\(\int \frac {\cos ^5(c+d x) \sin (c+d x)}{(a+b \sin (c+d x))^2} \, dx\) [1228]

Optimal result

Integrand size = 27, antiderivative size = 157 \[ \int \frac {\cos ^5(c+d x) \sin (c+d x)}{(a+b \sin (c+d x))^2} \, dx=\frac {\left (5 a^4-6 a^2 b^2+b^4\right ) \log (a+b \sin (c+d x))}{b^6 d}-\frac {4 a \left (a^2-b^2\right ) \sin (c+d x)}{b^5 d}+\frac {\left (3 a^2-2 b^2\right ) \sin ^2(c+d x)}{2 b^4 d}-\frac {2 a \sin ^3(c+d x)}{3 b^3 d}+\frac {\sin ^4(c+d x)}{4 b^2 d}+\frac {a \left (a^2-b^2\right )^2}{b^6 d (a+b \sin (c+d x))} \]

[Out]

(5*a^4-6*a^2*b^2+b^4)*ln(a+b*sin(d*x+c))/b^6/d-4*a*(a^2-b^2)*sin(d*x+c)/b^5/d+1/2*(3*a^2-2*b^2)*sin(d*x+c)^2/b
^4/d-2/3*a*sin(d*x+c)^3/b^3/d+1/4*sin(d*x+c)^4/b^2/d+a*(a^2-b^2)^2/b^6/d/(a+b*sin(d*x+c))

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {2916, 12, 786} \[ \int \frac {\cos ^5(c+d x) \sin (c+d x)}{(a+b \sin (c+d x))^2} \, dx=\frac {a \left (a^2-b^2\right )^2}{b^6 d (a+b \sin (c+d x))}-\frac {4 a \left (a^2-b^2\right ) \sin (c+d x)}{b^5 d}+\frac {\left (3 a^2-2 b^2\right ) \sin ^2(c+d x)}{2 b^4 d}+\frac {\left (5 a^4-6 a^2 b^2+b^4\right ) \log (a+b \sin (c+d x))}{b^6 d}-\frac {2 a \sin ^3(c+d x)}{3 b^3 d}+\frac {\sin ^4(c+d x)}{4 b^2 d} \]

[In]

Int[(Cos[c + d*x]^5*Sin[c + d*x])/(a + b*Sin[c + d*x])^2,x]

[Out]

((5*a^4 - 6*a^2*b^2 + b^4)*Log[a + b*Sin[c + d*x]])/(b^6*d) - (4*a*(a^2 - b^2)*Sin[c + d*x])/(b^5*d) + ((3*a^2
- 2*b^2)*Sin[c + d*x]^2)/(2*b^4*d) - (2*a*Sin[c + d*x]^3)/(3*b^3*d) + Sin[c + d*x]^4/(4*b^2*d) + (a*(a^2 - b^
2)^2)/(b^6*d*(a + b*Sin[c + d*x]))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 786

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegr
and[(d + e*x)^m*(f + g*x)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, m}, x] && IGtQ[p, 0]

Rule 2916

Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)
*(x_)])^(n_.), x_Symbol] :> Dist[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]

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {x \left (b^2-x^2\right )^2}{b (a+x)^2} \, dx,x,b \sin (c+d x)\right )}{b^5 d} \\ & = \frac {\text {Subst}\left (\int \frac {x \left (b^2-x^2\right )^2}{(a+x)^2} \, dx,x,b \sin (c+d x)\right )}{b^6 d} \\ & = \frac {\text {Subst}\left (\int \left (-4 \left (a^3-a b^2\right )+\left (3 a^2-2 b^2\right ) x-2 a x^2+x^3-\frac {a \left (a^2-b^2\right )^2}{(a+x)^2}+\frac {5 a^4-6 a^2 b^2+b^4}{a+x}\right ) \, dx,x,b \sin (c+d x)\right )}{b^6 d} \\ & = \frac {\left (5 a^4-6 a^2 b^2+b^4\right ) \log (a+b \sin (c+d x))}{b^6 d}-\frac {4 a \left (a^2-b^2\right ) \sin (c+d x)}{b^5 d}+\frac {\left (3 a^2-2 b^2\right ) \sin ^2(c+d x)}{2 b^4 d}-\frac {2 a \sin ^3(c+d x)}{3 b^3 d}+\frac {\sin ^4(c+d x)}{4 b^2 d}+\frac {a \left (a^2-b^2\right )^2}{b^6 d (a+b \sin (c+d x))} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.51 (sec) , antiderivative size = 188, normalized size of antiderivative = 1.20 \[ \int \frac {\cos ^5(c+d x) \sin (c+d x)}{(a+b \sin (c+d x))^2} \, dx=\frac {12 a \left (a^2-b^2\right ) \left (a^2-b^2+\left (5 a^2-b^2\right ) \log (a+b \sin (c+d x))\right )+12 b \left (-a^2+b^2\right ) \left (4 a^2+\left (-5 a^2+b^2\right ) \log (a+b \sin (c+d x))\right ) \sin (c+d x)-6 a b^2 \left (5 a^2-6 b^2\right ) \sin ^2(c+d x)+2 b^3 \left (5 a^2-6 b^2\right ) \sin ^3(c+d x)-5 a b^4 \sin ^4(c+d x)+3 b^5 \sin ^5(c+d x)}{12 b^6 d (a+b \sin (c+d x))} \]

