Integrand size = 27, antiderivative size = 148 \[ \int \frac {\cos ^5(c+d x) \sin (c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {a \left (a^2-b^2\right )^2 \log (a+b \sin (c+d x))}{b^6 d}+\frac {\left (a^2-b^2\right )^2 \sin (c+d x)}{b^5 d}-\frac {a \left (a^2-2 b^2\right ) \sin ^2(c+d x)}{2 b^4 d}+\frac {\left (a^2-2 b^2\right ) \sin ^3(c+d x)}{3 b^3 d}-\frac {a \sin ^4(c+d x)}{4 b^2 d}+\frac {\sin ^5(c+d x)}{5 b d} \] Output:
-a*(a^2-b^2)^2*ln(a+b*sin(d*x+c))/b^6/d+(a^2-b^2)^2*sin(d*x+c)/b^5/d-1/2*a *(a^2-2*b^2)*sin(d*x+c)^2/b^4/d+1/3*(a^2-2*b^2)*sin(d*x+c)^3/b^3/d-1/4*a*s in(d*x+c)^4/b^2/d+1/5*sin(d*x+c)^5/b/d
Time = 0.33 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.86 \[ \int \frac {\cos ^5(c+d x) \sin (c+d x)}{a+b \sin (c+d x)} \, dx=\frac {-60 a \left (a^2-b^2\right )^2 \log (a+b \sin (c+d x))+60 b \left (a^2-b^2\right )^2 \sin (c+d x)-30 a b^2 \left (a^2-2 b^2\right ) \sin ^2(c+d x)+20 b^3 \left (a^2-2 b^2\right ) \sin ^3(c+d x)-15 a b^4 \sin ^4(c+d x)+12 b^5 \sin ^5(c+d x)}{60 b^6 d} \] Input:
Integrate[(Cos[c + d*x]^5*Sin[c + d*x])/(a + b*Sin[c + d*x]),x]
Output:
(-60*a*(a^2 - b^2)^2*Log[a + b*Sin[c + d*x]] + 60*b*(a^2 - b^2)^2*Sin[c + d*x] - 30*a*b^2*(a^2 - 2*b^2)*Sin[c + d*x]^2 + 20*b^3*(a^2 - 2*b^2)*Sin[c + d*x]^3 - 15*a*b^4*Sin[c + d*x]^4 + 12*b^5*Sin[c + d*x]^5)/(60*b^6*d)
Time = 0.35 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.89, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {3042, 3316, 27, 522, 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.
| \(\displaystyle \int \frac {\sin (c+d x) \cos ^5(c+d x)}{a+b \sin (c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin (c+d x) \cos (c+d x)^5}{a+b \sin (c+d x)}dx\) |
| \(\Big \downarrow \) 3316 |
| \(\displaystyle \frac {\int \frac {\sin (c+d x) \left (b^2-b^2 \sin ^2(c+d x)\right )^2}{a+b \sin (c+d x)}d(b \sin (c+d x))}{b^5 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {b \sin (c+d x) \left (b^2-b^2 \sin ^2(c+d x)\right )^2}{a+b \sin (c+d x)}d(b \sin (c+d x))}{b^6 d}\) |
| \(\Big \downarrow \) 522 |
| \(\displaystyle \frac {\int \left (b^4 \sin ^4(c+d x)-a b^3 \sin ^3(c+d x)+b^2 \left (a^2-2 b^2\right ) \sin ^2(c+d x)-a b \left (a^2-2 b^2\right ) \sin (c+d x)+\left (a^2-b^2\right )^2-\frac {a \left (a^2-b^2\right )^2}{a+b \sin (c+d x)}\right )d(b \sin (c+d x))}{b^6 d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {-\frac {1}{2} a b^2 \left (a^2-2 b^2\right ) \sin ^2(c+d x)+b \left (a^2-b^2\right )^2 \sin (c+d x)-a \left (a^2-b^2\right )^2 \log (a+b \sin (c+d x))+\frac {1}{3} b^3 \left (a^2-2 b^2\right ) \sin ^3(c+d x)-\frac {1}{4} a b^4 \sin ^4(c+d x)+\frac {1}{5} b^5 \sin ^5(c+d x)}{b^6 d}\) |
Input:
Int[(Cos[c + d*x]^5*Sin[c + d*x])/(a + b*Sin[c + d*x]),x]
Output:
