Integrand size = 31, antiderivative size = 51 \[ \int \frac {\cos ^5(c+d x) (A+B \sin (c+d x))}{(a+a \sin (c+d x))^2} \, dx=-\frac {(A+B) (a-a \sin (c+d x))^3}{3 a^5 d}+\frac {B (a-a \sin (c+d x))^4}{4 a^6 d} \] Output:
-1/3*(A+B)*(a-a*sin(d*x+c))^3/a^5/d+1/4*B*(a-a*sin(d*x+c))^4/a^6/d
Time = 0.08 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.67 \[ \int \frac {\cos ^5(c+d x) (A+B \sin (c+d x))}{(a+a \sin (c+d x))^2} \, dx=\frac {(-1+\sin (c+d x))^3 (4 A+B+3 B \sin (c+d x))}{12 a^2 d} \] Input:
Integrate[(Cos[c + d*x]^5*(A + B*Sin[c + d*x]))/(a + a*Sin[c + d*x])^2,x]
Output:
((-1 + Sin[c + d*x])^3*(4*A + B + 3*B*Sin[c + d*x]))/(12*a^2*d)
Time = 0.30 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.92, 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, 49, 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.
\(\displaystyle \int \frac {\cos ^5(c+d x) (A+B \sin (c+d x))}{(a \sin (c+d x)+a)^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\cos (c+d x)^5 (A+B \sin (c+d x))}{(a \sin (c+d x)+a)^2}dx\) |
\(\Big \downarrow \) 3315 |
\(\displaystyle \frac {\int \frac {(a-a \sin (c+d x))^2 (a A+a B \sin (c+d x))}{a}d(a \sin (c+d x))}{a^5 d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int (a-a \sin (c+d x))^2 (a A+a B \sin (c+d x))d(a \sin (c+d x))}{a^6 d}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {\int \left (a (A+B) (a-a \sin (c+d x))^2-B (a-a \sin (c+d x))^3\right )d(a \sin (c+d x))}{a^6 d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {1}{4} B (a-a \sin (c+d x))^4-\frac {1}{3} a (A+B) (a-a \sin (c+d x))^3}{a^6 d}\) |
Input:
Int[(Cos[c + d*x]^5*(A + B*Sin[c + d*x]))/(a + a*Sin[c + d*x])^2,x]
Output:
(-1/3*(a*(A + B)*(a - a*Sin[c + d*x])^3) + (B*(a - a*Sin[c + d*x])^4)/4)/( 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_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int [ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && IGtQ[m, 0] && IGtQ[m + n + 2, 0]
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 = 0.64 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.14
method | result | size |
derivativedivides | \(\frac {\frac {B \sin \left (d x +c \right )^{4}}{4}+\frac {\left (A -2 B \right ) \sin \left (d x +c \right )^{3}}{3}+\frac {\left (B -2 A \right ) \sin \left (d x +c \right )^{2}}{2}+A \sin \left (d x +c \right )}{d \,a^{2}}\) | \(58\) |
default | \(\frac {\frac {B \sin \left (d x +c \right )^{4}}{4}+\frac {\left (A -2 B \right ) \sin \left (d x +c \right )^{3}}{3}+\frac {\left (B -2 A \right ) \sin \left (d x +c \right )^{2}}{2}+A \sin \left (d x +c \right )}{d \,a^{2}}\) | \(58\) |
parallelrisch | \(\frac {\left (48 A -36 B \right ) \cos \left (2 d x +2 c \right )+\left (-8 A +16 B \right ) \sin \left (3 d x +3 c \right )+3 B \cos \left (4 d x +4 c \right )+\left (120 A -48 B \right ) \sin \left (d x +c \right )-48 A +33 B}{96 d \,a^{2}}\) | \(76\) |
risch | \(\frac {5 \sin \left (d x +c \right ) A}{4 d \,a^{2}}-\frac {\sin \left (d x +c \right ) B}{2 d \,a^{2}}+\frac {\cos \left (4 d x +4 c \right ) B}{32 a^{2} d}-\frac {\sin \left (3 d x +3 c \right ) A}{12 d \,a^{2}}+\frac {\sin \left (3 d x +3 c \right ) B}{6 d \,a^{2}}+\frac {\cos \left (2 d x +2 c \right ) A}{2 a^{2} d}-\frac {3 \cos \left (2 d x +2 c \right ) B}{8 a^{2} d}\) | \(122\) |