[In]

Integrate[(Cos[c + d*x]^5*Sin[c + d*x])/(a + b*Sin[c + d*x])^2,x]

[Out]

(12*a*(a^2 - b^2)*(a^2 - b^2 + (5*a^2 - b^2)*Log[a + b*Sin[c + d*x]]) + 12*b*(-a^2 + b^2)*(4*a^2 + (-5*a^2 + b
^2)*Log[a + b*Sin[c + d*x]])*Sin[c + d*x] - 6*a*b^2*(5*a^2 - 6*b^2)*Sin[c + d*x]^2 + 2*b^3*(5*a^2 - 6*b^2)*Sin
[c + d*x]^3 - 5*a*b^4*Sin[c + d*x]^4 + 3*b^5*Sin[c + d*x]^5)/(12*b^6*d*(a + b*Sin[c + d*x]))

Maple [A] (verified)

Time = 0.88 (sec) , antiderivative size = 152, normalized size of antiderivative = 0.97

method	result	size
derivativedivides	\(\frac {-\frac {-\frac {\left (\sin ^{4}\left (d x +c \right )\right ) b^{3}}{4}+\frac {2 a \left (\sin ^{3}\left (d x +c \right )\right ) b^{2}}{3}-\frac {3 a^{2} b \left (\sin ^{2}\left (d x +c \right )\right )}{2}+b^{3} \left (\sin ^{2}\left (d x +c \right )\right )+4 a^{3} \sin \left (d x +c \right )-4 \sin \left (d x +c \right ) a \,b^{2}}{b^{5}}+\frac {\left (5 a^{4}-6 a^{2} b^{2}+b^{4}\right ) \ln \left (a +b \sin \left (d x +c \right )\right )}{b^{6}}+\frac {a \left (a^{4}-2 a^{2} b^{2}+b^{4}\right )}{b^{6} \left (a +b \sin \left (d x +c \right )\right )}}{d}\)	\(152\)
default	\(\frac {-\frac {-\frac {\left (\sin ^{4}\left (d x +c \right )\right ) b^{3}}{4}+\frac {2 a \left (\sin ^{3}\left (d x +c \right )\right ) b^{2}}{3}-\frac {3 a^{2} b \left (\sin ^{2}\left (d x +c \right )\right )}{2}+b^{3} \left (\sin ^{2}\left (d x +c \right )\right )+4 a^{3} \sin \left (d x +c \right )-4 \sin \left (d x +c \right ) a \,b^{2}}{b^{5}}+\frac {\left (5 a^{4}-6 a^{2} b^{2}+b^{4}\right ) \ln \left (a +b \sin \left (d x +c \right )\right )}{b^{6}}+\frac {a \left (a^{4}-2 a^{2} b^{2}+b^{4}\right )}{b^{6} \left (a +b \sin \left (d x +c \right )\right )}}{d}\)	\(152\)
parallelrisch	\(\frac {960 \left (a +b \right ) \left (a +b \sin \left (d x +c \right )\right ) \left (a -b \right ) \left (a^{2}-\frac {b^{2}}{5}\right ) \ln \left (2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )-960 \left (a +b \right ) \left (a +b \sin \left (d x +c \right )\right ) \left (a -b \right ) \left (a^{2}-\frac {b^{2}}{5}\right ) \ln \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (240 a^{3} b^{2}-248 a \,b^{4}\right ) \cos \left (2 d x +2 c \right )+\left (-40 a^{2} b^{3}+33 b^{5}\right ) \sin \left (3 d x +3 c \right )-10 b^{4} \cos \left (4 d x +4 c \right ) a +3 b^{5} \sin \left (5 d x +5 c \right )+\left (-960 a^{4} b +1272 a^{2} b^{3}-306 b^{5}\right ) \sin \left (d x +c \right )-240 a^{3} b^{2}+258 a \,b^{4}}{192 b^{6} d \left (a +b \sin \left (d x +c \right )\right )}\)	\(238\)
