\(\int \frac {1}{(a+b \tan ^2(c+d x))^3} \, dx\) [256]

Optimal result

Integrand size = 14, antiderivative size = 150 \[ \int \frac {1}{\left (a+b \tan ^2(c+d x)\right )^3} \, dx=\frac {x}{(a-b)^3}-\frac {\sqrt {b} \left (15 a^2-10 a b+3 b^2\right ) \arctan \left (\frac {\sqrt {b} \tan (c+d x)}{\sqrt {a}}\right )}{8 a^{5/2} (a-b)^3 d}-\frac {b \tan (c+d x)}{4 a (a-b) d \left (a+b \tan ^2(c+d x)\right )^2}-\frac {(7 a-3 b) b \tan (c+d x)}{8 a^2 (a-b)^2 d \left (a+b \tan ^2(c+d x)\right )} \] Output:

x/(a-b)^3-1/8*b^(1/2)*(15*a^2-10*a*b+3*b^2)*arctan(b^(1/2)*tan(d*x+c)/a^(1 
/2))/a^(5/2)/(a-b)^3/d-1/4*b*tan(d*x+c)/a/(a-b)/d/(a+b*tan(d*x+c)^2)^2-1/8 
*(7*a-3*b)*b*tan(d*x+c)/a^2/(a-b)^2/d/(a+b*tan(d*x+c)^2)
 

Mathematica [A] (verified)

Time = 1.28 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.92 \[ \int \frac {1}{\left (a+b \tan ^2(c+d x)\right )^3} \, dx=-\frac {-8 \arctan (\tan (c+d x))+\frac {\sqrt {b} \left (15 a^2-10 a b+3 b^2\right ) \arctan \left (\frac {\sqrt {b} \tan (c+d x)}{\sqrt {a}}\right )}{a^{5/2}}+\frac {2 (a-b)^2 b \tan (c+d x)}{a \left (a+b \tan ^2(c+d x)\right )^2}+\frac {(7 a-3 b) (a-b) b \tan (c+d x)}{a^2 \left (a+b \tan ^2(c+d x)\right )}}{8 (a-b)^3 d} \] Input:

Integrate[(a + b*Tan[c + d*x]^2)^(-3),x]
 

Output:

-1/8*(-8*ArcTan[Tan[c + d*x]] + (Sqrt[b]*(15*a^2 - 10*a*b + 3*b^2)*ArcTan[ 
(Sqrt[b]*Tan[c + d*x])/Sqrt[a]])/a^(5/2) + (2*(a - b)^2*b*Tan[c + d*x])/(a 
*(a + b*Tan[c + d*x]^2)^2) + ((7*a - 3*b)*(a - b)*b*Tan[c + d*x])/(a^2*(a 
+ b*Tan[c + d*x]^2)))/((a - b)^3*d)
 

Rubi [A] (verified)

Time = 0.55 (sec) , antiderivative size = 183, normalized size of antiderivative = 1.22, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {3042, 4144, 316, 402, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4144

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

\(\Big \downarrow \) 316

\(\displaystyle \frac {\frac {\int \frac {-3 b \tan ^2(c+d x)+4 a-3 b}{\left (\tan ^2(c+d x)+1\right ) \left (b \tan ^2(c+d x)+a\right )^2}d\tan (c+d x)}{4 a (a-b)}-\frac {b \tan (c+d x)}{4 a (a-b) \left (a+b \tan ^2(c+d x)\right )^2}}{d}\)

\(\Big \downarrow \) 402

\(\displaystyle \frac {\frac {\frac {\int \frac {8 a^2-7 b a+3 b^2-(7 a-3 b) b \tan ^2(c+d x)}{\left (\tan ^2(c+d x)+1\right ) \left (b \tan ^2(c+d x)+a\right )}d\tan (c+d x)}{2 a (a-b)}-\frac {b (7 a-3 b) \tan (c+d x)}{2 a (a-b) \left (a+b \tan ^2(c+d x)\right )}}{4 a (a-b)}-\frac {b \tan (c+d x)}{4 a (a-b) \left (a+b \tan ^2(c+d x)\right )^2}}{d}\)

