3.1.100 \(\int (c+d \tan (e+f x))^{3/2} (A+B \tan (e+f x)+C \tan ^2(e+f x)) \, dx\) [100]

3.1.100.1 Optimal result

Integrand size = 35, antiderivative size = 187 \[ \int (c+d \tan (e+f x))^{3/2} \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right ) \, dx=-\frac {(i A+B-i C) (c-i d)^{3/2} \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{f}-\frac {(B-i (A-C)) (c+i d)^{3/2} \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{f}+\frac {2 (B c+(A-C) d) \sqrt {c+d \tan (e+f x)}}{f}+\frac {2 B (c+d \tan (e+f x))^{3/2}}{3 f}+\frac {2 C (c+d \tan (e+f x))^{5/2}}{5 d f} \]

output

-(I*A+B-I*C)*(c-I*d)^(3/2)*arctanh((c+d*tan(f*x+e))^(1/2)/(c-I*d)^(1/2))/f 
-(B-I*(A-C))*(c+I*d)^(3/2)*arctanh((c+d*tan(f*x+e))^(1/2)/(c+I*d)^(1/2))/f 
+2*(B*c+(A-C)*d)*(c+d*tan(f*x+e))^(1/2)/f+2/3*B*(c+d*tan(f*x+e))^(3/2)/f+2 
/5*C*(c+d*tan(f*x+e))^(5/2)/d/f
 

3.1.100.2 Mathematica [A] (verified)

Time = 1.32 (sec) , antiderivative size = 202, normalized size of antiderivative = 1.08 \[ \int (c+d \tan (e+f x))^{3/2} \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right ) \, dx=\frac {\frac {6 C (c+d \tan (e+f x))^{5/2}}{d}+5 (i A+B-i C) \left (-3 (c-i d)^{3/2} \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )+\sqrt {c+d \tan (e+f x)} (4 c-3 i d+d \tan (e+f x))\right )+5 (-i A+B+i C) \left (-3 (c+i d)^{3/2} \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )+\sqrt {c+d \tan (e+f x)} (4 c+3 i d+d \tan (e+f x))\right )}{15 f} \]

input

Integrate[(c + d*Tan[e + f*x])^(3/2)*(A + B*Tan[e + f*x] + C*Tan[e + f*x]^ 
2),x]
 

output

((6*C*(c + d*Tan[e + f*x])^(5/2))/d + 5*(I*A + B - I*C)*(-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])) + 5*((-I)*A + B + I*C)*(-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])))/(15*f)
 

3.1.100.3 Rubi [A] (warning: unable to verify)

Time = 0.99 (sec) , antiderivative size = 166, normalized size of antiderivative = 0.89, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.371, Rules used = {3042, 4113, 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.

\(\displaystyle \int (c+d \tan (e+f x))^{3/2} \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right ) \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int (c+d \tan (e+f x))^{3/2} \left (A+B \tan (e+f x)+C \tan (e+f x)^2\right )dx\)

\(\Big \downarrow \) 4113

\(\displaystyle \int (A-C+B \tan (e+f x)) (c+d \tan (e+f x))^{3/2}dx+\frac {2 C (c+d \tan (e+f x))^{5/2}}{5 d f}\)

\(\Big \downarrow \) 3042

\(\displaystyle \int (A-C+B \tan (e+f x)) (c+d \tan (e+f x))^{3/2}dx+\frac {2 C (c+d \tan (e+f x))^{5/2}}{5 d f}\)

\(\Big \downarrow \) 4011

\(\displaystyle \int \sqrt {c+d \tan (e+f x)} (A c-C c-B d+(B c+(A-C) d) \tan (e+f x))dx+\frac {2 B (c+d \tan (e+f x))^{3/2}}{3 f}+\frac {2 C (c+d \tan (e+f x))^{5/2}}{5 d f}\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \sqrt {c+d \tan (e+f x)} (A c-C c-B d+(B c+(A-C) d) \tan (e+f x))dx+\frac {2 B (c+d \tan (e+f x))^{3/2}}{3 f}+\frac {2 C (c+d \tan (e+f x))^{5/2}}{5 d f}\)

