3.319 \(\int \frac{\cot ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx\)

Optimal. Leaf size=82 \[ -\frac{\cot (c+d x)}{a^3 d}+\frac{3 \tanh ^{-1}(\cos (c+d x))}{a^3 d}-\frac{13 \cot (c+d x)}{3 a^3 d (\csc (c+d x)+1)}+\frac{2 \cot (c+d x)}{3 a^3 d (\csc (c+d x)+1)^2} \]

[Out]

(3*ArcTanh[Cos[c + d*x]])/(a^3*d) - Cot[c + d*x]/(a^3*d) + (2*Cot[c + d*x])/(3*a^3*d*(1 + Csc[c + d*x])^2) - (
13*Cot[c + d*x])/(3*a^3*d*(1 + Csc[c + d*x]))

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Rubi [A]  time = 0.207704, antiderivative size = 82, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 7, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {2709, 3770, 3767, 8, 3777, 3919, 3794} \[ -\frac{\cot (c+d x)}{a^3 d}+\frac{3 \tanh ^{-1}(\cos (c+d x))}{a^3 d}-\frac{13 \cot (c+d x)}{3 a^3 d (\csc (c+d x)+1)}+\frac{2 \cot (c+d x)}{3 a^3 d (\csc (c+d x)+1)^2} \]

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]^2/(a + a*Sin[c + d*x])^3,x]

[Out]

(3*ArcTanh[Cos[c + d*x]])/(a^3*d) - Cot[c + d*x]/(a^3*d) + (2*Cot[c + d*x])/(3*a^3*d*(1 + Csc[c + d*x])^2) - (
13*Cot[c + d*x])/(3*a^3*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]

Rule 3919

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[(c*x)/a,
x] - Dist[(b*c - a*d)/a, Int[Csc[e + f*x]/(a + b*Csc[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[
b*c - a*d, 0]

Rule 3794

Int[csc[(e_.) + (f_.)*(x_)]/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> -Simp[Cot[e + f*x]/(f*(b + a*
Csc[e + f*x])), x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0]

Rubi steps

\begin{align*} \int \frac{\cot ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx &=\frac{\int \left (\frac{5}{a}-\frac{3 \csc (c+d x)}{a}+\frac{\csc ^2(c+d x)}{a}+\frac{2}{a (1+\csc (c+d x))^2}-\frac{7}{a (1+\csc (c+d x))}\right ) \, dx}{a^2}\\ &=\frac{5 x}{a^3}+\frac{\int \csc ^2(c+d x) \, dx}{a^3}+\frac{2 \int \frac{1}{(1+\csc (c+d x))^2} \, dx}{a^3}-\frac{3 \int \csc (c+d x) \, dx}{a^3}-\frac{7 \int \frac{1}{1+\csc (c+d x)} \, dx}{a^3}\\ &=\frac{5 x}{a^3}+\frac{3 \tanh ^{-1}(\cos (c+d x))}{a^3 d}+\frac{2 \cot (c+d x)}{3 a^3 d (1+\csc (c+d x))^2}-\frac{7 \cot (c+d x)}{a^3 d (1+\csc (c+d x))}-\frac{2 \int \frac{-3+\csc (c+d x)}{1+\csc (c+d x)} \, dx}{3 a^3}+\frac{7 \int -1 \, dx}{a^3}-\frac{\operatorname{Subst}(\int 1 \, dx,x,\cot (c+d x))}{a^3 d}\\ &=\frac{3 \tanh ^{-1}(\cos (c+d x))}{a^3 d}-\frac{\cot (c+d x)}{a^3 d}+\frac{2 \cot (c+d x)}{3 a^3 d (1+\csc (c+d x))^2}-\frac{7 \cot (c+d x)}{a^3 d (1+\csc (c+d x))}-\frac{8 \int \frac{\csc (c+d x)}{1+\csc (c+d x)} \, dx}{3 a^3}\\ &=\frac{3 \tanh ^{-1}(\cos (c+d x))}{a^3 d}-\frac{\cot (c+d x)}{a^3 d}+\frac{2 \cot (c+d x)}{3 a^3 d (1+\csc (c+d x))^2}-\frac{13 \cot (c+d x)}{3 a^3 d (1+\csc (c+d x))}\\ \end{align*}

