Integrand size = 25, antiderivative size = 357 \[ \int \frac {A+B \tan (c+d x)}{\sqrt [3]{a+b \tan (c+d x)}} \, dx=-\frac {(A-i B) x}{4 \sqrt [3]{a-i b}}-\frac {(A+i B) x}{4 \sqrt [3]{a+i b}}+\frac {\sqrt {3} (i A+B) \arctan \left (\frac {1+\frac {2 \sqrt [3]{a+b \tan (c+d x)}}{\sqrt [3]{a-i b}}}{\sqrt {3}}\right )}{2 \sqrt [3]{a-i b} d}-\frac {\sqrt {3} (i A-B) \arctan \left (\frac {1+\frac {2 \sqrt [3]{a+b \tan (c+d x)}}{\sqrt [3]{a+i b}}}{\sqrt {3}}\right )}{2 \sqrt [3]{a+i b} d}-\frac {(i A-B) \log (\cos (c+d x))}{4 \sqrt [3]{a+i b} d}+\frac {(i A+B) \log (\cos (c+d x))}{4 \sqrt [3]{a-i b} d}+\frac {3 (i A+B) \log \left (\sqrt [3]{a-i b}-\sqrt [3]{a+b \tan (c+d x)}\right )}{4 \sqrt [3]{a-i b} d}-\frac {3 (i A-B) \log \left (\sqrt [3]{a+i b}-\sqrt [3]{a+b \tan (c+d x)}\right )}{4 \sqrt [3]{a+i b} d} \]
-1/4*(A-I*B)*x/(a-I*b)^(1/3)-1/4*(A+I*B)*x/(a+I*b)^(1/3)-1/4*(I*A-B)*ln(co s(d*x+c))/(a+I*b)^(1/3)/d+1/4*(I*A+B)*ln(cos(d*x+c))/(a-I*b)^(1/3)/d+3/4*( I*A+B)*ln((a-I*b)^(1/3)-(a+b*tan(d*x+c))^(1/3))/(a-I*b)^(1/3)/d-3/4*(I*A-B )*ln((a+I*b)^(1/3)-(a+b*tan(d*x+c))^(1/3))/(a+I*b)^(1/3)/d+1/2*(I*A+B)*arc tan(1/3*(1+2*(a+b*tan(d*x+c))^(1/3)/(a-I*b)^(1/3))*3^(1/2))*3^(1/2)/(a-I*b )^(1/3)/d-1/2*(I*A-B)*arctan(1/3*(1+2*(a+b*tan(d*x+c))^(1/3)/(a+I*b)^(1/3) )*3^(1/2))*3^(1/2)/(a+I*b)^(1/3)/d
Time = 0.46 (sec) , antiderivative size = 227, normalized size of antiderivative = 0.64 \[ \int \frac {A+B \tan (c+d x)}{\sqrt [3]{a+b \tan (c+d x)}} \, dx=\frac {i \left (\frac {(A-i B) \left (2 \sqrt {3} \arctan \left (\frac {1+\frac {2 \sqrt [3]{a+b \tan (c+d x)}}{\sqrt [3]{a-i b}}}{\sqrt {3}}\right )-\log (i+\tan (c+d x))+3 \log \left (\sqrt [3]{a-i b}-\sqrt [3]{a+b \tan (c+d x)}\right )\right )}{\sqrt [3]{a-i b}}-\frac {(A+i B) \left (2 \sqrt {3} \arctan \left (\frac {1+\frac {2 \sqrt [3]{a+b \tan (c+d x)}}{\sqrt [3]{a+i b}}}{\sqrt {3}}\right )-\log (i-\tan (c+d x))+3 \log \left (\sqrt [3]{a+i b}-\sqrt [3]{a+b \tan (c+d x)}\right )\right )}{\sqrt [3]{a+i b}}\right )}{4 d} \]
((I/4)*(((A - I*B)*(2*Sqrt[3]*ArcTan[(1 + (2*(a + b*Tan[c + d*x])^(1/3))/( a - I*b)^(1/3))/Sqrt[3]] - Log[I + Tan[c + d*x]] + 3*Log[(a - I*b)^(1/3) - (a + b*Tan[c + d*x])^(1/3)]))/(a - I*b)^(1/3) - ((A + I*B)*(2*Sqrt[3]*Arc Tan[(1 + (2*(a + b*Tan[c + d*x])^(1/3))/(a + I*b)^(1/3))/Sqrt[3]] - Log[I - Tan[c + d*x]] + 3*Log[(a + I*b)^(1/3) - (a + b*Tan[c + d*x])^(1/3)]))/(a + I*b)^(1/3)))/d
Time = 0.54 (sec) , antiderivative size = 233, normalized size of antiderivative = 0.65, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.360, Rules used = {3042, 4022, 3042, 4020, 25, 67, 16, 1082, 217}
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.
