Integrand size = 23, antiderivative size = 148 \[ \int \frac {\cos ^2(c+d x)}{\left (a+b \tan ^2(c+d x)\right )^2} \, dx=\frac {(a-5 b) x}{2 (a-b)^3}+\frac {(5 a-b) b^{3/2} \arctan \left (\frac {\sqrt {b} \tan (c+d x)}{\sqrt {a}}\right )}{2 a^{3/2} (a-b)^3 d}+\frac {\cos (c+d x) \sin (c+d x)}{2 (a-b) d \left (a+b \tan ^2(c+d x)\right )}+\frac {b (a+b) \tan (c+d x)}{2 a (a-b)^2 d \left (a+b \tan ^2(c+d x)\right )} \]
1/2*(a-5*b)*x/(a-b)^3+1/2*(5*a-b)*b^(3/2)*arctan(b^(1/2)*tan(d*x+c)/a^(1/2 ))/a^(3/2)/(a-b)^3/d+1/2*cos(d*x+c)*sin(d*x+c)/(a-b)/d/(a+b*tan(d*x+c)^2)+ 1/2*b*(a+b)*tan(d*x+c)/a/(a-b)^2/d/(a+b*tan(d*x+c)^2)
Time = 1.80 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.78 \[ \int \frac {\cos ^2(c+d x)}{\left (a+b \tan ^2(c+d x)\right )^2} \, dx=\frac {2 (a-5 b) (c+d x)-\frac {2 b^{3/2} (-5 a+b) \arctan \left (\frac {\sqrt {b} \tan (c+d x)}{\sqrt {a}}\right )}{a^{3/2}}+(a-b) \sin (2 (c+d x))+\frac {2 (a-b) b^2 \sin (2 (c+d x))}{a (a+b+(a-b) \cos (2 (c+d x)))}}{4 (a-b)^3 d} \]
(2*(a - 5*b)*(c + d*x) - (2*b^(3/2)*(-5*a + b)*ArcTan[(Sqrt[b]*Tan[c + d*x ])/Sqrt[a]])/a^(3/2) + (a - b)*Sin[2*(c + d*x)] + (2*(a - b)*b^2*Sin[2*(c + d*x)])/(a*(a + b + (a - b)*Cos[2*(c + d*x)])))/(4*(a - b)^3*d)
Time = 0.37 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.16, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.391, Rules used = {3042, 4158, 316, 25, 402, 27, 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.
| \(\displaystyle \int \frac {\cos ^2(c+d x)}{\left (a+b \tan ^2(c+d x)\right )^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\sec (c+d x)^2 \left (a+b \tan (c+d x)^2\right )^2}dx\) |
| \(\Big \downarrow \) 4158 |
| \(\displaystyle \frac {\int \frac {1}{\left (\tan ^2(c+d x)+1\right )^2 \left (b \tan ^2(c+d x)+a\right )^2}d\tan (c+d x)}{d}\) |
| \(\Big \downarrow \) 316 |
| \(\displaystyle \frac {\frac {\tan (c+d x)}{2 (a-b) \left (\tan ^2(c+d x)+1\right ) \left (a+b \tan ^2(c+d x)\right )}-\frac {\int -\frac {3 b \tan ^2(c+d x)+a-2 b}{\left (\tan ^2(c+d x)+1\right ) \left (b \tan ^2(c+d x)+a\right )^2}d\tan (c+d x)}{2 (a-b)}}{d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {\int \frac {3 b \tan ^2(c+d x)+a-2 b}{\left (\tan ^2(c+d x)+1\right ) \left (b \tan ^2(c+d x)+a\right )^2}d\tan (c+d x)}{2 (a-b)}+\frac {\tan (c+d x)}{2 (a-b) \left (\tan ^2(c+d x)+1\right ) \left (a+b \tan ^2(c+d x)\right )}}{d}\) |
| \(\Big \downarrow \) 402 |
