Integrand size = 14, antiderivative size = 150 \[ \int \frac {1}{\left (a+b \cot ^2(c+d x)\right )^3} \, dx=\frac {x}{(a-b)^3}+\frac {\sqrt {b} \left (15 a^2-10 a b+3 b^2\right ) \arctan \left (\frac {\sqrt {b} \cot (c+d x)}{\sqrt {a}}\right )}{8 a^{5/2} (a-b)^3 d}+\frac {b \cot (c+d x)}{4 a (a-b) d \left (a+b \cot ^2(c+d x)\right )^2}+\frac {(7 a-3 b) b \cot (c+d x)}{8 a^2 (a-b)^2 d \left (a+b \cot ^2(c+d x)\right )} \]
x/(a-b)^3+1/4*b*cot(d*x+c)/a/(a-b)/d/(a+b*cot(d*x+c)^2)^2+1/8*(7*a-3*b)*b* cot(d*x+c)/a^2/(a-b)^2/d/(a+b*cot(d*x+c)^2)+1/8*(15*a^2-10*a*b+3*b^2)*arct an(cot(d*x+c)*b^(1/2)/a^(1/2))*b^(1/2)/a^(5/2)/(a-b)^3/d
Time = 0.35 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.92 \[ \int \frac {1}{\left (a+b \cot ^2(c+d x)\right )^3} \, dx=\frac {-8 \arctan (\cot (c+d x))+\frac {\sqrt {b} \left (15 a^2-10 a b+3 b^2\right ) \arctan \left (\frac {\sqrt {b} \cot (c+d x)}{\sqrt {a}}\right )}{a^{5/2}}+\frac {2 (a-b)^2 b \cot (c+d x)}{a \left (a+b \cot ^2(c+d x)\right )^2}+\frac {(7 a-3 b) (a-b) b \cot (c+d x)}{a^2 \left (a+b \cot ^2(c+d x)\right )}}{8 (a-b)^3 d} \]
(-8*ArcTan[Cot[c + d*x]] + (Sqrt[b]*(15*a^2 - 10*a*b + 3*b^2)*ArcTan[(Sqrt [b]*Cot[c + d*x])/Sqrt[a]])/a^(5/2) + (2*(a - b)^2*b*Cot[c + d*x])/(a*(a + b*Cot[c + d*x]^2)^2) + ((7*a - 3*b)*(a - b)*b*Cot[c + d*x])/(a^2*(a + b*C ot[c + d*x]^2)))/(8*(a - b)^3*d)
Time = 0.33 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.23, 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.
\(\displaystyle \int \frac {1}{\left (a+b \cot ^2(c+d x)\right )^3} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\left (a+b \tan \left (c+d x+\frac {\pi }{2}\right )^2\right )^3}dx\) |
\(\Big \downarrow \) 4144 |
\(\displaystyle -\frac {\int \frac {1}{\left (\cot ^2(c+d x)+1\right ) \left (b \cot ^2(c+d x)+a\right )^3}d\cot (c+d x)}{d}\) |
\(\Big \downarrow \) 316 |
\(\displaystyle -\frac {\frac {\int \frac {-3 b \cot ^2(c+d x)+4 a-3 b}{\left (\cot ^2(c+d x)+1\right ) \left (b \cot ^2(c+d x)+a\right )^2}d\cot (c+d x)}{4 a (a-b)}-\frac {b \cot (c+d x)}{4 a (a-b) \left (a+b \cot ^2(c+d x)\right )^2}}{d}\) |
\(\Big \downarrow \) 402 |
\(\displaystyle -\frac {\frac {\frac {\int \frac {8 a^2-7 b a+3 b^2-(7 a-3 b) b \cot ^2(c+d x)}{\left (\cot ^2(c+d x)+1\right ) \left (b \cot ^2(c+d x)+a\right )}d\cot (c+d x)}{2 a (a-b)}-\frac {b (7 a-3 b) \cot (c+d x)}{2 a (a-b) \left (a+b \cot ^2(c+d x)\right )}}{4 a (a-b)}-\frac {b \cot (c+d x)}{4 a (a-b) \left (a+b \cot ^2(c+d x)\right )^2}}{d}\) |
\(\Big \downarrow \) 397 |