(-(a*(a^2 - b^2)^2*Log[a + b*Sin[c + d*x]]) + b*(a^2 - b^2)^2*Sin[c + d*x] - (a*b^2*(a^2 - 2*b^2)*Sin[c + d*x]^2)/2 + (b^3*(a^2 - 2*b^2)*Sin[c + d*x ]^3)/3 - (a*b^4*Sin[c + d*x]^4)/4 + (b^5*Sin[c + d*x]^5)/5)/(b^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[((e_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_)^2)^(p_. ), x_Symbol] :> Int[ExpandIntegrand[(e*x)^m*(c + d*x)^n*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && IGtQ[p, 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*(c + (d/b)*x)^n*(b^2 - x^2)^((p - 1)/2), x], x, b* Sin[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && IntegerQ[(p - 1) /2] && NeQ[a^2 - b^2, 0]
Time = 0.98 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.08
| method | result | size |
| derivativedivides | \(\frac {\frac {\frac {\sin \left (d x +c \right )^{5} b^{4}}{5}-\frac {a \sin \left (d x +c \right )^{4} b^{3}}{4}+\frac {a^{2} b^{2} \sin \left (d x +c \right )^{3}}{3}-\frac {2 b^{4} \sin \left (d x +c \right )^{3}}{3}-\frac {a^{3} b \sin \left (d x +c \right )^{2}}{2}+a \,b^{3} \sin \left (d x +c \right )^{2}+a^{4} \sin \left (d x +c \right )-2 \sin \left (d x +c \right ) a^{2} b^{2}+b^{4} \sin \left (d x +c \right )}{b^{5}}-\frac {a \left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) \ln \left (a +b \sin \left (d x +c \right )\right )}{b^{6}}}{d}\) | \(160\) |
| default | \(\frac {\frac {\frac {\sin \left (d x +c \right )^{5} b^{4}}{5}-\frac {a \sin \left (d x +c \right )^{4} b^{3}}{4}+\frac {a^{2} b^{2} \sin \left (d x +c \right )^{3}}{3}-\frac {2 b^{4} \sin \left (d x +c \right )^{3}}{3}-\frac {a^{3} b \sin \left (d x +c \right )^{2}}{2}+a \,b^{3} \sin \left (d x +c \right )^{2}+a^{4} \sin \left (d x +c \right )-2 \sin \left (d x +c \right ) a^{2} b^{2}+b^{4} \sin \left (d x +c \right )}{b^{5}}-\frac {a \left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) \ln \left (a +b \sin \left (d x +c \right )\right )}{b^{6}}}{d}\) | \(160\) |
| parallelrisch | \(\frac {-480 a \left (a -b \right )^{2} \left (a +b \right )^{2} \ln \left (2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )+480 a \left (a -b \right )^{2} \left (a +b \right )^{2} \ln \left (\sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )+\left (120 a^{3} b^{2}-180 a \,b^{4}\right ) \cos \left (2 d x +2 c \right )+\left (-40 a^{2} b^{3}+50 b^{5}\right ) \sin \left (3 d x +3 c \right )-15 b^{4} a \cos \left (4 d x +4 c \right )+6 b^{5} \sin \left (5 d x +5 c \right )+\left (480 a^{4} b -840 a^{2} b^{3}+300 b^{5}\right ) \sin \left (d x +c \right )-120 a^{3} b^{2}+195 a \,b^{4}}{480 b^{6} d}\) | \(198\) |
