Integrand size = 31, antiderivative size = 79 \[ \int \frac {\cos ^7(c+d x) (A+B \sin (c+d x))}{(a+a \sin (c+d x))^2} \, dx=-\frac {(A+B) (a-a \sin (c+d x))^4}{2 a^6 d}+\frac {(A+3 B) (a-a \sin (c+d x))^5}{5 a^7 d}-\frac {B (a-a \sin (c+d x))^6}{6 a^8 d} \] Output:
-1/2*(A+B)*(a-a*sin(d*x+c))^4/a^6/d+1/5*(A+3*B)*(a-a*sin(d*x+c))^5/a^7/d-1 /6*B*(a-a*sin(d*x+c))^6/a^8/d
Time = 0.20 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.66 \[ \int \frac {\cos ^7(c+d x) (A+B \sin (c+d x))}{(a+a \sin (c+d x))^2} \, dx=-\frac {(-1+\sin (c+d x))^4 \left (9 A+2 B+(6 A+8 B) \sin (c+d x)+5 B \sin ^2(c+d x)\right )}{30 a^2 d} \] Input:
Integrate[(Cos[c + d*x]^7*(A + B*Sin[c + d*x]))/(a + a*Sin[c + d*x])^2,x]
Output:
-1/30*((-1 + Sin[c + d*x])^4*(9*A + 2*B + (6*A + 8*B)*Sin[c + d*x] + 5*B*S in[c + d*x]^2))/(a^2*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.
\(\displaystyle \int \frac {\cos ^7(c+d x) (A+B \sin (c+d x))}{(a \sin (c+d x)+a)^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\cos (c+d x)^7 (A+B \sin (c+d x))}{(a \sin (c+d x)+a)^2}dx\) |
\(\Big \downarrow \) 3315 |
\(\displaystyle \frac {\int \frac {(a-a \sin (c+d x))^3 (\sin (c+d x) a+a) (a A+a B \sin (c+d x))}{a}d(a \sin (c+d x))}{a^7 d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int (a-a \sin (c+d x))^3 (\sin (c+d x) a+a) (a A+a B \sin (c+d x))d(a \sin (c+d x))}{a^8 d}\) |
\(\Big \downarrow \) 86 |
\(\displaystyle \frac {\int \left (B (a-a \sin (c+d x))^5-a (A+3 B) (a-a \sin (c+d x))^4+2 a^2 (A+B) (a-a \sin (c+d x))^3\right )d(a \sin (c+d x))}{a^8 d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {-\frac {1}{2} a^2 (A+B) (a-a \sin (c+d x))^4+\frac {1}{5} a (A+3 B) (a-a \sin (c+d x))^5-\frac {1}{6} B (a-a \sin (c+d x))^6}{a^8 d}\) |
Input:
Int[(Cos[c + d*x]^7*(A + B*Sin[c + d*x]))/(a + a*Sin[c + d*x])^2,x]
Output:
(-1/2*(a^2*(A + B)*(a - a*Sin[c + d*x])^4) + (a*(A + 3*B)*(a - a*Sin[c + d *x])^5)/5 - (B*(a - a*Sin[c + d*x])^6)/6)/(a^8*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 = 1.27 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.06
method | result | size |
derivativedivides | \(-\frac {\frac {B \sin \left (d x +c \right )^{6}}{6}+\frac {\left (A -2 B \right ) \sin \left (d x +c \right )^{5}}{5}-\frac {\sin \left (d x +c \right )^{4} A}{2}+\frac {2 B \sin \left (d x +c \right )^{3}}{3}+\frac {\left (2 A -B \right ) \sin \left (d x +c \right )^{2}}{2}-A \sin \left (d x +c \right )}{d \,a^{2}}\) | \(84\) |
