Integrand size = 21, antiderivative size = 202 \[ \int \frac {\cos ^2(c+d x)}{(a+b \tan (c+d x))^3} \, dx=\frac {a \left (a^4+10 a^2 b^2-15 b^4\right ) x}{2 \left (a^2+b^2\right )^4}+\frac {2 b^3 \left (5 a^2-b^2\right ) \log (a \cos (c+d x)+b \sin (c+d x))}{\left (a^2+b^2\right )^4 d}+\frac {b \left (a^2-2 b^2\right )}{2 \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))^2}+\frac {\cos ^2(c+d x) (b+a \tan (c+d x))}{2 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}+\frac {a b \left (a^2-11 b^2\right )}{2 \left (a^2+b^2\right )^3 d (a+b \tan (c+d x))} \]
1/2*a*(a^4+10*a^2*b^2-15*b^4)*x/(a^2+b^2)^4+2*b^3*(5*a^2-b^2)*ln(a*cos(d*x +c)+b*sin(d*x+c))/(a^2+b^2)^4/d+1/2*b*(a^2-2*b^2)/(a^2+b^2)^2/d/(a+b*tan(d *x+c))^2+1/2*cos(d*x+c)^2*(b+a*tan(d*x+c))/(a^2+b^2)/d/(a+b*tan(d*x+c))^2+ 1/2*a*b*(a^2-11*b^2)/(a^2+b^2)^3/d/(a+b*tan(d*x+c))
Leaf count is larger than twice the leaf count of optimal. \(458\) vs. \(2(202)=404\).
Time = 6.37 (sec) , antiderivative size = 458, normalized size of antiderivative = 2.27 \[ \int \frac {\cos ^2(c+d x)}{(a+b \tan (c+d x))^3} \, dx=\frac {b^3 \left (\frac {\cos ^2(c+d x) \left (b^2+a b \tan (c+d x)\right )}{2 b^4 \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}-\frac {\left (2 a^2-4 b^2\right ) \left (-\frac {\left (3 a^2-b^2-\frac {a^3-3 a b^2}{\sqrt {-b^2}}\right ) \log \left (\sqrt {-b^2}-b \tan (c+d x)\right )}{2 \left (a^2+b^2\right )^3}+\frac {\left (3 a^2-b^2\right ) \log (a+b \tan (c+d x))}{\left (a^2+b^2\right )^3}-\frac {\left (3 a^2-b^2+\frac {a^3-3 a b^2}{\sqrt {-b^2}}\right ) \log \left (\sqrt {-b^2}+b \tan (c+d x)\right )}{2 \left (a^2+b^2\right )^3}-\frac {1}{2 \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}-\frac {2 a}{\left (a^2+b^2\right )^2 (a+b \tan (c+d x))}\right )-3 a \left (-\frac {\left (2 a-\frac {a^2-b^2}{\sqrt {-b^2}}\right ) \log \left (\sqrt {-b^2}-b \tan (c+d x)\right )}{2 \left (a^2+b^2\right )^2}+\frac {2 a \log (a+b \tan (c+d x))}{\left (a^2+b^2\right )^2}-\frac {\left (2 a+\frac {a^2-b^2}{\sqrt {-b^2}}\right ) \log \left (\sqrt {-b^2}+b \tan (c+d x)\right )}{2 \left (a^2+b^2\right )^2}-\frac {1}{\left (a^2+b^2\right ) (a+b \tan (c+d x))}\right )}{2 b^2 \left (a^2+b^2\right )}\right )}{d} \]
(b^3*((Cos[c + d*x]^2*(b^2 + a*b*Tan[c + d*x]))/(2*b^4*(a^2 + b^2)*(a + b* Tan[c + d*x])^2) - ((2*a^2 - 4*b^2)*(-1/2*((3*a^2 - b^2 - (a^3 - 3*a*b^2)/ Sqrt[-b^2])*Log[Sqrt[-b^2] - b*Tan[c + d*x]])/(a^2 + b^2)^3 + ((3*a^2 - b^ 2)*Log[a + b*Tan[c + d*x]])/(a^2 + b^2)^3 - ((3*a^2 - b^2 + (a^3 - 3*a*b^2 )/Sqrt[-b^2])*Log[Sqrt[-b^2] + b*Tan[c + d*x]])/(2*(a^2 + b^2)^3) - 1/(2*( a^2 + b^2)*(a + b*Tan[c + d*x])^2) - (2*a)/((a^2 + b^2)^2*(a + b*Tan[c + d *x]))) - 3*a*(-1/2*((2*a - (a^2 - b^2)/Sqrt[-b^2])*Log[Sqrt[-b^2] - b*Tan[ c + d*x]])/(a^2 + b^2)^2 + (2*a*Log[a + b*Tan[c + d*x]])/(a^2 + b^2)^2 - ( (2*a + (a^2 - b^2)/Sqrt[-b^2])*Log[Sqrt[-b^2] + b*Tan[c + d*x]])/(2*(a^2 + b^2)^2) - 1/((a^2 + b^2)*(a + b*Tan[c + d*x]))))/(2*b^2*(a^2 + b^2))))/d
Time = 0.48 (sec) , antiderivative size = 263, normalized size of antiderivative = 1.30, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 3987, 27, 496, 25, 657, 2009}
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.
