\(\int \frac {\cos ^5(c+d x) \sin ^2(c+d x)}{a+b \sin (c+d x)} \, dx\) [1311]

Optimal result

Integrand size = 29, antiderivative size = 180 \[ \int \frac {\cos ^5(c+d x) \sin ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {a^2 \left (a^2-b^2\right )^2 \log (a+b \sin (c+d x))}{b^7 d}-\frac {a \left (a^2-b^2\right )^2 \sin (c+d x)}{b^6 d}+\frac {\left (a^2-b^2\right )^2 \sin ^2(c+d x)}{2 b^5 d}-\frac {a \left (a^2-2 b^2\right ) \sin ^3(c+d x)}{3 b^4 d}+\frac {\left (a^2-2 b^2\right ) \sin ^4(c+d x)}{4 b^3 d}-\frac {a \sin ^5(c+d x)}{5 b^2 d}+\frac {\sin ^6(c+d x)}{6 b d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.28 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {2916, 12, 962} \[ \int \frac {\cos ^5(c+d x) \sin ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {a^2 \left (a^2-b^2\right )^2 \log (a+b \sin (c+d x))}{b^7 d}-\frac {a \left (a^2-b^2\right )^2 \sin (c+d x)}{b^6 d}+\frac {\left (a^2-b^2\right )^2 \sin ^2(c+d x)}{2 b^5 d}-\frac {a \left (a^2-2 b^2\right ) \sin ^3(c+d x)}{3 b^4 d}+\frac {\left (a^2-2 b^2\right ) \sin ^4(c+d x)}{4 b^3 d}-\frac {a \sin ^5(c+d x)}{5 b^2 d}+\frac {\sin ^6(c+d x)}{6 b d} \]

[In]

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

[Out]

(a^2*(a^2 - b^2)^2*Log[a + b*Sin[c + d*x]])/(b^7*d) - (a*(a^2 - b^2)^2*Sin[c + d*x])/(b^6*d) + ((a^2 - b^2)^2*
Sin[c + d*x]^2)/(2*b^5*d) - (a*(a^2 - 2*b^2)*Sin[c + d*x]^3)/(3*b^4*d) + ((a^2 - 2*b^2)*Sin[c + d*x]^4)/(4*b^3
*d) - (a*Sin[c + d*x]^5)/(5*b^2*d) + Sin[c + d*x]^6/(6*b*d)

Rule 12

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

Rule 962

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIn
tegrand[(d + e*x)^m*(f + g*x)^n*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[e*f - d*g, 0] &&
NeQ[c*d^2 + a*e^2, 0] && IGtQ[p, 0] && (IGtQ[m, 0] || (EqQ[m, -2] && EqQ[p, 1] && EqQ[d, 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^2 \left (b^2-x^2\right )^2}{b^2 (a+x)} \, dx,x,b \sin (c+d x)\right )}{b^5 d} \\ & = \frac {\text {Subst}\left (\int \frac {x^2 \left (b^2-x^2\right )^2}{a+x} \, dx,x,b \sin (c+d x)\right )}{b^7 d} \\ & = \frac {\text {Subst}\left (\int \left (-a \left (a^2-b^2\right )^2+\left (a^2-b^2\right )^2 x-a \left (a^2-2 b^2\right ) x^2+\left (a^2-2 b^2\right ) x^3-a x^4+x^5+\frac {\left (a^3-a b^2\right )^2}{a+x}\right ) \, dx,x,b \sin (c+d x)\right )}{b^7 d} \\ & = \frac {a^2 \left (a^2-b^2\right )^2 \log (a+b \sin (c+d x))}{b^7 d}-\frac {a \left (a^2-b^2\right )^2 \sin (c+d x)}{b^6 d}+\frac {\left (a^2-b^2\right )^2 \sin ^2(c+d x)}{2 b^5 d}-\frac {a \left (a^2-2 b^2\right ) \sin ^3(c+d x)}{3 b^4 d}+\frac {\left (a^2-2 b^2\right ) \sin ^4(c+d x)}{4 b^3 d}-\frac {a \sin ^5(c+d x)}{5 b^2 d}+\frac {\sin ^6(c+d x)}{6 b d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.41 (sec) , antiderivative size = 153, normalized size of antiderivative = 0.85 \[ \int \frac {\cos ^5(c+d x) \sin ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {60 \left (a^3-a b^2\right )^2 \log (a+b \sin (c+d x))-60 a b \left (a^2-b^2\right )^2 \sin (c+d x)+30 b^2 \left (a^2-b^2\right )^2 \sin ^2(c+d x)-20 a b^3 \left (a^2-2 b^2\right ) \sin ^3(c+d x)+15 b^4 \left (a^2-2 b^2\right ) \sin ^4(c+d x)-12 a b^5 \sin ^5(c+d x)+10 b^6 \sin ^6(c+d x)}{60 b^7 d} \]

