Integrand size = 23, antiderivative size = 212 \[ \int \frac {\cos ^4(c+d x)}{\left (a+b \tan ^2(c+d x)\right )^2} \, dx=\frac {\left (3 a^2-14 a b+35 b^2\right ) x}{8 (a-b)^4}-\frac {(7 a-b) b^{5/2} \arctan \left (\frac {\sqrt {b} \tan (c+d x)}{\sqrt {a}}\right )}{2 a^{3/2} (a-b)^4 d}+\frac {3 (a-3 b) \cos (c+d x) \sin (c+d x)}{8 (a-b)^2 d \left (a+b \tan ^2(c+d x)\right )}+\frac {\cos ^3(c+d x) \sin (c+d x)}{4 (a-b) d \left (a+b \tan ^2(c+d x)\right )}+\frac {(a-4 b) b (3 a+b) \tan (c+d x)}{8 a (a-b)^3 d \left (a+b \tan ^2(c+d x)\right )} \]
1/8*(3*a^2-14*a*b+35*b^2)*x/(a-b)^4-1/2*(7*a-b)*b^(5/2)*arctan(b^(1/2)*tan (d*x+c)/a^(1/2))/a^(3/2)/(a-b)^4/d+3/8*(a-3*b)*cos(d*x+c)*sin(d*x+c)/(a-b) ^2/d/(a+b*tan(d*x+c)^2)+1/4*cos(d*x+c)^3*sin(d*x+c)/(a-b)/d/(a+b*tan(d*x+c )^2)+1/8*(a-4*b)*b*(3*a+b)*tan(d*x+c)/a/(a-b)^3/d/(a+b*tan(d*x+c)^2)
Time = 3.26 (sec) , antiderivative size = 148, normalized size of antiderivative = 0.70 \[ \int \frac {\cos ^4(c+d x)}{\left (a+b \tan ^2(c+d x)\right )^2} \, dx=\frac {4 \left (3 a^2-14 a b+35 b^2\right ) (c+d x)+\frac {16 b^{5/2} (-7 a+b) \arctan \left (\frac {\sqrt {b} \tan (c+d x)}{\sqrt {a}}\right )}{a^{3/2}}+8 (a-3 b) (a-b) \sin (2 (c+d x))-\frac {16 (a-b) b^3 \sin (2 (c+d x))}{a (a+b+(a-b) \cos (2 (c+d x)))}+(a-b)^2 \sin (4 (c+d x))}{32 (a-b)^4 d} \]
(4*(3*a^2 - 14*a*b + 35*b^2)*(c + d*x) + (16*b^(5/2)*(-7*a + b)*ArcTan[(Sq rt[b]*Tan[c + d*x])/Sqrt[a]])/a^(3/2) + 8*(a - 3*b)*(a - b)*Sin[2*(c + d*x )] - (16*(a - b)*b^3*Sin[2*(c + d*x)])/(a*(a + b + (a - b)*Cos[2*(c + d*x) ])) + (a - b)^2*Sin[4*(c + d*x)])/(32*(a - b)^4*d)
Time = 0.44 (sec) , antiderivative size = 249, normalized size of antiderivative = 1.17, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.478, Rules used = {3042, 4158, 316, 25, 402, 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.
