Integrand size = 27, antiderivative size = 55 \[ \int \frac {\cos ^5(c+d x) \sin (c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {\cos ^4(c+d x)}{4 a d}-\frac {\sin ^3(c+d x)}{3 a d}+\frac {\sin ^5(c+d x)}{5 a d} \] Output:
-1/4*cos(d*x+c)^4/a/d-1/3*sin(d*x+c)^3/a/d+1/5*sin(d*x+c)^5/a/d
Time = 0.09 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.87 \[ \int \frac {\cos ^5(c+d x) \sin (c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\sin ^2(c+d x) \left (30-20 \sin (c+d x)-15 \sin ^2(c+d x)+12 \sin ^3(c+d x)\right )}{60 a d} \] Input:
Integrate[(Cos[c + d*x]^5*Sin[c + d*x])/(a + a*Sin[c + d*x]),x]
Output:
(Sin[c + d*x]^2*(30 - 20*Sin[c + d*x] - 15*Sin[c + d*x]^2 + 12*Sin[c + d*x ]^3))/(60*a*d)
Time = 0.40 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.95, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.296, Rules used = {3042, 3314, 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.
\(\displaystyle \int \frac {\sin (c+d x) \cos ^5(c+d x)}{a \sin (c+d x)+a} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sin (c+d x) \cos (c+d x)^5}{a \sin (c+d x)+a}dx\) |
\(\Big \downarrow \) 3314 |
\(\displaystyle \frac {\int \cos ^3(c+d x) \sin (c+d x)dx}{a}-\frac {\int \cos ^3(c+d x) \sin ^2(c+d x)dx}{a}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \cos (c+d x)^3 \sin (c+d x)dx}{a}-\frac {\int \cos (c+d x)^3 \sin (c+d x)^2dx}{a}\) |
\(\Big \downarrow \) 3044 |
\(\displaystyle \frac {\int \cos (c+d x)^3 \sin (c+d x)dx}{a}-\frac {\int \sin ^2(c+d x) \left (1-\sin ^2(c+d x)\right )d\sin (c+d x)}{a d}\) |
\(\Big \downarrow \) 244 |
\(\displaystyle \frac {\int \cos (c+d x)^3 \sin (c+d x)dx}{a}-\frac {\int \left (\sin ^2(c+d x)-\sin ^4(c+d x)\right )d\sin (c+d x)}{a d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\int \cos (c+d x)^3 \sin (c+d x)dx}{a}-\frac {\frac {1}{3} \sin ^3(c+d x)-\frac {1}{5} \sin ^5(c+d x)}{a d}\) |
\(\Big \downarrow \) 3045 |
\(\displaystyle -\frac {\int \cos ^3(c+d x)d\cos (c+d x)}{a d}-\frac {\frac {1}{3} \sin ^3(c+d x)-\frac {1}{5} \sin ^5(c+d x)}{a d}\) |
\(\Big \downarrow \) 15 |
\(\displaystyle -\frac {\frac {1}{3} \sin ^3(c+d x)-\frac {1}{5} \sin ^5(c+d x)}{a d}-\frac {\cos ^4(c+d x)}{4 a d}\) |
Input:
Int[(Cos[c + d*x]^5*Sin[c + d*x])/(a + a*Sin[c + d*x]),x]
Output:
-1/4*Cos[c + d*x]^4/(a*d) - (Sin[c + d*x]^3/3 - Sin[c + d*x]^5/5)/(a*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[1/a Int[Cos[e + f *x]^(p - 2)*(d*Sin[e + f*x])^n, x], x] - Simp[1/(b*d) Int[Cos[e + f*x]^(p - 2)*(d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n, p}, x] & & IntegerQ[(p - 1)/2] && EqQ[a^2 - b^2, 0] && IntegerQ[n] && (LtQ[0, n, (p + 1)/2] || (LeQ[p, -n] && LtQ[-n, 2*p - 3]) || (GtQ[n, 0] && LeQ[n, -p]))
Time = 0.53 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.89
method | result | size |
derivativedivides | \(\frac {\frac {\sin \left (d x +c \right )^{5}}{5}-\frac {\sin \left (d x +c \right )^{4}}{4}-\frac {\sin \left (d x +c \right )^{3}}{3}+\frac {\sin \left (d x +c \right )^{2}}{2}}{d a}\) | \(49\) |
