\(\int \frac {\tan (c+d x) (A+B \tan (c+d x))}{\sqrt {a+b \tan (c+d x)}} \, dx\) [345]

Optimal result

Integrand size = 31, antiderivative size = 124 \[ \int \frac {\tan (c+d x) (A+B \tan (c+d x))}{\sqrt {a+b \tan (c+d x)}} \, dx=-\frac {(A-i B) \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-i b}}\right )}{\sqrt {a-i b} d}-\frac {(A+i B) \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )}{\sqrt {a+i b} d}+\frac {2 B \sqrt {a+b \tan (c+d x)}}{b d} \] Output:

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

Mathematica [A] (verified)

Time = 0.37 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.95 \[ \int \frac {\tan (c+d x) (A+B \tan (c+d x))}{\sqrt {a+b \tan (c+d x)}} \, dx=-\frac {\frac {(A-i B) \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-i b}}\right )}{\sqrt {a-i b}}+\frac {(A+i B) \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )}{\sqrt {a+i b}}-\frac {2 B \sqrt {a+b \tan (c+d x)}}{b}}{d} \] Input:

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

Output:

-((((A - I*B)*ArcTanh[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a - I*b]])/Sqrt[a - I* 
b] + ((A + I*B)*ArcTanh[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a + I*b]])/Sqrt[a + 
I*b] - (2*B*Sqrt[a + b*Tan[c + d*x]])/b)/d)
 

Rubi [A] (warning: unable to verify)

Time = 0.57 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.88, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.290, Rules used = {3042, 4075, 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.

\(\displaystyle \int \frac {\tan (c+d x) (A+B \tan (c+d x))}{\sqrt {a+b \tan (c+d x)}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\tan (c+d x) (A+B \tan (c+d x))}{\sqrt {a+b \tan (c+d x)}}dx\)

\(\Big \downarrow \) 4075

\(\displaystyle \int \frac {A \tan (c+d x)-B}{\sqrt {a+b \tan (c+d x)}}dx+\frac {2 B \sqrt {a+b \tan (c+d x)}}{b d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {A \tan (c+d x)-B}{\sqrt {a+b \tan (c+d x)}}dx+\frac {2 B \sqrt {a+b \tan (c+d x)}}{b d}\)

\(\Big \downarrow \) 4022

\(\displaystyle \frac {1}{2} (-B+i A) \int \frac {1-i \tan (c+d x)}{\sqrt {a+b \tan (c+d x)}}dx-\frac {1}{2} (B+i A) \int \frac {i \tan (c+d x)+1}{\sqrt {a+b \tan (c+d x)}}dx+\frac {2 B \sqrt {a+b \tan (c+d x)}}{b d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{2} (-B+i A) \int \frac {1-i \tan (c+d x)}{\sqrt {a+b \tan (c+d x)}}dx-\frac {1}{2} (B+i A) \int \frac {i \tan (c+d x)+1}{\sqrt {a+b \tan (c+d x)}}dx+\frac {2 B \sqrt {a+b \tan (c+d x)}}{b d}\)

\(\Big \downarrow \) 4020

\(\displaystyle -\frac {i (B+i A) \int -\frac {1}{(1-i \tan (c+d x)) \sqrt {a+b \tan (c+d x)}}d(i \tan (c+d x))}{2 d}-\frac {i (-B+i A) \int -\frac {1}{(i \tan (c+d x)+1) \sqrt {a+b \tan (c+d x)}}d(-i \tan (c+d x))}{2 d}+\frac {2 B \sqrt {a+b \tan (c+d x)}}{b d}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {i (B+i A) \int \frac {1}{(1-i \tan (c+d x)) \sqrt {a+b \tan (c+d x)}}d(i \tan (c+d x))}{2 d}+\frac {i (-B+i A) \int \frac {1}{(i \tan (c+d x)+1) \sqrt {a+b \tan (c+d x)}}d(-i \tan (c+d x))}{2 d}+\frac {2 B \sqrt {a+b \tan (c+d x)}}{b d}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {(-B+i A) \int \frac {1}{-\frac {i \tan ^2(c+d x)}{b}-\frac {i a}{b}+1}d\sqrt {a+b \tan (c+d x)}}{b d}-\frac {(B+i A) \int \frac {1}{\frac {i \tan ^2(c+d x)}{b}+\frac {i a}{b}+1}d\sqrt {a+b \tan (c+d x)}}{b d}+\frac {2 B \sqrt {a+b \tan (c+d x)}}{b d}\)

\(\Big \downarrow \) 221

\(\displaystyle -\frac {(B+i A) \arctan \left (\frac {\tan (c+d x)}{\sqrt {a-i b}}\right )}{d \sqrt {a-i b}}+\frac {(-B+i A) \arctan \left (\frac {\tan (c+d x)}{\sqrt {a+i b}}\right )}{d \sqrt {a+i b}}+\frac {2 B \sqrt {a+b \tan (c+d x)}}{b d}\)

Input:

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

Output:

-(((I*A + B)*ArcTan[Tan[c + d*x]/Sqrt[a - I*b]])/(Sqrt[a - I*b]*d)) + ((I* 
A - B)*ArcTan[Tan[c + d*x]/Sqrt[a + I*b]])/(Sqrt[a + I*b]*d) + (2*B*Sqrt[a 
 + b*Tan[c + d*x]])/(b*d)
 

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 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 4075

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1918\) vs. \(2(104)=208\).

