Integrand size = 29, antiderivative size = 73 \[ \int \frac {\cos ^5(c+d x) \sin ^2(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\sin ^3(c+d x)}{3 a d}-\frac {\sin ^4(c+d x)}{4 a d}-\frac {\sin ^5(c+d x)}{5 a d}+\frac {\sin ^6(c+d x)}{6 a d} \] Output:
1/3*sin(d*x+c)^3/a/d-1/4*sin(d*x+c)^4/a/d-1/5*sin(d*x+c)^5/a/d+1/6*sin(d*x +c)^6/a/d
Time = 0.16 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.66 \[ \int \frac {\cos ^5(c+d x) \sin ^2(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\sin ^3(c+d x) \left (20-15 \sin (c+d x)-12 \sin ^2(c+d x)+10 \sin ^3(c+d x)\right )}{60 a d} \] Input:
Integrate[(Cos[c + d*x]^5*Sin[c + d*x]^2)/(a + a*Sin[c + d*x]),x]
Output:
(Sin[c + d*x]^3*(20 - 15*Sin[c + d*x] - 12*Sin[c + d*x]^2 + 10*Sin[c + d*x ]^3))/(60*a*d)
Time = 0.43 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.90, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.207, Rules used = {3042, 3314, 3042, 3044, 244, 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.
\(\displaystyle \int \frac {\sin ^2(c+d x) \cos ^5(c+d x)}{a \sin (c+d x)+a} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sin (c+d x)^2 \cos (c+d x)^5}{a \sin (c+d x)+a}dx\) |
\(\Big \downarrow \) 3314 |
\(\displaystyle \frac {\int \cos ^3(c+d x) \sin ^2(c+d x)dx}{a}-\frac {\int \cos ^3(c+d x) \sin ^3(c+d x)dx}{a}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \cos (c+d x)^3 \sin (c+d x)^2dx}{a}-\frac {\int \cos (c+d x)^3 \sin (c+d x)^3dx}{a}\) |
\(\Big \downarrow \) 3044 |
\(\displaystyle \frac {\int \sin ^2(c+d x) \left (1-\sin ^2(c+d x)\right )d\sin (c+d x)}{a d}-\frac {\int \sin ^3(c+d x) \left (1-\sin ^2(c+d x)\right )d\sin (c+d x)}{a d}\) |
\(\Big \downarrow \) 244 |
\(\displaystyle \frac {\int \left (\sin ^2(c+d x)-\sin ^4(c+d x)\right )d\sin (c+d x)}{a d}-\frac {\int \left (\sin ^3(c+d x)-\sin ^5(c+d x)\right )d\sin (c+d x)}{a d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {1}{3} \sin ^3(c+d x)-\frac {1}{5} \sin ^5(c+d x)}{a d}-\frac {\frac {1}{4} \sin ^4(c+d x)-\frac {1}{6} \sin ^6(c+d x)}{a d}\) |
Input:
Int[(Cos[c + d*x]^5*Sin[c + d*x]^2)/(a + a*Sin[c + d*x]),x]
Output:
(Sin[c + d*x]^3/3 - Sin[c + d*x]^5/5)/(a*d) - (Sin[c + d*x]^4/4 - Sin[c + d*x]^6/6)/(a*d)
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_)]^(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 = 0.58 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.67
method | result | size |
derivativedivides | \(\frac {\frac {\sin \left (d x +c \right )^{6}}{6}-\frac {\sin \left (d x +c \right )^{5}}{5}-\frac {\sin \left (d x +c \right )^{4}}{4}+\frac {\sin \left (d x +c \right )^{3}}{3}}{d a}\) | \(49\) |
default | \(\frac {\frac {\sin \left (d x +c \right )^{6}}{6}-\frac {\sin \left (d x +c \right )^{5}}{5}-\frac {\sin \left (d x +c \right )^{4}}{4}+\frac {\sin \left (d x +c \right )^{3}}{3}}{d a}\) | \(49\) |
