\(\int \frac {\cos (c+d x) (A+B \sin (c+d x))}{a+b \sin (c+d x)} \, dx\) [1547]

Optimal result

Integrand size = 29, antiderivative size = 41 \[ \int \frac {\cos (c+d x) (A+B \sin (c+d x))}{a+b \sin (c+d x)} \, dx=\frac {(A b-a B) \log (a+b \sin (c+d x))}{b^2 d}+\frac {B \sin (c+d x)}{b d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.069, Rules used = {2912, 45} \[ \int \frac {\cos (c+d x) (A+B \sin (c+d x))}{a+b \sin (c+d x)} \, dx=\frac {(A b-a B) \log (a+b \sin (c+d x))}{b^2 d}+\frac {B \sin (c+d x)}{b d} \]

[In]

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

[Out]

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

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

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.95 \[ \int \frac {\cos (c+d x) (A+B \sin (c+d x))}{a+b \sin (c+d x)} \, dx=\frac {\frac {(A b-a B) \log (a+b \sin (c+d x))}{b}+B \sin (c+d x)}{b d} \]

[In]

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

[Out]

(((A*b - a*B)*Log[a + b*Sin[c + d*x]])/b + B*Sin[c + d*x])/(b*d)

Maple [A] (verified)

Time = 0.36 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.98

method	result	size
derivativedivides	\(\frac {\frac {\sin \left (d x +c \right ) B}{b}+\frac {\left (A b -B a \right ) \ln \left (a +b \sin \left (d x +c \right )\right )}{b^{2}}}{d}\)	\(40\)
default	\(\frac {\frac {\sin \left (d x +c \right ) B}{b}+\frac {\left (A b -B a \right ) \ln \left (a +b \sin \left (d x +c \right )\right )}{b^{2}}}{d}\)	\(40\)
parallelrisch	\(\frac {-A \ln \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b +A \ln \left (2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right ) b +B \ln \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a -B \ln \left (2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right ) a +B b \sin \left (d x +c \right )}{d \,b^{2}}\)	\(110\)
norman	\(\frac {\frac {2 B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d b}+\frac {2 B \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d b}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+\frac {\left (A b -B a \right ) \ln \left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \right )}{b^{2} d}-\frac {\left (A b -B a \right ) \ln \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{2} d}\)	\(130\)
risch	\(-\frac {i x A}{b}+\frac {i x B a}{b^{2}}-\frac {i {\mathrm e}^{i \left (d x +c \right )} B}{2 b d}+\frac {i {\mathrm e}^{-i \left (d x +c \right )} B}{2 b d}-\frac {2 i A c}{d b}+\frac {2 i B a c}{d \,b^{2}}+\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}\right ) A}{d b}-\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}\right ) B a}{d \,b^{2}}\)	\(154\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.93 \[ \int \frac {\cos (c+d x) (A+B \sin (c+d x))}{a+b \sin (c+d x)} \, dx=\frac {B b \sin \left (d x + c\right ) - {\left (B a - A b\right )} \log \left (b \sin \left (d x + c\right ) + a\right )}{b^{2} d} \]

[In]

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

[Out]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 104 vs. \(2 (34) = 68\).

Time = 0.35 (sec) , antiderivative size = 104, normalized size of antiderivative = 2.54 \[ \int \frac {\cos (c+d x) (A+B \sin (c+d x))}{a+b \sin (c+d x)} \, dx=\begin {cases} \frac {x \left (A + B \sin {\left (c \right )}\right ) \cos {\left (c \right )}}{a} & \text {for}\: b = 0 \wedge d = 0 \\\frac {\frac {A \sin {\left (c + d x \right )}}{d} + \frac {B \sin ^{2}{\left (c + d x \right )}}{2 d}}{a} & \text {for}\: b = 0 \\\frac {x \left (A + B \sin {\left (c \right )}\right ) \cos {\left (c \right )}}{a + b \sin {\left (c \right )}} & \text {for}\: d = 0 \\\frac {A \log {\left (\frac {a}{b} + \sin {\left (c + d x \right )} \right )}}{b d} - \frac {B a \log {\left (\frac {a}{b} + \sin {\left (c + d x \right )} \right )}}{b^{2} d} + \frac {B \sin {\left (c + d x \right )}}{b d} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.98 \[ \int \frac {\cos (c+d x) (A+B \sin (c+d x))}{a+b \sin (c+d x)} \, dx=\frac {\frac {B \sin \left (d x + c\right )}{b} - \frac {{\left (B a - A b\right )} \log \left (b \sin \left (d x + c\right ) + a\right )}{b^{2}}}{d} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.00 \[ \int \frac {\cos (c+d x) (A+B \sin (c+d x))}{a+b \sin (c+d x)} \, dx=\frac {\frac {B \sin \left (d x + c\right )}{b} - \frac {{\left (B a - A b\right )} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{b^{2}}}{d} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.00 \[ \int \frac {\cos (c+d x) (A+B \sin (c+d x))}{a+b \sin (c+d x)} \, dx=\frac {B\,\sin \left (c+d\,x\right )}{b\,d}+\frac {\ln \left (a+b\,\sin \left (c+d\,x\right )\right )\,\left (A\,b-B\,a\right )}{b^2\,d} \]

[In]

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

[Out]

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