\(\int \frac {\sqrt {a+b \tan (e+f x)} (A+B \tan (e+f x)+C \tan ^2(e+f x))}{(c+d \tan (e+f x))^{3/2}} \, dx\) [155]

Optimal result

Integrand size = 49, antiderivative size = 299 \[ \int \frac {\sqrt {a+b \tan (e+f x)} \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right )}{(c+d \tan (e+f x))^{3/2}} \, dx=-\frac {\sqrt {a-i b} (i A+B-i C) \text {arctanh}\left (\frac {\sqrt {c-i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a-i b} \sqrt {c+d \tan (e+f x)}}\right )}{(c-i d)^{3/2} f}-\frac {\sqrt {a+i b} (B-i (A-C)) \text {arctanh}\left (\frac {\sqrt {c+i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a+i b} \sqrt {c+d \tan (e+f x)}}\right )}{(c+i d)^{3/2} f}+\frac {2 \sqrt {b} C \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b \tan (e+f x)}}{\sqrt {b} \sqrt {c+d \tan (e+f x)}}\right )}{d^{3/2} f}-\frac {2 \left (c^2 C-B c d+A d^2\right ) \sqrt {a+b \tan (e+f x)}}{d \left (c^2+d^2\right ) f \sqrt {c+d \tan (e+f x)}} \] Output:

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

Mathematica [A] (verified)

Time = 3.50 (sec) , antiderivative size = 403, normalized size of antiderivative = 1.35 \[ \int \frac {\sqrt {a+b \tan (e+f x)} \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right )}{(c+d \tan (e+f x))^{3/2}} \, dx=\frac {\frac {\sqrt {-a+i b} (i A+B-i C) \text {arctanh}\left (\frac {\sqrt {-c+i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {-a+i b} \sqrt {c+d \tan (e+f x)}}\right )}{(-c+i d)^{3/2}}+\frac {i \sqrt {a+i b} (A+i B-C) \text {arctanh}\left (\frac {\sqrt {c+i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a+i b} \sqrt {c+d \tan (e+f x)}}\right )}{(c+i d)^{3/2}}+\frac {(B+i (A-C)) \sqrt {a+b \tan (e+f x)}}{(c-i d) \sqrt {c+d \tan (e+f x)}}+\frac {(-i A+B+i C) \sqrt {a+b \tan (e+f x)}}{(c+i d) \sqrt {c+d \tan (e+f x)}}+\frac {2 C \left (-\sqrt {d} \sqrt {a+b \tan (e+f x)}+\sqrt {b c-a d} \text {arcsinh}\left (\frac {\sqrt {d} \sqrt {a+b \tan (e+f x)}}{\sqrt {b c-a d}}\right ) \sqrt {\frac {b (c+d \tan (e+f x))}{b c-a d}}\right )}{d^{3/2} \sqrt {c+d \tan (e+f x)}}}{f} \] Input:

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

Output:

((Sqrt[-a + I*b]*(I*A + B - I*C)*ArcTanh[(Sqrt[-c + I*d]*Sqrt[a + b*Tan[e 
+ f*x]])/(Sqrt[-a + I*b]*Sqrt[c + d*Tan[e + f*x]])])/(-c + I*d)^(3/2) + (I 
*Sqrt[a + I*b]*(A + I*B - C)*ArcTanh[(Sqrt[c + I*d]*Sqrt[a + b*Tan[e + f*x 
]])/(Sqrt[a + I*b]*Sqrt[c + d*Tan[e + f*x]])])/(c + I*d)^(3/2) + ((B + I*( 
A - C))*Sqrt[a + b*Tan[e + f*x]])/((c - I*d)*Sqrt[c + d*Tan[e + f*x]]) + ( 
((-I)*A + B + I*C)*Sqrt[a + b*Tan[e + f*x]])/((c + I*d)*Sqrt[c + d*Tan[e + 
 f*x]]) + (2*C*(-(Sqrt[d]*Sqrt[a + b*Tan[e + f*x]]) + Sqrt[b*c - a*d]*ArcS 
inh[(Sqrt[d]*Sqrt[a + b*Tan[e + f*x]])/Sqrt[b*c - a*d]]*Sqrt[(b*(c + d*Tan 
[e + f*x]))/(b*c - a*d)]))/(d^(3/2)*Sqrt[c + d*Tan[e + f*x]]))/f
 

Rubi [A] (verified)

Time = 2.46 (sec) , antiderivative size = 328, normalized size of antiderivative = 1.10, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3042, 4128, 27, 3042, 4138, 2348, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4128

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4138

\(\displaystyle \frac {\int \frac {b C \left (c^2+d^2\right ) \tan ^2(e+f x)+d ((A-C) (b c-a d)+B (a c+b d)) \tan (e+f x)+A d (a c+b d)+(b c-a d) (c C-B d)}{\sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)} \left (\tan ^2(e+f x)+1\right )}d\tan (e+f x)}{d f \left (c^2+d^2\right )}-\frac {2 \left (A d^2-B c d+c^2 C\right ) \sqrt {a+b \tan (e+f x)}}{d f \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}\)