\(\Big \downarrow \) 397

\(\displaystyle \frac {\frac {\frac {\frac {8 a^2 \int \frac {1}{\tan ^2(c+d x)+1}d\tan (c+d x)}{a-b}-\frac {b \left (15 a^2-10 a b+3 b^2\right ) \int \frac {1}{b \tan ^2(c+d x)+a}d\tan (c+d x)}{a-b}}{2 a (a-b)}-\frac {b (7 a-3 b) \tan (c+d x)}{2 a (a-b) \left (a+b \tan ^2(c+d x)\right )}}{4 a (a-b)}-\frac {b \tan (c+d x)}{4 a (a-b) \left (a+b \tan ^2(c+d x)\right )^2}}{d}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {\frac {\frac {\frac {8 a^2 \arctan (\tan (c+d x))}{a-b}-\frac {b \left (15 a^2-10 a b+3 b^2\right ) \int \frac {1}{b \tan ^2(c+d x)+a}d\tan (c+d x)}{a-b}}{2 a (a-b)}-\frac {b (7 a-3 b) \tan (c+d x)}{2 a (a-b) \left (a+b \tan ^2(c+d x)\right )}}{4 a (a-b)}-\frac {b \tan (c+d x)}{4 a (a-b) \left (a+b \tan ^2(c+d x)\right )^2}}{d}\)

\(\Big \downarrow \) 218

\(\displaystyle \frac {\frac {\frac {\frac {8 a^2 \arctan (\tan (c+d x))}{a-b}-\frac {\sqrt {b} \left (15 a^2-10 a b+3 b^2\right ) \arctan \left (\frac {\sqrt {b} \tan (c+d x)}{\sqrt {a}}\right )}{\sqrt {a} (a-b)}}{2 a (a-b)}-\frac {b (7 a-3 b) \tan (c+d x)}{2 a (a-b) \left (a+b \tan ^2(c+d x)\right )}}{4 a (a-b)}-\frac {b \tan (c+d x)}{4 a (a-b) \left (a+b \tan ^2(c+d x)\right )^2}}{d}\)

Input:

Int[(a + b*Tan[c + d*x]^2)^(-3),x]
 

Output:

(-1/4*(b*Tan[c + d*x])/(a*(a - b)*(a + b*Tan[c + d*x]^2)^2) + (((8*a^2*Arc 
Tan[Tan[c + d*x]])/(a - b) - (Sqrt[b]*(15*a^2 - 10*a*b + 3*b^2)*ArcTan[(Sq 
rt[b]*Tan[c + d*x])/Sqrt[a]])/(Sqrt[a]*(a - b)))/(2*a*(a - b)) - ((7*a - 3 
*b)*b*Tan[c + d*x])/(2*a*(a - b)*(a + b*Tan[c + d*x]^2)))/(4*a*(a - b)))/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 402

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_.)*((e_) + (f_.)*(x 
_)^2), x_Symbol] :> Simp[(-(b*e - a*f))*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^ 
(q + 1)/(a*2*(b*c - a*d)*(p + 1))), x] + Simp[1/(a*2*(b*c - a*d)*(p + 1)) 
 Int[(a + b*x^2)^(p + 1)*(c + d*x^2)^q*Simp[c*(b*e - a*f) + e*2*(b*c - a*d) 
*(p + 1) + d*(b*e - a*f)*(2*(p + q + 2) + 1)*x^2, x], x], x] /; FreeQ[{a, b 
, c, d, e, f, q}, x] && LtQ[p, -1]
 