| \(\displaystyle \int \frac {\cos ^2(c+d x)}{(a+b \tan (c+d x))^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\sec (c+d x)^2 (a+b \tan (c+d x))^3}dx\) |
| \(\Big \downarrow \) 3987 |
| \(\displaystyle \frac {\int \frac {b^4}{(a+b \tan (c+d x))^3 \left (\tan ^2(c+d x) b^2+b^2\right )^2}d(b \tan (c+d x))}{b d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {b^3 \int \frac {1}{(a+b \tan (c+d x))^3 \left (\tan ^2(c+d x) b^2+b^2\right )^2}d(b \tan (c+d x))}{d}\) |
| \(\Big \downarrow \) 496 |
| \(\displaystyle \frac {b^3 \left (\frac {a b \tan (c+d x)+b^2}{2 b^2 \left (a^2+b^2\right ) \left (b^2 \tan ^2(c+d x)+b^2\right ) (a+b \tan (c+d x))^2}-\frac {\int -\frac {a^2+3 b \tan (c+d x) a+4 b^2}{(a+b \tan (c+d x))^3 \left (\tan ^2(c+d x) b^2+b^2\right )}d(b \tan (c+d x))}{2 b^2 \left (a^2+b^2\right )}\right )}{d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {b^3 \left (\frac {\int \frac {a^2+3 b \tan (c+d x) a+4 b^2}{(a+b \tan (c+d x))^3 \left (\tan ^2(c+d x) b^2+b^2\right )}d(b \tan (c+d x))}{2 b^2 \left (a^2+b^2\right )}+\frac {a b \tan (c+d x)+b^2}{2 b^2 \left (a^2+b^2\right ) \left (b^2 \tan ^2(c+d x)+b^2\right ) (a+b \tan (c+d x))^2}\right )}{d}\) |
| \(\Big \downarrow \) 657 |
| \(\displaystyle \frac {b^3 \left (\frac {\int \left (-\frac {a \left (a^2-11 b^2\right )}{\left (a^2+b^2\right )^2 (a+b \tan (c+d x))^2}+\frac {4 \left (5 a^2 b^2-b^4\right )}{\left (a^2+b^2\right )^3 (a+b \tan (c+d x))}+\frac {a \left (a^4+10 b^2 a^2-15 b^4\right )-4 b^3 \left (5 a^2-b^2\right ) \tan (c+d x)}{\left (a^2+b^2\right )^3 \left (\tan ^2(c+d x) b^2+b^2\right )}-\frac {2 \left (a^2-2 b^2\right )}{\left (a^2+b^2\right ) (a+b \tan (c+d x))^3}\right )d(b \tan (c+d x))}{2 b^2 \left (a^2+b^2\right )}+\frac {a b \tan (c+d x)+b^2}{2 b^2 \left (a^2+b^2\right ) \left (b^2 \tan ^2(c+d x)+b^2\right ) (a+b \tan (c+d x))^2}\right )}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {b^3 \left (\frac {a b \tan (c+d x)+b^2}{2 b^2 \left (a^2+b^2\right ) \left (b^2 \tan ^2(c+d x)+b^2\right ) (a+b \tan (c+d x))^2}+\frac {\frac {a \left (a^2-11 b^2\right )}{\left (a^2+b^2\right )^2 (a+b \tan (c+d x))}+\frac {a^2-2 b^2}{\left (a^2+b^2\right ) (a+b \tan (c+d x))^2}-\frac {2 b^2 \left (5 a^2-b^2\right ) \log \left (b^2 \tan ^2(c+d x)+b^2\right )}{\left (a^2+b^2\right )^3}+\frac {4 b^2 \left (5 a^2-b^2\right ) \log (a+b \tan (c+d x))}{\left (a^2+b^2\right )^3}+\frac {a \left (a^4+10 a^2 b^2-15 b^4\right ) \arctan (\tan (c+d x))}{b \left (a^2+b^2\right )^3}}{2 b^2 \left (a^2+b^2\right )}\right )}{d}\) |
