\(\int \frac {\tan (c+d x)}{(a+b \tan (c+d x))^2} \, dx\) [472]

Optimal result

Integrand size = 19, antiderivative size = 82 \[ \int \frac {\tan (c+d x)}{(a+b \tan (c+d x))^2} \, dx=\frac {2 a b x}{\left (a^2+b^2\right )^2}-\frac {\left (a^2-b^2\right ) \log (a \cos (c+d x)+b \sin (c+d x))}{\left (a^2+b^2\right )^2 d}+\frac {a}{\left (a^2+b^2\right ) d (a+b \tan (c+d x))} \] Output:

2*a*b*x/(a^2+b^2)^2-(a^2-b^2)*ln(a*cos(d*x+c)+b*sin(d*x+c))/(a^2+b^2)^2/d+ 
a/(a^2+b^2)/d/(a+b*tan(d*x+c))
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.32 (sec) , antiderivative size = 181, normalized size of antiderivative = 2.21 \[ \int \frac {\tan (c+d x)}{(a+b \tan (c+d x))^2} \, dx=\frac {a \left ((a-i b)^2 \log (i-\tan (c+d x))+(a+i b)^2 \log (i+\tan (c+d x))+2 \left (a^2+b^2+\left (-a^2+b^2\right ) \log (a+b \tan (c+d x))\right )\right )+b \left ((a-i b)^2 \log (i-\tan (c+d x))+(a+i b)^2 \log (i+\tan (c+d x))+2 \left (-a^2+b^2\right ) \log (a+b \tan (c+d x))\right ) \tan (c+d x)}{2 \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))} \] Input:

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

Output:

(a*((a - I*b)^2*Log[I - Tan[c + d*x]] + (a + I*b)^2*Log[I + Tan[c + d*x]] 
+ 2*(a^2 + b^2 + (-a^2 + b^2)*Log[a + b*Tan[c + d*x]])) + b*((a - I*b)^2*L 
og[I - Tan[c + d*x]] + (a + I*b)^2*Log[I + Tan[c + d*x]] + 2*(-a^2 + b^2)* 
Log[a + b*Tan[c + d*x]])*Tan[c + d*x])/(2*(a^2 + b^2)^2*d*(a + b*Tan[c + d 
*x]))
 

Rubi [A] (verified)

Time = 0.48 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.13, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {3042, 4012, 3042, 4014, 3042, 4013}

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.

\(\displaystyle \int \frac {\tan (c+d x)}{(a+b \tan (c+d x))^2} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\tan (c+d x)}{(a+b \tan (c+d x))^2}dx\)

\(\Big \downarrow \) 4012

\(\displaystyle \frac {\int \frac {b+a \tan (c+d x)}{a+b \tan (c+d x)}dx}{a^2+b^2}+\frac {a}{d \left (a^2+b^2\right ) (a+b \tan (c+d x))}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {b+a \tan (c+d x)}{a+b \tan (c+d x)}dx}{a^2+b^2}+\frac {a}{d \left (a^2+b^2\right ) (a+b \tan (c+d x))}\)

\(\Big \downarrow \) 4014

\(\displaystyle \frac {\frac {2 a b x}{a^2+b^2}-\frac {\left (a^2-b^2\right ) \int \frac {b-a \tan (c+d x)}{a+b \tan (c+d x)}dx}{a^2+b^2}}{a^2+b^2}+\frac {a}{d \left (a^2+b^2\right ) (a+b \tan (c+d x))}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {2 a b x}{a^2+b^2}-\frac {\left (a^2-b^2\right ) \int \frac {b-a \tan (c+d x)}{a+b \tan (c+d x)}dx}{a^2+b^2}}{a^2+b^2}+\frac {a}{d \left (a^2+b^2\right ) (a+b \tan (c+d x))}\)

\(\Big \downarrow \) 4013

\(\displaystyle \frac {a}{d \left (a^2+b^2\right ) (a+b \tan (c+d x))}+\frac {\frac {2 a b x}{a^2+b^2}-\frac {\left (a^2-b^2\right ) \log (a \cos (c+d x)+b \sin (c+d x))}{d \left (a^2+b^2\right )}}{a^2+b^2}\)

Input:

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

Output:

((2*a*b*x)/(a^2 + b^2) - ((a^2 - b^2)*Log[a*Cos[c + d*x] + b*Sin[c + d*x]] 
)/((a^2 + b^2)*d))/(a^2 + b^2) + a/((a^2 + b^2)*d*(a + b*Tan[c + d*x]))
 

Defintions of rubi rules used

rule 3042

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

rule 4012

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

rule 4013

Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/((a_) + (b_.)*tan[(e_.) + (f_.)* 
(x_)]), x_Symbol] :> Simp[(c/(b*f))*Log[RemoveContent[a*Cos[e + f*x] + b*Si 
n[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] && EqQ[a*c + b*d, 0]
 

rule 4014

Int[((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])/((a_.) + (b_.)*tan[(e_.) + (f_. 
)*(x_)]), x_Symbol] :> Simp[(a*c + b*d)*(x/(a^2 + b^2)), x] + Simp[(b*c - a 
*d)/(a^2 + b^2)   Int[(b - a*Tan[e + f*x])/(a + b*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] && N 
eQ[a*c + b*d, 0]
 

Maple [A] (verified)

Time = 0.50 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.27

method	result	size
derivativedivides	\(\frac {\frac {\frac {\left (a^{2}-b^{2}\right ) \ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2}+2 a b \arctan \left (\tan \left (d x +c \right )\right )}{\left (a^{2}+b^{2}\right )^{2}}+\frac {a}{\left (a^{2}+b^{2}\right ) \left (a +b \tan \left (d x +c \right )\right )}-\frac {\left (a^{2}-b^{2}\right ) \ln \left (a +b \tan \left (d x +c \right )\right )}{\left (a^{2}+b^{2}\right )^{2}}}{d}\)	\(104\)
default	\(\frac {\frac {\frac {\left (a^{2}-b^{2}\right ) \ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2}+2 a b \arctan \left (\tan \left (d x +c \right )\right )}{\left (a^{2}+b^{2}\right )^{2}}+\frac {a}{\left (a^{2}+b^{2}\right ) \left (a +b \tan \left (d x +c \right )\right )}-\frac {\left (a^{2}-b^{2}\right ) \ln \left (a +b \tan \left (d x +c \right )\right )}{\left (a^{2}+b^{2}\right )^{2}}}{d}\)	\(104\)
norman	\(\frac {-\frac {b \tan \left (d x +c \right )}{d \left (a^{2}+b^{2}\right )}+\frac {2 b \,a^{2} x}{a^{4}+2 b^{2} a^{2}+b^{4}}+\frac {2 b^{2} a x \tan \left (d x +c \right )}{a^{4}+2 b^{2} a^{2}+b^{4}}}{a +b \tan \left (d x +c \right )}+\frac {\left (a^{2}-b^{2}\right ) \ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2 d \left (a^{4}+2 b^{2} a^{2}+b^{4}\right )}-\frac {\left (a^{2}-b^{2}\right ) \ln \left (a +b \tan \left (d x +c \right )\right )}{d \left (a^{4}+2 b^{2} a^{2}+b^{4}\right )}\)	\(175\)
parallelrisch	\(\frac {4 x \tan \left (d x +c \right ) a^{2} b^{2} d +\ln \left (1+\tan \left (d x +c \right )^{2}\right ) \tan \left (d x +c \right ) a^{3} b -\ln \left (1+\tan \left (d x +c \right )^{2}\right ) \tan \left (d x +c \right ) a \,b^{3}-2 \ln \left (a +b \tan \left (d x +c \right )\right ) \tan \left (d x +c \right ) a^{3} b +2 \ln \left (a +b \tan \left (d x +c \right )\right ) \tan \left (d x +c \right ) a \,b^{3}+4 a^{3} b d x +\ln \left (1+\tan \left (d x +c \right )^{2}\right ) a^{4}-\ln \left (1+\tan \left (d x +c \right )^{2}\right ) a^{2} b^{2}-2 a^{4} \ln \left (a +b \tan \left (d x +c \right )\right )+2 \ln \left (a +b \tan \left (d x +c \right )\right ) a^{2} b^{2}-2 a^{3} b \tan \left (d x +c \right )-2 a \,b^{3} \tan \left (d x +c \right )}{2 \left (a +b \tan \left (d x +c \right )\right ) d a \left (a^{2}+b^{2}\right )^{2}}\)	\(239\)
risch	\(\frac {i x}{2 i a b -a^{2}+b^{2}}+\frac {2 i a^{2} x}{a^{4}+2 b^{2} a^{2}+b^{4}}-\frac {2 i b^{2} x}{a^{4}+2 b^{2} a^{2}+b^{4}}+\frac {2 i a^{2} c}{\left (a^{4}+2 b^{2} a^{2}+b^{4}\right ) d}-\frac {2 i b^{2} c}{d \left (a^{4}+2 b^{2} a^{2}+b^{4}\right )}-\frac {2 i a b}{\left (i b +a \right ) d \left (-i b +a \right )^{2} \left (-i b \,{\mathrm e}^{2 i \left (d x +c \right )}+a \,{\mathrm e}^{2 i \left (d x +c \right )}+i b +a \right )}-\frac {a^{2} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {i b +a}{i b -a}\right )}{\left (a^{4}+2 b^{2} a^{2}+b^{4}\right ) d}+\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {i b +a}{i b -a}\right ) b^{2}}{d \left (a^{4}+2 b^{2} a^{2}+b^{4}\right )}\)	\(287\)