| \(\displaystyle \frac {\frac {\frac {\int \frac {2 \left (a^2-4 b a+b^2+b (a+b) \tan ^2(c+d x)\right )}{\left (\tan ^2(c+d x)+1\right ) \left (b \tan ^2(c+d x)+a\right )}d\tan (c+d x)}{2 a (a-b)}+\frac {b (a+b) \tan (c+d x)}{a (a-b) \left (a+b \tan ^2(c+d x)\right )}}{2 (a-b)}+\frac {\tan (c+d x)}{2 (a-b) \left (\tan ^2(c+d x)+1\right ) \left (a+b \tan ^2(c+d x)\right )}}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\frac {\int \frac {a^2-4 b a+b^2+b (a+b) \tan ^2(c+d x)}{\left (\tan ^2(c+d x)+1\right ) \left (b \tan ^2(c+d x)+a\right )}d\tan (c+d x)}{a (a-b)}+\frac {b (a+b) \tan (c+d x)}{a (a-b) \left (a+b \tan ^2(c+d x)\right )}}{2 (a-b)}+\frac {\tan (c+d x)}{2 (a-b) \left (\tan ^2(c+d x)+1\right ) \left (a+b \tan ^2(c+d x)\right )}}{d}\) |
| \(\Big \downarrow \) 397 |
| \(\displaystyle \frac {\frac {\frac {\frac {b^2 (5 a-b) \int \frac {1}{b \tan ^2(c+d x)+a}d\tan (c+d x)}{a-b}+\frac {a (a-5 b) \int \frac {1}{\tan ^2(c+d x)+1}d\tan (c+d x)}{a-b}}{a (a-b)}+\frac {b (a+b) \tan (c+d x)}{a (a-b) \left (a+b \tan ^2(c+d x)\right )}}{2 (a-b)}+\frac {\tan (c+d x)}{2 (a-b) \left (\tan ^2(c+d x)+1\right ) \left (a+b \tan ^2(c+d x)\right )}}{d}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle \frac {\frac {\frac {\frac {b^2 (5 a-b) \int \frac {1}{b \tan ^2(c+d x)+a}d\tan (c+d x)}{a-b}+\frac {a (a-5 b) \arctan (\tan (c+d x))}{a-b}}{a (a-b)}+\frac {b (a+b) \tan (c+d x)}{a (a-b) \left (a+b \tan ^2(c+d x)\right )}}{2 (a-b)}+\frac {\tan (c+d x)}{2 (a-b) \left (\tan ^2(c+d x)+1\right ) \left (a+b \tan ^2(c+d x)\right )}}{d}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {\frac {\frac {\frac {b^{3/2} (5 a-b) \arctan \left (\frac {\sqrt {b} \tan (c+d x)}{\sqrt {a}}\right )}{\sqrt {a} (a-b)}+\frac {a (a-5 b) \arctan (\tan (c+d x))}{a-b}}{a (a-b)}+\frac {b (a+b) \tan (c+d x)}{a (a-b) \left (a+b \tan ^2(c+d x)\right )}}{2 (a-b)}+\frac {\tan (c+d x)}{2 (a-b) \left (\tan ^2(c+d x)+1\right ) \left (a+b \tan ^2(c+d x)\right )}}{d}\) |
(Tan[c + d*x]/(2*(a - b)*(1 + Tan[c + d*x]^2)*(a + b*Tan[c + d*x]^2)) + (( (a*(a - 5*b)*ArcTan[Tan[c + d*x]])/(a - b) + ((5*a - b)*b^(3/2)*ArcTan[(Sq rt[b]*Tan[c + d*x])/Sqrt[a]])/(Sqrt[a]*(a - b)))/(a*(a - b)) + (b*(a + b)* Tan[c + d*x])/(a*(a - b)*(a + b*Tan[c + d*x]^2)))/(2*(a - b)))/d
3.5.72.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_ )])^(n_))^(p_.), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Sim p[ff/(c^(m - 1)*f) Subst[Int[(c^2 + ff^2*x^2)^(m/2 - 1)*(a + b*(ff*x)^n)^ p, x], x, c*(Tan[e + f*x]/ff)], x]] /; FreeQ[{a, b, c, e, f, n, p}, x] && I ntegerQ[m/2] && (IntegersQ[n, p] || IGtQ[m, 0] || IGtQ[p, 0] || EqQ[n^2, 4] || EqQ[n^2, 16])
Time = 3.63 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.86
| method | result | size |
