\(\int \frac {\cos ^2(c+d x) \sin (c+d x)}{a+a \sin (c+d x)} \, dx\) [300]

Optimal result

Integrand size = 27, antiderivative size = 45 \[ \int \frac {\cos ^2(c+d x) \sin (c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {x}{2 a}-\frac {\cos (c+d x)}{a d}+\frac {\cos (c+d x) \sin (c+d x)}{2 a d} \]

[Out]

-1/2*x/a-cos(d*x+c)/a/d+1/2*cos(d*x+c)*sin(d*x+c)/a/d

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {2918, 2718, 2715, 8} \[ \int \frac {\cos ^2(c+d x) \sin (c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {\cos (c+d x)}{a d}+\frac {\sin (c+d x) \cos (c+d x)}{2 a d}-\frac {x}{2 a} \]

[In]

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

[Out]

-1/2*x/a - Cos[c + d*x]/(a*d) + (Cos[c + d*x]*Sin[c + d*x])/(2*a*d)

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 2715

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d*x]*((b*Sin[c + d*x])^(n - 1)/(d*n))
, x] + Dist[b^2*((n - 1)/n), Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integ
erQ[2*n]

Rule 2718

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-Cos[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 2918

Int[((cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.))/((a_) + (b_.)*sin[(e_.) + (f_
.)*(x_)]), x_Symbol] :> Dist[g^2/a, Int[(g*Cos[e + f*x])^(p - 2)*(d*Sin[e + f*x])^n, x], x] - Dist[g^2/(b*d),
Int[(g*Cos[e + f*x])^(p - 2)*(d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, g, n, p}, x] && EqQ[a^2
- b^2, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {\int \sin (c+d x) \, dx}{a}-\frac {\int \sin ^2(c+d x) \, dx}{a} \\ & = -\frac {\cos (c+d x)}{a d}+\frac {\cos (c+d x) \sin (c+d x)}{2 a d}-\frac {\int 1 \, dx}{2 a} \\ & = -\frac {x}{2 a}-\frac {\cos (c+d x)}{a d}+\frac {\cos (c+d x) \sin (c+d x)}{2 a d} \\ \end{align*}

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(161\) vs. \(2(45)=90\).

Time = 0.38 (sec) , antiderivative size = 161, normalized size of antiderivative = 3.58 \[ \int \frac {\cos ^2(c+d x) \sin (c+d x)}{a+a \sin (c+d x)} \, dx=\frac {2 (c-2 d x) \cos \left (\frac {c}{2}\right )-4 \cos \left (\frac {c}{2}+d x\right )-4 \cos \left (\frac {3 c}{2}+d x\right )+\cos \left (\frac {3 c}{2}+2 d x\right )-\cos \left (\frac {5 c}{2}+2 d x\right )-4 \sin \left (\frac {c}{2}\right )+2 c \sin \left (\frac {c}{2}\right )-4 d x \sin \left (\frac {c}{2}\right )+4 \sin \left (\frac {c}{2}+d x\right )-4 \sin \left (\frac {3 c}{2}+d x\right )+\sin \left (\frac {3 c}{2}+2 d x\right )+\sin \left (\frac {5 c}{2}+2 d x\right )}{8 a d \left (\cos \left (\frac {c}{2}\right )+\sin \left (\frac {c}{2}\right )\right )} \]

[In]

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

[Out]

(2*(c - 2*d*x)*Cos[c/2] - 4*Cos[c/2 + d*x] - 4*Cos[(3*c)/2 + d*x] + Cos[(3*c)/2 + 2*d*x] - Cos[(5*c)/2 + 2*d*x
] - 4*Sin[c/2] + 2*c*Sin[c/2] - 4*d*x*Sin[c/2] + 4*Sin[c/2 + d*x] - 4*Sin[(3*c)/2 + d*x] + Sin[(3*c)/2 + 2*d*x
] + Sin[(5*c)/2 + 2*d*x])/(8*a*d*(Cos[c/2] + Sin[c/2]))

Maple [A] (verified)

Time = 0.14 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.71

method	result	size
parallelrisch	\(\frac {-2 d x -4 \cos \left (d x +c \right )+\sin \left (2 d x +2 c \right )+4}{4 d a}\)	\(32\)
risch	\(-\frac {x}{2 a}-\frac {\cos \left (d x +c \right )}{a d}+\frac {\sin \left (2 d x +2 c \right )}{4 d a}\)	\(39\)
derivativedivides	\(\frac {\frac {4 \left (-\frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4}-\frac {\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4}-\frac {1}{2}\right )}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}-\arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}\)	\(77\)
default	\(\frac {\frac {4 \left (-\frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4}-\frac {\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4}-\frac {1}{2}\right )}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}-\arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}\)	\(77\)
norman	\(\frac {-\frac {1}{a d}+\frac {\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )}{d a}-\frac {\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )}{d a}+\frac {\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )}{d a}-\frac {x}{2 a}-\frac {x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a}-\frac {3 x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 a}-\frac {3 x \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 a}-\frac {3 x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 a}-\frac {3 x \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 a}-\frac {x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 a}-\frac {x \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 a}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}\)	\(217\)