Mathematica [B]  time = 1.47682, size = 255, normalized size = 3.11 \[ \frac{\left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^3 \left (8 \sin \left (\frac{1}{2} (c+d x)\right )+44 \sin \left (\frac{1}{2} (c+d x)\right ) \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^2-4 \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )+18 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right ) \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^3-18 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right ) \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^3+3 \tan \left (\frac{1}{2} (c+d x)\right ) \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^3-3 \cot \left (\frac{1}{2} (c+d x)\right ) \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^3\right )}{6 d (a \sin (c+d x)+a)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[c + d*x]^2/(a + a*Sin[c + d*x])^3,x]

[Out]

((Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^3*(8*Sin[(c + d*x)/2] - 4*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]) + 44*Si
n[(c + d*x)/2]*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2 - 3*Cot[(c + d*x)/2]*(Cos[(c + d*x)/2] + Sin[(c + d*x)/
2])^3 + 18*Log[Cos[(c + d*x)/2]]*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^3 - 18*Log[Sin[(c + d*x)/2]]*(Cos[(c +
d*x)/2] + Sin[(c + d*x)/2])^3 + 3*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^3*Tan[(c + d*x)/2]))/(6*d*(a + a*Sin[c
+ d*x])^3)

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Maple [A]  time = 0.166, size = 119, normalized size = 1.5 \begin{align*}{\frac{1}{2\,d{a}^{3}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }-{\frac{8}{3\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-3}}+4\,{\frac{1}{d{a}^{3} \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) ^{2}}}-10\,{\frac{1}{d{a}^{3} \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) }}-{\frac{1}{2\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1}}-3\,{\frac{\ln \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) }{d{a}^{3}}} \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))^3,x)

[Out]

1/2/d/a^3*tan(1/2*d*x+1/2*c)-8/3/d/a^3/(tan(1/2*d*x+1/2*c)+1)^3+4/d/a^3/(tan(1/2*d*x+1/2*c)+1)^2-10/d/a^3/(tan
(1/2*d*x+1/2*c)+1)-1/2/d/a^3/tan(1/2*d*x+1/2*c)-3/d/a^3*ln(tan(1/2*d*x+1/2*c))

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Maxima [B]  time = 1.03531, size = 273, normalized size = 3.33 \begin{align*} -\frac{\frac{\frac{61 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{105 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{63 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + 3}{\frac{a^{3} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{3 \, a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{3 \, a^{3} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac{a^{3} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}} + \frac{18 \, \log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}} - \frac{3 \, \sin \left (d x + c\right )}{a^{3}{\left (\cos \left (d x + c\right ) + 1\right )}}}{6 \, 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))^3,x, algorithm="maxima")

[Out]

-1/6*((61*sin(d*x + c)/(cos(d*x + c) + 1) + 105*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 63*sin(d*x + c)^3/(cos(d
*x + c) + 1)^3 + 3)/(a^3*sin(d*x + c)/(cos(d*x + c) + 1) + 3*a^3*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 3*a^3*s
in(d*x + c)^3/(cos(d*x + c) + 1)^3 + a^3*sin(d*x + c)^4/(cos(d*x + c) + 1)^4) + 18*log(sin(d*x + c)/(cos(d*x +
c) + 1))/a^3 - 3*sin(d*x + c)/(a^3*(cos(d*x + c) + 1)))/d