| \(\displaystyle \int \frac {A+B \tan (c+d x)}{\sqrt [3]{a+b \tan (c+d x)}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {A+B \tan (c+d x)}{\sqrt [3]{a+b \tan (c+d x)}}dx\) |
| \(\Big \downarrow \) 4022 |
| \(\displaystyle \frac {1}{2} (A+i B) \int \frac {1-i \tan (c+d x)}{\sqrt [3]{a+b \tan (c+d x)}}dx+\frac {1}{2} (A-i B) \int \frac {i \tan (c+d x)+1}{\sqrt [3]{a+b \tan (c+d x)}}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{2} (A+i B) \int \frac {1-i \tan (c+d x)}{\sqrt [3]{a+b \tan (c+d x)}}dx+\frac {1}{2} (A-i B) \int \frac {i \tan (c+d x)+1}{\sqrt [3]{a+b \tan (c+d x)}}dx\) |
| \(\Big \downarrow \) 4020 |
| \(\displaystyle \frac {i (A-i B) \int -\frac {1}{(1-i \tan (c+d x)) \sqrt [3]{a+b \tan (c+d x)}}d(i \tan (c+d x))}{2 d}-\frac {i (A+i B) \int -\frac {1}{(i \tan (c+d x)+1) \sqrt [3]{a+b \tan (c+d x)}}d(-i \tan (c+d x))}{2 d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {i (A+i B) \int \frac {1}{(i \tan (c+d x)+1) \sqrt [3]{a+b \tan (c+d x)}}d(-i \tan (c+d x))}{2 d}-\frac {i (A-i B) \int \frac {1}{(1-i \tan (c+d x)) \sqrt [3]{a+b \tan (c+d x)}}d(i \tan (c+d x))}{2 d}\) |
| \(\Big \downarrow \) 67 |
| \(\displaystyle \frac {i (A-i B) \left (\frac {3}{2} \int \frac {1}{-\tan ^2(c+d x)+(a-i b)^{2/3}+\sqrt [3]{a-i b} \sqrt [3]{a+b \tan (c+d x)}}d\sqrt [3]{a+b \tan (c+d x)}-\frac {3 \int \frac {1}{\sqrt [3]{a-i b}-i \tan (c+d x)}d\sqrt [3]{a+b \tan (c+d x)}}{2 \sqrt [3]{a-i b}}-\frac {\log (1-i \tan (c+d x))}{2 \sqrt [3]{a-i b}}\right )}{2 d}-\frac {i (A+i B) \left (\frac {3}{2} \int \frac {1}{-\tan ^2(c+d x)+(a+i b)^{2/3}+\sqrt [3]{a+i b} \sqrt [3]{a+b \tan (c+d x)}}d\sqrt [3]{a+b \tan (c+d x)}-\frac {3 \int \frac {1}{i \tan (c+d x)+\sqrt [3]{a+i b}}d\sqrt [3]{a+b \tan (c+d x)}}{2 \sqrt [3]{a+i b}}-\frac {\log (1+i \tan (c+d x))}{2 \sqrt [3]{a+i b}}\right )}{2 d}\) |
| \(\Big \downarrow \) 16 |
| \(\displaystyle \frac {i (A-i B) \left (\frac {3}{2} \int \frac {1}{-\tan ^2(c+d x)+(a-i b)^{2/3}+\sqrt [3]{a-i b} \sqrt [3]{a+b \tan (c+d x)}}d\sqrt [3]{a+b \tan (c+d x)}-\frac {\log (1-i \tan (c+d x))}{2 \sqrt [3]{a-i b}}+\frac {3 \log \left (\sqrt [3]{a-i b}-i \tan (c+d x)\right )}{2 \sqrt [3]{a-i b}}\right )}{2 d}-\frac {i (A+i B) \left (\frac {3}{2} \int \frac {1}{-\tan ^2(c+d x)+(a+i b)^{2/3}+\sqrt [3]{a+i b} \sqrt [3]{a+b \tan (c+d x)}}d\sqrt [3]{a+b \tan (c+d x)}-\frac {\log (1+i \tan (c+d x))}{2 \sqrt [3]{a+i b}}+\frac {3 \log \left (\sqrt [3]{a+i b}+i \tan (c+d x)\right )}{2 \sqrt [3]{a+i b}}\right )}{2 d}\) |
