Integrand size = 21, antiderivative size = 50 \[ \int \cos ^3(c+d x) \left (A+C \cos ^2(c+d x)\right ) \, dx=\frac {(A+C) \sin (c+d x)}{d}-\frac {(A+2 C) \sin ^3(c+d x)}{3 d}+\frac {C \sin ^5(c+d x)}{5 d} \] Output:
(A+C)*sin(d*x+c)/d-1/3*(A+2*C)*sin(d*x+c)^3/d+1/5*C*sin(d*x+c)^5/d
Time = 0.03 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.42 \[ \int \cos ^3(c+d x) \left (A+C \cos ^2(c+d x)\right ) \, dx=\frac {A \sin (c+d x)}{d}+\frac {C \sin (c+d x)}{d}-\frac {A \sin ^3(c+d x)}{3 d}-\frac {2 C \sin ^3(c+d x)}{3 d}+\frac {C \sin ^5(c+d x)}{5 d} \] Input:
Integrate[Cos[c + d*x]^3*(A + C*Cos[c + d*x]^2),x]
Output:
(A*Sin[c + d*x])/d + (C*Sin[c + d*x])/d - (A*Sin[c + d*x]^3)/(3*d) - (2*C* Sin[c + d*x]^3)/(3*d) + (C*Sin[c + d*x]^5)/(5*d)
Time = 0.25 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.94, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {3042, 3492, 290, 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.
\(\displaystyle \int \cos ^3(c+d x) \left (A+C \cos ^2(c+d x)\right ) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sin \left (c+d x+\frac {\pi }{2}\right )^3 \left (A+C \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )dx\) |
\(\Big \downarrow \) 3492 |
\(\displaystyle -\frac {\int \left (1-\sin ^2(c+d x)\right ) \left (-C \sin ^2(c+d x)+A+C\right )d(-\sin (c+d x))}{d}\) |
\(\Big \downarrow \) 290 |
\(\displaystyle -\frac {\int \left (C \sin ^4(c+d x)-(A+2 C) \sin ^2(c+d x)+A \left (\frac {C}{A}+1\right )\right )d(-\sin (c+d x))}{d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {\frac {1}{3} (A+2 C) \sin ^3(c+d x)-(A+C) \sin (c+d x)-\frac {1}{5} C \sin ^5(c+d x)}{d}\) |
Input:
Int[Cos[c + d*x]^3*(A + C*Cos[c + d*x]^2),x]
Output:
-((-((A + C)*Sin[c + d*x]) + ((A + 2*C)*Sin[c + d*x]^3)/3 - (C*Sin[c + d*x ]^5)/5)/d)
Int[((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.), x_Symbol] :> I nt[ExpandIntegrand[(a + b*x^2)^p*(c + d*x^2)^q, x], x] /; FreeQ[{a, b, c, d }, x] && NeQ[b*c - a*d, 0] && IGtQ[p, 0] && IGtQ[q, 0]
Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((A_) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[-f^(-1) Subst[Int[(1 - x^2)^((m - 1)/2)*(A + C - C*x^2 ), x], x, Cos[e + f*x]], x] /; FreeQ[{e, f, A, C}, x] && IGtQ[(m + 1)/2, 0]
Time = 1.48 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.98
method | result | size |
parallelrisch | \(\frac {\left (20 A +25 C \right ) \sin \left (3 d x +3 c \right )+3 C \sin \left (5 d x +5 c \right )+180 \left (A +\frac {5 C}{6}\right ) \sin \left (d x +c \right )}{240 d}\) | \(49\) |
derivativedivides | \(\frac {\frac {C \left (\frac {8}{3}+\cos \left (d x +c \right )^{4}+\frac {4 \cos \left (d x +c \right )^{2}}{3}\right ) \sin \left (d x +c \right )}{5}+\frac {A \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}}{d}\) | \(54\) |