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 = 0.58 (sec) , antiderivative size = 142, normalized size of antiderivative = 0.95

method	result	size
derivativedivides	\(\frac {-\frac {b \left (\frac {\frac {b \left (7 a^{2}-10 a b +3 b^{2}\right ) \tan \left (d x +c \right )^{3}}{8 a^{2}}+\frac {\left (9 a^{2}-14 a b +5 b^{2}\right ) \tan \left (d x +c \right )}{8 a}}{\left (a +b \tan \left (d x +c \right )^{2}\right )^{2}}+\frac {\left (15 a^{2}-10 a b +3 b^{2}\right ) \arctan \left (\frac {b \tan \left (d x +c \right )}{\sqrt {a b}}\right )}{8 a^{2} \sqrt {a b}}\right )}{\left (a -b \right )^{3}}+\frac {\arctan \left (\tan \left (d x +c \right )\right )}{\left (a -b \right )^{3}}}{d}\)	\(142\)
default	\(\frac {-\frac {b \left (\frac {\frac {b \left (7 a^{2}-10 a b +3 b^{2}\right ) \tan \left (d x +c \right )^{3}}{8 a^{2}}+\frac {\left (9 a^{2}-14 a b +5 b^{2}\right ) \tan \left (d x +c \right )}{8 a}}{\left (a +b \tan \left (d x +c \right )^{2}\right )^{2}}+\frac {\left (15 a^{2}-10 a b +3 b^{2}\right ) \arctan \left (\frac {b \tan \left (d x +c \right )}{\sqrt {a b}}\right )}{8 a^{2} \sqrt {a b}}\right )}{\left (a -b \right )^{3}}+\frac {\arctan \left (\tan \left (d x +c \right )\right )}{\left (a -b \right )^{3}}}{d}\)	\(142\)
risch	\(\frac {x}{a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}}+\frac {i \left (9 a^{3} {\mathrm e}^{6 i \left (d x +c \right )}+a^{2} b \,{\mathrm e}^{6 i \left (d x +c \right )}-13 a \,b^{2} {\mathrm e}^{6 i \left (d x +c \right )}+3 b^{3} {\mathrm e}^{6 i \left (d x +c \right )}+27 a^{3} {\mathrm e}^{4 i \left (d x +c \right )}+9 a^{2} b \,{\mathrm e}^{4 i \left (d x +c \right )}+21 a \,b^{2} {\mathrm e}^{4 i \left (d x +c \right )}-9 b^{3} {\mathrm e}^{4 i \left (d x +c \right )}+27 a^{3} {\mathrm e}^{2 i \left (d x +c \right )}-13 a^{2} b \,{\mathrm e}^{2 i \left (d x +c \right )}-23 a \,b^{2} {\mathrm e}^{2 i \left (d x +c \right )}+9 b^{3} {\mathrm e}^{2 i \left (d x +c \right )}+9 a^{3}-21 a^{2} b +15 a \,b^{2}-3 b^{3}\right ) b}{4 \left (-a \,{\mathrm e}^{4 i \left (d x +c \right )}+b \,{\mathrm e}^{4 i \left (d x +c \right )}-2 a \,{\mathrm e}^{2 i \left (d x +c \right )}-2 b \,{\mathrm e}^{2 i \left (d x +c \right )}-a +b \right )^{2} d \left (-a +b \right )^{3} a^{2}}+\frac {15 \sqrt {-a b}\, \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 i \sqrt {-a b}+a +b}{a -b}\right )}{16 a \left (a -b \right )^{3} d}-\frac {5 \sqrt {-a b}\, \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 i \sqrt {-a b}+a +b}{a -b}\right ) b}{8 a^{2} \left (a -b \right )^{3} d}+\frac {3 \sqrt {-a b}\, \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 i \sqrt {-a b}+a +b}{a -b}\right ) b^{2}}{16 a^{3} \left (a -b \right )^{3} d}-\frac {15 \sqrt {-a b}\, \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {2 i \sqrt {-a b}-a -b}{a -b}\right )}{16 a \left (a -b \right )^{3} d}+\frac {5 \sqrt {-a b}\, \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {2 i \sqrt {-a b}-a -b}{a -b}\right ) b}{8 a^{2} \left (a -b \right )^{3} d}-\frac {3 \sqrt {-a b}\, \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {2 i \sqrt {-a b}-a -b}{a -b}\right ) b^{2}}{16 a^{3} \left (a -b \right )^{3} d}\)	\(629\)

