\(\int (a+b \tan (c+d x))^{2/3} (A+B \tan (c+d x)) \, dx\) [473]

Optimal result

Integrand size = 25, antiderivative size = 379 \[ \int (a+b \tan (c+d x))^{2/3} (A+B \tan (c+d x)) \, dx=-\frac {1}{4} (a-i b)^{2/3} (A-i B) x-\frac {1}{4} (a+i b)^{2/3} (A+i B) x+\frac {\sqrt {3} (a-i b)^{2/3} (i A+B) \arctan \left (\frac {1+\frac {2 \sqrt [3]{a+b \tan (c+d x)}}{\sqrt [3]{a-i b}}}{\sqrt {3}}\right )}{2 d}-\frac {\sqrt {3} (a+i b)^{2/3} (i A-B) \arctan \left (\frac {1+\frac {2 \sqrt [3]{a+b \tan (c+d x)}}{\sqrt [3]{a+i b}}}{\sqrt {3}}\right )}{2 d}-\frac {(a+i b)^{2/3} (i A-B) \log (\cos (c+d x))}{4 d}+\frac {(a-i b)^{2/3} (i A+B) \log (\cos (c+d x))}{4 d}+\frac {3 (a-i b)^{2/3} (i A+B) \log \left (\sqrt [3]{a-i b}-\sqrt [3]{a+b \tan (c+d x)}\right )}{4 d}-\frac {3 (a+i b)^{2/3} (i A-B) \log \left (\sqrt [3]{a+i b}-\sqrt [3]{a+b \tan (c+d x)}\right )}{4 d}+\frac {3 B (a+b \tan (c+d x))^{2/3}}{2 d} \]

[Out]

-1/4*(a-I*b)^(2/3)*(A-I*B)*x-1/4*(a+I*b)^(2/3)*(A+I*B)*x-1/4*(a+I*b)^(2/3)*(I*A-B)*ln(cos(d*x+c))/d+1/4*(a-I*b
)^(2/3)*(I*A+B)*ln(cos(d*x+c))/d+3/4*(a-I*b)^(2/3)*(I*A+B)*ln((a-I*b)^(1/3)-(a+b*tan(d*x+c))^(1/3))/d-3/4*(a+I
*b)^(2/3)*(I*A-B)*ln((a+I*b)^(1/3)-(a+b*tan(d*x+c))^(1/3))/d+1/2*(a-I*b)^(2/3)*(I*A+B)*arctan(1/3*(1+2*(a+b*ta
n(d*x+c))^(1/3)/(a-I*b)^(1/3))*3^(1/2))*3^(1/2)/d-1/2*(a+I*b)^(2/3)*(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)/d+3/2*B*(a+b*tan(d*x+c))^(2/3)/d

Rubi [A] (verified)

Time = 0.57 (sec) , antiderivative size = 379, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {3609, 3620, 3618, 57, 631, 210, 31} \[ \int (a+b \tan (c+d x))^{2/3} (A+B \tan (c+d x)) \, dx=\frac {\sqrt {3} (a-i b)^{2/3} (B+i A) \arctan \left (\frac {1+\frac {2 \sqrt [3]{a+b \tan (c+d x)}}{\sqrt [3]{a-i b}}}{\sqrt {3}}\right )}{2 d}-\frac {\sqrt {3} (a+i b)^{2/3} (-B+i A) \arctan \left (\frac {1+\frac {2 \sqrt [3]{a+b \tan (c+d x)}}{\sqrt [3]{a+i b}}}{\sqrt {3}}\right )}{2 d}+\frac {3 (a-i b)^{2/3} (B+i A) \log \left (-\sqrt [3]{a+b \tan (c+d x)}+\sqrt [3]{a-i b}\right )}{4 d}-\frac {3 (a+i b)^{2/3} (-B+i A) \log \left (-\sqrt [3]{a+b \tan (c+d x)}+\sqrt [3]{a+i b}\right )}{4 d}-\frac {(a+i b)^{2/3} (-B+i A) \log (\cos (c+d x))}{4 d}+\frac {(a-i b)^{2/3} (B+i A) \log (\cos (c+d x))}{4 d}-\frac {1}{4} x (a-i b)^{2/3} (A-i B)-\frac {1}{4} x (a+i b)^{2/3} (A+i B)+\frac {3 B (a+b \tan (c+d x))^{2/3}}{2 d} \]