| \(\displaystyle \int \frac {\cos ^4(c+d x)}{\left (a+b \tan ^2(c+d x)\right )^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\sec (c+d x)^4 \left (a+b \tan (c+d x)^2\right )^2}dx\) |
| \(\Big \downarrow \) 4158 |
| \(\displaystyle \frac {\int \frac {1}{\left (\tan ^2(c+d x)+1\right )^3 \left (b \tan ^2(c+d x)+a\right )^2}d\tan (c+d x)}{d}\) |
| \(\Big \downarrow \) 316 |
| \(\displaystyle \frac {\frac {\tan (c+d x)}{4 (a-b) \left (\tan ^2(c+d x)+1\right )^2 \left (a+b \tan ^2(c+d x)\right )}-\frac {\int -\frac {5 b \tan ^2(c+d x)+3 a-4 b}{\left (\tan ^2(c+d x)+1\right )^2 \left (b \tan ^2(c+d x)+a\right )^2}d\tan (c+d x)}{4 (a-b)}}{d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {\int \frac {5 b \tan ^2(c+d x)+3 a-4 b}{\left (\tan ^2(c+d x)+1\right )^2 \left (b \tan ^2(c+d x)+a\right )^2}d\tan (c+d x)}{4 (a-b)}+\frac {\tan (c+d x)}{4 (a-b) \left (\tan ^2(c+d x)+1\right )^2 \left (a+b \tan ^2(c+d x)\right )}}{d}\) |
| \(\Big \downarrow \) 402 |
| \(\displaystyle \frac {\frac {\frac {3 (a-3 b) \tan (c+d x)}{2 (a-b) \left (\tan ^2(c+d x)+1\right ) \left (a+b \tan ^2(c+d x)\right )}-\frac {\int -\frac {3 a^2-5 b a+8 b^2+9 (a-3 b) b \tan ^2(c+d x)}{\left (\tan ^2(c+d x)+1\right ) \left (b \tan ^2(c+d x)+a\right )^2}d\tan (c+d x)}{2 (a-b)}}{4 (a-b)}+\frac {\tan (c+d x)}{4 (a-b) \left (\tan ^2(c+d x)+1\right )^2 \left (a+b \tan ^2(c+d x)\right )}}{d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {\frac {\int \frac {3 a^2-5 b a+8 b^2+9 (a-3 b) b \tan ^2(c+d x)}{\left (\tan ^2(c+d x)+1\right ) \left (b \tan ^2(c+d x)+a\right )^2}d\tan (c+d x)}{2 (a-b)}+\frac {3 (a-3 b) \tan (c+d x)}{2 (a-b) \left (\tan ^2(c+d x)+1\right ) \left (a+b \tan ^2(c+d x)\right )}}{4 (a-b)}+\frac {\tan (c+d x)}{4 (a-b) \left (\tan ^2(c+d x)+1\right )^2 \left (a+b \tan ^2(c+d x)\right )}}{d}\) |
| \(\Big \downarrow \) 402 |
| \(\displaystyle \frac {\frac {\frac {\frac {\int \frac {2 \left (3 a^3-11 b a^2+24 b^2 a-4 b^3+(a-4 b) b (3 a+b) \tan ^2(c+d x)\right )}{\left (\tan ^2(c+d x)+1\right ) \left (b \tan ^2(c+d x)+a\right )}d\tan (c+d x)}{2 a (a-b)}+\frac {b (a-4 b) (3 a+b) \tan (c+d x)}{a (a-b) \left (a+b \tan ^2(c+d x)\right )}}{2 (a-b)}+\frac {3 (a-3 b) \tan (c+d x)}{2 (a-b) \left (\tan ^2(c+d x)+1\right ) \left (a+b \tan ^2(c+d x)\right )}}{4 (a-b)}+\frac {\tan (c+d x)}{4 (a-b) \left (\tan ^2(c+d x)+1\right )^2 \left (a+b \tan ^2(c+d x)\right )}}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\frac {\frac {\int \frac {3 a^3-11 b a^2+24 b^2 a-4 b^3+(a-4 b) b (3 a+b) \tan ^2(c+d x)}{\left (\tan ^2(c+d x)+1\right ) \left (b \tan ^2(c+d x)+a\right )}d\tan (c+d x)}{a (a-b)}+\frac {b (a-4 b) (3 a+b) \tan (c+d x)}{a (a-b) \left (a+b \tan ^2(c+d x)\right )}}{2 (a-b)}+\frac {3 (a-3 b) \tan (c+d x)}{2 (a-b) \left (\tan ^2(c+d x)+1\right ) \left (a+b \tan ^2(c+d x)\right )}}{4 (a-b)}+\frac {\tan (c+d x)}{4 (a-b) \left (\tan ^2(c+d x)+1\right )^2 \left (a+b \tan ^2(c+d x)\right )}}{d}\) |
| \(\Big \downarrow \) 397 |
| \(\displaystyle \frac {\frac {\frac {\frac {\frac {a \left (3 a^2-14 a b+35 b^2\right ) \int \frac {1}{\tan ^2(c+d x)+1}d\tan (c+d x)}{a-b}-\frac {4 b^3 (7 a-b) \int \frac {1}{b \tan ^2(c+d x)+a}d\tan (c+d x)}{a-b}}{a (a-b)}+\frac {b (a-4 b) (3 a+b) \tan (c+d x)}{a (a-b) \left (a+b \tan ^2(c+d x)\right )}}{2 (a-b)}+\frac {3 (a-3 b) \tan (c+d x)}{2 (a-b) \left (\tan ^2(c+d x)+1\right ) \left (a+b \tan ^2(c+d x)\right )}}{4 (a-b)}+\frac {\tan (c+d x)}{4 (a-b) \left (\tan ^2(c+d x)+1\right )^2 \left (a+b \tan ^2(c+d x)\right )}}{d}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle \frac {\frac {\frac {\frac {\frac {a \left (3 a^2-14 a b+35 b^2\right ) \arctan (\tan (c+d x))}{a-b}-\frac {4 b^3 (7 a-b) \int \frac {1}{b \tan ^2(c+d x)+a}d\tan (c+d x)}{a-b}}{a (a-b)}+\frac {b (a-4 b) (3 a+b) \tan (c+d x)}{a (a-b) \left (a+b \tan ^2(c+d x)\right )}}{2 (a-b)}+\frac {3 (a-3 b) \tan (c+d x)}{2 (a-b) \left (\tan ^2(c+d x)+1\right ) \left (a+b \tan ^2(c+d x)\right )}}{4 (a-b)}+\frac {\tan (c+d x)}{4 (a-b) \left (\tan ^2(c+d x)+1\right )^2 \left (a+b \tan ^2(c+d x)\right )}}{d}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {\frac {\frac {\frac {\frac {a \left (3 a^2-14 a b+35 b^2\right ) \arctan (\tan (c+d x))}{a-b}-\frac {4 b^{5/2} (7 a-b) \arctan \left (\frac {\sqrt {b} \tan (c+d x)}{\sqrt {a}}\right )}{\sqrt {a} (a-b)}}{a (a-b)}+\frac {b (a-4 b) (3 a+b) \tan (c+d x)}{a (a-b) \left (a+b \tan ^2(c+d x)\right )}}{2 (a-b)}+\frac {3 (a-3 b) \tan (c+d x)}{2 (a-b) \left (\tan ^2(c+d x)+1\right ) \left (a+b \tan ^2(c+d x)\right )}}{4 (a-b)}+\frac {\tan (c+d x)}{4 (a-b) \left (\tan ^2(c+d x)+1\right )^2 \left (a+b \tan ^2(c+d x)\right )}}{d}\) |
(Tan[c + d*x]/(4*(a - b)*(1 + Tan[c + d*x]^2)^2*(a + b*Tan[c + d*x]^2)) + ((3*(a - 3*b)*Tan[c + d*x])/(2*(a - b)*(1 + Tan[c + d*x]^2)*(a + b*Tan[c + d*x]^2)) + (((a*(3*a^2 - 14*a*b + 35*b^2)*ArcTan[Tan[c + d*x]])/(a - b) - (4*(7*a - b)*b^(5/2)*ArcTan[(Sqrt[b]*Tan[c + d*x])/Sqrt[a]])/(Sqrt[a]*(a - b)))/(a*(a - b)) + ((a - 4*b)*b*(3*a + b)*Tan[c + d*x])/(a*(a - b)*(a + b*Tan[c + d*x]^2)))/(2*(a - b)))/(4*(a - b)))/d
