Integrand size = 20, antiderivative size = 162 \[ \int \frac {x^3 \left (A+B x^3\right )}{a+b x^3} \, dx=\frac {(A b-a B) x}{b^2}+\frac {B x^4}{4 b}+\frac {\sqrt [3]{a} (A b-a B) \arctan \left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt {3} b^{7/3}}-\frac {\sqrt [3]{a} (A b-a B) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 b^{7/3}}+\frac {\sqrt [3]{a} (A b-a B) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{6 b^{7/3}} \]
(A*b-B*a)*x/b^2+1/4*B*x^4/b-1/3*a^(1/3)*(A*b-B*a)*ln(a^(1/3)+b^(1/3)*x)/b^ (7/3)+1/6*a^(1/3)*(A*b-B*a)*ln(a^(2/3)-a^(1/3)*b^(1/3)*x+b^(2/3)*x^2)/b^(7 /3)+1/3*a^(1/3)*(A*b-B*a)*arctan(1/3*(a^(1/3)-2*b^(1/3)*x)/a^(1/3)*3^(1/2) )/b^(7/3)*3^(1/2)
Time = 0.08 (sec) , antiderivative size = 152, normalized size of antiderivative = 0.94 \[ \int \frac {x^3 \left (A+B x^3\right )}{a+b x^3} \, dx=\frac {12 \sqrt [3]{b} (A b-a B) x+3 b^{4/3} B x^4-4 \sqrt {3} \sqrt [3]{a} (-A b+a B) \arctan \left (\frac {1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}}{\sqrt {3}}\right )+4 \sqrt [3]{a} (-A b+a B) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )-2 \sqrt [3]{a} (-A b+a B) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{12 b^{7/3}} \]
(12*b^(1/3)*(A*b - a*B)*x + 3*b^(4/3)*B*x^4 - 4*Sqrt[3]*a^(1/3)*(-(A*b) + a*B)*ArcTan[(1 - (2*b^(1/3)*x)/a^(1/3))/Sqrt[3]] + 4*a^(1/3)*(-(A*b) + a*B )*Log[a^(1/3) + b^(1/3)*x] - 2*a^(1/3)*(-(A*b) + a*B)*Log[a^(2/3) - a^(1/3 )*b^(1/3)*x + b^(2/3)*x^2])/(12*b^(7/3))
Time = 0.33 (sec) , antiderivative size = 147, normalized size of antiderivative = 0.91, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {959, 843, 750, 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.
\(\displaystyle \int \frac {x^3 \left (A+B x^3\right )}{a+b x^3} \, dx\) |
\(\Big \downarrow \) 959 |
\(\displaystyle \frac {(A b-a B) \int \frac {x^3}{b x^3+a}dx}{b}+\frac {B x^4}{4 b}\) |
\(\Big \downarrow \) 843 |
\(\displaystyle \frac {(A b-a B) \left (\frac {x}{b}-\frac {a \int \frac {1}{b x^3+a}dx}{b}\right )}{b}+\frac {B x^4}{4 b}\) |
\(\Big \downarrow \) 750 |
\(\displaystyle \frac {(A b-a B) \left (\frac {x}{b}-\frac {a \left (\frac {\int \frac {2 \sqrt [3]{a}-\sqrt [3]{b} x}{b^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3}}dx}{3 a^{2/3}}+\frac {\int \frac {1}{\sqrt [3]{b} x+\sqrt [3]{a}}dx}{3 a^{2/3}}\right )}{b}\right )}{b}+\frac {B x^4}{4 b}\) |
\(\Big \downarrow \) 16 |
\(\displaystyle \frac {(A b-a B) \left (\frac {x}{b}-\frac {a \left (\frac {\int \frac {2 \sqrt [3]{a}-\sqrt [3]{b} x}{b^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3}}dx}{3 a^{2/3}}+\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 a^{2/3} \sqrt [3]{b}}\right )}{b}\right )}{b}+\frac {B x^4}{4 b}\) |
\(\Big \downarrow \) 1142 |