[In]

Int[(a + b*Tan[c + d*x])^(2/3)*(A + B*Tan[c + d*x]),x]

[Out]

-1/4*((a - I*b)^(2/3)*(A - I*B)*x) - ((a + I*b)^(2/3)*(A + I*B)*x)/4 + (Sqrt[3]*(a - I*b)^(2/3)*(I*A + B)*ArcT
an[(1 + (2*(a + b*Tan[c + d*x])^(1/3))/(a - I*b)^(1/3))/Sqrt[3]])/(2*d) - (Sqrt[3]*(a + I*b)^(2/3)*(I*A - B)*A
rcTan[(1 + (2*(a + b*Tan[c + d*x])^(1/3))/(a + I*b)^(1/3))/Sqrt[3]])/(2*d) - ((a + I*b)^(2/3)*(I*A - B)*Log[Co
s[c + d*x]])/(4*d) + ((a - I*b)^(2/3)*(I*A + B)*Log[Cos[c + d*x]])/(4*d) + (3*(a - I*b)^(2/3)*(I*A + B)*Log[(a
- I*b)^(1/3) - (a + b*Tan[c + d*x])^(1/3)])/(4*d) - (3*(a + I*b)^(2/3)*(I*A - B)*Log[(a + I*b)^(1/3) - (a + b
*Tan[c + d*x])^(1/3)])/(4*d) + (3*B*(a + b*Tan[c + d*x])^(2/3))/(2*d)

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 57

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(1/3)), x_Symbol] :> With[{q = Rt[(b*c - a*d)/b, 3]}, Simp[-L
og[RemoveContent[a + b*x, x]]/(2*b*q), x] + (Dist[3/(2*b), Subst[Int[1/(q^2 + q*x + x^2), x], x, (c + d*x)^(1/
3)], x] - Dist[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]

Rule 210

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])

Rule 631

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[a*(c/b^2)]}, Dist[-2/b, Sub
st[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])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 3609

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[d*
((a + b*Tan[e + f*x])^m/(f*m)), x] + Int[(a + b*Tan[e + f*x])^(m - 1)*Simp[a*c - b*d + (b*c + a*d)*Tan[e + f*x
], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && GtQ[m, 0]

Rule 3618

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[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]

Rule 3620

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[(c
+ I*d)/2, Int[(a + b*Tan[e + f*x])^m*(1 - I*Tan[e + f*x]), x], x] + Dist[(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]