risch	\(-\frac {i x}{b^{2}}-\frac {7 i a \,{\mathrm e}^{i \left (d x +c \right )}}{4 b^{3} d}-\frac {10 i a^{4} c}{b^{6} d}-\frac {3 \,{\mathrm e}^{2 i \left (d x +c \right )} a^{2}}{8 b^{4} d}+\frac {3 \,{\mathrm e}^{2 i \left (d x +c \right )}}{16 b^{2} d}+\frac {12 i a^{2} c}{b^{4} d}-\frac {2 i c}{b^{2} d}+\frac {6 i a^{2} x}{b^{4}}-\frac {5 i a^{4} x}{b^{6}}-\frac {3 \,{\mathrm e}^{-2 i \left (d x +c \right )} a^{2}}{8 b^{4} d}+\frac {3 \,{\mathrm e}^{-2 i \left (d x +c \right )}}{16 b^{2} d}+\frac {2 i a^{3} {\mathrm e}^{i \left (d x +c \right )}}{b^{5} d}+\frac {7 i a \,{\mathrm e}^{-i \left (d x +c \right )}}{4 b^{3} d}-\frac {2 i a^{3} {\mathrm e}^{-i \left (d x +c \right )}}{b^{5} d}+\frac {2 a \left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) {\mathrm e}^{i \left (d x +c \right )}}{d \,b^{6} \left (-i b \,{\mathrm e}^{2 i \left (d x +c \right )}+i b +2 a \,{\mathrm e}^{i \left (d x +c \right )}\right )}+\frac {5 a^{4} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}\right )}{b^{6} d}-\frac {6 a^{2} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}\right )}{b^{4} d}+\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}\right )}{b^{2} d}+\frac {\cos \left (4 d x +4 c \right )}{32 d \,b^{2}}+\frac {a \sin \left (3 d x +3 c \right )}{6 b^{3} d}\)	\(432\)
norman	\(\frac {-\frac {\left (150 a^{3}-160 a \,b^{2}\right ) \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 b^{4} d}-\frac {\left (150 a^{3}-160 a \,b^{2}\right ) \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 b^{4} d}-\frac {\left (10 a^{3}-12 a \,b^{2}\right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{4} d}-\frac {\left (10 a^{3}-12 a \,b^{2}\right ) \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{4} d}-\frac {5 \left (20 a^{3}-20 a \,b^{2}\right ) \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{4} d}-\frac {5 \left (20 a^{3}-20 a \,b^{2}\right ) \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{4} d}-\frac {4 \left (45 a^{5}-59 a^{3} b^{2}+15 a \,b^{4}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 a \,b^{5} d}-\frac {4 \left (45 a^{5}-59 a^{3} b^{2}+15 a \,b^{4}\right ) \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 a \,b^{5} d}-\frac {2 \left (225 a^{5}-310 a^{3} b^{2}+81 a \,b^{4}\right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d \,b^{5} a}-\frac {2 \left (225 a^{5}-310 a^{3} b^{2}+81 a \,b^{4}\right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d \,b^{5} a}-\frac {8 \left (25 a^{5}-35 a^{3} b^{2}+9 a \,b^{4}\right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,b^{5} a}-\frac {2 \left (5 a^{5}-6 a^{3} b^{2}+a \,b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a \,b^{5} d}-\frac {2 \left (5 a^{5}-6 a^{3} b^{2}+a \,b^{4}\right ) \left (\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a \,b^{5} d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{6} \left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \right )}+\frac {\left (5 a^{4}-6 a^{2} b^{2}+b^{4}\right ) \ln \left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \right )}{d \,b^{6}}-\frac {\left (5 a^{4}-6 a^{2} b^{2}+b^{4}\right ) \ln \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,b^{6}}\)	\(615\)

