Integrand size = 20, antiderivative size = 244 \[ \int \frac {x^9 \left (A+B x^3\right )}{\left (a+b x^3\right )^3} \, dx=\frac {7 (2 A b-5 a B) x}{9 b^4}-\frac {7 (2 A b-5 a B) x^4}{36 a b^3}+\frac {(A b-a B) x^{10}}{6 a b \left (a+b x^3\right )^2}+\frac {(2 A b-5 a B) x^7}{9 a b^2 \left (a+b x^3\right )}+\frac {7 \sqrt [3]{a} (2 A b-5 a B) \arctan \left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right )}{9 \sqrt {3} b^{13/3}}-\frac {7 \sqrt [3]{a} (2 A b-5 a B) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{27 b^{13/3}}+\frac {7 \sqrt [3]{a} (2 A b-5 a B) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{54 b^{13/3}} \]
7/9*(2*A*b-5*B*a)*x/b^4-7/36*(2*A*b-5*B*a)*x^4/a/b^3+1/6*(A*b-B*a)*x^10/a/ b/(b*x^3+a)^2+1/9*(2*A*b-5*B*a)*x^7/a/b^2/(b*x^3+a)-7/27*a^(1/3)*(2*A*b-5* B*a)*ln(a^(1/3)+b^(1/3)*x)/b^(13/3)+7/54*a^(1/3)*(2*A*b-5*B*a)*ln(a^(2/3)- a^(1/3)*b^(1/3)*x+b^(2/3)*x^2)/b^(13/3)+7/27*a^(1/3)*(2*A*b-5*B*a)*arctan( 1/3*(a^(1/3)-2*b^(1/3)*x)/a^(1/3)*3^(1/2))/b^(13/3)*3^(1/2)
Time = 0.16 (sec) , antiderivative size = 210, normalized size of antiderivative = 0.86 \[ \int \frac {x^9 \left (A+B x^3\right )}{\left (a+b x^3\right )^3} \, dx=\frac {108 \sqrt [3]{b} (A b-3 a B) x+27 b^{4/3} B x^4+\frac {18 a^2 \sqrt [3]{b} (-A b+a B) x}{\left (a+b x^3\right )^2}+\frac {6 a \sqrt [3]{b} (13 A b-19 a B) x}{a+b x^3}-28 \sqrt {3} \sqrt [3]{a} (-2 A b+5 a B) \arctan \left (\frac {1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}}{\sqrt {3}}\right )+28 \sqrt [3]{a} (-2 A b+5 a B) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )-14 \sqrt [3]{a} (-2 A b+5 a B) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{108 b^{13/3}} \]
(108*b^(1/3)*(A*b - 3*a*B)*x + 27*b^(4/3)*B*x^4 + (18*a^2*b^(1/3)*(-(A*b) + a*B)*x)/(a + b*x^3)^2 + (6*a*b^(1/3)*(13*A*b - 19*a*B)*x)/(a + b*x^3) - 28*Sqrt[3]*a^(1/3)*(-2*A*b + 5*a*B)*ArcTan[(1 - (2*b^(1/3)*x)/a^(1/3))/Sqr t[3]] + 28*a^(1/3)*(-2*A*b + 5*a*B)*Log[a^(1/3) + b^(1/3)*x] - 14*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])/(108*b^(1 3/3))
Time = 0.34 (sec) , antiderivative size = 209, normalized size of antiderivative = 0.86, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {957, 817, 831, 2009}
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.