\(\displaystyle \frac {(A b-a B) \left (\frac {x}{b}-\frac {a \left (\frac {\frac {3}{2} \sqrt [3]{a} \int \frac {1}{b^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3}}dx-\frac {\int -\frac {\sqrt [3]{b} \left (\sqrt [3]{a}-2 \sqrt [3]{b} x\right )}{b^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3}}dx}{2 \sqrt [3]{b}}}{3 a^{2/3}}+\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 a^{2/3} \sqrt [3]{b}}\right )}{b}\right )}{b}+\frac {B x^4}{4 b}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {(A b-a B) \left (\frac {x}{b}-\frac {a \left (\frac {\frac {3}{2} \sqrt [3]{a} \int \frac {1}{b^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3}}dx+\frac {\int \frac {\sqrt [3]{b} \left (\sqrt [3]{a}-2 \sqrt [3]{b} x\right )}{b^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3}}dx}{2 \sqrt [3]{b}}}{3 a^{2/3}}+\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 a^{2/3} \sqrt [3]{b}}\right )}{b}\right )}{b}+\frac {B x^4}{4 b}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {(A b-a B) \left (\frac {x}{b}-\frac {a \left (\frac {\frac {3}{2} \sqrt [3]{a} \int \frac {1}{b^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3}}dx+\frac {1}{2} \int \frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{b^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3}}dx}{3 a^{2/3}}+\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 a^{2/3} \sqrt [3]{b}}\right )}{b}\right )}{b}+\frac {B x^4}{4 b}\) |
\(\Big \downarrow \) 1082 |
\(\displaystyle \frac {(A b-a B) \left (\frac {x}{b}-\frac {a \left (\frac {\frac {1}{2} \int \frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{b^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3}}dx+\frac {3 \int \frac {1}{-\left (1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}\right )^2-3}d\left (1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}\right )}{\sqrt [3]{b}}}{3 a^{2/3}}+\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 a^{2/3} \sqrt [3]{b}}\right )}{b}\right )}{b}+\frac {B x^4}{4 b}\) |
\(\Big \downarrow \) 217 |
\(\displaystyle \frac {(A b-a B) \left (\frac {x}{b}-\frac {a \left (\frac {\frac {1}{2} \int \frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{b^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3}}dx-\frac {\sqrt {3} \arctan \left (\frac {1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}}{\sqrt {3}}\right )}{\sqrt [3]{b}}}{3 a^{2/3}}+\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 a^{2/3} \sqrt [3]{b}}\right )}{b}\right )}{b}+\frac {B x^4}{4 b}\) |
\(\Big \downarrow \) 1103 |
\(\displaystyle \frac {(A b-a B) \left (\frac {x}{b}-\frac {a \left (\frac {-\frac {\log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{2 \sqrt [3]{b}}-\frac {\sqrt {3} \arctan \left (\frac {1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}}{\sqrt {3}}\right )}{\sqrt [3]{b}}}{3 a^{2/3}}+\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 a^{2/3} \sqrt [3]{b}}\right )}{b}\right )}{b}+\frac {B x^4}{4 b}\) |
(B*x^4)/(4*b) + ((A*b - a*B)*(x/b - (a*(Log[a^(1/3) + b^(1/3)*x]/(3*a^(2/3 )*b^(1/3)) + (-((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^(2 /3))))/b))/b
3.1.59.3.1 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[((a_) + (b_.)*(x_)^3)^(-1), x_Symbol] :> Simp[1/(3*Rt[a, 3]^2) Int[1/ (Rt[a, 3] + Rt[b, 3]*x), x], x] + Simp[1/(3*Rt[a, 3]^2) Int[(2*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[d*(e*x)^(m + 1)*((a + b*x^n)^(p + 1)/(b*e*(m + n*(p + 1) + 1))), x] - Simp[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(b*(m + n*(p + 1) + 1)) Int[(e*x)^m*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, m, n, p}, x] && NeQ[b*c - a*d, 0] && NeQ[m + n*(p + 1) + 1, 0]