[In]

int(cos(d*x+c)^5*sin(d*x+c)/(a+b*sin(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

1/d*(-1/b^5*(-1/4*sin(d*x+c)^4*b^3+2/3*a*sin(d*x+c)^3*b^2-3/2*a^2*b*sin(d*x+c)^2+b^3*sin(d*x+c)^2+4*a^3*sin(d*
x+c)-4*sin(d*x+c)*a*b^2)+1/b^6*(5*a^4-6*a^2*b^2+b^4)*ln(a+b*sin(d*x+c))+a*(a^4-2*a^2*b^2+b^4)/b^6/(a+b*sin(d*x
+c)))

Fricas [A] (verification not implemented)

none

Time = 0.37 (sec) , antiderivative size = 203, normalized size of antiderivative = 1.29 \[ \int \frac {\cos ^5(c+d x) \sin (c+d x)}{(a+b \sin (c+d x))^2} \, dx=-\frac {40 \, a b^{4} \cos \left (d x + c\right )^{4} - 96 \, a^{5} + 504 \, a^{3} b^{2} - 383 \, a b^{4} - 16 \, {\left (15 \, a^{3} b^{2} - 13 \, a b^{4}\right )} \cos \left (d x + c\right )^{2} - 96 \, {\left (5 \, a^{5} - 6 \, a^{3} b^{2} + a b^{4} + {\left (5 \, a^{4} b - 6 \, a^{2} b^{3} + b^{5}\right )} \sin \left (d x + c\right )\right )} \log \left (b \sin \left (d x + c\right ) + a\right ) - {\left (24 \, b^{5} \cos \left (d x + c\right )^{4} - 384 \, a^{4} b + 392 \, a^{2} b^{3} - 33 \, b^{5} - 16 \, {\left (5 \, a^{2} b^{3} - 3 \, b^{5}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{96 \, {\left (b^{7} d \sin \left (d x + c\right ) + a b^{6} d\right )}} \]

[In]

integrate(cos(d*x+c)^5*sin(d*x+c)/(a+b*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

-1/96*(40*a*b^4*cos(d*x + c)^4 - 96*a^5 + 504*a^3*b^2 - 383*a*b^4 - 16*(15*a^3*b^2 - 13*a*b^4)*cos(d*x + c)^2
- 96*(5*a^5 - 6*a^3*b^2 + a*b^4 + (5*a^4*b - 6*a^2*b^3 + b^5)*sin(d*x + c))*log(b*sin(d*x + c) + a) - (24*b^5*
cos(d*x + c)^4 - 384*a^4*b + 392*a^2*b^3 - 33*b^5 - 16*(5*a^2*b^3 - 3*b^5)*cos(d*x + c)^2)*sin(d*x + c))/(b^7*
d*sin(d*x + c) + a*b^6*d)

Sympy [F(-1)]

Timed out. \[ \int \frac {\cos ^5(c+d x) \sin (c+d x)}{(a+b \sin (c+d x))^2} \, dx=\text {Timed out} \]

[In]

integrate(cos(d*x+c)**5*sin(d*x+c)/(a+b*sin(d*x+c))**2,x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 148, normalized size of antiderivative = 0.94 \[ \int \frac {\cos ^5(c+d x) \sin (c+d x)}{(a+b \sin (c+d x))^2} \, dx=\frac {\frac {12 \, {\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )}}{b^{7} \sin \left (d x + c\right ) + a b^{6}} + \frac {3 \, b^{3} \sin \left (d x + c\right )^{4} - 8 \, a b^{2} \sin \left (d x + c\right )^{3} + 6 \, {\left (3 \, a^{2} b - 2 \, b^{3}\right )} \sin \left (d x + c\right )^{2} - 48 \, {\left (a^{3} - a b^{2}\right )} \sin \left (d x + c\right )}{b^{5}} + \frac {12 \, {\left (5 \, a^{4} - 6 \, a^{2} b^{2} + b^{4}\right )} \log \left (b \sin \left (d x + c\right ) + a\right )}{b^{6}}}{12 \, d} \]

[In]

integrate(cos(d*x+c)^5*sin(d*x+c)/(a+b*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