3.5.73.3.1 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 = 15.26 (sec) , antiderivative size = 173, normalized size of antiderivative = 0.82
| method | result | size |
| derivativedivides | \(\frac {-\frac {b^{3} \left (\frac {\left (a -b \right ) \tan \left (d x +c \right )}{2 a \left (a +b \tan \left (d x +c \right )^{2}\right )}+\frac {\left (7 a -b \right ) \arctan \left (\frac {b \tan \left (d x +c \right )}{\sqrt {a b}}\right )}{2 a \sqrt {a b}}\right )}{\left (a -b \right )^{4}}+\frac {\frac {\left (\frac {3}{8} a^{2}-\frac {7}{4} a b +\frac {11}{8} b^{2}\right ) \tan \left (d x +c \right )^{3}+\left (-\frac {9}{4} a b +\frac {13}{8} b^{2}+\frac {5}{8} a^{2}\right ) \tan \left (d x +c \right )}{\left (1+\tan \left (d x +c \right )^{2}\right )^{2}}+\frac {\left (3 a^{2}-14 a b +35 b^{2}\right ) \arctan \left (\tan \left (d x +c \right )\right )}{8}}{\left (a -b \right )^{4}}}{d}\) | \(173\) |
| default | \(\frac {-\frac {b^{3} \left (\frac {\left (a -b \right ) \tan \left (d x +c \right )}{2 a \left (a +b \tan \left (d x +c \right )^{2}\right )}+\frac {\left (7 a -b \right ) \arctan \left (\frac {b \tan \left (d x +c \right )}{\sqrt {a b}}\right )}{2 a \sqrt {a b}}\right )}{\left (a -b \right )^{4}}+\frac {\frac {\left (\frac {3}{8} a^{2}-\frac {7}{4} a b +\frac {11}{8} b^{2}\right ) \tan \left (d x +c \right )^{3}+\left (-\frac {9}{4} a b +\frac {13}{8} b^{2}+\frac {5}{8} a^{2}\right ) \tan \left (d x +c \right )}{\left (1+\tan \left (d x +c \right )^{2}\right )^{2}}+\frac {\left (3 a^{2}-14 a b +35 b^{2}\right ) \arctan \left (\tan \left (d x +c \right )\right )}{8}}{\left (a -b \right )^{4}}}{d}\) | \(173\) |
| risch | \(\frac {3 x \,a^{2}}{8 \left (a^{2}-2 a b +b^{2}\right ) \left (a -b \right )^{2}}-\frac {7 x a b}{4 \left (a^{2}-2 a b +b^{2}\right ) \left (a -b \right )^{2}}+\frac {35 x \,b^{2}}{8 \left (a^{2}-2 a b +b^{2}\right ) \left (a -b \right )^{2}}-\frac {i {\mathrm e}^{4 i \left (d x +c \right )}}{64 \left (a -b \right )^{2} d}-\frac {i {\mathrm e}^{2 i \left (d x +c \right )} a}{8 \left (a -b \right )^{3} d}+\frac {3 i {\mathrm e}^{2 i \left (d x +c \right )} b}{8 \left (a -b \right )^{3} d}+\frac {i {\mathrm e}^{-2 i \left (d x +c \right )} a}{8 d \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right )}-\frac {3 i {\mathrm