Time = 0.13 (sec) , antiderivative size = 1919, normalized size of antiderivative = 15.48

method	result	size
parts	\(\text {Expression too large to display}\)	\(1919\)
derivativedivides	\(\text {Expression too large to display}\)	\(3997\)
default	\(\text {Expression too large to display}\)	\(3997\)

Input:

int(tan(d*x+c)*(A+B*tan(d*x+c))/(a+b*tan(d*x+c))^(1/2),x,method=_RETURNVER 
BOSE)
 

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1706 vs. \(2 (98) = 196\).

Time = 0.12 (sec) , antiderivative size = 1706, normalized size of antiderivative = 13.76 \[ \int \frac {\tan (c+d x) (A+B \tan (c+d x))}{\sqrt {a+b \tan (c+d x)}} \, dx=\text {Too large to display} \] Input:

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

Output:

1/2*(b*d*sqrt(((a^2 + b^2)*d^2*sqrt(-(4*A^2*B^2*a^2 - 4*(A^3*B - A*B^3)*a* 
b + (A^4 - 2*A^2*B^2 + B^4)*b^2)/((a^4 + 2*a^2*b^2 + b^4)*d^4)) + 2*A*B*b 
+ (A^2 - B^2)*a)/((a^2 + b^2)*d^2))*log((2*(A^3*B + A*B^3)*a - (A^4 - B^4) 
*b)*sqrt(b*tan(d*x + c) + a) + ((B*a^3 - A*a^2*b + B*a*b^2 - A*b^3)*d^3*sq 
rt(-(4*A^2*B^2*a^2 - 4*(A^3*B - A*B^3)*a*b + (A^4 - 2*A^2*B^2 + B^4)*b^2)/ 
((a^4 + 2*a^2*b^2 + b^4)*d^4)) - (2*A^2*B*a^2 - (A^3 - 3*A*B^2)*a*b - (A^2 
*B - B^3)*b^2)*d)*sqrt(((a^2 + b^2)*d^2*sqrt(-(4*A^2*B^2*a^2 - 4*(A^3*B - 
A*B^3)*a*b + (A^4 - 2*A^2*B^2 + B^4)*b^2)/((a^4 + 2*a^2*b^2 + b^4)*d^4)) + 
 2*A*B*b + (A^2 - B^2)*a)/((a^2 + b^2)*d^2))) - b*d*sqrt(((a^2 + b^2)*d^2* 
sqrt(-(4*A^2*B^2*a^2 - 4*(A^3*B - A*B^3)*a*b + (A^4 - 2*A^2*B^2 + B^4)*b^2 
)/((a^4 + 2*a^2*b^2 + b^4)*d^4)) + 2*A*B*b + (A^2 - B^2)*a)/((a^2 + b^2)*d 
^2))*log((2*(A^3*B + A*B^3)*a - (A^4 - B^4)*b)*sqrt(b*tan(d*x + c) + a) - 
((B*a^3 - A*a^2*b + B*a*b^2 - A*b^3)*d^3*sqrt(-(4*A^2*B^2*a^2 - 4*(A^3*B - 
 A*B^3)*a*b + (A^4 - 2*A^2*B^2 + B^4)*b^2)/((a^4 + 2*a^2*b^2 + b^4)*d^4)) 
- (2*A^2*B*a^2 - (A^3 - 3*A*B^2)*a*b - (A^2*B - B^3)*b^2)*d)*sqrt(((a^2 + 
b^2)*d^2*sqrt(-(4*A^2*B^2*a^2 - 4*(A^3*B - A*B^3)*a*b + (A^4 - 2*A^2*B^2 + 
 B^4)*b^2)/((a^4 + 2*a^2*b^2 + b^4)*d^4)) + 2*A*B*b + (A^2 - B^2)*a)/((a^2 
 + b^2)*d^2))) - b*d*sqrt(-((a^2 + b^2)*d^2*sqrt(-(4*A^2*B^2*a^2 - 4*(A^3* 
B - A*B^3)*a*b + (A^4 - 2*A^2*B^2 + B^4)*b^2)/((a^4 + 2*a^2*b^2 + b^4)*d^4 
)) - 2*A*B*b - (A^2 - B^2)*a)/((a^2 + b^2)*d^2))*log((2*(A^3*B + A*B^3)...
 

