Integrand size = 31, antiderivative size = 66 \[ \int \frac {\cos ^3(c+d x) (A+B \sin (c+d x))}{(a+a \sin (c+d x))^2} \, dx=\frac {2 (A-B) \log (1+\sin (c+d x))}{a^2 d}-\frac {(A-B) \sin (c+d x)}{a^2 d}-\frac {B (a-a \sin (c+d x))^2}{2 a^4 d} \] Output:
2*(A-B)*ln(1+sin(d*x+c))/a^2/d-(A-B)*sin(d*x+c)/a^2/d-1/2*B*(a-a*sin(d*x+c ))^2/a^4/d
Time = 0.11 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.77 \[ \int \frac {\cos ^3(c+d x) (A+B \sin (c+d x))}{(a+a \sin (c+d x))^2} \, dx=-\frac {B-4 (A-B) \log (1+\sin (c+d x))+2 (A-2 B) \sin (c+d x)+B \sin ^2(c+d x)}{2 a^2 d} \] Input:
Integrate[(Cos[c + d*x]^3*(A + B*Sin[c + d*x]))/(a + a*Sin[c + d*x])^2,x]
Output:
-1/2*(B - 4*(A - B)*Log[1 + Sin[c + d*x]] + 2*(A - 2*B)*Sin[c + d*x] + B*S in[c + d*x]^2)/(a^2*d)
Time = 0.32 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.95, 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.
\(\displaystyle \int \frac {\cos ^3(c+d x) (A+B \sin (c+d x))}{(a \sin (c+d x)+a)^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\cos (c+d x)^3 (A+B \sin (c+d x))}{(a \sin (c+d x)+a)^2}dx\) |
\(\Big \downarrow \) 3315 |
\(\displaystyle \frac {\int \frac {(a-a \sin (c+d x)) (a A+a B \sin (c+d x))}{a (\sin (c+d x) a+a)}d(a \sin (c+d x))}{a^3 d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {(a-a \sin (c+d x)) (a A+a B \sin (c+d x))}{\sin (c+d x) a+a}d(a \sin (c+d x))}{a^4 d}\) |
\(\Big \downarrow \) 86 |
\(\displaystyle \frac {\int \left (\frac {2 (A-B) a^2}{\sin (c+d x) a+a}-(A-B) a+B (a-a \sin (c+d x))\right )d(a \sin (c+d x))}{a^4 d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {-a^2 (A-B) \sin (c+d x)+2 a^2 (A-B) \log (a \sin (c+d x)+a)-\frac {1}{2} B (a-a \sin (c+d x))^2}{a^4 d}\) |
Input:
Int[(Cos[c + d*x]^3*(A + B*Sin[c + d*x]))/(a + a*Sin[c + d*x])^2,x]
Output:
(2*a^2*(A - B)*Log[a + a*Sin[c + d*x]] - a^2*(A - B)*Sin[c + d*x] - (B*(a - a*Sin[c + d*x])^2)/2)/(a^4*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 = 0.70 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.83
method | result | size |
derivativedivides | \(\frac {-\frac {B \sin \left (d x +c \right )^{2}}{2}-A \sin \left (d x +c \right )+2 B \sin \left (d x +c \right )+\left (-2 B +2 A \right ) \ln \left (1+\sin \left (d x +c \right )\right )}{d \,a^{2}}\) | \(55\) |
default | \(\frac {-\frac {B \sin \left (d x +c \right )^{2}}{2}-A \sin \left (d x +c \right )+2 B \sin \left (d x +c \right )+\left (-2 B +2 A \right ) \ln \left (1+\sin \left (d x +c \right )\right )}{d \,a^{2}}\) | \(55\) |
parallelrisch | \(\frac {8 \left (-A +B \right ) \ln \left (\sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )+16 \left (A -B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+B \cos \left (2 d x +2 c \right )+4 \left (2 B -A \right ) \sin \left (d x +c \right )-B}{4 d \,a^{2}}\) | \(77\) |
risch | \(-\frac {2 i x A}{a^{2}}+\frac {2 i x B}{a^{2}}-\frac {i {\mathrm e}^{i \left (d x +c \right )} B}{d \,a^{2}}+\frac {i {\mathrm e}^{i \left (d x +c \right )} A}{2 d \,a^{2}}+\frac {i {\mathrm e}^{-i \left (d x +c \right )} B}{d \,a^{2}}-\frac {i {\mathrm e}^{-i \left (d x +c \right )} A}{2 d \,a^{2}}-\frac {4 i A c}{d \,a^{2}}+\frac {4 i B c}{d \,a^{2}}+\frac {4 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A}{d \,a^{2}}-\frac {4 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{d \,a^{2}}+\frac {\cos \left (2 d x +2 c \right ) B}{4 a^{2} d}\) | \(178\) |
