\(\int \frac {1}{(a+b \tan ^2(c+d x))^2} \, dx\) [255]

Optimal result

Integrand size = 14, antiderivative size = 97 \[ \int \frac {1}{\left (a+b \tan ^2(c+d x)\right )^2} \, dx=\frac {x}{(a-b)^2}-\frac {(3 a-b) \sqrt {b} \arctan \left (\frac {\sqrt {b} \tan (c+d x)}{\sqrt {a}}\right )}{2 a^{3/2} (a-b)^2 d}-\frac {b \tan (c+d x)}{2 a (a-b) d \left (a+b \tan ^2(c+d x)\right )} \] Output:

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

Mathematica [A] (verified)

Time = 0.70 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.91 \[ \int \frac {1}{\left (a+b \tan ^2(c+d x)\right )^2} \, dx=\frac {2 \arctan (\tan (c+d x))+\frac {\sqrt {b} (-3 a+b) \arctan \left (\frac {\sqrt {b} \tan (c+d x)}{\sqrt {a}}\right )}{a^{3/2}}+\frac {b (-a+b) \tan (c+d x)}{a \left (a+b \tan ^2(c+d x)\right )}}{2 (a-b)^2 d} \] Input:

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

Output:

(2*ArcTan[Tan[c + d*x]] + (Sqrt[b]*(-3*a + b)*ArcTan[(Sqrt[b]*Tan[c + d*x] 
)/Sqrt[a]])/a^(3/2) + (b*(-a + b)*Tan[c + d*x])/(a*(a + b*Tan[c + d*x]^2)) 
)/(2*(a - b)^2*d)
 

Rubi [A] (verified)

Time = 0.43 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.20, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {3042, 4144, 316, 397, 216, 218}

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.

\(\displaystyle \int \frac {1}{\left (a+b \tan ^2(c+d x)\right )^2} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\left (a+b \tan (c+d x)^2\right )^2}dx\)

\(\Big \downarrow \) 4144

\(\displaystyle \frac {\int \frac {1}{\left (\tan ^2(c+d x)+1\right ) \left (b \tan ^2(c+d x)+a\right )^2}d\tan (c+d x)}{d}\)

\(\Big \downarrow \) 316

\(\displaystyle \frac {\frac {\int \frac {-b \tan ^2(c+d x)+2 a-b}{\left (\tan ^2(c+d x)+1\right ) \left (b \tan ^2(c+d x)+a\right )}d\tan (c+d x)}{2 a (a-b)}-\frac {b \tan (c+d x)}{2 a (a-b) \left (a+b \tan ^2(c+d x)\right )}}{d}\)

\(\Big \downarrow \) 397

\(\displaystyle \frac {\frac {\frac {2 a \int \frac {1}{\tan ^2(c+d x)+1}d\tan (c+d x)}{a-b}-\frac {b (3 a-b) \int \frac {1}{b \tan ^2(c+d x)+a}d\tan (c+d x)}{a-b}}{2 a (a-b)}-\frac {b \tan (c+d x)}{2 a (a-b) \left (a+b \tan ^2(c+d x)\right )}}{d}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {\frac {\frac {2 a \arctan (\tan (c+d x))}{a-b}-\frac {b (3 a-b) \int \frac {1}{b \tan ^2(c+d x)+a}d\tan (c+d x)}{a-b}}{2 a (a-b)}-\frac {b \tan (c+d x)}{2 a (a-b) \left (a+b \tan ^2(c+d x)\right )}}{d}\)

\(\Big \downarrow \) 218

\(\displaystyle \frac {\frac {\frac {2 a \arctan (\tan (c+d x))}{a-b}-\frac {\sqrt {b} (3 a-b) \arctan \left (\frac {\sqrt {b} \tan (c+d x)}{\sqrt {a}}\right )}{\sqrt {a} (a-b)}}{2 a (a-b)}-\frac {b \tan (c+d x)}{2 a (a-b) \left (a+b \tan ^2(c+d x)\right )}}{d}\)

