Integrand size = 27, antiderivative size = 68 \[ \int \frac {\cos ^5(c+d x) \sin (c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {4 \log (1+\sin (c+d x))}{a^3 d}+\frac {4 \sin (c+d x)}{a^3 d}-\frac {3 \sin ^2(c+d x)}{2 a^3 d}+\frac {\sin ^3(c+d x)}{3 a^3 d} \] Output:
-4*ln(1+sin(d*x+c))/a^3/d+4*sin(d*x+c)/a^3/d-3/2*sin(d*x+c)^2/a^3/d+1/3*si n(d*x+c)^3/a^3/d
Time = 0.56 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.75 \[ \int \frac {\cos ^5(c+d x) \sin (c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {15-384 \log (1+\sin (c+d x))+384 \sin (c+d x)-144 \sin ^2(c+d x)+32 \sin ^3(c+d x)}{96 a^3 d} \] Input:
Integrate[(Cos[c + d*x]^5*Sin[c + d*x])/(a + a*Sin[c + d*x])^3,x]
Output:
(15 - 384*Log[1 + Sin[c + d*x]] + 384*Sin[c + d*x] - 144*Sin[c + d*x]^2 + 32*Sin[c + d*x]^3)/(96*a^3*d)
Time = 0.29 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.96, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, 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.
\(\displaystyle \int \frac {\sin (c+d x) \cos ^5(c+d x)}{(a \sin (c+d x)+a)^3} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sin (c+d x) \cos (c+d x)^5}{(a \sin (c+d x)+a)^3}dx\) |
\(\Big \downarrow \) 3315 |
\(\displaystyle \frac {\int \frac {\sin (c+d x) (a-a \sin (c+d x))^2}{\sin (c+d x) a+a}d(a \sin (c+d x))}{a^5 d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {a \sin (c+d x) (a-a \sin (c+d x))^2}{\sin (c+d x) a+a}d(a \sin (c+d x))}{a^6 d}\) |
\(\Big \downarrow \) 86 |
\(\displaystyle \frac {\int \left (-\frac {4 a^3}{\sin (c+d x) a+a}+\sin ^2(c+d x) a^2-3 \sin (c+d x) a^2+4 a^2\right )d(a \sin (c+d x))}{a^6 d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {1}{3} a^3 \sin ^3(c+d x)-\frac {3}{2} a^3 \sin ^2(c+d x)+4 a^3 \sin (c+d x)-4 a^3 \log (a \sin (c+d x)+a)}{a^6 d}\) |
Input:
Int[(Cos[c + d*x]^5*Sin[c + d*x])/(a + a*Sin[c + d*x])^3,x]
Output:
(-4*a^3*Log[a + a*Sin[c + d*x]] + 4*a^3*Sin[c + d*x] - (3*a^3*Sin[c + d*x] ^2)/2 + (a^3*Sin[c + d*x]^3)/3)/(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 = 0.67 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.71
method | result | size |
derivativedivides | \(\frac {\frac {\sin \left (d x +c \right )^{3}}{3}-\frac {3 \sin \left (d x +c \right )^{2}}{2}+4 \sin \left (d x +c \right )-4 \ln \left (1+\sin \left (d x +c \right )\right )}{d \,a^{3}}\) | \(48\) |
default | \(\frac {\frac {\sin \left (d x +c \right )^{3}}{3}-\frac {3 \sin \left (d x +c \right )^{2}}{2}+4 \sin \left (d x +c \right )-4 \ln \left (1+\sin \left (d x +c \right )\right )}{d \,a^{3}}\) | \(48\) |
parallelrisch | \(\frac {-96 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+48 \ln \left (\sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )-9-\sin \left (3 d x +3 c \right )+51 \sin \left (d x +c \right )+9 \cos \left (2 d x +2 c \right )}{12 d \,a^{3}}\) | \(69\) |
risch | \(\frac {4 i x}{a^{3}}-\frac {17 i {\mathrm e}^{i \left (d x +c \right )}}{8 d \,a^{3}}+\frac {17 i {\mathrm e}^{-i \left (d x +c \right )}}{8 d \,a^{3}}+\frac {8 i c}{d \,a^{3}}-\frac {8 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d \,a^{3}}-\frac {\sin \left (3 d x +3 c \right )}{12 d \,a^{3}}+\frac {3 \cos \left (2 d x +2 c \right )}{4 d \,a^{3}}\) | \(110\) |
