\(\int \frac {\cos ^3(c+d x) \sin (c+d x)}{a+b \sin (c+d x)} \, dx\) [1297]

Optimal result

Integrand size = 27, antiderivative size = 89 \[ \int \frac {\cos ^3(c+d x) \sin (c+d x)}{a+b \sin (c+d x)} \, dx=\frac {a \left (a^2-b^2\right ) \log (a+b \sin (c+d x))}{b^4 d}-\frac {\left (a^2-b^2\right ) \sin (c+d x)}{b^3 d}+\frac {a \sin ^2(c+d x)}{2 b^2 d}-\frac {\sin ^3(c+d x)}{3 b d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 89, 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} \[ \int \frac {\cos ^3(c+d x) \sin (c+d x)}{a+b \sin (c+d x)} \, dx=\frac {a \left (a^2-b^2\right ) \log (a+b \sin (c+d x))}{b^4 d}-\frac {\left (a^2-b^2\right ) \sin (c+d x)}{b^3 d}+\frac {a \sin ^2(c+d x)}{2 b^2 d}-\frac {\sin ^3(c+d x)}{3 b d} \]

[In]

Int[(Cos[c + d*x]^3*Sin[c + d*x])/(a + b*Sin[c + d*x]),x]

[Out]

(a*(a^2 - b^2)*Log[a + b*Sin[c + d*x]])/(b^4*d) - ((a^2 - b^2)*Sin[c + d*x])/(b^3*d) + (a*Sin[c + d*x]^2)/(2*b
^2*d) - Sin[c + d*x]^3/(3*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 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 )}{b (a+x)} \, dx,x,b \sin (c+d x)\right )}{b^3 d} \\ & = \frac {\text {Subst}\left (\int \frac {x \left (b^2-x^2\right )}{a+x} \, dx,x,b \sin (c+d x)\right )}{b^4 d} \\ & = \frac {\text {Subst}\left (\int \left (-a^2 \left (1-\frac {b^2}{a^2}\right )+a x-x^2+\frac {a^3-a b^2}{a+x}\right ) \, dx,x,b \sin (c+d x)\right )}{b^4 d} \\ & = \frac {a \left (a^2-b^2\right ) \log (a+b \sin (c+d x))}{b^4 d}-\frac {\left (a^2-b^2\right ) \sin (c+d x)}{b^3 d}+\frac {a \sin ^2(c+d x)}{2 b^2 d}-\frac {\sin ^3(c+d x)}{3 b d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.89 \[ \int \frac {\cos ^3(c+d x) \sin (c+d x)}{a+b \sin (c+d x)} \, dx=\frac {6 a \left (a^2-b^2\right ) \log (a+b \sin (c+d x))+6 b \left (-a^2+b^2\right ) \sin (c+d x)+3 a b^2 \sin ^2(c+d x)-2 b^3 \sin ^3(c+d x)}{6 b^4 d} \]

[In]

Integrate[(Cos[c + d*x]^3*Sin[c + d*x])/(a + b*Sin[c + d*x]),x]

[Out]

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

Maple [A] (verified)

