\(\int \cot (c+d x) \csc (c+d x) (a+a \sin (c+d x))^2 \, dx\) [200]

Optimal result

Integrand size = 25, antiderivative size = 43 \[ \int \cot (c+d x) \csc (c+d x) (a+a \sin (c+d x))^2 \, dx=-\frac {a^2 \csc (c+d x)}{d}+\frac {2 a^2 \log (\sin (c+d x))}{d}+\frac {a^2 \sin (c+d x)}{d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 43, 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} \[ \int \cot (c+d x) \csc (c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {a^2 \sin (c+d x)}{d}-\frac {a^2 \csc (c+d x)}{d}+\frac {2 a^2 \log (\sin (c+d x))}{d} \]

[In]

Int[Cot[c + d*x]*Csc[c + d*x]*(a + a*Sin[c + d*x])^2,x]

[Out]

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

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.88 \[ \int \cot (c+d x) \csc (c+d x) (a+a \sin (c+d x))^2 \, dx=a^2 \left (-\frac {\csc (c+d x)}{d}+\frac {2 \log (\sin (c+d x))}{d}+\frac {\sin (c+d x)}{d}\right ) \]

[In]

Integrate[Cot[c + d*x]*Csc[c + d*x]*(a + a*Sin[c + d*x])^2,x]

[Out]

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

Maple [A] (verified)

Time = 0.14 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.81

method	result	size
derivativedivides	\(-\frac {a^{2} \left (\csc \left (d x +c \right )+2 \ln \left (\csc \left (d x +c \right )\right )-\frac {1}{\csc \left (d x +c \right )}\right )}{d}\)	\(35\)
default	\(-\frac {a^{2} \left (\csc \left (d x +c \right )+2 \ln \left (\csc \left (d x +c \right )\right )-\frac {1}{\csc \left (d x +c \right )}\right )}{d}\)	\(35\)
parallelrisch	\(-\frac {a^{2} \left (-4 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )+4 \ln \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2 \cot \left (\frac {d x}{2}+\frac {c}{2}\right ) \cos \left (d x +c \right )+\sec \left (\frac {d x}{2}+\frac {c}{2}\right ) \csc \left (\frac {d x}{2}+\frac {c}{2}\right )-2 \cot \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}\)	\(83\)
risch	\(-2 i a^{2} x -\frac {i a^{2} {\mathrm e}^{i \left (d x +c \right )}}{2 d}+\frac {i a^{2} {\mathrm e}^{-i \left (d x +c \right )}}{2 d}-\frac {4 i a^{2} c}{d}-\frac {2 i a^{2} {\mathrm e}^{i \left (d x +c \right )}}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}+\frac {2 a^{2} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d}\)	\(106\)
norman	\(\frac {-\frac {a^{2}}{2 d}+\frac {a^{2} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}+\frac {a^{2} \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}-\frac {a^{2} \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+\frac {2 a^{2} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {2 a^{2} \ln \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}\)	\(135\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.07 \[ \int \cot (c+d x) \csc (c+d x) (a+a \sin (c+d x))^2 \, dx=-\frac {a^{2} \cos \left (d x + c\right )^{2} - 2 \, a^{2} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) \sin \left (d x + c\right )}{d \sin \left (d x + c\right )} \]

[In]

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

[Out]

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

Sympy [F]

\[ \int \cot (c+d x) \csc (c+d x) (a+a \sin (c+d x))^2 \, dx=a^{2} \left (\int \cos {\left (c + d x \right )} \csc ^{2}{\left (c + d x \right )}\, dx + \int 2 \sin {\left (c + d x \right )} \cos {\left (c + d x \right )} \csc ^{2}{\left (c + d x \right )}\, dx + \int \sin ^{2}{\left (c + d x \right )} \cos {\left (c + d x \right )} \csc ^{2}{\left (c + d x \right )}\, dx\right ) \]

[In]

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

[Out]

a**2*(Integral(cos(c + d*x)*csc(c + d*x)**2, x) + Integral(2*sin(c + d*x)*cos(c + d*x)*csc(c + d*x)**2, x) + I
ntegral(sin(c + d*x)**2*cos(c + d*x)*csc(c + d*x)**2, x))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.93 \[ \int \cot (c+d x) \csc (c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {2 \, a^{2} \log \left (\sin \left (d x + c\right )\right ) + a^{2} \sin \left (d x + c\right ) - \frac {a^{2}}{\sin \left (d x + c\right )}}{d} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.95 \[ \int \cot (c+d x) \csc (c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {2 \, a^{2} \log \left ({\left | \sin \left (d x + c\right ) \right |}\right ) + a^{2} \sin \left (d x + c\right ) - \frac {a^{2}}{\sin \left (d x + c\right )}}{d} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 9.40 (sec) , antiderivative size = 111, normalized size of antiderivative = 2.58 \[ \int \cot (c+d x) \csc (c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {2\,a^2\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}+\frac {3\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-a^2}{d\,\left (2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}-\frac {a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,d}-\frac {2\,a^2\,\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{d} \]

[In]

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

[Out]

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