\(\displaystyle -\frac {\frac {\frac {\frac {8 a^2 \int \frac {1}{\cot ^2(c+d x)+1}d\cot (c+d x)}{a-b}-\frac {b \left (15 a^2-10 a b+3 b^2\right ) \int \frac {1}{b \cot ^2(c+d x)+a}d\cot (c+d x)}{a-b}}{2 a (a-b)}-\frac {b (7 a-3 b) \cot (c+d x)}{2 a (a-b) \left (a+b \cot ^2(c+d x)\right )}}{4 a (a-b)}-\frac {b \cot (c+d x)}{4 a (a-b) \left (a+b \cot ^2(c+d x)\right )^2}}{d}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle -\frac {\frac {\frac {\frac {8 a^2 \arctan (\cot (c+d x))}{a-b}-\frac {b \left (15 a^2-10 a b+3 b^2\right ) \int \frac {1}{b \cot ^2(c+d x)+a}d\cot (c+d x)}{a-b}}{2 a (a-b)}-\frac {b (7 a-3 b) \cot (c+d x)}{2 a (a-b) \left (a+b \cot ^2(c+d x)\right )}}{4 a (a-b)}-\frac {b \cot (c+d x)}{4 a (a-b) \left (a+b \cot ^2(c+d x)\right )^2}}{d}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle -\frac {\frac {\frac {\frac {8 a^2 \arctan (\cot (c+d x))}{a-b}-\frac {\sqrt {b} \left (15 a^2-10 a b+3 b^2\right ) \arctan \left (\frac {\sqrt {b} \cot (c+d x)}{\sqrt {a}}\right )}{\sqrt {a} (a-b)}}{2 a (a-b)}-\frac {b (7 a-3 b) \cot (c+d x)}{2 a (a-b) \left (a+b \cot ^2(c+d x)\right )}}{4 a (a-b)}-\frac {b \cot (c+d x)}{4 a (a-b) \left (a+b \cot ^2(c+d x)\right )^2}}{d}\) |
-((-1/4*(b*Cot[c + d*x])/(a*(a - b)*(a + b*Cot[c + d*x]^2)^2) + (((8*a^2*A rcTan[Cot[c + d*x]])/(a - b) - (Sqrt[b]*(15*a^2 - 10*a*b + 3*b^2)*ArcTan[( Sqrt[b]*Cot[c + d*x])/Sqrt[a]])/(Sqrt[a]*(a - b)))/(2*a*(a - b)) - ((7*a - 3*b)*b*Cot[c + d*x])/(2*a*(a - b)*(a + b*Cot[c + d*x]^2)))/(4*a*(a - b))) /d)
3.1.7.3.1 Defintions of rubi rules used
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])
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]
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]
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]
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]
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])
Time = 0.21 (sec) , antiderivative size = 148, normalized size of antiderivative = 0.99
method | result | size |
derivativedivides | \(\frac {-\frac {\frac {\pi }{2}-\operatorname {arccot}\left (\cot \left (d x +c \right )\right )}{\left (a -b \right )^{3}}+\frac {b \left (\frac {\frac {b \left (7 a^{2}-10 a b +3 b^{2}\right ) \cot \left (d x +c \right )^{3}}{8 a^{2}}+\frac {\left (9 a^{2}-14 a b +5 b^{2}\right ) \cot \left (d x +c \right )}{8 a}}{\left (a +b \cot \left (d x +c \right )^{2}\right )^{2}}+\frac {\left (15 a^{2}-10 a b +3 b^{2}\right ) \arctan \left (\frac {b \cot \left (d x +c \right )}{\sqrt {a b}}\right )}{8 a^{2} \sqrt {a b}}\right )}{\left (a -b \right )^{3}}}{d}\) | \(148\) |