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 = 4.28 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.37
method | result | size |
risch | \(\frac {B \,x^{4}}{4 b}+\frac {A x}{b}-\frac {B a x}{b^{2}}+\frac {a \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (b \,\textit {\_Z}^{3}+a \right )}{\sum }\frac {\left (-A b +B a \right ) \ln \left (x -\textit {\_R} \right )}{\textit {\_R}^{2}}\right )}{3 b^{3}}\) | \(60\) |
default | \(\frac {\frac {1}{4} b B \,x^{4}+A b x -B a x}{b^{2}}-\frac {\left (\frac {\ln \left (x +\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}-\frac {\ln \left (x^{2}-\left (\frac {a}{b}\right )^{\frac {1}{3}} x +\left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}+\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {a}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}\right ) a \left (A b -B a \right )}{b^{2}}\) | \(127\) |
Time = 0.37 (sec) , antiderivative size = 145, normalized size of antiderivative = 0.90 \[ \int \frac {x^3 \left (A+B x^3\right )}{a+b x^3} \, dx=\frac {3 \, B b x^{4} - 4 \, \sqrt {3} {\left (B a - A b\right )} \left (-\frac {a}{b}\right )^{\frac {1}{3}} \arctan \left (\frac {2 \, \sqrt {3} b x \left (-\frac {a}{b}\right )^{\frac {2}{3}} - \sqrt {3} a}{3 \, a}\right ) + 2 \, {\left (B a - A b\right )} \left (-\frac {a}{b}\right )^{\frac {1}{3}} \log \left (x^{2} + x \left (-\frac {a}{b}\right )^{\frac {1}{3}} + \left (-\frac {a}{b}\right )^{\frac {2}{3}}\right ) - 4 \, {\left (B a - A b\right )} \left (-\frac {a}{b}\right )^{\frac {1}{3}} \log \left (x - \left (-\frac {a}{b}\right )^{\frac {1}{3}}\right ) - 12 \, {\left (B a - A b\right )} x}{12 \, b^{2}} \]
1/12*(3*B*b*x^4 - 4*sqrt(3)*(B*a - A*b)*(-a/b)^(1/3)*arctan(1/3*(2*sqrt(3) *b*x*(-a/b)^(2/3) - sqrt(3)*a)/a) + 2*(B*a - A*b)*(-a/b)^(1/3)*log(x^2 + x *(-a/b)^(1/3) + (-a/b)^(2/3)) - 4*(B*a - A*b)*(-a/b)^(1/3)*log(x - (-a/b)^ (1/3)) - 12*(B*a - A*b)*x)/b^2
Time = 0.27 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.54 \[ \int \frac {x^3 \left (A+B x^3\right )}{a+b x^3} \, dx=\frac {B x^{4}}{4 b} + x \left (\frac {A}{b} - \frac {B a}{b^{2}}\right ) + \operatorname {RootSum} {\left (27 t^{3} b^{7} + A^{3} a b^{3} - 3 A^{2} B a^{2} b^{2} + 3 A B^{2} a^{3} b - B^{3} a^{4}, \left ( t \mapsto t \log {\left (\frac {3 t b^{2}}{- A b + B a} + x \right )} \right )\right )} \]
B*x**4/(4*b) + x*(A/b - B*a/b**2) + RootSum(27*_t**3*b**7 + A**3*a*b**3 - 3*A**2*B*a**2*b**2 + 3*A*B**2*a**3*b - B**3*a**4, Lambda(_t, _t*log(3*_t*b **2/(-A*b + B*a) + x)))
Time = 0.30 (sec) , antiderivative size = 154, normalized size of antiderivative = 0.95 \[ \int \frac {x^3 \left (A+B x^3\right )}{a+b x^3} \, dx=\frac {B b x^{4} - 4 \, {\left (B a - A b\right )} x}{4 \, b^{2}} + \frac {\sqrt {3} {\left (B a^{2} - A a b\right )} \arctan \left (\frac {\sqrt {3} {\left (2 \, x - \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{3 \, b^{3} \left (\frac {a}{b}\right )^{\frac {2}{3}}} - \frac {{\left (B a^{2} - A a b\right )} \log \left (x^{2} - x \left (\frac {a}{b}\right )^{\frac {1}{3}} + \left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 \, b^{3} \left (\frac {a}{b}\right )^{\frac {2}{3}}} + \frac {{\left (B a^{2} - A a b\right )} \log \left (x + \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 \, b^{3} \left (\frac {a}{b}\right )^{\frac {2}{3}}} \]