1/12*(12*(a^5 - 2*a^3*b^2 + a*b^4)/(b^7*sin(d*x + c) + a*b^6) + (3*b^3*sin(d*x + c)^4 - 8*a*b^2*sin(d*x + c)^3
+ 6*(3*a^2*b - 2*b^3)*sin(d*x + c)^2 - 48*(a^3 - a*b^2)*sin(d*x + c))/b^5 + 12*(5*a^4 - 6*a^2*b^2 + b^4)*log(
b*sin(d*x + c) + a)/b^6)/d

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 194, normalized size of antiderivative = 1.24 \[ \int \frac {\cos ^5(c+d x) \sin (c+d x)}{(a+b \sin (c+d x))^2} \, dx=\frac {\frac {12 \, {\left (5 \, a^{4} - 6 \, a^{2} b^{2} + b^{4}\right )} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{b^{6}} - \frac {12 \, {\left (5 \, a^{4} b \sin \left (d x + c\right ) - 6 \, a^{2} b^{3} \sin \left (d x + c\right ) + b^{5} \sin \left (d x + c\right ) + 4 \, a^{5} - 4 \, a^{3} b^{2}\right )}}{{\left (b \sin \left (d x + c\right ) + a\right )} b^{6}} + \frac {3 \, b^{6} \sin \left (d x + c\right )^{4} - 8 \, a b^{5} \sin \left (d x + c\right )^{3} + 18 \, a^{2} b^{4} \sin \left (d x + c\right )^{2} - 12 \, b^{6} \sin \left (d x + c\right )^{2} - 48 \, a^{3} b^{3} \sin \left (d x + c\right ) + 48 \, a b^{5} \sin \left (d x + c\right )}{b^{8}}}{12 \, d} \]

[In]

integrate(cos(d*x+c)^5*sin(d*x+c)/(a+b*sin(d*x+c))^2,x, algorithm="giac")

[Out]

1/12*(12*(5*a^4 - 6*a^2*b^2 + b^4)*log(abs(b*sin(d*x + c) + a))/b^6 - 12*(5*a^4*b*sin(d*x + c) - 6*a^2*b^3*sin
(d*x + c) + b^5*sin(d*x + c) + 4*a^5 - 4*a^3*b^2)/((b*sin(d*x + c) + a)*b^6) + (3*b^6*sin(d*x + c)^4 - 8*a*b^5
*sin(d*x + c)^3 + 18*a^2*b^4*sin(d*x + c)^2 - 12*b^6*sin(d*x + c)^2 - 48*a^3*b^3*sin(d*x + c) + 48*a*b^5*sin(d
*x + c))/b^8)/d

Mupad [B] (verification not implemented)

Time = 12.96 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.03 \[ \int \frac {\cos ^5(c+d x) \sin (c+d x)}{(a+b \sin (c+d x))^2} \, dx=\frac {\frac {{\sin \left (c+d\,x\right )}^4}{4\,b^2}-{\sin \left (c+d\,x\right )}^2\,\left (\frac {1}{b^2}-\frac {3\,a^2}{2\,b^4}\right )+\sin \left (c+d\,x\right )\,\left (\frac {2\,a^3}{b^5}+\frac {2\,a\,\left (\frac {2}{b^2}-\frac {3\,a^2}{b^4}\right )}{b}\right )-\frac {2\,a\,{\sin \left (c+d\,x\right )}^3}{3\,b^3}+\frac {a^5-2\,a^3\,b^2+a\,b^4}{b\,\left (\sin \left (c+d\,x\right )\,b^6+a\,b^5\right )}+\frac {\ln \left (a+b\,\sin \left (c+d\,x\right )\right )\,\left (5\,a^4-6\,a^2\,b^2+b^4\right )}{b^6}}{d} \]

[In]

int((cos(c + d*x)^5*sin(c + d*x))/(a + b*sin(c + d*x))^2,x)

[Out]

(sin(c + d*x)^4/(4*b^2) - sin(c + d*x)^2*(1/b^2 - (3*a^2)/(2*b^4)) + sin(c + d*x)*((2*a^3)/b^5 + (2*a*(2/b^2 -
(3*a^2)/b^4))/b) - (2*a*sin(c + d*x)^3)/(3*b^3) + (a*b^4 + a^5 - 2*a^3*b^2)/(b*(a*b^5 + b^6*sin(c + d*x))) +
(log(a + b*sin(c + d*x))*(5*a^4 + b^4 - 6*a^2*b^2))/b^6)/d