B x^{2}}{2 \, b^{2}} - \frac {\sqrt {3} {\left (5 \, B a - 2 \, A b\right )} \arctan \left (\frac {\sqrt {3} {\left (2 \, x + \left (-\frac {a}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (-\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{9 \, \left (-a b^{2}\right )^{\frac {1}{3}} b^{2}} + \frac {{\left (5 \, B a - 2 \, A b\right )} \log \left (x^{2} + x \left (-\frac {a}{b}\right )^{\frac {1}{3}} + \left (-\frac {a}{b}\right )^{\frac {2}{3}}\right )}{18 \, \left (-a b^{2}\right )^{\frac {1}{3}} b^{2}} + \frac {{\left (5 \, B a \left (-\frac {a}{b}\right )^{\frac {1}{3}} - 2 \, A b \left (-\frac {a}{b}\right )^{\frac {1}{3}}\right )} \left (-\frac {a}{b}\right )^{\frac {1}{3}} \log \left ({\left | x - \left (-\frac {a}{b}\right )^{\frac {1}{3}} \right |}\right )}{9 \, a b^{2}} + \frac {B a x^{2} - A b x^{2}}{3 \, {\left (b x^{3} + a\right )} b^{2}} \]
1/2*B*x^2/b^2 - 1/9*sqrt(3)*(5*B*a - 2*A*b)*arctan(1/3*sqrt(3)*(2*x + (-a/ b)^(1/3))/(-a/b)^(1/3))/((-a*b^2)^(1/3)*b^2) + 1/18*(5*B*a - 2*A*b)*log(x^ 2 + x*(-a/b)^(1/3) + (-a/b)^(2/3))/((-a*b^2)^(1/3)*b^2) + 1/9*(5*B*a*(-a/b )^(1/3) - 2*A*b*(-a/b)^(1/3))*(-a/b)^(1/3)*log(abs(x - (-a/b)^(1/3)))/(a*b ^2) + 1/3*(B*a*x^2 - A*b*x^2)/((b*x^3 + a)*b^2)
Time = 6.83 (sec) , antiderivative size = 158, normalized size of antiderivative = 0.81 \[ \int \frac {x^4 \left (A+B x^3\right )}{\left (a+b x^3\right )^2} \, dx=\frac {B\,x^2}{2\,b^2}-\frac {x^2\,\left (\frac {A\,b}{3}-\frac {B\,a}{3}\right )}{b^3\,x^3+a\,b^2}-\frac {\ln \left (b^{1/3}\,x+a^{1/3}\right )\,\left (2\,A\,b-5\,B\,a\right )}{9\,a^{1/3}\,b^{8/3}}-\frac {\ln \left (a^{1/3}-2\,b^{1/3}\,x+\sqrt {3}\,a^{1/3}\,1{}\mathrm {i}\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (2\,A\,b-5\,B\,a\right )}{9\,a^{1/3}\,b^{8/3}}+\frac {\ln \left (2\,b^{1/3}\,x-a^{1/3}+\sqrt {3}\,a^{1/3}\,1{}\mathrm {i}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (2\,A\,b-5\,B\,a\right )}{9\,a^{1/3}\,b^{8/3}} \]
(B*x^2)/(2*b^2) - (x^2*((A*b)/3 - (B*a)/3))/(a*b^2 + b^3*x^3) - (log(b^(1/ 3)*x + a^(1/3))*(2*A*b - 5*B*a))/(9*a^(1/3)*b^(8/3)) - (log(3^(1/2)*a^(1/3 )*1i - 2*b^(1/3)*x + a^(1/3))*((3^(1/2)*1i)/2 - 1/2)*(2*A*b - 5*B*a))/(9*a ^(1/3)*b^(8/3)) + (log(3^(1/2)*a^(1/3)*1i + 2*b^(1/3)*x - a^(1/3))*((3^(1/ 2)*1i)/2 + 1/2)*(2*A*b - 5*B*a))/(9*a^(1/3)*b^(8/3))