| risch | \(\frac {i a^{5} x}{b^{6}}+\frac {2 a^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}\right )}{b^{4} d}-\frac {3 a \,{\mathrm e}^{2 i \left (d x +c \right )}}{16 b^{2} d}-\frac {3 a \,{\mathrm e}^{-2 i \left (d x +c \right )}}{16 b^{2} d}-\frac {a \cos \left (4 d x +4 c \right )}{32 b^{2} d}-\frac {\sin \left (3 d x +3 c \right ) a^{2}}{12 b^{3} d}+\frac {i x a}{b^{2}}-\frac {5 i {\mathrm e}^{i \left (d x +c \right )}}{16 b d}+\frac {2 i a c}{b^{2} d}+\frac {2 i a^{5} c}{b^{6} d}-\frac {7 i {\mathrm e}^{-i \left (d x +c \right )} a^{2}}{8 b^{3} d}+\frac {i {\mathrm e}^{-i \left (d x +c \right )} a^{4}}{2 b^{5} d}-\frac {i {\mathrm e}^{i \left (d x +c \right )} a^{4}}{2 b^{5} d}+\frac {7 i {\mathrm e}^{i \left (d x +c \right )} a^{2}}{8 b^{3} d}-\frac {2 i a^{3} x}{b^{4}}+\frac {5 i {\mathrm e}^{-i \left (d x +c \right )}}{16 b d}-\frac {a^{5} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}\right )}{b^{6} d}-\frac {a \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}\right )}{b^{2} d}-\frac {4 i a^{3} c}{b^{4} d}+\frac {a^{3} {\mathrm e}^{2 i \left (d x +c \right )}}{8 b^{4} d}+\frac {a^{3} {\mathrm e}^{-2 i \left (d x +c \right )}}{8 b^{4} d}+\frac {\sin \left (5 d x +5 c \right )}{80 b d}+\frac {5 \sin \left (3 d x +3 c \right )}{48 b d}\) | \(450\) |
| norman | \(\frac {\frac {2 \left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{b^{5} d}+\frac {2 \left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{b^{5} d}+\frac {2 \left (15 a^{4}-26 a^{2} b^{2}+7 b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 b^{5} d}+\frac {2 \left (15 a^{4}-26 a^{2} b^{2}+7 b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{3 b^{5} d}+\frac {4 \left (25 a^{4}-40 a^{2} b^{2}+13 b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{5 b^{5} d}+\frac {4 \left (25 a^{4}-40 a^{2} b^{2}+13 b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{5 b^{5} d}-\frac {4 \left (3 a^{3}-4 a \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{d \,b^{4}}-\frac {\left (8 a^{3}-12 a \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{b^{4} d}-\frac {\left (8 a^{3}-12 a \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{b^{4} d}-\frac {2 \left (a^{3}-2 a \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{b^{4} d}-\frac {2 \left (a^{3}-2 a \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}}{b^{4} d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{6}}+\frac {a \left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) \ln \left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}{b^{6} d}-\frac {a \left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \right )}{b^{6} d}\) | \(477\) |
Input:
int(cos(d*x+c)^5*sin(d*x+c)/(a+b*sin(d*x+c)),x,method=_RETURNVERBOSE)
Output:
1/d*(1/b^5*(1/5*sin(d*x+c)^5*b^4-1/4*a*sin(d*x+c)^4*b^3+1/3*a^2*b^2*sin(d* x+c)^3-2/3*b^4*sin(d*x+c)^3-1/2*a^3*b*sin(d*x+c)^2+a*b^3*sin(d*x+c)^2+a^4* sin(d*x+c)-2*sin(d*x+c)*a^2*b^2+b^4*sin(d*x+c))-a*(a^4-2*a^2*b^2+b^4)/b^6* ln(a+b*sin(d*x+c)))
