\(\int \frac {\cos ^5(c+d x) (A+B \sin (c+d x))}{a+b \sin (c+d x)} \, dx\) [1545]

Optimal result

Integrand size = 31, antiderivative size = 202 \[ \int \frac {\cos ^5(c+d x) (A+B \sin (c+d x))}{a+b \sin (c+d x)} \, dx=\frac {\left (a^2-b^2\right )^2 (A b-a B) \log (a+b \sin (c+d x))}{b^6 d}-\frac {\left (a^3 A b-2 a A b^3-a^4 B+2 a^2 b^2 B-b^4 B\right ) \sin (c+d x)}{b^5 d}+\frac {\left (a^2-2 b^2\right ) (A b-a B) \sin ^2(c+d x)}{2 b^4 d}-\frac {\left (a A b-a^2 B+2 b^2 B\right ) \sin ^3(c+d x)}{3 b^3 d}+\frac {(A b-a B) \sin ^4(c+d x)}{4 b^2 d}+\frac {B \sin ^5(c+d x)}{5 b d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.17 (sec) , antiderivative size = 202, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.065, Rules used = {2916, 786} \[ \int \frac {\cos ^5(c+d x) (A+B \sin (c+d x))}{a+b \sin (c+d x)} \, dx=\frac {\left (a^2-b^2\right )^2 (A b-a B) \log (a+b \sin (c+d x))}{b^6 d}+\frac {\left (a^2-2 b^2\right ) (A b-a B) \sin ^2(c+d x)}{2 b^4 d}-\frac {\left (a^2 (-B)+a A b+2 b^2 B\right ) \sin ^3(c+d x)}{3 b^3 d}-\frac {\left (a^4 (-B)+a^3 A b+2 a^2 b^2 B-2 a A b^3-b^4 B\right ) \sin (c+d x)}{b^5 d}+\frac {(A b-a B) \sin ^4(c+d x)}{4 b^2 d}+\frac {B \sin ^5(c+d x)}{5 b d} \]

[In]

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

[Out]

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

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

Mathematica [A] (verified)

Time = 0.31 (sec) , antiderivative size = 148, normalized size of antiderivative = 0.73 \[ \int \frac {\cos ^5(c+d x) (A+B \sin (c+d x))}{a+b \sin (c+d x)} \, dx=\frac {20 (A b-a B) \left (3 b^4 \cos ^4(c+d x)+12 \left (a^2-b^2\right )^2 \log (a+b \sin (c+d x))-12 a b \left (a^2-2 b^2\right ) \sin (c+d x)+6 b^2 \left (a^2-b^2\right ) \sin ^2(c+d x)-4 a b^3 \sin ^3(c+d x)\right )+b^5 B (150 \sin (c+d x)+25 \sin (3 (c+d x))+3 \sin (5 (c+d x)))}{240 b^6 d} \]

[In]

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

[Out]

(20*(A*b - a*B)*(3*b^4*Cos[c + d*x]^4 + 12*(a^2 - b^2)^2*Log[a + b*Sin[c + d*x]] - 12*a*b*(a^2 - 2*b^2)*Sin[c
+ d*x] + 6*b^2*(a^2 - b^2)*Sin[c + d*x]^2 - 4*a*b^3*Sin[c + d*x]^3) + b^5*B*(150*Sin[c + d*x] + 25*Sin[3*(c +
d*x)] + 3*Sin[5*(c + d*x)]))/(240*b^6*d)

Maple [A] (verified)

Time = 0.80 (sec) , antiderivative size = 254, normalized size of antiderivative = 1.26