Input:

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

Output:

1/d*(1/(a^2+b^2)^2*(1/2*(a^2-b^2)*ln(1+tan(d*x+c)^2)+2*a*b*arctan(tan(d*x+ 
c)))+a/(a^2+b^2)/(a+b*tan(d*x+c))-(a^2-b^2)/(a^2+b^2)^2*ln(a+b*tan(d*x+c)) 
)
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.91 \[ \int \frac {\tan (c+d x)}{(a+b \tan (c+d x))^2} \, dx=\frac {4 \, a^{2} b d x + 2 \, a b^{2} - {\left (a^{3} - a b^{2} + {\left (a^{2} b - b^{3}\right )} \tan \left (d x + c\right )\right )} \log \left (\frac {b^{2} \tan \left (d x + c\right )^{2} + 2 \, a b \tan \left (d x + c\right ) + a^{2}}{\tan \left (d x + c\right )^{2} + 1}\right ) + 2 \, {\left (2 \, a b^{2} d x - a^{2} b\right )} \tan \left (d x + c\right )}{2 \, {\left ({\left (a^{4} b + 2 \, a^{2} b^{3} + b^{5}\right )} d \tan \left (d x + c\right ) + {\left (a^{5} + 2 \, a^{3} b^{2} + a b^{4}\right )} d\right )}} \] Input:

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

Output:

1/2*(4*a^2*b*d*x + 2*a*b^2 - (a^3 - a*b^2 + (a^2*b - b^3)*tan(d*x + c))*lo 
g((b^2*tan(d*x + c)^2 + 2*a*b*tan(d*x + c) + a^2)/(tan(d*x + c)^2 + 1)) + 
2*(2*a*b^2*d*x - a^2*b)*tan(d*x + c))/((a^4*b + 2*a^2*b^3 + b^5)*d*tan(d*x 
 + c) + (a^5 + 2*a^3*b^2 + a*b^4)*d)
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.78 (sec) , antiderivative size = 1476, normalized size of antiderivative = 18.00 \[ \int \frac {\tan (c+d x)}{(a+b \tan (c+d x))^2} \, dx=\text {Too large to display} \] Input:

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

Output:

Piecewise((zoo*x/tan(c), Eq(a, 0) & Eq(b, 0) & Eq(d, 0)), (log(tan(c + d*x 
)**2 + 1)/(2*a**2*d), Eq(b, 0)), (I*d*x*tan(c + d*x)**2/(4*b**2*d*tan(c + 
d*x)**2 - 8*I*b**2*d*tan(c + d*x) - 4*b**2*d) + 2*d*x*tan(c + d*x)/(4*b**2 
*d*tan(c + d*x)**2 - 8*I*b**2*d*tan(c + d*x) - 4*b**2*d) - I*d*x/(4*b**2*d 
*tan(c + d*x)**2 - 8*I*b**2*d*tan(c + d*x) - 4*b**2*d) + I*tan(c + d*x)/(4 
*b**2*d*tan(c + d*x)**2 - 8*I*b**2*d*tan(c + d*x) - 4*b**2*d), Eq(a, -I*b) 
), (-I*d*x*tan(c + d*x)**2/(4*b**2*d*tan(c + d*x)**2 + 8*I*b**2*d*tan(c + 
d*x) - 4*b**2*d) + 2*d*x*tan(c + d*x)/(4*b**2*d*tan(c + d*x)**2 + 8*I*b**2 
*d*tan(c + d*x) - 4*b**2*d) + I*d*x/(4*b**2*d*tan(c + d*x)**2 + 8*I*b**2*d 
*tan(c + d*x) - 4*b**2*d) - I*tan(c + d*x)/(4*b**2*d*tan(c + d*x)**2 + 8*I 
*b**2*d*tan(c + d*x) - 4*b**2*d), Eq(a, I*b)), (x*tan(c)/(a + b*tan(c))**2 
, Eq(d, 0)), (-2*a**3*log(a/b + tan(c + d*x))/(2*a**5*d + 2*a**4*b*d*tan(c 
 + d*x) + 4*a**3*b**2*d + 4*a**2*b**3*d*tan(c + d*x) + 2*a*b**4*d + 2*b**5 
*d*tan(c + d*x)) + a**3*log(tan(c + d*x)**2 + 1)/(2*a**5*d + 2*a**4*b*d*ta 
n(c + d*x) + 4*a**3*b**2*d + 4*a**2*b**3*d*tan(c + d*x) + 2*a*b**4*d + 2*b 
**5*d*tan(c + d*x)) + 2*a**3/(2*a**5*d + 2*a**4*b*d*tan(c + d*x) + 4*a**3* 
b**2*d + 4*a**2*b**3*d*tan(c + d*x) + 2*a*b**4*d + 2*b**5*d*tan(c + d*x)) 
+ 4*a**2*b*d*x/(2*a**5*d + 2*a**4*b*d*tan(c + d*x) + 4*a**3*b**2*d + 4*a** 
2*b**3*d*tan(c + d*x) + 2*a*b**4*d + 2*b**5*d*tan(c + d*x)) - 2*a**2*b*log 
(a/b + tan(c + d*x))*tan(c + d*x)/(2*a**5*d + 2*a**4*b*d*tan(c + d*x) +...
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.70 \[ \int \frac {\tan (c+d x)}{(a+b \tan (c+d x))^2} \, dx=\frac {\frac {4 \, {\left (d x + c\right )} a b}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} - \frac {2 \, {\left (a^{2} - b^{2}\right )} \log \left (b \tan \left (d x + c\right ) + a\right )}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} + \frac {{\left (a^{2} - b^{2}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} + \frac {2 \, a}{a^{3} + a b^{2} + {\left (a^{2} b + b^{3}\right )} \tan \left (d x + c\right )}}{2 \, d} \] Input:

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

Output:

1/2*(4*(d*x + c)*a*b/(a^4 + 2*a^2*b^2 + b^4) - 2*(a^2 - b^2)*log(b*tan(d*x 
 + c) + a)/(a^4 + 2*a^2*b^2 + b^4) + (a^2 - b^2)*log(tan(d*x + c)^2 + 1)/( 
a^4 + 2*a^2*b^2 + b^4) + 2*a/(a^3 + a*b^2 + (a^2*b + b^3)*tan(d*x + c)))/d
 

Giac [A] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.93 \[ \int \frac {\tan (c+d x)}{(a+b \tan (c+d x))^2} \, dx=\frac {2 \, {\left (d x + c\right )} a b}{a^{4} d + 2 \, a^{2} b^{2} d + b^{4} d} + \frac {{\left (a^{2} - b^{2}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{2 \, {\left (a^{4} d + 2 \, a^{2} b^{2} d + b^{4} d\right )}} - \frac {{\left (a^{2} b - b^{3}\right )} \log \left ({\left | b \tan \left (d x + c\right ) + a \right |}\right )}{a^{4} b d + 2 \, a^{2} b^{3} d + b^{5} d} + \frac {a^{3} + a b^{2}}{{\left (a^{2} + b^{2}\right )}^{2} {\left (b \tan \left (d x + c\right ) + a\right )} d} \] Input:

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

Output:

2*(d*x + c)*a*b/(a^4*d + 2*a^2*b^2*d + b^4*d) + 1/2*(a^2 - b^2)*log(tan(d* 
x + c)^2 + 1)/(a^4*d + 2*a^2*b^2*d + b^4*d) - (a^2*b - b^3)*log(abs(b*tan( 
d*x + c) + a))/(a^4*b*d + 2*a^2*b^3*d + b^5*d) + (a^3 + a*b^2)/((a^2 + b^2 
)^2*(b*tan(d*x + c) + a)*d)
 

Mupad [B] (verification not implemented)

Time = 1.18 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.62 \[ \int \frac {\tan (c+d x)}{(a+b \tan (c+d x))^2} \, dx=\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )}{2\,d\,\left (a^2+a\,b\,2{}\mathrm {i}-b^2\right )}-\frac {\ln \left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )\,\left (\frac {1}{a^2+b^2}-\frac {2\,b^2}{{\left (a^2+b^2\right )}^2}\right )}{d}+\frac {a}{d\,\left (a^2+b^2\right )\,\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,1{}\mathrm {i}}{2\,d\,\left (a^2\,1{}\mathrm {i}+2\,a\,b-b^2\,1{}\mathrm {i}\right )} \] Input:

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

Output:

log(tan(c + d*x) - 1i)/(2*d*(a*b*2i + a^2 - b^2)) + (log(tan(c + d*x) + 1i 
)*1i)/(2*d*(2*a*b + a^2*1i - b^2*1i)) - (log(a + b*tan(c + d*x))*(1/(a^2 + 
 b^2) - (2*b^2)/(a^2 + b^2)^2))/d + a/(d*(a^2 + b^2)*(a + b*tan(c + d*x)))
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 259, normalized size of antiderivative = 3.16 \[ \int \frac {\tan (c+d x)}{(a+b \tan (c+d x))^2} \, dx=\frac {\mathrm {log}\left (\tan \left (d x +c \right )^{2}+1\right ) \tan \left (d x +c \right ) a^{2} b -\mathrm {log}\left (\tan \left (d x +c \right )^{2}+1\right ) \tan \left (d x +c \right ) b^{3}+\mathrm {log}\left (\tan \left (d x +c \right )^{2}+1\right ) a^{3}-\mathrm {log}\left (\tan \left (d x +c \right )^{2}+1\right ) a \,b^{2}-2 \,\mathrm {log}\left (a +\tan \left (d x +c \right ) b \right ) \tan \left (d x +c \right ) a^{2} b +2 \,\mathrm {log}\left (a +\tan \left (d x +c \right ) b \right ) \tan \left (d x +c \right ) b^{3}-2 \,\mathrm {log}\left (a +\tan \left (d x +c \right ) b \right ) a^{3}+2 \,\mathrm {log}\left (a +\tan \left (d x +c \right ) b \right ) a \,b^{2}-2 \tan \left (d x +c \right ) a^{2} b +4 \tan \left (d x +c \right ) a \,b^{2} d x -2 \tan \left (d x +c \right ) b^{3}+4 a^{2} b d x}{2 d \left (\tan \left (d x +c \right ) a^{4} b +2 \tan \left (d x +c \right ) a^{2} b^{3}+\tan \left (d x +c \right ) b^{5}+a^{5}+2 a^{3} b^{2}+a \,b^{4}\right )} \] Input:

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

Output:

(log(tan(c + d*x)**2 + 1)*tan(c + d*x)*a**2*b - log(tan(c + d*x)**2 + 1)*t 
an(c + d*x)*b**3 + log(tan(c + d*x)**2 + 1)*a**3 - log(tan(c + d*x)**2 + 1 
)*a*b**2 - 2*log(tan(c + d*x)*b + a)*tan(c + d*x)*a**2*b + 2*log(tan(c + d 
*x)*b + a)*tan(c + d*x)*b**3 - 2*log(tan(c + d*x)*b + a)*a**3 + 2*log(tan( 
c + d*x)*b + a)*a*b**2 - 2*tan(c + d*x)*a**2*b + 4*tan(c + d*x)*a*b**2*d*x 
 - 2*tan(c + d*x)*b**3 + 4*a**2*b*d*x)/(2*d*(tan(c + d*x)*a**4*b + 2*tan(c 
 + d*x)*a**2*b**3 + tan(c + d*x)*b**5 + a**5 + 2*a**3*b**2 + a*b**4))