Integrand size = 28, antiderivative size = 120 \[ \int \cos ^2(c+d x) (a \sin (c+d x)+b \tan (c+d x))^3 \, dx=-\frac {3 a b^2 \cos (c+d x)}{d}-\frac {b \left (3 a^2-b^2\right ) \cos ^2(c+d x)}{2 d}-\frac {a \left (a^2-3 b^2\right ) \cos ^3(c+d x)}{3 d}+\frac {3 a^2 b \cos ^4(c+d x)}{4 d}+\frac {a^3 \cos ^5(c+d x)}{5 d}-\frac {b^3 \log (\cos (c+d x))}{d} \]
-3*a*b^2*cos(d*x+c)/d-1/2*b*(3*a^2-b^2)*cos(d*x+c)^2/d-1/3*a*(a^2-3*b^2)*c os(d*x+c)^3/d+3/4*a^2*b*cos(d*x+c)^4/d+1/5*a^3*cos(d*x+c)^5/d-b^3*ln(cos(d *x+c))/d
Time = 0.22 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.88 \[ \int \cos ^2(c+d x) (a \sin (c+d x)+b \tan (c+d x))^3 \, dx=-\frac {3 a b^2 \cos (c+d x)+\frac {1}{2} b \left (3 a^2-b^2\right ) \cos ^2(c+d x)+\frac {1}{3} a \left (a^2-3 b^2\right ) \cos ^3(c+d x)-\frac {3}{4} a^2 b \cos ^4(c+d x)-\frac {1}{5} a^3 \cos ^5(c+d x)+b^3 \log (\cos (c+d x))}{d} \]
-((3*a*b^2*Cos[c + d*x] + (b*(3*a^2 - b^2)*Cos[c + d*x]^2)/2 + (a*(a^2 - 3 *b^2)*Cos[c + d*x]^3)/3 - (3*a^2*b*Cos[c + d*x]^4)/4 - (a^3*Cos[c + d*x]^5 )/5 + b^3*Log[Cos[c + d*x]])/d)
Time = 0.44 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.01, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3042, 4897, 3042, 25, 3316, 27, 522, 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.
\(\displaystyle \int \cos ^2(c+d x) (a \sin (c+d x)+b \tan (c+d x))^3 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \cos (c+d x)^2 (a \sin (c+d x)+b \tan (c+d x))^3dx\) |
\(\Big \downarrow \) 4897 |
\(\displaystyle \int \sin ^2(c+d x) \tan (c+d x) (a \cos (c+d x)+b)^3dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int -\frac {\cos \left (c+d x+\frac {\pi }{2}\right )^3 \left (a \sin \left (c+d x+\frac {\pi }{2}\right )+b\right )^3}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\int \frac {\cos \left (\frac {1}{2} (2 c+\pi )+d x\right )^3 \left (b+a \sin \left (\frac {1}{2} (2 c+\pi )+d x\right )\right )^3}{\sin \left (\frac {1}{2} (2 c+\pi )+d x\right )}dx\) |
\(\Big \downarrow \) 3316 |
\(\displaystyle -\frac {\int (b+a \cos (c+d x))^3 \left (a^2-a^2 \cos ^2(c+d x)\right ) \sec (c+d x)d(a \cos (c+d x))}{a^3 d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {\int \frac {(b+a \cos (c+d x))^3 \left (a^2-a^2 \cos ^2(c+d x)\right ) \sec (c+d x)}{a}d(a \cos (c+d x))}{a^2 d}\) |
\(\Big \downarrow \) 522 |
\(\displaystyle -\frac {\int \left (-a^4 \cos ^4(c+d x)-3 a^3 b \cos ^3(c+d x)+a^2 \left (a^2-3 b^2\right ) \cos ^2(c+d x)+a b \left (3 a^2-b^2\right ) \cos (c+d x)+3 a^2 b^2+a b^3 \sec (c+d x)\right )d(a \cos (c+d x))}{a^2 d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {-\frac {1}{5} a^5 \cos ^5(c+d x)-\frac {3}{4} a^4 b \cos ^4(c+d x)+3 a^3 b^2 \cos (c+d x)+a^2 b^3 \log (a \cos (c+d x))+\frac {1}{2} a^2 b \left (3 a^2-b^2\right ) \cos ^2(c+d x)+\frac {1}{3} a^3 \left (a^2-3 b^2\right ) \cos ^3(c+d x)}{a^2 d}\) |