default | \(\frac {\frac {C \left (\frac {8}{3}+\cos \left (d x +c \right )^{4}+\frac {4 \cos \left (d x +c \right )^{2}}{3}\right ) \sin \left (d x +c \right )}{5}+\frac {A \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}}{d}\) | \(54\) |
parts | \(\frac {A \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3 d}+\frac {C \left (\frac {8}{3}+\cos \left (d x +c \right )^{4}+\frac {4 \cos \left (d x +c \right )^{2}}{3}\right ) \sin \left (d x +c \right )}{5 d}\) | \(56\) |
risch | \(\frac {3 \sin \left (d x +c \right ) A}{4 d}+\frac {5 C \sin \left (d x +c \right )}{8 d}+\frac {\sin \left (5 d x +5 c \right ) C}{80 d}+\frac {\sin \left (3 d x +3 c \right ) A}{12 d}+\frac {5 \sin \left (3 d x +3 c \right ) C}{48 d}\) | \(71\) |
norman | \(\frac {\frac {2 \left (A +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {2 \left (A +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{d}+\frac {8 \left (2 A +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 d}+\frac {8 \left (2 A +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{3 d}+\frac {4 \left (25 A +29 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{15 d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{5}}\) | \(119\) |
orering | \(-\frac {259 \left (-3 \cos \left (d x +c \right )^{2} \left (A +C \cos \left (d x +c \right )^{2}\right ) d \sin \left (d x +c \right )-2 \cos \left (d x +c \right )^{4} C d \sin \left (d x +c \right )\right )}{225 d^{2}}-\frac {7 \left (-6 d^{3} \sin \left (d x +c \right )^{3} \left (A +C \cos \left (d x +c \right )^{2}\right )-54 \cos \left (d x +c \right )^{2} C \,d^{3} \sin \left (d x +c \right )^{3}+21 \cos \left (d x +c \right )^{2} \left (A +C \cos \left (d x +c \right )^{2}\right ) d^{3} \sin \left (d x +c \right )+44 \cos \left (d x +c \right )^{4} C \,d^{3} \sin \left (d x +c \right )\right )}{45 d^{4}}-\frac {-183 d^{5} \sin \left (d x +c \right ) \left (A +C \cos \left (d x +c \right )^{2}\right ) \cos \left (d x +c \right )^{2}+1740 d^{5} \sin \left (d x +c \right )^{3} C \cos \left (d x +c \right )^{2}+60 d^{5} \sin \left (d x +c \right )^{3} \left (A +C \cos \left (d x +c \right )^{2}\right )-120 d^{5} \sin \left (d x +c \right )^{5} C -1022 \cos \left (d x +c \right )^{4} C \,d^{5} \sin \left (d x +c \right )}{225 d^{6}}\) | \(277\) |
Input:
int(cos(d*x+c)^3*(A+C*cos(d*x+c)^2),x,method=_RETURNVERBOSE)
Output:
1/240*((20*A+25*C)*sin(3*d*x+3*c)+3*C*sin(5*d*x+5*c)+180*(A+5/6*C)*sin(d*x +c))/d
Time = 0.11 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.90 \[ \int \cos ^3(c+d x) \left (A+C \cos ^2(c+d x)\right ) \, dx=\frac {{\left (3 \, C \cos \left (d x + c\right )^{4} + {\left (5 \, A + 4 \, C\right )} \cos \left (d x + c\right )^{2} + 10 \, A + 8 \, C\right )} \sin \left (d x + c\right )}{15 \, d} \] Input:
integrate(cos(d*x+c)^3*(A+C*cos(d*x+c)^2),x, algorithm="fricas")
Output:
1/15*(3*C*cos(d*x + c)^4 + (5*A + 4*C)*cos(d*x + c)^2 + 10*A + 8*C)*sin(d* x + c)/d
Leaf count of result is larger than twice the leaf count of optimal. 105 vs. \(2 (42) = 84\).
