3.57 \(\int \frac{\cot ^2(c+d x)}{a+a \sin (c+d x)} \, dx\)

Optimal. Leaf size=29 \[ \frac{\tanh ^{-1}(\cos (c+d x))}{a d}-\frac{\cot (c+d x)}{a d} \]

[Out]

ArcTanh[Cos[c + d*x]]/(a*d) - Cot[c + d*x]/(a*d)

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Rubi [A]  time = 0.0512723, antiderivative size = 29, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {2706, 3767, 8, 3770} \[ \frac{\tanh ^{-1}(\cos (c+d x))}{a d}-\frac{\cot (c+d x)}{a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

ArcTanh[Cos[c + d*x]]/(a*d) - Cot[c + d*x]/(a*d)

Rule 2706

Int[((g_.)*tan[(e_.) + (f_.)*(x_)])^(p_.)/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[1/a, Int[S
ec[e + f*x]^2*(g*Tan[e + f*x])^p, x], x] - Dist[1/(b*g), Int[Sec[e + f*x]*(g*Tan[e + f*x])^(p + 1), x], x] /;
FreeQ[{a, b, e, f, g, p}, x] && EqQ[a^2 - b^2, 0] && NeQ[p, -1]

Rule 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

\begin{align*} \int \frac{\cot ^2(c+d x)}{a+a \sin (c+d x)} \, dx &=-\frac{\int \csc (c+d x) \, dx}{a}+\frac{\int \csc ^2(c+d x) \, dx}{a}\\ &=\frac{\tanh ^{-1}(\cos (c+d x))}{a d}-\frac{\operatorname{Subst}(\int 1 \, dx,x,\cot (c+d x))}{a d}\\ &=\frac{\tanh ^{-1}(\cos (c+d x))}{a d}-\frac{\cot (c+d x)}{a d}\\ \end{align*}

Mathematica [B]  time = 0.218073, size = 69, normalized size = 2.38 \[ -\frac{\csc \left (\frac{1}{2} (c+d x)\right ) \sec \left (\frac{1}{2} (c+d x)\right ) \left (\cos (c+d x)+\sin (c+d x) \left (\log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )-\log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )\right )}{2 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.068, size = 56, normalized size = 1.9 \begin{align*}{\frac{1}{2\,da}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }-{\frac{1}{2\,da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1}}-{\frac{1}{da}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 1.08722, size = 95, normalized size = 3.28 \begin{align*} -\frac{\frac{2 \, \log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a} + \frac{\cos \left (d x + c\right ) + 1}{a \sin \left (d x + c\right )} - \frac{\sin \left (d x + c\right )}{a{\left (\cos \left (d x + c\right ) + 1\right )}}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 1.53889, size = 173, normalized size = 5.97 \begin{align*} \frac{\log \left (\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) \sin \left (d x + c\right ) - \log \left (-\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) \sin \left (d x + c\right ) - 2 \, \cos \left (d x + c\right )}{2 \, a d \sin \left (d x + c\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{\cot ^{2}{\left (c + d x \right )}}{\sin{\left (c + d x \right )} + 1}\, dx}{a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(cot(c + d*x)**2/(sin(c + d*x) + 1), x)/a

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Giac [B]  time = 1.2855, size = 88, normalized size = 3.03 \begin{align*} -\frac{\frac{2 \, \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right )}{a} - \frac{\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{a} - \frac{2 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1}{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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