| \(\displaystyle \int \frac {x^9 \left (A+B x^3\right )}{\left (a+b x^3\right )^3} \, dx\) |
| \(\Big \downarrow \) 957 |
| \(\displaystyle \frac {x^{10} (A b-a B)}{6 a b \left (a+b x^3\right )^2}-\frac {(2 A b-5 a B) \int \frac {x^9}{\left (b x^3+a\right )^2}dx}{3 a b}\) |
| \(\Big \downarrow \) 817 |
| \(\displaystyle \frac {x^{10} (A b-a B)}{6 a b \left (a+b x^3\right )^2}-\frac {(2 A b-5 a B) \left (\frac {7 \int \frac {x^6}{b x^3+a}dx}{3 b}-\frac {x^7}{3 b \left (a+b x^3\right )}\right )}{3 a b}\) |
| \(\Big \downarrow \) 831 |
| \(\displaystyle \frac {x^{10} (A b-a B)}{6 a b \left (a+b x^3\right )^2}-\frac {(2 A b-5 a B) \left (\frac {7 \int \left (\frac {x^3}{b}+\frac {a^2}{b^2 \left (b x^3+a\right )}-\frac {a}{b^2}\right )dx}{3 b}-\frac {x^7}{3 b \left (a+b x^3\right )}\right )}{3 a b}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {x^{10} (A b-a B)}{6 a b \left (a+b x^3\right )^2}-\frac {(2 A b-5 a B) \left (\frac {7 \left (-\frac {a^{4/3} \arctan \left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt {3} b^{7/3}}-\frac {a^{4/3} \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{6 b^{7/3}}+\frac {a^{4/3} \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 b^{7/3}}-\frac {a x}{b^2}+\frac {x^4}{4 b}\right )}{3 b}-\frac {x^7}{3 b \left (a+b x^3\right )}\right )}{3 a b}\) |
((A*b - a*B)*x^10)/(6*a*b*(a + b*x^3)^2) - ((2*A*b - 5*a*B)*(-1/3*x^7/(b*( a + b*x^3)) + (7*(-((a*x)/b^2) + x^4/(4*b) - (a^(4/3)*ArcTan[(a^(1/3) - 2* b^(1/3)*x)/(Sqrt[3]*a^(1/3))])/(Sqrt[3]*b^(7/3)) + (a^(4/3)*Log[a^(1/3) + b^(1/3)*x])/(3*b^(7/3)) - (a^(4/3)*Log[a^(2/3) - a^(1/3)*b^(1/3)*x + b^(2/ 3)*x^2])/(6*b^(7/3))))/(3*b)))/(3*a*b)
3.1.96.3.1 Defintions of rubi rules used
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*n*(p + 1))), x] - Simp[c^n *((m - n + 1)/(b*n*(p + 1))) Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x], x ] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] && ! ILtQ[(m + n*(p + 1) + 1)/n, 0] && IntBinomialQ[a, b, c, n, m, p, x]
Int[(x_)^(m_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[PolynomialDivide[x ^m, a + b*x^n, x], x] /; FreeQ[{a, b}, x] && IGtQ[m, 0] && IGtQ[n, 0] && Gt Q[m, 2*n - 1]
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)]))
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 4.19 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.45
| method | result | size |
| risch | \(\frac {B \,x^{4}}{4 b^{3}}+\frac {A x}{b^{3}}-\frac {3 B a x}{b^{4}}+\frac {\left (\frac {13}{18} a \,b^{2} A -\frac {19}{18} a^{2} b B \right ) x^{4}+\frac {a^{2} \left (5 A b -8 B a \right ) x}{9}}{b^{4} \left (b \,x^{3}+a \right )^{2}}-\frac {7 a \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (b \,\textit {\_Z}^{3}+a \right )}{\sum }\frac {\left (2 A b -5 B a \right ) \ln \left (x -\textit {\_R} \right )}{\textit {\_R}^{2}}\right )}{27 b^{5}}\) | \(109\) |
