Integrand size = 24, antiderivative size = 290 \[ \int \frac {x^{14}}{\sqrt [3]{a+b x^3} \left (c+d x^3\right )} \, dx=-\frac {(b c+a d) \left (b^2 c^2+a^2 d^2\right ) \left (a+b x^3\right )^{2/3}}{2 b^4 d^4}+\frac {\left (b^2 c^2+2 a b c d+3 a^2 d^2\right ) \left (a+b x^3\right )^{5/3}}{5 b^4 d^3}-\frac {(b c+3 a d) \left (a+b x^3\right )^{8/3}}{8 b^4 d^2}+\frac {\left (a+b x^3\right )^{11/3}}{11 b^4 d}-\frac {c^4 \arctan \left (\frac {1-\frac {2 \sqrt [3]{d} \sqrt [3]{a+b x^3}}{\sqrt [3]{b c-a d}}}{\sqrt {3}}\right )}{\sqrt {3} d^{14/3} \sqrt [3]{b c-a d}}+\frac {c^4 \log \left (c+d x^3\right )}{6 d^{14/3} \sqrt [3]{b c-a d}}-\frac {c^4 \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{2 d^{14/3} \sqrt [3]{b c-a d}} \]
-1/2*(a*d+b*c)*(a^2*d^2+b^2*c^2)*(b*x^3+a)^(2/3)/b^4/d^4+1/5*(3*a^2*d^2+2* a*b*c*d+b^2*c^2)*(b*x^3+a)^(5/3)/b^4/d^3-1/8*(3*a*d+b*c)*(b*x^3+a)^(8/3)/b ^4/d^2+1/11*(b*x^3+a)^(11/3)/b^4/d+1/6*c^4*ln(d*x^3+c)/d^(14/3)/(-a*d+b*c) ^(1/3)-1/2*c^4*ln((-a*d+b*c)^(1/3)+d^(1/3)*(b*x^3+a)^(1/3))/d^(14/3)/(-a*d +b*c)^(1/3)-1/3*c^4*arctan(1/3*(1-2*d^(1/3)*(b*x^3+a)^(1/3)/(-a*d+b*c)^(1/ 3))*3^(1/2))/d^(14/3)/(-a*d+b*c)^(1/3)*3^(1/2)
Time = 1.01 (sec) , antiderivative size = 307, normalized size of antiderivative = 1.06 \[ \int \frac {x^{14}}{\sqrt [3]{a+b x^3} \left (c+d x^3\right )} \, dx=\frac {-3 d^{2/3} \sqrt [3]{b c-a d} \left (a+b x^3\right )^{2/3} \left (81 a^3 d^3+9 a^2 b d^2 \left (11 c-6 d x^3\right )+3 a b^2 d \left (44 c^2-22 c d x^3+15 d^2 x^6\right )+b^3 \left (220 c^3-88 c^2 d x^3+55 c d^2 x^6-40 d^3 x^9\right )\right )-440 \sqrt {3} b^4 c^4 \arctan \left (\frac {1-\frac {2 \sqrt [3]{d} \sqrt [3]{a+b x^3}}{\sqrt [3]{b c-a d}}}{\sqrt {3}}\right )-440 b^4 c^4 \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )+220 b^4 c^4 \log \left ((b c-a d)^{2/3}-\sqrt [3]{d} \sqrt [3]{b c-a d} \sqrt [3]{a+b x^3}+d^{2/3} \left (a+b x^3\right )^{2/3}\right )}{1320 b^4 d^{14/3} \sqrt [3]{b c-a d}} \]
(-3*d^(2/3)*(b*c - a*d)^(1/3)*(a + b*x^3)^(2/3)*(81*a^3*d^3 + 9*a^2*b*d^2* (11*c - 6*d*x^3) + 3*a*b^2*d*(44*c^2 - 22*c*d*x^3 + 15*d^2*x^6) + b^3*(220 *c^3 - 88*c^2*d*x^3 + 55*c*d^2*x^6 - 40*d^3*x^9)) - 440*Sqrt[3]*b^4*c^4*Ar cTan[(1 - (2*d^(1/3)*(a + b*x^3)^(1/3))/(b*c - a*d)^(1/3))/Sqrt[3]] - 440* b^4*c^4*Log[(b*c - a*d)^(1/3) + d^(1/3)*(a + b*x^3)^(1/3)] + 220*b^4*c^4*L og[(b*c - a*d)^(2/3) - d^(1/3)*(b*c - a*d)^(1/3)*(a + b*x^3)^(1/3) + d^(2/ 3)*(a + b*x^3)^(2/3)])/(1320*b^4*d^(14/3)*(b*c - a*d)^(1/3))
Time = 0.47 (sec) , antiderivative size = 294, normalized size of antiderivative = 1.01, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {948, 99, 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.