Rubi steps \begin{align*} \text {integral}& = \frac {3 B (a+b \tan (c+d x))^{2/3}}{2 d}+\int \frac {a A-b B+(A b+a B) \tan (c+d x)}{\sqrt [3]{a+b \tan (c+d x)}} \, dx \\ & = \frac {3 B (a+b \tan (c+d x))^{2/3}}{2 d}+\frac {1}{2} ((a-i b) (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) (A+i B)) \int \frac {1-i \tan (c+d x)}{\sqrt [3]{a+b \tan (c+d x)}} \, dx \\ & = \frac {3 B (a+b \tan (c+d x))^{2/3}}{2 d}+\frac {(i (a-i b) (A-i B)) \text {Subst}\left (\int \frac {1}{(-1+x) \sqrt [3]{a-i b x}} \, dx,x,i \tan (c+d x)\right )}{2 d}-\frac {((i a-b) (A+i B)) \text {Subst}\left (\int \frac {1}{(-1+x) \sqrt [3]{a+i b x}} \, dx,x,-i \tan (c+d x)\right )}{2 d} \\ & = -\frac {1}{4} (a-i b)^{2/3} (A-i B) x-\frac {1}{4} (a+i b)^{2/3} (A+i B) x-\frac {(a+i b)^{2/3} (i A-B) \log (\cos (c+d x))}{4 d}+\frac {(a-i b)^{2/3} (i A+B) \log (\cos (c+d x))}{4 d}+\frac {3 B (a+b \tan (c+d x))^{2/3}}{2 d}+\frac {\left (3 (a+i b)^{2/3} (i A-B)\right ) \text {Subst}\left (\int \frac {1}{\sqrt [3]{a+i b}-x} \, dx,x,\sqrt [3]{a+b \tan (c+d x)}\right )}{4 d}+\frac {(3 i (a-i b) (A-i B)) \text {Subst}\left (\int \frac {1}{(a-i b)^{2/3}+\sqrt [3]{a-i b} x+x^2} \, dx,x,\sqrt [3]{a+b \tan (c+d x)}\right )}{4 d}-\frac {(3 (i a-b) (A+i B)) \text {Subst}\left (\int \frac {1}{(a+i b)^{2/3}+\sqrt [3]{a+i b} x+x^2} \, dx,x,\sqrt [3]{a+b \tan (c+d x)}\right )}{4 d}-\frac {\left (3 (a-i b)^{2/3} (i A+B)\right ) \text {Subst}\left (\int \frac {1}{\sqrt [3]{a-i b}-x} \, dx,x,\sqrt [3]{a+b \tan (c+d x)}\right )}{4 d} \\ & = -\frac {1}{4} (a-i b)^{2/3} (A-i B) x-\frac {1}{4} (a+i b)^{2/3} (A+i B) x-\frac {(a+i b)^{2/3} (i A-B) \log (\cos (c+d x))}{4 d}+\frac {(a-i b)^{2/3} (i A+B) \log (\cos (c+d x))}{4 d}+\frac {3 (a-i b)^{2/3} (i A+B) \log \left (\sqrt [3]{a-i b}-\sqrt [3]{a+b \tan (c+d x)}\right )}{4 d}-\frac {3 (a+i b)^{2/3} (i A-B) \log \left (\sqrt [3]{a+i b}-\sqrt [3]{a+b \tan (c+d x)}\right )}{4 d}+\frac {3 B (a+b \tan (c+d x))^{2/3}}{2 d}+\frac {\left (3 (a+i b)^{2/3} (i A-B)\right ) \text {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1+\frac {2 \sqrt [3]{a+b \tan (c+d x)}}{\sqrt [3]{a+i b}}\right )}{2 d}-\frac {\left (3 (a-i b)^{2/3} (i A+B)\right ) \text {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1+\frac {2 \sqrt [3]{a+b \tan (c+d x)}}{\sqrt [3]{a-i b}}\right )}{2 d} \\ & = -\frac {1}{4} (a-i b)^{2/3} (A-i B) x-\frac {1}{4} (a+i b)^{2/3} (A+i B) x+\frac {\sqrt {3} (a-i b)^{2/3} (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 d}-\frac {\sqrt {3} (a+i b)^{2/3} (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 d}-\frac {(a+i b)^{2/3} (i A-B) \log (\cos (c+d x))}{4 d}+\frac {(a-i b)^{2/3} (i A+B) \log (\cos (c+d x))}{4 d}+\frac {3 (a-i b)^{2/3} (i A+B) \log \left (\sqrt [3]{a-i b}-\sqrt [3]{a+b \tan (c+d x)}\right )}{4 d}-\frac {3 (a+i b)^{2/3} (i A-B) \log \left (\sqrt [3]{a+i b}-\sqrt [3]{a+b \tan (c+d x)}\right )}{4 d}+\frac {3 B (a+b \tan (c+d x))^{2/3}}{2 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.07 (sec) , antiderivative size = 263, normalized size of antiderivative = 0.69 \[ \int (a+b \tan (c+d x))^{2/3} (A+B \tan (c+d x)) \, dx=\frac {i \left ((A-i B) \left ((a-i b)^{2/3} \left (2 \sqrt {3} \arctan \left (\frac {1+\frac {2 \sqrt [3]{a+b \tan (c+d x)}}{\sqrt [3]{a-i b}}}{\sqrt {3}}\right )-\log (i+\tan (c+d x))+3 \log \left (\sqrt [3]{a-i b}-\sqrt [3]{a+b \tan (c+d x)}\right )\right )+3 (a+b \tan (c+d x))^{2/3}\right )-(A+i B) \left ((a+i b)^{2/3} \left (2 \sqrt {3} \arctan \left (\frac {1+\frac {2 \sqrt [3]{a+b \tan (c+d x)}}{\sqrt [3]{a+i b}}}{\sqrt {3}}\right )-\log (i-\tan (c+d x))+3 \log \left (\sqrt [3]{a+i b}-\sqrt [3]{a+b \tan (c+d x)}\right )\right )+3 (a+b \tan (c+d x))^{2/3}\right )\right )}{4 d} \]

