Integrand size = 21, antiderivative size = 77 \[ \int \cos ^3(c+d x) (a+b \sin (c+d x))^2 \, dx=-\frac {\left (a^2-b^2\right ) (a+b \sin (c+d x))^3}{3 b^3 d}+\frac {a (a+b \sin (c+d x))^4}{2 b^3 d}-\frac {(a+b \sin (c+d x))^5}{5 b^3 d} \] Output:
-1/3*(a^2-b^2)*(a+b*sin(d*x+c))^3/b^3/d+1/2*a*(a+b*sin(d*x+c))^4/b^3/d-1/5 *(a+b*sin(d*x+c))^5/b^3/d
Time = 0.13 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.73 \[ \int \cos ^3(c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {(a+b \sin (c+d x))^3 \left (-a^2+7 b^2+3 b^2 \cos (2 (c+d x))+3 a b \sin (c+d x)\right )}{30 b^3 d} \] Input:
Integrate[Cos[c + d*x]^3*(a + b*Sin[c + d*x])^2,x]
Output:
((a + b*Sin[c + d*x])^3*(-a^2 + 7*b^2 + 3*b^2*Cos[2*(c + d*x)] + 3*a*b*Sin [c + d*x]))/(30*b^3*d)
Time = 0.27 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.86, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {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.
\(\displaystyle \int \cos ^3(c+d x) (a+b \sin (c+d x))^2 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \cos (c+d x)^3 (a+b \sin (c+d x))^2dx\) |
\(\Big \downarrow \) 3147 |
\(\displaystyle \frac {\int (a+b \sin (c+d x))^2 \left (b^2-b^2 \sin ^2(c+d x)\right )d(b \sin (c+d x))}{b^3 d}\) |
\(\Big \downarrow \) 476 |
\(\displaystyle \frac {\int \left (-(a+b \sin (c+d x))^4+2 a (a+b \sin (c+d x))^3+\left (b^2-a^2\right ) (a+b \sin (c+d x))^2\right )d(b \sin (c+d x))}{b^3 d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {-\frac {1}{3} \left (a^2-b^2\right ) (a+b \sin (c+d x))^3-\frac {1}{5} (a+b \sin (c+d x))^5+\frac {1}{2} a (a+b \sin (c+d x))^4}{b^3 d}\) |
Input:
Int[Cos[c + d*x]^3*(a + b*Sin[c + d*x])^2,x]
Output:
(-1/3*((a^2 - b^2)*(a + b*Sin[c + d*x])^3) + (a*(a + b*Sin[c + d*x])^4)/2 - (a + b*Sin[c + d*x])^5/5)/(b^3*d)
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 = 6.35 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.96
method | result | size |
derivativedivides | \(-\frac {\frac {b^{2} \sin \left (d x +c \right )^{5}}{5}+\frac {a b \sin \left (d x +c \right )^{4}}{2}+\frac {\left (a^{2}-b^{2}\right ) \sin \left (d x +c \right )^{3}}{3}-a b \sin \left (d x +c \right )^{2}-a^{2} \sin \left (d x +c \right )}{d}\) | \(74\) |
default | \(-\frac {\frac {b^{2} \sin \left (d x +c \right )^{5}}{5}+\frac {a b \sin \left (d x +c \right )^{4}}{2}+\frac {\left (a^{2}-b^{2}\right ) \sin \left (d x +c \right )^{3}}{3}-a b \sin \left (d x +c \right )^{2}-a^{2} \sin \left (d x +c \right )}{d}\) | \(74\) |
parallelrisch | \(\frac {180 a^{2} \sin \left (d x +c \right )+20 \sin \left (3 d x +3 c \right ) a^{2}+30 \sin \left (d x +c \right ) b^{2}-5 \sin \left (3 d x +3 c \right ) b^{2}+75 a b -60 a b \cos \left (2 d x +2 c \right )-3 \sin \left (5 d x +5 c \right ) b^{2}-15 a b \cos \left (4 d x +4 c \right )}{240 d}\) | \(101\) |
