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

Optimal. Leaf size=58 \[ \frac{\cot (c+d x)}{a d}+\frac{\tanh ^{-1}(\cos (c+d x))}{2 a d}-\frac{\cot (c+d x) \csc (c+d x)}{2 a d}+\frac{x}{a} \]

[Out]

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

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Rubi [A]  time = 0.105861, antiderivative size = 58, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.185, Rules used = {2839, 2611, 3770, 3473, 8} \[ \frac{\cot (c+d x)}{a d}+\frac{\tanh ^{-1}(\cos (c+d x))}{2 a d}-\frac{\cot (c+d x) \csc (c+d x)}{2 a d}+\frac{x}{a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2839

Int[((cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.))/((a_) + (b_.)*sin[(e_.) + (f_
.)*(x_)]), x_Symbol] :> Dist[g^2/a, Int[(g*Cos[e + f*x])^(p - 2)*(d*Sin[e + f*x])^n, x], x] - Dist[g^2/(b*d),
Int[(g*Cos[e + f*x])^(p - 2)*(d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, g, n, p}, x] && EqQ[a^2
- b^2, 0]

Rule 2611

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(a*Sec[e
+ f*x])^m*(b*Tan[e + f*x])^(n - 1))/(f*(m + n - 1)), x] - Dist[(b^2*(n - 1))/(m + n - 1), Int[(a*Sec[e + f*x])
^m*(b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && GtQ[n, 1] && NeQ[m + n - 1, 0] && Integers
Q[2*m, 2*n]

Rule 3770

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

Rule 3473

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(b*Tan[c + d*x])^(n - 1))/(d*(n - 1)), x] - Dis
t[b^2, Int[(b*Tan[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1]

Rule 8

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

Rubi steps

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

Mathematica [A]  time = 0.461766, size = 102, normalized size = 1.76 \[ \frac{\left (\csc \left (\frac{1}{2} (c+d x)\right )+\sec \left (\frac{1}{2} (c+d x)\right )\right )^2 \left ((2 \sin (c+d x)-1) \cos (c+d x)+\sin ^2(c+d x) \left (-\log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )+\log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )+2 c+2 d x\right )\right )}{8 a d (\sin (c+d x)+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((Csc[(c + d*x)/2] + Sec[(c + d*x)/2])^2*((2*c + 2*d*x + Log[Cos[(c + d*x)/2]] - Log[Sin[(c + d*x)/2]])*Sin[c
+ d*x]^2 + Cos[c + d*x]*(-1 + 2*Sin[c + d*x])))/(8*a*d*(1 + Sin[c + d*x]))

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Maple [B]  time = 0.112, size = 112, normalized size = 1.9 \begin{align*}{\frac{1}{8\,da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}}-{\frac{1}{2\,da}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }+2\,{\frac{\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) }{da}}-{\frac{1}{8\,da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-2}}+{\frac{1}{2\,da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1}}-{\frac{1}{2\,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(cos(d*x+c)^4*csc(d*x+c)^3/(a+a*sin(d*x+c)),x)

[Out]

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

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Maxima [B]  time = 1.6215, size = 186, normalized size = 3.21 \begin{align*} -\frac{\frac{\frac{4 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac{\sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}}{a} - \frac{16 \, \arctan \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a} + \frac{4 \, \log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a} - \frac{{\left (\frac{4 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}{a \sin \left (d x + c\right )^{2}}}{8 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*((4*sin(d*x + c)/(cos(d*x + c) + 1) - sin(d*x + c)^2/(cos(d*x + c) + 1)^2)/a - 16*arctan(sin(d*x + c)/(co
s(d*x + c) + 1))/a + 4*log(sin(d*x + c)/(cos(d*x + c) + 1))/a - (4*sin(d*x + c)/(cos(d*x + c) + 1) - 1)*(cos(d
*x + c) + 1)^2/(a*sin(d*x + c)^2))/d

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Fricas [A]  time = 1.12652, size = 288, normalized size = 4.97 \begin{align*} \frac{4 \, d x \cos \left (d x + c\right )^{2} - 4 \, d x +{\left (\cos \left (d x + c\right )^{2} - 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} - 1\right )} \log \left (-\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) - 4 \, \cos \left (d x + c\right ) \sin \left (d x + c\right ) + 2 \, \cos \left (d x + c\right )}{4 \,{\left (a d \cos \left (d x + c\right )^{2} - a d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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)**3/(a+a*sin(d*x+c)),x)

[Out]

Timed out

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Giac [A]  time = 1.42191, size = 139, normalized size = 2.4 \begin{align*} \frac{\frac{8 \,{\left (d x + c\right )}}{a} - \frac{4 \, \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right )}{a} + \frac{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 4 \, a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{a^{2}} + \frac{6 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 4 \, \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}}}{8 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(8*(d*x + c)/a - 4*log(abs(tan(1/2*d*x + 1/2*c)))/a + (a*tan(1/2*d*x + 1/2*c)^2 - 4*a*tan(1/2*d*x + 1/2*c)
)/a^2 + (6*tan(1/2*d*x + 1/2*c)^2 + 4*tan(1/2*d*x + 1/2*c) - 1)/(a*tan(1/2*d*x + 1/2*c)^2))/d