default | \(-\frac {\frac {B \sin \left (d x +c \right )^{6}}{6}+\frac {\left (A -2 B \right ) \sin \left (d x +c \right )^{5}}{5}-\frac {\sin \left (d x +c \right )^{4} A}{2}+\frac {2 B \sin \left (d x +c \right )^{3}}{3}+\frac {\left (2 A -B \right ) \sin \left (d x +c \right )^{2}}{2}-A \sin \left (d x +c \right )}{d \,a^{2}}\) | \(84\) |
parallelrisch | \(\frac {\left (240 A -165 B \right ) \cos \left (2 d x +2 c \right )+\left (60 A -30 B \right ) \cos \left (4 d x +4 c \right )+\left (60 A +40 B \right ) \sin \left (3 d x +3 c \right )+\left (-12 A +24 B \right ) \sin \left (5 d x +5 c \right )+5 B \cos \left (6 d x +6 c \right )+\left (840 A -240 B \right ) \sin \left (d x +c \right )-300 A +190 B}{960 d \,a^{2}}\) | \(110\) |
risch | \(\frac {7 \sin \left (d x +c \right ) A}{8 d \,a^{2}}-\frac {\sin \left (d x +c \right ) B}{4 d \,a^{2}}+\frac {B \cos \left (6 d x +6 c \right )}{192 a^{2} d}-\frac {\sin \left (5 d x +5 c \right ) A}{80 d \,a^{2}}+\frac {\sin \left (5 d x +5 c \right ) B}{40 d \,a^{2}}+\frac {\cos \left (4 d x +4 c \right ) A}{16 a^{2} d}-\frac {\cos \left (4 d x +4 c \right ) B}{32 a^{2} d}+\frac {\sin \left (3 d x +3 c \right ) A}{16 d \,a^{2}}+\frac {\sin \left (3 d x +3 c \right ) B}{24 d \,a^{2}}+\frac {\cos \left (2 d x +2 c \right ) A}{4 a^{2} d}-\frac {11 \cos \left (2 d x +2 c \right ) B}{64 a^{2} d}\) | \(194\) |
Input:
int(cos(d*x+c)^7*(A+B*sin(d*x+c))/(a+a*sin(d*x+c))^2,x,method=_RETURNVERBO SE)
Output:
-1/d/a^2*(1/6*B*sin(d*x+c)^6+1/5*(A-2*B)*sin(d*x+c)^5-1/2*sin(d*x+c)^4*A+2 /3*B*sin(d*x+c)^3+1/2*(2*A-B)*sin(d*x+c)^2-A*sin(d*x+c))
Time = 0.08 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.04 \[ \int \frac {\cos ^7(c+d x) (A+B \sin (c+d x))}{(a+a \sin (c+d x))^2} \, dx=\frac {5 \, B \cos \left (d x + c\right )^{6} + 15 \, {\left (A - B\right )} \cos \left (d x + c\right )^{4} - 2 \, {\left (3 \, {\left (A - 2 \, B\right )} \cos \left (d x + c\right )^{4} - 2 \, {\left (3 \, A - B\right )} \cos \left (d x + c\right )^{2} - 12 \, A + 4 \, B\right )} \sin \left (d x + c\right )}{30 \, a^{2} d} \] Input:
integrate(cos(d*x+c)^7*(A+B*sin(d*x+c))/(a+a*sin(d*x+c))^2,x, algorithm="f ricas")
Output:
1/30*(5*B*cos(d*x + c)^6 + 15*(A - B)*cos(d*x + c)^4 - 2*(3*(A - 2*B)*cos( d*x + c)^4 - 2*(3*A - B)*cos(d*x + c)^2 - 12*A + 4*B)*sin(d*x + c))/(a^2*d )
Leaf count of result is larger than twice the leaf count of optimal. 2705 vs. \(2 (70) = 140\).