Time = 0.33 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.93

method	result	size
derivativedivides	\(\frac {-\frac {\frac {\left (\sin ^{3}\left (d x +c \right )\right ) b^{2}}{3}-\frac {b a \left (\sin ^{2}\left (d x +c \right )\right )}{2}+a^{2} \sin \left (d x +c \right )-\sin \left (d x +c \right ) b^{2}}{b^{3}}+\frac {a \left (a^{2}-b^{2}\right ) \ln \left (a +b \sin \left (d x +c \right )\right )}{b^{4}}}{d}\)	\(83\)
default	\(\frac {-\frac {\frac {\left (\sin ^{3}\left (d x +c \right )\right ) b^{2}}{3}-\frac {b a \left (\sin ^{2}\left (d x +c \right )\right )}{2}+a^{2} \sin \left (d x +c \right )-\sin \left (d x +c \right ) b^{2}}{b^{3}}+\frac {a \left (a^{2}-b^{2}\right ) \ln \left (a +b \sin \left (d x +c \right )\right )}{b^{4}}}{d}\)	\(83\)
parallelrisch	\(\frac {12 a^{3} \ln \left (2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )-12 \ln \left (2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right ) a \,b^{2}-12 a^{3} \ln \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+12 \ln \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a \,b^{2}-3 a \,b^{2} \cos \left (2 d x +2 c \right )-12 \sin \left (d x +c \right ) a^{2} b +9 b^{3} \sin \left (d x +c \right )+b^{3} \sin \left (3 d x +3 c \right )+3 a \,b^{2}}{12 d \,b^{4}}\)	\(167\)
risch	\(-\frac {i a^{3} x}{b^{4}}+\frac {i a x}{b^{2}}-\frac {a \,{\mathrm e}^{2 i \left (d x +c \right )}}{8 d \,b^{2}}+\frac {i {\mathrm e}^{i \left (d x +c \right )} a^{2}}{2 b^{3} d}-\frac {3 i {\mathrm e}^{i \left (d x +c \right )}}{8 b d}-\frac {i {\mathrm e}^{-i \left (d x +c \right )} a^{2}}{2 b^{3} d}+\frac {3 i {\mathrm e}^{-i \left (d x +c \right )}}{8 b d}-\frac {a \,{\mathrm e}^{-2 i \left (d x +c \right )}}{8 d \,b^{2}}-\frac {2 i a^{3} c}{b^{4} d}+\frac {2 i a c}{b^{2} d}+\frac {a^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}\right )}{b^{4} d}-\frac {a \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}\right )}{b^{2} d}+\frac {\sin \left (3 d x +3 c \right )}{12 b d}\)	\(250\)
norman	\(\frac {-\frac {2 \left (a^{2}-b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{b^{3} d}-\frac {2 \left (a^{2}-b^{2}\right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{3} d}-\frac {2 \left (9 a^{2}-5 b^{2}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 b^{3} d}-\frac {2 \left (9 a^{2}-5 b^{2}\right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 b^{3} d}+\frac {2 a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{2} d}+\frac {2 a \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{2} d}+\frac {4 a \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,b^{2}}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}+\frac {a \left (a^{2}-b^{2}\right ) \ln \left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \right )}{b^{4} d}-\frac {a \left (a^{2}-b^{2}\right ) \ln \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{4} d}\)	\(270\)

[In]

int(cos(d*x+c)^3*sin(d*x+c)/(a+b*sin(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.41 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.88 \[ \int \frac {\cos ^3(c+d x) \sin (c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {3 \, a b^{2} \cos \left (d x + c\right )^{2} - 6 \, {\left (a^{3} - a b^{2}\right )} \log \left (b \sin \left (d x + c\right ) + a\right ) - 2 \, {\left (b^{3} \cos \left (d x + c\right )^{2} - 3 \, a^{2} b + 2 \, b^{3}\right )} \sin \left (d x + c\right )}{6 \, b^{4} d} \]

[In]

integrate(cos(d*x+c)^3*sin(d*x+c)/(a+b*sin(d*x+c)),x, algorithm="fricas")

[Out]

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

Sympy [F(-1)]

Timed out. \[ \int \frac {\cos ^3(c+d x) \sin (c+d x)}{a+b \sin (c+d x)} \, dx=\text {Timed out} \]

[In]

integrate(cos(d*x+c)**3*sin(d*x+c)/(a+b*sin(d*x+c)),x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.89 \[ \int \frac {\cos ^3(c+d x) \sin (c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {\frac {2 \, b^{2} \sin \left (d x + c\right )^{3} - 3 \, a b \sin \left (d x + c\right )^{2} + 6 \, {\left (a^{2} - b^{2}\right )} \sin \left (d x + c\right )}{b^{3}} - \frac {6 \, {\left (a^{3} - a b^{2}\right )} \log \left (b \sin \left (d x + c\right ) + a\right )}{b^{4}}}{6 \, d} \]

[In]

integrate(cos(d*x+c)^3*sin(d*x+c)/(a+b*sin(d*x+c)),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.96 \[ \int \frac {\cos ^3(c+d x) \sin (c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {\frac {2 \, b^{2} \sin \left (d x + c\right )^{3} - 3 \, a b \sin \left (d x + c\right )^{2} + 6 \, a^{2} \sin \left (d x + c\right ) - 6 \, b^{2} \sin \left (d x + c\right )}{b^{3}} - \frac {6 \, {\left (a^{3} - a b^{2}\right )} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{b^{4}}}{6 \, d} \]

[In]

integrate(cos(d*x+c)^3*sin(d*x+c)/(a+b*sin(d*x+c)),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 11.54 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.88 \[ \int \frac {\cos ^3(c+d x) \sin (c+d x)}{a+b \sin (c+d x)} \, dx=\frac {\sin \left (c+d\,x\right )\,\left (\frac {1}{b}-\frac {a^2}{b^3}\right )-\frac {{\sin \left (c+d\,x\right )}^3}{3\,b}+\frac {a\,{\sin \left (c+d\,x\right )}^2}{2\,b^2}-\frac {\ln \left (a+b\,\sin \left (c+d\,x\right )\right )\,\left (a\,b^2-a^3\right )}{b^4}}{d} \]

[In]

int((cos(c + d*x)^3*sin(c + d*x))/(a + b*sin(c + d*x)),x)

[Out]

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