norman | \(\frac {\frac {8 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d a}+\frac {8 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{16}}{d a}+\frac {34 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{d a}+\frac {34 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{15}}{d a}+\frac {278 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 d a}+\frac {278 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{14}}{3 d a}+\frac {628 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{3 d a}+\frac {628 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{13}}{3 d a}+\frac {1124 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{3 d a}+\frac {1124 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}}{3 d a}+\frac {1706 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{3 d a}+\frac {1706 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{3 d a}+\frac {2266 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{3 d a}+\frac {2266 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}}{3 d a}+\frac {2576 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{3 d a}+\frac {2576 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{3 d a}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{6} a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}-\frac {8 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d \,a^{3}}+\frac {4 \ln \left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}{d \,a^{3}}\) | \(379\) |
Input:
int(cos(d*x+c)^5*sin(d*x+c)/(a+a*sin(d*x+c))^3,x,method=_RETURNVERBOSE)
Output:
1/d/a^3*(1/3*sin(d*x+c)^3-3/2*sin(d*x+c)^2+4*sin(d*x+c)-4*ln(1+sin(d*x+c)) )
Time = 0.08 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.71 \[ \int \frac {\cos ^5(c+d x) \sin (c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {9 \, \cos \left (d x + c\right )^{2} - 2 \, {\left (\cos \left (d x + c\right )^{2} - 13\right )} \sin \left (d x + c\right ) - 24 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{6 \, a^{3} d} \] Input:
integrate(cos(d*x+c)^5*sin(d*x+c)/(a+a*sin(d*x+c))^3,x, algorithm="fricas" )
Output:
1/6*(9*cos(d*x + c)^2 - 2*(cos(d*x + c)^2 - 13)*sin(d*x + c) - 24*log(sin( d*x + c) + 1))/(a^3*d)
Leaf count of result is larger than twice the leaf count of optimal. 1102 vs. \(2 (61) = 122\).
Time = 38.75 (sec) , antiderivative size = 1102, normalized size of antiderivative = 16.21 \[ \int \frac {\cos ^5(c+d x) \sin (c+d x)}{(a+a \sin (c+d x))^3} \, dx=\text {Too large to display} \] Input:
integrate(cos(d*x+c)**5*sin(d*x+c)/(a+a*sin(d*x+c))**3,x)
Output:
Piecewise((-24*log(tan(c/2 + d*x/2) + 1)*tan(c/2 + d*x/2)**6/(3*a**3*d*tan (c/2 + d*x/2)**6 + 9*a**3*d*tan(c/2 + d*x/2)**4 + 9*a**3*d*tan(c/2 + d*x/2 )**2 + 3*a**3*d) - 72*log(tan(c/2 + d*x/2) + 1)*tan(c/2 + d*x/2)**4/(3*a** 3*d*tan(c/2 + d*x/2)**6 + 9*a**3*d*tan(c/2 + d*x/2)**4 + 9*a**3*d*tan(c/2 + d*x/2)**2 + 3*a**3*d) - 72*log(tan(c/2 + d*x/2) + 1)*tan(c/2 + d*x/2)**2 /(3*a**3*d*tan(c/2 + d*x/2)**6 + 9*a**3*d*tan(c/2 + d*x/2)**4 + 9*a**3*d*t an(c/2 + d*x/2)**2 + 3*a**3*d) - 24*log(tan(c/2 + d*x/2) + 1)/(3*a**3*d*ta n(c/2 + d*x/2)**6 + 9*a**3*d*tan(c/2 + d*x/2)**4 + 9*a**3*d*tan(c/2 + d*x/ 2)**2 + 3*a**3*d) + 12*log(tan(c/2 + d*x/2)**2 + 1)*tan(c/2 + d*x/2)**6/(3 *a**3*d*tan(c/2 + d*x/2)**6 + 9*a**3*d*tan(c/2 + d*x/2)**4 + 9*a**3*d*tan( c/2 + d*x/2)**2 + 3*a**3*d) + 36*log(tan(c/2 + d*x/2)**2 + 1)*tan(c/2 + d* x/2)**4/(3*a**3*d*tan(c/2 + d*x/2)**6 + 9*a**3*d*tan(c/2 + d*x/2)**4 + 9*a **3*d*tan(c/2 + d*x/2)**2 + 3*a**3*d) + 36*log(tan(c/2 + d*x/2)**2 + 1)*ta n(c/2 + d*x/2)**2/(3*a**3*d*tan(c/2 + d*x/2)**6 + 9*a**3*d*tan(c/2 + d*x/2 )**4 + 9*a**3*d*tan(c/2 + d*x/2)**2 + 3*a**3*d) + 12*log(tan(c/2 + d*x/2)* *2 + 1)/(3*a**3*d*tan(c/2 + d*x/2)**6 + 9*a**3*d*tan(c/2 + d*x/2)**4 + 9*a **3*d*tan(c/2 + d*x/2)**2 + 3*a**3*d) + 24*tan(c/2 + d*x/2)**5/(3*a**3*d*t an(c/2 + d*x/2)**6 + 9*a**3*d*tan(c/2 + d*x/2)**4 + 9*a**3*d*tan(c/2 + d*x /2)**2 + 3*a**3*d) - 18*tan(c/2 + d*x/2)**4/(3*a**3*d*tan(c/2 + d*x/2)**6 + 9*a**3*d*tan(c/2 + d*x/2)**4 + 9*a**3*d*tan(c/2 + d*x/2)**2 + 3*a**3*...