Time = 56.90 (sec) , antiderivative size = 2705, normalized size of antiderivative = 34.24 \[ \int \frac {\cos ^7(c+d x) (A+B \sin (c+d x))}{(a+a \sin (c+d x))^2} \, dx=\text {Too large to display} \] Input:
integrate(cos(d*x+c)**7*(A+B*sin(d*x+c))/(a+a*sin(d*x+c))**2,x)
Output:
Piecewise((30*A*tan(c/2 + d*x/2)**11/(15*a**2*d*tan(c/2 + d*x/2)**12 + 90* a**2*d*tan(c/2 + d*x/2)**10 + 225*a**2*d*tan(c/2 + d*x/2)**8 + 300*a**2*d* tan(c/2 + d*x/2)**6 + 225*a**2*d*tan(c/2 + d*x/2)**4 + 90*a**2*d*tan(c/2 + d*x/2)**2 + 15*a**2*d) - 60*A*tan(c/2 + d*x/2)**10/(15*a**2*d*tan(c/2 + d *x/2)**12 + 90*a**2*d*tan(c/2 + d*x/2)**10 + 225*a**2*d*tan(c/2 + d*x/2)** 8 + 300*a**2*d*tan(c/2 + d*x/2)**6 + 225*a**2*d*tan(c/2 + d*x/2)**4 + 90*a **2*d*tan(c/2 + d*x/2)**2 + 15*a**2*d) + 150*A*tan(c/2 + d*x/2)**9/(15*a** 2*d*tan(c/2 + d*x/2)**12 + 90*a**2*d*tan(c/2 + d*x/2)**10 + 225*a**2*d*tan (c/2 + d*x/2)**8 + 300*a**2*d*tan(c/2 + d*x/2)**6 + 225*a**2*d*tan(c/2 + d *x/2)**4 + 90*a**2*d*tan(c/2 + d*x/2)**2 + 15*a**2*d) - 120*A*tan(c/2 + d* x/2)**8/(15*a**2*d*tan(c/2 + d*x/2)**12 + 90*a**2*d*tan(c/2 + d*x/2)**10 + 225*a**2*d*tan(c/2 + d*x/2)**8 + 300*a**2*d*tan(c/2 + d*x/2)**6 + 225*a** 2*d*tan(c/2 + d*x/2)**4 + 90*a**2*d*tan(c/2 + d*x/2)**2 + 15*a**2*d) + 204 *A*tan(c/2 + d*x/2)**7/(15*a**2*d*tan(c/2 + d*x/2)**12 + 90*a**2*d*tan(c/2 + d*x/2)**10 + 225*a**2*d*tan(c/2 + d*x/2)**8 + 300*a**2*d*tan(c/2 + d*x/ 2)**6 + 225*a**2*d*tan(c/2 + d*x/2)**4 + 90*a**2*d*tan(c/2 + d*x/2)**2 + 1 5*a**2*d) - 120*A*tan(c/2 + d*x/2)**6/(15*a**2*d*tan(c/2 + d*x/2)**12 + 90 *a**2*d*tan(c/2 + d*x/2)**10 + 225*a**2*d*tan(c/2 + d*x/2)**8 + 300*a**2*d *tan(c/2 + d*x/2)**6 + 225*a**2*d*tan(c/2 + d*x/2)**4 + 90*a**2*d*tan(c/2 + d*x/2)**2 + 15*a**2*d) + 204*A*tan(c/2 + d*x/2)**5/(15*a**2*d*tan(c/2...