Sympy [F]

\[ \int \frac {\tan (c+d x) (A+B \tan (c+d x))}{\sqrt {a+b \tan (c+d x)}} \, dx=\int \frac {\left (A + B \tan {\left (c + d x \right )}\right ) \tan {\left (c + d x \right )}}{\sqrt {a + b \tan {\left (c + d x \right )}}}\, dx \] Input:

integrate(tan(d*x+c)*(A+B*tan(d*x+c))/(a+b*tan(d*x+c))**(1/2),x)
 

Output:

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

Maxima [F]

\[ \int \frac {\tan (c+d x) (A+B \tan (c+d x))}{\sqrt {a+b \tan (c+d x)}} \, dx=\int { \frac {{\left (B \tan \left (d x + c\right ) + A\right )} \tan \left (d x + c\right )}{\sqrt {b \tan \left (d x + c\right ) + a}} \,d x } \] Input:

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

Output:

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

Giac [F(-2)]

Exception generated. \[ \int \frac {\tan (c+d x) (A+B \tan (c+d x))}{\sqrt {a+b \tan (c+d x)}} \, dx=\text {Exception raised: TypeError} \] Input:

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

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Unable to divide, perhaps due to ro 
unding error%%%{%%%{1,[0,9,3]%%%}+%%%{4,[0,7,3]%%%}+%%%{6,[0,5,3]%%%}+%%%{ 
4,[0,3,3]
 

Mupad [B] (verification not implemented)

Time = 6.33 (sec) , antiderivative size = 2930, normalized size of antiderivative = 23.63 \[ \int \frac {\tan (c+d x) (A+B \tan (c+d x))}{\sqrt {a+b \tan (c+d x)}} \, dx=\text {Too large to display} \] Input:

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

Output:

2*atanh((32*B^2*b^2*((-16*B^4*b^2*d^4)^(1/2)/(16*(a^2*d^4 + b^2*d^4)) - (B 
^2*a*d^2)/(4*(a^2*d^4 + b^2*d^4)))^(1/2)*(a + b*tan(c + d*x))^(1/2))/((16* 
B^3*a*b^3*d^3)/(a^2*d^4 + b^2*d^4) - (4*B*b^3*d^2*(-16*B^4*b^2*d^4)^(1/2)) 
/(a^2*d^5 + b^2*d^5)) + (8*a*b^2*((-16*B^4*b^2*d^4)^(1/2)/(16*(a^2*d^4 + b 
^2*d^4)) - (B^2*a*d^2)/(4*(a^2*d^4 + b^2*d^4)))^(1/2)*(a + b*tan(c + d*x)) 
^(1/2)*(-16*B^4*b^2*d^4)^(1/2))/((16*B^3*a*b^5*d^5)/(a^2*d^4 + b^2*d^4) - 
(4*B*b^5*d^4*(-16*B^4*b^2*d^4)^(1/2))/(a^2*d^5 + b^2*d^5) + (16*B^3*a^3*b^ 
3*d^5)/(a^2*d^4 + b^2*d^4) - (4*B*a^2*b^3*d^4*(-16*B^4*b^2*d^4)^(1/2))/(a^ 
2*d^5 + b^2*d^5)) - (32*B^2*a^2*b^2*d^2*((-16*B^4*b^2*d^4)^(1/2)/(16*(a^2* 
d^4 + b^2*d^4)) - (B^2*a*d^2)/(4*(a^2*d^4 + b^2*d^4)))^(1/2)*(a + b*tan(c 
+ d*x))^(1/2))/((16*B^3*a*b^5*d^5)/(a^2*d^4 + b^2*d^4) - (4*B*b^5*d^4*(-16 
*B^4*b^2*d^4)^(1/2))/(a^2*d^5 + b^2*d^5) + (16*B^3*a^3*b^3*d^5)/(a^2*d^4 + 
 b^2*d^4) - (4*B*a^2*b^3*d^4*(-16*B^4*b^2*d^4)^(1/2))/(a^2*d^5 + b^2*d^5)) 
)*((-16*B^4*b^2*d^4)^(1/2)/(16*(a^2*d^4 + b^2*d^4)) - (B^2*a*d^2)/(4*(a^2* 
d^4 + b^2*d^4)))^(1/2) - 2*atanh((8*a*b^2*((-16*A^4*b^2*d^4)^(1/2)/(16*(a^ 
2*d^4 + b^2*d^4)) + (A^2*a*d^2)/(4*(a^2*d^4 + b^2*d^4)))^(1/2)*(a + b*tan( 
c + d*x))^(1/2)*(-16*A^4*b^2*d^4)^(1/2))/((16*A^3*a^2*b^4*d^5)/(a^2*d^4 + 
b^2*d^4) - 16*A^3*a^2*b^2*d - 16*A^3*b^4*d + (16*A^3*a^4*b^2*d^5)/(a^2*d^4 
 + b^2*d^4) + (4*A*a^3*b^2*d^4*(-16*A^4*b^2*d^4)^(1/2))/(a^2*d^5 + b^2*d^5 
) + (4*A*a*b^4*d^4*(-16*A^4*b^2*d^4)^(1/2))/(a^2*d^5 + b^2*d^5)) - (32*...
 

Reduce [F]

\[ \int \frac {\tan (c+d x) (A+B \tan (c+d x))}{\sqrt {a+b \tan (c+d x)}} \, dx=\int \sqrt {a +\tan \left (d x +c \right ) b}\, \tan \left (d x +c \right )d x \] Input:

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

Output:

int(sqrt(tan(c + d*x)*b + a)*tan(c + d*x),x)