\(\Big \downarrow \) 2348

\(\displaystyle -\frac {2 \left (A d^2-B c d+c^2 C\right ) \sqrt {a+b \tan (e+f x)}}{d f \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}+\frac {\int \left (\frac {b C \left (c^2+d^2\right )}{\sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}+\frac {a A d^2-b B d^2-a C d^2-A b c d-a B c d+b c C d+i \left (A b d^2+a B d^2-b C d^2+a A c d-b B c d-a c C d\right )}{2 (i-\tan (e+f x)) \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}+\frac {-a A d^2+b B d^2+a C d^2+A b c d+a B c d-b c C d+i \left (A b d^2+a B d^2-b C d^2+a A c d-b B c d-a c C d\right )}{2 (\tan (e+f x)+i) \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}\right )d\tan (e+f x)}{d f \left (c^2+d^2\right )}\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {2 \left (A d^2-B c d+c^2 C\right ) \sqrt {a+b \tan (e+f x)}}{d f \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}+\frac {-\frac {d \sqrt {a-i b} (c+i d) (i A+B-i C) \text {arctanh}\left (\frac {\sqrt {c-i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a-i b} \sqrt {c+d \tan (e+f x)}}\right )}{\sqrt {c-i d}}+\frac {d \sqrt {a+i b} (d+i c) (A+i B-C) \text {arctanh}\left (\frac {\sqrt {c+i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a+i b} \sqrt {c+d \tan (e+f x)}}\right )}{\sqrt {c+i d}}+\frac {2 \sqrt {b} C \left (c^2+d^2\right ) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b \tan (e+f x)}}{\sqrt {b} \sqrt {c+d \tan (e+f x)}}\right )}{\sqrt {d}}}{d f \left (c^2+d^2\right )}\)

Input:

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

Output:

(-((Sqrt[a - I*b]*(I*A + B - I*C)*(c + I*d)*d*ArcTanh[(Sqrt[c - I*d]*Sqrt[ 
a + b*Tan[e + f*x]])/(Sqrt[a - I*b]*Sqrt[c + d*Tan[e + f*x]])])/Sqrt[c - I 
*d]) + (Sqrt[a + I*b]*(A + I*B - C)*d*(I*c + d)*ArcTanh[(Sqrt[c + I*d]*Sqr 
t[a + b*Tan[e + f*x]])/(Sqrt[a + I*b]*Sqrt[c + d*Tan[e + f*x]])])/Sqrt[c + 
 I*d] + (2*Sqrt[b]*C*(c^2 + d^2)*ArcTanh[(Sqrt[d]*Sqrt[a + b*Tan[e + f*x]] 
)/(Sqrt[b]*Sqrt[c + d*Tan[e + f*x]])])/Sqrt[d])/(d*(c^2 + d^2)*f) - (2*(c^ 
2*C - B*c*d + A*d^2)*Sqrt[a + b*Tan[e + f*x]])/(d*(c^2 + d^2)*f*Sqrt[c + d 
*Tan[e + f*x]])
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2348

Int[(Px_)*((c_) + (d_.)*(x_))^(m_.)*((e_) + (f_.)*(x_))^(n_.)*((a_.) + (b_. 
)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[Px*(c + d*x)^m*(e + f*x)^ 
n*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && PolyQ[P 
x, x] && (IntegerQ[p] || (IntegerQ[2*p] && IntegerQ[m] && ILtQ[n, 0])) && 
!(IGtQ[m, 0] && IGtQ[n, 0])
 

rule 3042

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

rule 4128

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

rule 4138

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + 
 (f_.)*(x_)])^(n_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) 
 + (f_.)*(x_)]^2), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, S 
imp[ff/f   Subst[Int[(a + b*ff*x)^m*(c + d*ff*x)^n*((A + B*ff*x + C*ff^2*x^ 
2)/(1 + ff^2*x^2)), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, c, d, e, f 
, A, B, C, m, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + 
 d^2, 0]
 

Maple [F(-1)]

Timed out.

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

Input:

int((a+b*tan(f*x+e))^(1/2)*(A+B*tan(f*x+e)+C*tan(f*x+e)^2)/(c+d*tan(f*x+e) 
)^(3/2),x)
 

Output:

int((a+b*tan(f*x+e))^(1/2)*(A+B*tan(f*x+e)+C*tan(f*x+e)^2)/(c+d*tan(f*x+e) 
)^(3/2),x)
 

Fricas [F(-1)]

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

integrate((a+b*tan(f*x+e))^(1/2)*(A+B*tan(f*x+e)+C*tan(f*x+e)^2)/(c+d*tan( 
f*x+e))^(3/2),x, algorithm="fricas")
 

Output:

Timed out
 

Sympy [F]