\(\Big \downarrow \) 4011

\(\displaystyle \int \frac {-C c^2-2 B d c+C d^2+A \left (c^2-d^2\right )+\left (2 c (A-C) d+B \left (c^2-d^2\right )\right ) \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx+\frac {2 (d (A-C)+B c) \sqrt {c+d \tan (e+f x)}}{f}+\frac {2 B (c+d \tan (e+f x))^{3/2}}{3 f}+\frac {2 C (c+d \tan (e+f x))^{5/2}}{5 d f}\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {-C c^2-2 B d c+C d^2+A \left (c^2-d^2\right )+\left (2 c (A-C) d+B \left (c^2-d^2\right )\right ) \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx+\frac {2 (d (A-C)+B c) \sqrt {c+d \tan (e+f x)}}{f}+\frac {2 B (c+d \tan (e+f x))^{3/2}}{3 f}+\frac {2 C (c+d \tan (e+f x))^{5/2}}{5 d f}\)

\(\Big \downarrow \) 4022

\(\displaystyle \frac {1}{2} (c+i d)^2 (A+i B-C) \int \frac {1-i \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx+\frac {1}{2} (c-i d)^2 (A-i B-C) \int \frac {i \tan (e+f x)+1}{\sqrt {c+d \tan (e+f x)}}dx+\frac {2 (d (A-C)+B c) \sqrt {c+d \tan (e+f x)}}{f}+\frac {2 B (c+d \tan (e+f x))^{3/2}}{3 f}+\frac {2 C (c+d \tan (e+f x))^{5/2}}{5 d f}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{2} (c+i d)^2 (A+i B-C) \int \frac {1-i \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx+\frac {1}{2} (c-i d)^2 (A-i B-C) \int \frac {i \tan (e+f x)+1}{\sqrt {c+d \tan (e+f x)}}dx+\frac {2 (d (A-C)+B c) \sqrt {c+d \tan (e+f x)}}{f}+\frac {2 B (c+d \tan (e+f x))^{3/2}}{3 f}+\frac {2 C (c+d \tan (e+f x))^{5/2}}{5 d f}\)

\(\Big \downarrow \) 4020

\(\displaystyle \frac {i (c-i d)^2 (A-i B-C) \int -\frac {1}{(1-i \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}d(i \tan (e+f x))}{2 f}-\frac {i (c+i d)^2 (A+i B-C) \int -\frac {1}{(i \tan (e+f x)+1) \sqrt {c+d \tan (e+f x)}}d(-i \tan (e+f x))}{2 f}+\frac {2 (d (A-C)+B c) \sqrt {c+d \tan (e+f x)}}{f}+\frac {2 B (c+d \tan (e+f x))^{3/2}}{3 f}+\frac {2 C (c+d \tan (e+f x))^{5/2}}{5 d f}\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {i (c-i d)^2 (A-i B-C) \int \frac {1}{(1-i \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}d(i \tan (e+f x))}{2 f}+\frac {i (c+i d)^2 (A+i B-C) \int \frac {1}{(i \tan (e+f x)+1) \sqrt {c+d \tan (e+f x)}}d(-i \tan (e+f x))}{2 f}+\frac {2 (d (A-C)+B c) \sqrt {c+d \tan (e+f x)}}{f}+\frac {2 B (c+d \tan (e+f x))^{3/2}}{3 f}+\frac {2 C (c+d \tan (e+f x))^{5/2}}{5 d f}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {(c+i d)^2 (A+i B-C) \int \frac {1}{-\frac {i \tan ^2(e+f x)}{d}-\frac {i c}{d}+1}d\sqrt {c+d \tan (e+f x)}}{d f}+\frac {(c-i d)^2 (A-i B-C) \int \frac {1}{\frac {i \tan ^2(e+f x)}{d}+\frac {i c}{d}+1}d\sqrt {c+d \tan (e+f x)}}{d f}+\frac {2 (d (A-C)+B c) \sqrt {c+d \tan (e+f x)}}{f}+\frac {2 B (c+d \tan (e+f x))^{3/2}}{3 f}+\frac {2 C (c+d \tan (e+f x))^{5/2}}{5 d f}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {(c-i d)^{3/2} (A-i B-C) \arctan \left (\frac {\tan (e+f x)}{\sqrt {c-i d}}\right )}{f}+\frac {(c+i d)^{3/2} (A+i B-C) \arctan \left (\frac {\tan (e+f x)}{\sqrt {c+i d}}\right )}{f}+\frac {2 (d (A-C)+B c) \sqrt {c+d \tan (e+f x)}}{f}+\frac {2 B (c+d \tan (e+f x))^{3/2}}{3 f}+\frac {2 C (c+d \tan (e+f x))^{5/2}}{5 d f}\)

