Integrand size = 35, antiderivative size = 229 \[ \int (c+d \tan (e+f x))^{5/2} \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right ) \, dx=-\frac {(i A+B-i C) (c-i d)^{5/2} \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{f}-\frac {(B-i (A-C)) (c+i d)^{5/2} \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{f}+\frac {2 \left (2 c (A-C) d+B \left (c^2-d^2\right )\right ) \sqrt {c+d \tan (e+f x)}}{f}+\frac {2 (B c+(A-C) d) (c+d \tan (e+f x))^{3/2}}{3 f}+\frac {2 B (c+d \tan (e+f x))^{5/2}}{5 f}+\frac {2 C (c+d \tan (e+f x))^{7/2}}{7 d f} \]
-(I*A+B-I*C)*(c-I*d)^(5/2)*arctanh((c+d*tan(f*x+e))^(1/2)/(c-I*d)^(1/2))/f -(B-I*(A-C))*(c+I*d)^(5/2)*arctanh((c+d*tan(f*x+e))^(1/2)/(c+I*d)^(1/2))/f +2*(2*c*(A-C)*d+B*(c^2-d^2))*(c+d*tan(f*x+e))^(1/2)/f+2/3*(B*c+(A-C)*d)*(c +d*tan(f*x+e))^(3/2)/f+2/5*B*(c+d*tan(f*x+e))^(5/2)/f+2/7*C*(c+d*tan(f*x+e ))^(7/2)/d/f
Time = 2.20 (sec) , antiderivative size = 262, normalized size of antiderivative = 1.14 \[ \int (c+d \tan (e+f x))^{5/2} \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right ) \, dx=\frac {\frac {4 C (c+d \tan (e+f x))^{7/2}}{d}+7 i (A-i B-C) \left (\frac {2}{5} (c+d \tan (e+f x))^{5/2}+\frac {2}{3} (c-i d) \left (-3 (c-i d)^{3/2} \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )+\sqrt {c+d \tan (e+f x)} (4 c-3 i d+d \tan (e+f x))\right )\right )-7 i (A+i B-C) \left (\frac {2}{5} (c+d \tan (e+f x))^{5/2}+\frac {2}{3} (c+i d) \left (-3 (c+i d)^{3/2} \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )+\sqrt {c+d \tan (e+f x)} (4 c+3 i d+d \tan (e+f x))\right )\right )}{14 f} \]
((4*C*(c + d*Tan[e + f*x])^(7/2))/d + (7*I)*(A - I*B - C)*((2*(c + d*Tan[e + f*x])^(5/2))/5 + (2*(c - I*d)*(-3*(c - I*d)^(3/2)*ArcTanh[Sqrt[c + d*Ta n[e + f*x]]/Sqrt[c - I*d]] + Sqrt[c + d*Tan[e + f*x]]*(4*c - (3*I)*d + d*T an[e + f*x])))/3) - (7*I)*(A + I*B - C)*((2*(c + d*Tan[e + f*x])^(5/2))/5 + (2*(c + I*d)*(-3*(c + I*d)^(3/2)*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c + I*d]] + Sqrt[c + d*Tan[e + f*x]]*(4*c + (3*I)*d + d*Tan[e + f*x])))/3)) /(14*f)
Time = 1.37 (sec) , antiderivative size = 208, normalized size of antiderivative = 0.91, number of steps used = 16, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {3042, 4113, 3042, 4011, 3042, 4011, 3042, 4011, 3042, 4022, 3042, 4020, 25, 73, 221}
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.
