Integrand size = 26, antiderivative size = 112 \[ \int \cos (c+d x) (a \sin (c+d x)+b \tan (c+d x))^3 \, dx=-\frac {b \left (3 a^2-b^2\right ) \cos (c+d x)}{d}-\frac {a \left (a^2-3 b^2\right ) \cos ^2(c+d x)}{2 d}+\frac {a^2 b \cos ^3(c+d x)}{d}+\frac {a^3 \cos ^4(c+d x)}{4 d}-\frac {3 a b^2 \log (\cos (c+d x))}{d}+\frac {b^3 \sec (c+d x)}{d} \]
-b*(3*a^2-b^2)*cos(d*x+c)/d-1/2*a*(a^2-3*b^2)*cos(d*x+c)^2/d+a^2*b*cos(d*x +c)^3/d+1/4*a^3*cos(d*x+c)^4/d-3*a*b^2*ln(cos(d*x+c))/d+b^3*sec(d*x+c)/d
Time = 1.09 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.88 \[ \int \cos (c+d x) (a \sin (c+d x)+b \tan (c+d x))^3 \, dx=\frac {8 b \left (-9 a^2+4 b^2\right ) \cos (c+d x)-4 \left (a^3-6 a b^2\right ) \cos (2 (c+d x))+8 a^2 b \cos (3 (c+d x))+a^3 \cos (4 (c+d x))-96 a b^2 \log (\cos (c+d x))+32 b^3 \sec (c+d x)}{32 d} \]
(8*b*(-9*a^2 + 4*b^2)*Cos[c + d*x] - 4*(a^3 - 6*a*b^2)*Cos[2*(c + d*x)] + 8*a^2*b*Cos[3*(c + d*x)] + a^3*Cos[4*(c + d*x)] - 96*a*b^2*Log[Cos[c + d*x ]] + 32*b^3*Sec[c + d*x])/(32*d)
Time = 0.42 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.99, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, 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.
\(\displaystyle \int \cos (c+d x) (a \sin (c+d x)+b \tan (c+d x))^3 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \cos (c+d x) (a \sin (c+d x)+b \tan (c+d x))^3dx\) |
\(\Big \downarrow \) 4897 |
\(\displaystyle \int \sin (c+d x) \tan ^2(c+d x) (a \cos (c+d x)+b)^3dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int -\frac {\cos \left (c+d x+\frac {\pi }{2}\right )^3 \left (a \sin \left (c+d x+\frac {\pi }{2}\right )+b\right )^3}{\sin \left (c+d x+\frac {\pi }{2}\right )^2}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\int \frac {\cos \left (\frac {1}{2} (2 c+\pi )+d x\right )^3 \left (b+a \sin \left (\frac {1}{2} (2 c+\pi )+d x\right )\right )^3}{\sin \left (\frac {1}{2} (2 c+\pi )+d x\right )^2}dx\) |
\(\Big \downarrow \) 3316 |
\(\displaystyle -\frac {\int (b+a \cos (c+d x))^3 \left (a^2-a^2 \cos ^2(c+d x)\right ) \sec ^2(c+d x)d(a \cos (c+d x))}{a^3 d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {\int \frac {(b+a \cos (c+d x))^3 \left (a^2-a^2 \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{a^2}d(a \cos (c+d x))}{a d}\) |
\(\Big \downarrow \) 522 |
\(\displaystyle -\frac {\int \left (\sec ^2(c+d x) b^3+3 a \sec (c+d x) b^2-3 a^2 \cos ^2(c+d x) b+3 a^2 \left (1-\frac {b^2}{3 a^2}\right ) b-a^3 \cos ^3(c+d x)+a \left (a^2-3 b^2\right ) \cos (c+d x)\right )d(a \cos (c+d x))}{a d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {-\frac {1}{4} a^4 \cos ^4(c+d x)-a^3 b \cos ^3(c+d x)+\frac {1}{2} a^2 \left (a^2-3 b^2\right ) \cos ^2(c+d x)+a b \left (3 a^2-b^2\right ) \cos (c+d x)+3 a^2 b^2 \log (a \cos (c+d x))-a b^3 \sec (c+d x)}{a d}\) |