| derivativedivides | \(\frac {\frac {\frac {\left (\frac {a}{2}-\frac {b}{2}\right ) \tan \left (d x +c \right )}{1+\tan \left (d x +c \right )^{2}}+\frac {\left (a -5 b \right ) \arctan \left (\tan \left (d x +c \right )\right )}{2}}{\left (a -b \right )^{3}}+\frac {b^{2} \left (\frac {\left (a -b \right ) \tan \left (d x +c \right )}{2 a \left (a +b \tan \left (d x +c \right )^{2}\right )}+\frac {\left (5 a -b \right ) \arctan \left (\frac {b \tan \left (d x +c \right )}{\sqrt {a b}}\right )}{2 a \sqrt {a b}}\right )}{\left (a -b \right )^{3}}}{d}\) | \(128\) |
| default | \(\frac {\frac {\frac {\left (\frac {a}{2}-\frac {b}{2}\right ) \tan \left (d x +c \right )}{1+\tan \left (d x +c \right )^{2}}+\frac {\left (a -5 b \right ) \arctan \left (\tan \left (d x +c \right )\right )}{2}}{\left (a -b \right )^{3}}+\frac {b^{2} \left (\frac {\left (a -b \right ) \tan \left (d x +c \right )}{2 a \left (a +b \tan \left (d x +c \right )^{2}\right )}+\frac {\left (5 a -b \right ) \arctan \left (\frac {b \tan \left (d x +c \right )}{\sqrt {a b}}\right )}{2 a \sqrt {a b}}\right )}{\left (a -b \right )^{3}}}{d}\) | \(128\) |
| risch | \(\frac {x a}{2 \left (a^{2}-2 a b +b^{2}\right ) \left (a -b \right )}-\frac {5 x b}{2 \left (a^{2}-2 a b +b^{2}\right ) \left (a -b \right )}-\frac {i {\mathrm e}^{2 i \left (d x +c \right )}}{8 d \left (a^{2}-2 a b +b^{2}\right )}+\frac {i {\mathrm e}^{-2 i \left (d x +c \right )}}{8 d \left (a^{2}-2 a b +b^{2}\right )}+\frac {i b^{2} \left (a \,{\mathrm e}^{2 i \left (d x +c \right )}+b \,{\mathrm e}^{2 i \left (d x +c \right )}+a -b \right )}{d a \left (-a +b \right )^{3} \left (-a \,{\mathrm e}^{4 i \left (d x +c \right )}+b \,{\mathrm e}^{4 i \left (d x +c \right )}-2 a \,{\mathrm e}^{2 i \left (d x +c \right )}-2 b \,{\mathrm e}^{2 i \left (d x +c \right )}-a +b \right )}+\frac {5 \sqrt {-a b}\, b \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {2 i \sqrt {-a b}-a -b}{a -b}\right )}{4 a \left (a -b \right )^{3} d}-\frac {\sqrt {-a b}\, b^{2} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {2 i \sqrt {-a b}-a -b}{a -b}\right )}{4 a^{2} \left (a -b \right )^{3} d}-\frac {5 \sqrt {-a b}\, b \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 i \sqrt {-a b}+a +b}{a -b}\right )}{4 a \left (a -b \right )^{3} d}+\frac {\sqrt {-a b}\, b^{2} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 i \sqrt {-a b}+a +b}{a -b}\right )}{4 a^{2} \left (a -b \right )^{3} d}\) | \(432\) |
