\(\int \frac {\cos (c+d x) (A+B \sin (c+d x))}{(a+b \sin (c+d x))^2} \, dx\) [1555]

Optimal result

Integrand size = 29, antiderivative size = 48 \[ \int \frac {\cos (c+d x) (A+B \sin (c+d x))}{(a+b \sin (c+d x))^2} \, dx=\frac {B \log (a+b \sin (c+d x))}{b^2 d}-\frac {A b-a B}{b^2 d (a+b \sin (c+d x))} \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 48, 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} \[ \int \frac {\cos (c+d x) (A+B \sin (c+d x))}{(a+b \sin (c+d x))^2} \, dx=\frac {B \log (a+b \sin (c+d x))}{b^2 d}-\frac {A b-a B}{b^2 d (a+b \sin (c+d x))} \]

[In]

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

[Out]

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

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.88 \[ \int \frac {\cos (c+d x) (A+B \sin (c+d x))}{(a+b \sin (c+d x))^2} \, dx=\frac {B \log (a+b \sin (c+d x))+\frac {-A b+a B}{a+b \sin (c+d x)}}{b^2 d} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.43 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.98

method	result	size
derivativedivides	\(\frac {-\frac {A b -B a}{b^{2} \left (a +b \sin \left (d x +c \right )\right )}+\frac {B \ln \left (a +b \sin \left (d x +c \right )\right )}{b^{2}}}{d}\)	\(47\)
default	\(\frac {-\frac {A b -B a}{b^{2} \left (a +b \sin \left (d x +c \right )\right )}+\frac {B \ln \left (a +b \sin \left (d x +c \right )\right )}{b^{2}}}{d}\)	\(47\)
parallelrisch	\(\frac {B \left (a +b \sin \left (d x +c \right )\right ) \ln \left (2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )-B \left (a +b \sin \left (d x +c \right )\right ) \ln \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-A b +B a}{b^{2} d \left (a +b \sin \left (d x +c \right )\right )}\)	\(92\)
risch	\(-\frac {i x B}{b^{2}}-\frac {2 i B c}{b^{2} d}-\frac {2 \left (A b -B a \right ) {\mathrm e}^{i \left (d x +c \right )}}{b^{2} d \left (-i b \,{\mathrm e}^{2 i \left (d x +c \right )}+i b +2 a \,{\mathrm e}^{i \left (d x +c \right )}\right )}+\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}\right ) B}{b^{2} d}\)	\(114\)
norman	\(\frac {\frac {2 \left (A b -B a \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d b a}+\frac {4 \left (A b -B a \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d b a}+\frac {2 \left (A b -B a \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d b a}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2} \left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \right )}+\frac {B \ln \left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \right )}{b^{2} d}-\frac {B \ln \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{2} d}\)	\(195\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.12 \[ \int \frac {\cos (c+d x) (A+B \sin (c+d x))}{(a+b \sin (c+d x))^2} \, dx=\frac {B a - A b + {\left (B b \sin \left (d x + c\right ) + B a\right )} \log \left (b \sin \left (d x + c\right ) + a\right )}{b^{3} d \sin \left (d x + c\right ) + a b^{2} d} \]

[In]

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

[Out]

(B*a - A*b + (B*b*sin(d*x + c) + B*a)*log(b*sin(d*x + c) + a))/(b^3*d*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. 178 vs. \(2 (39) = 78\).

Time = 0.54 (sec) , antiderivative size = 178, normalized size of antiderivative = 3.71 \[ \int \frac {\cos (c+d x) (A+B \sin (c+d x))}{(a+b \sin (c+d x))^2} \, dx=\begin {cases} \frac {x \left (A + B \sin {\left (c \right )}\right ) \cos {\left (c \right )}}{a^{2}} & \text {for}\: b = 0 \wedge d = 0 \\\frac {\frac {A \sin {\left (c + d x \right )}}{d} + \frac {B \sin ^{2}{\left (c + d x \right )}}{2 d}}{a^{2}} & \text {for}\: b = 0 \\\frac {x \left (A + B \sin {\left (c \right )}\right ) \cos {\left (c \right )}}{\left (a + b \sin {\left (c \right )}\right )^{2}} & \text {for}\: d = 0 \\- \frac {A b}{a b^{2} d + b^{3} d \sin {\left (c + d x \right )}} + \frac {B a \log {\left (\frac {a}{b} + \sin {\left (c + d x \right )} \right )}}{a b^{2} d + b^{3} d \sin {\left (c + d x \right )}} + \frac {B a}{a b^{2} d + b^{3} d \sin {\left (c + d x \right )}} + \frac {B b \log {\left (\frac {a}{b} + \sin {\left (c + d x \right )} \right )} \sin {\left (c + d x \right )}}{a b^{2} d + b^{3} d \sin {\left (c + d x \right )}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((x*(A + B*sin(c))*cos(c)/a**2, Eq(b, 0) & Eq(d, 0)), ((A*sin(c + d*x)/d + B*sin(c + d*x)**2/(2*d))/a
**2, Eq(b, 0)), (x*(A + B*sin(c))*cos(c)/(a + b*sin(c))**2, Eq(d, 0)), (-A*b/(a*b**2*d + b**3*d*sin(c + d*x))
+ B*a*log(a/b + sin(c + d*x))/(a*b**2*d + b**3*d*sin(c + d*x)) + B*a/(a*b**2*d + b**3*d*sin(c + d*x)) + B*b*lo
g(a/b + sin(c + d*x))*sin(c + d*x)/(a*b**2*d + b**3*d*sin(c + d*x)), True))

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.00 \[ \int \frac {\cos (c+d x) (A+B \sin (c+d x))}{(a+b \sin (c+d x))^2} \, dx=\frac {\frac {B a - A b}{b^{3} \sin \left (d x + c\right ) + a b^{2}} + \frac {B \log \left (b \sin \left (d x + c\right ) + a\right )}{b^{2}}}{d} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.67 \[ \int \frac {\cos (c+d x) (A+B \sin (c+d x))}{(a+b \sin (c+d x))^2} \, dx=-\frac {\frac {B {\left (\frac {\log \left (\frac {{\left | b \sin \left (d x + c\right ) + a \right |}}{{\left (b \sin \left (d x + c\right ) + a\right )}^{2} {\left | b \right |}}\right )}{b} - \frac {a}{{\left (b \sin \left (d x + c\right ) + a\right )} b}\right )}}{b} + \frac {A}{{\left (b \sin \left (d x + c\right ) + a\right )} b}}{d} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.00 \[ \int \frac {\cos (c+d x) (A+B \sin (c+d x))}{(a+b \sin (c+d x))^2} \, dx=\frac {B\,\ln \left (a+b\,\sin \left (c+d\,x\right )\right )}{b^2\,d}-\frac {A\,b-B\,a}{b^2\,d\,\left (a+b\,\sin \left (c+d\,x\right )\right )} \]

[In]

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

[Out]

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