Input:

int(1/(a+b*tan(d*x+c)^2)^3,x,method=_RETURNVERBOSE)
 

Output:

1/d*(-b/(a-b)^3*((1/8*b*(7*a^2-10*a*b+3*b^2)/a^2*tan(d*x+c)^3+1/8*(9*a^2-1 
4*a*b+5*b^2)/a*tan(d*x+c))/(a+b*tan(d*x+c)^2)^2+1/8*(15*a^2-10*a*b+3*b^2)/ 
a^2/(a*b)^(1/2)*arctan(b*tan(d*x+c)/(a*b)^(1/2)))+1/(a-b)^3*arctan(tan(d*x 
+c)))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 341 vs. \(2 (136) = 272\).

Time = 0.12 (sec) , antiderivative size = 742, normalized size of antiderivative = 4.95 \[ \int \frac {1}{\left (a+b \tan ^2(c+d x)\right )^3} \, dx =\text {Too large to display} \] Input:

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

Output:

[1/32*(32*a^2*b^2*d*x*tan(d*x + c)^4 + 64*a^3*b*d*x*tan(d*x + c)^2 + 32*a^ 
4*d*x - 4*(7*a^2*b^2 - 10*a*b^3 + 3*b^4)*tan(d*x + c)^3 - ((15*a^2*b^2 - 1 
0*a*b^3 + 3*b^4)*tan(d*x + c)^4 + 15*a^4 - 10*a^3*b + 3*a^2*b^2 + 2*(15*a^ 
3*b - 10*a^2*b^2 + 3*a*b^3)*tan(d*x + c)^2)*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*(9* 
a^3*b - 14*a^2*b^2 + 5*a*b^3)*tan(d*x + c))/((a^5*b^2 - 3*a^4*b^3 + 3*a^3* 
b^4 - a^2*b^5)*d*tan(d*x + c)^4 + 2*(a^6*b - 3*a^5*b^2 + 3*a^4*b^3 - a^3*b 
^4)*d*tan(d*x + c)^2 + (a^7 - 3*a^6*b + 3*a^5*b^2 - a^4*b^3)*d), 1/16*(16* 
a^2*b^2*d*x*tan(d*x + c)^4 + 32*a^3*b*d*x*tan(d*x + c)^2 + 16*a^4*d*x - 2* 
(7*a^2*b^2 - 10*a*b^3 + 3*b^4)*tan(d*x + c)^3 - ((15*a^2*b^2 - 10*a*b^3 + 
3*b^4)*tan(d*x + c)^4 + 15*a^4 - 10*a^3*b + 3*a^2*b^2 + 2*(15*a^3*b - 10*a 
^2*b^2 + 3*a*b^3)*tan(d*x + c)^2)*sqrt(b/a)*arctan(1/2*(b*tan(d*x + c)^2 - 
 a)*sqrt(b/a)/(b*tan(d*x + c))) - 2*(9*a^3*b - 14*a^2*b^2 + 5*a*b^3)*tan(d 
*x + c))/((a^5*b^2 - 3*a^4*b^3 + 3*a^3*b^4 - a^2*b^5)*d*tan(d*x + c)^4 + 2 
*(a^6*b - 3*a^5*b^2 + 3*a^4*b^3 - a^3*b^4)*d*tan(d*x + c)^2 + (a^7 - 3*a^6 
*b + 3*a^5*b^2 - a^4*b^3)*d)]
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 8964 vs. \(2 (129) = 258\).