Time = 0.20 (sec) , antiderivative size = 142, normalized size of antiderivative = 0.96 \[ \int \frac {\cos ^5(c+d x) \sin (c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {15 \, a b^{4} \cos \left (d x + c\right )^{4} - 30 \, {\left (a^{3} b^{2} - a b^{4}\right )} \cos \left (d x + c\right )^{2} + 60 \, {\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} \log \left (b \sin \left (d x + c\right ) + a\right ) - 4 \, {\left (3 \, b^{5} \cos \left (d x + c\right )^{4} + 15 \, a^{4} b - 25 \, a^{2} b^{3} + 8 \, b^{5} - {\left (5 \, a^{2} b^{3} - 4 \, b^{5}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{60 \, b^{6} d} \] Input:
integrate(cos(d*x+c)^5*sin(d*x+c)/(a+b*sin(d*x+c)),x, algorithm="fricas")
Output:
-1/60*(15*a*b^4*cos(d*x + c)^4 - 30*(a^3*b^2 - a*b^4)*cos(d*x + c)^2 + 60* (a^5 - 2*a^3*b^2 + a*b^4)*log(b*sin(d*x + c) + a) - 4*(3*b^5*cos(d*x + c)^ 4 + 15*a^4*b - 25*a^2*b^3 + 8*b^5 - (5*a^2*b^3 - 4*b^5)*cos(d*x + c)^2)*si n(d*x + c))/(b^6*d)
Timed out. \[ \int \frac {\cos ^5(c+d x) \sin (c+d x)}{a+b \sin (c+d x)} \, dx=\text {Timed out} \] Input:
integrate(cos(d*x+c)**5*sin(d*x+c)/(a+b*sin(d*x+c)),x)
Output:
Timed out
Time = 0.03 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.94 \[ \int \frac {\cos ^5(c+d x) \sin (c+d x)}{a+b \sin (c+d x)} \, dx=\frac {\frac {12 \, b^{4} \sin \left (d x + c\right )^{5} - 15 \, a b^{3} \sin \left (d x + c\right )^{4} + 20 \, {\left (a^{2} b^{2} - 2 \, b^{4}\right )} \sin \left (d x + c\right )^{3} - 30 \, {\left (a^{3} b - 2 \, a b^{3}\right )} \sin \left (d x + c\right )^{2} + 60 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sin \left (d x + c\right )}{b^{5}} - \frac {60 \, {\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} \log \left (b \sin \left (d x + c\right ) + a\right )}{b^{6}}}{60 \, d} \] Input:
integrate(cos(d*x+c)^5*sin(d*x+c)/(a+b*sin(d*x+c)),x, algorithm="maxima")
Output:
1/60*((12*b^4*sin(d*x + c)^5 - 15*a*b^3*sin(d*x + c)^4 + 20*(a^2*b^2 - 2*b ^4)*sin(d*x + c)^3 - 30*(a^3*b - 2*a*b^3)*sin(d*x + c)^2 + 60*(a^4 - 2*a^2 *b^2 + b^4)*sin(d*x + c))/b^5 - 60*(a^5 - 2*a^3*b^2 + a*b^4)*log(b*sin(d*x + c) + a)/b^6)/d
Time = 0.24 (sec) , antiderivative size = 194, normalized size of antiderivative = 1.31 \[ \int \frac {\cos ^5(c+d x) \sin (c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {{\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{b^{6} d} + \frac {12 \, b^{4} d^{4} \sin \left (d x + c\right )^{5} - 15 \, a b^{3} d^{4} \sin \left (d x + c\right )^{4} + 20 \, a^{2} b^{2} d^{4} \sin \left (d x + c\right )^{3} - 40 \, b^{4} d^{4} \sin \left (d x + c\right )^{3} - 30 \, a^{3} b d^{4} \sin \left (d x + c\right )^{2} + 60 \, a b^{3} d^{4} \sin \left (d x + c\right )^{2} + 60 \, a^{4} d^{4} \sin \left (d x + c\right ) - 120 \, a^{2} b^{2} d^{4} \sin \left (d x + c\right ) + 60 \, b^{4} d^{4} \sin \left (d x + c\right )}{60 \, b^{5} d^{5}} \] Input:
integrate(cos(d*x+c)^5*sin(d*x+c)/(a+b*sin(d*x+c)),x, algorithm="giac")
Output:
-(a^5 - 2*a^3*b^2 + a*b^4)*log(abs(b*sin(d*x + c) + a))/(b^6*d) + 1/60*(12 *b^4*d^4*sin(d*x + c)^5 - 15*a*b^3*d^4*sin(d*x + c)^4 + 20*a^2*b^2*d^4*sin (d*x + c)^3 - 40*b^4*d^4*sin(d*x + c)^3 - 30*a^3*b*d^4*sin(d*x + c)^2 + 60 *a*b^3*d^4*sin(d*x + c)^2 + 60*a^4*d^4*sin(d*x + c) - 120*a^2*b^2*d^4*sin( d*x + c) + 60*b^4*d^4*sin(d*x + c))/(b^5*d^5)
Time = 0.04 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.01 \[ \int \frac {\cos ^5(c+d x) \sin (c+d x)}{a+b \sin (c+d x)} \, dx=\frac {\sin \left (c+d\,x\right )\,\left (\frac {1}{b}-\frac {a^2\,\left (\frac {2}{b}-\frac {a^2}{b^3}\right )}{b^2}\right )-{\sin \left (c+d\,x\right )}^3\,\left (\frac {2}{3\,b}-\frac {a^2}{3\,b^3}\right )+\frac {{\sin \left (c+d\,x\right )}^5}{5\,b}-\frac {\ln \left (a+b\,\sin \left (c+d\,x\right )\right )\,\left (a^5-2\,a^3\,b^2+a\,b^4\right )}{b^6}-\frac {a\,{\sin \left (c+d\,x\right )}^4}{4\,b^2}+\frac {a\,{\sin \left (c+d\,x\right )}^2\,\left (\frac {2}{b}-\frac {a^2}{b^3}\right )}{2\,b}}{d} \] Input:
int((cos(c + d*x)^5*sin(c + d*x))/(a + b*sin(c + d*x)),x)
Output:
(sin(c + d*x)*(1/b - (a^2*(2/b - a^2/b^3))/b^2) - sin(c + d*x)^3*(2/(3*b) - a^2/(3*b^3)) + sin(c + d*x)^5/(5*b) - (log(a + b*sin(c + d*x))*(a*b^4 + a^5 - 2*a^3*b^2))/b^6 - (a*sin(c + d*x)^4)/(4*b^2) + (a*sin(c + d*x)^2*(2/ b - a^2/b^3))/(2*b))/d
Time = 0.20 (sec) , antiderivative size = 310, normalized size of antiderivative = 2.09 \[ \int \frac {\cos ^5(c+d x) \sin (c+d x)}{a+b \sin (c+d x)} \, dx=\frac {60 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) a^{5}-120 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) a^{3} b^{2}+60 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) a \,b^{4}-60 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b +a \right ) a^{5}+120 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b +a \right ) a^{3} b^{2}-60 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b +a \right ) a \,b^{4}+12 \sin \left (d x +c \right )^{5} b^{5}-15 \sin \left (d x +c \right )^{4} a \,b^{4}+20 \sin \left (d x +c \right )^{3} a^{2} b^{3}-40 \sin \left (d x +c \right )^{3} b^{5}-30 \sin \left (d x +c \right )^{2} a^{3} b^{2}+60 \sin \left (d x +c \right )^{2} a \,b^{4}+60 \sin \left (d x +c \right ) a^{4} b -120 \sin \left (d x +c \right ) a^{2} b^{3}+60 \sin \left (d x +c \right ) b^{5}+24 a^{3} b^{2}-48 a \,b^{4}}{60 b^{6} d} \] Input:
int(cos(d*x+c)^5*sin(d*x+c)/(a+b*sin(d*x+c)),x)
Output:
(60*log(tan((c + d*x)/2)**2 + 1)*a**5 - 120*log(tan((c + d*x)/2)**2 + 1)*a **3*b**2 + 60*log(tan((c + d*x)/2)**2 + 1)*a*b**4 - 60*log(tan((c + d*x)/2 )**2*a + 2*tan((c + d*x)/2)*b + a)*a**5 + 120*log(tan((c + d*x)/2)**2*a + 2*tan((c + d*x)/2)*b + a)*a**3*b**2 - 60*log(tan((c + d*x)/2)**2*a + 2*tan ((c + d*x)/2)*b + a)*a*b**4 + 12*sin(c + d*x)**5*b**5 - 15*sin(c + d*x)**4 *a*b**4 + 20*sin(c + d*x)**3*a**2*b**3 - 40*sin(c + d*x)**3*b**5 - 30*sin( c + d*x)**2*a**3*b**2 + 60*sin(c + d*x)**2*a*b**4 + 60*sin(c + d*x)*a**4*b - 120*sin(c + d*x)*a**2*b**3 + 60*sin(c + d*x)*b**5 + 24*a**3*b**2 - 48*a *b**4)/(60*b**6*d)