risch | \(\frac {\sin \left (d x +c \right )}{8 a d}-\frac {\cos \left (6 d x +6 c \right )}{192 a d}-\frac {\sin \left (5 d x +5 c \right )}{80 d a}-\frac {\sin \left (3 d x +3 c \right )}{48 d a}+\frac {3 \cos \left (2 d x +2 c \right )}{64 a d}\) | \(84\) |
parallelrisch | \(\frac {\left (-\sin \left (\frac {3 d x}{2}+\frac {3 c}{2}\right )+3 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \left (28+12 \cos \left (2 d x +2 c \right )-5 \sin \left (3 d x +3 c \right )-15 \sin \left (d x +c \right )\right ) \left (\cos \left (\frac {3 d x}{2}+\frac {3 c}{2}\right )+3 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{240 a d}\) | \(85\) |
norman | \(\frac {\frac {8 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 d a}+\frac {8 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}}{3 d a}-\frac {4 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{3 d a}-\frac {4 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{3 d a}+\frac {28 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{15 d a}+\frac {28 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{15 d a}+\frac {4 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{15 d a}+\frac {4 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}}{15 d a}+\frac {44 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{15 d a}+\frac {44 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{15 d a}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{7} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}\) | \(221\) |
Input:
int(cos(d*x+c)^5*sin(d*x+c)^2/(a+a*sin(d*x+c)),x,method=_RETURNVERBOSE)
Output:
1/d/a*(1/6*sin(d*x+c)^6-1/5*sin(d*x+c)^5-1/4*sin(d*x+c)^4+1/3*sin(d*x+c)^3 )
Time = 0.08 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.81 \[ \int \frac {\cos ^5(c+d x) \sin ^2(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {10 \, \cos \left (d x + c\right )^{6} - 15 \, \cos \left (d x + c\right )^{4} + 4 \, {\left (3 \, \cos \left (d x + c\right )^{4} - \cos \left (d x + c\right )^{2} - 2\right )} \sin \left (d x + c\right )}{60 \, a d} \] Input:
integrate(cos(d*x+c)^5*sin(d*x+c)^2/(a+a*sin(d*x+c)),x, algorithm="fricas" )
Output:
-1/60*(10*cos(d*x + c)^6 - 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. 862 vs. \(2 (53) = 106\).
Time = 21.29 (sec) , antiderivative size = 862, normalized size of antiderivative = 11.81 \[ \int \frac {\cos ^5(c+d x) \sin ^2(c+d x)}{a+a \sin (c+d x)} \, dx=\text {Too large to display} \] Input:
integrate(cos(d*x+c)**5*sin(d*x+c)**2/(a+a*sin(d*x+c)),x)
Output:
Piecewise((40*tan(c/2 + d*x/2)**9/(15*a*d*tan(c/2 + d*x/2)**12 + 90*a*d*ta n(c/2 + d*x/2)**10 + 225*a*d*tan(c/2 + d*x/2)**8 + 300*a*d*tan(c/2 + d*x/2 )**6 + 225*a*d*tan(c/2 + d*x/2)**4 + 90*a*d*tan(c/2 + d*x/2)**2 + 15*a*d) - 60*tan(c/2 + d*x/2)**8/(15*a*d*tan(c/2 + d*x/2)**12 + 90*a*d*tan(c/2 + d *x/2)**10 + 225*a*d*tan(c/2 + d*x/2)**8 + 300*a*d*tan(c/2 + d*x/2)**6 + 22 5*a*d*tan(c/2 + d*x/2)**4 + 90*a*d*tan(c/2 + d*x/2)**2 + 15*a*d) + 24*tan( c/2 + d*x/2)**7/(15*a*d*tan(c/2 + d*x/2)**12 + 90*a*d*tan(c/2 + d*x/2)**10 + 225*a*d*tan(c/2 + d*x/2)**8 + 300*a*d*tan(c/2 + d*x/2)**6 + 225*a*d*tan (c/2 + d*x/2)**4 + 90*a*d*tan(c/2 + d*x/2)**2 + 15*a*d) + 40*tan(c/2 + d*x /2)**6/(15*a*d*tan(c/2 + d*x/2)**12 + 90*a*d*tan(c/2 + d*x/2)**10 + 225*a* d*tan(c/2 + d*x/2)**8 + 300*a*d*tan(c/2 + d*x/2)**6 + 225*a*d*tan(c/2 + d* x/2)**4 + 90*a*d*tan(c/2 + d*x/2)**2 + 15*a*d) + 24*tan(c/2 + d*x/2)**5/(1 5*a*d*tan(c/2 + d*x/2)**12 + 90*a*d*tan(c/2 + d*x/2)**10 + 225*a*d*tan(c/2 + d*x/2)**8 + 300*a*d*tan(c/2 + d*x/2)**6 + 225*a*d*tan(c/2 + d*x/2)**4 + 90*a*d*tan(c/2 + d*x/2)**2 + 15*a*d) - 60*tan(c/2 + d*x/2)**4/(15*a*d*tan (c/2 + d*x/2)**12 + 90*a*d*tan(c/2 + d*x/2)**10 + 225*a*d*tan(c/2 + d*x/2) **8 + 300*a*d*tan(c/2 + d*x/2)**6 + 225*a*d*tan(c/2 + d*x/2)**4 + 90*a*d*t an(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)**12 + 90*a*d*tan(c/2 + d*x/2)**10 + 225*a*d*tan(c/2 + d*x/2)**8 + 300 *a*d*tan(c/2 + d*x/2)**6 + 225*a*d*tan(c/2 + d*x/2)**4 + 90*a*d*tan(c/2...
Time = 0.03 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.67 \[ \int \frac {\cos ^5(c+d x) \sin ^2(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {10 \, \sin \left (d x + c\right )^{6} - 12 \, \sin \left (d x + c\right )^{5} - 15 \, \sin \left (d x + c\right )^{4} + 20 \, \sin \left (d x + c\right )^{3}}{60 \, a d} \] Input:
integrate(cos(d*x+c)^5*sin(d*x+c)^2/(a+a*sin(d*x+c)),x, algorithm="maxima" )
Output:
1/60*(10*sin(d*x + c)^6 - 12*sin(d*x + c)^5 - 15*sin(d*x + c)^4 + 20*sin(d *x + c)^3)/(a*d)
Time = 0.17 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.67 \[ \int \frac {\cos ^5(c+d x) \sin ^2(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {10 \, \sin \left (d x + c\right )^{6} - 12 \, \sin \left (d x + c\right )^{5} - 15 \, \sin \left (d x + c\right )^{4} + 20 \, \sin \left (d x + c\right )^{3}}{60 \, a d} \] Input:
integrate(cos(d*x+c)^5*sin(d*x+c)^2/(a+a*sin(d*x+c)),x, algorithm="giac")
Output:
1/60*(10*sin(d*x + c)^6 - 12*sin(d*x + c)^5 - 15*sin(d*x + c)^4 + 20*sin(d *x + c)^3)/(a*d)
Time = 0.07 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.78 \[ \int \frac {\cos ^5(c+d x) \sin ^2(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\frac {{\sin \left (c+d\,x\right )}^3}{3\,a}-\frac {{\sin \left (c+d\,x\right )}^4}{4\,a}-\frac {{\sin \left (c+d\,x\right )}^5}{5\,a}+\frac {{\sin \left (c+d\,x\right )}^6}{6\,a}}{d} \] Input:
int((cos(c + d*x)^5*sin(c + d*x)^2)/(a + a*sin(c + d*x)),x)
Output:
(sin(c + d*x)^3/(3*a) - sin(c + d*x)^4/(4*a) - sin(c + d*x)^5/(5*a) + sin( c + d*x)^6/(6*a))/d
Time = 0.16 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.63 \[ \int \frac {\cos ^5(c+d x) \sin ^2(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\sin \left (d x +c \right )^{3} \left (10 \sin \left (d x +c \right )^{3}-12 \sin \left (d x +c \right )^{2}-15 \sin \left (d x +c \right )+20\right )}{60 a d} \] Input:
int(cos(d*x+c)^5*sin(d*x+c)^2/(a+a*sin(d*x+c)),x)
Output:
(sin(c + d*x)**3*(10*sin(c + d*x)**3 - 12*sin(c + d*x)**2 - 15*sin(c + d*x ) + 20))/(60*a*d)