| default | \(\frac {\frac {1}{4} b B \,x^{4}+A b x -3 B a x}{b^{4}}-\frac {a \left (\frac {\left (-\frac {13}{18} b^{2} A +\frac {19}{18} a b B \right ) x^{4}-\frac {a \left (5 A b -8 B a \right ) x}{9}}{\left (b \,x^{3}+a \right )^{2}}+\frac {7 \left (2 A b -5 B a \right ) \left (\frac {\ln \left (x +\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}-\frac {\ln \left (x^{2}-\left (\frac {a}{b}\right )^{\frac {1}{3}} x +\left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}+\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {a}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}\right )}{9}\right )}{b^{4}}\) | \(171\) |
1/4*B*x^4/b^3+1/b^3*A*x-3/b^4*B*a*x+((13/18*a*b^2*A-19/18*a^2*b*B)*x^4+1/9 *a^2*(5*A*b-8*B*a)*x)/b^4/(b*x^3+a)^2-7/27/b^5*a*sum((2*A*b-5*B*a)/_R^2*ln (x-_R),_R=RootOf(_Z^3*b+a))
Time = 0.29 (sec) , antiderivative size = 347, normalized size of antiderivative = 1.42 \[ \int \frac {x^9 \left (A+B x^3\right )}{\left (a+b x^3\right )^3} \, dx=\frac {27 \, B b^{3} x^{10} - 54 \, {\left (5 \, B a b^{2} - 2 \, A b^{3}\right )} x^{7} - 147 \, {\left (5 \, B a^{2} b - 2 \, A a b^{2}\right )} x^{4} - 28 \, \sqrt {3} {\left ({\left (5 \, B a b^{2} - 2 \, A b^{3}\right )} x^{6} + 5 \, B a^{3} - 2 \, A a^{2} b + 2 \, {\left (5 \, B a^{2} b - 2 \, A a b^{2}\right )} x^{3}\right )} \left (-\frac {a}{b}\right )^{\frac {1}{3}} \arctan \left (\frac {2 \, \sqrt {3} b x \left (-\frac {a}{b}\right )^{\frac {2}{3}} - \sqrt {3} a}{3 \, a}\right ) + 14 \, {\left ({\left (5 \, B a b^{2} - 2 \, A b^{3}\right )} x^{6} + 5 \, B a^{3} - 2 \, A a^{2} b + 2 \, {\left (5 \, B a^{2} b - 2 \, A a b^{2}\right )} x^{3}\right )} \left (-\frac {a}{b}\right )^{\frac {1}{3}} \log \left (x^{2} + x \left (-\frac {a}{b}\right )^{\frac {1}{3}} + \left (-\frac {a}{b}\right )^{\frac {2}{3}}\right ) - 28 \, {\left ({\left (5 \, B a b^{2} - 2 \, A b^{3}\right )} x^{6} + 5 \, B a^{3} - 2 \, A a^{2} b + 2 \, {\left (5 \, B a^{2} b - 2 \, A a b^{2}\right )} x^{3}\right )} \left (-\frac {a}{b}\right )^{\frac {1}{3}} \log \left (x - \left (-\frac {a}{b}\right )^{\frac {1}{3}}\right ) - 84 \, {\left (5 \, B a^{3} - 2 \, A a^{2} b\right )} x}{108 \, {\left (b^{6} x^{6} + 2 \, a b^{5} x^{3} + a^{2} b^{4}\right )}} \]
1/108*(27*B*b^3*x^10 - 54*(5*B*a*b^2 - 2*A*b^3)*x^7 - 147*(5*B*a^2*b - 2*A *a*b^2)*x^4 - 28*sqrt(3)*((5*B*a*b^2 - 2*A*b^3)*x^6 + 5*B*a^3 - 2*A*a^2*b + 2*(5*B*a^2*b - 2*A*a*b^2)*x^3)*(-a/b)^(1/3)*arctan(1/3*(2*sqrt(3)*b*x*(- a/b)^(2/3) - sqrt(3)*a)/a) + 14*((5*B*a*b^2 - 2*A*b^3)*x^6 + 5*B*a^3 - 2*A *a^2*b + 2*(5*B*a^2*b - 2*A*a*b^2)*x^3)*(-a/b)^(1/3)*log(x^2 + x*(-a/b)^(1 /3) + (-a/b)^(2/3)) - 28*((5*B*a*b^2 - 2*A*b^3)*x^6 + 5*B*a^3 - 2*A*a^2*b + 2*(5*B*a^2*b - 2*A*a*b^2)*x^3)*(-a/b)^(1/3)*log(x - (-a/b)^(1/3)) - 84*( 5*B*a^3 - 2*A*a^2*b)*x)/(b^6*x^6 + 2*a*b^5*x^3 + a^2*b^4)