[In]

Integrate[(a + b*Tan[c + d*x])^(2/3)*(A + B*Tan[c + d*x]),x]

[Out]

((I/4)*((A - I*B)*((a - I*b)^(2/3)*(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)]) + 3*(a + b*Tan[c + d*x])^(
2/3)) - (A + I*B)*((a + I*b)^(2/3)*(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)]) + 3*(a + b*Tan[c + d*x])^(
2/3))))/d

Maple [C] (verified)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 0.56 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.26

method	result	size
derivativedivides	\(\frac {\frac {3 \left (a +b \tan \left (d x +c \right )\right )^{\frac {2}{3}} B}{2}+\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{6}-2 a \,\textit {\_Z}^{3}+a^{2}+b^{2}\right )}{\sum }\frac {\left (\left (A b +B a \right ) \textit {\_R}^{4}+B \left (-a^{2}-b^{2}\right ) \textit {\_R} \right ) \ln \left (\left (a +b \tan \left (d x +c \right )\right )^{\frac {1}{3}}-\textit {\_R} \right )}{\textit {\_R}^{5}-\textit {\_R}^{2} a}\right )}{2}}{d}\)	\(99\)
default	\(\frac {\frac {3 \left (a +b \tan \left (d x +c \right )\right )^{\frac {2}{3}} B}{2}+\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{6}-2 a \,\textit {\_Z}^{3}+a^{2}+b^{2}\right )}{\sum }\frac {\left (\left (A b +B a \right ) \textit {\_R}^{4}+B \left (-a^{2}-b^{2}\right ) \textit {\_R} \right ) \ln \left (\left (a +b \tan \left (d x +c \right )\right )^{\frac {1}{3}}-\textit {\_R} \right )}{\textit {\_R}^{5}-\textit {\_R}^{2} a}\right )}{2}}{d}\)	\(99\)
parts	\(\frac {A b \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{6}-2 a \,\textit {\_Z}^{3}+a^{2}+b^{2}\right )}{\sum }\frac {\textit {\_R}^{2} \ln \left (\left (a +b \tan \left (d x +c \right )\right )^{\frac {1}{3}}-\textit {\_R} \right )}{\textit {\_R}^{3}-a}\right )}{2 d}+\frac {B \left (\frac {3 \left (a +b \tan \left (d x +c \right )\right )^{\frac {2}{3}}}{2}+\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{6}-2 a \,\textit {\_Z}^{3}+a^{2}+b^{2}\right )}{\sum }\frac {\left (-\textit {\_R}^{3} a +a^{2}+b^{2}\right ) \ln \left (\left (a +b \tan \left (d x +c \right )\right )^{\frac {1}{3}}-\textit {\_R} \right )}{\textit {\_R} \left (-\textit {\_R}^{3}+a \right )}\right )}{2}\right )}{d}\)	\(144\)

[In]

int((a+b*tan(d*x+c))^(2/3)*(A+B*tan(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

1/d*(3/2*(a+b*tan(d*x+c))^(2/3)*B+1/2*sum(((A*b+B*a)*_R^4+B*(-a^2-b^2)*_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)))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 5405 vs. \(2 (281) = 562\).

Time = 1.89 (sec) , antiderivative size = 5405, normalized size of antiderivative = 14.26 \[ \int (a+b \tan (c+d x))^{2/3} (A+B \tan (c+d x)) \, dx=\text {Too large to display} \]

[In]

integrate((a+b*tan(d*x+c))^(2/3)*(A+B*tan(d*x+c)),x, algorithm="fricas")

[Out]

Too large to include

Sympy [F]

\[ \int (a+b \tan (c+d x))^{2/3} (A+B \tan (c+d x)) \, dx=\int \left (A + B \tan {\left (c + d x \right )}\right ) \left (a + b \tan {\left (c + d x \right )}\right )^{\frac {2}{3}}\, dx \]

[In]

integrate((a+b*tan(d*x+c))**(2/3)*(A+B*tan(d*x+c)),x)

[Out]