| \(\displaystyle \int (c+d \tan (e+f x))^{5/2} \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int (c+d \tan (e+f x))^{5/2} \left (A+B \tan (e+f x)+C \tan (e+f x)^2\right )dx\) |
| \(\Big \downarrow \) 4113 |
| \(\displaystyle \int (A-C+B \tan (e+f x)) (c+d \tan (e+f x))^{5/2}dx+\frac {2 C (c+d \tan (e+f x))^{7/2}}{7 d f}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int (A-C+B \tan (e+f x)) (c+d \tan (e+f x))^{5/2}dx+\frac {2 C (c+d \tan (e+f x))^{7/2}}{7 d f}\) |
| \(\Big \downarrow \) 4011 |
| \(\displaystyle \int (c+d \tan (e+f x))^{3/2} (A c-C c-B d+(B c+(A-C) d) \tan (e+f x))dx+\frac {2 B (c+d \tan (e+f x))^{5/2}}{5 f}+\frac {2 C (c+d \tan (e+f x))^{7/2}}{7 d f}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int (c+d \tan (e+f x))^{3/2} (A c-C c-B d+(B c+(A-C) d) \tan (e+f x))dx+\frac {2 B (c+d \tan (e+f x))^{5/2}}{5 f}+\frac {2 C (c+d \tan (e+f x))^{7/2}}{7 d f}\) |
| \(\Big \downarrow \) 4011 |
| \(\displaystyle \int \sqrt {c+d \tan (e+f x)} \left (-C c^2-2 B d c+C d^2+A \left (c^2-d^2\right )+\left (2 c (A-C) d+B \left (c^2-d^2\right )\right ) \tan (e+f x)\right )dx+\frac {2 (d (A-C)+B c) (c+d \tan (e+f x))^{3/2}}{3 f}+\frac {2 B (c+d \tan (e+f x))^{5/2}}{5 f}+\frac {2 C (c+d \tan (e+f x))^{7/2}}{7 d f}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \sqrt {c+d \tan (e+f x)} \left (-C c^2-2 B d c+C d^2+A \left (c^2-d^2\right )+\left (2 c (A-C) d+B \left (c^2-d^2\right )\right ) \tan (e+f x)\right )dx+\frac {2 (d (A-C)+B c) (c+d \tan (e+f x))^{3/2}}{3 f}+\frac {2 B (c+d \tan (e+f x))^{5/2}}{5 f}+\frac {2 C (c+d \tan (e+f x))^{7/2}}{7 d f}\) |
| \(\Big \downarrow \) 4011 |
| \(\displaystyle \int \frac {-C c^3-3 B d c^2+3 C d^2 c+B d^3+A \left (c^3-3 c d^2\right )+\left ((A-C) d \left (3 c^2-d^2\right )+B \left (c^3-3 c d^2\right )\right ) \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx+\frac {2 \left (2 c d (A-C)+B \left (c^2-d^2\right )\right ) \sqrt {c+d \tan (e+f x)}}{f}+\frac {2 (d (A-C)+B c) (c+d \tan (e+f x))^{3/2}}{3 f}+\frac {2 B (c+d \tan (e+f x))^{5/2}}{5 f}+\frac {2 C (c+d \tan (e+f x))^{7/2}}{7 d f}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {-C c^3-3 B d c^2+3 C d^2 c+B d^3+A \left (c^3-3 c d^2\right )+\left ((A-C) d \left (3 c^2-d^2\right )+B \left (c^3-3 c d^2\right )\right ) \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx+\frac {2 \left (2 c d (A-C)+B \left (c^2-d^2\right )\right ) \sqrt {c+d \tan (e+f x)}}{f}+\frac {2 (d (A-C)+B c) (c+d \tan (e+f x))^{3/2}}{3 f}+\frac {2 B (c+d \tan (e+f x))^{5/2}}{5 f}+\frac {2 C (c+d \tan (e+f x))^{7/2}}{7 d f}\) |
| \(\Big \downarrow \) 4022 |
| \(\displaystyle \frac {1}{2} (c+i d)^3 (A+i B-C) \int \frac {1-i \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx+\frac {1}{2} (c-i d)^3 (A-i B-C) \int \frac {i \tan (e+f x)+1}{\sqrt {c+d \tan (e+f x)}}dx+\frac {2 \left (2 c d (A-C)+B \left (c^2-d^2\right )\right ) \sqrt {c+d \tan (e+f x)}}{f}+\frac {2 (d (A-C)+B c) (c+d \tan (e+f x))^{3/2}}{3 f}+\frac {2 B (c+d \tan (e+f x))^{5/2}}{5 f}+\frac {2 C (c+d \tan (e+f x))^{7/2}}{7 d f}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{2} (c+i d)^3 (A+i B-C) \int \frac {1-i \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx+\frac {1}{2} (c-i d)^3 (A-i B-C) \int \frac {i \tan (e+f x)+1}{\sqrt {c+d \tan (e+f x)}}dx+\frac {2 \left (2 c d (A-C)+B \left (c^2-d^2\right )\right ) \sqrt {c+d \tan (e+f x)}}{f}+\frac {2 (d (A-C)+B c) (c+d \tan (e+f x))^{3/2}}{3 f}+\frac {2 B (c+d \tan (e+f x))^{5/2}}{5 f}+\frac {2 C (c+d \tan (e+f x))^{7/2}}{7 d f}\) |