Time = 1.39 (sec) , antiderivative size = 163, normalized size of antiderivative = 0.67 \[ \int \frac {x^9 \left (A+B x^3\right )}{\left (a+b x^3\right )^3} \, dx=\frac {B x^{4}}{4 b^{3}} + x \left (\frac {A}{b^{3}} - \frac {3 B a}{b^{4}}\right ) + \frac {x^{4} \cdot \left (13 A a b^{2} - 19 B a^{2} b\right ) + x \left (10 A a^{2} b - 16 B a^{3}\right )}{18 a^{2} b^{4} + 36 a b^{5} x^{3} + 18 b^{6} x^{6}} + \operatorname {RootSum} {\left (19683 t^{3} b^{13} + 2744 A^{3} a b^{3} - 20580 A^{2} B a^{2} b^{2} + 51450 A B^{2} a^{3} b - 42875 B^{3} a^{4}, \left ( t \mapsto t \log {\left (\frac {27 t b^{4}}{- 14 A b + 35 B a} + x \right )} \right )\right )} \]
B*x**4/(4*b**3) + x*(A/b**3 - 3*B*a/b**4) + (x**4*(13*A*a*b**2 - 19*B*a**2 *b) + x*(10*A*a**2*b - 16*B*a**3))/(18*a**2*b**4 + 36*a*b**5*x**3 + 18*b** 6*x**6) + RootSum(19683*_t**3*b**13 + 2744*A**3*a*b**3 - 20580*A**2*B*a**2 *b**2 + 51450*A*B**2*a**3*b - 42875*B**3*a**4, Lambda(_t, _t*log(27*_t*b** 4/(-14*A*b + 35*B*a) + x)))
Time = 0.27 (sec) , antiderivative size = 223, normalized size of antiderivative = 0.91 \[ \int \frac {x^9 \left (A+B x^3\right )}{\left (a+b x^3\right )^3} \, dx=-\frac {{\left (19 \, B a^{2} b - 13 \, A a b^{2}\right )} x^{4} + 2 \, {\left (8 \, B a^{3} - 5 \, A a^{2} b\right )} x}{18 \, {\left (b^{6} x^{6} + 2 \, a b^{5} x^{3} + a^{2} b^{4}\right )}} + \frac {B b x^{4} - 4 \, {\left (3 \, B a - A b\right )} x}{4 \, b^{4}} + \frac {7 \, \sqrt {3} {\left (5 \, B a^{2} - 2 \, A a b\right )} \arctan \left (\frac {\sqrt {3} {\left (2 \, x - \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{27 \, b^{5} \left (\frac {a}{b}\right )^{\frac {2}{3}}} - \frac {7 \, {\left (5 \, B a^{2} - 2 \, A a b\right )} \log \left (x^{2} - x \left (\frac {a}{b}\right )^{\frac {1}{3}} + \left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{54 \, b^{5} \left (\frac {a}{b}\right )^{\frac {2}{3}}} + \frac {7 \, {\left (5 \, B a^{2} - 2 \, A a b\right )} \log \left (x + \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{27 \, b^{5} \left (\frac {a}{b}\right )^{\frac {2}{3}}} \]
-1/18*((19*B*a^2*b - 13*A*a*b^2)*x^4 + 2*(8*B*a^3 - 5*A*a^2*b)*x)/(b^6*x^6 + 2*a*b^5*x^3 + a^2*b^4) + 1/4*(B*b*x^4 - 4*(3*B*a - A*b)*x)/b^4 + 7/27*s qrt(3)*(5*B*a^2 - 2*A*a*b)*arctan(1/3*sqrt(3)*(2*x - (a/b)^(1/3))/(a/b)^(1 /3))/(b^5*(a/b)^(2/3)) - 7/54*(5*B*a^2 - 2*A*a*b)*log(x^2 - x*(a/b)^(1/3) + (a/b)^(2/3))/(b^5*(a/b)^(2/3)) + 7/27*(5*B*a^2 - 2*A*a*b)*log(x + (a/b)^ (1/3))/(b^5*(a/b)^(2/3))