| \(\Big \downarrow \) 1082 |
| \(\displaystyle \frac {i (A-i B) \left (-\frac {3 \int \frac {1}{\tan ^2(c+d x)-3}d\left (\frac {2 i \tan (c+d x)}{\sqrt [3]{a-i b}}+1\right )}{\sqrt [3]{a-i b}}-\frac {\log (1-i \tan (c+d x))}{2 \sqrt [3]{a-i b}}+\frac {3 \log \left (\sqrt [3]{a-i b}-i \tan (c+d x)\right )}{2 \sqrt [3]{a-i b}}\right )}{2 d}-\frac {i (A+i B) \left (-\frac {3 \int \frac {1}{\tan ^2(c+d x)-3}d\left (1-\frac {2 i \tan (c+d x)}{\sqrt [3]{a+i b}}\right )}{\sqrt [3]{a+i b}}-\frac {\log (1+i \tan (c+d x))}{2 \sqrt [3]{a+i b}}+\frac {3 \log \left (\sqrt [3]{a+i b}+i \tan (c+d x)\right )}{2 \sqrt [3]{a+i b}}\right )}{2 d}\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle \frac {i (A-i B) \left (\frac {i \sqrt {3} \text {arctanh}\left (\frac {\tan (c+d x)}{\sqrt {3}}\right )}{\sqrt [3]{a-i b}}-\frac {\log (1-i \tan (c+d x))}{2 \sqrt [3]{a-i b}}+\frac {3 \log \left (\sqrt [3]{a-i b}-i \tan (c+d x)\right )}{2 \sqrt [3]{a-i b}}\right )}{2 d}-\frac {i (A+i B) \left (-\frac {i \sqrt {3} \text {arctanh}\left (\frac {\tan (c+d x)}{\sqrt {3}}\right )}{\sqrt [3]{a+i b}}-\frac {\log (1+i \tan (c+d x))}{2 \sqrt [3]{a+i b}}+\frac {3 \log \left (\sqrt [3]{a+i b}+i \tan (c+d x)\right )}{2 \sqrt [3]{a+i b}}\right )}{2 d}\) |
((I/2)*(A - I*B)*((I*Sqrt[3]*ArcTanh[Tan[c + d*x]/Sqrt[3]])/(a - I*b)^(1/3 ) - Log[1 - I*Tan[c + d*x]]/(2*(a - I*b)^(1/3)) + (3*Log[(a - I*b)^(1/3) - I*Tan[c + d*x]])/(2*(a - I*b)^(1/3))))/d - ((I/2)*(A + I*B)*(((-I)*Sqrt[3 ]*ArcTanh[Tan[c + d*x]/Sqrt[3]])/(a + I*b)^(1/3) - Log[1 + I*Tan[c + d*x]] /(2*(a + I*b)^(1/3)) + (3*Log[(a + I*b)^(1/3) + I*Tan[c + d*x]])/(2*(a + I *b)^(1/3))))/d
3.5.75.3.1 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[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(1/3)), x_Symbol] :> With[ {q = Rt[(b*c - a*d)/b, 3]}, Simp[-Log[RemoveContent[a + b*x, x]]/(2*b*q), x ] + (Simp[3/(2*b) Subst[Int[1/(q^2 + q*x + x^2), x], x, (c + d*x)^(1/3)], x] - Simp[3/(2*b*q) Subst[Int[1/(q - x), x], x, (c + d*x)^(1/3)], x])] / ; FreeQ[{a, b, c, d}, x] && PosQ[(b*c - a*d)/b]