Time = 49.24 (sec) , antiderivative size = 8964, normalized size of antiderivative = 59.76 \[ \int \frac {1}{\left (a+b \tan ^2(c+d x)\right )^3} \, dx=\text {Too large to display} \] Input:

integrate(1/(a+b*tan(d*x+c)**2)**3,x)
 

Output:

Piecewise((zoo*x/tan(c)**6, Eq(a, 0) & Eq(b, 0) & Eq(d, 0)), (x/a**3, Eq(b 
, 0)), ((-x - 1/(d*tan(c + d*x)) + 1/(3*d*tan(c + d*x)**3) - 1/(5*d*tan(c 
+ d*x)**5))/b**3, Eq(a, 0)), (15*d*x*tan(c + d*x)**6/(48*b**3*d*tan(c + d* 
x)**6 + 144*b**3*d*tan(c + d*x)**4 + 144*b**3*d*tan(c + d*x)**2 + 48*b**3* 
d) + 45*d*x*tan(c + d*x)**4/(48*b**3*d*tan(c + d*x)**6 + 144*b**3*d*tan(c 
+ d*x)**4 + 144*b**3*d*tan(c + d*x)**2 + 48*b**3*d) + 45*d*x*tan(c + d*x)* 
*2/(48*b**3*d*tan(c + d*x)**6 + 144*b**3*d*tan(c + d*x)**4 + 144*b**3*d*ta 
n(c + d*x)**2 + 48*b**3*d) + 15*d*x/(48*b**3*d*tan(c + d*x)**6 + 144*b**3* 
d*tan(c + d*x)**4 + 144*b**3*d*tan(c + d*x)**2 + 48*b**3*d) + 15*tan(c + d 
*x)**5/(48*b**3*d*tan(c + d*x)**6 + 144*b**3*d*tan(c + d*x)**4 + 144*b**3* 
d*tan(c + d*x)**2 + 48*b**3*d) + 40*tan(c + d*x)**3/(48*b**3*d*tan(c + d*x 
)**6 + 144*b**3*d*tan(c + d*x)**4 + 144*b**3*d*tan(c + d*x)**2 + 48*b**3*d 
) + 33*tan(c + d*x)/(48*b**3*d*tan(c + d*x)**6 + 144*b**3*d*tan(c + d*x)** 
4 + 144*b**3*d*tan(c + d*x)**2 + 48*b**3*d), Eq(a, b)), (x/(a + b*tan(c)** 
2)**3, Eq(d, 0)), (16*a**4*d*x*sqrt(-a/b)/(16*a**7*d*sqrt(-a/b) + 32*a**6* 
b*d*sqrt(-a/b)*tan(c + d*x)**2 - 48*a**6*b*d*sqrt(-a/b) + 16*a**5*b**2*d*s 
qrt(-a/b)*tan(c + d*x)**4 - 96*a**5*b**2*d*sqrt(-a/b)*tan(c + d*x)**2 + 48 
*a**5*b**2*d*sqrt(-a/b) - 48*a**4*b**3*d*sqrt(-a/b)*tan(c + d*x)**4 + 96*a 
**4*b**3*d*sqrt(-a/b)*tan(c + d*x)**2 - 16*a**4*b**3*d*sqrt(-a/b) + 48*a** 
3*b**4*d*sqrt(-a/b)*tan(c + d*x)**4 - 32*a**3*b**4*d*sqrt(-a/b)*tan(c +...
 

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 227, normalized size of antiderivative = 1.51 \[ \int \frac {1}{\left (a+b \tan ^2(c+d x)\right )^3} \, dx=-\frac {\frac {{\left (15 \, a^{2} b - 10 \, a b^{2} + 3 \, b^{3}\right )} \arctan \left (\frac {b \tan \left (d x + c\right )}{\sqrt {a b}}\right )}{{\left (a^{5} - 3 \, a^{4} b + 3 \, a^{3} b^{2} - a^{2} b^{3}\right )} \sqrt {a b}} + \frac {{\left (7 \, a b^{2} - 3 \, b^{3}\right )} \tan \left (d x + c\right )^{3} + {\left (9 \, a^{2} b - 5 \, a b^{2}\right )} \tan \left (d x + c\right )}{a^{6} - 2 \, a^{5} b + a^{4} b^{2} + {\left (a^{4} b^{2} - 2 \, a^{3} b^{3} + a^{2} b^{4}\right )} \tan \left (d x + c\right )^{4} + 2 \, {\left (a^{5} b - 2 \, a^{4} b^{2} + a^{3} b^{3}\right )} \tan \left (d x + c\right )^{2}} - \frac {8 \, {\left (d x + c\right )}}{a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}}}{8 \, d} \] Input:

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

Output:

-1/8*((15*a^2*b - 10*a*b^2 + 3*b^3)*arctan(b*tan(d*x + c)/sqrt(a*b))/((a^5 
 - 3*a^4*b + 3*a^3*b^2 - a^2*b^3)*sqrt(a*b)) + ((7*a*b^2 - 3*b^3)*tan(d*x 
+ c)^3 + (9*a^2*b - 5*a*b^2)*tan(d*x + c))/(a^6 - 2*a^5*b + a^4*b^2 + (a^4 
*b^2 - 2*a^3*b^3 + a^2*b^4)*tan(d*x + c)^4 + 2*(a^5*b - 2*a^4*b^2 + a^3*b^ 
3)*tan(d*x + c)^2) - 8*(d*x + c)/(a^3 - 3*a^2*b + 3*a*b^2 - b^3))/d
 

Giac [A] (verification not implemented)

Time = 0.65 (sec) , antiderivative size = 198, normalized size of antiderivative = 1.32 \[ \int \frac {1}{\left (a+b \tan ^2(c+d x)\right )^3} \, dx=-\frac {{\left (15 \, a^{2} b - 10 \, a b^{2} + 3 \, b^{3}\right )} \arctan \left (\frac {b \tan \left (d x + c\right )}{\sqrt {a b}}\right )}{8 \, {\left (a^{5} d - 3 \, a^{4} b d + 3 \, a^{3} b^{2} d - a^{2} b^{3} d\right )} \sqrt {a b}} + \frac {d x + c}{a^{3} d - 3 \, a^{2} b d + 3 \, a b^{2} d - b^{3} d} - \frac {7 \, a b^{2} \tan \left (d x + c\right )^{3} - 3 \, b^{3} \tan \left (d x + c\right )^{3} + 9 \, a^{2} b \tan \left (d x + c\right ) - 5 \, a b^{2} \tan \left (d x + c\right )}{8 \, {\left (a^{4} d - 2 \, a^{3} b d + a^{2} b^{2} d\right )} {\left (b \tan \left (d x + c\right )^{2} + a\right )}^{2}} \] Input:

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

Output:

-1/8*(15*a^2*b - 10*a*b^2 + 3*b^3)*arctan(b*tan(d*x + c)/sqrt(a*b))/((a^5* 
d - 3*a^4*b*d + 3*a^3*b^2*d - a^2*b^3*d)*sqrt(a*b)) + (d*x + c)/(a^3*d - 3 
*a^2*b*d + 3*a*b^2*d - b^3*d) - 1/8*(7*a*b^2*tan(d*x + c)^3 - 3*b^3*tan(d* 
x + c)^3 + 9*a^2*b*tan(d*x + c) - 5*a*b^2*tan(d*x + c))/((a^4*d - 2*a^3*b* 
d + a^2*b^2*d)*(b*tan(d*x + c)^2 + a)^2)
 

Mupad [B] (verification not implemented)

Time = 9.76 (sec) , antiderivative size = 3901, normalized size of antiderivative = 26.01 \[ \int \frac {1}{\left (a+b \tan ^2(c+d x)\right )^3} \, dx=\text {Too large to display} \] Input:

int(1/(a + b*tan(c + d*x)^2)^3,x)
 

Output:

(atan((((-a^5*b)^(1/2)*((tan(c + d*x)*(9*b^7 - 60*a*b^6 + 190*a^2*b^5 - 30 
0*a^3*b^4 + 289*a^4*b^3))/(32*(a^8 - 4*a^7*b + a^4*b^4 - 4*a^5*b^3 + 6*a^6 
*b^2)) - (((96*a^2*b^10 - 800*a^3*b^9 + 3040*a^4*b^8 - 6816*a^5*b^7 + 9760 
*a^6*b^6 - 9056*a^7*b^5 + 5280*a^8*b^4 - 1760*a^9*b^3 + 256*a^10*b^2)/(64* 
(a^10 - 6*a^9*b + a^4*b^6 - 6*a^5*b^5 + 15*a^6*b^4 - 20*a^7*b^3 + 15*a^8*b 
^2)) - (tan(c + d*x)*(-a^5*b)^(1/2)*(15*a^2 - 10*a*b + 3*b^2)*(256*a^4*b^9 
 - 1280*a^5*b^8 + 2304*a^6*b^7 - 1280*a^7*b^6 - 1280*a^8*b^5 + 2304*a^9*b^ 
4 - 1280*a^10*b^3 + 256*a^11*b^2))/(512*(3*a^7*b - a^8 + a^5*b^3 - 3*a^6*b 
^2)*(a^8 - 4*a^7*b + a^4*b^4 - 4*a^5*b^3 + 6*a^6*b^2)))*(-a^5*b)^(1/2)*(15 
*a^2 - 10*a*b + 3*b^2))/(16*(3*a^7*b - a^8 + a^5*b^3 - 3*a^6*b^2)))*(15*a^ 
2 - 10*a*b + 3*b^2)*1i)/(16*(3*a^7*b - a^8 + a^5*b^3 - 3*a^6*b^2)) + ((-a^ 
5*b)^(1/2)*((tan(c + d*x)*(9*b^7 - 60*a*b^6 + 190*a^2*b^5 - 300*a^3*b^4 + 
289*a^4*b^3))/(32*(a^8 - 4*a^7*b + a^4*b^4 - 4*a^5*b^3 + 6*a^6*b^2)) + ((( 
96*a^2*b^10 - 800*a^3*b^9 + 3040*a^4*b^8 - 6816*a^5*b^7 + 9760*a^6*b^6 - 9 
056*a^7*b^5 + 5280*a^8*b^4 - 1760*a^9*b^3 + 256*a^10*b^2)/(64*(a^10 - 6*a^ 
9*b + a^4*b^6 - 6*a^5*b^5 + 15*a^6*b^4 - 20*a^7*b^3 + 15*a^8*b^2)) + (tan( 
c + d*x)*(-a^5*b)^(1/2)*(15*a^2 - 10*a*b + 3*b^2)*(256*a^4*b^9 - 1280*a^5* 
b^8 + 2304*a^6*b^7 - 1280*a^7*b^6 - 1280*a^8*b^5 + 2304*a^9*b^4 - 1280*a^1 
0*b^3 + 256*a^11*b^2))/(512*(3*a^7*b - a^8 + a^5*b^3 - 3*a^6*b^2)*(a^8 - 4 
*a^7*b + a^4*b^4 - 4*a^5*b^3 + 6*a^6*b^2)))*(-a^5*b)^(1/2)*(15*a^2 - 10...
 