| \(\Big \downarrow \) 4020 |
| \(\displaystyle \frac {i (c-i d)^3 (A-i B-C) \int -\frac {1}{(1-i \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}d(i \tan (e+f x))}{2 f}-\frac {i (c+i d)^3 (A+i B-C) \int -\frac {1}{(i \tan (e+f x)+1) \sqrt {c+d \tan (e+f x)}}d(-i \tan (e+f x))}{2 f}+\frac {2 \left (2 c d (A-C)+B \left (c^2-d^2\right )\right ) \sqrt {c+d \tan (e+f x)}}{f}+\frac {2 (d (A-C)+B c) (c+d \tan (e+f x))^{3/2}}{3 f}+\frac {2 B (c+d \tan (e+f x))^{5/2}}{5 f}+\frac {2 C (c+d \tan (e+f x))^{7/2}}{7 d f}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {i (c-i d)^3 (A-i B-C) \int \frac {1}{(1-i \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}d(i \tan (e+f x))}{2 f}+\frac {i (c+i d)^3 (A+i B-C) \int \frac {1}{(i \tan (e+f x)+1) \sqrt {c+d \tan (e+f x)}}d(-i \tan (e+f x))}{2 f}+\frac {2 \left (2 c d (A-C)+B \left (c^2-d^2\right )\right ) \sqrt {c+d \tan (e+f x)}}{f}+\frac {2 (d (A-C)+B c) (c+d \tan (e+f x))^{3/2}}{3 f}+\frac {2 B (c+d \tan (e+f x))^{5/2}}{5 f}+\frac {2 C (c+d \tan (e+f x))^{7/2}}{7 d f}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {(c+i d)^3 (A+i B-C) \int \frac {1}{-\frac {i \tan ^2(e+f x)}{d}-\frac {i c}{d}+1}d\sqrt {c+d \tan (e+f x)}}{d f}+\frac {(c-i d)^3 (A-i B-C) \int \frac {1}{\frac {i \tan ^2(e+f x)}{d}+\frac {i c}{d}+1}d\sqrt {c+d \tan (e+f x)}}{d f}+\frac {2 \left (2 c d (A-C)+B \left (c^2-d^2\right )\right ) \sqrt {c+d \tan (e+f x)}}{f}+\frac {2 (d (A-C)+B c) (c+d \tan (e+f x))^{3/2}}{3 f}+\frac {2 B (c+d \tan (e+f x))^{5/2}}{5 f}+\frac {2 C (c+d \tan (e+f x))^{7/2}}{7 d f}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {(c-i d)^{5/2} (A-i B-C) \arctan \left (\frac {\tan (e+f x)}{\sqrt {c-i d}}\right )}{f}+\frac {(c+i d)^{5/2} (A+i B-C) \arctan \left (\frac {\tan (e+f x)}{\sqrt {c+i d}}\right )}{f}+\frac {2 \left (2 c d (A-C)+B \left (c^2-d^2\right )\right ) \sqrt {c+d \tan (e+f x)}}{f}+\frac {2 (d (A-C)+B c) (c+d \tan (e+f x))^{3/2}}{3 f}+\frac {2 B (c+d \tan (e+f x))^{5/2}}{5 f}+\frac {2 C (c+d \tan (e+f x))^{7/2}}{7 d f}\) |
((A - I*B - C)*(c - I*d)^(5/2)*ArcTan[Tan[e + f*x]/Sqrt[c - I*d]])/f + ((A + I*B - C)*(c + I*d)^(5/2)*ArcTan[Tan[e + f*x]/Sqrt[c + I*d]])/f + (2*(2* c*(A - C)*d + B*(c^2 - d^2))*Sqrt[c + d*Tan[e + f*x]])/f + (2*(B*c + (A - C)*d)*(c + d*Tan[e + f*x])^(3/2))/(3*f) + (2*B*(c + d*Tan[e + f*x])^(5/2)) /(5*f) + (2*C*(c + d*Tan[e + f*x])^(7/2))/(7*d*f)
3.2.6.3.1 Defintions of rubi rules used
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
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]
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]
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[C*((a + b*Tan[e + f*x])^(m + 1)/(b*f*(m + 1))), x] + Int[(a + b*Tan[e + f*x])^m*Si mp[A - C + B*Tan[e + f*x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] && NeQ[A*b^2 - a*b*B + a^2*C, 0] && !LeQ[m, -1]
Leaf count of result is larger than twice the leaf count of optimal. \(3561\) vs. \(2(196)=392\).