Time = 0.03 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.78 \[ \int \frac {\cos ^5(c+d x) \sin (c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {\frac {2 \, \sin \left (d x + c\right )^{3} - 9 \, \sin \left (d x + c\right )^{2} + 24 \, \sin \left (d x + c\right )}{a^{3}} - \frac {24 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{3}}}{6 \, d} \] Input:
integrate(cos(d*x+c)^5*sin(d*x+c)/(a+a*sin(d*x+c))^3,x, algorithm="maxima" )
Output:
1/6*((2*sin(d*x + c)^3 - 9*sin(d*x + c)^2 + 24*sin(d*x + c))/a^3 - 24*log( sin(d*x + c) + 1)/a^3)/d
Time = 0.14 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.09 \[ \int \frac {\cos ^5(c+d x) \sin (c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {4 \, \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a^{3} d} + \frac {2 \, a^{6} d^{2} \sin \left (d x + c\right )^{3} - 9 \, a^{6} d^{2} \sin \left (d x + c\right )^{2} + 24 \, a^{6} d^{2} \sin \left (d x + c\right )}{6 \, a^{9} d^{3}} \] Input:
integrate(cos(d*x+c)^5*sin(d*x+c)/(a+a*sin(d*x+c))^3,x, algorithm="giac")
Output:
-4*log(abs(sin(d*x + c) + 1))/(a^3*d) + 1/6*(2*a^6*d^2*sin(d*x + c)^3 - 9* a^6*d^2*sin(d*x + c)^2 + 24*a^6*d^2*sin(d*x + c))/(a^9*d^3)
Time = 0.06 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.84 \[ \int \frac {\cos ^5(c+d x) \sin (c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {\frac {4\,\ln \left (\sin \left (c+d\,x\right )+1\right )}{a^3}-\frac {4\,\sin \left (c+d\,x\right )}{a^3}+\frac {3\,{\sin \left (c+d\,x\right )}^2}{2\,a^3}-\frac {{\sin \left (c+d\,x\right )}^3}{3\,a^3}}{d} \] Input:
int((cos(c + d*x)^5*sin(c + d*x))/(a + a*sin(c + d*x))^3,x)
Output:
-((4*log(sin(c + d*x) + 1))/a^3 - (4*sin(c + d*x))/a^3 + (3*sin(c + d*x)^2 )/(2*a^3) - sin(c + d*x)^3/(3*a^3))/d
Time = 0.15 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.00 \[ \int \frac {\cos ^5(c+d x) \sin (c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {24 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right )-48 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+2 \sin \left (d x +c \right )^{3}-9 \sin \left (d x +c \right )^{2}+24 \sin \left (d x +c \right )+12}{6 a^{3} d} \] Input:
int(cos(d*x+c)^5*sin(d*x+c)/(a+a*sin(d*x+c))^3,x)
Output:
(24*log(tan((c + d*x)/2)**2 + 1) - 48*log(tan((c + d*x)/2) + 1) + 2*sin(c + d*x)**3 - 9*sin(c + d*x)**2 + 24*sin(c + d*x) + 12)/(6*a**3*d)