Integral((A + B*tan(c + d*x))*(a + b*tan(c + d*x))**(2/3), x)

Maxima [F]

\[ \int (a+b \tan (c+d x))^{2/3} (A+B \tan (c+d x)) \, dx=\int { {\left (B \tan \left (d x + c\right ) + A\right )} {\left (b \tan \left (d x + c\right ) + a\right )}^{\frac {2}{3}} \,d x } \]

[In]

integrate((a+b*tan(d*x+c))^(2/3)*(A+B*tan(d*x+c)),x, algorithm="maxima")

[Out]

integrate((B*tan(d*x + c) + A)*(b*tan(d*x + c) + a)^(2/3), x)

Giac [F]

\[ \int (a+b \tan (c+d x))^{2/3} (A+B \tan (c+d x)) \, dx=\int { {\left (B \tan \left (d x + c\right ) + A\right )} {\left (b \tan \left (d x + c\right ) + a\right )}^{\frac {2}{3}} \,d x } \]

[In]

integrate((a+b*tan(d*x+c))^(2/3)*(A+B*tan(d*x+c)),x, algorithm="giac")

[Out]

undef

Mupad [B] (verification not implemented)

Time = 25.43 (sec) , antiderivative size = 3945, normalized size of antiderivative = 10.41 \[ \int (a+b \tan (c+d x))^{2/3} (A+B \tan (c+d x)) \, dx=\text {Too large to display} \]

[In]

int((A + B*tan(c + d*x))*(a + b*tan(c + d*x))^(2/3),x)

[Out]