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_) + (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[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[c*(d/f) Subst[Int[(a + (b/d)*x)^m/(d^2 + c*x), x], x, d*Tan[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[ b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && EqQ[c^2 + d^2, 0]
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(c + I*d)/2 Int[(a + b*Tan[e + f*x])^m*( 1 - I*Tan[e + f*x]), x], x] + Simp[(c - I*d)/2 Int[(a + b*Tan[e + f*x])^m *(1 + I*Tan[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && !IntegerQ[m]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.43 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.20
| method | result | size |
| derivativedivides | \(\frac {\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{6}-2 a \,\textit {\_Z}^{3}+a^{2}+b^{2}\right )}{\sum }\frac {\left (B \,\textit {\_R}^{4}+\left (A b -B a \right ) \textit {\_R} \right ) \ln \left (\left (a +b \tan \left (d x +c \right )\right )^{\frac {1}{3}}-\textit {\_R} \right )}{\textit {\_R}^{5}-\textit {\_R}^{2} a}}{2 d}\) | \(72\) |
| default | \(\frac {\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{6}-2 a \,\textit {\_Z}^{3}+a^{2}+b^{2}\right )}{\sum }\frac {\left (B \,\textit {\_R}^{4}+\left (A b -B a \right ) \textit {\_R} \right ) \ln \left (\left (a +b \tan \left (d x +c \right )\right )^{\frac {1}{3}}-\textit {\_R} \right )}{\textit {\_R}^{5}-\textit {\_R}^{2} a}}{2 d}\) | \(72\) |
| parts | \(\frac {A b \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{6}-2 a \,\textit {\_Z}^{3}+a^{2}+b^{2}\right )}{\sum }\frac {\ln \left (\left (a +b \tan \left (d x +c \right )\right )^{\frac {1}{3}}-\textit {\_R} \right )}{\textit {\_R} \left (\textit {\_R}^{3}-a \right )}\right )}{2 d}+\frac {B \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{6}-2 a \,\textit {\_Z}^{3}+a^{2}+b^{2}\right )}{\sum }\frac {\ln \left (\left (a +b \tan \left (d x +c \right )\right )^{\frac {1}{3}}-\textit {\_R} \right )}{\textit {\_R}}\right )}{2 d}\) | \(106\) |
1/2/d*sum((B*_R^4+(A*b-B*a)*_R)/(_R^5-_R^2*a)*ln((a+b*tan(d*x+c))^(1/3)-_R ),_R=RootOf(_Z^6-2*_Z^3*a+a^2+b^2))
Leaf count of result is larger than twice the leaf count of optimal. 4121 vs. \(2 (263) = 526\).
