Integrand size = 28, antiderivative size = 77 \[ \int \cos ^3(c+d x) (a \sin (c+d x)+b \tan (c+d x))^3 \, dx=-\frac {\left (a^2-b^2\right ) (b+a \cos (c+d x))^4}{4 a^3 d}-\frac {2 b (b+a \cos (c+d x))^5}{5 a^3 d}+\frac {(b+a \cos (c+d x))^6}{6 a^3 d} \]
-1/4*(a^2-b^2)*(b+a*cos(d*x+c))^4/a^3/d-2/5*b*(b+a*cos(d*x+c))^5/a^3/d+1/6 *(b+a*cos(d*x+c))^6/a^3/d
Time = 0.97 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.48 \[ \int \cos ^3(c+d x) (a \sin (c+d x)+b \tan (c+d x))^3 \, dx=\frac {-360 b \left (a^2+2 b^2\right ) \cos (c+d x)-45 \left (a^3+8 a b^2\right ) \cos (2 (c+d x))-60 a^2 b \cos (3 (c+d x))+80 b^3 \cos (3 (c+d x))+90 a b^2 \cos (4 (c+d x))+36 a^2 b \cos (5 (c+d x))+5 a^3 \cos (6 (c+d x))}{960 d} \]
(-360*b*(a^2 + 2*b^2)*Cos[c + d*x] - 45*(a^3 + 8*a*b^2)*Cos[2*(c + d*x)] - 60*a^2*b*Cos[3*(c + d*x)] + 80*b^3*Cos[3*(c + d*x)] + 90*a*b^2*Cos[4*(c + d*x)] + 36*a^2*b*Cos[5*(c + d*x)] + 5*a^3*Cos[6*(c + d*x)])/(960*d)
Time = 0.40 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.87, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {3042, 4897, 3042, 3147, 476, 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.
\(\displaystyle \int \cos ^3(c+d x) (a \sin (c+d x)+b \tan (c+d x))^3 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \cos (c+d x)^3 (a \sin (c+d x)+b \tan (c+d x))^3dx\) |
\(\Big \downarrow \) 4897 |
\(\displaystyle \int \sin ^3(c+d x) (a \cos (c+d x)+b)^3dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \cos \left (c+d x-\frac {\pi }{2}\right )^3 \left (b-a \sin \left (c+d x-\frac {\pi }{2}\right )\right )^3dx\) |
\(\Big \downarrow \) 3147 |
\(\displaystyle -\frac {\int (b+a \cos (c+d x))^3 \left (a^2-a^2 \cos ^2(c+d x)\right )d(a \cos (c+d x))}{a^3 d}\) |
\(\Big \downarrow \) 476 |
\(\displaystyle -\frac {\int \left (-(b+a \cos (c+d x))^5+2 b (b+a \cos (c+d x))^4+\left (a^2-b^2\right ) (b+a \cos (c+d x))^3\right )d(a \cos (c+d x))}{a^3 d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {\frac {1}{4} \left (a^2-b^2\right ) (a \cos (c+d x)+b)^4-\frac {1}{6} (a \cos (c+d x)+b)^6+\frac {2}{5} b (a \cos (c+d x)+b)^5}{a^3 d}\) |
-((((a^2 - b^2)*(b + a*Cos[c + d*x])^4)/4 + (2*b*(b + a*Cos[c + d*x])^5)/5 - (b + a*Cos[c + d*x])^6/6)/(a^3*d))
3.3.43.3.1 Defintions of rubi rules used
Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ ExpandIntegrand[(c + d*x)^n*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, n}, x] && IGtQ[p, 0]
Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m _.), x_Symbol] :> Simp[1/(b^p*f) Subst[Int[(a + x)^m*(b^2 - x^2)^((p - 1) /2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && IntegerQ[(p - 1)/2] && NeQ[a^2 - b^2, 0]
Time = 30.63 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.30
method | result | size |
derivativedivides | \(\frac {\frac {a^{3} \cos \left (d x +c \right )^{6}}{6}+\frac {3 a^{2} b \cos \left (d x +c \right )^{5}}{5}+\frac {\left (-a^{3}+3 a \,b^{2}\right ) \cos \left (d x +c \right )^{4}}{4}+\frac {\left (-3 a^{2} b +b^{3}\right ) \cos \left (d x +c \right )^{3}}{3}-\frac {3 a \,b^{2} \cos \left (d x +c \right )^{2}}{2}-\cos \left (d x +c \right ) b^{3}}{d}\) | \(100\) |