norman | \(\frac {\frac {2 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d a}+\frac {2 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{14}}{d a}+\frac {\left (2 A +2 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{d a}+\frac {\left (2 A +2 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{13}}{d a}+\frac {\left (12 A +2 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{d a}+\frac {\left (12 A +2 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{d a}+\frac {\left (14 A +6 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{d a}+\frac {\left (14 A +6 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}}{d a}+\frac {\left (24 A +8 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{d a}+\frac {\left (24 A +8 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{d a}+\frac {\left (20 A +2 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 d a}+\frac {\left (20 A +2 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}}{3 d a}+\frac {\left (74 A +8 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{3 d a}+\frac {\left (74 A +8 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{3 d a}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{6} a \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}\) | \(376\) |
Input:
int(cos(d*x+c)^5*(A+B*sin(d*x+c))/(a+a*sin(d*x+c))^2,x,method=_RETURNVERBO SE)
Output:
1/d/a^2*(1/4*B*sin(d*x+c)^4+1/3*(A-2*B)*sin(d*x+c)^3+1/2*(B-2*A)*sin(d*x+c )^2+A*sin(d*x+c))
Time = 0.08 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.25 \[ \int \frac {\cos ^5(c+d x) (A+B \sin (c+d x))}{(a+a \sin (c+d x))^2} \, dx=\frac {3 \, B \cos \left (d x + c\right )^{4} + 12 \, {\left (A - B\right )} \cos \left (d x + c\right )^{2} - 4 \, {\left ({\left (A - 2 \, B\right )} \cos \left (d x + c\right )^{2} - 4 \, A + 2 \, B\right )} \sin \left (d x + c\right )}{12 \, a^{2} d} \] Input:
integrate(cos(d*x+c)^5*(A+B*sin(d*x+c))/(a+a*sin(d*x+c))^2,x, algorithm="f ricas")
Output:
1/12*(3*B*cos(d*x + c)^4 + 12*(A - B)*cos(d*x + c)^2 - 4*((A - 2*B)*cos(d* x + c)^2 - 4*A + 2*B)*sin(d*x + c))/(a^2*d)
Leaf count of result is larger than twice the leaf count of optimal. 1182 vs. \(2 (44) = 88\).
Time = 22.06 (sec) , antiderivative size = 1182, normalized size of antiderivative = 23.18 \[ \int \frac {\cos ^5(c+d x) (A+B \sin (c+d x))}{(a+a \sin (c+d x))^2} \, dx=\text {Too large to display} \] Input:
integrate(cos(d*x+c)**5*(A+B*sin(d*x+c))/(a+a*sin(d*x+c))**2,x)
Output:
Piecewise((6*A*tan(c/2 + d*x/2)**7/(3*a**2*d*tan(c/2 + d*x/2)**8 + 12*a**2 *d*tan(c/2 + d*x/2)**6 + 18*a**2*d*tan(c/2 + d*x/2)**4 + 12*a**2*d*tan(c/2 + d*x/2)**2 + 3*a**2*d) - 12*A*tan(c/2 + d*x/2)**6/(3*a**2*d*tan(c/2 + d* x/2)**8 + 12*a**2*d*tan(c/2 + d*x/2)**6 + 18*a**2*d*tan(c/2 + d*x/2)**4 + 12*a**2*d*tan(c/2 + d*x/2)**2 + 3*a**2*d) + 26*A*tan(c/2 + d*x/2)**5/(3*a* *2*d*tan(c/2 + d*x/2)**8 + 12*a**2*d*tan(c/2 + d*x/2)**6 + 18*a**2*d*tan(c /2 + d*x/2)**4 + 12*a**2*d*tan(c/2 + d*x/2)**2 + 3*a**2*d) - 24*A*tan(c/2 + d*x/2)**4/(3*a**2*d*tan(c/2 + d*x/2)**8 + 12*a**2*d*tan(c/2 + d*x/2)**6 + 18*a**2*d*tan(c/2 + d*x/2)**4 + 12*a**2*d*tan(c/2 + d*x/2)**2 + 3*a**2*d ) + 26*A*tan(c/2 + d*x/2)**3/(3*a**2*d*tan(c/2 + d*x/2)**8 + 12*a**2*d*tan (c/2 + d*x/2)**6 + 18*a**2*d*tan(c/2 + d*x/2)**4 + 12*a**2*d*tan(c/2 + d*x /2)**2 + 3*a**2*d) - 12*A*tan(c/2 + d*x/2)**2/(3*a**2*d*tan(c/2 + d*x/2)** 8 + 12*a**2*d*tan(c/2 + d*x/2)**6 + 18*a**2*d*tan(c/2 + d*x/2)**4 + 12*a** 2*d*tan(c/2 + d*x/2)**2 + 3*a**2*d) + 6*A*tan(c/2 + d*x/2)/(3*a**2*d*tan(c /2 + d*x/2)**8 + 12*a**2*d*tan(c/2 + d*x/2)**6 + 18*a**2*d*tan(c/2 + d*x/2 )**4 + 12*a**2*d*tan(c/2 + d*x/2)**2 + 3*a**2*d) + 6*B*tan(c/2 + d*x/2)**6 /(3*a**2*d*tan(c/2 + d*x/2)**8 + 12*a**2*d*tan(c/2 + d*x/2)**6 + 18*a**2*d *tan(c/2 + d*x/2)**4 + 12*a**2*d*tan(c/2 + d*x/2)**2 + 3*a**2*d) - 16*B*ta n(c/2 + d*x/2)**5/(3*a**2*d*tan(c/2 + d*x/2)**8 + 12*a**2*d*tan(c/2 + d*x/ 2)**6 + 18*a**2*d*tan(c/2 + d*x/2)**4 + 12*a**2*d*tan(c/2 + d*x/2)**2 +...