Time = 0.44 (sec) , antiderivative size = 4121, normalized size of antiderivative = 11.54 \[ \int \frac {A+B \tan (c+d x)}{\sqrt [3]{a+b \tan (c+d x)}} \, dx=\text {Too large to display} \]
1/4*(sqrt(-3) - 1)*(-((a^2 + b^2)*d^3*sqrt(-((A^6 - 6*A^4*B^2 + 9*A^2*B^4) *a^2 + 2*(3*A^5*B - 10*A^3*B^3 + 3*A*B^5)*a*b + (9*A^4*B^2 - 6*A^2*B^4 + B ^6)*b^2)/((a^4 + 2*a^2*b^2 + b^4)*d^6)) + (3*A^2*B - B^3)*a - (A^3 - 3*A*B ^2)*b)/((a^2 + b^2)*d^3))^(1/3)*log(1/2*(sqrt(-3)*((A^5 - 4*A^3*B^2 + 3*A* B^4)*a^2 + (5*A^4*B - 10*A^2*B^3 + B^5)*a*b + 2*(3*A^3*B^2 - A*B^4)*b^2)*d ^2 + ((A^5 - 4*A^3*B^2 + 3*A*B^4)*a^2 + (5*A^4*B - 10*A^2*B^3 + B^5)*a*b + 2*(3*A^3*B^2 - A*B^4)*b^2)*d^2 + (sqrt(-3)*(2*A*B*a^3 + 2*A*B*a*b^2 - (A^ 2 - B^2)*a^2*b - (A^2 - B^2)*b^3)*d^5 + (2*A*B*a^3 + 2*A*B*a*b^2 - (A^2 - B^2)*a^2*b - (A^2 - B^2)*b^3)*d^5)*sqrt(-((A^6 - 6*A^4*B^2 + 9*A^2*B^4)*a^ 2 + 2*(3*A^5*B - 10*A^3*B^3 + 3*A*B^5)*a*b + (9*A^4*B^2 - 6*A^2*B^4 + B^6) *b^2)/((a^4 + 2*a^2*b^2 + b^4)*d^6)))*(-((a^2 + b^2)*d^3*sqrt(-((A^6 - 6*A ^4*B^2 + 9*A^2*B^4)*a^2 + 2*(3*A^5*B - 10*A^3*B^3 + 3*A*B^5)*a*b + (9*A^4* B^2 - 6*A^2*B^4 + B^6)*b^2)/((a^4 + 2*a^2*b^2 + b^4)*d^6)) + (3*A^2*B - B^ 3)*a - (A^3 - 3*A*B^2)*b)/((a^2 + b^2)*d^3))^(2/3) - ((A^7 - A^5*B^2 - 5*A ^3*B^4 - 3*A*B^6)*a + (3*A^6*B + 5*A^4*B^3 + A^2*B^5 - B^7)*b)*(b*tan(d*x + c) + a)^(1/3)) - 1/4*(sqrt(-3) + 1)*(-((a^2 + b^2)*d^3*sqrt(-((A^6 - 6*A ^4*B^2 + 9*A^2*B^4)*a^2 + 2*(3*A^5*B - 10*A^3*B^3 + 3*A*B^5)*a*b + (9*A^4* B^2 - 6*A^2*B^4 + B^6)*b^2)/((a^4 + 2*a^2*b^2 + b^4)*d^6)) + (3*A^2*B - B^ 3)*a - (A^3 - 3*A*B^2)*b)/((a^2 + b^2)*d^3))^(1/3)*log(-1/2*(sqrt(-3)*((A^ 5 - 4*A^3*B^2 + 3*A*B^4)*a^2 + (5*A^4*B - 10*A^2*B^3 + B^5)*a*b + 2*(3*...
\[ \int \frac {A+B \tan (c+d x)}{\sqrt [3]{a+b \tan (c+d x)}} \, dx=\int \frac {A + B \tan {\left (c + d x \right )}}{\sqrt [3]{a + b \tan {\left (c + d x \right )}}}\, dx \]
\[ \int \frac {A+B \tan (c+d x)}{\sqrt [3]{a+b \tan (c+d x)}} \, dx=\int { \frac {B \tan \left (d x + c\right ) + A}{{\left (b \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}}} \,d x } \]