default | \(\frac {\frac {\sin \left (d x +c \right )^{5}}{5}-\frac {\sin \left (d x +c \right )^{4}}{4}-\frac {\sin \left (d x +c \right )^{3}}{3}+\frac {\sin \left (d x +c \right )^{2}}{2}}{d a}\) | \(49\) |
parallelrisch | \(\frac {6 \sin \left (5 d x +5 c \right )-60 \sin \left (d x +c \right )+10 \sin \left (3 d x +3 c \right )-15 \cos \left (4 d x +4 c \right )+75-60 \cos \left (2 d x +2 c \right )}{480 d a}\) | \(63\) |
risch | \(-\frac {\sin \left (d x +c \right )}{8 a d}+\frac {\sin \left (5 d x +5 c \right )}{80 d a}-\frac {\cos \left (4 d x +4 c \right )}{32 a d}+\frac {\sin \left (3 d x +3 c \right )}{48 d a}-\frac {\cos \left (2 d x +2 c \right )}{8 a d}\) | \(84\) |
norman | \(\frac {\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{d a}+\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{d a}-\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 d a}-\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}}{3 d a}+\frac {12 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{5 d a}+\frac {12 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{5 d a}+\frac {12 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{5 d a}+\frac {12 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{5 d a}+\frac {4 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{3 d a}+\frac {4 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{3 d a}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{6} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}\) | \(221\) |
Input:
int(cos(d*x+c)^5*sin(d*x+c)/(a+a*sin(d*x+c)),x,method=_RETURNVERBOSE)
Output:
1/d/a*(1/5*sin(d*x+c)^5-1/4*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 = 49, normalized size of antiderivative = 0.89 \[ \int \frac {\cos ^5(c+d x) \sin (c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {15 \, \cos \left (d x + c\right )^{4} - 4 \, {\left (3 \, \cos \left (d x + c\right )^{4} - \cos \left (d x + c\right )^{2} - 2\right )} \sin \left (d x + c\right )}{60 \, a d} \] Input:
integrate(cos(d*x+c)^5*sin(d*x+c)/(a+a*sin(d*x+c)),x, algorithm="fricas")
Output:
-1/60*(15*cos(d*x + c)^4 - 4*(3*cos(d*x + c)^4 - cos(d*x + c)^2 - 2)*sin(d *x + c))/(a*d)
Leaf count of result is larger than twice the leaf count of optimal. 741 vs. \(2 (39) = 78\).
Time = 12.78 (sec) , antiderivative size = 741, normalized size of antiderivative = 13.47 \[ \int \frac {\cos ^5(c+d x) \sin (c+d x)}{a+a \sin (c+d x)} \, dx =\text {Too large to display} \] Input:
integrate(cos(d*x+c)**5*sin(d*x+c)/(a+a*sin(d*x+c)),x)
Output:
Piecewise((30*tan(c/2 + d*x/2)**8/(15*a*d*tan(c/2 + d*x/2)**10 + 75*a*d*ta n(c/2 + d*x/2)**8 + 150*a*d*tan(c/2 + d*x/2)**6 + 150*a*d*tan(c/2 + d*x/2) **4 + 75*a*d*tan(c/2 + d*x/2)**2 + 15*a*d) - 40*tan(c/2 + d*x/2)**7/(15*a* d*tan(c/2 + d*x/2)**10 + 75*a*d*tan(c/2 + d*x/2)**8 + 150*a*d*tan(c/2 + d* x/2)**6 + 150*a*d*tan(c/2 + d*x/2)**4 + 75*a*d*tan(c/2 + d*x/2)**2 + 15*a* d) + 30*tan(c/2 + d*x/2)**6/(15*a*d*tan(c/2 + d*x/2)**10 + 75*a*d*tan(c/2 + d*x/2)**8 + 150*a*d*tan(c/2 + d*x/2)**6 + 150*a*d*tan(c/2 + d*x/2)**4 + 75*a*d*tan(c/2 + d*x/2)**2 + 15*a*d) + 16*tan(c/2 + d*x/2)**5/(15*a*d*tan( c/2 + d*x/2)**10 + 75*a*d*tan(c/2 + d*x/2)**8 + 150*a*d*tan(c/2 + d*x/2)** 6 + 150*a*d*tan(c/2 + d*x/2)**4 + 75*a*d*tan(c/2 + d*x/2)**2 + 15*a*d) + 3 0*tan(c/2 + d*x/2)**4/(15*a*d*tan(c/2 + d*x/2)**10 + 75*a*d*tan(c/2 + d*x/ 2)**8 + 150*a*d*tan(c/2 + d*x/2)**6 + 150*a*d*tan(c/2 + d*x/2)**4 + 75*a*d *tan(c/2 + d*x/2)**2 + 15*a*d) - 40*tan(c/2 + d*x/2)**3/(15*a*d*tan(c/2 + d*x/2)**10 + 75*a*d*tan(c/2 + d*x/2)**8 + 150*a*d*tan(c/2 + d*x/2)**6 + 15 0*a*d*tan(c/2 + d*x/2)**4 + 75*a*d*tan(c/2 + d*x/2)**2 + 15*a*d) + 30*tan( c/2 + d*x/2)**2/(15*a*d*tan(c/2 + d*x/2)**10 + 75*a*d*tan(c/2 + d*x/2)**8 + 150*a*d*tan(c/2 + d*x/2)**6 + 150*a*d*tan(c/2 + d*x/2)**4 + 75*a*d*tan(c /2 + d*x/2)**2 + 15*a*d), Ne(d, 0)), (x*sin(c)*cos(c)**5/(a*sin(c) + a), T rue))
Time = 0.03 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.89 \[ \int \frac {\cos ^5(c+d x) \sin (c+d x)}{a+a \sin (c+d x)} \, dx=\frac {12 \, \sin \left (d x + c\right )^{5} - 15 \, \sin \left (d x + c\right )^{4} - 20 \, \sin \left (d x + c\right )^{3} + 30 \, \sin \left (d x + c\right )^{2}}{60 \, a d} \] Input:
integrate(cos(d*x+c)^5*sin(d*x+c)/(a+a*sin(d*x+c)),x, algorithm="maxima")
Output:
1/60*(12*sin(d*x + c)^5 - 15*sin(d*x + c)^4 - 20*sin(d*x + c)^3 + 30*sin(d *x + c)^2)/(a*d)
Time = 0.14 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.89 \[ \int \frac {\cos ^5(c+d x) \sin (c+d x)}{a+a \sin (c+d x)} \, dx=\frac {12 \, \sin \left (d x + c\right )^{5} - 15 \, \sin \left (d x + c\right )^{4} - 20 \, \sin \left (d x + c\right )^{3} + 30 \, \sin \left (d x + c\right )^{2}}{60 \, a d} \] Input:
integrate(cos(d*x+c)^5*sin(d*x+c)/(a+a*sin(d*x+c)),x, algorithm="giac")
Output:
1/60*(12*sin(d*x + c)^5 - 15*sin(d*x + c)^4 - 20*sin(d*x + c)^3 + 30*sin(d *x + c)^2)/(a*d)
Time = 0.07 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.04 \[ \int \frac {\cos ^5(c+d x) \sin (c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\frac {{\sin \left (c+d\,x\right )}^2}{2\,a}-\frac {{\sin \left (c+d\,x\right )}^3}{3\,a}-\frac {{\sin \left (c+d\,x\right )}^4}{4\,a}+\frac {{\sin \left (c+d\,x\right )}^5}{5\,a}}{d} \] Input:
int((cos(c + d*x)^5*sin(c + d*x))/(a + a*sin(c + d*x)),x)
Output:
(sin(c + d*x)^2/(2*a) - sin(c + d*x)^3/(3*a) - sin(c + d*x)^4/(4*a) + sin( c + d*x)^5/(5*a))/d
Time = 0.16 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.84 \[ \int \frac {\cos ^5(c+d x) \sin (c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\sin \left (d x +c \right )^{2} \left (12 \sin \left (d x +c \right )^{3}-15 \sin \left (d x +c \right )^{2}-20 \sin \left (d x +c \right )+30\right )}{60 a d} \] Input:
int(cos(d*x+c)^5*sin(d*x+c)/(a+a*sin(d*x+c)),x)
Output:
(sin(c + d*x)**2*(12*sin(c + d*x)**3 - 15*sin(c + d*x)**2 - 20*sin(c + d*x ) + 30))/(60*a*d)