e}^{-2 i \left (d x +c \right )} b}{8 d \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right )}+\frac {i {\mathrm e}^{-4 i \left (d x +c \right )}}{64 \left (a^{2}-2 a b +b^{2}\right ) d}+\frac {i b^{3} \left (a \,{\mathrm e}^{2 i \left (d x +c \right )}+b \,{\mathrm e}^{2 i \left (d x +c \right )}+a -b \right )}{d \left (-a +b \right )^{2} a \left (a^{2}-2 a b +b^{2}\right ) \left (-a \,{\mathrm e}^{4 i \left (d x +c \right )}+b \,{\mathrm e}^{4 i \left (d x +c \right )}-2 a \,{\mathrm e}^{2 i \left (d x +c \right )}-2 b \,{\mathrm e}^{2 i \left (d x +c \right )}-a +b \right )}+\frac {7 \sqrt {-a b}\, b^{2} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 i \sqrt {-a b}+a +b}{a -b}\right )}{4 a \left (a -b \right )^{4} d}-\frac {\sqrt {-a b}\, b^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 i \sqrt {-a b}+a +b}{a -b}\right )}{4 a^{2} \left (a -b \right )^{4} d}-\frac {7 \sqrt {-a b}\, b^{2} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {2 i \sqrt {-a b}-a -b}{a -b}\right )}{4 a \left (a -b \right )^{4} d}+\frac {\sqrt {-a b}\, b^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {2 i \sqrt {-a b}-a -b}{a -b}\right )}{4 a^{2} \left (a -b \right )^{4} d}\) | \(596\) |
1/d*(-b^3/(a-b)^4*(1/2/a*(a-b)*tan(d*x+c)/(a+b*tan(d*x+c)^2)+1/2*(7*a-b)/a /(a*b)^(1/2)*arctan(b*tan(d*x+c)/(a*b)^(1/2)))+1/(a-b)^4*(((3/8*a^2-7/4*a* b+11/8*b^2)*tan(d*x+c)^3+(-9/4*a*b+13/8*b^2+5/8*a^2)*tan(d*x+c))/(1+tan(d* x+c)^2)^2+1/8*(3*a^2-14*a*b+35*b^2)*arctan(tan(d*x+c))))
Time = 0.34 (sec) , antiderivative size = 801, normalized size of antiderivative = 3.78 \[ \int \frac {\cos ^4(c+d x)}{\left (a+b \tan ^2(c+d x)\right )^2} \, dx=\left [\frac {{\left (3 \, a^{4} - 17 \, a^{3} b + 49 \, a^{2} b^{2} - 35 \, a b^{3}\right )} d x \cos \left (d x + c\right )^{2} + {\left (3 \, a^{3} b - 14 \, a^{2} b^{2} + 35 \, a b^{3}\right )} d x - {\left (7 \, a b^{3} - b^{4} + {\left (7 \, a^{2} b^{2} - 8 \, a b^{3} + b^{4}\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 ) + {\left (2 \, {\left (a^{4} - 3 \, a^{3} b + 3 \, a^{2} b^{2} - a b^{3}\right )} \cos \left (d x + c\right )^{5} + 3 \, {\left (a^{4} - 5 \, a^{3} b + 7 \, a^{2} b^{2} - 3 \, a b^{3}\right )} \cos \left (d x + c\right )^{3} + {\left (3 \, a^{3} b - 14 \, a^{2} b^{2} + 7 \, a b^{3} + 4 \, b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{8 \, {\left ({\left (a^{6} - 5 \, a^{5} b + 10 \, a^{4} b^{2} - 10 \, a^{3} b^{3} + 5 \, a^{2} b^{4} - a