Time = 0.15 (sec) , antiderivative size = 3562, normalized size of antiderivative = 15.55
| method | result | size |
| parts | \(\text {Expression too large to display}\) | \(3562\) |
| derivativedivides | \(\text {Expression too large to display}\) | \(3614\) |
| default | \(\text {Expression too large to display}\) | \(3614\) |
A*(2/3/f*d*(c+d*tan(f*x+e))^(3/2)+4/f*d*(c+d*tan(f*x+e))^(1/2)*c+1/4/f/d*l n(d*tan(f*x+e)+c-(c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)+(c^2 +d^2)^(1/2))*(c^2+d^2)^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*c^2-1/4/f*d*ln( d*tan(f*x+e)+c-(c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)+(c^2+d ^2)^(1/2))*(c^2+d^2)^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)-1/4/f/d*ln(d*tan( f*x+e)+c-(c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)+(c^2+d^2)^(1 /2))*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*c^3+3/4/f*d*ln(d*tan(f*x+e)+c-(c+d*tan( f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)+(c^2+d^2)^(1/2))*(2*(c^2+d^2)^ (1/2)+2*c)^(1/2)*c+3/f*d/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arctan((2*(c+d*tan( f*x+e))^(1/2)-(2*(c^2+d^2)^(1/2)+2*c)^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2) )*c^2-1/f*d^3/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arctan((2*(c+d*tan(f*x+e))^(1/ 2)-(2*(c^2+d^2)^(1/2)+2*c)^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))-2/f*d/(2* (c^2+d^2)^(1/2)-2*c)^(1/2)*arctan((2*(c+d*tan(f*x+e))^(1/2)-(2*(c^2+d^2)^( 1/2)+2*c)^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))*(c^2+d^2)^(1/2)*c-1/4/f/d* ln(d*tan(f*x+e)+c+(c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)+(c^ 2+d^2)^(1/2))*(c^2+d^2)^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*c^2+1/4/f*d*ln (d*tan(f*x+e)+c+(c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)+(c^2+ d^2)^(1/2))*(c^2+d^2)^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)+1/4/f/d*ln(d*tan (f*x+e)+c+(c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)+(c^2+d^2)^( 1/2))*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*c^3-3/4/f*d*ln(d*tan(f*x+e)+c+(c+d*...
Leaf count of result is larger than twice the leaf count of optimal. 10840 vs. \(2 (189) = 378\).