| \(\displaystyle \int \frac {x^{14}}{\sqrt [3]{a+b x^3} \left (c+d x^3\right )} \, dx\) |
| \(\Big \downarrow \) 948 |
| \(\displaystyle \frac {1}{3} \int \frac {x^{12}}{\sqrt [3]{b x^3+a} \left (d x^3+c\right )}dx^3\) |
| \(\Big \downarrow \) 99 |
| \(\displaystyle \frac {1}{3} \int \left (\frac {c^4}{d^4 \sqrt [3]{b x^3+a} \left (d x^3+c\right )}+\frac {\left (b x^3+a\right )^{8/3}}{b^3 d}+\frac {(-b c-3 a d) \left (b x^3+a\right )^{5/3}}{b^3 d^2}+\frac {\left (b^2 c^2+2 a b d c+3 a^2 d^2\right ) \left (b x^3+a\right )^{2/3}}{b^3 d^3}+\frac {(b c+a d) \left (-b^2 c^2-a^2 d^2\right )}{b^3 d^4 \sqrt [3]{b x^3+a}}\right )dx^3\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {1}{3} \left (-\frac {3 \left (a+b x^3\right )^{2/3} (a d+b c) \left (a^2 d^2+b^2 c^2\right )}{2 b^4 d^4}+\frac {3 \left (a+b x^3\right )^{5/3} \left (3 a^2 d^2+2 a b c d+b^2 c^2\right )}{5 b^4 d^3}-\frac {\sqrt {3} c^4 \arctan \left (\frac {1-\frac {2 \sqrt [3]{d} \sqrt [3]{a+b x^3}}{\sqrt [3]{b c-a d}}}{\sqrt {3}}\right )}{d^{14/3} \sqrt [3]{b c-a d}}-\frac {3 \left (a+b x^3\right )^{8/3} (3 a d+b c)}{8 b^4 d^2}+\frac {3 \left (a+b x^3\right )^{11/3}}{11 b^4 d}+\frac {c^4 \log \left (c+d x^3\right )}{2 d^{14/3} \sqrt [3]{b c-a d}}-\frac {3 c^4 \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{2 d^{14/3} \sqrt [3]{b c-a d}}\right )\) |
((-3*(b*c + a*d)*(b^2*c^2 + a^2*d^2)*(a + b*x^3)^(2/3))/(2*b^4*d^4) + (3*( b^2*c^2 + 2*a*b*c*d + 3*a^2*d^2)*(a + b*x^3)^(5/3))/(5*b^4*d^3) - (3*(b*c + 3*a*d)*(a + b*x^3)^(8/3))/(8*b^4*d^2) + (3*(a + b*x^3)^(11/3))/(11*b^4*d ) - (Sqrt[3]*c^4*ArcTan[(1 - (2*d^(1/3)*(a + b*x^3)^(1/3))/(b*c - a*d)^(1/ 3))/Sqrt[3]])/(d^(14/3)*(b*c - a*d)^(1/3)) + (c^4*Log[c + d*x^3])/(2*d^(14 /3)*(b*c - a*d)^(1/3)) - (3*c^4*Log[(b*c - a*d)^(1/3) + d^(1/3)*(a + b*x^3 )^(1/3)])/(2*d^(14/3)*(b*c - a*d)^(1/3)))/3