risch | \(\frac {3 a^{2} \sin \left (d x +c \right )}{4 d}+\frac {b^{2} \sin \left (d x +c \right )}{8 d}-\frac {\sin \left (5 d x +5 c \right ) b^{2}}{80 d}-\frac {a b \cos \left (4 d x +4 c \right )}{16 d}+\frac {a^{2} \sin \left (3 d x +3 c \right )}{12 d}-\frac {\sin \left (3 d x +3 c \right ) b^{2}}{48 d}-\frac {a b \cos \left (2 d x +2 c \right )}{4 d}\) | \(113\) |
norman | \(\frac {\frac {4 a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{d}+\frac {4 a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{d}+\frac {2 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {2 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{d}+\frac {8 \left (2 a^{2}+b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 d}+\frac {8 \left (2 a^{2}+b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{3 d}+\frac {4 \left (25 a^{2}-4 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{15 d}+\frac {4 a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{d}+\frac {4 a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{5}}\) | \(203\) |
orering | \(-\frac {5269 \left (-3 \cos \left (d x +c \right )^{2} \left (a +b \sin \left (d x +c \right )\right )^{2} \sin \left (d x +c \right ) d +2 \cos \left (d x +c \right )^{4} \left (a +b \sin \left (d x +c \right )\right ) b d \right )}{3600 d^{2}}-\frac {1529 \left (-6 \sin \left (d x +c \right )^{3} d^{3} \left (a +b \sin \left (d x +c \right )\right )^{2}+54 \cos \left (d x +c \right )^{2} \left (a +b \sin \left (d x +c \right )\right ) \sin \left (d x +c \right )^{2} d^{3} b +21 \cos \left (d x +c \right )^{2} \left (a +b \sin \left (d x +c \right )\right )^{2} \sin \left (d x +c \right ) d^{3}-24 b^{2} \cos \left (d x +c \right )^{4} \sin \left (d x +c \right ) d^{3}-20 \cos \left (d x +c \right )^{4} \left (a +b \sin \left (d x +c \right )\right ) d^{3} b \right )}{2880 d^{4}}-\frac {341 \left (-183 \sin \left (d x +c \right ) d^{5} \left (a +b \sin \left (d x +c \right )\right )^{2} \cos \left (d x +c \right )^{2}-1170 \sin \left (d x +c \right )^{2} d^{5} \left (a +b \sin \left (d x +c \right )\right ) \cos \left (d x +c \right )^{2} b +60 \sin \left (d x +c \right )^{3} d^{5} \left (a +b \sin \left (d x +c \right )\right )^{2}-570 b^{2} \cos \left (d x +c \right )^{2} \sin \left (d x +c \right )^{3} d^{5}+120 \sin \left (d x +c \right )^{4} d^{5} \left (a +b \sin \left (d x +c \right )\right ) b +750 b^{2} \cos \left (d x +c \right )^{4} \sin \left (d x +c \right ) d^{5}+272 \cos \left (d x +c \right )^{4} \left (a +b \sin \left (d x +c \right )\right ) d^{5} b \right )}{4800 d^{6}}-\frac {11 \left (1641 d^{7} \cos \left (d x +c \right )^{2} \left (a +b \sin \left (d x +c \right )\right )^{2} \sin \left (d