b^{5}\right )} d \cos \left (d x + c\right )^{2} + {\left (a^{5} b - 4 \, a^{4} b^{2} + 6 \, a^{3} b^{3} - 4 \, a^{2} b^{4} + a b^{5}\right )} d\right )}}, \frac {{\left (3 \, a^{4} - 17 \, a^{3} b + 49 \, a^{2} b^{2} - 35 \, a b^{3}\right )} d x \cos \left (d x + c\right )^{2} + {\left (3 \, a^{3} b - 14 \, a^{2} b^{2} + 35 \, a b^{3}\right )} d x + 2 \, {\left (7 \, a b^{3} - b^{4} + {\left (7 \, a^{2} b^{2} - 8 \, a b^{3} + b^{4}\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 ) + {\left (2 \, {\left (a^{4} - 3 \, a^{3} b + 3 \, a^{2} b^{2} - a b^{3}\right )} \cos \left (d x + c\right )^{5} + 3 \, {\left (a^{4} - 5 \, a^{3} b + 7 \, a^{2} b^{2} - 3 \, a b^{3}\right )} \cos \left (d x + c\right )^{3} + {\left (3 \, a^{3} b - 14 \, a^{2} b^{2} + 7 \, a b^{3} + 4 \, b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{8 \, {\left ({\left (a^{6} - 5 \, a^{5} b + 10 \, a^{4} b^{2} - 10 \, a^{3} b^{3} + 5 \, a^{2} b^{4} - a b^{5}\right )} d \cos \left (d x + c\right )^{2} + {\left (a^{5} b - 4 \, a^{4} b^{2} + 6 \, a^{3} b^{3} - 4 \, a^{2} b^{4} + a b^{5}\right )} d\right )}}\right ] \]
[1/8*((3*a^4 - 17*a^3*b + 49*a^2*b^2 - 35*a*b^3)*d*x*cos(d*x + c)^2 + (3*a ^3*b - 14*a^2*b^2 + 35*a*b^3)*d*x - (7*a*b^3 - b^4 + (7*a^2*b^2 - 8*a*b^3 + b^4)*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)) + (2*(a^4 - 3*a^3*b + 3*a^2*b^2 - a*b^3)*cos(d*x + c)^5 + 3*(a^4 - 5*a^3*b + 7*a^2*b^2 - 3*a*b^3)*cos(d*x + c)^3 + (3*a^3*b - 14*a^2*b^2 + 7*a*b^3 + 4*b^4)*cos(d*x + c))*sin(d*x + c))/((a^6 - 5*a^5*b + 10*a^4*b^2 - 10*a^3*b^3 + 5*a^2*b^4 - a*b^5)*d*cos( d*x + c)^2 + (a^5*b - 4*a^4*b^2 + 6*a^3*b^3 - 4*a^2*b^4 + a*b^5)*d), 1/8*( (3*a^4 - 17*a^3*b + 49*a^2*b^2 - 35*a*b^3)*d*x*cos(d*x + c)^2 + (3*a^3*b - 14*a^2*b^2 + 35*a*b^3)*d*x + 2*(7*a*b^3 - b^4 + (7*a^2*b^2 - 8*a*b^3 + b^ 4)*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^4 - 3*a^3*b + 3*a^2*b^2 - a*b^ 3)*cos(d*x + c)^5 + 3*(a^4 - 5*a^3*b + 7*a^2*b^2 - 3*a*b^3)*cos(d*x + c)^3 + (3*a^3*b - 14*a^2*b^2 + 7*a*b^3 + 4*b^4)*cos(d*x + c))*sin(d*x + c))/(( a^6 - 5*a^5*b + 10*a^4*b^2 - 10*a^3*b^3 + 5*a^2*b^4 - a*b^5)*d*cos(d*x + c )^2 + (a^5*b - 4*a^4*b^2 + 6*a^3*b^3 - 4*a^2*b^4 + a*b^5)*d)]
Timed out. \[ \int \frac {\cos ^4(c+d x)}{\left (a+b \tan ^2(c+d x)\right )^2} \, dx=\text {Timed out} \]