3.8.14.3.1 Defintions of rubi rules used
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] && (IntegerQ[p] | | (GtQ[m, 0] && GeQ[n, -1]))
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_. ), x_Symbol] :> Simp[1/n Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(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] && IntegerQ[Simplify[(m + 1)/n]]
Time = 4.80 (sec) , antiderivative size = 275, normalized size of antiderivative = 0.95
| method | result | size |
| pseudoelliptic | \(-\frac {\frac {243 \left (\frac {a d -b c}{d}\right )^{\frac {1}{3}} \left (\frac {\left (-40 d^{3} x^{9}+55 c \,d^{2} x^{6}-88 c^{2} d \,x^{3}+220 c^{3}\right ) b^{3}}{81}+\frac {44 d a \left (\frac {15}{44} d^{2} x^{6}-\frac {1}{2} c d \,x^{3}+c^{2}\right ) b^{2}}{27}+\frac {11 \left (-\frac {6 d \,x^{3}}{11}+c \right ) d^{2} a^{2} b}{9}+a^{3} d^{3}\right ) d \left (b \,x^{3}+a \right )^{\frac {2}{3}}}{220}+b^{4} c^{4} \left (-2 \arctan \left (\frac {\sqrt {3}\, \left (2 \left (b \,x^{3}+a \right )^{\frac {1}{3}}+\left (\frac {a d -b c}{d}\right )^{\frac {1}{3}}\right )}{3 \left (\frac {a d -b c}{d}\right )^{\frac {1}{3}}}\right ) \sqrt {3}+\ln \left (\left (b \,x^{3}+a \right )^{\frac {2}{3}}+\left (\frac {a d -b c}{d}\right )^{\frac {1}{3}} \left (b \,x^{3}+a \right )^{\frac {1}{3}}+\left (\frac {a d -b c}{d}\right )^{\frac {2}{3}}\right )-2 \ln \left (\left (b \,x^{3}+a \right )^{\frac {1}{3}}-\left (\frac {a d -b c}{d}\right )^{\frac {1}{3}}\right )\right )}{6 \left (\frac {a d -b c}{d}\right )^{\frac {1}{3}} d^{5} b^{4}}\) | \(275\) |
-1/6/(1/d*(a*d-b*c))^(1/3)*(243/220*(1/d*(a*d-b*c))^(1/3)*(1/81*(-40*d^3*x ^9+55*c*d^2*x^6-88*c^2*d*x^3+220*c^3)*b^3+44/27*d*a*(15/44*d^2*x^6-1/2*c*d *x^3+c^2)*b^2+11/9*(-6/11*d*x^3+c)*d^2*a^2*b+a^3*d^3)*d*(b*x^3+a)^(2/3)+b^ 4*c^4*(-2*arctan(1/3*3^(1/2)*(2*(b*x^3+a)^(1/3)+(1/d*(a*d-b*c))^(1/3))/(1/ d*(a*d-b*c))^(1/3))*3^(1/2)+ln((b*x^3+a)^(2/3)+(1/d*(a*d-b*c))^(1/3)*(b*x^ 3+a)^(1/3)+(1/d*(a*d-b*c))^(2/3))-2*ln((b*x^3+a)^(1/3)-(1/d*(a*d-b*c))^(1/ 3))))/d^5/b^4