Time = 0.04 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.05 \[ \int \frac {\cos ^7(c+d x) (A+B \sin (c+d x))}{(a+a \sin (c+d x))^2} \, dx=-\frac {5 \, B \sin \left (d x + c\right )^{6} + 6 \, {\left (A - 2 \, B\right )} \sin \left (d x + c\right )^{5} - 15 \, A \sin \left (d x + c\right )^{4} + 20 \, B \sin \left (d x + c\right )^{3} + 15 \, {\left (2 \, A - B\right )} \sin \left (d x + c\right )^{2} - 30 \, A \sin \left (d x + c\right )}{30 \, a^{2} d} \] Input:
integrate(cos(d*x+c)^7*(A+B*sin(d*x+c))/(a+a*sin(d*x+c))^2,x, algorithm="m axima")
Output:
-1/30*(5*B*sin(d*x + c)^6 + 6*(A - 2*B)*sin(d*x + c)^5 - 15*A*sin(d*x + c) ^4 + 20*B*sin(d*x + c)^3 + 15*(2*A - B)*sin(d*x + c)^2 - 30*A*sin(d*x + c) )/(a^2*d)
Time = 0.14 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.20 \[ \int \frac {\cos ^7(c+d x) (A+B \sin (c+d x))}{(a+a \sin (c+d x))^2} \, dx=-\frac {5 \, B \sin \left (d x + c\right )^{6} + 6 \, A \sin \left (d x + c\right )^{5} - 12 \, B \sin \left (d x + c\right )^{5} - 15 \, A \sin \left (d x + c\right )^{4} + 20 \, B \sin \left (d x + c\right )^{3} + 30 \, A \sin \left (d x + c\right )^{2} - 15 \, B \sin \left (d x + c\right )^{2} - 30 \, A \sin \left (d x + c\right )}{30 \, a^{2} d} \] Input:
integrate(cos(d*x+c)^7*(A+B*sin(d*x+c))/(a+a*sin(d*x+c))^2,x, algorithm="g iac")
Output:
-1/30*(5*B*sin(d*x + c)^6 + 6*A*sin(d*x + c)^5 - 12*B*sin(d*x + c)^5 - 15* A*sin(d*x + c)^4 + 20*B*sin(d*x + c)^3 + 30*A*sin(d*x + c)^2 - 15*B*sin(d* x + c)^2 - 30*A*sin(d*x + c))/(a^2*d)
Time = 0.09 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.24 \[ \int \frac {\cos ^7(c+d x) (A+B \sin (c+d x))}{(a+a \sin (c+d x))^2} \, dx=-\frac {\frac {{\sin \left (c+d\,x\right )}^5\,\left (A-2\,B\right )}{5\,a^2}-\frac {A\,{\sin \left (c+d\,x\right )}^4}{2\,a^2}+\frac {2\,B\,{\sin \left (c+d\,x\right )}^3}{3\,a^2}+\frac {B\,{\sin \left (c+d\,x\right )}^6}{6\,a^2}+\frac {{\sin \left (c+d\,x\right )}^2\,\left (2\,A-B\right )}{2\,a^2}-\frac {A\,\sin \left (c+d\,x\right )}{a^2}}{d} \] Input:
int((cos(c + d*x)^7*(A + B*sin(c + d*x)))/(a + a*sin(c + d*x))^2,x)
Output:
-((sin(c + d*x)^5*(A - 2*B))/(5*a^2) - (A*sin(c + d*x)^4)/(2*a^2) + (2*B*s in(c + d*x)^3)/(3*a^2) + (B*sin(c + d*x)^6)/(6*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 = 91, normalized size of antiderivative = 1.15 \[ \int \frac {\cos ^7(c+d x) (A+B \sin (c+d x))}{(a+a \sin (c+d x))^2} \, dx=\frac {\sin \left (d x +c \right ) \left (-5 \sin \left (d x +c \right )^{5} b -6 \sin \left (d x +c \right )^{4} a +12 \sin \left (d x +c \right )^{4} b +15 \sin \left (d x +c \right )^{3} a -20 \sin \left (d x +c \right )^{2} b -30 \sin \left (d x +c \right ) a +15 \sin \left (d x +c \right ) b +30 a \right )}{30 a^{2} d} \] Input:
int(cos(d*x+c)^7*(A+B*sin(d*x+c))/(a+a*sin(d*x+c))^2,x)
Output:
(sin(c + d*x)*( - 5*sin(c + d*x)**5*b - 6*sin(c + d*x)**4*a + 12*sin(c + d *x)**4*b + 15*sin(c + d*x)**3*a - 20*sin(c + d*x)**2*b - 30*sin(c + d*x)*a + 15*sin(c + d*x)*b + 30*a))/(30*a**2*d)