[In]

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

[Out]

(60*(a^3 - a*b^2)^2*Log[a + b*Sin[c + d*x]] - 60*a*b*(a^2 - b^2)^2*Sin[c + d*x] + 30*b^2*(a^2 - b^2)^2*Sin[c +
d*x]^2 - 20*a*b^3*(a^2 - 2*b^2)*Sin[c + d*x]^3 + 15*b^4*(a^2 - 2*b^2)*Sin[c + d*x]^4 - 12*a*b^5*Sin[c + d*x]^
5 + 10*b^6*Sin[c + d*x]^6)/(60*b^7*d)

Maple [A] (verified)

Time = 0.67 (sec) , antiderivative size = 242, normalized size of antiderivative = 1.34

method	result	size
parallelrisch	\(\frac {960 a^{2} \left (a -b \right )^{2} \left (a +b \right )^{2} \ln \left (2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )-960 a^{2} \left (a -b \right )^{2} \left (a +b \right )^{2} \ln \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-240 a^{4} b^{2}+360 a^{2} b^{4}-75 b^{6}\right ) \cos \left (2 d x +2 c \right )+\left (30 a^{2} b^{4}-30 b^{6}\right ) \cos \left (4 d x +4 c \right )+\left (80 a^{3} b^{3}-100 a \,b^{5}\right ) \sin \left (3 d x +3 c \right )-5 b^{6} \cos \left (6 d x +6 c \right )-12 a \,b^{5} \sin \left (5 d x +5 c \right )+\left (-960 a^{5} b +1680 a^{3} b^{3}-600 a \,b^{5}\right ) \sin \left (d x +c \right )+240 a^{4} b^{2}-390 a^{2} b^{4}+110 b^{6}}{960 b^{7} d}\)	\(242\)
derivativedivides	\(\frac {\sin ^{6}\left (d x +c \right )}{6 b d}-\frac {a \left (\sin ^{5}\left (d x +c \right )\right )}{5 b^{2} d}-\frac {\sin ^{4}\left (d x +c \right )}{2 b d}+\frac {\left (\sin ^{4}\left (d x +c \right )\right ) a^{2}}{4 d \,b^{3}}-\frac {\left (\sin ^{3}\left (d x +c \right )\right ) a^{3}}{3 d \,b^{4}}+\frac {2 a \left (\sin ^{3}\left (d x +c \right )\right )}{3 b^{2} d}+\frac {\left (\sin ^{2}\left (d x +c \right )\right ) a^{4}}{2 d \,b^{5}}-\frac {\left (\sin ^{2}\left (d x +c \right )\right ) a^{2}}{d \,b^{3}}+\frac {\sin ^{2}\left (d x +c \right )}{2 b d}-\frac {a^{5} \sin \left (d x +c \right )}{d \,b^{6}}+\frac {2 a^{3} \sin \left (d x +c \right )}{d \,b^{4}}-\frac {a \sin \left (d x +c \right )}{b^{2} d}+\frac {a^{6} \ln \left (a +b \sin \left (d x +c \right )\right )}{d \,b^{7}}-\frac {2 a^{4} \ln \left (a +b \sin \left (d x +c \right )\right )}{d \,b^{5}}+\frac {a^{2} \ln \left (a +b \sin \left (d x +c \right )\right )}{b^{3} d}\)	\(273\)
default	\(\frac {\sin ^{6}\left (d x +c \right )}{6 b d}-\frac {a \left (\sin ^{5}\left (d x +c \right )\right )}{5 b^{2} d}-\frac {\sin ^{4}\left (d x +c \right )}{2 b d}+\frac {\left (\sin ^{4}\left (d x +c \right )\right ) a^{2}}{4 d \,b^{3}}-\frac {\left (\sin ^{3}\left (d x +c \right )\right ) a^{3}}{3 d \,b^{4}}+\frac {2 a \left (\sin ^{3}\left (d x +c \right )\right )}{3 b^{2} d}+\frac {\left (\sin ^{2}\left (d x +c \right )\right ) a^{4}}{2 d \,b^{5}}-\frac {\left (\sin ^{2}\left (d x +c \right )\right ) a^{2}}{d \,b^{3}}+\frac {\sin ^{2}\left (d x +c \right )}{2 b d}-\frac {a^{5} \sin \left (d x +c \right )}{d \,b^{6}}+\frac {2 a^{3} \sin \left (d x +c \right )}{d \,b^{4}}-\frac {a \sin \left (d x +c \right )}{b^{2} d}+\frac {a^{6} \ln \left (a +b \sin \left (d x +c \right )\right )}{d \,b^{7}}-\frac {2 a^{4} \ln \left (a +b \sin \left (d x +c \right )\right )}{d \,b^{5}}+\frac {a^{2} \ln \left (a +b \sin \left (d x +c \right )\right )}{b^{3} d}\)	\(273\)