norman | \(\frac {-\frac {\left (12 A -18 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{d a}-\frac {\left (12 A -18 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{d a}-\frac {2 \left (12 A -17 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{d a}-\frac {2 \left (12 A -17 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{d a}-\frac {2 \left (10 A -15 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{d a}-\frac {2 \left (10 A -15 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{d a}-\frac {\left (6 A -10 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{d a}-\frac {\left (6 A -10 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{d a}-\frac {\left (2 A -4 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d a}-\frac {\left (2 A -4 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}}{d a}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{4} a \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}+\frac {4 \left (A -B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d \,a^{2}}-\frac {2 \left (A -B \right ) \ln \left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}{d \,a^{2}}\) | \(345\) |
Input:
int(cos(d*x+c)^3*(A+B*sin(d*x+c))/(a+a*sin(d*x+c))^2,x,method=_RETURNVERBO SE)
Output:
1/d/a^2*(-1/2*B*sin(d*x+c)^2-A*sin(d*x+c)+2*B*sin(d*x+c)+(-2*B+2*A)*ln(1+s in(d*x+c)))
Time = 0.09 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.73 \[ \int \frac {\cos ^3(c+d x) (A+B \sin (c+d x))}{(a+a \sin (c+d x))^2} \, dx=\frac {B \cos \left (d x + c\right )^{2} + 4 \, {\left (A - B\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (A - 2 \, B\right )} \sin \left (d x + c\right )}{2 \, a^{2} d} \] Input:
integrate(cos(d*x+c)^3*(A+B*sin(d*x+c))/(a+a*sin(d*x+c))^2,x, algorithm="f ricas")
Output:
1/2*(B*cos(d*x + c)^2 + 4*(A - B)*log(sin(d*x + c) + 1) - 2*(A - 2*B)*sin( d*x + c))/(a^2*d)
Leaf count of result is larger than twice the leaf count of optimal. 1096 vs. \(2 (56) = 112\).
Time = 7.54 (sec) , antiderivative size = 1096, normalized size of antiderivative = 16.61 \[ \int \frac {\cos ^3(c+d x) (A+B \sin (c+d x))}{(a+a \sin (c+d x))^2} \, dx=\text {Too large to display} \] Input:
integrate(cos(d*x+c)**3*(A+B*sin(d*x+c))/(a+a*sin(d*x+c))**2,x)
Output:
Piecewise((4*A*log(tan(c/2 + d*x/2) + 1)*tan(c/2 + d*x/2)**4/(a**2*d*tan(c /2 + d*x/2)**4 + 2*a**2*d*tan(c/2 + d*x/2)**2 + a**2*d) + 8*A*log(tan(c/2 + d*x/2) + 1)*tan(c/2 + d*x/2)**2/(a**2*d*tan(c/2 + d*x/2)**4 + 2*a**2*d*t an(c/2 + d*x/2)**2 + a**2*d) + 4*A*log(tan(c/2 + d*x/2) + 1)/(a**2*d*tan(c /2 + d*x/2)**4 + 2*a**2*d*tan(c/2 + d*x/2)**2 + a**2*d) - 2*A*log(tan(c/2 + d*x/2)**2 + 1)*tan(c/2 + d*x/2)**4/(a**2*d*tan(c/2 + d*x/2)**4 + 2*a**2* d*tan(c/2 + d*x/2)**2 + a**2*d) - 4*A*log(tan(c/2 + d*x/2)**2 + 1)*tan(c/2 + d*x/2)**2/(a**2*d*tan(c/2 + d*x/2)**4 + 2*a**2*d*tan(c/2 + d*x/2)**2 + a**2*d) - 2*A*log(tan(c/2 + d*x/2)**2 + 1)/(a**2*d*tan(c/2 + d*x/2)**4 + 2 *a**2*d*tan(c/2 + d*x/2)**2 + a**2*d) - 2*A*tan(c/2 + d*x/2)**3/(a**2*d*ta n(c/2 + d*x/2)**4 + 2*a**2*d*tan(c/2 + d*x/2)**2 + a**2*d) - 2*A*tan(c/2 + d*x/2)/(a**2*d*tan(c/2 + d*x/2)**4 + 2*a**2*d*tan(c/2 + d*x/2)**2 + a**2* d) - 4*B*log(tan(c/2 + d*x/2) + 1)*tan(c/2 + d*x/2)**4/(a**2*d*tan(c/2 + d *x/2)**4 + 2*a**2*d*tan(c/2 + d*x/2)**2 + a**2*d) - 8*B*log(tan(c/2 + d*x/ 2) + 1)*tan(c/2 + d*x/2)**2/(a**2*d*tan(c/2 + d*x/2)**4 + 2*a**2*d*tan(c/2 + d*x/2)**2 + a**2*d) - 4*B*log(tan(c/2 + d*x/2) + 1)/(a**2*d*tan(c/2 + d *x/2)**4 + 2*a**2*d*tan(c/2 + d*x/2)**2 + a**2*d) + 2*B*log(tan(c/2 + d*x/ 2)**2 + 1)*tan(c/2 + d*x/2)**4/(a**2*d*tan(c/2 + d*x/2)**4 + 2*a**2*d*tan( c/2 + d*x/2)**2 + a**2*d) + 4*B*log(tan(c/2 + d*x/2)**2 + 1)*tan(c/2 + d*x /2)**2/(a**2*d*tan(c/2 + d*x/2)**4 + 2*a**2*d*tan(c/2 + d*x/2)**2 + a**...