Input:

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

Output:

(((2*a*ArcTan[Tan[c + d*x]])/(a - b) - ((3*a - b)*Sqrt[b]*ArcTan[(Sqrt[b]* 
Tan[c + d*x])/Sqrt[a]])/(Sqrt[a]*(a - b)))/(2*a*(a - b)) - (b*Tan[c + d*x] 
)/(2*a*(a - b)*(a + b*Tan[c + d*x]^2)))/d
 

Defintions of rubi rules used

rule 216

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A 
rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a 
, 0] || GtQ[b, 0])
 

rule 218

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R 
t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
 

rule 316

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim 
p[(-b)*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q + 1)/(2*a*(p + 1)*(b*c - a*d)) 
), x] + Simp[1/(2*a*(p + 1)*(b*c - a*d))   Int[(a + b*x^2)^(p + 1)*(c + d*x 
^2)^q*Simp[b*c + 2*(p + 1)*(b*c - a*d) + d*b*(2*(p + q + 2) + 1)*x^2, x], x 
], x] /; FreeQ[{a, b, c, d, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] &&  ! 
( !IntegerQ[p] && IntegerQ[q] && LtQ[q, -1]) && IntBinomialQ[a, b, c, d, 2, 
 p, q, x]
 

rule 397

Int[((e_) + (f_.)*(x_)^2)/(((a_) + (b_.)*(x_)^2)*((c_) + (d_.)*(x_)^2)), x_ 
Symbol] :> Simp[(b*e - a*f)/(b*c - a*d)   Int[1/(a + b*x^2), x], x] - Simp[ 
(d*e - c*f)/(b*c - a*d)   Int[1/(c + d*x^2), x], x] /; FreeQ[{a, b, c, d, e 
, f}, x]
 

rule 3042

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

rule 4144

Int[((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_), x_Symbol] :> 
With[{ff = FreeFactors[Tan[e + f*x], x]}, Simp[c*(ff/f)   Subst[Int[(a + b* 
(ff*x)^n)^p/(c^2 + ff^2*x^2), x], x, c*(Tan[e + f*x]/ff)], x]] /; FreeQ[{a, 
 b, c, e, f, n, p}, x] && (IntegersQ[n, p] || IGtQ[p, 0] || EqQ[n^2, 4] || 
EqQ[n^2, 16])
 

Maple [A] (verified)

Time = 0.83 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.96

method	result	size
derivativedivides	\(\frac {-\frac {b \left (\frac {\left (a -b \right ) \tan \left (d x +c \right )}{2 a \left (a +b \tan \left (d x +c \right )^{2}\right )}+\frac {\left (3 a -b \right ) \arctan \left (\frac {b \tan \left (d x +c \right )}{\sqrt {a b}}\right )}{2 a \sqrt {a b}}\right )}{\left (a -b \right )^{2}}+\frac {\arctan \left (\tan \left (d x +c \right )\right )}{\left (a -b \right )^{2}}}{d}\)	\(93\)
default	\(\frac {-\frac {b \left (\frac {\left (a -b \right ) \tan \left (d x +c \right )}{2 a \left (a +b \tan \left (d x +c \right )^{2}\right )}+\frac {\left (3 a -b \right ) \arctan \left (\frac {b \tan \left (d x +c \right )}{\sqrt {a b}}\right )}{2 a \sqrt {a b}}\right )}{\left (a -b \right )^{2}}+\frac {\arctan \left (\tan \left (d x +c \right )\right )}{\left (a -b \right )^{2}}}{d}\)	\(93\)
risch	\(\frac {x}{a^{2}-2 a b +b^{2}}+\frac {i b \left (a \,{\mathrm e}^{2 i \left (d x +c \right )}+b \,{\mathrm e}^{2 i \left (d x +c \right )}+a -b \right )}{d a \left (-a +b \right )^{2} \left (-a \,{\mathrm e}^{4 i \left (d x +c \right )}+b \,{\mathrm e}^{4 i \left (d x +c \right )}-2 a \,{\mathrm e}^{2 i \left (d x +c \right )}-2 b \,{\mathrm e}^{2 i \left (d x +c \right )}-a +b \right )}+\frac {3 \sqrt {-a b}\, \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 i \sqrt {-a b}+a +b}{a -b}\right )}{4 a \left (a -b \right )^{2} d}-\frac {\sqrt {-a b}\, \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 i \sqrt {-a b}+a +b}{a -b}\right ) b}{4 a^{2} \left (a -b \right )^{2} d}-\frac {3 \sqrt {-a b}\, \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {2 i \sqrt {-a b}-a -b}{a -b}\right )}{4 a \left (a -b \right )^{2} d}+\frac {\sqrt {-a b}\, \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {2 i \sqrt {-a b}-a -b}{a -b}\right ) b}{4 a^{2} \left (a -b \right )^{2} d}\)	\(335\)