Time = 0.30 (sec) , antiderivative size = 1004, normalized size of antiderivative = 3.46 \[ \int \frac {x^{14}}{\sqrt [3]{a+b x^3} \left (c+d x^3\right )} \, dx=\text {Too large to display} \]
[1/1320*(220*(-b*c*d^2 + a*d^3)^(2/3)*b^4*c^4*log((b*x^3 + a)^(2/3)*d^2 + (-b*c*d^2 + a*d^3)^(1/3)*(b*x^3 + a)^(1/3)*d + (-b*c*d^2 + a*d^3)^(2/3)) - 440*(-b*c*d^2 + a*d^3)^(2/3)*b^4*c^4*log((b*x^3 + a)^(1/3)*d - (-b*c*d^2 + a*d^3)^(1/3)) + 660*sqrt(1/3)*(b^5*c^5*d - a*b^4*c^4*d^2)*sqrt((-b*c*d^2 + a*d^3)^(1/3)/(b*c - a*d))*log((2*b*d^2*x^3 - b*c*d + 3*a*d^2 + 3*sqrt(1 /3)*(2*(-b*c*d^2 + a*d^3)^(2/3)*(b*x^3 + a)^(2/3) + (b*x^3 + a)^(1/3)*(b*c *d - a*d^2) + (-b*c*d^2 + a*d^3)^(1/3)*(b*c - a*d))*sqrt((-b*c*d^2 + a*d^3 )^(1/3)/(b*c - a*d)) - 3*(-b*c*d^2 + a*d^3)^(2/3)*(b*x^3 + a)^(1/3))/(d*x^ 3 + c)) - 3*(220*b^4*c^4*d^2 - 88*a*b^3*c^3*d^3 - 33*a^2*b^2*c^2*d^4 - 18* a^3*b*c*d^5 - 81*a^4*d^6 - 40*(b^4*c*d^5 - a*b^3*d^6)*x^9 + 5*(11*b^4*c^2* d^4 - 2*a*b^3*c*d^5 - 9*a^2*b^2*d^6)*x^6 - 2*(44*b^4*c^3*d^3 - 11*a*b^3*c^ 2*d^4 - 6*a^2*b^2*c*d^5 - 27*a^3*b*d^6)*x^3)*(b*x^3 + a)^(2/3))/(b^5*c*d^6 - a*b^4*d^7), 1/1320*(220*(-b*c*d^2 + a*d^3)^(2/3)*b^4*c^4*log((b*x^3 + a )^(2/3)*d^2 + (-b*c*d^2 + a*d^3)^(1/3)*(b*x^3 + a)^(1/3)*d + (-b*c*d^2 + a *d^3)^(2/3)) - 440*(-b*c*d^2 + a*d^3)^(2/3)*b^4*c^4*log((b*x^3 + a)^(1/3)* d - (-b*c*d^2 + a*d^3)^(1/3)) + 1320*sqrt(1/3)*(b^5*c^5*d - a*b^4*c^4*d^2) *sqrt(-(-b*c*d^2 + a*d^3)^(1/3)/(b*c - a*d))*arctan(sqrt(1/3)*(2*(b*x^3 + a)^(1/3)*d + (-b*c*d^2 + a*d^3)^(1/3))*sqrt(-(-b*c*d^2 + a*d^3)^(1/3)/(b*c - a*d))/d) - 3*(220*b^4*c^4*d^2 - 88*a*b^3*c^3*d^3 - 33*a^2*b^2*c^2*d^4 - 18*a^3*b*c*d^5 - 81*a^4*d^6 - 40*(b^4*c*d^5 - a*b^3*d^6)*x^9 + 5*(11*b...