default | \(\frac {\frac {a^{3} \cos \left (d x +c \right )^{6}}{6}+\frac {3 a^{2} b \cos \left (d x +c \right )^{5}}{5}+\frac {\left (-a^{3}+3 a \,b^{2}\right ) \cos \left (d x +c \right )^{4}}{4}+\frac {\left (-3 a^{2} b +b^{3}\right ) \cos \left (d x +c \right )^{3}}{3}-\frac {3 a \,b^{2} \cos \left (d x +c \right )^{2}}{2}-\cos \left (d x +c \right ) b^{3}}{d}\) | \(100\) |
risch | \(-\frac {3 a^{2} b \cos \left (d x +c \right )}{8 d}-\frac {3 b^{3} \cos \left (d x +c \right )}{4 d}+\frac {a^{3} \cos \left (6 d x +6 c \right )}{192 d}+\frac {3 b \,a^{2} \cos \left (5 d x +5 c \right )}{80 d}+\frac {3 a \,b^{2} \cos \left (4 d x +4 c \right )}{32 d}-\frac {b \cos \left (3 d x +3 c \right ) a^{2}}{16 d}+\frac {b^{3} \cos \left (3 d x +3 c \right )}{12 d}-\frac {3 a^{3} \cos \left (2 d x +2 c \right )}{64 d}-\frac {3 a \cos \left (2 d x +2 c \right ) b^{2}}{8 d}\) | \(154\) |
1/d*(1/6*a^3*cos(d*x+c)^6+3/5*a^2*b*cos(d*x+c)^5+1/4*(-a^3+3*a*b^2)*cos(d* x+c)^4+1/3*(-3*a^2*b+b^3)*cos(d*x+c)^3-3/2*a*b^2*cos(d*x+c)^2-cos(d*x+c)*b ^3)
Time = 0.25 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.30 \[ \int \cos ^3(c+d x) (a \sin (c+d x)+b \tan (c+d x))^3 \, dx=\frac {10 \, a^{3} \cos \left (d x + c\right )^{6} + 36 \, a^{2} b \cos \left (d x + c\right )^{5} - 90 \, a b^{2} \cos \left (d x + c\right )^{2} - 15 \, {\left (a^{3} - 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{4} - 60 \, b^{3} \cos \left (d x + c\right ) - 20 \, {\left (3 \, a^{2} b - b^{3}\right )} \cos \left (d x + c\right )^{3}}{60 \, d} \]
1/60*(10*a^3*cos(d*x + c)^6 + 36*a^2*b*cos(d*x + c)^5 - 90*a*b^2*cos(d*x + c)^2 - 15*(a^3 - 3*a*b^2)*cos(d*x + c)^4 - 60*b^3*cos(d*x + c) - 20*(3*a^ 2*b - b^3)*cos(d*x + c)^3)/d
\[ \int \cos ^3(c+d x) (a \sin (c+d x)+b \tan (c+d x))^3 \, dx=\int \left (a \sin {\left (c + d x \right )} + b \tan {\left (c + d x \right )}\right )^{3} \cos ^{3}{\left (c + d x \right )}\, dx \]
Time = 0.21 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.23 \[ \int \cos ^3(c+d x) (a \sin (c+d x)+b \tan (c+d x))^3 \, dx=\frac {45 \, a b^{2} \sin \left (d x + c\right )^{4} - 5 \, {\left (2 \, \sin \left (d x + c\right )^{6} - 3 \, \sin \left (d x + c\right )^{4}\right )} a^{3} + 12 \, {\left (3 \, \cos \left (d x + c\right )^{5} - 5 \, \cos \left (d x + c\right )^{3}\right )} a^{2} b + 20 \, {\left (\cos \left (d x + c\right )^{3} - 3 \, \cos \left (d x + c\right )\right )} b^{3}}{60 \, d} \]
1/60*(45*a*b^2*sin(d*x + c)^4 - 5*(2*sin(d*x + c)^6 - 3*sin(d*x + c)^4)*a^ 3 + 12*(3*cos(d*x + c)^5 - 5*cos(d*x + c)^3)*a^2*b + 20*(cos(d*x + c)^3 - 3*cos(d*x + c))*b^3)/d
Timed out. \[ \int \cos ^3(c+d x) (a \sin (c+d x)+b \tan (c+d x))^3 \, dx=\text {Timed out} \]
Time = 22.71 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.94 \[ \int \cos ^3(c+d x) (a \sin (c+d x)+b \tan (c+d x))^3 \, dx=\frac {32\,a^3}{3\,d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^6}+\frac {4\,{\left (a-b\right )}^3}{d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^2}-\frac {32\,a^2\,\left (5\,a-3\,b\right )}{5\,d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^5}-\frac {8\,{\left (a-b\right )}^2\,\left (7\,a-b\right )}{3\,d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^3}+\frac {12\,a\,\left (3\,a^2-4\,a\,b+b^2\right )}{d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^4} \]