Time = 1.98 (sec) , antiderivative size = 10840, normalized size of antiderivative = 47.34 \[ \int (c+d \tan (e+f x))^{5/2} \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right ) \, dx=\text {Too large to display} \]
\[ \int (c+d \tan (e+f x))^{5/2} \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right ) \, dx=\int \left (c + d \tan {\left (e + f x \right )}\right )^{\frac {5}{2}} \left (A + B \tan {\left (e + f x \right )} + C \tan ^{2}{\left (e + f x \right )}\right )\, dx \]
\[ \int (c+d \tan (e+f x))^{5/2} \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right ) \, dx=\int { {\left (C \tan \left (f x + e\right )^{2} + B \tan \left (f x + e\right ) + A\right )} {\left (d \tan \left (f x + e\right ) + c\right )}^{\frac {5}{2}} \,d x } \]
Timed out. \[ \int (c+d \tan (e+f x))^{5/2} \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right ) \, dx=\text {Timed out} \]
Time = 114.33 (sec) , antiderivative size = 5863, normalized size of antiderivative = 25.60 \[ \int (c+d \tan (e+f x))^{5/2} \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right ) \, dx=\text {Too large to display} \]
((2*C*c^2)/(3*d*f) - (2*C*(d^3*f + c^2*d*f))/(3*d^2*f^2))*(c + d*tan(e + f *x))^(3/2) - log(((((-B^4*d^2*f^4*(5*c^4 + d^4 - 10*c^2*d^2)^2)^(1/2) + B^ 2*c^5*f^2 - 10*B^2*c^3*d^2*f^2 + 5*B^2*c*d^4*f^2)/f^4)^(1/2)*(((((-B^4*d^2 *f^4*(5*c^4 + d^4 - 10*c^2*d^2)^2)^(1/2) + B^2*c^5*f^2 - 10*B^2*c^3*d^2*f^ 2 + 5*B^2*c*d^4*f^2)/f^4)^(1/2)*(32*B*c^4*d^2 - 32*B*d^6 + 32*c*d^2*f*(((- B^4*d^2*f^4*(5*c^4 + d^4 - 10*c^2*d^2)^2)^(1/2) + B^2*c^5*f^2 - 10*B^2*c^3 *d^2*f^2 + 5*B^2*c*d^4*f^2)/f^4)^(1/2)*(c + d*tan(e + f*x))^(1/2)))/(2*f) - (16*B^2*d^2*(c + d*tan(e + f*x))^(1/2)*(c^6 - d^6 + 15*c^2*d^4 - 15*c^4* d^2))/f^2))/2 - (8*B^3*c*d^2*(c^2 - 3*d^2)*(c^2 + d^2)^3)/f^3)*(((20*B^4*c ^2*d^8*f^4 - B^4*d^10*f^4 - 110*B^4*c^4*d^6*f^4 + 100*B^4*c^6*d^4*f^4 - 25 *B^4*c^8*d^2*f^4)^(1/2) + B^2*c^5*f^2 - 10*B^2*c^3*d^2*f^2 + 5*B^2*c*d^4*f ^2)/(4*f^4))^(1/2) + log(- ((((-B^4*d^2*f^4*(5*c^4 + d^4 - 10*c^2*d^2)^2)^ (1/2) + B^2*c^5*f^2 - 10*B^2*c^3*d^2*f^2 + 5*B^2*c*d^4*f^2)/f^4)^(1/2)*((( ((-B^4*d^2*f^4*(5*c^4 + d^4 - 10*c^2*d^2)^2)^(1/2) + B^2*c^5*f^2 - 10*B^2* c^3*d^2*f^2 + 5*B^2*c*d^4*f^2)/f^4)^(1/2)*(32*B*d^6 - 32*B*c^4*d^2 + 32*c* d^2*f*(((-B^4*d^2*f^4*(5*c^4 + d^4 - 10*c^2*d^2)^2)^(1/2) + B^2*c^5*f^2 - 10*B^2*c^3*d^2*f^2 + 5*B^2*c*d^4*f^2)/f^4)^(1/2)*(c + d*tan(e + f*x))^(1/2 )))/(2*f) - (16*B^2*d^2*(c + d*tan(e + f*x))^(1/2)*(c^6 - d^6 + 15*c^2*d^4 - 15*c^4*d^2))/f^2))/2 - (8*B^3*c*d^2*(c^2 - 3*d^2)*(c^2 + d^2)^3)/f^3)*( (20*B^4*c^2*d^8*f^4 - B^4*d^10*f^4 - 110*B^4*c^4*d^6*f^4 + 100*B^4*c^6*...