risch	\(-\frac {{\mathrm e}^{-2 i \left (d x +c \right )} a^{4}}{8 b^{5} d}+\frac {3 \,{\mathrm e}^{-2 i \left (d x +c \right )} a^{2}}{16 b^{3} d}-\frac {i x \,a^{6}}{b^{7}}+\frac {2 i x \,a^{4}}{b^{5}}-\frac {i a^{2} x}{b^{3}}-\frac {{\mathrm e}^{2 i \left (d x +c \right )} a^{4}}{8 b^{5} d}+\frac {3 \,{\mathrm e}^{2 i \left (d x +c \right )} a^{2}}{16 b^{3} d}+\frac {a^{6} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}\right )}{b^{7} d}-\frac {2 a^{4} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}\right )}{b^{5} d}+\frac {a^{2} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}\right )}{b^{3} d}+\frac {i a^{5} {\mathrm e}^{i \left (d x +c \right )}}{2 b^{6} d}-\frac {7 i a^{3} {\mathrm e}^{i \left (d x +c \right )}}{8 b^{4} d}+\frac {5 i a \,{\mathrm e}^{i \left (d x +c \right )}}{16 b^{2} d}-\frac {i a^{5} {\mathrm e}^{-i \left (d x +c \right )}}{2 b^{6} d}-\frac {\cos \left (6 d x +6 c \right )}{192 b d}+\frac {7 i a^{3} {\mathrm e}^{-i \left (d x +c \right )}}{8 d \,b^{4}}-\frac {\cos \left (4 d x +4 c \right )}{32 b d}+\frac {a^{3} \sin \left (3 d x +3 c \right )}{12 b^{4} d}-\frac {5 a \sin \left (3 d x +3 c \right )}{48 b^{2} d}-\frac {a \sin \left (5 d x +5 c \right )}{80 b^{2} d}+\frac {\cos \left (4 d x +4 c \right ) a^{2}}{32 b^{3} d}-\frac {5 i a \,{\mathrm e}^{-i \left (d x +c \right )}}{16 b^{2} d}-\frac {2 i a^{6} c}{b^{7} d}+\frac {4 i a^{4} c}{b^{5} d}-\frac {2 i a^{2} c}{b^{3} d}-\frac {5 \,{\mathrm e}^{2 i \left (d x +c \right )}}{128 b d}-\frac {5 \,{\mathrm e}^{-2 i \left (d x +c \right )}}{128 b d}\)	\(532\)
norman	\(\frac {\frac {\left (10 a^{4}-16 a^{2} b^{2}+2 b^{4}\right ) \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,b^{5}}+\frac {\left (10 a^{4}-16 a^{2} b^{2}+2 b^{4}\right ) \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,b^{5}}+\frac {\left (2 a^{4}-4 a^{2} b^{2}+2 b^{4}\right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{5} d}+\frac {\left (2 a^{4}-4 a^{2} b^{2}+2 b^{4}\right ) \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{5} d}+\frac {\left (60 a^{4}-84 a^{2} b^{2}+20 b^{4}\right ) \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d \,b^{5}}+\frac {\left (60 a^{4}-84 a^{2} b^{2}+20 b^{4}\right ) \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d \,b^{5}}-\frac {2 a \left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{b^{6} d}-\frac {2 a \left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) \left (\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{6} d}-\frac {4 a \left (9 a^{4}-16 a^{2} b^{2}+5 b^{4}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 b^{6} d}-\frac {4 a \left (9 a^{4}-16 a^{2} b^{2}+5 b^{4}\right ) \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 b^{6} d}-\frac {8 a \left (25 a^{4}-40 a^{2} b^{2}+13 b^{4}\right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5 b^{6} d}-\frac {2 a \left (225 a^{4}-370 a^{2} b^{2}+113 b^{4}\right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{15 b^{6} d}-\frac {2 a \left (225 a^{4}-370 a^{2} b^{2}+113 b^{4}\right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{15 b^{6} d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{7}}+\frac {a^{2} \left (a^{4}-2 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 )}{b^{7} d}-\frac {a^{2} \left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) \ln \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{7} d}\)	\(599\)

