Integrand size = 31, antiderivative size = 79 \[ \int \frac {\cos ^5(c+d x) (A+B \sin (c+d x))}{a+a \sin (c+d x)} \, dx=-\frac {2 (A+B) (a-a \sin (c+d x))^3}{3 a^4 d}+\frac {(A+3 B) (a-a \sin (c+d x))^4}{4 a^5 d}-\frac {B (a-a \sin (c+d x))^5}{5 a^6 d} \] Output:
-2/3*(A+B)*(a-a*sin(d*x+c))^3/a^4/d+1/4*(A+3*B)*(a-a*sin(d*x+c))^4/a^5/d-1 /5*B*(a-a*sin(d*x+c))^5/a^6/d
Time = 0.18 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.91 \[ \int \frac {\cos ^5(c+d x) (A+B \sin (c+d x))}{a+a \sin (c+d x)} \, dx=\frac {\sin (c+d x) \left (60 A-30 (A-B) \sin (c+d x)-20 (A+B) \sin ^2(c+d x)+15 (A-B) \sin ^3(c+d x)+12 B \sin ^4(c+d x)\right )}{60 a d} \] Input:
Integrate[(Cos[c + d*x]^5*(A + B*Sin[c + d*x]))/(a + a*Sin[c + d*x]),x]
Output:
(Sin[c + d*x]*(60*A - 30*(A - B)*Sin[c + d*x] - 20*(A + B)*Sin[c + d*x]^2 + 15*(A - B)*Sin[c + d*x]^3 + 12*B*Sin[c + d*x]^4))/(60*a*d)
Time = 0.33 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.91, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.161, Rules used = {3042, 3315, 27, 86, 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.
\(\displaystyle \int \frac {\cos ^5(c+d x) (A+B \sin (c+d x))}{a \sin (c+d x)+a} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\cos (c+d x)^5 (A+B \sin (c+d x))}{a \sin (c+d x)+a}dx\) |
\(\Big \downarrow \) 3315 |
\(\displaystyle \frac {\int \frac {(a-a \sin (c+d x))^2 (\sin (c+d x) a+a) (a A+a B \sin (c+d x))}{a}d(a \sin (c+d x))}{a^5 d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int (a-a \sin (c+d x))^2 (\sin (c+d x) a+a) (a A+a B \sin (c+d x))d(a \sin (c+d x))}{a^6 d}\) |
\(\Big \downarrow \) 86 |
\(\displaystyle \frac {\int \left (B (a-a \sin (c+d x))^4-a (A+3 B) (a-a \sin (c+d x))^3+2 a^2 (A+B) (a-a \sin (c+d x))^2\right )d(a \sin (c+d x))}{a^6 d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {-\frac {2}{3} a^2 (A+B) (a-a \sin (c+d x))^3+\frac {1}{4} a (A+3 B) (a-a \sin (c+d x))^4-\frac {1}{5} B (a-a \sin (c+d x))^5}{a^6 d}\) |
Input:
Int[(Cos[c + d*x]^5*(A + B*Sin[c + d*x]))/(a + a*Sin[c + d*x]),x]
Output:
((-2*a^2*(A + B)*(a - a*Sin[c + d*x])^3)/3 + (a*(A + 3*B)*(a - a*Sin[c + d *x])^4)/4 - (B*(a - a*Sin[c + d*x])^5)/5)/(a^6*d)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_ .), x_] :> Int[ExpandIntegrand[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && ((ILtQ[n, 0] && ILtQ[p, 0]) || EqQ[p, 1 ] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))
Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_ .)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Simp[1/(b^p* f) Subst[Int[(a + x)^(m + (p - 1)/2)*(a - x)^((p - 1)/2)*(c + (d/b)*x)^n, x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, c, d, m, n}, x] && Intege rQ[(p - 1)/2] && EqQ[a^2 - b^2, 0]
Time = 0.72 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.95
method | result | size |
derivativedivides | \(\frac {\frac {B \sin \left (d x +c \right )^{5}}{5}+\frac {\left (A -B \right ) \sin \left (d x +c \right )^{4}}{4}+\frac {\left (-A -B \right ) \sin \left (d x +c \right )^{3}}{3}+\frac {\left (-A +B \right ) \sin \left (d x +c \right )^{2}}{2}+A \sin \left (d x +c \right )}{d a}\) | \(75\) |
default | \(\frac {\frac {B \sin \left (d x +c \right )^{5}}{5}+\frac {\left (A -B \right ) \sin \left (d x +c \right )^{4}}{4}+\frac {\left (-A -B \right ) \sin \left (d x +c \right )^{3}}{3}+\frac {\left (-A +B \right ) \sin \left (d x +c \right )^{2}}{2}+A \sin \left (d x +c \right )}{d a}\) | \(75\) |
