Integrand size = 25, antiderivative size = 81 \[ \int \cos ^7(c+d x) \sin (c+d x) (a+a \sin (c+d x)) \, dx=-\frac {a \cos ^8(c+d x)}{8 d}+\frac {a \sin ^3(c+d x)}{3 d}-\frac {3 a \sin ^5(c+d x)}{5 d}+\frac {3 a \sin ^7(c+d x)}{7 d}-\frac {a \sin ^9(c+d x)}{9 d} \] Output:
-1/8*a*cos(d*x+c)^8/d+1/3*a*sin(d*x+c)^3/d-3/5*a*sin(d*x+c)^5/d+3/7*a*sin( d*x+c)^7/d-1/9*a*sin(d*x+c)^9/d
Time = 0.38 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.74 \[ \int \cos ^7(c+d x) \sin (c+d x) (a+a \sin (c+d x)) \, dx=\frac {a \left (-1260 \cos ^8(c+d x)+(1606+1389 \cos (2 (c+d x))+330 \cos (4 (c+d x))+35 \cos (6 (c+d x))) \sin ^3(c+d x)\right )}{10080 d} \] Input:
Integrate[Cos[c + d*x]^7*Sin[c + d*x]*(a + a*Sin[c + d*x]),x]
Output:
(a*(-1260*Cos[c + d*x]^8 + (1606 + 1389*Cos[2*(c + d*x)] + 330*Cos[4*(c + d*x)] + 35*Cos[6*(c + d*x)])*Sin[c + d*x]^3))/(10080*d)
Time = 0.38 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.88, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {3042, 3313, 3042, 3044, 244, 2009, 3045, 15}
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.
\(\displaystyle \int \sin (c+d x) \cos ^7(c+d x) (a \sin (c+d x)+a) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sin (c+d x) \cos (c+d x)^7 (a \sin (c+d x)+a)dx\) |
\(\Big \downarrow \) 3313 |
\(\displaystyle a \int \cos ^7(c+d x) \sin ^2(c+d x)dx+a \int \cos ^7(c+d x) \sin (c+d x)dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle a \int \cos (c+d x)^7 \sin (c+d x)dx+a \int \cos (c+d x)^7 \sin (c+d x)^2dx\) |
\(\Big \downarrow \) 3044 |
\(\displaystyle \frac {a \int \sin ^2(c+d x) \left (1-\sin ^2(c+d x)\right )^3d\sin (c+d x)}{d}+a \int \cos (c+d x)^7 \sin (c+d x)dx\) |
\(\Big \downarrow \) 244 |
\(\displaystyle \frac {a \int \left (-\sin ^8(c+d x)+3 \sin ^6(c+d x)-3 \sin ^4(c+d x)+\sin ^2(c+d x)\right )d\sin (c+d x)}{d}+a \int \cos (c+d x)^7 \sin (c+d x)dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle a \int \cos (c+d x)^7 \sin (c+d x)dx+\frac {a \left (-\frac {1}{9} \sin ^9(c+d x)+\frac {3}{7} \sin ^7(c+d x)-\frac {3}{5} \sin ^5(c+d x)+\frac {1}{3} \sin ^3(c+d x)\right )}{d}\) |
\(\Big \downarrow \) 3045 |
\(\displaystyle \frac {a \left (-\frac {1}{9} \sin ^9(c+d x)+\frac {3}{7} \sin ^7(c+d x)-\frac {3}{5} \sin ^5(c+d x)+\frac {1}{3} \sin ^3(c+d x)\right )}{d}-\frac {a \int \cos ^7(c+d x)d\cos (c+d x)}{d}\) |
\(\Big \downarrow \) 15 |
\(\displaystyle \frac {a \left (-\frac {1}{9} \sin ^9(c+d x)+\frac {3}{7} \sin ^7(c+d x)-\frac {3}{5} \sin ^5(c+d x)+\frac {1}{3} \sin ^3(c+d x)\right )}{d}-\frac {a \cos ^8(c+d x)}{8 d}\) |
Input:
Int[Cos[c + d*x]^7*Sin[c + d*x]*(a + a*Sin[c + d*x]),x]
Output:
-1/8*(a*Cos[c + d*x]^8)/d + (a*(Sin[c + d*x]^3/3 - (3*Sin[c + d*x]^5)/5 + (3*Sin[c + d*x]^7)/7 - Sin[c + d*x]^9/9))/d
Int[(a_.)*(x_)^(m_.), x_Symbol] :> Simp[a*(x^(m + 1)/(m + 1)), x] /; FreeQ[ {a, m}, x] && NeQ[m, -1]
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[Expand Integrand[(c*x)^m*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, m}, x] && IGtQ[p , 0]