(b^3*((b^2 + a*b*Tan[c + d*x])/(2*b^2*(a^2 + b^2)*(a + b*Tan[c + d*x])^2*( b^2 + b^2*Tan[c + d*x]^2)) + ((a*(a^4 + 10*a^2*b^2 - 15*b^4)*ArcTan[Tan[c + d*x]])/(b*(a^2 + b^2)^3) + (4*b^2*(5*a^2 - b^2)*Log[a + b*Tan[c + d*x]]) /(a^2 + b^2)^3 - (2*b^2*(5*a^2 - b^2)*Log[b^2 + b^2*Tan[c + d*x]^2])/(a^2 + b^2)^3 + (a^2 - 2*b^2)/((a^2 + b^2)*(a + b*Tan[c + d*x])^2) + (a*(a^2 - 11*b^2))/((a^2 + b^2)^2*(a + b*Tan[c + d*x])))/(2*b^2*(a^2 + b^2))))/d
3.6.70.3.1 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[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ (-(a*d + b*c*x))*(c + d*x)^(n + 1)*((a + b*x^2)^(p + 1)/(2*a*(p + 1)*(b*c^2 + a*d^2))), x] + Simp[1/(2*a*(p + 1)*(b*c^2 + a*d^2)) Int[(c + d*x)^n*(a + b*x^2)^(p + 1)*Simp[b*c^2*(2*p + 3) + a*d^2*(n + 2*p + 3) + b*c*d*(n + 2 *p + 4)*x, x], x], x] /; FreeQ[{a, b, c, d, n}, x] && LtQ[p, -1] && IntQuad raticQ[a, 0, b, c, d, n, p, x]
Int[(((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.))/((a_) + (c_.)*( x_)^2), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*((f + g*x)^n/(a + c*x^ 2)), x], x] /; FreeQ[{a, c, d, e, f, g, m}, x] && IntegersQ[n]
Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_ ), x_Symbol] :> Simp[1/(b*f) Subst[Int[(a + x)^n*(1 + x^2/b^2)^(m/2 - 1), x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b, e, f, n}, x] && NeQ[a^2 + b^2, 0] && IntegerQ[m/2]
Time = 10.26 (sec) , antiderivative size = 219, normalized size of antiderivative = 1.08
| method | result | size |
| derivativedivides | \(\frac {\frac {\frac {\left (\frac {1}{2} a^{5}-a^{3} b^{2}-\frac {3}{2} a \,b^{4}\right ) \tan \left (d x +c \right )+\frac {3 a^{4} b}{2}+a^{2} b^{3}-\frac {b^{5}}{2}}{1+\tan ^{2}\left (d x +c \right )}+\frac {\left (-20 a^{2} b^{3}+4 b^{5}\right ) \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{4}+\frac {\left (a^{5}+10 a^{3} b^{2}-15 a \,b^{4}\right ) \arctan \left (\tan \left (d x +c \right )\right )}{2}}{\left (a^{2}+b^{2}\right )^{4}}-\frac {b^{3}}{2 \left (a^{2}+b^{2}\right )^{2} \left (a +b \tan \left (d x +c \right )\right )^{2}}-\frac {4 b^{3} a}{\left (a^{2}+b^{2}\right )^{3} \left (a +b \tan \left (d x +c \right )\right )}+\frac {2 b^{3} \left (5 a^{2}-b^{2}\right ) \ln \left (a +b \tan \left (d x +c \right )\right )}{\left (a^{2}+b^{2}\right )^{4}}}{d}\) | \(219\) |