parallelrisch | \(\frac {60 \left (A -B \right ) \cos \left (2 d x +2 c \right )+15 \left (A -B \right ) \cos \left (4 d x +4 c \right )+10 \left (4 A +B \right ) \sin \left (3 d x +3 c \right )+6 B \sin \left (5 d x +5 c \right )+60 \left (6 A -B \right ) \sin \left (d x +c \right )-75 A +75 B}{480 d a}\) | \(91\) |
risch | \(\frac {3 A \sin \left (d x +c \right )}{4 a d}-\frac {B \sin \left (d x +c \right )}{8 a d}+\frac {\sin \left (5 d x +5 c \right ) B}{80 d a}+\frac {\cos \left (4 d x +4 c \right ) A}{32 a d}-\frac {\cos \left (4 d x +4 c \right ) B}{32 a d}+\frac {\sin \left (3 d x +3 c \right ) A}{12 d a}+\frac {\sin \left (3 d x +3 c \right ) B}{48 d a}+\frac {\cos \left (2 d x +2 c \right ) A}{8 a d}-\frac {\cos \left (2 d x +2 c \right ) B}{8 a d}\) | \(158\) |
norman | \(\frac {\frac {2 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d a}+\frac {2 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}}{d a}+\frac {2 B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{d a}+\frac {2 B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{d a}+\frac {2 \left (8 A -B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 d a}+\frac {2 \left (8 A -B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}}{3 d a}+\frac {4 \left (10 A +3 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{5 d a}+\frac {4 \left (10 A +3 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{5 d a}+\frac {\left (10 A +4 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{3 d a}+\frac {\left (10 A +4 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{3 d a}+\frac {\left (40 A +12 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{5 d a}+\frac {\left (40 A +12 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{5 d a}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{6} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}\) | \(317\) |
Input:
int(cos(d*x+c)^5*(A+B*sin(d*x+c))/(a+a*sin(d*x+c)),x,method=_RETURNVERBOSE )
Output:
1/d/a*(1/5*B*sin(d*x+c)^5+1/4*(A-B)*sin(d*x+c)^4+1/3*(-A-B)*sin(d*x+c)^3+1 /2*(-A+B)*sin(d*x+c)^2+A*sin(d*x+c))
Time = 0.09 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.84 \[ \int \frac {\cos ^5(c+d x) (A+B \sin (c+d x))}{a+a \sin (c+d x)} \, dx=\frac {15 \, {\left (A - B\right )} \cos \left (d x + c\right )^{4} + 4 \, {\left (3 \, B \cos \left (d x + c\right )^{4} + {\left (5 \, A - B\right )} \cos \left (d x + c\right )^{2} + 10 \, A - 2 \, B\right )} \sin \left (d x + c\right )}{60 \, a d} \] Input:
integrate(cos(d*x+c)^5*(A+B*sin(d*x+c))/(a+a*sin(d*x+c)),x, algorithm="fri cas")
Output:
1/60*(15*(A - B)*cos(d*x + c)^4 + 4*(3*B*cos(d*x + c)^4 + (5*A - B)*cos(d* x + c)^2 + 10*A - 2*B)*sin(d*x + c))/(a*d)
Leaf count of result is larger than twice the leaf count of optimal. 1703 vs. \(2 (73) = 146\).
Time = 12.86 (sec) , antiderivative size = 1703, normalized size of antiderivative = 21.56 \[ \int \frac {\cos ^5(c+d x) (A+B \sin (c+d x))}{a+a \sin (c+d x)} \, dx=\text {Too large to display} \] Input:
integrate(cos(d*x+c)**5*(A+B*sin(d*x+c))/(a+a*sin(d*x+c)),x)
Output:
Piecewise((30*A*tan(c/2 + d*x/2)**9/(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*A*tan(c/2 + d*x/2)**8/(1 5*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 + 1 5*a*d) + 80*A*tan(c/2 + d*x/2)**7/(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) - 30*A*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) + 100*A*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) - 30*A*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) + 80*A*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 + 150*a*d*tan(c/2 + d*x/2)**4 + 75*a*d*tan(c/2 + d*x/2)**2 + 15*a*d) - 30*A*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*...