-((a*b*(3*a^2 - b^2)*Cos[c + d*x] + (a^2*(a^2 - 3*b^2)*Cos[c + d*x]^2)/2 - a^3*b*Cos[c + d*x]^3 - (a^4*Cos[c + d*x]^4)/4 + 3*a^2*b^2*Log[a*Cos[c + d *x]] - a*b^3*Sec[c + d*x])/(a*d))
3.3.45.3.1 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 = 9.20 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.95
method | result | size |
derivativedivides | \(\frac {\frac {a^{3} \sin \left (d x +c \right )^{4}}{4}-a^{2} b \left (2+\sin \left (d x +c \right )^{2}\right ) \cos \left (d x +c \right )+3 a \,b^{2} \left (-\frac {\sin \left (d x +c \right )^{2}}{2}-\ln \left (\cos \left (d x +c \right )\right )\right )+b^{3} \left (\frac {\sin \left (d x +c \right )^{4}}{\cos \left (d x +c \right )}+\left (2+\sin \left (d x +c \right )^{2}\right ) \cos \left (d x +c \right )\right )}{d}\) | \(106\) |
default | \(\frac {\frac {a^{3} \sin \left (d x +c \right )^{4}}{4}-a^{2} b \left (2+\sin \left (d x +c \right )^{2}\right ) \cos \left (d x +c \right )+3 a \,b^{2} \left (-\frac {\sin \left (d x +c \right )^{2}}{2}-\ln \left (\cos \left (d x +c \right )\right )\right )+b^{3} \left (\frac {\sin \left (d x +c \right )^{4}}{\cos \left (d x +c \right )}+\left (2+\sin \left (d x +c \right )^{2}\right ) \cos \left (d x +c \right )\right )}{d}\) | \(106\) |
risch | \(3 i x a \,b^{2}-\frac {a^{3} {\mathrm e}^{2 i \left (d x +c \right )}}{16 d}+\frac {3 a \,{\mathrm e}^{2 i \left (d x +c \right )} b^{2}}{8 d}-\frac {9 b \,{\mathrm e}^{i \left (d x +c \right )} a^{2}}{8 d}+\frac {b^{3} {\mathrm e}^{i \left (d x +c \right )}}{2 d}-\frac {9 b \,{\mathrm e}^{-i \left (d x +c \right )} a^{2}}{8 d}+\frac {b^{3} {\mathrm e}^{-i \left (d x +c \right )}}{2 d}-\frac {a^{3} {\mathrm e}^{-2 i \left (d x +c \right )}}{16 d}+\frac {3 a \,{\mathrm e}^{-2 i \left (d x +c \right )} b^{2}}{8 d}+\frac {6 i a \,b^{2} c}{d}+\frac {2 b^{3} {\mathrm e}^{i \left (d x +c \right )}}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}-\frac {3 a \,b^{2} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{d}+\frac {a^{3} \cos \left (4 d x +4 c \right )}{32 d}+\frac {b \cos \left (3 d x +3 c \right ) a^{2}}{4 d}\) | \(247\) |
1/d*(1/4*a^3*sin(d*x+c)^4-a^2*b*(2+sin(d*x+c)^2)*cos(d*x+c)+3*a*b^2*(-1/2* sin(d*x+c)^2-ln(cos(d*x+c)))+b^3*(sin(d*x+c)^4/cos(d*x+c)+(2+sin(d*x+c)^2) *cos(d*x+c)))