| default | \(\frac {\frac {\frac {\left (\frac {1}{2} a^{5}-a^{3} b^{2}-\frac {3}{2} a \,b^{4}\right ) \tan \left (d x +c \right )+\frac {3 a^{4} b}{2}+a^{2} b^{3}-\frac {b^{5}}{2}}{1+\tan ^{2}\left (d x +c \right )}+\frac {\left (-20 a^{2} b^{3}+4 b^{5}\right ) \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{4}+\frac {\left (a^{5}+10 a^{3} b^{2}-15 a \,b^{4}\right ) \arctan \left (\tan \left (d x +c \right )\right )}{2}}{\left (a^{2}+b^{2}\right )^{4}}-\frac {b^{3}}{2 \left (a^{2}+b^{2}\right )^{2} \left (a +b \tan \left (d x +c \right )\right )^{2}}-\frac {4 b^{3} a}{\left (a^{2}+b^{2}\right )^{3} \left (a +b \tan \left (d x +c \right )\right )}+\frac {2 b^{3} \left (5 a^{2}-b^{2}\right ) \ln \left (a +b \tan \left (d x +c \right )\right )}{\left (a^{2}+b^{2}\right )^{4}}}{d}\) | \(219\) |
| risch | \(\frac {4 i x b}{8 i a^{3} b -8 i a \,b^{3}-2 a^{4}+12 a^{2} b^{2}-2 b^{4}}-\frac {x a}{8 i a^{3} b -8 i a \,b^{3}-2 a^{4}+12 a^{2} b^{2}-2 b^{4}}-\frac {i {\mathrm e}^{2 i \left (d x +c \right )}}{8 \left (-3 i b \,a^{2}+i b^{3}+a^{3}-3 a \,b^{2}\right ) d}+\frac {i {\mathrm e}^{-2 i \left (d x +c \right )}}{8 \left (3 i b \,a^{2}-i b^{3}+a^{3}-3 a \,b^{2}\right ) d}-\frac {20 i b^{3} a^{2} x}{a^{8}+4 a^{6} b^{2}+6 a^{4} b^{4}+4 b^{6} a^{2}+b^{8}}+\frac {4 i b^{5} x}{a^{8}+4 a^{6} b^{2}+6 a^{4} b^{4}+4 b^{6} a^{2}+b^{8}}-\frac {20 i b^{3} a^{2} c}{d \left (a^{8}+4 a^{6} b^{2}+6 a^{4} b^{4}+4 b^{6} a^{2}+b^{8}\right )}+\frac {4 i b^{5} c}{d \left (a^{8}+4 a^{6} b^{2}+6 a^{4} b^{4}+4 b^{6} a^{2}+b^{8}\right )}+\frac {2 b^{4} \left (-4 i a b \,{\mathrm e}^{2 i \left (d x +c \right )}+5 a^{2} {\mathrm e}^{2 i \left (d x +c \right )}+b^{2} {\mathrm e}^{2 i \left (d x +c \right )}+5 i a b +5 a^{2}\right )}{\left (-i a +b \right )^{3} \left (b \,{\mathrm e}^{2 i \left (d x +c \right )}+i a \,{\mathrm e}^{2 i \left (d x +c \right )}-b +i a \right )^{2} d \left (i a +b \right )^{4}}+\frac {10 b^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {i b +a}{i b -a}\right ) a^{2}}{d \left (a^{8}+4 a^{6} b^{2}+6 a^{4} b^{4}+4 b^{6} a^{2}+b^{8}\right )}-\frac {2 b^{5} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {i b +a}{i b -a}\right )}{d \left (a^{8}+4 a^{6} b^{2}+6 a^{4} b^{4}+4 b^{6} a^{2}+b^{8}\right )}\) | \(586\) |
1/d*(1/(a^2+b^2)^4*(((1/2*a^5-a^3*b^2-3/2*a*b^4)*tan(d*x+c)+3/2*a^4*b+a^2* b^3-1/2*b^5)/(1+tan(d*x+c)^2)+1/4*(-20*a^2*b^3+4*b^5)*ln(1+tan(d*x+c)^2)+1 /2*(a^5+10*a^3*b^2-15*a*b^4)*arctan(tan(d*x+c)))-1/2*b^3/(a^2+b^2)^2/(a+b* tan(d*x+c))^2-4*b^3/(a^2+b^2)^3*a/(a+b*tan(d*x+c))+2*b^3*(5*a^2-b^2)/(a^2+ b^2)^4*ln(a+b*tan(d*x+c)))
Leaf count of result is larger than twice the leaf count of optimal. 503 vs. \(2 (194) = 388\).