Time = 0.05 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.20 \[ \int \frac {\cos ^5(c+d x) (A+B \sin (c+d x))}{(a+a \sin (c+d x))^2} \, dx=\frac {3 \, B \sin \left (d x + c\right )^{4} + 4 \, {\left (A - 2 \, B\right )} \sin \left (d x + c\right )^{3} - 6 \, {\left (2 \, A - B\right )} \sin \left (d x + c\right )^{2} + 12 \, A \sin \left (d x + c\right )}{12 \, a^{2} d} \] Input:
integrate(cos(d*x+c)^5*(A+B*sin(d*x+c))/(a+a*sin(d*x+c))^2,x, algorithm="m axima")
Output:
1/12*(3*B*sin(d*x + c)^4 + 4*(A - 2*B)*sin(d*x + c)^3 - 6*(2*A - B)*sin(d* x + c)^2 + 12*A*sin(d*x + c))/(a^2*d)
Time = 0.14 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.43 \[ \int \frac {\cos ^5(c+d x) (A+B \sin (c+d x))}{(a+a \sin (c+d x))^2} \, dx=\frac {3 \, B \sin \left (d x + c\right )^{4} + 4 \, A \sin \left (d x + c\right )^{3} - 8 \, B \sin \left (d x + c\right )^{3} - 12 \, A \sin \left (d x + c\right )^{2} + 6 \, B \sin \left (d x + c\right )^{2} + 12 \, A \sin \left (d x + c\right )}{12 \, a^{2} d} \] Input:
integrate(cos(d*x+c)^5*(A+B*sin(d*x+c))/(a+a*sin(d*x+c))^2,x, algorithm="g iac")
Output:
1/12*(3*B*sin(d*x + c)^4 + 4*A*sin(d*x + c)^3 - 8*B*sin(d*x + c)^3 - 12*A* sin(d*x + c)^2 + 6*B*sin(d*x + c)^2 + 12*A*sin(d*x + c))/(a^2*d)
Time = 0.10 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.33 \[ \int \frac {\cos ^5(c+d x) (A+B \sin (c+d x))}{(a+a \sin (c+d x))^2} \, dx=\frac {\frac {{\sin \left (c+d\,x\right )}^3\,\left (A-2\,B\right )}{3\,a^2}+\frac {B\,{\sin \left (c+d\,x\right )}^4}{4\,a^2}-\frac {{\sin \left (c+d\,x\right )}^2\,\left (2\,A-B\right )}{2\,a^2}+\frac {A\,\sin \left (c+d\,x\right )}{a^2}}{d} \] Input:
int((cos(c + d*x)^5*(A + B*sin(c + d*x)))/(a + a*sin(c + d*x))^2,x)
Output:
((sin(c + d*x)^3*(A - 2*B))/(3*a^2) + (B*sin(c + d*x)^4)/(4*a^2) - (sin(c + d*x)^2*(2*A - B))/(2*a^2) + (A*sin(c + d*x))/a^2)/d
Time = 0.18 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.35 \[ \int \frac {\cos ^5(c+d x) (A+B \sin (c+d x))}{(a+a \sin (c+d x))^2} \, dx=\frac {\sin \left (d x +c \right ) \left (3 \sin \left (d x +c \right )^{3} b +4 \sin \left (d x +c \right )^{2} a -8 \sin \left (d x +c \right )^{2} b -12 \sin \left (d x +c \right ) a +6 \sin \left (d x +c \right ) b +12 a \right )}{12 a^{2} d} \] Input:
int(cos(d*x+c)^5*(A+B*sin(d*x+c))/(a+a*sin(d*x+c))^2,x)
Output:
(sin(c + d*x)*(3*sin(c + d*x)**3*b + 4*sin(c + d*x)**2*a - 8*sin(c + d*x)* *2*b - 12*sin(c + d*x)*a + 6*sin(c + d*x)*b + 12*a))/(12*a**2*d)