log((((2*(-B^6*a^2*b^2*d^6)^(1/2) + B^3*a^2*d^3 - B^3*b^2*d^3)/d^6)^(2/3)*((((2*(-B^6*a^2*b^2*d^6)^(1/2) + B^3
*a^2*d^3 - B^3*b^2*d^3)/d^6)^(1/3)*(1944*a*b^4*((2*(-B^6*a^2*b^2*d^6)^(1/2) + B^3*a^2*d^3 - B^3*b^2*d^3)/d^6)^
(2/3)*(a^2 + b^2) - (1944*B^2*b^4*(a^2 + b^2)^2*(a + b*tan(c + d*x))^(1/3))/d^2))/2 + (972*B^3*a*b^4*(3*b^4 -
a^4 + 2*a^2*b^2))/d^3))/4 + (243*B^5*b^4*(a^2 - b^2)*(a^2 + b^2)^2*(a + b*tan(c + d*x))^(1/3))/d^5)*(((-4*B^6*
a^2*b^2*d^6)^(1/2) + B^3*a^2*d^3 - B^3*b^2*d^3)/(8*d^6))^(1/3) + log(((-(2*(-B^6*a^2*b^2*d^6)^(1/2) - B^3*a^2*
d^3 + B^3*b^2*d^3)/d^6)^(2/3)*(((-(2*(-B^6*a^2*b^2*d^6)^(1/2) - B^3*a^2*d^3 + B^3*b^2*d^3)/d^6)^(1/3)*(1944*a*
b^4*(-(2*(-B^6*a^2*b^2*d^6)^(1/2) - B^3*a^2*d^3 + B^3*b^2*d^3)/d^6)^(2/3)*(a^2 + b^2) - (1944*B^2*b^4*(a^2 + b
^2)^2*(a + b*tan(c + d*x))^(1/3))/d^2))/2 + (972*B^3*a*b^4*(3*b^4 - a^4 + 2*a^2*b^2))/d^3))/4 + (243*B^5*b^4*(
a^2 - b^2)*(a^2 + b^2)^2*(a + b*tan(c + d*x))^(1/3))/d^5)*(-((-4*B^6*a^2*b^2*d^6)^(1/2) - B^3*a^2*d^3 + B^3*b^
2*d^3)/(8*d^6))^(1/3) + log(((((-A^6*d^6*(a^2 - b^2)^2)^(1/2) - 2*A^3*a*b*d^3)/d^6)^(2/3)*(((1944*a*b^4*(((-A^
6*d^6*(a^2 - b^2)^2)^(1/2) - 2*A^3*a*b*d^3)/d^6)^(2/3)*(a^2 + b^2) + (1944*A^2*b^4*(a^2 + b^2)^2*(a + b*tan(c
+ d*x))^(1/3))/d^2)*(((-A^6*d^6*(a^2 - b^2)^2)^(1/2) - 2*A^3*a*b*d^3)/d^6)^(1/3))/2 + (972*A^3*b^5*(3*a^4 - b^
4 + 2*a^2*b^2))/d^3))/4 + (486*A^5*a*b^5*(a^2 + b^2)^2*(a + b*tan(c + d*x))^(1/3))/d^5)*((2*A^6*a^2*b^2*d^6 -
A^6*b^4*d^6 - A^6*a^4*d^6)^(1/2)/(8*d^6) - (A^3*a*b)/(4*d^3))^(1/3) + log(((-((-A^6*d^6*(a^2 - b^2)^2)^(1/2) +
2*A^3*a*b*d^3)/d^6)^(2/3)*(((1944*a*b^4*(-((-A^6*d^6*(a^2 - b^2)^2)^(1/2) + 2*A^3*a*b*d^3)/d^6)^(2/3)*(a^2 +
b^2) + (1944*A^2*b^4*(a^2 + b^2)^2*(a + b*tan(c + d*x))^(1/3))/d^2)*(-((-A^6*d^6*(a^2 - b^2)^2)^(1/2) + 2*A^3*
a*b*d^3)/d^6)^(1/3))/2 + (972*A^3*b^5*(3*a^4 - b^4 + 2*a^2*b^2))/d^3))/4 + (486*A^5*a*b^5*(a^2 + b^2)^2*(a + b
*tan(c + d*x))^(1/3))/d^5)*(- (2*A^6*a^2*b^2*d^6 - A^6*b^4*d^6 - A^6*a^4*d^6)^(1/2)/(8*d^6) - (A^3*a*b)/(4*d^3
))^(1/3) + (log((243*B^5*b^4*(a^2 - b^2)*(a^2 + b^2)^2*(a + b*tan(c + d*x))^(1/3))/d^5 - ((3^(1/2)*1i - 1)^2*(
((3^(1/2)*1i - 1)*((1944*B^2*b^4*(a^2 + b^2)^2*(a + b*tan(c + d*x))^(1/3))/d^2 - 486*a*b^4*(3^(1/2)*1i - 1)^2*