Time = 0.29 (sec) , antiderivative size = 234, normalized size of antiderivative = 0.96 \[ \int \frac {x^9 \left (A+B x^3\right )}{\left (a+b x^3\right )^3} \, dx=\frac {7 \, \sqrt {3} {\left (5 \, \left (-a b^{2}\right )^{\frac {1}{3}} B a - 2 \, \left (-a b^{2}\right )^{\frac {1}{3}} A b\right )} \arctan \left (\frac {\sqrt {3} {\left (2 \, x + \left (-\frac {a}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (-\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{27 \, b^{5}} - \frac {7 \, {\left (5 \, B a^{2} - 2 \, A a b\right )} \left (-\frac {a}{b}\right )^{\frac {1}{3}} \log \left ({\left | x - \left (-\frac {a}{b}\right )^{\frac {1}{3}} \right |}\right )}{27 \, a b^{4}} + \frac {7 \, {\left (5 \, \left (-a b^{2}\right )^{\frac {1}{3}} B a - 2 \, \left (-a b^{2}\right )^{\frac {1}{3}} A b\right )} \log \left (x^{2} + x \left (-\frac {a}{b}\right )^{\frac {1}{3}} + \left (-\frac {a}{b}\right )^{\frac {2}{3}}\right )}{54 \, b^{5}} - \frac {19 \, B a^{2} b x^{4} - 13 \, A a b^{2} x^{4} + 16 \, B a^{3} x - 10 \, A a^{2} b x}{18 \, {\left (b x^{3} + a\right )}^{2} b^{4}} + \frac {B b^{9} x^{4} - 12 \, B a b^{8} x + 4 \, A b^{9} x}{4 \, b^{12}} \]
7/27*sqrt(3)*(5*(-a*b^2)^(1/3)*B*a - 2*(-a*b^2)^(1/3)*A*b)*arctan(1/3*sqrt (3)*(2*x + (-a/b)^(1/3))/(-a/b)^(1/3))/b^5 - 7/27*(5*B*a^2 - 2*A*a*b)*(-a/ b)^(1/3)*log(abs(x - (-a/b)^(1/3)))/(a*b^4) + 7/54*(5*(-a*b^2)^(1/3)*B*a - 2*(-a*b^2)^(1/3)*A*b)*log(x^2 + x*(-a/b)^(1/3) + (-a/b)^(2/3))/b^5 - 1/18 *(19*B*a^2*b*x^4 - 13*A*a*b^2*x^4 + 16*B*a^3*x - 10*A*a^2*b*x)/((b*x^3 + a )^2*b^4) + 1/4*(B*b^9*x^4 - 12*B*a*b^8*x + 4*A*b^9*x)/b^12
Time = 0.34 (sec) , antiderivative size = 227, normalized size of antiderivative = 0.93 \[ \int \frac {x^9 \left (A+B x^3\right )}{\left (a+b x^3\right )^3} \, dx=\frac {x^4\,\left (\frac {13\,A\,a\,b^2}{18}-\frac {19\,B\,a^2\,b}{18}\right )-x\,\left (\frac {8\,B\,a^3}{9}-\frac {5\,A\,a^2\,b}{9}\right )}{a^2\,b^4+2\,a\,b^5\,x^3+b^6\,x^6}+x\,\left (\frac {A}{b^3}-\frac {3\,B\,a}{b^4}\right )+\frac {B\,x^4}{4\,b^3}+\frac {7\,{\left (-a\right )}^{1/3}\,\ln \left ({\left (-a\right )}^{4/3}+a\,b^{1/3}\,x\right )\,\left (2\,A\,b-5\,B\,a\right )}{27\,b^{13/3}}-\frac {7\,{\left (-a\right )}^{1/3}\,\ln \left ({\left (-a\right )}^{4/3}-2\,a\,b^{1/3}\,x+\sqrt {3}\,{\left (-a\right )}^{4/3}\,1{}\mathrm {i}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (2\,A\,b-5\,B\,a\right )}{27\,b^{13/3}}+\frac {7\,{\left (-a\right )}^{1/3}\,\ln \left (2\,a\,b^{1/3}\,x-{\left (-a\right )}^{4/3}+\sqrt {3}\,{\left (-a\right )}^{4/3}\,1{}\mathrm {i}\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (2\,A\,b-5\,B\,a\right )}{27\,b^{13/3}} \]
(x^4*((13*A*a*b^2)/18 - (19*B*a^2*b)/18) - x*((8*B*a^3)/9 - (5*A*a^2*b)/9) )/(a^2*b^4 + b^6*x^6 + 2*a*b^5*x^3) + x*(A/b^3 - (3*B*a)/b^4) + (B*x^4)/(4 *b^3) + (7*(-a)^(1/3)*log((-a)^(4/3) + a*b^(1/3)*x)*(2*A*b - 5*B*a))/(27*b ^(13/3)) - (7*(-a)^(1/3)*log((-a)^(4/3) + 3^(1/2)*(-a)^(4/3)*1i - 2*a*b^(1 /3)*x)*((3^(1/2)*1i)/2 + 1/2)*(2*A*b - 5*B*a))/(27*b^(13/3)) + (7*(-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)*(2*A*b - 5*B*a))/(27*b^(13/3))