\(\int \frac {\cos (c+d x) \sin ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx\) [234]

Optimal result

Integrand size = 27, antiderivative size = 52 \[ \int \frac {\cos (c+d x) \sin ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {2 \log (1+\sin (c+d x))}{a^2 d}+\frac {\sin (c+d x)}{a^2 d}-\frac {1}{d \left (a^2+a^2 \sin (c+d x)\right )} \]

[Out]

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

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {2912, 12, 45} \[ \int \frac {\cos (c+d x) \sin ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {\sin (c+d x)}{a^2 d}-\frac {1}{d \left (a^2 \sin (c+d x)+a^2\right )}-\frac {2 \log (\sin (c+d x)+1)}{a^2 d} \]

[In]

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

[Out]

(-2*Log[1 + Sin[c + d*x]])/(a^2*d) + 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^2}{a^2 (a+x)^2} \, dx,x,a \sin (c+d x)\right )}{a d} \\ & = \frac {\text {Subst}\left (\int \frac {x^2}{(a+x)^2} \, dx,x,a \sin (c+d x)\right )}{a^3 d} \\ & = \frac {\text {Subst}\left (\int \left (1+\frac {a^2}{(a+x)^2}-\frac {2 a}{a+x}\right ) \, dx,x,a \sin (c+d x)\right )}{a^3 d} \\ & = -\frac {2 \log (1+\sin (c+d x))}{a^2 d}+\frac {\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 = 0.04 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.12 \[ \int \frac {\cos (c+d x) \sin ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {-1-2 \log (1+\sin (c+d x))+(1-2 \log (1+\sin (c+d x))) \sin (c+d x)+\sin ^2(c+d x)}{a^2 d (1+\sin (c+d x))} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.20 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.73

method	result	size
derivativedivides	\(\frac {\sin \left (d x +c \right )-2 \ln \left (1+\sin \left (d x +c \right )\right )-\frac {1}{1+\sin \left (d x +c \right )}}{d \,a^{2}}\)	\(38\)
default	\(\frac {\sin \left (d x +c \right )-2 \ln \left (1+\sin \left (d x +c \right )\right )-\frac {1}{1+\sin \left (d x +c \right )}}{d \,a^{2}}\)	\(38\)
parallelrisch	\(\frac {\left (4 \sin \left (d x +c \right )+4\right ) \ln \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-8 \sin \left (d x +c \right )-8\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-\cos \left (2 d x +2 c \right )+4 \sin \left (d x +c \right )+1}{2 d \,a^{2} \left (1+\sin \left (d x +c \right )\right )}\)	\(86\)
risch	\(\frac {2 i x}{a^{2}}-\frac {i {\mathrm e}^{i \left (d x +c \right )}}{2 d \,a^{2}}+\frac {i {\mathrm e}^{-i \left (d x +c \right )}}{2 d \,a^{2}}+\frac {4 i c}{d \,a^{2}}-\frac {2 i {\mathrm e}^{i \left (d x +c \right )}}{d \,a^{2} \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{2}}-\frac {4 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d \,a^{2}}\)	\(108\)
norman	\(\frac {\frac {4 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d a}+\frac {4 \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}+\frac {8 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}+\frac {8 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}+\frac {20 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}+\frac {20 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}+\frac {16 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}+\frac {16 \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3} a \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}-\frac {4 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d \,a^{2}}+\frac {2 \ln \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,a^{2}}\)	\(227\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.10 \[ \int \frac {\cos (c+d x) \sin ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {\cos \left (d x + c\right )^{2} + 2 \, {\left (\sin \left (d x + c\right ) + 1\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - \sin \left (d x + c\right )}{a^{2} d \sin \left (d x + c\right ) + a^{2} d} \]

[In]

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

[Out]

-(cos(d*x + c)^2 + 2*(sin(d*x + c) + 1)*log(sin(d*x + c) + 1) - sin(d*x + c))/(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. 126 vs. \(2 (44) = 88\).

Time = 0.47 (sec) , antiderivative size = 126, normalized size of antiderivative = 2.42 \[ \int \frac {\cos (c+d x) \sin ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\begin {cases} - \frac {2 \log {\left (\sin {\left (c + d x \right )} + 1 \right )} \sin {\left (c + d x \right )}}{a^{2} d \sin {\left (c + d x \right )} + a^{2} d} - \frac {2 \log {\left (\sin {\left (c + d x \right )} + 1 \right )}}{a^{2} d \sin {\left (c + d x \right )} + a^{2} d} + \frac {\sin ^{2}{\left (c + d x \right )}}{a^{2} d \sin {\left (c + d x \right )} + a^{2} d} - \frac {2}{a^{2} d \sin {\left (c + d x \right )} + a^{2} d} & \text {for}\: d \neq 0 \\\frac {x \sin ^{2}{\left (c \right )} \cos {\left (c \right )}}{\left (a \sin {\left (c \right )} + a\right )^{2}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.90 \[ \int \frac {\cos (c+d x) \sin ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {\frac {1}{a^{2} \sin \left (d x + c\right ) + a^{2}} + \frac {2 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{2}} - \frac {\sin \left (d x + c\right )}{a^{2}}}{d} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.35 \[ \int \frac {\cos (c+d x) \sin ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {\frac {2 \, \log \left (\frac {{\left | a \sin \left (d x + c\right ) + a \right |}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{2} {\left | a \right |}}\right )}{a^{2}} + \frac {a \sin \left (d x + c\right ) + a}{a^{3}} - \frac {1}{{\left (a \sin \left (d x + c\right ) + a\right )} a}}{d} \]

[In]

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

[Out]

(2*log(abs(a*sin(d*x + c) + a)/((a*sin(d*x + c) + a)^2*abs(a)))/a^2 + (a*sin(d*x + c) + a)/a^3 - 1/((a*sin(d*x
+ c) + a)*a))/d

Mupad [B] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.87 \[ \int \frac {\cos (c+d x) \sin ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {{\sin \left (c+d\,x\right )}^2-2}{a^2\,d\,\left (\sin \left (c+d\,x\right )+1\right )}-\frac {2\,\ln \left (\sin \left (c+d\,x\right )+1\right )}{a^2\,d} \]

[In]

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

[Out]

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