Time = 0.03 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.91 \[ \int \frac {\cos ^5(c+d x) (A+B \sin (c+d x))}{a+a \sin (c+d x)} \, dx=\frac {12 \, B \sin \left (d x + c\right )^{5} + 15 \, {\left (A - B\right )} \sin \left (d x + c\right )^{4} - 20 \, {\left (A + B\right )} \sin \left (d x + c\right )^{3} - 30 \, {\left (A - B\right )} \sin \left (d x + c\right )^{2} + 60 \, A \sin \left (d x + c\right )}{60 \, a d} \] Input:
integrate(cos(d*x+c)^5*(A+B*sin(d*x+c))/(a+a*sin(d*x+c)),x, algorithm="max ima")
Output:
1/60*(12*B*sin(d*x + c)^5 + 15*(A - B)*sin(d*x + c)^4 - 20*(A + B)*sin(d*x + c)^3 - 30*(A - B)*sin(d*x + c)^2 + 60*A*sin(d*x + c))/(a*d)
Time = 0.15 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.20 \[ \int \frac {\cos ^5(c+d x) (A+B \sin (c+d x))}{a+a \sin (c+d x)} \, dx=\frac {12 \, B \sin \left (d x + c\right )^{5} + 15 \, A \sin \left (d x + c\right )^{4} - 15 \, B \sin \left (d x + c\right )^{4} - 20 \, A \sin \left (d x + c\right )^{3} - 20 \, B \sin \left (d x + c\right )^{3} - 30 \, A \sin \left (d x + c\right )^{2} + 30 \, B \sin \left (d x + c\right )^{2} + 60 \, A \sin \left (d x + c\right )}{60 \, a d} \] Input:
integrate(cos(d*x+c)^5*(A+B*sin(d*x+c))/(a+a*sin(d*x+c)),x, algorithm="gia c")
Output:
1/60*(12*B*sin(d*x + c)^5 + 15*A*sin(d*x + c)^4 - 15*B*sin(d*x + c)^4 - 20 *A*sin(d*x + c)^3 - 20*B*sin(d*x + c)^3 - 30*A*sin(d*x + c)^2 + 30*B*sin(d *x + c)^2 + 60*A*sin(d*x + c))/(a*d)
Time = 34.37 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.04 \[ \int \frac {\cos ^5(c+d x) (A+B \sin (c+d x))}{a+a \sin (c+d x)} \, dx=\frac {\frac {{\sin \left (c+d\,x\right )}^4\,\left (A-B\right )}{4\,a}-\frac {{\sin \left (c+d\,x\right )}^2\,\left (A-B\right )}{2\,a}+\frac {B\,{\sin \left (c+d\,x\right )}^5}{5\,a}+\frac {A\,\sin \left (c+d\,x\right )}{a}-\frac {{\sin \left (c+d\,x\right )}^3\,\left (A+B\right )}{3\,a}}{d} \] Input:
int((cos(c + d*x)^5*(A + B*sin(c + d*x)))/(a + a*sin(c + d*x)),x)
Output:
((sin(c + d*x)^4*(A - B))/(4*a) - (sin(c + d*x)^2*(A - B))/(2*a) + (B*sin( c + d*x)^5)/(5*a) + (A*sin(c + d*x))/a - (sin(c + d*x)^3*(A + B))/(3*a))/d
Time = 0.18 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.15 \[ \int \frac {\cos ^5(c+d x) (A+B \sin (c+d x))}{a+a \sin (c+d x)} \, dx=\frac {\sin \left (d x +c \right ) \left (12 \sin \left (d x +c \right )^{4} b +15 \sin \left (d x +c \right )^{3} a -15 \sin \left (d x +c \right )^{3} b -20 \sin \left (d x +c \right )^{2} a -20 \sin \left (d x +c \right )^{2} b -30 \sin \left (d x +c \right ) a +30 \sin \left (d x +c \right ) b +60 a \right )}{60 a d} \] Input:
int(cos(d*x+c)^5*(A+B*sin(d*x+c))/(a+a*sin(d*x+c)),x)
Output:
(sin(c + d*x)*(12*sin(c + d*x)**4*b + 15*sin(c + d*x)**3*a - 15*sin(c + d* x)**3*b - 20*sin(c + d*x)**2*a - 20*sin(c + d*x)**2*b - 30*sin(c + d*x)*a + 30*sin(c + d*x)*b + 60*a))/(60*a*d)