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

integrate((a+b*tan(f*x+e))**(1/2)*(A+B*tan(f*x+e)+C*tan(f*x+e)**2)/(c+d*ta 
n(f*x+e))**(3/2),x)
 

Output:

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

Maxima [F(-1)]

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

integrate((a+b*tan(f*x+e))^(1/2)*(A+B*tan(f*x+e)+C*tan(f*x+e)^2)/(c+d*tan( 
f*x+e))^(3/2),x, algorithm="maxima")
 

Output:

Timed out
 

Giac [F]

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

integrate((a+b*tan(f*x+e))^(1/2)*(A+B*tan(f*x+e)+C*tan(f*x+e)^2)/(c+d*tan( 
f*x+e))^(3/2),x, algorithm="giac")
 

Output:

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

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt {a+b \tan (e+f x)} \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right )}{(c+d \tan (e+f x))^{3/2}} \, dx=\int \frac {\sqrt {a+b\,\mathrm {tan}\left (e+f\,x\right )}\,\left (C\,{\mathrm {tan}\left (e+f\,x\right )}^2+B\,\mathrm {tan}\left (e+f\,x\right )+A\right )}{{\left (c+d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}} \,d x \] Input:

int(((a + b*tan(e + f*x))^(1/2)*(A + B*tan(e + f*x) + C*tan(e + f*x)^2))/( 
c + d*tan(e + f*x))^(3/2),x)
 

Output:

int(((a + b*tan(e + f*x))^(1/2)*(A + B*tan(e + f*x) + C*tan(e + f*x)^2))/( 
c + d*tan(e + f*x))^(3/2), x)
 

Reduce [F]

\[ \int \frac {\sqrt {a+b \tan (e+f x)} \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right )}{(c+d \tan (e+f x))^{3/2}} \, dx =\text {Too large to display} \] Input:

int((a+b*tan(f*x+e))^(1/2)*(A+B*tan(f*x+e)+C*tan(f*x+e)^2)/(c+d*tan(f*x+e) 
)^(3/2),x)
 

Output:

( - 2*sqrt(tan(e + f*x)*d + c)*sqrt(tan(e + f*x)*b + a)*b**2 + int((sqrt(t 
an(e + f*x)*d + c)*sqrt(tan(e + f*x)*b + a)*tan(e + f*x)**2)/(tan(e + f*x) 
**2*d**2 + 2*tan(e + f*x)*c*d + c**2),x)*tan(e + f*x)*a*c*d**2*f - int((sq 
rt(tan(e + f*x)*d + c)*sqrt(tan(e + f*x)*b + a)*tan(e + f*x)**2)/(tan(e + 
f*x)**2*d**2 + 2*tan(e + f*x)*c*d + c**2),x)*tan(e + f*x)*b*c**2*d*f + int 
((sqrt(tan(e + f*x)*d + c)*sqrt(tan(e + f*x)*b + a)*tan(e + f*x)**2)/(tan( 
e + f*x)**2*d**2 + 2*tan(e + f*x)*c*d + c**2),x)*a*c**2*d*f - int((sqrt(ta 
n(e + f*x)*d + c)*sqrt(tan(e + f*x)*b + a)*tan(e + f*x)**2)/(tan(e + f*x)* 
*2*d**2 + 2*tan(e + f*x)*c*d + c**2),x)*b*c**3*f + int((sqrt(tan(e + f*x)* 
d + c)*sqrt(tan(e + f*x)*b + a)*tan(e + f*x))/(tan(e + f*x)**3*b*d**2 + ta 
n(e + f*x)**2*a*d**2 + 2*tan(e + f*x)**2*b*c*d + 2*tan(e + f*x)*a*c*d + ta 
n(e + f*x)*b*c**2 + a*c**2),x)*tan(e + f*x)*a**2*b*d**2*f - int((sqrt(tan( 
e + f*x)*d + c)*sqrt(tan(e + f*x)*b + a)*tan(e + f*x))/(tan(e + f*x)**3*b* 
d**2 + tan(e + f*x)**2*a*d**2 + 2*tan(e + f*x)**2*b*c*d + 2*tan(e + f*x)*a 
*c*d + tan(e + f*x)*b*c**2 + a*c**2),x)*tan(e + f*x)*a*b**2*c*d*f + int((s 
qrt(tan(e + f*x)*d + c)*sqrt(tan(e + f*x)*b + a)*tan(e + f*x))/(tan(e + f* 
x)**3*b*d**2 + tan(e + f*x)**2*a*d**2 + 2*tan(e + f*x)**2*b*c*d + 2*tan(e 
+ f*x)*a*c*d + tan(e + f*x)*b*c**2 + a*c**2),x)*a**2*b*c*d*f - int((sqrt(t 
an(e + f*x)*d + c)*sqrt(tan(e + f*x)*b + a)*tan(e + f*x))/(tan(e + f*x)**3 
*b*d**2 + tan(e + f*x)**2*a*d**2 + 2*tan(e + f*x)**2*b*c*d + 2*tan(e + ...