Time = 0.22 (sec) , antiderivative size = 105, normalized size of antiderivative = 2.10 \[ \int \cos ^3(c+d x) \left (A+C \cos ^2(c+d x)\right ) \, dx=\begin {cases} \frac {2 A \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac {A \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} + \frac {8 C \sin ^{5}{\left (c + d x \right )}}{15 d} + \frac {4 C \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{3 d} + \frac {C \sin {\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{d} & \text {for}\: d \neq 0 \\x \left (A + C \cos ^{2}{\left (c \right )}\right ) \cos ^{3}{\left (c \right )} & \text {otherwise} \end {cases} \] Input:
integrate(cos(d*x+c)**3*(A+C*cos(d*x+c)**2),x)
Output:
Piecewise((2*A*sin(c + d*x)**3/(3*d) + A*sin(c + d*x)*cos(c + d*x)**2/d + 8*C*sin(c + d*x)**5/(15*d) + 4*C*sin(c + d*x)**3*cos(c + d*x)**2/(3*d) + C *sin(c + d*x)*cos(c + d*x)**4/d, Ne(d, 0)), (x*(A + C*cos(c)**2)*cos(c)**3 , True))
Time = 0.03 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.86 \[ \int \cos ^3(c+d x) \left (A+C \cos ^2(c+d x)\right ) \, dx=\frac {3 \, C \sin \left (d x + c\right )^{5} - 5 \, {\left (A + 2 \, C\right )} \sin \left (d x + c\right )^{3} + 15 \, {\left (A + C\right )} \sin \left (d x + c\right )}{15 \, d} \] Input:
integrate(cos(d*x+c)^3*(A+C*cos(d*x+c)^2),x, algorithm="maxima")
Output:
1/15*(3*C*sin(d*x + c)^5 - 5*(A + 2*C)*sin(d*x + c)^3 + 15*(A + C)*sin(d*x + c))/d
Time = 0.19 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.14 \[ \int \cos ^3(c+d x) \left (A+C \cos ^2(c+d x)\right ) \, dx=\frac {3 \, C \sin \left (d x + c\right )^{5} - 5 \, A \sin \left (d x + c\right )^{3} - 10 \, C \sin \left (d x + c\right )^{3} + 15 \, A \sin \left (d x + c\right ) + 15 \, C \sin \left (d x + c\right )}{15 \, d} \] Input:
integrate(cos(d*x+c)^3*(A+C*cos(d*x+c)^2),x, algorithm="giac")
Output:
1/15*(3*C*sin(d*x + c)^5 - 5*A*sin(d*x + c)^3 - 10*C*sin(d*x + c)^3 + 15*A *sin(d*x + c) + 15*C*sin(d*x + c))/d
Time = 40.83 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.86 \[ \int \cos ^3(c+d x) \left (A+C \cos ^2(c+d x)\right ) \, dx=\frac {\frac {C\,{\sin \left (c+d\,x\right )}^5}{5}+\left (-\frac {A}{3}-\frac {2\,C}{3}\right )\,{\sin \left (c+d\,x\right )}^3+\left (A+C\right )\,\sin \left (c+d\,x\right )}{d} \] Input:
int(cos(c + d*x)^3*(A + C*cos(c + d*x)^2),x)
Output:
((C*sin(c + d*x)^5)/5 + sin(c + d*x)*(A + C) - sin(c + d*x)^3*(A/3 + (2*C) /3))/d
Time = 0.16 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.02 \[ \int \cos ^3(c+d x) \left (A+C \cos ^2(c+d x)\right ) \, dx=\frac {\sin \left (d x +c \right ) \left (3 \sin \left (d x +c \right )^{4} c -5 \sin \left (d x +c \right )^{2} a -10 \sin \left (d x +c \right )^{2} c +15 a +15 c \right )}{15 d} \] Input:
int(cos(d*x+c)^3*(A+C*cos(d*x+c)^2),x)
Output:
(sin(c + d*x)*(3*sin(c + d*x)**4*c - 5*sin(c + d*x)**2*a - 10*sin(c + d*x) **2*c + 15*a + 15*c))/(15*d)