1/4*(B*b*x^4 - 4*(B*a - A*b)*x)/b^2 + 1/3*sqrt(3)*(B*a^2 - A*a*b)*arctan(1 /3*sqrt(3)*(2*x - (a/b)^(1/3))/(a/b)^(1/3))/(b^3*(a/b)^(2/3)) - 1/6*(B*a^2 - A*a*b)*log(x^2 - x*(a/b)^(1/3) + (a/b)^(2/3))/(b^3*(a/b)^(2/3)) + 1/3*( B*a^2 - A*a*b)*log(x + (a/b)^(1/3))/(b^3*(a/b)^(2/3))
Time = 0.29 (sec) , antiderivative size = 186, normalized size of antiderivative = 1.15 \[ \int \frac {x^3 \left (A+B x^3\right )}{a+b x^3} \, dx=\frac {\sqrt {3} {\left (\left (-a b^{2}\right )^{\frac {1}{3}} B a - \left (-a b^{2}\right )^{\frac {1}{3}} A b\right )} \arctan \left (\frac {\sqrt {3} {\left (2 \, x + \left (-\frac {a}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (-\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{3 \, b^{3}} + \frac {{\left (\left (-a b^{2}\right )^{\frac {1}{3}} B a - \left (-a b^{2}\right )^{\frac {1}{3}} A b\right )} \log \left (x^{2} + x \left (-\frac {a}{b}\right )^{\frac {1}{3}} + \left (-\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 \, b^{3}} - \frac {{\left (B a^{2} b^{2} - A a b^{3}\right )} \left (-\frac {a}{b}\right )^{\frac {1}{3}} \log \left ({\left | x - \left (-\frac {a}{b}\right )^{\frac {1}{3}} \right |}\right )}{3 \, a b^{4}} + \frac {B b^{3} x^{4} - 4 \, B a b^{2} x + 4 \, A b^{3} x}{4 \, b^{4}} \]
1/3*sqrt(3)*((-a*b^2)^(1/3)*B*a - (-a*b^2)^(1/3)*A*b)*arctan(1/3*sqrt(3)*( 2*x + (-a/b)^(1/3))/(-a/b)^(1/3))/b^3 + 1/6*((-a*b^2)^(1/3)*B*a - (-a*b^2) ^(1/3)*A*b)*log(x^2 + x*(-a/b)^(1/3) + (-a/b)^(2/3))/b^3 - 1/3*(B*a^2*b^2 - A*a*b^3)*(-a/b)^(1/3)*log(abs(x - (-a/b)^(1/3)))/(a*b^4) + 1/4*(B*b^3*x^ 4 - 4*B*a*b^2*x + 4*A*b^3*x)/b^4
Time = 7.05 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.00 \[ \int \frac {x^3 \left (A+B x^3\right )}{a+b x^3} \, dx=x\,\left (\frac {A}{b}-\frac {B\,a}{b^2}\right )+\frac {B\,x^4}{4\,b}+\frac {{\left (-a\right )}^{1/3}\,\ln \left ({\left (-a\right )}^{4/3}+a\,b^{1/3}\,x\right )\,\left (A\,b-B\,a\right )}{3\,b^{7/3}}-\frac {{\left (-a\right )}^{1/3}\,\ln \left (2\,a\,b^{1/3}\,x-{\left (-a\right )}^{4/3}-\sqrt {3}\,{\left (-a\right )}^{4/3}\,1{}\mathrm {i}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (A\,b-B\,a\right )}{3\,b^{7/3}}+\frac {{\left (-a\right )}^{1/3}\,\ln \left (2\,a\,b^{1/3}\,x-{\left (-a\right )}^{4/3}+\sqrt {3}\,{\left (-a\right )}^{4/3}\,1{}\mathrm {i}\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (A\,b-B\,a\right )}{3\,b^{7/3}} \]
x*(A/b - (B*a)/b^2) + (B*x^4)/(4*b) + ((-a)^(1/3)*log((-a)^(4/3) + a*b^(1/ 3)*x)*(A*b - B*a))/(3*b^(7/3)) - ((-a)^(1/3)*log(2*a*b^(1/3)*x - 3^(1/2)*( -a)^(4/3)*1i - (-a)^(4/3))*((3^(1/2)*1i)/2 + 1/2)*(A*b - B*a))/(3*b^(7/3)) + ((-a)^(1/3)*log(3^(1/2)*(-a)^(4/3)*1i - (-a)^(4/3) + 2*a*b^(1/3)*x)*((3 ^(1/2)*1i)/2 - 1/2)*(A*b - B*a))/(3*b^(7/3))