input

Int[(c + d*Tan[e + f*x])^(3/2)*(A + B*Tan[e + f*x] + C*Tan[e + f*x]^2),x]
 

output

((A - I*B - C)*(c - I*d)^(3/2)*ArcTan[Tan[e + f*x]/Sqrt[c - I*d]])/f + ((A 
 + I*B - C)*(c + I*d)^(3/2)*ArcTan[Tan[e + f*x]/Sqrt[c + I*d]])/f + (2*(B* 
c + (A - C)*d)*Sqrt[c + d*Tan[e + f*x]])/f + (2*B*(c + d*Tan[e + f*x])^(3/ 
2))/(3*f) + (2*C*(c + d*Tan[e + f*x])^(5/2))/(5*d*f)
 

3.1.100.3.1 Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 73

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]
 

rule 221

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]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4011

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 4020

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]
 

rule 4022

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]
 

rule 4113

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]
 

3.1.100.4 Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(2499\) vs. \(2(158)=316\).

Time = 0.14 (sec) , antiderivative size = 2500, normalized size of antiderivative = 13.37

method	result	size
parts	\(\text {Expression too large to display}\)	\(2500\)
derivativedivides	\(\text {Expression too large to display}\)	\(2517\)
default	\(\text {Expression too large to display}\)	\(2517\)

input

int((c+d*tan(f*x+e))^(3/2)*(A+B*tan(f*x+e)+C*tan(f*x+e)^2),x,method=_RETUR 
NVERBOSE)
 

output

A*(2/f*d*(c+d*tan(f*x+e))^(1/2)+1/4/f/d*ln((c+d*tan(f*x+e))^(1/2)*(2*(c^2+ 
d^2)^(1/2)+2*c)^(1/2)-d*tan(f*x+e)-c-(c^2+d^2)^(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((c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2) 
^(1/2)+2*c)^(1/2)-d*tan(f*x+e)-c-(c^2+d^2)^(1/2))*(2*(c^2+d^2)^(1/2)+2*c)^ 
(1/2)*c^2+1/4/f*d*ln((c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)- 
d*tan(f*x+e)-c-(c^2+d^2)^(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^2+d^2)^(1/2)+2*c)^(1/2)-2*(c+d*tan(f 
*x+e))^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))*c+1/f*d/(2*(c^2+d^2)^(1/2)-2* 
c)^(1/2)*arctan(((2*(c^2+d^2)^(1/2)+2*c)^(1/2)-2*(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)-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^2+d^2)^(1/2)*c+1/4/f/d*ln(d*tan(f*x+e)+c+(c+d*t 
an(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^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 
)+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-1/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))+B*(2/3/f*(c+d 
*tan(f*x+e))^(3/2)+2/f*(c+d*tan(f*x+e))^(1/2)*c-1/4/f*ln((c+d*tan(f*x+e...
 

3.1.100.5 Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 6846 vs. \(2 (151) = 302\).

Time = 0.95 (sec) , antiderivative size = 6846, normalized size of antiderivative = 36.61 \[ \int (c+d \tan (e+f x))^{3/2} \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right ) \, dx=\text {Too large to display} \]

input

integrate((c+d*tan(f*x+e))^(3/2)*(A+B*tan(f*x+e)+C*tan(f*x+e)^2),x, algori 
thm="fricas")
 

output

Too large to include
 

3.1.100.6 Sympy [F]

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

input

integrate((c+d*tan(f*x+e))**(3/2)*(A+B*tan(f*x+e)+C*tan(f*x+e)**2),x)
 

output

Integral((c + d*tan(e + f*x))**(3/2)*(A + B*tan(e + f*x) + C*tan(e + f*x)* 
*2), x)
 

3.1.100.7 Maxima [F]

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

input

integrate((c+d*tan(f*x+e))^(3/2)*(A+B*tan(f*x+e)+C*tan(f*x+e)^2),x, algori 
thm="maxima")
 

output

integrate((C*tan(f*x + e)^2 + B*tan(f*x + e) + A)*(d*tan(f*x + e) + c)^(3/ 
2), x)
 