Time = 0.31 (sec) , antiderivative size = 503, normalized size of antiderivative = 2.49 \[ \int \frac {\cos ^2(c+d x)}{(a+b \tan (c+d x))^3} \, dx=-\frac {3 \, a^{4} b^{3} - 16 \, a^{2} b^{5} + b^{7} - 2 \, {\left (a^{6} b + 3 \, a^{4} b^{3} + 3 \, a^{2} b^{5} + b^{7}\right )} \cos \left (d x + c\right )^{4} - 2 \, {\left (a^{5} b^{2} + 10 \, a^{3} b^{4} - 15 \, a b^{6}\right )} d x - {\left (a^{6} b - a^{4} b^{3} - 45 \, a^{2} b^{5} - 3 \, b^{7} + 2 \, {\left (a^{7} + 9 \, a^{5} b^{2} - 25 \, a^{3} b^{4} + 15 \, a b^{6}\right )} d x\right )} \cos \left (d x + c\right )^{2} - 4 \, {\left (5 \, a^{2} b^{5} - b^{7} + {\left (5 \, a^{4} b^{3} - 6 \, a^{2} b^{5} + b^{7}\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (5 \, a^{3} b^{4} - a b^{6}\right )} \cos \left (d x + c\right ) \sin \left (d x + c\right )\right )} \log \left (2 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right ) + {\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + b^{2}\right ) - 2 \, {\left ({\left (a^{7} + 3 \, a^{5} b^{2} + 3 \, a^{3} b^{4} + a b^{6}\right )} \cos \left (d x + c\right )^{3} - 2 \, {\left (a^{5} b^{2} - 3 \, a^{3} b^{4} + 6 \, a b^{6} - {\left (a^{6} b + 10 \, a^{4} b^{3} - 15 \, a^{2} b^{5}\right )} d x\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{4 \, {\left ({\left (a^{10} + 3 \, a^{8} b^{2} + 2 \, a^{6} b^{4} - 2 \, a^{4} b^{6} - 3 \, a^{2} b^{8} - b^{10}\right )} d \cos \left (d x + c\right )^{2} + 2 \, {\left (a^{9} b + 4 \, a^{7} b^{3} + 6 \, a^{5} b^{5} + 4 \, a^{3} b^{7} + a b^{9}\right )} d \cos \left (d x + c\right ) \sin \left (d x + c\right ) + {\left (a^{8} b^{2} + 4 \, a^{6} b^{4} + 6 \, a^{4} b^{6} + 4 \, a^{2} b^{8} + b^{10}\right )} d\right )}} \]
-1/4*(3*a^4*b^3 - 16*a^2*b^5 + b^7 - 2*(a^6*b + 3*a^4*b^3 + 3*a^2*b^5 + b^ 7)*cos(d*x + c)^4 - 2*(a^5*b^2 + 10*a^3*b^4 - 15*a*b^6)*d*x - (a^6*b - a^4 *b^3 - 45*a^2*b^5 - 3*b^7 + 2*(a^7 + 9*a^5*b^2 - 25*a^3*b^4 + 15*a*b^6)*d* x)*cos(d*x + c)^2 - 4*(5*a^2*b^5 - b^7 + (5*a^4*b^3 - 6*a^2*b^5 + b^7)*cos (d*x + c)^2 + 2*(5*a^3*b^4 - a*b^6)*cos(d*x + c)*sin(d*x + c))*log(2*a*b*c os(d*x + c)*sin(d*x + c) + (a^2 - b^2)*cos(d*x + c)^2 + b^2) - 2*((a^7 + 3 *a^5*b^2 + 3*a^3*b^4 + a*b^6)*cos(d*x + c)^3 - 2*(a^5*b^2 - 3*a^3*b^4 + 6* a*b^6 - (a^6*b + 10*a^4*b^3 - 15*a^2*b^5)*d*x)*cos(d*x + c))*sin(d*x + c)) /((a^10 + 3*a^8*b^2 + 2*a^6*b^4 - 2*a^4*b^6 - 3*a^2*b^8 - b^10)*d*cos(d*x + c)^2 + 2*(a^9*b + 4*a^7*b^3 + 6*a^5*b^5 + 4*a^3*b^7 + a*b^9)*d*cos(d*x + c)*sin(d*x + c) + (a^8*b^2 + 4*a^6*b^4 + 6*a^4*b^6 + 4*a^2*b^8 + b^10)*d)
Exception generated. \[ \int \frac {\cos ^2(c+d x)}{(a+b \tan (c+d x))^3} \, dx=\text {Exception raised: AttributeError} \]
Leaf count of result is larger than twice the leaf count of optimal. 458 vs. \(2 (194) = 388\).