((2*(-B^6*a^2*b^2*d^6)^(1/2) + B^3*a^2*d^3 - B^3*b^2*d^3)/d^6)^(2/3)*(a^2 + b^2))*((2*(-B^6*a^2*b^2*d^6)^(1/2)
+ B^3*a^2*d^3 - B^3*b^2*d^3)/d^6)^(1/3))/4 - (972*B^3*a*b^4*(3*b^4 - a^4 + 2*a^2*b^2))/d^3)*((2*(-B^6*a^2*b^2
*d^6)^(1/2) + B^3*a^2*d^3 - B^3*b^2*d^3)/d^6)^(2/3))/16)*(3^(1/2)*1i - 1)*((-4*B^6*a^2*b^2*d^6)^(1/2)/(8*d^6)
+ (B^3*a^2)/(8*d^3) - (B^3*b^2)/(8*d^3))^(1/3))/2 - (log(((3^(1/2)*1i + 1)^2*(((3^(1/2)*1i + 1)*((1944*B^2*b^4
*(a^2 + b^2)^2*(a + b*tan(c + d*x))^(1/3))/d^2 - 486*a*b^4*(3^(1/2)*1i + 1)^2*((2*(-B^6*a^2*b^2*d^6)^(1/2) + B
^3*a^2*d^3 - B^3*b^2*d^3)/d^6)^(2/3)*(a^2 + b^2))*((2*(-B^6*a^2*b^2*d^6)^(1/2) + B^3*a^2*d^3 - B^3*b^2*d^3)/d^
6)^(1/3))/4 + (972*B^3*a*b^4*(3*b^4 - a^4 + 2*a^2*b^2))/d^3)*((2*(-B^6*a^2*b^2*d^6)^(1/2) + B^3*a^2*d^3 - B^3*
b^2*d^3)/d^6)^(2/3))/16 + (243*B^5*b^4*(a^2 - b^2)*(a^2 + b^2)^2*(a + b*tan(c + d*x))^(1/3))/d^5)*(3^(1/2)*1i
+ 1)*((-4*B^6*a^2*b^2*d^6)^(1/2)/(8*d^6) + (B^3*a^2)/(8*d^3) - (B^3*b^2)/(8*d^3))^(1/3))/2 + (log((243*B^5*b^4
*(a^2 - b^2)*(a^2 + b^2)^2*(a + b*tan(c + d*x))^(1/3))/d^5 - ((3^(1/2)*1i - 1)^2*(((3^(1/2)*1i - 1)*((1944*B^2
*b^4*(a^2 + b^2)^2*(a + b*tan(c + d*x))^(1/3))/d^2 - 486*a*b^4*(3^(1/2)*1i - 1)^2*(-(2*(-B^6*a^2*b^2*d^6)^(1/2
) - B^3*a^2*d^3 + B^3*b^2*d^3)/d^6)^(2/3)*(a^2 + b^2))*(-(2*(-B^6*a^2*b^2*d^6)^(1/2) - B^3*a^2*d^3 + B^3*b^2*d
^3)/d^6)^(1/3))/4 - (972*B^3*a*b^4*(3*b^4 - a^4 + 2*a^2*b^2))/d^3)*(-(2*(-B^6*a^2*b^2*d^6)^(1/2) - B^3*a^2*d^3
+ B^3*b^2*d^3)/d^6)^(2/3))/16)*(3^(1/2)*1i - 1)*((B^3*a^2)/(8*d^3) - (-4*B^6*a^2*b^2*d^6)^(1/2)/(8*d^6) - (B^
3*b^2)/(8*d^3))^(1/3))/2 - (log(((3^(1/2)*1i + 1)^2*(((3^(1/2)*1i + 1)*((1944*B^2*b^4*(a^2 + b^2)^2*(a + b*tan
(c + d*x))^(1/3))/d^2 - 486*a*b^4*(3^(1/2)*1i + 1)^2*(-(2*(-B^6*a^2*b^2*d^6)^(1/2) - B^3*a^2*d^3 + B^3*b^2*d^3
)/d^6)^(2/3)*(a^2 + b^2))*(-(2*(-B^6*a^2*b^2*d^6)^(1/2) - B^3*a^2*d^3 + B^3*b^2*d^3)/d^6)^(1/3))/4 + (972*B^3*
a*b^4*(3*b^4 - a^4 + 2*a^2*b^2))/d^3)*(-(2*(-B^6*a^2*b^2*d^6)^(1/2) - B^3*a^2*d^3 + B^3*b^2*d^3)/d^6)^(2/3))/1
6 + (243*B^5*b^4*(a^2 - b^2)*(a^2 + b^2)^2*(a + b*tan(c + d*x))^(1/3))/d^5)*(3^(1/2)*1i + 1)*((B^3*a^2)/(8*d^3
) - (-4*B^6*a^2*b^2*d^6)^(1/2)/(8*d^6) - (B^3*b^2)/(8*d^3))^(1/3))/2 - log((486*A^5*a*b^5*(a^2 + b^2)^2*(a + b
*tan(c + d*x))^(1/3))/d^5 - (((3^(1/2)*1i)/2 - 1/2)*((((3^(1/2)*1i)/2 + 1/2)*((1944*A^2*b^4*(a^2 + b^2)^2*(a +