3.1.100.8 Giac [F(-1)]

Timed out. \[ \int (c+d \tan (e+f x))^{3/2} \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right ) \, dx=\text {Timed out} \]

input

integrate((c+d*tan(f*x+e))^(3/2)*(A+B*tan(f*x+e)+C*tan(f*x+e)^2),x, algori 
thm="giac")
 

output

Timed out
 

3.1.100.9 Mupad [B] (verification not implemented)

Time = 42.33 (sec) , antiderivative size = 4260, normalized size of antiderivative = 22.78 \[ \int (c+d \tan (e+f x))^{3/2} \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right ) \, dx=\text {Too large to display} \]

input

int((c + d*tan(e + f*x))^(3/2)*(A + B*tan(e + f*x) + C*tan(e + f*x)^2),x)
 

output

((2*C*c^2)/(d*f) - (2*C*(d^3*f + c^2*d*f))/(d^2*f^2))*(c + d*tan(e + f*x)) 
^(1/2) - log((((16*c*d^2*(((-B^4*d^2*f^4*(3*c^2 - d^2)^2)^(1/2) + B^2*c^3* 
f^2 - 3*B^2*c*d^2*f^2)/f^4)^(1/2)*(B*c^2 + B*d^2 + f*(((-B^4*d^2*f^4*(3*c^ 
2 - d^2)^2)^(1/2) + B^2*c^3*f^2 - 3*B^2*c*d^2*f^2)/f^4)^(1/2)*(c + d*tan(e 
 + f*x))^(1/2)))/f - (16*B^2*d^2*(c + d*tan(e + f*x))^(1/2)*(c^4 + d^4 - 6 
*c^2*d^2))/f^2)*(((-B^4*d^2*f^4*(3*c^2 - d^2)^2)^(1/2) + B^2*c^3*f^2 - 3*B 
^2*c*d^2*f^2)/f^4)^(1/2))/2 - (8*B^3*d^2*(c^2 - d^2)*(c^2 + d^2)^2)/f^3)*( 
((6*B^4*c^2*d^4*f^4 - B^4*d^6*f^4 - 9*B^4*c^4*d^2*f^4)^(1/2) + B^2*c^3*f^2 
 - 3*B^2*c*d^2*f^2)/(4*f^4))^(1/2) - log((((16*c*d^2*(-((-B^4*d^2*f^4*(3*c 
^2 - d^2)^2)^(1/2) - B^2*c^3*f^2 + 3*B^2*c*d^2*f^2)/f^4)^(1/2)*(B*c^2 + B* 
d^2 + f*(-((-B^4*d^2*f^4*(3*c^2 - d^2)^2)^(1/2) - B^2*c^3*f^2 + 3*B^2*c*d^ 
2*f^2)/f^4)^(1/2)*(c + d*tan(e + f*x))^(1/2)))/f - (16*B^2*d^2*(c + d*tan( 
e + f*x))^(1/2)*(c^4 + d^4 - 6*c^2*d^2))/f^2)*(-((-B^4*d^2*f^4*(3*c^2 - d^ 
2)^2)^(1/2) - B^2*c^3*f^2 + 3*B^2*c*d^2*f^2)/f^4)^(1/2))/2 - (8*B^3*d^2*(c 
^2 - d^2)*(c^2 + d^2)^2)/f^3)*(-((6*B^4*c^2*d^4*f^4 - B^4*d^6*f^4 - 9*B^4* 
c^4*d^2*f^4)^(1/2) - B^2*c^3*f^2 + 3*B^2*c*d^2*f^2)/(4*f^4))^(1/2) + log(( 
((16*c*d^2*(((-B^4*d^2*f^4*(3*c^2 - d^2)^2)^(1/2) + B^2*c^3*f^2 - 3*B^2*c* 
d^2*f^2)/f^4)^(1/2)*(B*c^2 + B*d^2 - f*(((-B^4*d^2*f^4*(3*c^2 - d^2)^2)^(1 
/2) + B^2*c^3*f^2 - 3*B^2*c*d^2*f^2)/f^4)^(1/2)*(c + d*tan(e + f*x))^(1/2) 
))/f + (16*B^2*d^2*(c + d*tan(e + f*x))^(1/2)*(c^4 + d^4 - 6*c^2*d^2))/...