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

Optimal. Leaf size=54 \[ -\frac{\cot (c+d x)}{a^2 d}+\frac{2 \tanh ^{-1}(\cos (c+d x))}{a^2 d}-\frac{2 \cot (c+d x)}{a^2 d (\csc (c+d x)+1)} \]

[Out]

(2*ArcTanh[Cos[c + d*x]])/(a^2*d) - Cot[c + d*x]/(a^2*d) - (2*Cot[c + d*x])/(a^2*d*(1 + Csc[c + d*x]))

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Rubi [A]  time = 0.0962669, antiderivative size = 54, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.238, Rules used = {2709, 3770, 3767, 8, 3777} \[ -\frac{\cot (c+d x)}{a^2 d}+\frac{2 \tanh ^{-1}(\cos (c+d x))}{a^2 d}-\frac{2 \cot (c+d x)}{a^2 d (\csc (c+d x)+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*ArcTanh[Cos[c + d*x]])/(a^2*d) - Cot[c + d*x]/(a^2*d) - (2*Cot[c + d*x])/(a^2*d*(1 + Csc[c + d*x]))

Rule 2709

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*tan[(e_.) + (f_.)*(x_)]^(p_), x_Symbol] :> Dist[a^p, Int[Expan
dIntegrand[(Sin[e + f*x]^p*(a + b*Sin[e + f*x])^(m - p/2))/(a - b*Sin[e + f*x])^(p/2), x], x], x] /; FreeQ[{a,
b, e, f}, x] && EqQ[a^2 - b^2, 0] && IntegersQ[m, p/2] && (LtQ[p, 0] || GtQ[m - p/2, 0])

Rule 3770

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

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 3777

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> -Simp[(Cot[c + d*x]*(a + b*Csc[c + d*x])^n)/(d*(
2*n + 1)), x] + Dist[1/(a^2*(2*n + 1)), Int[(a + b*Csc[c + d*x])^(n + 1)*(a*(2*n + 1) - b*(n + 1)*Csc[c + d*x]
), x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0] && LeQ[n, -1] && IntegerQ[2*n]

Rubi steps

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

Mathematica [B]  time = 0.761603, size = 216, normalized size = 4. \[ -\frac{\csc \left (\frac{1}{2} (c+d x)\right ) \sec \left (\frac{1}{2} (c+d x)\right ) \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^3 \left (\cos \left (\frac{3}{2} (c+d x)\right ) \left (-2 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )+2 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )+5\right )+\cos \left (\frac{1}{2} (c+d x)\right ) \left (2 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )-2 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )-3\right )+2 \sin \left (\frac{1}{2} (c+d x)\right ) \left (2 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )-2 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )+\cos (c+d x) \left (2 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )-2 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )+1\right )\right )\right )}{4 a^2 d (\sin (c+d x)+1)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Csc[(c + d*x)/2]*Sec[(c + d*x)/2]*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^3*(Cos[(3*(c + d*x))/2]*(5 + 2*Log[C
os[(c + d*x)/2]] - 2*Log[Sin[(c + d*x)/2]]) + Cos[(c + d*x)/2]*(-3 - 2*Log[Cos[(c + d*x)/2]] + 2*Log[Sin[(c +
d*x)/2]]) + 2*(-2*Log[Cos[(c + d*x)/2]] + 2*Log[Sin[(c + d*x)/2]] + Cos[c + d*x]*(1 - 2*Log[Cos[(c + d*x)/2]]
+ 2*Log[Sin[(c + d*x)/2]]))*Sin[(c + d*x)/2]))/(4*a^2*d*(1 + Sin[c + d*x])^2)

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Maple [A]  time = 0.145, size = 77, normalized size = 1.4 \begin{align*}{\frac{1}{2\,d{a}^{2}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }-4\,{\frac{1}{d{a}^{2} \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) }}-{\frac{1}{2\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1}}-2\,{\frac{\ln \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) }{d{a}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 1.16115, size = 157, normalized size = 2.91 \begin{align*} -\frac{\frac{\frac{9 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1}{\frac{a^{2} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}} + \frac{4 \, \log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}} - \frac{\sin \left (d x + c\right )}{a^{2}{\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(cos(d*x+c)^2*csc(d*x+c)^2/(a+a*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

-1/2*((9*sin(d*x + c)/(cos(d*x + c) + 1) + 1)/(a^2*sin(d*x + c)/(cos(d*x + c) + 1) + a^2*sin(d*x + c)^2/(cos(d
*x + c) + 1)^2) + 4*log(sin(d*x + c)/(cos(d*x + c) + 1))/a^2 - sin(d*x + c)/(a^2*(cos(d*x + c) + 1)))/d

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Fricas [B]  time = 1.70917, size = 433, normalized size = 8.02 \begin{align*} \frac{3 \, \cos \left (d x + c\right )^{2} +{\left (\cos \left (d x + c\right )^{2} -{\left (\cos \left (d x + c\right ) + 1\right )} \sin \left (d x + c\right ) - 1\right )} \log \left (\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) -{\left (\cos \left (d x + c\right )^{2} -{\left (\cos \left (d x + c\right ) + 1\right )} \sin \left (d x + c\right ) - 1\right )} \log \left (-\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) +{\left (3 \, \cos \left (d x + c\right ) + 2\right )} \sin \left (d x + c\right ) + \cos \left (d x + c\right ) - 2}{a^{2} d \cos \left (d x + c\right )^{2} - a^{2} d -{\left (a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )} \sin \left (d x + c\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.37124, size = 122, normalized size = 2.26 \begin{align*} -\frac{\frac{4 \, \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right )}{a^{2}} - \frac{\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{a^{2}} - \frac{2 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 7 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )} a^{2}}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(4*log(abs(tan(1/2*d*x + 1/2*c)))/a^2 - tan(1/2*d*x + 1/2*c)/a^2 - (2*tan(1/2*d*x + 1/2*c)^2 - 7*tan(1/2*
d*x + 1/2*c) - 1)/((tan(1/2*d*x + 1/2*c)^2 + tan(1/2*d*x + 1/2*c))*a^2))/d