b*tan(c + d*x))^(1/3))/d^2 + 1944*a*b^4*((3^(1/2)*1i)/2 - 1/2)*(((-A^6*d^6*(a^2 - b^2)^2)^(1/2) - 2*A^3*a*b*d
^3)/d^6)^(2/3)*(a^2 + b^2))*(((-A^6*d^6*(a^2 - b^2)^2)^(1/2) - 2*A^3*a*b*d^3)/d^6)^(1/3))/2 - (972*A^3*b^5*(3*
a^4 - b^4 + 2*a^2*b^2))/d^3)*(((-A^6*d^6*(a^2 - b^2)^2)^(1/2) - 2*A^3*a*b*d^3)/d^6)^(2/3))/4)*((3^(1/2)*1i)/2
+ 1/2)*(((4*A^6*a^2*b^2*d^6 - d^6*(A^6*a^4 + A^6*b^4 + 2*A^6*a^2*b^2))^(1/2) - 2*A^3*a*b*d^3)/(8*d^6))^(1/3) +
log((486*A^5*a*b^5*(a^2 + b^2)^2*(a + b*tan(c + d*x))^(1/3))/d^5 - (((3^(1/2)*1i)/2 + 1/2)*((((3^(1/2)*1i)/2
- 1/2)*((1944*A^2*b^4*(a^2 + b^2)^2*(a + b*tan(c + d*x))^(1/3))/d^2 - 1944*a*b^4*((3^(1/2)*1i)/2 + 1/2)*(((-A^
6*d^6*(a^2 - b^2)^2)^(1/2) - 2*A^3*a*b*d^3)/d^6)^(2/3)*(a^2 + b^2))*(((-A^6*d^6*(a^2 - b^2)^2)^(1/2) - 2*A^3*a
*b*d^3)/d^6)^(1/3))/2 + (972*A^3*b^5*(3*a^4 - b^4 + 2*a^2*b^2))/d^3)*(((-A^6*d^6*(a^2 - b^2)^2)^(1/2) - 2*A^3*
a*b*d^3)/d^6)^(2/3))/4)*((3^(1/2)*1i)/2 - 1/2)*(((4*A^6*a^2*b^2*d^6 - d^6*(A^6*a^4 + A^6*b^4 + 2*A^6*a^2*b^2))
^(1/2) - 2*A^3*a*b*d^3)/(8*d^6))^(1/3) - log((486*A^5*a*b^5*(a^2 + b^2)^2*(a + b*tan(c + d*x))^(1/3))/d^5 - ((
(3^(1/2)*1i)/2 - 1/2)*((((3^(1/2)*1i)/2 + 1/2)*((1944*A^2*b^4*(a^2 + b^2)^2*(a + b*tan(c + d*x))^(1/3))/d^2 +
1944*a*b^4*((3^(1/2)*1i)/2 - 1/2)*(-((-A^6*d^6*(a^2 - b^2)^2)^(1/2) + 2*A^3*a*b*d^3)/d^6)^(2/3)*(a^2 + b^2))*(
-((-A^6*d^6*(a^2 - b^2)^2)^(1/2) + 2*A^3*a*b*d^3)/d^6)^(1/3))/2 - (972*A^3*b^5*(3*a^4 - b^4 + 2*a^2*b^2))/d^3)
*(-((-A^6*d^6*(a^2 - b^2)^2)^(1/2) + 2*A^3*a*b*d^3)/d^6)^(2/3))/4)*((3^(1/2)*1i)/2 + 1/2)*(-((4*A^6*a^2*b^2*d^
6 - d^6*(A^6*a^4 + A^6*b^4 + 2*A^6*a^2*b^2))^(1/2) + 2*A^3*a*b*d^3)/(8*d^6))^(1/3) + log((486*A^5*a*b^5*(a^2 +
b^2)^2*(a + b*tan(c + d*x))^(1/3))/d^5 - (((3^(1/2)*1i)/2 + 1/2)*((((3^(1/2)*1i)/2 - 1/2)*((1944*A^2*b^4*(a^2
+ b^2)^2*(a + b*tan(c + d*x))^(1/3))/d^2 - 1944*a*b^4*((3^(1/2)*1i)/2 + 1/2)*(-((-A^6*d^6*(a^2 - b^2)^2)^(1/2
) + 2*A^3*a*b*d^3)/d^6)^(2/3)*(a^2 + b^2))*(-((-A^6*d^6*(a^2 - b^2)^2)^(1/2) + 2*A^3*a*b*d^3)/d^6)^(1/3))/2 +
(972*A^3*b^5*(3*a^4 - b^4 + 2*a^2*b^2))/d^3)*(-((-A^6*d^6*(a^2 - b^2)^2)^(1/2) + 2*A^3*a*b*d^3)/d^6)^(2/3))/4)
*((3^(1/2)*1i)/2 - 1/2)*(-((4*A^6*a^2*b^2*d^6 - d^6*(A^6*a^4 + A^6*b^4 + 2*A^6*a^2*b^2))^(1/2) + 2*A^3*a*b*d^3
)/(8*d^6))^(1/3) + (3*B*(a + b*tan(c + d*x))^(2/3))/(2*d)