1/d*(1/(a-b)^3*((1/2*a-1/2*b)*tan(d*x+c)/(1+tan(d*x+c)^2)+1/2*(a-5*b)*arct an(tan(d*x+c)))+b^2/(a-b)^3*(1/2/a*(a-b)*tan(d*x+c)/(a+b*tan(d*x+c)^2)+1/2 *(5*a-b)/a/(a*b)^(1/2)*arctan(b*tan(d*x+c)/(a*b)^(1/2))))
Time = 0.34 (sec) , antiderivative size = 614, normalized size of antiderivative = 4.15 \[ \int \frac {\cos ^2(c+d x)}{\left (a+b \tan ^2(c+d x)\right )^2} \, dx=\left [\frac {4 \, {\left (a^{3} - 6 \, a^{2} b + 5 \, a b^{2}\right )} d x \cos \left (d x + c\right )^{2} + 4 \, {\left (a^{2} b - 5 \, a b^{2}\right )} d x + {\left (5 \, a b^{2} - b^{3} + {\left (5 \, a^{2} b - 6 \, a b^{2} + b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \sqrt {-\frac {b}{a}} \log \left (\frac {{\left (a^{2} + 6 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{4} - 2 \, {\left (3 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{2} - 4 \, {\left ({\left (a^{2} + a b\right )} \cos \left (d x + c\right )^{3} - a b \cos \left (d x + c\right )\right )} \sqrt {-\frac {b}{a}} \sin \left (d x + c\right ) + b^{2}}{{\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{4} + 2 \, {\left (a b - b^{2}\right )} \cos \left (d x + c\right )^{2} + b^{2}}\right ) + 4 \, {\left ({\left (a^{3} - 2 \, a^{2} b + a b^{2}\right )} \cos \left (d x + c\right )^{3} + {\left (a^{2} b - b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{8 \, {\left ({\left (a^{5} - 4 \, a^{4} b + 6 \, a^{3} b^{2} - 4 \, a^{2} b^{3} + a b^{4}\right )} d \cos \left (d x + c\right )^{2} + {\left (a^{4} b - 3 \, a^{3} b^{2} + 3 \, a^{2} b^{3} - a b^{4}\right )} d\right )}}, \frac {2 \, {\left (a^{3} - 6 \, a^{2} b + 5 \, a b^{2}\right )} d x \cos \left (d x + c\right )^{2} + 2 \, {\left (a^{2} b - 5 \, a b^{2}\right )} d x - {\left (5 \, a b^{2} - b^{3} + {\left (5 \, a^{2} b - 6 \, a b^{2} + b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \sqrt {\frac {b}{a}} \arctan \left (\frac {{\left ({\left (a + b\right )} \cos \left (d x + c\right )^{2} - b\right )} \sqrt {\frac {b}{a}}}{2 \, b \cos \left (d x + c\right ) \sin \left (d x + c\right )}\right ) + 2 \, {\left ({\left (a^{3} - 2 \, a^{2} b + a b^{2}\right )} \cos \left (d x + c\right )^{3} + {\left (a^{2} b - b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{4 \, {\left ({\left (a^{5} - 4 \, a^{4} b + 6 \, a^{3} b^{2} - 4 \, a^{2} b^{3} + a b^{4}\right )} d \cos \left (d x + c\right )^{2} + {\left (a^{4} b - 3 \, a^{3} b^{2} + 3 \, a^{2} b^{3} - a b^{4}\right )} d\right )}}\right ] \]