method	result	size
parallelrisch	\(\frac {\left (a -b \right )^{2} \left (a +b \right )^{2} \left (A b -B a \right ) \ln \left (2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )-\left (a -b \right )^{2} \left (a +b \right )^{2} \left (A b -B a \right ) \ln \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-b \left (\frac {\left (a^{2}-\frac {3 b^{2}}{2}\right ) b \left (A b -B a \right ) \cos \left (2 d x +2 c \right )}{4}-\frac {b^{2} \left (A a b -B \,a^{2}+\frac {5}{4} B \,b^{2}\right ) \sin \left (3 d x +3 c \right )}{12}-\frac {b^{3} \left (A b -B a \right ) \cos \left (4 d x +4 c \right )}{32}-\frac {B \,b^{4} \sin \left (5 d x +5 c \right )}{80}+\left (A \,a^{3} b -\frac {7}{4} A a \,b^{3}-B \,a^{4}+\frac {7}{4} B \,a^{2} b^{2}-\frac {5}{8} B \,b^{4}\right ) \sin \left (d x +c \right )-\frac {\left (a^{2}-\frac {13 b^{2}}{8}\right ) b \left (A b -B a \right )}{4}\right )}{b^{6} d}\)	\(254\)
derivativedivides	\(\frac {-\frac {-\frac {B \left (\sin ^{5}\left (d x +c \right )\right ) b^{4}}{5}-\frac {A \,b^{4} \left (\sin ^{4}\left (d x +c \right )\right )}{4}+\frac {B a \,b^{3} \left (\sin ^{4}\left (d x +c \right )\right )}{4}+\frac {A a \,b^{3} \left (\sin ^{3}\left (d x +c \right )\right )}{3}-\frac {B \,a^{2} b^{2} \left (\sin ^{3}\left (d x +c \right )\right )}{3}+\frac {2 B \,b^{4} \left (\sin ^{3}\left (d x +c \right )\right )}{3}-\frac {A \,a^{2} b^{2} \left (\sin ^{2}\left (d x +c \right )\right )}{2}+A \,b^{4} \left (\sin ^{2}\left (d x +c \right )\right )+\frac {B \,a^{3} b \left (\sin ^{2}\left (d x +c \right )\right )}{2}-B a \,b^{3} \left (\sin ^{2}\left (d x +c \right )\right )+A \,a^{3} b \sin \left (d x +c \right )-2 A a \,b^{3} \sin \left (d x +c \right )-B \,a^{4} \sin \left (d x +c \right )+2 B \,a^{2} b^{2} \sin \left (d x +c \right )-B \,b^{4} \sin \left (d x +c \right )}{b^{5}}+\frac {\left (A \,a^{4} b -2 A \,a^{2} b^{3}+A \,b^{5}-B \,a^{5}+2 B \,a^{3} b^{2}-B a \,b^{4}\right ) \ln \left (a +b \sin \left (d x +c \right )\right )}{b^{6}}}{d}\)	\(283\)
default	\(\frac {-\frac {-\frac {B \left (\sin ^{5}\left (d x +c \right )\right ) b^{4}}{5}-\frac {A \,b^{4} \left (\sin ^{4}\left (d x +c \right )\right )}{4}+\frac {B a \,b^{3} \left (\sin ^{4}\left (d x +c \right )\right )}{4}+\frac {A a \,b^{3} \left (\sin ^{3}\left (d x +c \right )\right )}{3}-\frac {B \,a^{2} b^{2} \left (\sin ^{3}\left (d x +c \right )\right )}{3}+\frac {2 B \,b^{4} \left (\sin ^{3}\left (d x +c \right )\right )}{3}-\frac {A \,a^{2} b^{2} \left (\sin ^{2}\left (d x +c \right )\right )}{2}+A \,b^{4} \left (\sin ^{2}\left (d x +c \right )\right )+\frac {B \,a^{3} b \left (\sin ^{2}\left (d x +c \right )\right )}{2}-B a \,b^{3} \left (\sin ^{2}\left (d x +c \right )\right )+A \,a^{3} b \sin \left (d x +c \right )-2 A a \,b^{3} \sin \left (d x +c \right )-B \,a^{4} \sin \left (d x +c \right )+2 B \,a^{2} b^{2} \sin \left (d x +c \right )-B \,b^{4} \sin \left (d x +c \right )}{b^{5}}+\frac {\left (A \,a^{4} b -2 A \,a^{2} b^{3}+A \,b^{5}-B \,a^{5}+2 B \,a^{3} b^{2}-B a \,b^{4}\right ) \ln \left (a +b \sin \left (d x +c \right )\right )}{b^{6}}}{d}\)	\(283\)
norman	\(\frac {\frac {\left (8 A \,a^{2} b -12 A \,b^{3}-8 B \,a^{3}+12 B a \,b^{2}\right ) \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{4} d}+\frac {\left (8 A \,a^{2} b -12 A \,b^{3}-8 B \,a^{3}+12 B a \,b^{2}\right ) \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{4} d}+\frac {4 \left (3 A \,a^{2} b -4 A \,b^{3}-3 B \,a^{3}+4 B a \,b^{2}\right ) \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{4} d}-\frac {2 \left (15 A \,a^{3} b -26 A a \,b^{3}-15 B \,a^{4}+26 B \,a^{2} b^{2}-7 B \,b^{4}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d \,b^{5}}-\frac {2 \left (15 A \,a^{3} b -26 A a \,b^{3}-15 B \,a^{4}+26 B \,a^{2} b^{2}-7 B \,b^{4}\right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d \,b^{5}}-\frac {4 \left (25 A \,a^{3} b -40 A a \,b^{3}-25 B \,a^{4}+40 B \,a^{2} b^{2}-13 B \,b^{4}\right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5 d \,b^{5}}-\frac {4 \left (25 A \,a^{3} b -40 A a \,b^{3}-25 B \,a^{4}+40 B \,a^{2} b^{2}-13 B \,b^{4}\right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5 d \,b^{5}}-\frac {2 \left (A \,a^{3} b -2 A a \,b^{3}-B \,a^{4}+2 B \,a^{2} b^{2}-B \,b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{b^{5} d}-\frac {2 \left (A \,a^{3} b -2 A a \,b^{3}-B \,a^{4}+2 B \,a^{2} b^{2}-B \,b^{4}\right ) \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{5} d}+\frac {2 \left (A \,a^{2} b -2 A \,b^{3}-B \,a^{3}+2 B a \,b^{2}\right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{4} d}+\frac {2 \left (A \,a^{2} b -2 A \,b^{3}-B \,a^{3}+2 B a \,b^{2}\right ) \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{4} d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{6}}+\frac {\left (A \,a^{4} b -2 A \,a^{2} b^{3}+A \,b^{5}-B \,a^{5}+2 B \,a^{3} b^{2}-B a \,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^{6} d}-\frac {\left (A \,a^{4} b -2 A \,a^{2} b^{3}+A \,b^{5}-B \,a^{5}+2 B \,a^{3} b^{2}-B a \,b^{4}\right ) \ln \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{6} d}\)	\(714\)
risch	\(\frac {\sin \left (3 d x +3 c \right ) A a}{12 b^{2} d}-\frac {\sin \left (3 d x +3 c \right ) B \,a^{2}}{12 b^{3} d}-\frac {\cos \left (4 d x +4 c \right ) B a}{32 b^{2} d}+\frac {\sin \left (5 d x +5 c \right ) B}{80 b d}+\frac {\cos \left (4 d x +4 c \right ) A}{32 b d}+\frac {5 \sin \left (3 d x +3 c \right ) B}{48 b d}+\frac {7 i {\mathrm e}^{i \left (d x +c \right )} B \,a^{2}}{8 b^{3} d}-\frac {i {\mathrm e}^{-i \left (d x +c \right )} A \,a^{3}}{2 b^{4} d}+\frac {7 i {\mathrm e}^{-i \left (d x +c \right )} A a}{8 b^{2} d}+\frac {i {\mathrm e}^{-i \left (d x +c \right )} B \,a^{4}}{2 b^{5} d}-\frac {7 i {\mathrm e}^{-i \left (d x +c \right )} B \,a^{2}}{8 b^{3} d}-\frac {2 i A \,a^{4} c}{b^{5} d}+\frac {4 i A \,a^{2} c}{b^{3} d}+\frac {2 i B \,a^{5} c}{b^{6} d}-\frac {4 i B \,a^{3} c}{b^{4} d}+\frac {i {\mathrm e}^{i \left (d x +c \right )} A \,a^{3}}{2 b^{4} d}-\frac {7 i {\mathrm e}^{i \left (d x +c \right )} A a}{8 b^{2} d}+\frac {2 i B a c}{d \,b^{2}}-\frac {i {\mathrm e}^{i \left (d x +c \right )} B \,a^{4}}{2 b^{5} d}+\frac {i x B a}{b^{2}}+\frac {2 i x A \,a^{2}}{b^{3}}-\frac {2 i x B \,a^{3}}{b^{4}}-\frac {i x A \,a^{4}}{b^{5}}-\frac {{\mathrm e}^{-2 i \left (d x +c \right )} A \,a^{2}}{8 b^{3} d}+\frac {{\mathrm e}^{-2 i \left (d x +c \right )} B \,a^{3}}{8 b^{4} d}-\frac {3 \,{\mathrm e}^{-2 i \left (d x +c \right )} B a}{16 b^{2} d}-\frac {{\mathrm e}^{2 i \left (d x +c \right )} A \,a^{2}}{8 b^{3} d}+\frac {{\mathrm e}^{2 i \left (d x +c \right )} B \,a^{3}}{8 b^{4} d}-\frac {3 \,{\mathrm e}^{2 i \left (d x +c \right )} B a}{16 b^{2} d}+\frac {3 \,{\mathrm e}^{-2 i \left (d x +c \right )} A}{16 b 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 ) A}{d b}-\frac {i x A}{b}+\frac {3 \,{\mathrm e}^{2 i \left (d x +c \right )} A}{16 b d}+\frac {i x B \,a^{5}}{b^{6}}-\frac {5 i {\mathrm e}^{i \left (d x +c \right )} B}{16 b d}+\frac {5 i {\mathrm e}^{-i \left (d x +c \right )} B}{16 b 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 \,a^{5}}{d \,b^{6}}+\frac {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 \,a^{3}}{d \,b^{4}}-\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 a}{d \,b^{2}}-\frac {2 i A c}{d b}+\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 ) A \,a^{4}}{d \,b^{5}}-\frac {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 ) A \,a^{2}}{d \,b^{3}}\)	\(856\)