x +c \right )-4160 d^{7} \cos \left (d x +c \right )^{4} \left (a +b \sin \left (d x +c \right )\right ) b -20664 b^{2} \cos \left (d x +c \right )^{4} \sin \left (d x +c \right ) d^{7}+21294 \sin \left (d x +c \right )^{2} d^{7} \left (a +b \sin \left (d x +c \right )\right ) \cos \left (d x +c \right )^{2} b -546 \sin \left (d x +c \right )^{3} d^{7} \left (a +b \sin \left (d x +c \right )\right )^{2}+25620 b^{2} \sin \left (d x +c \right )^{3} d^{7} \cos \left (d x +c \right )^{2}-2940 \sin \left (d x +c \right )^{4} d^{7} \left (a +b \sin \left (d x +c \right )\right ) b -1260 b^{2} \sin \left (d x +c \right )^{5} d^{7}\right )}{2880 d^{8}}-\frac {4920 d^{9} \sin \left (d x +c \right )^{3} \left (a +b \sin \left (d x +c \right )\right )^{2}-363690 d^{9} \cos \left (d x +c \right )^{2} \left (a +b \sin \left (d x +c \right )\right ) \sin \left (d x +c \right )^{2} b -14763 d^{9} \cos \left (d x +c \right )^{2} \left (a +b \sin \left (d x +c \right )\right )^{2} \sin \left (d x +c \right )+548250 d^{9} \cos \left (d x +c \right )^{4} b^{2} \sin \left (d x +c \right )+65792 d^{9} \cos \left (d x +c \right )^{4} \left (a +b \sin \left (d x +c \right )\right ) b -839790 b^{2} \cos \left (d x +c \right )^{2} \sin \left (d x +c \right )^{3} d^{9}+55440 \sin \left (d x +c \right )^{4} d^{9} \left (a +b \sin \left (d x +c \right )\right ) b +60480 b^{2} \sin \left (d x +c \right )^{5} d^{9}}{14400 d^{10}}\) | \(795\) |
Input:
int(cos(d*x+c)^3*(a+b*sin(d*x+c))^2,x,method=_RETURNVERBOSE)
Output:
-1/d*(1/5*b^2*sin(d*x+c)^5+1/2*a*b*sin(d*x+c)^4+1/3*(a^2-b^2)*sin(d*x+c)^3 -a*b*sin(d*x+c)^2-a^2*sin(d*x+c))
Time = 0.08 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.90 \[ \int \cos ^3(c+d x) (a+b \sin (c+d x))^2 \, dx=-\frac {15 \, a b \cos \left (d x + c\right )^{4} + 2 \, {\left (3 \, b^{2} \cos \left (d x + c\right )^{4} - {\left (5 \, a^{2} + b^{2}\right )} \cos \left (d x + c\right )^{2} - 10 \, a^{2} - 2 \, b^{2}\right )} \sin \left (d x + c\right )}{30 \, d} \] Input:
integrate(cos(d*x+c)^3*(a+b*sin(d*x+c))^2,x, algorithm="fricas")
Output:
-1/30*(15*a*b*cos(d*x + c)^4 + 2*(3*b^2*cos(d*x + c)^4 - (5*a^2 + b^2)*cos (d*x + c)^2 - 10*a^2 - 2*b^2)*sin(d*x + c))/d
Time = 0.27 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.39 \[ \int \cos ^3(c+d x) (a+b \sin (c+d x))^2 \, dx=\begin {cases} \frac {2 a^{2} \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac {a^{2} \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} - \frac {a b \cos ^{4}{\left (c + d x \right )}}{2 d} + \frac {2 b^{2} \sin ^{5}{\left (c + d x \right )}}{15 d} + \frac {b^{2} \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{3 d} & \text {for}\: d \neq 0 \\x \left (a + b \sin {\left (c \right )}\right )^{2} \cos ^{3}{\left (c \right )} & \text {otherwise} \end {cases} \] Input:
integrate(cos(d*x+c)**3*(a+b*sin(d*x+c))**2,x)
Output:
Piecewise((2*a**2*sin(c + d*x)**3/(3*d) + a**2*sin(c + d*x)*cos(c + d*x)** 2/d - a*b*cos(c + d*x)**4/(2*d) + 2*b**2*sin(c + d*x)**5/(15*d) + b**2*sin (c + d*x)**3*cos(c + d*x)**2/(3*d), Ne(d, 0)), (x*(a + b*sin(c))**2*cos(c) **3, True))
Time = 0.03 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.95 \[ \int \cos ^3(c+d x) (a+b \sin (c+d x))^2 \, dx=-\frac {6 \, b^{2} \sin \left (d x + c\right )^{5} + 15 \, a b \sin \left (d x + c\right )^{4} - 30 \, a b \sin \left (d x + c\right )^{2} + 10 \, {\left (a^{2} - b^{2}\right )} \sin \left (d x + c\right )^{3} - 30 \, a^{2} \sin \left (d x + c\right )}{30 \, d} \] Input:
integrate(cos(d*x+c)^3*(a+b*sin(d*x+c))^2,x, algorithm="maxima")
Output:
-1/30*(6*b^2*sin(d*x + c)^5 + 15*a*b*sin(d*x + c)^4 - 30*a*b*sin(d*x + c)^ 2 + 10*(a^2 - b^2)*sin(d*x + c)^3 - 30*a^2*sin(d*x + c))/d
Time = 0.14 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.04 \[ \int \cos ^3(c+d x) (a+b \sin (c+d x))^2 \, dx=-\frac {6 \, b^{2} \sin \left (d x + c\right )^{5} + 15 \, a b \sin \left (d x + c\right )^{4} + 10 \, a^{2} \sin \left (d x + c\right )^{3} - 10 \, b^{2} \sin \left (d x + c\right )^{3} - 30 \, a b \sin \left (d x + c\right )^{2} - 30 \, a^{2} \sin \left (d x + c\right )}{30 \, d} \] Input:
integrate(cos(d*x+c)^3*(a+b*sin(d*x+c))^2,x, algorithm="giac")
Output:
-1/30*(6*b^2*sin(d*x + c)^5 + 15*a*b*sin(d*x + c)^4 + 10*a^2*sin(d*x + c)^ 3 - 10*b^2*sin(d*x + c)^3 - 30*a*b*sin(d*x + c)^2 - 30*a^2*sin(d*x + c))/d
Time = 0.03 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.96 \[ \int \cos ^3(c+d x) (a+b \sin (c+d x))^2 \, dx=-\frac {{\sin \left (c+d\,x\right )}^3\,\left (\frac {a^2}{3}-\frac {b^2}{3}\right )-a^2\,\sin \left (c+d\,x\right )+\frac {b^2\,{\sin \left (c+d\,x\right )}^5}{5}-a\,b\,{\sin \left (c+d\,x\right )}^2+\frac {a\,b\,{\sin \left (c+d\,x\right )}^4}{2}}{d} \] Input:
int(cos(c + d*x)^3*(a + b*sin(c + d*x))^2,x)
Output:
-(sin(c + d*x)^3*(a^2/3 - b^2/3) - a^2*sin(c + d*x) + (b^2*sin(c + d*x)^5) /5 - a*b*sin(c + d*x)^2 + (a*b*sin(c + d*x)^4)/2)/d
Time = 0.17 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.01 \[ \int \cos ^3(c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {\sin \left (d x +c \right ) \left (-6 \sin \left (d x +c \right )^{4} b^{2}-15 \sin \left (d x +c \right )^{3} a b -10 \sin \left (d x +c \right )^{2} a^{2}+10 \sin \left (d x +c \right )^{2} b^{2}+30 \sin \left (d x +c \right ) a b +30 a^{2}\right )}{30 d} \] Input:
int(cos(d*x+c)^3*(a+b*sin(d*x+c))^2,x)
Output:
(sin(c + d*x)*( - 6*sin(c + d*x)**4*b**2 - 15*sin(c + d*x)**3*a*b - 10*sin (c + d*x)**2*a**2 + 10*sin(c + d*x)**2*b**2 + 30*sin(c + d*x)*a*b + 30*a** 2))/(30*d)