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Fricas [B]  time = 1.73355, size = 745, normalized size = 9.09 \begin{align*} -\frac{28 \, \cos \left (d x + c\right )^{3} - 10 \, \cos \left (d x + c\right )^{2} - 9 \,{\left (\cos \left (d x + c\right )^{3} + 2 \, \cos \left (d x + c\right )^{2} +{\left (\cos \left (d x + c\right )^{2} - \cos \left (d x + c\right ) - 2\right )} \sin \left (d x + c\right ) - \cos \left (d x + c\right ) - 2\right )} \log \left (\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) + 9 \,{\left (\cos \left (d x + c\right )^{3} + 2 \, \cos \left (d x + c\right )^{2} +{\left (\cos \left (d x + c\right )^{2} - \cos \left (d x + c\right ) - 2\right )} \sin \left (d x + c\right ) - \cos \left (d x + c\right ) - 2\right )} \log \left (-\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) - 2 \,{\left (14 \, \cos \left (d x + c\right )^{2} + 19 \, \cos \left (d x + c\right ) + 2\right )} \sin \left (d x + c\right ) - 34 \, \cos \left (d x + c\right ) + 4}{6 \,{\left (a^{3} d \cos \left (d x + c\right )^{3} + 2 \, a^{3} d \cos \left (d x + c\right )^{2} - a^{3} d \cos \left (d x + c\right ) - 2 \, a^{3} d +{\left (a^{3} d \cos \left (d x + c\right )^{2} - a^{3} d \cos \left (d x + c\right ) - 2 \, a^{3} d\right )} \sin \left (d x + c\right )\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))^3,x, algorithm="fricas")

[Out]

-1/6*(28*cos(d*x + c)^3 - 10*cos(d*x + c)^2 - 9*(cos(d*x + c)^3 + 2*cos(d*x + c)^2 + (cos(d*x + c)^2 - cos(d*x
+ c) - 2)*sin(d*x + c) - cos(d*x + c) - 2)*log(1/2*cos(d*x + c) + 1/2) + 9*(cos(d*x + c)^3 + 2*cos(d*x + c)^2
+ (cos(d*x + c)^2 - cos(d*x + c) - 2)*sin(d*x + c) - cos(d*x + c) - 2)*log(-1/2*cos(d*x + c) + 1/2) - 2*(14*c
os(d*x + c)^2 + 19*cos(d*x + c) + 2)*sin(d*x + c) - 34*cos(d*x + c) + 4)/(a^3*d*cos(d*x + c)^3 + 2*a^3*d*cos(d
*x + c)^2 - a^3*d*cos(d*x + c) - 2*a^3*d + (a^3*d*cos(d*x + c)^2 - a^3*d*cos(d*x + c) - 2*a^3*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 ^{3}{\left (c + d x \right )} + 3 \sin ^{2}{\left (c + d x \right )} + 3 \sin{\left (c + d x \right )} + 1}\, dx}{a^{3}} \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))**3,x)

[Out]

Integral(cos(c + d*x)**2*csc(c + d*x)**2/(sin(c + d*x)**3 + 3*sin(c + d*x)**2 + 3*sin(c + d*x) + 1), x)/a**3

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Giac [A]  time = 1.44403, size = 147, normalized size = 1.79 \begin{align*} -\frac{\frac{18 \, \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right )}{a^{3}} - \frac{3 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{a^{3}} - \frac{3 \,{\left (6 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1\right )}}{a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )} + \frac{4 \,{\left (15 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 24 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 13\right )}}{a^{3}{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1\right )}^{3}}}{6 \, 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))^3,x, algorithm="giac")

[Out]

-1/6*(18*log(abs(tan(1/2*d*x + 1/2*c)))/a^3 - 3*tan(1/2*d*x + 1/2*c)/a^3 - 3*(6*tan(1/2*d*x + 1/2*c) - 1)/(a^3
*tan(1/2*d*x + 1/2*c)) + 4*(15*tan(1/2*d*x + 1/2*c)^2 + 24*tan(1/2*d*x + 1/2*c) + 13)/(a^3*(tan(1/2*d*x + 1/2*
c) + 1)^3))/d