-((3*a^3*b^2*Cos[c + d*x] + (a^2*b*(3*a^2 - b^2)*Cos[c + d*x]^2)/2 + (a^3* (a^2 - 3*b^2)*Cos[c + d*x]^3)/3 - (3*a^4*b*Cos[c + d*x]^4)/4 - (a^5*Cos[c + d*x]^5)/5 + a^2*b^3*Log[a*Cos[c + d*x]])/(a^2*d))
3.3.44.3.1 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[((e_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_)^2)^(p_. ), x_Symbol] :> Int[ExpandIntegrand[(e*x)^m*(c + d*x)^n*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && IGtQ[p, 0]
Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_ .)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Simp[1/(b^p* f) Subst[Int[(a + x)^m*(c + (d/b)*x)^n*(b^2 - x^2)^((p - 1)/2), x], x, b* Sin[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && IntegerQ[(p - 1) /2] && NeQ[a^2 - b^2, 0]
Time = 17.40 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.82
method | result | size |
derivativedivides | \(\frac {a^{3} \left (-\frac {\cos \left (d x +c \right )^{3} \sin \left (d x +c \right )^{2}}{5}-\frac {2 \cos \left (d x +c \right )^{3}}{15}\right )+\frac {3 a^{2} b \sin \left (d x +c \right )^{4}}{4}-a \,b^{2} \left (2+\sin \left (d x +c \right )^{2}\right ) \cos \left (d x +c \right )+b^{3} \left (-\frac {\sin \left (d x +c \right )^{2}}{2}-\ln \left (\cos \left (d x +c \right )\right )\right )}{d}\) | \(99\) |
default | \(\frac {a^{3} \left (-\frac {\cos \left (d x +c \right )^{3} \sin \left (d x +c \right )^{2}}{5}-\frac {2 \cos \left (d x +c \right )^{3}}{15}\right )+\frac {3 a^{2} b \sin \left (d x +c \right )^{4}}{4}-a \,b^{2} \left (2+\sin \left (d x +c \right )^{2}\right ) \cos \left (d x +c \right )+b^{3} \left (-\frac {\sin \left (d x +c \right )^{2}}{2}-\ln \left (\cos \left (d x +c \right )\right )\right )}{d}\) | \(99\) |
risch | \(i x \,b^{3}-\frac {3 b \,{\mathrm e}^{2 i \left (d x +c \right )} a^{2}}{16 d}+\frac {b^{3} {\mathrm e}^{2 i \left (d x +c \right )}}{8 d}-\frac {3 b \,{\mathrm e}^{-2 i \left (d x +c \right )} a^{2}}{16 d}+\frac {b^{3} {\mathrm e}^{-2 i \left (d x +c \right )}}{8 d}+\frac {2 i b^{3} c}{d}-\frac {b^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{d}-\frac {a^{3} \cos \left (d x +c \right )}{8 d}-\frac {9 a \,b^{2} \cos \left (d x +c \right )}{4 d}+\frac {a^{3} \cos \left (5 d x +5 c \right )}{80 d}+\frac {3 b \cos \left (4 d x +4 c \right ) a^{2}}{32 d}-\frac {a^{3} \cos \left (3 d x +3 c \right )}{48 d}+\frac {a \cos \left (3 d x +3 c \right ) b^{2}}{4 d}\) | \(208\) |
1/d*(a^3*(-1/5*cos(d*x+c)^3*sin(d*x+c)^2-2/15*cos(d*x+c)^3)+3/4*a^2*b*sin( d*x+c)^4-a*b^2*(2+sin(d*x+c)^2)*cos(d*x+c)+b^3*(-1/2*sin(d*x+c)^2-ln(cos(d *x+c))))
Time = 0.25 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.84 \[ \int \cos ^2(c+d x) (a \sin (c+d x)+b \tan (c+d x))^3 \, dx=\frac {12 \, a^{3} \cos \left (d x + c\right )^{5} + 45 \, a^{2} b \cos \left (d x + c\right )^{4} - 180 \, a b^{2} \cos \left (d x + c\right ) - 20 \, {\left (a^{3} - 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{3} - 60 \, b^{3} \log \left (-\cos \left (d x + c\right )\right ) - 30 \, {\left (3 \, a^{2} b - b^{3}\right )} \cos \left (d x + c\right )^{2}}{60 \, d} \]