[In]

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

[Out]

((a-b)^2*(a+b)^2*(A*b-B*a)*ln(2*b*tan(1/2*d*x+1/2*c)+a*sec(1/2*d*x+1/2*c)^2)-(a-b)^2*(a+b)^2*(A*b-B*a)*ln(sec(
1/2*d*x+1/2*c)^2)-b*(1/4*(a^2-3/2*b^2)*b*(A*b-B*a)*cos(2*d*x+2*c)-1/12*b^2*(A*a*b-B*a^2+5/4*B*b^2)*sin(3*d*x+3
*c)-1/32*b^3*(A*b-B*a)*cos(4*d*x+4*c)-1/80*B*b^4*sin(5*d*x+5*c)+(A*a^3*b-7/4*A*a*b^3-B*a^4+7/4*B*a^2*b^2-5/8*B
*b^4)*sin(d*x+c)-1/4*(a^2-13/8*b^2)*b*(A*b-B*a)))/b^6/d

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 222, normalized size of antiderivative = 1.10 \[ \int \frac {\cos ^5(c+d x) (A+B \sin (c+d x))}{a+b \sin (c+d x)} \, dx=-\frac {15 \, {\left (B a b^{4} - A b^{5}\right )} \cos \left (d x + c\right )^{4} - 30 \, {\left (B a^{3} b^{2} - A a^{2} b^{3} - B a b^{4} + A b^{5}\right )} \cos \left (d x + c\right )^{2} + 60 \, {\left (B a^{5} - A a^{4} b - 2 \, B a^{3} b^{2} + 2 \, A a^{2} b^{3} + B a b^{4} - A b^{5}\right )} \log \left (b \sin \left (d x + c\right ) + a\right ) - 4 \, {\left (3 \, B b^{5} \cos \left (d x + c\right )^{4} + 15 \, B a^{4} b - 15 \, A a^{3} b^{2} - 25 \, B a^{2} b^{3} + 25 \, A a b^{4} + 8 \, B b^{5} - {\left (5 \, B a^{2} b^{3} - 5 \, A a b^{4} - 4 \, B b^{5}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{60 \, b^{6} d} \]

[In]

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

[Out]

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

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 220, normalized size of antiderivative = 1.09 \[ \int \frac {\cos ^5(c+d x) (A+B \sin (c+d x))}{a+b \sin (c+d x)} \, dx=\frac {\frac {12 \, B b^{4} \sin \left (d x + c\right )^{5} - 15 \, {\left (B a b^{3} - A b^{4}\right )} \sin \left (d x + c\right )^{4} + 20 \, {\left (B a^{2} b^{2} - A a b^{3} - 2 \, B b^{4}\right )} \sin \left (d x + c\right )^{3} - 30 \, {\left (B a^{3} b - A a^{2} b^{2} - 2 \, B a b^{3} + 2 \, A b^{4}\right )} \sin \left (d x + c\right )^{2} + 60 \, {\left (B a^{4} - A a^{3} b - 2 \, B a^{2} b^{2} + 2 \, A a b^{3} + B b^{4}\right )} \sin \left (d x + c\right )}{b^{5}} - \frac {60 \, {\left (B a^{5} - A a^{4} b - 2 \, B a^{3} b^{2} + 2 \, A a^{2} b^{3} + B a b^{4} - A b^{5}\right )} \log \left (b \sin \left (d x + c\right ) + a\right )}{b^{6}}}{60 \, d} \]

[In]

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

[Out]

1/60*((12*B*b^4*sin(d*x + c)^5 - 15*(B*a*b^3 - A*b^4)*sin(d*x + c)^4 + 20*(B*a^2*b^2 - A*a*b^3 - 2*B*b^4)*sin(
d*x + c)^3 - 30*(B*a^3*b - A*a^2*b^2 - 2*B*a*b^3 + 2*A*b^4)*sin(d*x + c)^2 + 60*(B*a^4 - A*a^3*b - 2*B*a^2*b^2
+ 2*A*a*b^3 + B*b^4)*sin(d*x + c))/b^5 - 60*(B*a^5 - A*a^4*b - 2*B*a^3*b^2 + 2*A*a^2*b^3 + B*a*b^4 - A*b^5)*l
og(b*sin(d*x + c) + a)/b^6)/d