[1/8*(4*(a^3 - 6*a^2*b + 5*a*b^2)*d*x*cos(d*x + c)^2 + 4*(a^2*b - 5*a*b^2) *d*x + (5*a*b^2 - b^3 + (5*a^2*b - 6*a*b^2 + b^3)*cos(d*x + c)^2)*sqrt(-b/ a)*log(((a^2 + 6*a*b + b^2)*cos(d*x + c)^4 - 2*(3*a*b + b^2)*cos(d*x + c)^ 2 - 4*((a^2 + a*b)*cos(d*x + c)^3 - a*b*cos(d*x + c))*sqrt(-b/a)*sin(d*x + c) + b^2)/((a^2 - 2*a*b + b^2)*cos(d*x + c)^4 + 2*(a*b - b^2)*cos(d*x + c )^2 + b^2)) + 4*((a^3 - 2*a^2*b + a*b^2)*cos(d*x + c)^3 + (a^2*b - b^3)*co s(d*x + c))*sin(d*x + c))/((a^5 - 4*a^4*b + 6*a^3*b^2 - 4*a^2*b^3 + a*b^4) *d*cos(d*x + c)^2 + (a^4*b - 3*a^3*b^2 + 3*a^2*b^3 - a*b^4)*d), 1/4*(2*(a^ 3 - 6*a^2*b + 5*a*b^2)*d*x*cos(d*x + c)^2 + 2*(a^2*b - 5*a*b^2)*d*x - (5*a *b^2 - b^3 + (5*a^2*b - 6*a*b^2 + b^3)*cos(d*x + c)^2)*sqrt(b/a)*arctan(1/ 2*((a + b)*cos(d*x + c)^2 - b)*sqrt(b/a)/(b*cos(d*x + c)*sin(d*x + c))) + 2*((a^3 - 2*a^2*b + a*b^2)*cos(d*x + c)^3 + (a^2*b - b^3)*cos(d*x + c))*si n(d*x + c))/((a^5 - 4*a^4*b + 6*a^3*b^2 - 4*a^2*b^3 + a*b^4)*d*cos(d*x + c )^2 + (a^4*b - 3*a^3*b^2 + 3*a^2*b^3 - a*b^4)*d)]
Timed out. \[ \int \frac {\cos ^2(c+d x)}{\left (a+b \tan ^2(c+d x)\right )^2} \, dx=\text {Timed out} \]
Time = 0.30 (sec) , antiderivative size = 209, normalized size of antiderivative = 1.41 \[ \int \frac {\cos ^2(c+d x)}{\left (a+b \tan ^2(c+d x)\right )^2} \, dx=\frac {\frac {{\left (d x + c\right )} {\left (a - 5 \, b\right )}}{a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}} + \frac {{\left (5 \, a b^{2} - b^{3}\right )} \arctan \left (\frac {b \tan \left (d x + c\right )}{\sqrt {a b}}\right )}{{\left (a^{4} - 3 \, a^{3} b + 3 \, a^{2} b^{2} - a b^{3}\right )} \sqrt {a b}} + \frac {{\left (a b + b^{2}\right )} \tan \left (d x + c\right )^{3} + {\left (a^{2} + b^{2}\right )} \tan \left (d x + c\right )}{{\left (a^{3} b - 2 \, a^{2} b^{2} + a b^{3}\right )} \tan \left (d x + c\right )^{4} + a^{4} - 2 \, a^{3} b + a^{2} b^{2} + {\left (a^{4} - a^{3} b - a^{2} b^{2} + a b^{3}\right )} \tan \left (d x + c\right )^{2}}}{2 \, d} \]
1/2*((d*x + c)*(a - 5*b)/(a^3 - 3*a^2*b + 3*a*b^2 - b^3) + (5*a*b^2 - b^3) *arctan(b*tan(d*x + c)/sqrt(a*b))/((a^4 - 3*a^3*b + 3*a^2*b^2 - a*b^3)*sqr t(a*b)) + ((a*b + b^2)*tan(d*x + c)^3 + (a^2 + b^2)*tan(d*x + c))/((a^3*b - 2*a^2*b^2 + a*b^3)*tan(d*x + c)^4 + a^4 - 2*a^3*b + a^2*b^2 + (a^4 - a^3 *b - a^2*b^2 + a*b^3)*tan(d*x + c)^2))/d
Time = 0.64 (sec) , antiderivative size = 211, normalized size of antiderivative = 1.43 \[ \int \frac {\cos ^2(c+d x)}{\left (a+b \tan ^2(c+d x)\right )^2} \, dx=\frac {\frac {{\left (d x + c\right )} {\left (a - 5 \, b\right )}}{a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}} + \frac {{\left (5 \, a b^{2} - b^{3}\right )} {\left (\pi \left \lfloor \frac {d x + c}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (b\right ) + \arctan \left (\frac {b \tan \left (d x + c\right )}{\sqrt {a b}}\right )\right )}}{{\left (a^{4} - 3 \, a^{3} b + 3 \, a^{2} b^{2} - a b^{3}\right )} \sqrt {a b}} + \frac {a b \tan \left (d x + c\right )^{3} + b^{2} \tan \left (d x + c\right )^{3} + a^{2} \tan \left (d x + c\right ) + b^{2} \tan \left (d x + c\right )}{{\left (b \tan \left (d x + c\right )^{4} + a \tan \left (d x + c\right )^{2} + b \tan \left (d x + c\right )^{2} + a\right )} {\left (a^{3} - 2 \, a^{2} b + a b^{2}\right )}}}{2 \, d} \]
1/2*((d*x + c)*(a - 5*b)/(a^3 - 3*a^2*b + 3*a*b^2 - b^3) + (5*a*b^2 - b^3) *(pi*floor((d*x + c)/pi + 1/2)*sgn(b) + arctan(b*tan(d*x + c)/sqrt(a*b)))/ ((a^4 - 3*a^3*b + 3*a^2*b^2 - a*b^3)*sqrt(a*b)) + (a*b*tan(d*x + c)^3 + b^ 2*tan(d*x + c)^3 + a^2*tan(d*x + c) + b^2*tan(d*x + c))/((b*tan(d*x + c)^4 + a*tan(d*x + c)^2 + b*tan(d*x + c)^2 + a)*(a^3 - 2*a^2*b + a*b^2)))/d
Time = 15.92 (sec) , antiderivative size = 3843, normalized size of antiderivative = 25.97 \[ \int \frac {\cos ^2(c+d x)}{\left (a+b \tan ^2(c+d x)\right )^2} \, dx=\text {Too large to display} \]
((tan(c + d*x)*(a^2 + b^2))/(2*a*(a^2 - 2*a*b + b^2)) + (b*tan(c + d*x)^3* (a + b))/(2*a*(a^2 - 2*a*b + b^2)))/(d*(a + tan(c + d*x)^2*(a + b) + b*tan (c + d*x)^4)) - (atan(((((((2*a*b^10 - 20*a^2*b^9 + 80*a^3*b^8 - 172*a^4*b ^7 + 220*a^5*b^6 - 172*a^6*b^5 + 80*a^7*b^4 - 20*a^8*b^3 + 2*a^9*b^2)/(a^8 - 6*a^7*b + a^2*b^6 - 6*a^3*b^5 + 15*a^4*b^4 - 20*a^5*b^3 + 15*a^6*b^2) - (tan(c + d*x)*(a - 5*b)*(16*a^2*b^9 - 80*a^3*b^8 + 144*a^4*b^7 - 80*a^5*b ^6 - 80*a^6*b^5 + 144*a^7*b^4 - 80*a^8*b^3 + 16*a^9*b^2))/(8*(a*b^2*3i - a ^2*b*3i + a^3*1i - b^3*1i)*(a^6 - 4*a^5*b + a^2*b^4 - 4*a^3*b^3 + 6*a^4*b^ 2)))*(a - 5*b))/(4*(a*b^2*3i - a^2*b*3i + a^3*1i - b^3*1i)) - (tan(c + d*x )*(b^7 - 10*a*b^6 + 50*a^2*b^5 - 10*a^3*b^4 + a^4*b^3))/(2*(a^6 - 4*a^5*b + a^2*b^4 - 4*a^3*b^3 + 6*a^4*b^2)))*(a - 5*b)*1i)/(4*(a*b^2*3i - a^2*b*3i + a^3*1i - b^3*1i)) - (((((2*a*b^10 - 20*a^2*b^9 + 80*a^3*b^8 - 172*a^4*b ^7 + 220*a^5*b^6 - 172*a^6*b^5 + 80*a^7*b^4 - 20*a^8*b^3 + 2*a^9*b^2)/(a^8 - 6*a^7*b + a^2*b^6 - 6*a^3*b^5 + 15*a^4*b^4 - 20*a^5*b^3 + 15*a^6*b^2) + (tan(c + d*x)*(a - 5*b)*(16*a^2*b^9 - 80*a^3*b^8 + 144*a^4*b^7 - 80*a^5*b ^6 - 80*a^6*b^5 + 144*a^7*b^4 - 80*a^8*b^3 + 16*a^9*b^2))/(8*(a*b^2*3i - a ^2*b*3i + a^3*1i - b^3*1i)*(a^6 - 4*a^5*b + a^2*b^4 - 4*a^3*b^3 + 6*a^4*b^ 2)))*(a - 5*b))/(4*(a*b^2*3i - a^2*b*3i + a^3*1i - b^3*1i)) + (tan(c + d*x )*(b^7 - 10*a*b^6 + 50*a^2*b^5 - 10*a^3*b^4 + a^4*b^3))/(2*(a^6 - 4*a^5*b + a^2*b^4 - 4*a^3*b^3 + 6*a^4*b^2)))*(a - 5*b)*1i)/(4*(a*b^2*3i - a^2*b...