1/60*(12*a^3*cos(d*x + c)^5 + 45*a^2*b*cos(d*x + c)^4 - 180*a*b^2*cos(d*x + c) - 20*(a^3 - 3*a*b^2)*cos(d*x + c)^3 - 60*b^3*log(-cos(d*x + c)) - 30* (3*a^2*b - b^3)*cos(d*x + c)^2)/d
\[ \int \cos ^2(c+d x) (a \sin (c+d x)+b \tan (c+d x))^3 \, dx=\int \left (a \sin {\left (c + d x \right )} + b \tan {\left (c + d x \right )}\right )^{3} \cos ^{2}{\left (c + d x \right )}\, dx \]
Time = 0.21 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.78 \[ \int \cos ^2(c+d x) (a \sin (c+d x)+b \tan (c+d x))^3 \, dx=\frac {45 \, a^{2} b \sin \left (d x + c\right )^{4} + 4 \, {\left (3 \, \cos \left (d x + c\right )^{5} - 5 \, \cos \left (d x + c\right )^{3}\right )} a^{3} + 60 \, {\left (\cos \left (d x + c\right )^{3} - 3 \, \cos \left (d x + c\right )\right )} a b^{2} - 30 \, {\left (\sin \left (d x + c\right )^{2} + \log \left (\sin \left (d x + c\right )^{2} - 1\right )\right )} b^{3}}{60 \, d} \]
1/60*(45*a^2*b*sin(d*x + c)^4 + 4*(3*cos(d*x + c)^5 - 5*cos(d*x + c)^3)*a^ 3 + 60*(cos(d*x + c)^3 - 3*cos(d*x + c))*a*b^2 - 30*(sin(d*x + c)^2 + log( sin(d*x + c)^2 - 1))*b^3)/d
Exception generated. \[ \int \cos ^2(c+d x) (a \sin (c+d x)+b \tan (c+d x))^3 \, dx=\text {Exception raised: TypeError} \]
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Modgcd: no suitable evaluation poin tindex.cc index_m operator + Error: Bad Argument ValueDone
Time = 23.73 (sec) , antiderivative size = 237, normalized size of antiderivative = 1.98 \[ \int \cos ^2(c+d x) (a \sin (c+d x)+b \tan (c+d x))^3 \, dx=\frac {40\,a^3\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{3\,d}-\frac {4\,a^3\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{d}-\frac {16\,a^3\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8}{d}+\frac {32\,a^3\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}}{5\,d}-\frac {2\,b^3\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{d}+\frac {2\,b^3\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{d}+\frac {2\,b^3\,\mathrm {atanh}\left (\frac {1}{{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}-1\right )}{d}-\frac {12\,a\,b^2\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{d}+\frac {12\,a^2\,b\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{d}+\frac {8\,a\,b^2\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{d}-\frac {24\,a^2\,b\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{d}+\frac {12\,a^2\,b\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8}{d} \]
(40*a^3*cos(c/2 + (d*x)/2)^6)/(3*d) - (4*a^3*cos(c/2 + (d*x)/2)^4)/d - (16 *a^3*cos(c/2 + (d*x)/2)^8)/d + (32*a^3*cos(c/2 + (d*x)/2)^10)/(5*d) - (2*b ^3*cos(c/2 + (d*x)/2)^2)/d + (2*b^3*cos(c/2 + (d*x)/2)^4)/d + (2*b^3*atanh (1/cos(c/2 + (d*x)/2)^2 - 1))/d - (12*a*b^2*cos(c/2 + (d*x)/2)^4)/d + (12* a^2*b*cos(c/2 + (d*x)/2)^4)/d + (8*a*b^2*cos(c/2 + (d*x)/2)^6)/d - (24*a^2 *b*cos(c/2 + (d*x)/2)^6)/d + (12*a^2*b*cos(c/2 + (d*x)/2)^8)/d