\[ \int \frac {x^{14}}{\sqrt [3]{a+b x^3} \left (c+d x^3\right )} \, dx=\int \frac {x^{14}}{\sqrt [3]{a + b x^{3}} \left (c + d x^{3}\right )}\, dx \]
Exception generated. \[ \int \frac {x^{14}}{\sqrt [3]{a+b x^3} \left (c+d x^3\right )} \, dx=\text {Exception raised: ValueError} \]
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 = 0.31 (sec) , antiderivative size = 454, normalized size of antiderivative = 1.57 \[ \int \frac {x^{14}}{\sqrt [3]{a+b x^3} \left (c+d x^3\right )} \, dx=-\frac {b^{48} c^{4} d^{7} \left (-\frac {b c - a d}{d}\right )^{\frac {2}{3}} \log \left ({\left | {\left (b x^{3} + a\right )}^{\frac {1}{3}} - \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}} \right |}\right )}{3 \, {\left (b^{49} c d^{11} - a b^{48} d^{12}\right )}} - \frac {{\left (-b c d^{2} + a d^{3}\right )}^{\frac {2}{3}} c^{4} \arctan \left (\frac {\sqrt {3} {\left (2 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}} + \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}}\right )}}{3 \, \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}}}\right )}{\sqrt {3} b c d^{6} - \sqrt {3} a d^{7}} + \frac {{\left (-b c d^{2} + a d^{3}\right )}^{\frac {2}{3}} c^{4} \log \left ({\left (b x^{3} + a\right )}^{\frac {2}{3}} + {\left (b x^{3} + a\right )}^{\frac {1}{3}} \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}} + \left (-\frac {b c - a d}{d}\right )^{\frac {2}{3}}\right )}{6 \, {\left (b c d^{6} - a d^{7}\right )}} - \frac {220 \, {\left (b x^{3} + a\right )}^{\frac {2}{3}} b^{43} c^{3} d^{7} - 88 \, {\left (b x^{3} + a\right )}^{\frac {5}{3}} b^{42} c^{2} d^{8} + 220 \, {\left (b x^{3} + a\right )}^{\frac {2}{3}} a b^{42} c^{2} d^{8} + 55 \, {\left (b x^{3} + a\right )}^{\frac {8}{3}} b^{41} c d^{9} - 176 \, {\left (b x^{3} + a\right )}^{\frac {5}{3}} a b^{41} c d^{9} + 220 \, {\left (b x^{3} + a\right )}^{\frac {2}{3}} a^{2} b^{41} c d^{9} - 40 \, {\left (b x^{3} + a\right )}^{\frac {11}{3}} b^{40} d^{10} + 165 \, {\left (b x^{3} + a\right )}^{\frac {8}{3}} a b^{40} d^{10} - 264 \, {\left (b x^{3} + a\right )}^{\frac {5}{3}} a^{2} b^{40} d^{10} + 220 \, {\left (b x^{3} + a\right )}^{\frac {2}{3}} a^{3} b^{40} d^{10}}{440 \, b^{44} d^{11}} \]
-1/3*b^48*c^4*d^7*(-(b*c - a*d)/d)^(2/3)*log(abs((b*x^3 + a)^(1/3) - (-(b* c - a*d)/d)^(1/3)))/(b^49*c*d^11 - a*b^48*d^12) - (-b*c*d^2 + a*d^3)^(2/3) *c^4*arctan(1/3*sqrt(3)*(2*(b*x^3 + a)^(1/3) + (-(b*c - a*d)/d)^(1/3))/(-( b*c - a*d)/d)^(1/3))/(sqrt(3)*b*c*d^6 - sqrt(3)*a*d^7) + 1/6*(-b*c*d^2 + a *d^3)^(2/3)*c^4*log((b*x^3 + a)^(2/3) + (b*x^3 + a)^(1/3)*(-(b*c - a*d)/d) ^(1/3) + (-(b*c - a*d)/d)^(2/3))/(b*c*d^6 - a*d^7) - 1/440*(220*(b*x^3 + a )^(2/3)*b^43*c^3*d^7 - 88*(b*x^3 + a)^(5/3)*b^42*c^2*d^8 + 220*(b*x^3 + a) ^(2/3)*a*b^42*c^2*d^8 + 55*(b*x^3 + a)^(8/3)*b^41*c*d^9 - 176*(b*x^3 + a)^ (5/3)*a*b^41*c*d^9 + 220*(b*x^3 + a)^(2/3)*a^2*b^41*c*d^9 - 40*(b*x^3 + a) ^(11/3)*b^40*d^10 + 165*(b*x^3 + a)^(8/3)*a*b^40*d^10 - 264*(b*x^3 + a)^(5 /3)*a^2*b^40*d^10 + 220*(b*x^3 + a)^(2/3)*a^3*b^40*d^10)/(b^44*d^11)