Giac [A] (verification not implemented)

none

Time = 0.36 (sec) , antiderivative size = 286, normalized size of antiderivative = 1.42 \[ \int \frac {\cos ^5(c+d x) (A+B \sin (c+d x))}{a+b \sin (c+d x)} \, dx=\frac {\frac {12 \, B b^{4} \sin \left (d x + c\right )^{5} - 15 \, B a b^{3} \sin \left (d x + c\right )^{4} + 15 \, A b^{4} \sin \left (d x + c\right )^{4} + 20 \, B a^{2} b^{2} \sin \left (d x + c\right )^{3} - 20 \, A a b^{3} \sin \left (d x + c\right )^{3} - 40 \, B b^{4} \sin \left (d x + c\right )^{3} - 30 \, B a^{3} b \sin \left (d x + c\right )^{2} + 30 \, A a^{2} b^{2} \sin \left (d x + c\right )^{2} + 60 \, B a b^{3} \sin \left (d x + c\right )^{2} - 60 \, A b^{4} \sin \left (d x + c\right )^{2} + 60 \, B a^{4} \sin \left (d x + c\right ) - 60 \, A a^{3} b \sin \left (d x + c\right ) - 120 \, B a^{2} b^{2} \sin \left (d x + c\right ) + 120 \, A a b^{3} \sin \left (d x + c\right ) + 60 \, B b^{4} \sin \left (d x + c\right )}{b^{5}} - \frac {60 \, {\left (B a^{5} - A a^{4} b - 2 \, B a^{3} b^{2} + 2 \, A a^{2} b^{3} + B a b^{4} - A b^{5}\right )} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{b^{6}}}{60 \, d} \]

[In]

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

[Out]

1/60*((12*B*b^4*sin(d*x + c)^5 - 15*B*a*b^3*sin(d*x + c)^4 + 15*A*b^4*sin(d*x + c)^4 + 20*B*a^2*b^2*sin(d*x +
c)^3 - 20*A*a*b^3*sin(d*x + c)^3 - 40*B*b^4*sin(d*x + c)^3 - 30*B*a^3*b*sin(d*x + c)^2 + 30*A*a^2*b^2*sin(d*x
+ c)^2 + 60*B*a*b^3*sin(d*x + c)^2 - 60*A*b^4*sin(d*x + c)^2 + 60*B*a^4*sin(d*x + c) - 60*A*a^3*b*sin(d*x + c)
- 120*B*a^2*b^2*sin(d*x + c) + 120*A*a*b^3*sin(d*x + c) + 60*B*b^4*sin(d*x + c))/b^5 - 60*(B*a^5 - A*a^4*b -
2*B*a^3*b^2 + 2*A*a^2*b^3 + B*a*b^4 - A*b^5)*log(abs(b*sin(d*x + c) + a))/b^6)/d

Mupad [B] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 253, normalized size of antiderivative = 1.25 \[ \int \frac {\cos ^5(c+d x) (A+B \sin (c+d x))}{a+b \sin (c+d x)} \, dx=\frac {{\sin \left (c+d\,x\right )}^4\,\left (\frac {A}{4\,b}-\frac {B\,a}{4\,b^2}\right )}{d}-\frac {{\sin \left (c+d\,x\right )}^2\,\left (\frac {A}{b}-\frac {a\,\left (\frac {2\,B}{b}+\frac {a\,\left (\frac {A}{b}-\frac {B\,a}{b^2}\right )}{b}\right )}{2\,b}\right )}{d}-\frac {{\sin \left (c+d\,x\right )}^3\,\left (\frac {2\,B}{3\,b}+\frac {a\,\left (\frac {A}{b}-\frac {B\,a}{b^2}\right )}{3\,b}\right )}{d}+\frac {\sin \left (c+d\,x\right )\,\left (\frac {B}{b}+\frac {a\,\left (\frac {2\,A}{b}-\frac {a\,\left (\frac {2\,B}{b}+\frac {a\,\left (\frac {A}{b}-\frac {B\,a}{b^2}\right )}{b}\right )}{b}\right )}{b}\right )}{d}+\frac {\ln \left (a+b\,\sin \left (c+d\,x\right )\right )\,\left (-B\,a^5+A\,a^4\,b+2\,B\,a^3\,b^2-2\,A\,a^2\,b^3-B\,a\,b^4+A\,b^5\right )}{b^6\,d}+\frac {B\,{\sin \left (c+d\,x\right )}^5}{5\,b\,d} \]

[In]

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

[Out]

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