\(\int \cot (c+d x) \csc ^2(c+d x) (a+a \sin (c+d x))^2 \, dx\) [201]

Optimal result

Integrand size = 27, antiderivative size = 47 \[ \int \cot (c+d x) \csc ^2(c+d x) (a+a \sin (c+d x))^2 \, dx=-\frac {2 a^2 \csc (c+d x)}{d}-\frac {a^2 \csc ^2(c+d x)}{2 d}+\frac {a^2 \log (\sin (c+d x))}{d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 47, 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} \[ \int \cot (c+d x) \csc ^2(c+d x) (a+a \sin (c+d x))^2 \, dx=-\frac {a^2 \csc ^2(c+d x)}{2 d}-\frac {2 a^2 \csc (c+d x)}{d}+\frac {a^2 \log (\sin (c+d x))}{d} \]

[In]

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

[Out]

(-2*a^2*Csc[c + d*x])/d - (a^2*Csc[c + d*x]^2)/(2*d) + (a^2*Log[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^3 (a+x)^2}{x^3} \, dx,x,a \sin (c+d x)\right )}{a d} \\ & = \frac {a^2 \text {Subst}\left (\int \frac {(a+x)^2}{x^3} \, dx,x,a \sin (c+d x)\right )}{d} \\ & = \frac {a^2 \text {Subst}\left (\int \left (\frac {a^2}{x^3}+\frac {2 a}{x^2}+\frac {1}{x}\right ) \, dx,x,a \sin (c+d x)\right )}{d} \\ & = -\frac {2 a^2 \csc (c+d x)}{d}-\frac {a^2 \csc ^2(c+d x)}{2 d}+\frac {a^2 \log (\sin (c+d x))}{d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.89 \[ \int \cot (c+d x) \csc ^2(c+d x) (a+a \sin (c+d x))^2 \, dx=a^2 \left (-\frac {2 \csc (c+d x)}{d}-\frac {\csc ^2(c+d x)}{2 d}+\frac {\log (\sin (c+d x))}{d}\right ) \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.14 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.74

method	result	size
derivativedivides	\(-\frac {a^{2} \left (\frac {\left (\csc ^{2}\left (d x +c \right )\right )}{2}+2 \csc \left (d x +c \right )+\ln \left (\csc \left (d x +c \right )\right )\right )}{d}\)	\(35\)
default	\(-\frac {a^{2} \left (\frac {\left (\csc ^{2}\left (d x +c \right )\right )}{2}+2 \csc \left (d x +c \right )+\ln \left (\csc \left (d x +c \right )\right )\right )}{d}\)	\(35\)
parallelrisch	\(-\frac {a^{2} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )+\cot ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )+8 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+8 \cot \left (\frac {d x}{2}+\frac {c}{2}\right )+8 \ln \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-8 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )}{8 d}\)	\(80\)
risch	\(-i a^{2} x -\frac {2 i a^{2} c}{d}-\frac {2 i a^{2} \left (i {\mathrm e}^{2 i \left (d x +c \right )}+2 \,{\mathrm e}^{3 i \left (d x +c \right )}-2 \,{\mathrm e}^{i \left (d x +c \right )}\right )}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{2}}+\frac {a^{2} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d}\)	\(95\)
norman	\(\frac {-\frac {a^{2}}{8 d}-\frac {a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {3 a^{2} \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {3 a^{2} \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {a^{2} \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {a^{2} \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d}+\frac {a^{2} \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+\frac {a^{2} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {a^{2} \ln \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}\)	\(189\)

[In]

int(cos(d*x+c)*csc(d*x+c)^3*(a+a*sin(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.32 \[ \int \cot (c+d x) \csc ^2(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {4 \, a^{2} \sin \left (d x + c\right ) + a^{2} + 2 \, {\left (a^{2} \cos \left (d x + c\right )^{2} - a^{2}\right )} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right )}{2 \, {\left (d \cos \left (d x + c\right )^{2} - d\right )}} \]

[In]

integrate(cos(d*x+c)*csc(d*x+c)^3*(a+a*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

1/2*(4*a^2*sin(d*x + c) + a^2 + 2*(a^2*cos(d*x + c)^2 - a^2)*log(1/2*sin(d*x + c)))/(d*cos(d*x + c)^2 - d)

Sympy [F]

\[ \int \cot (c+d x) \csc ^2(c+d x) (a+a \sin (c+d x))^2 \, dx=a^{2} \left (\int \cos {\left (c + d x \right )} \csc ^{3}{\left (c + d x \right )}\, dx + \int 2 \sin {\left (c + d x \right )} \cos {\left (c + d x \right )} \csc ^{3}{\left (c + d x \right )}\, dx + \int \sin ^{2}{\left (c + d x \right )} \cos {\left (c + d x \right )} \csc ^{3}{\left (c + d x \right )}\, dx\right ) \]

[In]

integrate(cos(d*x+c)*csc(d*x+c)**3*(a+a*sin(d*x+c))**2,x)

[Out]

a**2*(Integral(cos(c + d*x)*csc(c + d*x)**3, x) + Integral(2*sin(c + d*x)*cos(c + d*x)*csc(c + d*x)**3, x) + I
ntegral(sin(c + d*x)**2*cos(c + d*x)*csc(c + d*x)**3, x))

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.91 \[ \int \cot (c+d x) \csc ^2(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {2 \, a^{2} \log \left (\sin \left (d x + c\right )\right ) - \frac {4 \, a^{2} \sin \left (d x + c\right ) + a^{2}}{\sin \left (d x + c\right )^{2}}}{2 \, d} \]

[In]

integrate(cos(d*x+c)*csc(d*x+c)^3*(a+a*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

1/2*(2*a^2*log(sin(d*x + c)) - (4*a^2*sin(d*x + c) + a^2)/sin(d*x + c)^2)/d

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.94 \[ \int \cot (c+d x) \csc ^2(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {2 \, a^{2} \log \left ({\left | \sin \left (d x + c\right ) \right |}\right ) - \frac {4 \, a^{2} \sin \left (d x + c\right ) + a^{2}}{\sin \left (d x + c\right )^{2}}}{2 \, d} \]

[In]

integrate(cos(d*x+c)*csc(d*x+c)^3*(a+a*sin(d*x+c))^2,x, algorithm="giac")

[Out]

1/2*(2*a^2*log(abs(sin(d*x + c))) - (4*a^2*sin(d*x + c) + a^2)/sin(d*x + c)^2)/d

Mupad [B] (verification not implemented)

Time = 9.17 (sec) , antiderivative size = 111, normalized size of antiderivative = 2.36 \[ \int \cot (c+d x) \csc ^2(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {a^2\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}-\frac {a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{8\,d}-\frac {{\mathrm {cot}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (\frac {a^2}{8}+a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}-\frac {a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d}-\frac {a^2\,\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{d} \]

[In]

int((cos(c + d*x)*(a + a*sin(c + d*x))^2)/sin(c + d*x)^3,x)

[Out]

(a^2*log(tan(c/2 + (d*x)/2)))/d - (a^2*tan(c/2 + (d*x)/2)^2)/(8*d) - (cot(c/2 + (d*x)/2)^2*(a^2/8 + a^2*tan(c/
2 + (d*x)/2)))/d - (a^2*tan(c/2 + (d*x)/2))/d - (a^2*log(tan(c/2 + (d*x)/2)^2 + 1))/d