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

Optimal. Leaf size=36 \[ -\frac{\cos (c+d x)}{a^2 d}-\frac{\tanh ^{-1}(\cos (c+d x))}{a^2 d}-\frac{2 x}{a^2} \]

[Out]

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

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Rubi [A]  time = 0.129302, antiderivative size = 36, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.148, Rules used = {2869, 2746, 2735, 3770} \[ -\frac{\cos (c+d x)}{a^2 d}-\frac{\tanh ^{-1}(\cos (c+d x))}{a^2 d}-\frac{2 x}{a^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2869

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

Rule 2746

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

Rule 2735

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*x)/d
, x] - Dist[(b*c - a*d)/d, Int[1/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d
, 0]

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

Mathematica [A]  time = 0.133405, size = 46, normalized size = 1.28 \[ -\frac{\cos (c+d x)-\log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )+\log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )+2 c+2 d x}{a^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.121, size = 60, normalized size = 1.7 \begin{align*} -2\,{\frac{1}{d{a}^{2} \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) }}-4\,{\frac{\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) }{d{a}^{2}}}+{\frac{1}{d{a}^{2}}\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(cos(d*x+c)^4*csc(d*x+c)/(a+a*sin(d*x+c))^2,x)

[Out]

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

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Maxima [B]  time = 1.73861, size = 111, normalized size = 3.08 \begin{align*} -\frac{\frac{2}{a^{2} + \frac{a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}} + \frac{4 \, \arctan \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}} - \frac{\log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{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)/(a+a*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.15881, size = 135, normalized size = 3.75 \begin{align*} -\frac{4 \, d x + 2 \, \cos \left (d x + c\right ) + \log \left (\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) - \log \left (-\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right )}{2 \, a^{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)/(a+a*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{\cos ^{4}{\left (c + d x \right )} \csc{\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)**4*csc(d*x+c)/(a+a*sin(d*x+c))**2,x)

[Out]

Integral(cos(c + d*x)**4*csc(c + d*x)/(sin(c + d*x)**2 + 2*sin(c + d*x) + 1), x)/a**2

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Giac [A]  time = 1.31897, size = 70, normalized size = 1.94 \begin{align*} -\frac{\frac{2 \,{\left (d x + c\right )}}{a^{2}} - \frac{\log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right )}{a^{2}} + \frac{2}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1\right )} a^{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)/(a+a*sin(d*x+c))^2,x, algorithm="giac")

[Out]

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