Time = 0.32 (sec) , antiderivative size = 355, normalized size of antiderivative = 1.67 \[ \int \frac {\cos ^4(c+d x)}{\left (a+b \tan ^2(c+d x)\right )^2} \, dx=\frac {\frac {{\left (3 \, a^{2} - 14 \, a b + 35 \, b^{2}\right )} {\left (d x + c\right )}}{a^{4} - 4 \, a^{3} b + 6 \, a^{2} b^{2} - 4 \, a b^{3} + b^{4}} - \frac {4 \, {\left (7 \, a b^{3} - b^{4}\right )} \arctan \left (\frac {b \tan \left (d x + c\right )}{\sqrt {a b}}\right )}{{\left (a^{5} - 4 \, a^{4} b + 6 \, a^{3} b^{2} - 4 \, a^{2} b^{3} + a b^{4}\right )} \sqrt {a b}} + \frac {{\left (3 \, a^{2} b - 11 \, a b^{2} - 4 \, b^{3}\right )} \tan \left (d x + c\right )^{5} + {\left (3 \, a^{3} - 6 \, a^{2} b - 13 \, a b^{2} - 8 \, b^{3}\right )} \tan \left (d x + c\right )^{3} + {\left (5 \, a^{3} - 13 \, a^{2} b - 4 \, b^{3}\right )} \tan \left (d x + c\right )}{{\left (a^{4} b - 3 \, a^{3} b^{2} + 3 \, a^{2} b^{3} - a b^{4}\right )} \tan \left (d x + c\right )^{6} + a^{5} - 3 \, a^{4} b + 3 \, a^{3} b^{2} - a^{2} b^{3} + {\left (a^{5} - a^{4} b - 3 \, a^{3} b^{2} + 5 \, a^{2} b^{3} - 2 \, a b^{4}\right )} \tan \left (d x + c\right )^{4} + {\left (2 \, a^{5} - 5 \, a^{4} b + 3 \, a^{3} b^{2} + a^{2} b^{3} - a b^{4}\right )} \tan \left (d x + c\right )^{2}}}{8 \, d} \]
1/8*((3*a^2 - 14*a*b + 35*b^2)*(d*x + c)/(a^4 - 4*a^3*b + 6*a^2*b^2 - 4*a* b^3 + b^4) - 4*(7*a*b^3 - b^4)*arctan(b*tan(d*x + c)/sqrt(a*b))/((a^5 - 4* a^4*b + 6*a^3*b^2 - 4*a^2*b^3 + a*b^4)*sqrt(a*b)) + ((3*a^2*b - 11*a*b^2 - 4*b^3)*tan(d*x + c)^5 + (3*a^3 - 6*a^2*b - 13*a*b^2 - 8*b^3)*tan(d*x + c) ^3 + (5*a^3 - 13*a^2*b - 4*b^3)*tan(d*x + c))/((a^4*b - 3*a^3*b^2 + 3*a^2* b^3 - a*b^4)*tan(d*x + c)^6 + a^5 - 3*a^4*b + 3*a^3*b^2 - a^2*b^3 + (a^5 - a^4*b - 3*a^3*b^2 + 5*a^2*b^3 - 2*a*b^4)*tan(d*x + c)^4 + (2*a^5 - 5*a^4* b + 3*a^3*b^2 + a^2*b^3 - a*b^4)*tan(d*x + c)^2))/d
Time = 0.64 (sec) , antiderivative size = 269, normalized size of antiderivative = 1.27 \[ \int \frac {\cos ^4(c+d x)}{\left (a+b \tan ^2(c+d x)\right )^2} \, dx=-\frac {\frac {4 \, b^{3} \tan \left (d x + c\right )}{{\left (a^{4} - 3 \, a^{3} b + 3 \, a^{2} b^{2} - a b^{3}\right )} {\left (b \tan \left (d x + c\right )^{2} + a\right )}} - \frac {{\left (3 \, a^{2} - 14 \, a b + 35 \, b^{2}\right )} {\left (d x + c\right )}}{a^{4} - 4 \, a^{3} b + 6 \, a^{2} b^{2} - 4 \, a b^{3} + b^{4}} + \frac {4 \, {\left (7 \, a b^{3} - b^{4}\right )} {\left (\pi \left \lfloor \frac {d x + c}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (b\right ) + \arctan \left (\frac {b \tan \left (d x + c\right )}{\sqrt {a b}}\right )\right )}}{{\left (a^{5} - 4 \, a^{4} b + 6 \, a^{3} b^{2} - 4 \, a^{2} b^{3} + a b^{4}\right )} \sqrt {a b}} - \frac {3 \, a \tan \left (d x + c\right )^{3} - 11 \, b \tan \left (d x + c\right )^{3} + 5 \, a \tan \left (d x + c\right ) - 13 \, b \tan \left (d x + c\right )}{{\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} {\left (\tan \left (d x + c\right )^{2} + 1\right )}^{2}}}{8 \, d} \]