Int[cos[(e_.) + (f_.)*(x_)]^(n_.)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_ Symbol] :> Simp[1/(a*f) Subst[Int[x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a *Sin[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2] && !(I ntegerQ[(m - 1)/2] && LtQ[0, m, n])
Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_.)*sin[(e_.) + (f_.)*(x_)]^(n_.), x_ Symbol] :> Simp[-(a*f)^(-1) Subst[Int[x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a*Cos[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2] && !(IntegerQ[(m - 1)/2] && GtQ[m, 0] && LeQ[m, n])
Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((a_ ) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[a Int[Cos[e + f*x]^ p*(d*Sin[e + f*x])^n, x], x] + Simp[b/d Int[Cos[e + f*x]^p*(d*Sin[e + f*x ])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n, p}, x] && IntegerQ[(p - 1)/2 ] && IntegerQ[n] && ((LtQ[p, 0] && NeQ[a^2 - b^2, 0]) || LtQ[0, n, p - 1] | | LtQ[p + 1, -n, 2*p + 1])
Time = 40.96 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.09
method | result | size |
derivativedivides | \(-\frac {a \left (\frac {\sin \left (d x +c \right )^{9}}{9}+\frac {\sin \left (d x +c \right )^{8}}{8}-\frac {3 \sin \left (d x +c \right )^{7}}{7}-\frac {\sin \left (d x +c \right )^{6}}{2}+\frac {3 \sin \left (d x +c \right )^{5}}{5}+\frac {3 \sin \left (d x +c \right )^{4}}{4}-\frac {\sin \left (d x +c \right )^{3}}{3}-\frac {\sin \left (d x +c \right )^{2}}{2}\right )}{d}\) | \(88\) |
default | \(-\frac {a \left (\frac {\sin \left (d x +c \right )^{9}}{9}+\frac {\sin \left (d x +c \right )^{8}}{8}-\frac {3 \sin \left (d x +c \right )^{7}}{7}-\frac {\sin \left (d x +c \right )^{6}}{2}+\frac {3 \sin \left (d x +c \right )^{5}}{5}+\frac {3 \sin \left (d x +c \right )^{4}}{4}-\frac {\sin \left (d x +c \right )^{3}}{3}-\frac {\sin \left (d x +c \right )^{2}}{2}\right )}{d}\) | \(88\) |
parallelrisch | \(-\frac {a \left (2016 \sin \left (5 d x +5 c \right )+315 \cos \left (8 d x +8 c \right )+2520 \cos \left (6 d x +6 c \right )+900 \sin \left (7 d x +7 c \right )+17640 \cos \left (2 d x +2 c \right )-17640 \sin \left (d x +c \right )-29295+8820 \cos \left (4 d x +4 c \right )+140 \sin \left (9 d x +9 c \right )\right )}{322560 d}\) | \(94\) |
risch | \(\frac {7 a \sin \left (d x +c \right )}{128 d}-\frac {a \sin \left (9 d x +9 c \right )}{2304 d}-\frac {a \cos \left (8 d x +8 c \right )}{1024 d}-\frac {5 a \sin \left (7 d x +7 c \right )}{1792 d}-\frac {a \cos \left (6 d x +6 c \right )}{128 d}-\frac {a \sin \left (5 d x +5 c \right )}{160 d}-\frac {7 a \cos \left (4 d x +4 c \right )}{256 d}-\frac {7 a \cos \left (2 d x +2 c \right )}{128 d}\) | \(119\) |