Time = 0.31 (sec) , antiderivative size = 458, normalized size of antiderivative = 2.27 \[ \int \frac {\cos ^2(c+d x)}{(a+b \tan (c+d x))^3} \, dx=\frac {\frac {{\left (a^{5} + 10 \, a^{3} b^{2} - 15 \, a b^{4}\right )} {\left (d x + c\right )}}{a^{8} + 4 \, a^{6} b^{2} + 6 \, a^{4} b^{4} + 4 \, a^{2} b^{6} + b^{8}} + \frac {4 \, {\left (5 \, a^{2} b^{3} - b^{5}\right )} \log \left (b \tan \left (d x + c\right ) + a\right )}{a^{8} + 4 \, a^{6} b^{2} + 6 \, a^{4} b^{4} + 4 \, a^{2} b^{6} + b^{8}} - \frac {2 \, {\left (5 \, a^{2} b^{3} - b^{5}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{8} + 4 \, a^{6} b^{2} + 6 \, a^{4} b^{4} + 4 \, a^{2} b^{6} + b^{8}} + \frac {3 \, a^{4} b - 10 \, a^{2} b^{3} - b^{5} + {\left (a^{3} b^{2} - 11 \, a b^{4}\right )} \tan \left (d x + c\right )^{3} + 2 \, {\left (a^{4} b - 6 \, a^{2} b^{3} - b^{5}\right )} \tan \left (d x + c\right )^{2} + {\left (a^{5} + 3 \, a^{3} b^{2} - 10 \, a b^{4}\right )} \tan \left (d x + c\right )}{a^{8} + 3 \, a^{6} b^{2} + 3 \, a^{4} b^{4} + a^{2} b^{6} + {\left (a^{6} b^{2} + 3 \, a^{4} b^{4} + 3 \, a^{2} b^{6} + b^{8}\right )} \tan \left (d x + c\right )^{4} + 2 \, {\left (a^{7} b + 3 \, a^{5} b^{3} + 3 \, a^{3} b^{5} + a b^{7}\right )} \tan \left (d x + c\right )^{3} + {\left (a^{8} + 4 \, a^{6} b^{2} + 6 \, a^{4} b^{4} + 4 \, a^{2} b^{6} + b^{8}\right )} \tan \left (d x + c\right )^{2} + 2 \, {\left (a^{7} b + 3 \, a^{5} b^{3} + 3 \, a^{3} b^{5} + a b^{7}\right )} \tan \left (d x + c\right )}}{2 \, d} \]
1/2*((a^5 + 10*a^3*b^2 - 15*a*b^4)*(d*x + c)/(a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8) + 4*(5*a^2*b^3 - b^5)*log(b*tan(d*x + c) + a)/(a^8 + 4* a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8) - 2*(5*a^2*b^3 - b^5)*log(tan(d*x + c)^2 + 1)/(a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8) + (3*a^4*b - 10 *a^2*b^3 - b^5 + (a^3*b^2 - 11*a*b^4)*tan(d*x + c)^3 + 2*(a^4*b - 6*a^2*b^ 3 - b^5)*tan(d*x + c)^2 + (a^5 + 3*a^3*b^2 - 10*a*b^4)*tan(d*x + c))/(a^8 + 3*a^6*b^2 + 3*a^4*b^4 + a^2*b^6 + (a^6*b^2 + 3*a^4*b^4 + 3*a^2*b^6 + b^8 )*tan(d*x + c)^4 + 2*(a^7*b + 3*a^5*b^3 + 3*a^3*b^5 + a*b^7)*tan(d*x + c)^ 3 + (a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)*tan(d*x + c)^2 + 2*(a^ 7*b + 3*a^5*b^3 + 3*a^3*b^5 + a*b^7)*tan(d*x + c)))/d
Leaf count of result is larger than twice the leaf count of optimal. 439 vs. \(2 (194) = 388\).