-1/8*(4*b^3*tan(d*x + c)/((a^4 - 3*a^3*b + 3*a^2*b^2 - a*b^3)*(b*tan(d*x + c)^2 + a)) - (3*a^2 - 14*a*b + 35*b^2)*(d*x + c)/(a^4 - 4*a^3*b + 6*a^2*b ^2 - 4*a*b^3 + b^4) + 4*(7*a*b^3 - b^4)*(pi*floor((d*x + c)/pi + 1/2)*sgn( b) + arctan(b*tan(d*x + c)/sqrt(a*b)))/((a^5 - 4*a^4*b + 6*a^3*b^2 - 4*a^2 *b^3 + a*b^4)*sqrt(a*b)) - (3*a*tan(d*x + c)^3 - 11*b*tan(d*x + c)^3 + 5*a *tan(d*x + c) - 13*b*tan(d*x + c))/((a^3 - 3*a^2*b + 3*a*b^2 - b^3)*(tan(d *x + c)^2 + 1)^2))/d
Time = 17.19 (sec) , antiderivative size = 5272, normalized size of antiderivative = 24.87 \[ \int \frac {\cos ^4(c+d x)}{\left (a+b \tan ^2(c+d x)\right )^2} \, dx=\text {Too large to display} \]
- ((tan(c + d*x)^5*(11*a*b^2 - 3*a^2*b + 4*b^3))/(8*a*(3*a*b^2 - 3*a^2*b + a^3 - b^3)) + (tan(c + d*x)^3*(13*a*b^2 + 6*a^2*b - 3*a^3 + 8*b^3))/(8*a* (a - b)*(a^2 - 2*a*b + b^2)) + (tan(c + d*x)*(13*a^2*b - 5*a^3 + 4*b^3))/( 8*a*(a - b)*(a^2 - 2*a*b + b^2)))/(d*(a + b*tan(c + d*x)^6 + tan(c + d*x)^ 2*(2*a + b) + tan(c + d*x)^4*(a + 2*b))) - (atan(((((((2*a*b^13 - 28*a^2*b ^12 + (315*a^3*b^11)/2 - (987*a^4*b^10)/2 + 978*a^5*b^9 - 1302*a^6*b^8 + 1 197*a^7*b^7 - 765*a^8*b^6 + 336*a^9*b^5 - 98*a^10*b^4 + (35*a^11*b^3)/2 - (3*a^12*b^2)/2)/(9*a^10*b - a^11 + a^2*b^9 - 9*a^3*b^8 + 36*a^4*b^7 - 84*a ^5*b^6 + 126*a^6*b^5 - 126*a^7*b^4 + 84*a^8*b^3 - 36*a^9*b^2) - (tan(c + d *x)*(a^2*3i - a*b*14i + b^2*35i)*(256*a^2*b^11 - 1792*a^3*b^10 + 5120*a^4* b^9 - 7168*a^5*b^8 + 3584*a^6*b^7 + 3584*a^7*b^6 - 7168*a^8*b^5 + 5120*a^9 *b^4 - 1792*a^10*b^3 + 256*a^11*b^2))/(512*(a^4 - 4*a^3*b - 4*a*b^3 + b^4 + 6*a^2*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)))*(a^2*3i - a*b*14i + b^2*35i))/(16*(a^4 - 4*a^3*b - 4*a*b ^3 + b^4 + 6*a^2*b^2)) - (tan(c + d*x)*(16*b^9 - 224*a*b^8 + 2009*a^2*b^7 - 980*a^3*b^6 + 406*a^4*b^5 - 84*a^5*b^4 + 9*a^6*b^3))/(32*(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)))*(a^2*3i - a*b*14i + b^2*35i)*1i)/(16*(a^4 - 4*a^3*b - 4*a*b^3 + b^4 + 6*a^2*b^2)) - (((((2*a*b^13 - 28*a^2*b^12 + (315*a^3*b^11)/2 - (987*a^4*b^10)/2 + 978*a^ 5*b^9 - 1302*a^6*b^8 + 1197*a^7*b^7 - 765*a^8*b^6 + 336*a^9*b^5 - 98*a^...