norman | \(\frac {\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{d}+\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{16}}{d}+\frac {8 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 d}-\frac {16 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{5 d}+\frac {632 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{35 d}-\frac {2848 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{315 d}+\frac {632 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{35 d}-\frac {16 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{13}}{5 d}+\frac {8 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{15}}{3 d}+\frac {14 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{d}+\frac {14 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}}{d}+\frac {14 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{d}+\frac {14 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}}{d}+\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{d}+\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{14}}{d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{9}}\) | \(273\) |
orering | \(\text {Expression too large to display}\) | \(1604\) |
Input:
int(cos(d*x+c)^7*sin(d*x+c)*(a+a*sin(d*x+c)),x,method=_RETURNVERBOSE)
Output:
-a/d*(1/9*sin(d*x+c)^9+1/8*sin(d*x+c)^8-3/7*sin(d*x+c)^7-1/2*sin(d*x+c)^6+ 3/5*sin(d*x+c)^5+3/4*sin(d*x+c)^4-1/3*sin(d*x+c)^3-1/2*sin(d*x+c)^2)
Time = 0.09 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.90 \[ \int \cos ^7(c+d x) \sin (c+d x) (a+a \sin (c+d x)) \, dx=-\frac {315 \, a \cos \left (d x + c\right )^{8} + 8 \, {\left (35 \, a \cos \left (d x + c\right )^{8} - 5 \, a \cos \left (d x + c\right )^{6} - 6 \, a \cos \left (d x + c\right )^{4} - 8 \, a \cos \left (d x + c\right )^{2} - 16 \, a\right )} \sin \left (d x + c\right )}{2520 \, d} \] Input:
integrate(cos(d*x+c)^7*sin(d*x+c)*(a+a*sin(d*x+c)),x, algorithm="fricas")
Output:
-1/2520*(315*a*cos(d*x + c)^8 + 8*(35*a*cos(d*x + c)^8 - 5*a*cos(d*x + c)^ 6 - 6*a*cos(d*x + c)^4 - 8*a*cos(d*x + c)^2 - 16*a)*sin(d*x + c))/d
Time = 0.93 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.41 \[ \int \cos ^7(c+d x) \sin (c+d x) (a+a \sin (c+d x)) \, dx=\begin {cases} \frac {16 a \sin ^{9}{\left (c + d x \right )}}{315 d} + \frac {8 a \sin ^{7}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{35 d} + \frac {2 a \sin ^{5}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{5 d} + \frac {a \sin ^{3}{\left (c + d x \right )} \cos ^{6}{\left (c + d x \right )}}{3 d} - \frac {a \cos ^{8}{\left (c + d x \right )}}{8 d} & \text {for}\: d \neq 0 \\x \left (a \sin {\left (c \right )} + a\right ) \sin {\left (c \right )} \cos ^{7}{\left (c \right )} & \text {otherwise} \end {cases} \] Input:
integrate(cos(d*x+c)**7*sin(d*x+c)*(a+a*sin(d*x+c)),x)
Output:
Piecewise((16*a*sin(c + d*x)**9/(315*d) + 8*a*sin(c + d*x)**7*cos(c + d*x) **2/(35*d) + 2*a*sin(c + d*x)**5*cos(c + d*x)**4/(5*d) + a*sin(c + d*x)**3 *cos(c + d*x)**6/(3*d) - a*cos(c + d*x)**8/(8*d), Ne(d, 0)), (x*(a*sin(c) + a)*sin(c)*cos(c)**7, True))