Time = 0.64 (sec) , antiderivative size = 439, normalized size of antiderivative = 2.17 \[ \int \frac {\cos ^2(c+d x)}{(a+b \tan (c+d x))^3} \, dx=\frac {\frac {{\left (a^{5} + 10 \, a^{3} b^{2} - 15 \, a b^{4}\right )} {\left (d x + c\right )}}{a^{8} + 4 \, a^{6} b^{2} + 6 \, a^{4} b^{4} + 4 \, a^{2} b^{6} + b^{8}} - \frac {2 \, {\left (5 \, a^{2} b^{3} - b^{5}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{8} + 4 \, a^{6} b^{2} + 6 \, a^{4} b^{4} + 4 \, a^{2} b^{6} + b^{8}} + \frac {4 \, {\left (5 \, a^{2} b^{4} - b^{6}\right )} \log \left ({\left | b \tan \left (d x + c\right ) + a \right |}\right )}{a^{8} b + 4 \, a^{6} b^{3} + 6 \, a^{4} b^{5} + 4 \, a^{2} b^{7} + b^{9}} + \frac {10 \, a^{2} b^{3} \tan \left (d x + c\right )^{2} - 2 \, b^{5} \tan \left (d x + c\right )^{2} + a^{5} \tan \left (d x + c\right ) - 2 \, a^{3} b^{2} \tan \left (d x + c\right ) - 3 \, a b^{4} \tan \left (d x + c\right ) + 3 \, a^{4} b + 12 \, a^{2} b^{3} - 3 \, b^{5}}{{\left (a^{8} + 4 \, a^{6} b^{2} + 6 \, a^{4} b^{4} + 4 \, a^{2} b^{6} + b^{8}\right )} {\left (\tan \left (d x + c\right )^{2} + 1\right )}} - \frac {30 \, a^{2} b^{5} \tan \left (d x + c\right )^{2} - 6 \, b^{7} \tan \left (d x + c\right )^{2} + 68 \, a^{3} b^{4} \tan \left (d x + c\right ) - 4 \, a b^{6} \tan \left (d x + c\right ) + 39 \, a^{4} b^{3} + 4 \, a^{2} b^{5} + b^{7}}{{\left (a^{8} + 4 \, a^{6} b^{2} + 6 \, a^{4} b^{4} + 4 \, a^{2} b^{6} + b^{8}\right )} {\left (b \tan \left (d x + c\right ) + a\right )}^{2}}}{2 \, d} \]
1/2*((a^5 + 10*a^3*b^2 - 15*a*b^4)*(d*x + c)/(a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8) - 2*(5*a^2*b^3 - b^5)*log(tan(d*x + c)^2 + 1)/(a^8 + 4* a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8) + 4*(5*a^2*b^4 - b^6)*log(abs(b*tan (d*x + c) + a))/(a^8*b + 4*a^6*b^3 + 6*a^4*b^5 + 4*a^2*b^7 + b^9) + (10*a^ 2*b^3*tan(d*x + c)^2 - 2*b^5*tan(d*x + c)^2 + a^5*tan(d*x + c) - 2*a^3*b^2 *tan(d*x + c) - 3*a*b^4*tan(d*x + c) + 3*a^4*b + 12*a^2*b^3 - 3*b^5)/((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)*(tan(d*x + c)^2 + 1)) - (30*a^ 2*b^5*tan(d*x + c)^2 - 6*b^7*tan(d*x + c)^2 + 68*a^3*b^4*tan(d*x + c) - 4* a*b^6*tan(d*x + c) + 39*a^4*b^3 + 4*a^2*b^5 + b^7)/((a^8 + 4*a^6*b^2 + 6*a ^4*b^4 + 4*a^2*b^6 + b^8)*(b*tan(d*x + c) + a)^2))/d
