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

Optimal. Leaf size=54 \[ -\frac{\cot (c+d x)}{a^3 d}-\frac{4 \cos (c+d x)}{a^3 d (\sin (c+d x)+1)}+\frac{3 \tanh ^{-1}(\cos (c+d x))}{a^3 d} \]

[Out]

(3*ArcTanh[Cos[c + d*x]])/(a^3*d) - Cot[c + d*x]/(a^3*d) - (4*Cos[c + d*x])/(a^3*d*(1 + Sin[c + d*x]))

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Rubi [A]  time = 0.235539, antiderivative size = 54, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.207, Rules used = {2875, 2872, 3770, 3767, 8, 2648} \[ -\frac{\cot (c+d x)}{a^3 d}-\frac{4 \cos (c+d x)}{a^3 d (\sin (c+d x)+1)}+\frac{3 \tanh ^{-1}(\cos (c+d x))}{a^3 d} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[c + d*x]^2*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) - (4*Cos[c + d*x])/(a^3*d*(1 + Sin[c + d*x]))

Rule 2875

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*
(x_)])^(m_), x_Symbol] :> Dist[(a/g)^(2*m), Int[((g*Cos[e + f*x])^(2*m + p)*(d*Sin[e + f*x])^n)/(a - b*Sin[e +
f*x])^m, x], x] /; FreeQ[{a, b, d, e, f, g, n, p}, x] && EqQ[a^2 - b^2, 0] && ILtQ[m, 0]

Rule 2872

Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(
m_), x_Symbol] :> Dist[1/a^p, Int[ExpandTrig[(d*sin[e + f*x])^n*(a - b*sin[e + f*x])^(p/2)*(a + b*sin[e + f*x]
)^(m + p/2), x], x], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 - b^2, 0] && IntegersQ[m, n, p/2] && ((GtQ[m,
0] && GtQ[p, 0] && LtQ[-m - p, n, -1]) || (GtQ[m, 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 2648

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> -Simp[Cos[c + d*x]/(d*(b + a*Sin[c + d*x])), x]
/; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rubi steps

\begin{align*} \int \frac{\cos ^2(c+d x) \cot ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx &=\frac{\int \csc ^2(c+d x) \sec ^2(c+d x) (a-a \sin (c+d x))^3 \, dx}{a^6}\\ &=\frac{\int \left (-3 a \csc (c+d x)+a \csc ^2(c+d x)+\frac{4 a}{1+\sin (c+d x)}\right ) \, dx}{a^4}\\ &=\frac{\int \csc ^2(c+d x) \, dx}{a^3}-\frac{3 \int \csc (c+d x) \, dx}{a^3}+\frac{4 \int \frac{1}{1+\sin (c+d x)} \, dx}{a^3}\\ &=\frac{3 \tanh ^{-1}(\cos (c+d x))}{a^3 d}-\frac{4 \cos (c+d x)}{a^3 d (1+\sin (c+d x))}-\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{4 \cos (c+d x)}{a^3 d (1+\sin (c+d x))}\\ \end{align*}

Mathematica [B]  time = 0.611471, size = 156, normalized size = 2.89 \[ -\frac{\tan \left (\frac{1}{2} (c+d x)\right ) \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^5 \left (\cos \left (\frac{1}{2} (c+d x)\right ) \left (\cot ^2\left (\frac{1}{2} (c+d x)\right )+6 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )-6 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )+\cot \left (\frac{1}{2} (c+d x)\right ) \left (6 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )-6 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )+1\right )-17\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )}{2 a^3 d (\sin (c+d x)+1)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((Cos[(c + d*x)/2]*(-17 + Cot[(c + d*x)/2]^2 - 6*Log[Cos[(c + d*x)/2]] + 6*Log[Sin[(c + d*x)/2]] + Cot[(c + d
*x)/2]*(1 - 6*Log[Cos[(c + d*x)/2]] + 6*Log[Sin[(c + d*x)/2]])) - Sin[(c + d*x)/2])*(Cos[(c + d*x)/2] + Sin[(c
+ d*x)/2])^5*Tan[(c + d*x)/2])/(2*a^3*d*(1 + Sin[c + d*x])^3)

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Maple [A]  time = 0.16, size = 77, normalized size = 1.4 \begin{align*}{\frac{1}{2\,d{a}^{3}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }-8\,{\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)^4*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/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 = 1.13214, size = 157, normalized size = 2.91 \begin{align*} -\frac{\frac{\frac{17 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1}{\frac{a^{3} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}} + \frac{6 \, \log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}} - \frac{\sin \left (d x + c\right )}{a^{3}{\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)^4*csc(d*x+c)^2/(a+a*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

-1/2*((17*sin(d*x + c)/(cos(d*x + c) + 1) + 1)/(a^3*sin(d*x + c)/(cos(d*x + c) + 1) + a^3*sin(d*x + c)^2/(cos(
d*x + c) + 1)^2) + 6*log(sin(d*x + c)/(cos(d*x + c) + 1))/a^3 - sin(d*x + c)/(a^3*(cos(d*x + c) + 1)))/d

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Fricas [B]  time = 1.13199, size = 451, normalized size = 8.35 \begin{align*} \frac{10 \, \cos \left (d x + c\right )^{2} + 3 \,{\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 ) - 3 \,{\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 ) + 2 \,{\left (5 \, \cos \left (d x + c\right ) + 4\right )} \sin \left (d x + c\right ) + 2 \, \cos \left (d x + c\right ) - 8}{2 \,{\left (a^{3} d \cos \left (d x + c\right )^{2} - a^{3} d -{\left (a^{3} d \cos \left (d x + c\right ) + 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)^4*csc(d*x+c)^2/(a+a*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

1/2*(10*cos(d*x + c)^2 + 3*(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 - (cos(d*x + c) + 1)*sin(d*x + c) - 1)*log(-1/2*cos(d*x + c) + 1/2) + 2*(5*cos(d*x + c) +
4)*sin(d*x + c) + 2*cos(d*x + c) - 8)/(a^3*d*cos(d*x + c)^2 - a^3*d - (a^3*d*cos(d*x + c) + a^3*d)*sin(d*x + c
))

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**4*csc(d*x+c)**2/(a+a*sin(d*x+c))**3,x)

[Out]

Timed out

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Giac [A]  time = 1.42827, size = 122, normalized size = 2.26 \begin{align*} -\frac{\frac{6 \, \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right )}{a^{3}} - \frac{\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{a^{3}} - \frac{3 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 14 \, \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^{3}}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*csc(d*x+c)^2/(a+a*sin(d*x+c))^3,x, algorithm="giac")

[Out]

-1/2*(6*log(abs(tan(1/2*d*x + 1/2*c)))/a^3 - tan(1/2*d*x + 1/2*c)/a^3 - (3*tan(1/2*d*x + 1/2*c)^2 - 14*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^3))/d