Time = 0.04 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.16 \[ \int \cos ^7(c+d x) \sin (c+d x) (a+a \sin (c+d x)) \, dx=-\frac {280 \, a \sin \left (d x + c\right )^{9} + 315 \, a \sin \left (d x + c\right )^{8} - 1080 \, a \sin \left (d x + c\right )^{7} - 1260 \, a \sin \left (d x + c\right )^{6} + 1512 \, a \sin \left (d x + c\right )^{5} + 1890 \, a \sin \left (d x + c\right )^{4} - 840 \, a \sin \left (d x + c\right )^{3} - 1260 \, a \sin \left (d x + c\right )^{2}}{2520 \, d} \] Input:
integrate(cos(d*x+c)^7*sin(d*x+c)*(a+a*sin(d*x+c)),x, algorithm="maxima")
Output:
-1/2520*(280*a*sin(d*x + c)^9 + 315*a*sin(d*x + c)^8 - 1080*a*sin(d*x + c) ^7 - 1260*a*sin(d*x + c)^6 + 1512*a*sin(d*x + c)^5 + 1890*a*sin(d*x + c)^4 - 840*a*sin(d*x + c)^3 - 1260*a*sin(d*x + c)^2)/d
Time = 0.14 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.16 \[ \int \cos ^7(c+d x) \sin (c+d x) (a+a \sin (c+d x)) \, dx=-\frac {280 \, a \sin \left (d x + c\right )^{9} + 315 \, a \sin \left (d x + c\right )^{8} - 1080 \, a \sin \left (d x + c\right )^{7} - 1260 \, a \sin \left (d x + c\right )^{6} + 1512 \, a \sin \left (d x + c\right )^{5} + 1890 \, a \sin \left (d x + c\right )^{4} - 840 \, a \sin \left (d x + c\right )^{3} - 1260 \, a \sin \left (d x + c\right )^{2}}{2520 \, d} \] Input:
integrate(cos(d*x+c)^7*sin(d*x+c)*(a+a*sin(d*x+c)),x, algorithm="giac")
Output:
-1/2520*(280*a*sin(d*x + c)^9 + 315*a*sin(d*x + c)^8 - 1080*a*sin(d*x + c) ^7 - 1260*a*sin(d*x + c)^6 + 1512*a*sin(d*x + c)^5 + 1890*a*sin(d*x + c)^4 - 840*a*sin(d*x + c)^3 - 1260*a*sin(d*x + c)^2)/d
Time = 34.08 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.15 \[ \int \cos ^7(c+d x) \sin (c+d x) (a+a \sin (c+d x)) \, dx=\frac {-\frac {a\,{\sin \left (c+d\,x\right )}^9}{9}-\frac {a\,{\sin \left (c+d\,x\right )}^8}{8}+\frac {3\,a\,{\sin \left (c+d\,x\right )}^7}{7}+\frac {a\,{\sin \left (c+d\,x\right )}^6}{2}-\frac {3\,a\,{\sin \left (c+d\,x\right )}^5}{5}-\frac {3\,a\,{\sin \left (c+d\,x\right )}^4}{4}+\frac {a\,{\sin \left (c+d\,x\right )}^3}{3}+\frac {a\,{\sin \left (c+d\,x\right )}^2}{2}}{d} \] Input:
int(cos(c + d*x)^7*sin(c + d*x)*(a + a*sin(c + d*x)),x)
Output:
((a*sin(c + d*x)^2)/2 + (a*sin(c + d*x)^3)/3 - (3*a*sin(c + d*x)^4)/4 - (3 *a*sin(c + d*x)^5)/5 + (a*sin(c + d*x)^6)/2 + (3*a*sin(c + d*x)^7)/7 - (a* sin(c + d*x)^8)/8 - (a*sin(c + d*x)^9)/9)/d
Time = 0.16 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.04 \[ \int \cos ^7(c+d x) \sin (c+d x) (a+a \sin (c+d x)) \, dx=\frac {\sin \left (d x +c \right )^{2} a \left (-280 \sin \left (d x +c \right )^{7}-315 \sin \left (d x +c \right )^{6}+1080 \sin \left (d x +c \right )^{5}+1260 \sin \left (d x +c \right )^{4}-1512 \sin \left (d x +c \right )^{3}-1890 \sin \left (d x +c \right )^{2}+840 \sin \left (d x +c \right )+1260\right )}{2520 d} \] Input:
int(cos(d*x+c)^7*sin(d*x+c)*(a+a*sin(d*x+c)),x)
Output:
(sin(c + d*x)**2*a*( - 280*sin(c + d*x)**7 - 315*sin(c + d*x)**6 + 1080*si n(c + d*x)**5 + 1260*sin(c + d*x)**4 - 1512*sin(c + d*x)**3 - 1890*sin(c + d*x)**2 + 840*sin(c + d*x) + 1260))/(2520*d)