Reduce [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 577, normalized size of antiderivative = 3.85 \[ \int \frac {1}{\left (a+b \tan ^2(c+d x)\right )^3} \, dx=\frac {-15 \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 )^{4} a^{2} b^{2}+10 \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 )^{4} a \,b^{3}-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 )^{4} b^{4}-30 \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^{3} b +20 \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^{2} b^{2}-6 \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^{3}-15 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {\tan \left (d x +c \right ) b}{\sqrt {b}\, \sqrt {a}}\right ) a^{4}+10 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {\tan \left (d x +c \right ) b}{\sqrt {b}\, \sqrt {a}}\right ) a^{3} b -3 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {\tan \left (d x +c \right ) b}{\sqrt {b}\, \sqrt {a}}\right ) a^{2} b^{2}+8 \tan \left (d x +c \right )^{4} a^{3} b^{2} d x -7 \tan \left (d x +c \right )^{3} a^{3} b^{2}+10 \tan \left (d x +c \right )^{3} a^{2} b^{3}-3 \tan \left (d x +c \right )^{3} a \,b^{4}+16 \tan \left (d x +c \right )^{2} a^{4} b d x -9 \tan \left (d x +c \right ) a^{4} b +14 \tan \left (d x +c \right ) a^{3} b^{2}-5 \tan \left (d x +c \right ) a^{2} b^{3}+8 a^{5} d x}{8 a^{3} d \left (\tan \left (d x +c \right )^{4} a^{3} b^{2}-3 \tan \left (d x +c \right )^{4} a^{2} b^{3}+3 \tan \left (d x +c \right )^{4} a \,b^{4}-\tan \left (d x +c \right )^{4} b^{5}+2 \tan \left (d x +c \right )^{2} a^{4} b -6 \tan \left (d x +c \right )^{2} a^{3} b^{2}+6 \tan \left (d x +c \right )^{2} a^{2} b^{3}-2 \tan \left (d x +c \right )^{2} a \,b^{4}+a^{5}-3 a^{4} b +3 a^{3} b^{2}-a^{2} b^{3}\right )} \] Input:

int(1/(a+b*tan(d*x+c)^2)^3,x)
 

Output:

( - 15*sqrt(b)*sqrt(a)*atan((tan(c + d*x)*b)/(sqrt(b)*sqrt(a)))*tan(c + d* 
x)**4*a**2*b**2 + 10*sqrt(b)*sqrt(a)*atan((tan(c + d*x)*b)/(sqrt(b)*sqrt(a 
)))*tan(c + d*x)**4*a*b**3 - 3*sqrt(b)*sqrt(a)*atan((tan(c + d*x)*b)/(sqrt 
(b)*sqrt(a)))*tan(c + d*x)**4*b**4 - 30*sqrt(b)*sqrt(a)*atan((tan(c + d*x) 
*b)/(sqrt(b)*sqrt(a)))*tan(c + d*x)**2*a**3*b + 20*sqrt(b)*sqrt(a)*atan((t 
an(c + d*x)*b)/(sqrt(b)*sqrt(a)))*tan(c + d*x)**2*a**2*b**2 - 6*sqrt(b)*sq 
rt(a)*atan((tan(c + d*x)*b)/(sqrt(b)*sqrt(a)))*tan(c + d*x)**2*a*b**3 - 15 
*sqrt(b)*sqrt(a)*atan((tan(c + d*x)*b)/(sqrt(b)*sqrt(a)))*a**4 + 10*sqrt(b 
)*sqrt(a)*atan((tan(c + d*x)*b)/(sqrt(b)*sqrt(a)))*a**3*b - 3*sqrt(b)*sqrt 
(a)*atan((tan(c + d*x)*b)/(sqrt(b)*sqrt(a)))*a**2*b**2 + 8*tan(c + d*x)**4 
*a**3*b**2*d*x - 7*tan(c + d*x)**3*a**3*b**2 + 10*tan(c + d*x)**3*a**2*b** 
3 - 3*tan(c + d*x)**3*a*b**4 + 16*tan(c + d*x)**2*a**4*b*d*x - 9*tan(c + d 
*x)*a**4*b + 14*tan(c + d*x)*a**3*b**2 - 5*tan(c + d*x)*a**2*b**3 + 8*a**5 
*d*x)/(8*a**3*d*(tan(c + d*x)**4*a**3*b**2 - 3*tan(c + d*x)**4*a**2*b**3 + 
 3*tan(c + d*x)**4*a*b**4 - tan(c + d*x)**4*b**5 + 2*tan(c + d*x)**2*a**4* 
b - 6*tan(c + d*x)**2*a**3*b**2 + 6*tan(c + d*x)**2*a**2*b**3 - 2*tan(c + 
d*x)**2*a*b**4 + a**5 - 3*a**4*b + 3*a**3*b**2 - a**2*b**3))