Input:

int(1/(a+b*tan(d*x+c)^2)^2,x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 390, normalized size of antiderivative = 4.02 \[ \int \frac {1}{\left (a+b \tan ^2(c+d x)\right )^2} \, dx=\left [\frac {8 \, a b d x \tan \left (d x + c\right )^{2} + 8 \, a^{2} d x - {\left ({\left (3 \, a b - b^{2}\right )} \tan \left (d x + c\right )^{2} + 3 \, a^{2} - a b\right )} \sqrt {-\frac {b}{a}} \log \left (\frac {b^{2} \tan \left (d x + c\right )^{4} - 6 \, a b \tan \left (d x + c\right )^{2} + a^{2} + 4 \, {\left (a b \tan \left (d x + c\right )^{3} - a^{2} \tan \left (d x + c\right )\right )} \sqrt {-\frac {b}{a}}}{b^{2} \tan \left (d x + c\right )^{4} + 2 \, a b \tan \left (d x + c\right )^{2} + a^{2}}\right ) - 4 \, {\left (a b - b^{2}\right )} \tan \left (d x + c\right )}{8 \, {\left ({\left (a^{3} b - 2 \, a^{2} b^{2} + a b^{3}\right )} d \tan \left (d x + c\right )^{2} + {\left (a^{4} - 2 \, a^{3} b + a^{2} b^{2}\right )} d\right )}}, \frac {4 \, a b d x \tan \left (d x + c\right )^{2} + 4 \, a^{2} d x - {\left ({\left (3 \, a b - b^{2}\right )} \tan \left (d x + c\right )^{2} + 3 \, a^{2} - a b\right )} \sqrt {\frac {b}{a}} \arctan \left (\frac {{\left (b \tan \left (d x + c\right )^{2} - a\right )} \sqrt {\frac {b}{a}}}{2 \, b \tan \left (d x + c\right )}\right ) - 2 \, {\left (a b - b^{2}\right )} \tan \left (d x + c\right )}{4 \, {\left ({\left (a^{3} b - 2 \, a^{2} b^{2} + a b^{3}\right )} d \tan \left (d x + c\right )^{2} + {\left (a^{4} - 2 \, a^{3} b + a^{2} b^{2}\right )} d\right )}}\right ] \] Input:

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

Output:

[1/8*(8*a*b*d*x*tan(d*x + c)^2 + 8*a^2*d*x - ((3*a*b - b^2)*tan(d*x + c)^2 
 + 3*a^2 - a*b)*sqrt(-b/a)*log((b^2*tan(d*x + c)^4 - 6*a*b*tan(d*x + c)^2 
+ a^2 + 4*(a*b*tan(d*x + c)^3 - a^2*tan(d*x + c))*sqrt(-b/a))/(b^2*tan(d*x 
 + c)^4 + 2*a*b*tan(d*x + c)^2 + a^2)) - 4*(a*b - b^2)*tan(d*x + c))/((a^3 
*b - 2*a^2*b^2 + a*b^3)*d*tan(d*x + c)^2 + (a^4 - 2*a^3*b + a^2*b^2)*d), 1 
/4*(4*a*b*d*x*tan(d*x + c)^2 + 4*a^2*d*x - ((3*a*b - b^2)*tan(d*x + c)^2 + 
 3*a^2 - a*b)*sqrt(b/a)*arctan(1/2*(b*tan(d*x + c)^2 - a)*sqrt(b/a)/(b*tan 
(d*x + c))) - 2*(a*b - b^2)*tan(d*x + c))/((a^3*b - 2*a^2*b^2 + a*b^3)*d*t 
an(d*x + c)^2 + (a^4 - 2*a^3*b + a^2*b^2)*d)]
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 2125 vs. \(2 (78) = 156\).

Time = 10.00 (sec) , antiderivative size = 2125, normalized size of antiderivative = 21.91 \[ \int \frac {1}{\left (a+b \tan ^2(c+d x)\right )^2} \, dx=\text {Too large to display} \] Input:

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

Output:

Piecewise((zoo*x/tan(c)**4, Eq(a, 0) & Eq(b, 0) & Eq(d, 0)), (x/a**2, Eq(b 
, 0)), ((x + 1/(d*tan(c + d*x)) - 1/(3*d*tan(c + d*x)**3))/b**2, Eq(a, 0)) 
, (3*d*x*tan(c + d*x)**4/(8*b**2*d*tan(c + d*x)**4 + 16*b**2*d*tan(c + d*x 
)**2 + 8*b**2*d) + 6*d*x*tan(c + d*x)**2/(8*b**2*d*tan(c + d*x)**4 + 16*b* 
*2*d*tan(c + d*x)**2 + 8*b**2*d) + 3*d*x/(8*b**2*d*tan(c + d*x)**4 + 16*b* 
*2*d*tan(c + d*x)**2 + 8*b**2*d) + 3*tan(c + d*x)**3/(8*b**2*d*tan(c + d*x 
)**4 + 16*b**2*d*tan(c + d*x)**2 + 8*b**2*d) + 5*tan(c + d*x)/(8*b**2*d*ta 
n(c + d*x)**4 + 16*b**2*d*tan(c + d*x)**2 + 8*b**2*d), Eq(a, b)), (x/(a + 
b*tan(c)**2)**2, Eq(d, 0)), (4*a**2*d*x*sqrt(-a/b)/(4*a**4*d*sqrt(-a/b) + 
4*a**3*b*d*sqrt(-a/b)*tan(c + d*x)**2 - 8*a**3*b*d*sqrt(-a/b) - 8*a**2*b** 
2*d*sqrt(-a/b)*tan(c + d*x)**2 + 4*a**2*b**2*d*sqrt(-a/b) + 4*a*b**3*d*sqr 
t(-a/b)*tan(c + d*x)**2) - 3*a**2*log(-sqrt(-a/b) + tan(c + d*x))/(4*a**4* 
d*sqrt(-a/b) + 4*a**3*b*d*sqrt(-a/b)*tan(c + d*x)**2 - 8*a**3*b*d*sqrt(-a/ 
b) - 8*a**2*b**2*d*sqrt(-a/b)*tan(c + d*x)**2 + 4*a**2*b**2*d*sqrt(-a/b) + 
 4*a*b**3*d*sqrt(-a/b)*tan(c + d*x)**2) + 3*a**2*log(sqrt(-a/b) + tan(c + 
d*x))/(4*a**4*d*sqrt(-a/b) + 4*a**3*b*d*sqrt(-a/b)*tan(c + d*x)**2 - 8*a** 
3*b*d*sqrt(-a/b) - 8*a**2*b**2*d*sqrt(-a/b)*tan(c + d*x)**2 + 4*a**2*b**2* 
d*sqrt(-a/b) + 4*a*b**3*d*sqrt(-a/b)*tan(c + d*x)**2) + 4*a*b*d*x*sqrt(-a/ 
b)*tan(c + d*x)**2/(4*a**4*d*sqrt(-a/b) + 4*a**3*b*d*sqrt(-a/b)*tan(c + d* 
x)**2 - 8*a**3*b*d*sqrt(-a/b) - 8*a**2*b**2*d*sqrt(-a/b)*tan(c + d*x)**...
 