Time = 8.95 (sec) , antiderivative size = 438, normalized size of antiderivative = 1.51 \[ \int \frac {x^{14}}{\sqrt [3]{a+b x^3} \left (c+d x^3\right )} \, dx=\left (\frac {6\,a^2}{5\,b^4\,d}+\frac {\left (\frac {4\,a}{b^4\,d}+\frac {b^5\,c-a\,b^4\,d}{b^8\,d^2}\right )\,\left (b^5\,c-a\,b^4\,d\right )}{5\,b^4\,d}\right )\,{\left (b\,x^3+a\right )}^{5/3}-\left (\frac {a}{2\,b^4\,d}+\frac {b^5\,c-a\,b^4\,d}{8\,b^8\,d^2}\right )\,{\left (b\,x^3+a\right )}^{8/3}-{\left (b\,x^3+a\right )}^{2/3}\,\left (\frac {2\,a^3}{b^4\,d}+\frac {\left (\frac {6\,a^2}{b^4\,d}+\frac {\left (\frac {4\,a}{b^4\,d}+\frac {b^5\,c-a\,b^4\,d}{b^8\,d^2}\right )\,\left (b^5\,c-a\,b^4\,d\right )}{b^4\,d}\right )\,\left (b^5\,c-a\,b^4\,d\right )}{2\,b^4\,d}\right )+\frac {{\left (b\,x^3+a\right )}^{11/3}}{11\,b^4\,d}+\frac {c^4\,\ln \left (\frac {c^8\,{\left (b\,x^3+a\right )}^{1/3}}{d^7}-\frac {c^8\,{\left (a\,d-b\,c\right )}^{1/3}}{d^{22/3}}\right )}{3\,d^{14/3}\,{\left (a\,d-b\,c\right )}^{1/3}}-\frac {\ln \left (\frac {c^8\,{\left (b\,x^3+a\right )}^{1/3}}{d^7}-\frac {c^8\,{\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}^2\,{\left (a\,d-b\,c\right )}^{1/3}}{4\,d^{22/3}}\right )\,\left (c^4+\sqrt {3}\,c^4\,1{}\mathrm {i}\right )}{6\,d^{14/3}\,{\left (a\,d-b\,c\right )}^{1/3}}+\frac {c^4\,\ln \left (\frac {c^8\,{\left (b\,x^3+a\right )}^{1/3}}{d^7}-\frac {c^8\,{\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}^2\,{\left (a\,d-b\,c\right )}^{1/3}}{4\,d^{22/3}}\right )\,\left (-\frac {1}{6}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right )}{d^{14/3}\,{\left (a\,d-b\,c\right )}^{1/3}} \]
((6*a^2)/(5*b^4*d) + (((4*a)/(b^4*d) + (b^5*c - a*b^4*d)/(b^8*d^2))*(b^5*c - a*b^4*d))/(5*b^4*d))*(a + b*x^3)^(5/3) - (a/(2*b^4*d) + (b^5*c - a*b^4* d)/(8*b^8*d^2))*(a + b*x^3)^(8/3) - (a + b*x^3)^(2/3)*((2*a^3)/(b^4*d) + ( ((6*a^2)/(b^4*d) + (((4*a)/(b^4*d) + (b^5*c - a*b^4*d)/(b^8*d^2))*(b^5*c - a*b^4*d))/(b^4*d))*(b^5*c - a*b^4*d))/(2*b^4*d)) + (a + b*x^3)^(11/3)/(11 *b^4*d) + (c^4*log((c^8*(a + b*x^3)^(1/3))/d^7 - (c^8*(a*d - b*c)^(1/3))/d ^(22/3)))/(3*d^(14/3)*(a*d - b*c)^(1/3)) - (log((c^8*(a + b*x^3)^(1/3))/d^ 7 - (c^8*(3^(1/2)*1i + 1)^2*(a*d - b*c)^(1/3))/(4*d^(22/3)))*(3^(1/2)*c^4* 1i + c^4))/(6*d^(14/3)*(a*d - b*c)^(1/3)) + (c^4*log((c^8*(a + b*x^3)^(1/3 ))/d^7 - (c^8*(3^(1/2)*1i - 1)^2*(a*d - b*c)^(1/3))/(4*d^(22/3)))*((3^(1/2 )*1i)/6 - 1/6))/(d^(14/3)*(a*d - b*c)^(1/3))