default | \(\frac {-\frac {\frac {\pi }{2}-\operatorname {arccot}\left (\cot \left (d x +c \right )\right )}{\left (a -b \right )^{3}}+\frac {b \left (\frac {\frac {b \left (7 a^{2}-10 a b +3 b^{2}\right ) \cot \left (d x +c \right )^{3}}{8 a^{2}}+\frac {\left (9 a^{2}-14 a b +5 b^{2}\right ) \cot \left (d x +c \right )}{8 a}}{\left (a +b \cot \left (d x +c \right )^{2}\right )^{2}}+\frac {\left (15 a^{2}-10 a b +3 b^{2}\right ) \arctan \left (\frac {b \cot \left (d x +c \right )}{\sqrt {a b}}\right )}{8 a^{2} \sqrt {a b}}\right )}{\left (a -b \right )^{3}}}{d}\) | \(148\) |
risch | \(\frac {x}{a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}}-\frac {i b \left (9 a^{3} {\mathrm e}^{6 i \left (d x +c \right )}+a^{2} b \,{\mathrm e}^{6 i \left (d x +c \right )}-13 a \,b^{2} {\mathrm e}^{6 i \left (d x +c \right )}+3 b^{3} {\mathrm e}^{6 i \left (d x +c \right )}-27 a^{3} {\mathrm e}^{4 i \left (d x +c \right )}-9 a^{2} b \,{\mathrm e}^{4 i \left (d x +c \right )}-21 a \,b^{2} {\mathrm e}^{4 i \left (d x +c \right )}+9 b^{3} {\mathrm e}^{4 i \left (d x +c \right )}+27 a^{3} {\mathrm e}^{2 i \left (d x +c \right )}-13 a^{2} b \,{\mathrm e}^{2 i \left (d x +c \right )}-23 a \,b^{2} {\mathrm e}^{2 i \left (d x +c \right )}+9 b^{3} {\mathrm e}^{2 i \left (d x +c \right )}-9 a^{3}+21 a^{2} b -15 a \,b^{2}+3 b^{3}\right )}{4 \left (-a \,{\mathrm e}^{4 i \left (d x +c \right )}+b \,{\mathrm e}^{4 i \left (d x +c \right )}+2 \,{\mathrm e}^{2 i \left (d x +c \right )} a +2 \,{\mathrm e}^{2 i \left (d x +c \right )} b -a +b \right )^{2} \left (-a +b \right ) \left (a^{2}-2 a b +b^{2}\right ) a^{2} d}+\frac {15 \sqrt {-a b}\, \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {2 i \sqrt {-a b}+a +b}{a -b}\right )}{16 a \left (a -b \right )^{3} d}-\frac {5 \sqrt {-a b}\, \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {2 i \sqrt {-a b}+a +b}{a -b}\right ) b}{8 a^{2} \left (a -b \right )^{3} d}+\frac {3 \sqrt {-a b}\, \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {2 i \sqrt {-a b}+a +b}{a -b}\right ) b^{2}}{16 a^{3} \left (a -b \right )^{3} d}-\frac {15 \sqrt {-a b}\, \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 i \sqrt {-a b}-a -b}{a -b}\right )}{16 a \left (a -b \right )^{3} d}+\frac {5 \sqrt {-a b}\, \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 i \sqrt {-a b}-a -b}{a -b}\right ) b}{8 a^{2} \left (a -b \right )^{3} d}-\frac {3 \sqrt {-a b}\, \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 i \sqrt {-a b}-a -b}{a -b}\right ) b^{2}}{16 a^{3} \left (a -b \right )^{3} d}\) | \(642\) |
1/d*(-1/(a-b)^3*(1/2*Pi-arccot(cot(d*x+c)))+b/(a-b)^3*((1/8*b*(7*a^2-10*a* b+3*b^2)/a^2*cot(d*x+c)^3+1/8*(9*a^2-14*a*b+5*b^2)/a*cot(d*x+c))/(a+b*cot( d*x+c)^2)^2+1/8*(15*a^2-10*a*b+3*b^2)/a^2/(a*b)^(1/2)*arctan(b*cot(d*x+c)/ (a*b)^(1/2))))
Leaf count of result is larger than twice the leaf count of optimal. 475 vs. \(2 (136) = 272\).