Time = 0.03 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.82 \[ \int \frac {\cos ^3(c+d x) (A+B \sin (c+d x))}{(a+a \sin (c+d x))^2} \, dx=\frac {\frac {4 \, {\left (A - B\right )} \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{2}} - \frac {B \sin \left (d x + c\right )^{2} + 2 \, {\left (A - 2 \, B\right )} \sin \left (d x + c\right )}{a^{2}}}{2 \, d} \] Input:
integrate(cos(d*x+c)^3*(A+B*sin(d*x+c))/(a+a*sin(d*x+c))^2,x, algorithm="m axima")
Output:
1/2*(4*(A - B)*log(sin(d*x + c) + 1)/a^2 - (B*sin(d*x + c)^2 + 2*(A - 2*B) *sin(d*x + c))/a^2)/d
Time = 0.14 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.11 \[ \int \frac {\cos ^3(c+d x) (A+B \sin (c+d x))}{(a+a \sin (c+d x))^2} \, dx=\frac {2 \, {\left (A - B\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a^{2} d} - \frac {B a^{2} d \sin \left (d x + c\right )^{2} + 2 \, A a^{2} d \sin \left (d x + c\right ) - 4 \, B a^{2} d \sin \left (d x + c\right )}{2 \, a^{4} d^{2}} \] Input:
integrate(cos(d*x+c)^3*(A+B*sin(d*x+c))/(a+a*sin(d*x+c))^2,x, algorithm="g iac")
Output:
2*(A - B)*log(abs(sin(d*x + c) + 1))/(a^2*d) - 1/2*(B*a^2*d*sin(d*x + c)^2 + 2*A*a^2*d*sin(d*x + c) - 4*B*a^2*d*sin(d*x + c))/(a^4*d^2)
Time = 0.08 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.92 \[ \int \frac {\cos ^3(c+d x) (A+B \sin (c+d x))}{(a+a \sin (c+d x))^2} \, dx=-\frac {2\,A\,\sin \left (c+d\,x\right )-4\,B\,\sin \left (c+d\,x\right )+B\,{\sin \left (c+d\,x\right )}^2-4\,A\,\ln \left (\sin \left (c+d\,x\right )+1\right )+4\,B\,\ln \left (\sin \left (c+d\,x\right )+1\right )}{2\,a^2\,d} \] Input:
int((cos(c + d*x)^3*(A + B*sin(c + d*x)))/(a + a*sin(c + d*x))^2,x)
Output:
-(2*A*sin(c + d*x) - 4*B*sin(c + d*x) + B*sin(c + d*x)^2 - 4*A*log(sin(c + d*x) + 1) + 4*B*log(sin(c + d*x) + 1))/(2*a^2*d)
Time = 0.18 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.59 \[ \int \frac {\cos ^3(c+d x) (A+B \sin (c+d x))}{(a+a \sin (c+d x))^2} \, dx=\frac {-4 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) a +4 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) b +8 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a -8 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) b -\sin \left (d x +c \right )^{2} b -2 \sin \left (d x +c \right ) a +4 \sin \left (d x +c \right ) b +2 b}{2 a^{2} d} \] Input:
int(cos(d*x+c)^3*(A+B*sin(d*x+c))/(a+a*sin(d*x+c))^2,x)
Output:
( - 4*log(tan((c + d*x)/2)**2 + 1)*a + 4*log(tan((c + d*x)/2)**2 + 1)*b + 8*log(tan((c + d*x)/2) + 1)*a - 8*log(tan((c + d*x)/2) + 1)*b - sin(c + d* x)**2*b - 2*sin(c + d*x)*a + 4*sin(c + d*x)*b + 2*b)/(2*a**2*d)