\(\int \frac {\cos (c+d x) (A+B \sin (c+d x))}{(a+a \sin (c+d x))^2} \, dx\) [1014]

Optimal result

Integrand size = 29, antiderivative size = 44 \[ \int \frac {\cos (c+d x) (A+B \sin (c+d x))}{(a+a \sin (c+d x))^2} \, dx=\frac {B \log (1+\sin (c+d x))}{a^2 d}-\frac {A-B}{d \left (a^2+a^2 \sin (c+d x)\right )} \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 44, 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} \[ \int \frac {\cos (c+d x) (A+B \sin (c+d x))}{(a+a \sin (c+d x))^2} \, dx=\frac {B \log (\sin (c+d x)+1)}{a^2 d}-\frac {A-B}{d \left (a^2 \sin (c+d x)+a^2\right )} \]

[In]

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

[Out]

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

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.93 \[ \int \frac {\cos (c+d x) (A+B \sin (c+d x))}{(a+a \sin (c+d x))^2} \, dx=\frac {\frac {B \log (1+\sin (c+d x))}{a}-\frac {A-B}{a+a \sin (c+d x)}}{a d} \]

[In]

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

[Out]

((B*Log[1 + Sin[c + d*x]])/a - (A - B)/(a + a*Sin[c + d*x]))/(a*d)

Maple [A] (verified)

Time = 0.24 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.84

method	result	size
derivativedivides	\(\frac {B \ln \left (1+\sin \left (d x +c \right )\right )-\frac {A -B}{1+\sin \left (d x +c \right )}}{d \,a^{2}}\)	\(37\)
default	\(\frac {B \ln \left (1+\sin \left (d x +c \right )\right )-\frac {A -B}{1+\sin \left (d x +c \right )}}{d \,a^{2}}\)	\(37\)
parallelrisch	\(\frac {-B \left (1+\sin \left (d x +c \right )\right ) \ln \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2 B \left (1+\sin \left (d x +c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\sin \left (d x +c \right ) \left (A -B \right )}{\left (1+\sin \left (d x +c \right )\right ) a^{2} d}\)	\(77\)
risch	\(-\frac {i x B}{a^{2}}-\frac {2 i B c}{a^{2} d}-\frac {2 i {\mathrm e}^{i \left (d x +c \right )} \left (A -B \right )}{d \,a^{2} \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{2}}+\frac {2 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{a^{2} d}\)	\(80\)
norman	\(\frac {\frac {\left (-2 B +2 A \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a d}+\frac {\left (-2 B +2 A \right ) \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a d}+\frac {\left (-2 B +2 A \right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a d}+\frac {\left (-2 B +2 A \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a d}+\frac {\left (4 A -4 B \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a d}+\frac {\left (4 A -4 B \right ) \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2} a \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}+\frac {2 B \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{a^{2} d}-\frac {B \ln \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{2} d}\)	\(227\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.02 \[ \int \frac {\cos (c+d x) (A+B \sin (c+d x))}{(a+a \sin (c+d x))^2} \, dx=\frac {{\left (B \sin \left (d x + c\right ) + B\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - A + B}{a^{2} d \sin \left (d x + c\right ) + a^{2} d} \]

[In]

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

[Out]

((B*sin(d*x + c) + B)*log(sin(d*x + c) + 1) - A + B)/(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. 121 vs. \(2 (34) = 68\).

Time = 0.41 (sec) , antiderivative size = 121, normalized size of antiderivative = 2.75 \[ \int \frac {\cos (c+d x) (A+B \sin (c+d x))}{(a+a \sin (c+d x))^2} \, dx=\begin {cases} - \frac {A}{a^{2} d \sin {\left (c + d x \right )} + a^{2} d} + \frac {B \log {\left (\sin {\left (c + d x \right )} + 1 \right )} \sin {\left (c + d x \right )}}{a^{2} d \sin {\left (c + d x \right )} + a^{2} d} + \frac {B \log {\left (\sin {\left (c + d x \right )} + 1 \right )}}{a^{2} d \sin {\left (c + d x \right )} + a^{2} d} + \frac {B}{a^{2} d \sin {\left (c + d x \right )} + a^{2} d} & \text {for}\: d \neq 0 \\\frac {x \left (A + B \sin {\left (c \right )}\right ) \cos {\left (c \right )}}{\left (a \sin {\left (c \right )} + a\right )^{2}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((-A/(a**2*d*sin(c + d*x) + a**2*d) + B*log(sin(c + d*x) + 1)*sin(c + d*x)/(a**2*d*sin(c + d*x) + a**
2*d) + B*log(sin(c + d*x) + 1)/(a**2*d*sin(c + d*x) + a**2*d) + B/(a**2*d*sin(c + d*x) + a**2*d), Ne(d, 0)), (
x*(A + B*sin(c))*cos(c)/(a*sin(c) + a)**2, True))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.98 \[ \int \frac {\cos (c+d x) (A+B \sin (c+d x))}{(a+a \sin (c+d x))^2} \, dx=-\frac {\frac {A - B}{a^{2} \sin \left (d x + c\right ) + a^{2}} - \frac {B \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{2}}}{d} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.73 \[ \int \frac {\cos (c+d x) (A+B \sin (c+d x))}{(a+a \sin (c+d x))^2} \, dx=-\frac {\frac {B {\left (\frac {\log \left (\frac {{\left | a \sin \left (d x + c\right ) + a \right |}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{2} {\left | a \right |}}\right )}{a} - \frac {1}{a \sin \left (d x + c\right ) + a}\right )}}{a} + \frac {A}{{\left (a \sin \left (d x + c\right ) + a\right )} a}}{d} \]

[In]

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

[Out]

-(B*(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 + A/((a*sin(d*
x + c) + a)*a))/d

Mupad [B] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.93 \[ \int \frac {\cos (c+d x) (A+B \sin (c+d x))}{(a+a \sin (c+d x))^2} \, dx=\frac {B\,\ln \left (\sin \left (c+d\,x\right )+1\right )}{a^2\,d}-\frac {A-B}{a^2\,d\,\left (\sin \left (c+d\,x\right )+1\right )} \]

[In]

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

[Out]

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