\(\int \frac {\cos (c+d x) \sin (c+d x)}{a+b \sin (c+d x)} \, dx\) [1281]

Optimal result

Integrand size = 25, antiderivative size = 34 \[ \int \frac {\cos (c+d x) \sin (c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {a \log (a+b \sin (c+d x))}{b^2 d}+\frac {\sin (c+d x)}{b d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 34, 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} \[ \int \frac {\cos (c+d x) \sin (c+d x)}{a+b \sin (c+d x)} \, dx=\frac {\sin (c+d x)}{b d}-\frac {a \log (a+b \sin (c+d x))}{b^2 d} \]

[In]

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

[Out]

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.15 \[ \int \frac {\cos (c+d x) \sin (c+d x)}{a+b \sin (c+d x)} \, dx=\frac {1}{2} \left (-\frac {2 a \log (a+b \sin (c+d x))}{b^2 d}+\frac {2 \sin (c+d x)}{b d}\right ) \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.18 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.97

method	result	size
derivativedivides	\(\frac {\frac {\sin \left (d x +c \right )}{b}-\frac {a \ln \left (a +b \sin \left (d x +c \right )\right )}{b^{2}}}{d}\)	\(33\)
default	\(\frac {\frac {\sin \left (d x +c \right )}{b}-\frac {a \ln \left (a +b \sin \left (d x +c \right )\right )}{b^{2}}}{d}\)	\(33\)
parallelrisch	\(\frac {b \sin \left (d x +c \right )+a \ln \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-a \ln \left (2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )}{b^{2} d}\)	\(61\)
risch	\(\frac {i a x}{b^{2}}-\frac {i {\mathrm e}^{i \left (d x +c \right )}}{2 b d}+\frac {i {\mathrm e}^{-i \left (d x +c \right )}}{2 b d}+\frac {2 i a c}{b^{2} d}-\frac {a \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}\right )}{b^{2} d}\)	\(94\)
norman	\(\frac {\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{b d}+\frac {2 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+\frac {a \ln \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{2} d}-\frac {a \ln \left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \right )}{b^{2} d}\)	\(114\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.91 \[ \int \frac {\cos (c+d x) \sin (c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {a \log \left (b \sin \left (d x + c\right ) + a\right ) - b \sin \left (d x + c\right )}{b^{2} d} \]

[In]

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

[Out]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 65 vs. \(2 (27) = 54\).

Time = 0.34 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.91 \[ \int \frac {\cos (c+d x) \sin (c+d x)}{a+b \sin (c+d x)} \, dx=\begin {cases} \frac {x \sin {\left (c \right )} \cos {\left (c \right )}}{a} & \text {for}\: b = 0 \wedge d = 0 \\\frac {\sin ^{2}{\left (c + d x \right )}}{2 a d} & \text {for}\: b = 0 \\\frac {x \sin {\left (c \right )} \cos {\left (c \right )}}{a + b \sin {\left (c \right )}} & \text {for}\: d = 0 \\- \frac {a \log {\left (\frac {a}{b} + \sin {\left (c + d x \right )} \right )}}{b^{2} d} + \frac {\sin {\left (c + d x \right )}}{b d} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((x*sin(c)*cos(c)/a, Eq(b, 0) & Eq(d, 0)), (sin(c + d*x)**2/(2*a*d), Eq(b, 0)), (x*sin(c)*cos(c)/(a +
b*sin(c)), Eq(d, 0)), (-a*log(a/b + sin(c + d*x))/(b**2*d) + sin(c + d*x)/(b*d), True))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.97 \[ \int \frac {\cos (c+d x) \sin (c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {\frac {a \log \left (b \sin \left (d x + c\right ) + a\right )}{b^{2}} - \frac {\sin \left (d x + c\right )}{b}}{d} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.43 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.00 \[ \int \frac {\cos (c+d x) \sin (c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {\frac {a \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{b^{2}} - \frac {\sin \left (d x + c\right )}{b}}{d} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.91 \[ \int \frac {\cos (c+d x) \sin (c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {a\,\ln \left (a+b\,\sin \left (c+d\,x\right )\right )-b\,\sin \left (c+d\,x\right )}{b^2\,d} \]

[In]

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

[Out]

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