[In]

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

[Out]

1/960*(960*a^2*(a-b)^2*(a+b)^2*ln(2*b*tan(1/2*d*x+1/2*c)+a*sec(1/2*d*x+1/2*c)^2)-960*a^2*(a-b)^2*(a+b)^2*ln(se
c(1/2*d*x+1/2*c)^2)+(-240*a^4*b^2+360*a^2*b^4-75*b^6)*cos(2*d*x+2*c)+(30*a^2*b^4-30*b^6)*cos(4*d*x+4*c)+(80*a^
3*b^3-100*a*b^5)*sin(3*d*x+3*c)-5*b^6*cos(6*d*x+6*c)-12*a*b^5*sin(5*d*x+5*c)+(-960*a^5*b+1680*a^3*b^3-600*a*b^
5)*sin(d*x+c)+240*a^4*b^2-390*a^2*b^4+110*b^6)/b^7/d

Fricas [A] (verification not implemented)

none

Time = 0.37 (sec) , antiderivative size = 164, normalized size of antiderivative = 0.91 \[ \int \frac {\cos ^5(c+d x) \sin ^2(c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {10 \, b^{6} \cos \left (d x + c\right )^{6} - 15 \, a^{2} b^{4} \cos \left (d x + c\right )^{4} + 30 \, {\left (a^{4} b^{2} - a^{2} b^{4}\right )} \cos \left (d x + c\right )^{2} - 60 \, {\left (a^{6} - 2 \, a^{4} b^{2} + a^{2} b^{4}\right )} \log \left (b \sin \left (d x + c\right ) + a\right ) + 4 \, {\left (3 \, a b^{5} \cos \left (d x + c\right )^{4} + 15 \, a^{5} b - 25 \, a^{3} b^{3} + 8 \, a b^{5} - {\left (5 \, a^{3} b^{3} - 4 \, a b^{5}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{60 \, b^{7} d} \]

[In]

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

[Out]

-1/60*(10*b^6*cos(d*x + c)^6 - 15*a^2*b^4*cos(d*x + c)^4 + 30*(a^4*b^2 - a^2*b^4)*cos(d*x + c)^2 - 60*(a^6 - 2
*a^4*b^2 + a^2*b^4)*log(b*sin(d*x + c) + a) + 4*(3*a*b^5*cos(d*x + c)^4 + 15*a^5*b - 25*a^3*b^3 + 8*a*b^5 - (5
*a^3*b^3 - 4*a*b^5)*cos(d*x + c)^2)*sin(d*x + c))/(b^7*d)

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 172, normalized size of antiderivative = 0.96 \[ \int \frac {\cos ^5(c+d x) \sin ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {\frac {10 \, b^{5} \sin \left (d x + c\right )^{6} - 12 \, a b^{4} \sin \left (d x + c\right )^{5} + 15 \, {\left (a^{2} b^{3} - 2 \, b^{5}\right )} \sin \left (d x + c\right )^{4} - 20 \, {\left (a^{3} b^{2} - 2 \, a b^{4}\right )} \sin \left (d x + c\right )^{3} + 30 \, {\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} \sin \left (d x + c\right )^{2} - 60 \, {\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} \sin \left (d x + c\right )}{b^{6}} + \frac {60 \, {\left (a^{6} - 2 \, a^{4} b^{2} + a^{2} b^{4}\right )} \log \left (b \sin \left (d x + c\right ) + a\right )}{b^{7}}}{60 \, d} \]

