\(\int \frac {\cos (c+d x) \sin (c+d x)}{(a+a \sin (c+d x))^2} \, dx\) [235]

Optimal result

Integrand size = 25, antiderivative size = 37 \[ \int \frac {\cos (c+d x) \sin (c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {\log (1+\sin (c+d x))}{a^2 d}+\frac {1}{d \left (a^2+a^2 \sin (c+d x)\right )} \]

[Out]

ln(1+sin(d*x+c))/a^2/d+1/d/(a^2+a^2*sin(d*x+c))

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {2912, 12, 45} \[ \int \frac {\cos (c+d x) \sin (c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {1}{d \left (a^2 \sin (c+d x)+a^2\right )}+\frac {\log (\sin (c+d x)+1)}{a^2 d} \]

[In]

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

[Out]

Log[1 + Sin[c + d*x]]/(a^2*d) + 1/(d*(a^2 + a^2*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 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 2912

Int[cos[(e_.) + (f_.)*(x_)]*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)
])^(n_.), x_Symbol] :> Dist[1/(b*f), Subst[Int[(a + x)^m*(c + (d/b)*x)^n, x], x, b*Sin[e + f*x]], x] /; FreeQ[
{a, b, c, d, e, f, m, n}, x]

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

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.73 \[ \int \frac {\cos (c+d x) \sin (c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {\log (1+\sin (c+d x))+\frac {1}{1+\sin (c+d x)}}{a^2 d} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.14 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.76

method	result	size
derivativedivides	\(\frac {\ln \left (1+\sin \left (d x +c \right )\right )+\frac {1}{1+\sin \left (d x +c \right )}}{d \,a^{2}}\)	\(28\)
default	\(\frac {\ln \left (1+\sin \left (d x +c \right )\right )+\frac {1}{1+\sin \left (d x +c \right )}}{d \,a^{2}}\)	\(28\)
risch	\(-\frac {i x}{a^{2}}-\frac {2 i c}{d \,a^{2}}+\frac {2 i {\mathrm e}^{i \left (d x +c \right )}}{d \,a^{2} \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{2}}+\frac {2 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d \,a^{2}}\)	\(72\)
parallelrisch	\(\frac {\left (-1-\sin \left (d x +c \right )\right ) \ln \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (2 \sin \left (d x +c \right )+2\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-\sin \left (d x +c \right )}{d \,a^{2} \left (1+\sin \left (d x +c \right )\right )}\)	\(73\)
norman	\(\frac {-\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d a}-\frac {2 \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}-\frac {2 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}-\frac {2 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}-\frac {4 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}-\frac {4 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2} a \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}+\frac {2 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d \,a^{2}}-\frac {\ln \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,a^{2}}\)	\(189\)

[In]

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

[Out]

1/d/a^2*(ln(1+sin(d*x+c))+1/(1+sin(d*x+c)))

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.08 \[ \int \frac {\cos (c+d x) \sin (c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {{\left (\sin \left (d x + c\right ) + 1\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + 1}{a^{2} d \sin \left (d x + c\right ) + a^{2} d} \]

[In]

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

[Out]

((sin(d*x + c) + 1)*log(sin(d*x + c) + 1) + 1)/(a^2*d*sin(d*x + c) + a^2*d)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 95 vs. \(2 (31) = 62\).

Time = 0.42 (sec) , antiderivative size = 95, normalized size of antiderivative = 2.57 \[ \int \frac {\cos (c+d x) \sin (c+d x)}{(a+a \sin (c+d x))^2} \, dx=\begin {cases} \frac {\log {\left (\sin {\left (c + d x \right )} + 1 \right )} \sin {\left (c + d x \right )}}{a^{2} d \sin {\left (c + d x \right )} + a^{2} d} + \frac {\log {\left (\sin {\left (c + d x \right )} + 1 \right )}}{a^{2} d \sin {\left (c + d x \right )} + a^{2} d} + \frac {1}{a^{2} d \sin {\left (c + d x \right )} + a^{2} d} & \text {for}\: d \neq 0 \\\frac {x \sin {\left (c \right )} \cos {\left (c \right )}}{\left (a \sin {\left (c \right )} + a\right )^{2}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((log(sin(c + d*x) + 1)*sin(c + d*x)/(a**2*d*sin(c + d*x) + a**2*d) + log(sin(c + d*x) + 1)/(a**2*d*s
in(c + d*x) + a**2*d) + 1/(a**2*d*sin(c + d*x) + a**2*d), Ne(d, 0)), (x*sin(c)*cos(c)/(a*sin(c) + a)**2, True)
)

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.92 \[ \int \frac {\cos (c+d x) \sin (c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {\frac {1}{a^{2} \sin \left (d x + c\right ) + a^{2}} + \frac {\log \left (\sin \left (d x + c\right ) + 1\right )}{a^{2}}}{d} \]

[In]

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

[Out]

(1/(a^2*sin(d*x + c) + a^2) + log(sin(d*x + c) + 1)/a^2)/d

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.51 \[ \int \frac {\cos (c+d x) \sin (c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {\frac {\log \left (\frac {{\left | a \sin \left (d x + c\right ) + a \right |}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{2} {\left | a \right |}}\right )}{a} - \frac {1}{a \sin \left (d x + c\right ) + a}}{a d} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 9.34 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.92 \[ \int \frac {\cos (c+d x) \sin (c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {1}{a^2\,d\,\left (\sin \left (c+d\,x\right )+1\right )}+\frac {\ln \left (\sin \left (c+d\,x\right )+1\right )}{a^2\,d} \]

[In]

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

[Out]

1/(a^2*d*(sin(c + d*x) + 1)) + log(sin(c + d*x) + 1)/(a^2*d)