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.18 \[ \int \frac {1}{\left (a+b \tan ^2(c+d x)\right )^2} \, dx=-\frac {\frac {b \tan \left (d x + c\right )}{a^{3} - a^{2} b + {\left (a^{2} b - a b^{2}\right )} \tan \left (d x + c\right )^{2}} + \frac {{\left (3 \, a b - b^{2}\right )} \arctan \left (\frac {b \tan \left (d x + c\right )}{\sqrt {a b}}\right )}{{\left (a^{3} - 2 \, a^{2} b + a b^{2}\right )} \sqrt {a b}} - \frac {2 \, {\left (d x + c\right )}}{a^{2} - 2 \, a b + b^{2}}}{2 \, d} \] Input:

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.56 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.16 \[ \int \frac {1}{\left (a+b \tan ^2(c+d x)\right )^2} \, dx=-\frac {{\left (3 \, a b - b^{2}\right )} \arctan \left (\frac {b \tan \left (d x + c\right )}{\sqrt {a b}}\right )}{2 \, {\left (a^{3} d - 2 \, a^{2} b d + a b^{2} d\right )} \sqrt {a b}} + \frac {d x + c}{a^{2} d - 2 \, a b d + b^{2} d} - \frac {b \tan \left (d x + c\right )}{2 \, {\left (a^{2} d - a b d\right )} {\left (b \tan \left (d x + c\right )^{2} + a\right )}} \] Input:

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

Output:

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

Mupad [B] (verification not implemented)

Time = 8.88 (sec) , antiderivative size = 2489, normalized size of antiderivative = 25.66 \[ \int \frac {1}{\left (a+b \tan ^2(c+d x)\right )^2} \, dx=\text {Too large to display} \] Input:

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

Output:

(2*atan((((((2*a*b^7 - 12*a^2*b^6 + 28*a^3*b^5 - 32*a^4*b^4 + 18*a^5*b^3 - 
 4*a^6*b^2)*1i)/(3*a^4*b - a^5 + a^2*b^3 - 3*a^3*b^2) - (tan(c + d*x)*(16* 
a^2*b^7 - 48*a^3*b^6 + 32*a^4*b^5 + 32*a^5*b^4 - 48*a^6*b^3 + 16*a^7*b^2)) 
/(2*(a^4 - 2*a^3*b + a^2*b^2)*(2*a^2 - 4*a*b + 2*b^2)))/(2*a^2 - 4*a*b + 2 
*b^2) + (tan(c + d*x)*(b^5 - 6*a*b^4 + 13*a^2*b^3))/(2*(a^4 - 2*a^3*b + a^ 
2*b^2)))/(2*a^2 - 4*a*b + 2*b^2) - ((((2*a*b^7 - 12*a^2*b^6 + 28*a^3*b^5 - 
 32*a^4*b^4 + 18*a^5*b^3 - 4*a^6*b^2)*1i)/(3*a^4*b - a^5 + a^2*b^3 - 3*a^3 
*b^2) + (tan(c + d*x)*(16*a^2*b^7 - 48*a^3*b^6 + 32*a^4*b^5 + 32*a^5*b^4 - 
 48*a^6*b^3 + 16*a^7*b^2))/(2*(a^4 - 2*a^3*b + a^2*b^2)*(2*a^2 - 4*a*b + 2 
*b^2)))/(2*a^2 - 4*a*b + 2*b^2) - (tan(c + d*x)*(b^5 - 6*a*b^4 + 13*a^2*b^ 
3))/(2*(a^4 - 2*a^3*b + a^2*b^2)))/(2*a^2 - 4*a*b + 2*b^2))/((((((2*a*b^7 
- 12*a^2*b^6 + 28*a^3*b^5 - 32*a^4*b^4 + 18*a^5*b^3 - 4*a^6*b^2)*1i)/(3*a^ 
4*b - a^5 + a^2*b^3 - 3*a^3*b^2) - (tan(c + d*x)*(16*a^2*b^7 - 48*a^3*b^6 
+ 32*a^4*b^5 + 32*a^5*b^4 - 48*a^6*b^3 + 16*a^7*b^2))/(2*(a^4 - 2*a^3*b + 
a^2*b^2)*(2*a^2 - 4*a*b + 2*b^2)))*1i)/(2*a^2 - 4*a*b + 2*b^2) + (tan(c + 
d*x)*(b^5 - 6*a*b^4 + 13*a^2*b^3)*1i)/(2*(a^4 - 2*a^3*b + a^2*b^2)))/(2*a^ 
2 - 4*a*b + 2*b^2) + (((((2*a*b^7 - 12*a^2*b^6 + 28*a^3*b^5 - 32*a^4*b^4 + 
 18*a^5*b^3 - 4*a^6*b^2)*1i)/(3*a^4*b - a^5 + a^2*b^3 - 3*a^3*b^2) + (tan( 
c + d*x)*(16*a^2*b^7 - 48*a^3*b^6 + 32*a^4*b^5 + 32*a^5*b^4 - 48*a^6*b^3 + 
 16*a^7*b^2))/(2*(a^4 - 2*a^3*b + a^2*b^2)*(2*a^2 - 4*a*b + 2*b^2)))*1i...
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 227, normalized size of antiderivative = 2.34 \[ \int \frac {1}{\left (a+b \tan ^2(c+d x)\right )^2} \, dx=\frac {-3 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {\tan \left (d x +c \right ) b}{\sqrt {b}\, \sqrt {a}}\right ) \tan \left (d x +c \right )^{2} a b +\sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {\tan \left (d x +c \right ) b}{\sqrt {b}\, \sqrt {a}}\right ) \tan \left (d x +c \right )^{2} b^{2}-3 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {\tan \left (d x +c \right ) b}{\sqrt {b}\, \sqrt {a}}\right ) a^{2}+\sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {\tan \left (d x +c \right ) b}{\sqrt {b}\, \sqrt {a}}\right ) a b +2 \tan \left (d x +c \right )^{2} a^{2} b d x -\tan \left (d x +c \right ) a^{2} b +\tan \left (d x +c \right ) a \,b^{2}+2 a^{3} d x}{2 a^{2} d \left (\tan \left (d x +c \right )^{2} a^{2} b -2 \tan \left (d x +c \right )^{2} a \,b^{2}+\tan \left (d x +c \right )^{2} b^{3}+a^{3}-2 a^{2} b +a \,b^{2}\right )} \] Input:

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

Output:

( - 3*sqrt(b)*sqrt(a)*atan((tan(c + d*x)*b)/(sqrt(b)*sqrt(a)))*tan(c + d*x 
)**2*a*b + sqrt(b)*sqrt(a)*atan((tan(c + d*x)*b)/(sqrt(b)*sqrt(a)))*tan(c 
+ d*x)**2*b**2 - 3*sqrt(b)*sqrt(a)*atan((tan(c + d*x)*b)/(sqrt(b)*sqrt(a)) 
)*a**2 + sqrt(b)*sqrt(a)*atan((tan(c + d*x)*b)/(sqrt(b)*sqrt(a)))*a*b + 2* 
tan(c + d*x)**2*a**2*b*d*x - tan(c + d*x)*a**2*b + tan(c + d*x)*a*b**2 + 2 
*a**3*d*x)/(2*a**2*d*(tan(c + d*x)**2*a**2*b - 2*tan(c + d*x)**2*a*b**2 + 
tan(c + d*x)**2*b**3 + a**3 - 2*a**2*b + a*b**2))