[In]

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

[Out]

1/60*((10*b^5*sin(d*x + c)^6 - 12*a*b^4*sin(d*x + c)^5 + 15*(a^2*b^3 - 2*b^5)*sin(d*x + c)^4 - 20*(a^3*b^2 - 2
*a*b^4)*sin(d*x + c)^3 + 30*(a^4*b - 2*a^2*b^3 + b^5)*sin(d*x + c)^2 - 60*(a^5 - 2*a^3*b^2 + a*b^4)*sin(d*x +
c))/b^6 + 60*(a^6 - 2*a^4*b^2 + a^2*b^4)*log(b*sin(d*x + c) + a)/b^7)/d

Giac [A] (verification not implemented)

none

Time = 0.43 (sec) , antiderivative size = 213, normalized size of antiderivative = 1.18 \[ \int \frac {\cos ^5(c+d x) \sin ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {\frac {10 \, b^{5} \sin \left (d x + c\right )^{6} - 12 \, a b^{4} \sin \left (d x + c\right )^{5} + 15 \, a^{2} b^{3} \sin \left (d x + c\right )^{4} - 30 \, b^{5} \sin \left (d x + c\right )^{4} - 20 \, a^{3} b^{2} \sin \left (d x + c\right )^{3} + 40 \, a b^{4} \sin \left (d x + c\right )^{3} + 30 \, a^{4} b \sin \left (d x + c\right )^{2} - 60 \, a^{2} b^{3} \sin \left (d x + c\right )^{2} + 30 \, b^{5} \sin \left (d x + c\right )^{2} - 60 \, a^{5} \sin \left (d x + c\right ) + 120 \, a^{3} b^{2} \sin \left (d x + c\right ) - 60 \, a b^{4} \sin \left (d x + c\right )}{b^{6}} + \frac {60 \, {\left (a^{6} - 2 \, a^{4} b^{2} + a^{2} b^{4}\right )} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{b^{7}}}{60 \, d} \]

[In]

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

[Out]

1/60*((10*b^5*sin(d*x + c)^6 - 12*a*b^4*sin(d*x + c)^5 + 15*a^2*b^3*sin(d*x + c)^4 - 30*b^5*sin(d*x + c)^4 - 2
0*a^3*b^2*sin(d*x + c)^3 + 40*a*b^4*sin(d*x + c)^3 + 30*a^4*b*sin(d*x + c)^2 - 60*a^2*b^3*sin(d*x + c)^2 + 30*
b^5*sin(d*x + c)^2 - 60*a^5*sin(d*x + c) + 120*a^3*b^2*sin(d*x + c) - 60*a*b^4*sin(d*x + c))/b^6 + 60*(a^6 - 2
*a^4*b^2 + a^2*b^4)*log(abs(b*sin(d*x + c) + a))/b^7)/d

Mupad [B] (verification not implemented)

Time = 11.69 (sec) , antiderivative size = 191, normalized size of antiderivative = 1.06 \[ \int \frac {\cos ^5(c+d x) \sin ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {{\sin \left (c+d\,x\right )}^2\,\left (\frac {1}{2\,b}-\frac {a^2\,\left (\frac {1}{b}-\frac {a^2}{2\,b^3}\right )}{b^2}\right )-{\sin \left (c+d\,x\right )}^4\,\left (\frac {1}{2\,b}-\frac {a^2}{4\,b^3}\right )+\frac {{\sin \left (c+d\,x\right )}^6}{6\,b}-\frac {a\,{\sin \left (c+d\,x\right )}^5}{5\,b^2}+\frac {\ln \left (a+b\,\sin \left (c+d\,x\right )\right )\,\left (a^6-2\,a^4\,b^2+a^2\,b^4\right )}{b^7}-\frac {a\,\sin \left (c+d\,x\right )\,\left (\frac {1}{b}-\frac {a^2\,\left (\frac {2}{b}-\frac {a^2}{b^3}\right )}{b^2}\right )}{b}+\frac {a\,{\sin \left (c+d\,x\right )}^3\,\left (\frac {2}{b}-\frac {a^2}{b^3}\right )}{3\,b}}{d} \]

[In]

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

[Out]

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