\(\int \cot (c+d x) (a+a \sin (c+d x))^2 \, dx\) [199]

Optimal result

Integrand size = 19, antiderivative size = 47 \[ \int \cot (c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {a^2 \log (\sin (c+d x))}{d}+\frac {2 a^2 \sin (c+d x)}{d}+\frac {a^2 \sin ^2(c+d x)}{2 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {2786, 45} \[ \int \cot (c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {a^2 \sin ^2(c+d x)}{2 d}+\frac {2 a^2 \sin (c+d x)}{d}+\frac {a^2 \log (\sin (c+d x))}{d} \]

[In]

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

[Out]

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

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 2786

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(p_.), x_Symbol] :> Dist[1/f, Subst[I
nt[x^p*((a + x)^(m - (p + 1)/2)/(a - x)^((p + 1)/2)), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x]
&& EqQ[a^2 - b^2, 0] && IntegerQ[(p + 1)/2]

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {(a+x)^2}{x} \, dx,x,a \sin (c+d x)\right )}{d} \\ & = \frac {\text {Subst}\left (\int \left (2 a+\frac {a^2}{x}+x\right ) \, dx,x,a \sin (c+d x)\right )}{d} \\ & = \frac {a^2 \log (\sin (c+d x))}{d}+\frac {2 a^2 \sin (c+d x)}{d}+\frac {a^2 \sin ^2(c+d x)}{2 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.00 \[ \int \cot (c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {a^2 \log (\sin (c+d x))}{d}+\frac {2 a^2 \sin (c+d x)}{d}+\frac {a^2 \sin ^2(c+d x)}{2 d} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.20 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.79

method	result	size
derivativedivides	\(-\frac {a^{2} \left (\ln \left (\csc \left (d x +c \right )\right )-\frac {2}{\csc \left (d x +c \right )}-\frac {1}{2 \csc \left (d x +c \right )^{2}}\right )}{d}\)	\(37\)
default	\(-\frac {a^{2} \left (\ln \left (\csc \left (d x +c \right )\right )-\frac {2}{\csc \left (d x +c \right )}-\frac {1}{2 \csc \left (d x +c \right )^{2}}\right )}{d}\)	\(37\)
risch	\(-i a^{2} x -\frac {a^{2} {\mathrm e}^{2 i \left (d x +c \right )}}{8 d}-\frac {a^{2} {\mathrm e}^{-2 i \left (d x +c \right )}}{8 d}-\frac {2 i a^{2} c}{d}+\frac {a^{2} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d}+\frac {2 a^{2} \sin \left (d x +c \right )}{d}\)	\(86\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.91 \[ \int \cot (c+d x) (a+a \sin (c+d x))^2 \, dx=-\frac {a^{2} \cos \left (d x + c\right )^{2} - 2 \, a^{2} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) - 4 \, a^{2} \sin \left (d x + c\right )}{2 \, d} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

a**2*(Integral(cos(c + d*x)*csc(c + d*x), x) + Integral(2*sin(c + d*x)*cos(c + d*x)*csc(c + d*x), x) + Integra
l(sin(c + d*x)**2*cos(c + d*x)*csc(c + d*x), x))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.87 \[ \int \cot (c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {a^{2} \sin \left (d x + c\right )^{2} + 2 \, a^{2} \log \left (\sin \left (d x + c\right )\right ) + 4 \, a^{2} \sin \left (d x + c\right )}{2 \, d} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.89 \[ \int \cot (c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {a^{2} \sin \left (d x + c\right )^{2} + 2 \, a^{2} \log \left ({\left | \sin \left (d x + c\right ) \right |}\right ) + 4 \, a^{2} \sin \left (d x + c\right )}{2 \, d} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 9.30 (sec) , antiderivative size = 119, normalized size of antiderivative = 2.53 \[ \int \cot (c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {a^2\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}+\frac {4\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+2\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+4\,a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}-\frac {a^2\,\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{d} \]

[In]

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

[Out]

(a^2*log(tan(c/2 + (d*x)/2)))/d + (2*a^2*tan(c/2 + (d*x)/2)^2 + 4*a^2*tan(c/2 + (d*x)/2)^3 + 4*a^2*tan(c/2 + (
d*x)/2))/(d*(2*tan(c/2 + (d*x)/2)^2 + tan(c/2 + (d*x)/2)^4 + 1)) - (a^2*log(tan(c/2 + (d*x)/2)^2 + 1))/d