Time = 0.33 (sec) , antiderivative size = 1068, normalized size of antiderivative = 7.12 \[ \int \frac {1}{\left (a+b \cot ^2(c+d x)\right )^3} \, dx=\text {Too large to display} \]
[1/32*(32*(a^4 - 2*a^3*b + a^2*b^2)*d*x*cos(2*d*x + 2*c)^2 - 64*(a^4 - a^2 *b^2)*d*x*cos(2*d*x + 2*c) + 32*(a^4 + 2*a^3*b + a^2*b^2)*d*x - (15*a^4 + 20*a^3*b - 2*a^2*b^2 - 4*a*b^3 + 3*b^4 + (15*a^4 - 40*a^3*b + 38*a^2*b^2 - 16*a*b^3 + 3*b^4)*cos(2*d*x + 2*c)^2 - 2*(15*a^4 - 10*a^3*b - 12*a^2*b^2 + 10*a*b^3 - 3*b^4)*cos(2*d*x + 2*c))*sqrt(-b/a)*log(((a^2 + 6*a*b + b^2)* cos(2*d*x + 2*c)^2 + 4*(a^2 - a*b - (a^2 + a*b)*cos(2*d*x + 2*c))*sqrt(-b/ a)*sin(2*d*x + 2*c) + a^2 - 6*a*b + b^2 - 2*(a^2 - b^2)*cos(2*d*x + 2*c))/ ((a^2 - 2*a*b + b^2)*cos(2*d*x + 2*c)^2 + a^2 + 2*a*b + b^2 - 2*(a^2 - b^2 )*cos(2*d*x + 2*c))) + 4*(9*a^3*b - 7*a^2*b^2 - 5*a*b^3 + 3*b^4 - 3*(3*a^3 *b - 7*a^2*b^2 + 5*a*b^3 - b^4)*cos(2*d*x + 2*c))*sin(2*d*x + 2*c))/((a^7 - 5*a^6*b + 10*a^5*b^2 - 10*a^4*b^3 + 5*a^3*b^4 - a^2*b^5)*d*cos(2*d*x + 2 *c)^2 - 2*(a^7 - 3*a^6*b + 2*a^5*b^2 + 2*a^4*b^3 - 3*a^3*b^4 + a^2*b^5)*d* cos(2*d*x + 2*c) + (a^7 - a^6*b - 2*a^5*b^2 + 2*a^4*b^3 + a^3*b^4 - a^2*b^ 5)*d), 1/16*(16*(a^4 - 2*a^3*b + a^2*b^2)*d*x*cos(2*d*x + 2*c)^2 - 32*(a^4 - a^2*b^2)*d*x*cos(2*d*x + 2*c) + 16*(a^4 + 2*a^3*b + a^2*b^2)*d*x + (15* a^4 + 20*a^3*b - 2*a^2*b^2 - 4*a*b^3 + 3*b^4 + (15*a^4 - 40*a^3*b + 38*a^2 *b^2 - 16*a*b^3 + 3*b^4)*cos(2*d*x + 2*c)^2 - 2*(15*a^4 - 10*a^3*b - 12*a^ 2*b^2 + 10*a*b^3 - 3*b^4)*cos(2*d*x + 2*c))*sqrt(b/a)*arctan(1/2*((a + b)* cos(2*d*x + 2*c) - a + b)*sqrt(b/a)/(b*sin(2*d*x + 2*c))) + 2*(9*a^3*b - 7 *a^2*b^2 - 5*a*b^3 + 3*b^4 - 3*(3*a^3*b - 7*a^2*b^2 + 5*a*b^3 - b^4)*co...
Leaf count of result is larger than twice the leaf count of optimal. 8964 vs. \(2 (133) = 266\).
Time = 48.20 (sec) , antiderivative size = 8964, normalized size of antiderivative = 59.76 \[ \int \frac {1}{\left (a+b \cot ^2(c+d x)\right )^3} \, dx=\text {Too large to display} \]
Piecewise((zoo*x/cot(c)**6, Eq(a, 0) & Eq(b, 0) & Eq(d, 0)), (x/a**3, Eq(b , 0)), ((-x + 1/(d*cot(c + d*x)) - 1/(3*d*cot(c + d*x)**3) + 1/(5*d*cot(c + d*x)**5))/b**3, Eq(a, 0)), (15*d*x*cot(c + d*x)**6/(48*b**3*d*cot(c + d* x)**6 + 144*b**3*d*cot(c + d*x)**4 + 144*b**3*d*cot(c + d*x)**2 + 48*b**3* d) + 45*d*x*cot(c + d*x)**4/(48*b**3*d*cot(c + d*x)**6 + 144*b**3*d*cot(c + d*x)**4 + 144*b**3*d*cot(c + d*x)**2 + 48*b**3*d) + 45*d*x*cot(c + d*x)* *2/(48*b**3*d*cot(c + d*x)**6 + 144*b**3*d*cot(c + d*x)**4 + 144*b**3*d*co t(c + d*x)**2 + 48*b**3*d) + 15*d*x/(48*b**3*d*cot(c + d*x)**6 + 144*b**3* d*cot(c + d*x)**4 + 144*b**3*d*cot(c + d*x)**2 + 48*b**3*d) - 15*cot(c + d *x)**5/(48*b**3*d*cot(c + d*x)**6 + 144*b**3*d*cot(c + d*x)**4 + 144*b**3* d*cot(c + d*x)**2 + 48*b**3*d) - 40*cot(c + d*x)**3/(48*b**3*d*cot(c + d*x )**6 + 144*b**3*d*cot(c + d*x)**4 + 144*b**3*d*cot(c + d*x)**2 + 48*b**3*d ) - 33*cot(c + d*x)/(48*b**3*d*cot(c + d*x)**6 + 144*b**3*d*cot(c + d*x)** 4 + 144*b**3*d*cot(c + d*x)**2 + 48*b**3*d), Eq(a, b)), (x/(a + b*cot(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)*cot(c + d*x)**2 - 48*a**6*b*d*sqrt(-a/b) + 16*a**5*b**2*d*s qrt(-a/b)*cot(c + d*x)**4 - 96*a**5*b**2*d*sqrt(-a/b)*cot(c + d*x)**2 + 48 *a**5*b**2*d*sqrt(-a/b) - 48*a**4*b**3*d*sqrt(-a/b)*cot(c + d*x)**4 + 96*a **4*b**3*d*sqrt(-a/b)*cot(c + d*x)**2 - 16*a**4*b**3*d*sqrt(-a/b) + 48*a** 3*b**4*d*sqrt(-a/b)*cot(c + d*x)**4 - 32*a**3*b**4*d*sqrt(-a/b)*cot(c +...
Time = 0.32 (sec) , antiderivative size = 228, normalized size of antiderivative = 1.52 \[ \int \frac {1}{\left (a+b \cot ^2(c+d x)\right )^3} \, dx=-\frac {\frac {{\left (15 \, a^{2} b - 10 \, a b^{2} + 3 \, b^{3}\right )} \arctan \left (\frac {a \tan \left (d x + c\right )}{\sqrt {a b}}\right )}{{\left (a^{5} - 3 \, a^{4} b + 3 \, a^{3} b^{2} - a^{2} b^{3}\right )} \sqrt {a b}} - \frac {{\left (9 \, a^{2} b - 5 \, a b^{2}\right )} \tan \left (d x + c\right )^{3} + {\left (7 \, a b^{2} - 3 \, b^{3}\right )} \tan \left (d x + c\right )}{a^{4} b^{2} - 2 \, a^{3} b^{3} + a^{2} b^{4} + {\left (a^{6} - 2 \, a^{5} b + a^{4} b^{2}\right )} \tan \left (d x + c\right )^{4} + 2 \, {\left (a^{5} b - 2 \, a^{4} b^{2} + a^{3} b^{3}\right )} \tan \left (d x + c\right )^{2}} - \frac {8 \, {\left (d x + c\right )}}{a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}}}{8 \, d} \]
-1/8*((15*a^2*b - 10*a*b^2 + 3*b^3)*arctan(a*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)) - ((9*a^2*b - 5*a*b^2)*tan(d* x + c)^3 + (7*a*b^2 - 3*b^3)*tan(d*x + c))/(a^4*b^2 - 2*a^3*b^3 + a^2*b^4 + (a^6 - 2*a^5*b + a^4*b^2)*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
Time = 0.40 (sec) , antiderivative size = 206, normalized size of antiderivative = 1.37 \[ \int \frac {1}{\left (a+b \cot ^2(c+d x)\right )^3} \, dx=-\frac {\frac {{\left (15 \, a^{2} b - 10 \, a b^{2} + 3 \, b^{3}\right )} {\left (\pi \left \lfloor \frac {d x + c}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (a\right ) + \arctan \left (\frac {a \tan \left (d x + c\right )}{\sqrt {a b}}\right )\right )}}{{\left (a^{5} - 3 \, a^{4} b + 3 \, a^{3} b^{2} - a^{2} b^{3}\right )} \sqrt {a b}} - \frac {8 \, {\left (d x + c\right )}}{a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}} - \frac {9 \, a^{2} b \tan \left (d x + c\right )^{3} - 5 \, a b^{2} \tan \left (d x + c\right )^{3} + 7 \, a b^{2} \tan \left (d x + c\right ) - 3 \, b^{3} \tan \left (d x + c\right )}{{\left (a^{4} - 2 \, a^{3} b + a^{2} b^{2}\right )} {\left (a \tan \left (d x + c\right )^{2} + b\right )}^{2}}}{8 \, d} \]
-1/8*((15*a^2*b - 10*a*b^2 + 3*b^3)*(pi*floor((d*x + c)/pi + 1/2)*sgn(a) + arctan(a*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)) - 8*(d*x + c)/(a^3 - 3*a^2*b + 3*a*b^2 - b^3) - (9*a^2*b*tan(d* x + c)^3 - 5*a*b^2*tan(d*x + c)^3 + 7*a*b^2*tan(d*x + c) - 3*b^3*tan(d*x + c))/((a^4 - 2*a^3*b + a^2*b^2)*(a*tan(d*x + c)^2 + b)^2))/d
Time = 16.09 (sec) , antiderivative size = 4866, normalized size of antiderivative = 32.44 \[ \int \frac {1}{\left (a+b \cot ^2(c+d x)\right )^3} \, dx=\text {Too large to display} \]
((cot(c + d*x)^3*(7*a*b^2 - 3*b^3))/(8*a^2*(a^2 - 2*a*b + b^2)) + (cot(c + d*x)*(9*a*b - 5*b^2))/(8*a*(a^2 - 2*a*b + b^2)))/(a^2*d + b^2*d*cot(c + d *x)^4 + 2*a*b*d*cot(c + d*x)^2) + (2*atan((((((96*a^2*b^10*d^2 - 800*a^3*b ^9*d^2 + 3040*a^4*b^8*d^2 - 6816*a^5*b^7*d^2 + 9760*a^6*b^6*d^2 - 9056*a^7 *b^5*d^2 + 5280*a^8*b^4*d^2 - 1760*a^9*b^3*d^2 + 256*a^10*b^2*d^2)/(64*(a^ 10*d^3 - 6*a^9*b*d^3 + a^4*b^6*d^3 - 6*a^5*b^5*d^3 + 15*a^6*b^4*d^3 - 20*a ^7*b^3*d^3 + 15*a^8*b^2*d^3)) - (cot(c + d*x)*(256*a^4*b^9*d^2 - 1280*a^5* b^8*d^2 + 2304*a^6*b^7*d^2 - 1280*a^7*b^6*d^2 - 1280*a^8*b^5*d^2 + 2304*a^ 9*b^4*d^2 - 1280*a^10*b^3*d^2 + 256*a^11*b^2*d^2)*1i)/(32*(2*a^3*d - 2*b^3 *d + 6*a*b^2*d - 6*a^2*b*d)*(a^8*d^2 - 4*a^7*b*d^2 + a^4*b^4*d^2 - 4*a^5*b ^3*d^2 + 6*a^6*b^2*d^2)))*1i)/(2*a^3*d - 2*b^3*d + 6*a*b^2*d - 6*a^2*b*d) - (cot(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*d^2 - 4*a^7*b*d^2 + a^4*b^4*d^2 - 4*a^5*b^3*d^2 + 6*a^6*b^2*d ^2)))/(2*a^3*d - 2*b^3*d + 6*a*b^2*d - 6*a^2*b*d) - ((((96*a^2*b^10*d^2 - 800*a^3*b^9*d^2 + 3040*a^4*b^8*d^2 - 6816*a^5*b^7*d^2 + 9760*a^6*b^6*d^2 - 9056*a^7*b^5*d^2 + 5280*a^8*b^4*d^2 - 1760*a^9*b^3*d^2 + 256*a^10*b^2*d^2 )/(64*(a^10*d^3 - 6*a^9*b*d^3 + a^4*b^6*d^3 - 6*a^5*b^5*d^3 + 15*a^6*b^4*d ^3 - 20*a^7*b^3*d^3 + 15*a^8*b^2*d^3)) + (cot(c + d*x)*(256*a^4*b^9*d^2 - 1280*a^5*b^8*d^2 + 2304*a^6*b^7*d^2 - 1280*a^7*b^6*d^2 - 1280*a^8*b^5*d^2 + 2304*a^9*b^4*d^2 - 1280*a^10*b^3*d^2 + 256*a^11*b^2*d^2)*1i)/(32*(2*a...