Time = 3.43 (sec) , antiderivative size = 147, normalized size of antiderivative = 0.41 \[ \int \frac {A+B \tan (c+d x)}{\sqrt [3]{a+b \tan (c+d x)}} \, dx=\frac {\left (\frac {A^{3} - 3 i \, A^{2} B - 3 \, A B^{2} + i \, B^{3}}{8 i \, a + 8 \, b}\right )^{\frac {1}{3}} \log \left (-a + i \, b - {\left (-a^{2} + 2 i \, a b + b^{2}\right )}^{\frac {1}{3}} {\left (b \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}}\right ) + \left (-\frac {A^{3} + 3 i \, A^{2} B - 3 \, A B^{2} - i \, B^{3}}{8 i \, a - 8 \, b}\right )^{\frac {1}{3}} \log \left (-a - i \, b + {\left (a^{2} + 2 i \, a b - b^{2}\right )}^{\frac {1}{3}} {\left (b \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}}\right )}{d} \]
(((A^3 - 3*I*A^2*B - 3*A*B^2 + I*B^3)/(8*I*a + 8*b))^(1/3)*log(-a + I*b - (-a^2 + 2*I*a*b + b^2)^(1/3)*(b*tan(d*x + c) + a)^(1/3)) + (-(A^3 + 3*I*A^ 2*B - 3*A*B^2 - I*B^3)/(8*I*a - 8*b))^(1/3)*log(-a - I*b + (a^2 + 2*I*a*b - b^2)^(1/3)*(b*tan(d*x + c) + a)^(1/3)))/d
Time = 20.96 (sec) , antiderivative size = 3228, normalized size of antiderivative = 9.04 \[ \int \frac {A+B \tan (c+d x)}{\sqrt [3]{a+b \tan (c+d x)}} \, dx=\text {Too large to display} \]
log(((((-B^6*b^2*d^6)^(1/2) + B^3*a*d^3)/(d^6*(a^2 + b^2)))^(2/3)*(972*a*b ^4*(-B^6*b^2*d^6)^(1/2) - 972*B^3*b^6*d^3 + 972*B^2*b^6*d^4*(a + b*tan(c + d*x))^(1/3)*(((-B^6*b^2*d^6)^(1/2) + B^3*a*d^3)/(d^6*(a^2 + b^2)))^(1/3) - 972*B^2*a^2*b^4*d^4*(a + b*tan(c + d*x))^(1/3)*(((-B^6*b^2*d^6)^(1/2) + B^3*a*d^3)/(d^6*(a^2 + b^2)))^(1/3)))/(4*d^6) + (243*B^5*a*b^4*(a + b*tan( c + d*x))^(1/3))/d^5)*((-64*B^6*b^2*d^6)^(1/2)/(64*(a^2*d^6 + b^2*d^6)) + (B^3*a*d^3)/(8*(a^2*d^6 + b^2*d^6)))^(1/3) + log(((1944*a*b^4*(a^2 + b^2)* (((-A^6*a^2*d^6)^(1/2) + A^3*b*d^3)/(d^6*(a^2 + b^2)))^(2/3) + (1944*A^2*b ^4*(a^2 - b^2)*(a + b*tan(c + d*x))^(1/3))/d^2)*((-A^6*a^2*d^6)^(1/2) + A^ 3*b*d^3))/(8*d^6*(a^2 + b^2)) + (243*A^5*b^5*(a + b*tan(c + d*x))^(1/3))/d ^5)*((-64*A^6*a^2*d^6)^(1/2)/(64*(a^2*d^6 + b^2*d^6)) + (A^3*b*d^3)/(8*(a^ 2*d^6 + b^2*d^6)))^(1/3) + log((243*B^5*a*b^4*(a + b*tan(c + d*x))^(1/3))/ d^5 - ((-(8*(-B^6*b^2*d^6)^(1/2) - 8*B^3*a*d^3)/(d^6*(a^2 + b^2)))^(2/3)*( 972*a*b^4*(-B^6*b^2*d^6)^(1/2) + 972*B^3*b^6*d^3 - 486*B^2*b^6*d^4*(a + b* tan(c + d*x))^(1/3)*(-(8*(-B^6*b^2*d^6)^(1/2) - 8*B^3*a*d^3)/(d^6*(a^2 + b ^2)))^(1/3) + 486*B^2*a^2*b^4*d^4*(a + b*tan(c + d*x))^(1/3)*(-(8*(-B^6*b^ 2*d^6)^(1/2) - 8*B^3*a*d^3)/(d^6*(a^2 + b^2)))^(1/3)))/(16*d^6))*((B^3*a*d ^3)/(8*(a^2*d^6 + b^2*d^6)) - (-64*B^6*b^2*d^6)^(1/2)/(64*(a^2*d^6 + b^2*d ^6)))^(1/3) + log((243*A^5*b^5*(a + b*tan(c + d*x))^(1/3))/d^5 - ((486*a*b ^4*(a^2 + b^2)*(-(8*(-A^6*a^2*d^6)^(1/2) - 8*A^3*b*d^3)/(d^6*(a^2 + b^2...