Time = 0.27 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.14 \[ \int \cos (c+d x) (a \sin (c+d x)+b \tan (c+d x))^3 \, dx=\frac {8 \, a^{3} \cos \left (d x + c\right )^{5} + 32 \, a^{2} b \cos \left (d x + c\right )^{4} - 96 \, a b^{2} \cos \left (d x + c\right ) \log \left (-\cos \left (d x + c\right )\right ) - 16 \, {\left (a^{3} - 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{3} + 32 \, b^{3} - 32 \, {\left (3 \, a^{2} b - b^{3}\right )} \cos \left (d x + c\right )^{2} + {\left (5 \, a^{3} - 24 \, a b^{2}\right )} \cos \left (d x + c\right )}{32 \, d \cos \left (d x + c\right )} \]
1/32*(8*a^3*cos(d*x + c)^5 + 32*a^2*b*cos(d*x + c)^4 - 96*a*b^2*cos(d*x + c)*log(-cos(d*x + c)) - 16*(a^3 - 3*a*b^2)*cos(d*x + c)^3 + 32*b^3 - 32*(3 *a^2*b - b^3)*cos(d*x + c)^2 + (5*a^3 - 24*a*b^2)*cos(d*x + c))/(d*cos(d*x + c))
\[ \int \cos (c+d x) (a \sin (c+d x)+b \tan (c+d x))^3 \, dx=\int \left (a \sin {\left (c + d x \right )} + b \tan {\left (c + d x \right )}\right )^{3} \cos {\left (c + d x \right )}\, dx \]
Time = 0.21 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.78 \[ \int \cos (c+d x) (a \sin (c+d x)+b \tan (c+d x))^3 \, dx=\frac {a^{3} \sin \left (d x + c\right )^{4} + 4 \, {\left (\cos \left (d x + c\right )^{3} - 3 \, \cos \left (d x + c\right )\right )} a^{2} b - 6 \, {\left (\sin \left (d x + c\right )^{2} + \log \left (\sin \left (d x + c\right )^{2} - 1\right )\right )} a b^{2} + 4 \, b^{3} {\left (\frac {1}{\cos \left (d x + c\right )} + \cos \left (d x + c\right )\right )}}{4 \, d} \]
1/4*(a^3*sin(d*x + c)^4 + 4*(cos(d*x + c)^3 - 3*cos(d*x + c))*a^2*b - 6*(s in(d*x + c)^2 + log(sin(d*x + c)^2 - 1))*a*b^2 + 4*b^3*(1/cos(d*x + c) + c os(d*x + c)))/d
Exception generated. \[ \int \cos (c+d x) (a \sin (c+d x)+b \tan (c+d x))^3 \, dx=\text {Exception raised: TypeError} \]
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 = 25.90 (sec) , antiderivative size = 225, normalized size of antiderivative = 2.01 \[ \int \cos (c+d x) (a \sin (c+d x)+b \tan (c+d x))^3 \, dx=\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (4\,a^3+4\,a^2\,b-6\,a\,b^2+12\,b^3\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (12\,a^2\,b+6\,a\,b^2-12\,b^3\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (-4\,a^3+12\,a^2\,b+6\,a\,b^2+4\,b^3\right )-4\,a^2\,b+4\,b^3+6\,a\,b^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8}{d\,\left (-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}+\frac {6\,a\,b^2\,\mathrm {atanh}\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\right )}{d} \]
(tan(c/2 + (d*x)/2)^4*(4*a^2*b - 6*a*b^2 + 4*a^3 + 12*b^3) - tan(c/2 + (d* x)/2)^2*(6*a*b^2 + 12*a^2*b - 12*b^3) + tan(c/2 + (d*x)/2)^6*(6*a*b^2 + 12 *a^2*b - 4*a^3 + 4*b^3) - 4*a^2*b + 4*b^3 + 6*a*b^2*tan(c/2 + (d*x)/2)^8)/ (d*(3*tan(c/2 + (d*x)/2)^2 + 2*tan(c/2 + (d*x)/2)^4 - 2*tan(c/2 + (d*x)/2) ^6 - 3*tan(c/2 + (d*x)/2)^8 - tan(c/2 + (d*x)/2)^10 + 1)) + (6*a*b^2*atanh (tan(c/2 + (d*x)/2)^2))/d