Time = 5.26 (sec) , antiderivative size = 419, normalized size of antiderivative = 2.07 \[ \int \frac {\cos ^2(c+d x)}{(a+b \tan (c+d x))^3} \, dx=\frac {\ln \left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )\,\left (\frac {10\,b^3}{{\left (a^2+b^2\right )}^3}-\frac {12\,b^5}{{\left (a^2+b^2\right )}^4}\right )}{d}-\frac {\frac {-3\,a^4\,b+10\,a^2\,b^3+b^5}{2\,\left (a^2+b^2\right )\,\left (a^4+2\,a^2\,b^2+b^4\right )}+\frac {{\mathrm {tan}\left (c+d\,x\right )}^3\,\left (11\,a\,b^4-a^3\,b^2\right )}{2\,\left (a^6+3\,a^4\,b^2+3\,a^2\,b^4+b^6\right )}+\frac {{\mathrm {tan}\left (c+d\,x\right )}^2\,\left (-a^4\,b+6\,a^2\,b^3+b^5\right )}{\left (a^2+b^2\right )\,\left (a^4+2\,a^2\,b^2+b^4\right )}-\frac {a\,\mathrm {tan}\left (c+d\,x\right )\,\left (a^4+3\,a^2\,b^2-10\,b^4\right )}{2\,\left (a^2+b^2\right )\,\left (a^4+2\,a^2\,b^2+b^4\right )}}{d\,\left ({\mathrm {tan}\left (c+d\,x\right )}^2\,\left (a^2+b^2\right )+a^2+b^2\,{\mathrm {tan}\left (c+d\,x\right )}^4+2\,a\,b\,\mathrm {tan}\left (c+d\,x\right )+2\,a\,b\,{\mathrm {tan}\left (c+d\,x\right )}^3\right )}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,\left (b+\frac {a\,1{}\mathrm {i}}{4}\right )}{d\,\left (a^4-a^3\,b\,4{}\mathrm {i}-6\,a^2\,b^2+a\,b^3\,4{}\mathrm {i}+b^4\right )}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,\left (a+b\,4{}\mathrm {i}\right )}{4\,d\,\left (a^4\,1{}\mathrm {i}-4\,a^3\,b-a^2\,b^2\,6{}\mathrm {i}+4\,a\,b^3+b^4\,1{}\mathrm {i}\right )} \]
(log(a + b*tan(c + d*x))*((10*b^3)/(a^2 + b^2)^3 - (12*b^5)/(a^2 + b^2)^4) )/d - ((b^5 - 3*a^4*b + 10*a^2*b^3)/(2*(a^2 + b^2)*(a^4 + b^4 + 2*a^2*b^2) ) + (tan(c + d*x)^3*(11*a*b^4 - a^3*b^2))/(2*(a^6 + b^6 + 3*a^2*b^4 + 3*a^ 4*b^2)) + (tan(c + d*x)^2*(b^5 - a^4*b + 6*a^2*b^3))/((a^2 + b^2)*(a^4 + b ^4 + 2*a^2*b^2)) - (a*tan(c + d*x)*(a^4 - 10*b^4 + 3*a^2*b^2))/(2*(a^2 + b ^2)*(a^4 + b^4 + 2*a^2*b^2)))/(d*(tan(c + d*x)^2*(a^2 + b^2) + a^2 + b^2*t an(c + d*x)^4 + 2*a*b*tan(c + d*x) + 2*a*b*tan(c + d*x)^3)) + (log(tan(c + d*x) + 1i)*((a*1i)/4 + b))/(d*(a*b^3*4i - a^3*b*4i + a^4 + b^4 - 6*a^2*b^ 2)) + (log(tan(c + d*x) - 1i)*(a + b*4i))/(4*d*(4*a*b^3 - 4*a^3*b + a^4*1i + b^4*1i - a^2*b^2*6i))