[In]

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

[Out]

1/4*(-2*d*x-4*cos(d*x+c)+sin(2*d*x+2*c)+4)/d/a

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.76 \[ \int \frac {\cos ^2(c+d x) \sin (c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {d x - \cos \left (d x + c\right ) \sin \left (d x + c\right ) + 2 \, \cos \left (d x + c\right )}{2 \, a d} \]

[In]

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

[Out]

-1/2*(d*x - cos(d*x + c)*sin(d*x + c) + 2*cos(d*x + c))/(a*d)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 366 vs. \(2 (32) = 64\).

Time = 1.89 (sec) , antiderivative size = 366, normalized size of antiderivative = 8.13 \[ \int \frac {\cos ^2(c+d x) \sin (c+d x)}{a+a \sin (c+d x)} \, dx=\begin {cases} - \frac {d x \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{2 a d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 4 a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 2 a d} - \frac {2 d x \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{2 a d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 4 a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 2 a d} - \frac {d x}{2 a d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 4 a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 2 a d} - \frac {2 \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{2 a d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 4 a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 2 a d} - \frac {4 \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{2 a d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 4 a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 2 a d} + \frac {2 \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{2 a d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 4 a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 2 a d} - \frac {4}{2 a d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 4 a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 2 a d} & \text {for}\: d \neq 0 \\\frac {x \sin {\left (c \right )} \cos ^{2}{\left (c \right )}}{a \sin {\left (c \right )} + a} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((-d*x*tan(c/2 + d*x/2)**4/(2*a*d*tan(c/2 + d*x/2)**4 + 4*a*d*tan(c/2 + d*x/2)**2 + 2*a*d) - 2*d*x*ta
n(c/2 + d*x/2)**2/(2*a*d*tan(c/2 + d*x/2)**4 + 4*a*d*tan(c/2 + d*x/2)**2 + 2*a*d) - d*x/(2*a*d*tan(c/2 + d*x/2
)**4 + 4*a*d*tan(c/2 + d*x/2)**2 + 2*a*d) - 2*tan(c/2 + d*x/2)**3/(2*a*d*tan(c/2 + d*x/2)**4 + 4*a*d*tan(c/2 +
d*x/2)**2 + 2*a*d) - 4*tan(c/2 + d*x/2)**2/(2*a*d*tan(c/2 + d*x/2)**4 + 4*a*d*tan(c/2 + d*x/2)**2 + 2*a*d) +
2*tan(c/2 + d*x/2)/(2*a*d*tan(c/2 + d*x/2)**4 + 4*a*d*tan(c/2 + d*x/2)**2 + 2*a*d) - 4/(2*a*d*tan(c/2 + d*x/2)
**4 + 4*a*d*tan(c/2 + d*x/2)**2 + 2*a*d), Ne(d, 0)), (x*sin(c)*cos(c)**2/(a*sin(c) + a), True))

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 133 vs. \(2 (41) = 82\).

Time = 0.29 (sec) , antiderivative size = 133, normalized size of antiderivative = 2.96 \[ \int \frac {\cos ^2(c+d x) \sin (c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\frac {\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {2 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - 2}{a + \frac {2 \, a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {a \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}} - \frac {\arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a}}{d} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.39 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.60 \[ \int \frac {\cos ^2(c+d x) \sin (c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {\frac {d x + c}{a} + \frac {2 \, {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2} a}}{2 \, d} \]

[In]

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

[Out]

-1/2*((d*x + c)/a + 2*(tan(1/2*d*x + 1/2*c)^3 + 2*tan(1/2*d*x + 1/2*c)^2 - tan(1/2*d*x + 1/2*c) + 2)/((tan(1/2
*d*x + 1/2*c)^2 + 1)^2*a))/d

Mupad [B] (verification not implemented)

Time = 9.62 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.73 \[ \int \frac {\cos ^2(c+d x) \sin (c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {x}{2\,a}-\frac {\cos \left (c+d\,x\right )-\frac {\sin \left (2\,c+2\,d\,x\right )}{4}}{a\,d} \]

[In]

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

[Out]

- x/(2*a) - (cos(c + d*x) - sin(2*c + 2*d*x)/4)/(a*d)