3.58 \(\int \frac{\cot ^4(c+d x)}{a+a \sin (c+d x)} \, dx\)

Optimal. Leaf size=58 \[ -\frac{\cot ^3(c+d x)}{3 a d}-\frac{\tanh ^{-1}(\cos (c+d x))}{2 a d}+\frac{\cot (c+d x) \csc (c+d x)}{2 a d} \]

[Out]

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

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Rubi [A]  time = 0.088263, antiderivative size = 58, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.238, Rules used = {2706, 2607, 30, 2611, 3770} \[ -\frac{\cot ^3(c+d x)}{3 a d}-\frac{\tanh ^{-1}(\cos (c+d x))}{2 a d}+\frac{\cot (c+d x) \csc (c+d x)}{2 a d} \]

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]^4/(a + a*Sin[c + d*x]),x]

[Out]

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

Rule 2706

Int[((g_.)*tan[(e_.) + (f_.)*(x_)])^(p_.)/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[1/a, Int[S
ec[e + f*x]^2*(g*Tan[e + f*x])^p, x], x] - Dist[1/(b*g), Int[Sec[e + f*x]*(g*Tan[e + f*x])^(p + 1), x], x] /;
FreeQ[{a, b, e, f, g, p}, x] && EqQ[a^2 - b^2, 0] && NeQ[p, -1]

Rule 2607

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[1/f, Subst[Int[(b*x)
^n*(1 + x^2)^(m/2 - 1), x], x, Tan[e + f*x]], x] /; FreeQ[{b, e, f, n}, x] && IntegerQ[m/2] &&  !(IntegerQ[(n
- 1)/2] && LtQ[0, n, m - 1])

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

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]

Rubi steps

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

Mathematica [B]  time = 0.491465, size = 124, normalized size = 2.14 \[ -\frac{\csc \left (\frac{1}{2} (c+d x)\right ) \sec \left (\frac{1}{2} (c+d x)\right ) \left (\csc \left (\frac{1}{2} (c+d x)\right )+\sec \left (\frac{1}{2} (c+d x)\right )\right )^2 \left (\cos (3 (c+d x))+(3-6 \sin (c+d x)) \cos (c+d x)+6 \sin ^3(c+d x) \left (\log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )\right )\right )}{96 a d (\sin (c+d x)+1)} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[c + d*x]^4/(a + a*Sin[c + d*x]),x]

[Out]

-(Csc[(c + d*x)/2]*Sec[(c + d*x)/2]*(Csc[(c + d*x)/2] + Sec[(c + d*x)/2])^2*(Cos[3*(c + d*x)] + Cos[c + d*x]*(
3 - 6*Sin[c + d*x]) + 6*(Log[Cos[(c + d*x)/2]] - Log[Sin[(c + d*x)/2]])*Sin[c + d*x]^3))/(96*a*d*(1 + Sin[c +
d*x]))

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Maple [B]  time = 0.075, size = 132, normalized size = 2.3 \begin{align*}{\frac{1}{24\,da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}-{\frac{1}{8\,da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}}-{\frac{1}{8\,da}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }+{\frac{1}{8\,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 ) }-{\frac{1}{24\,da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-3}}+{\frac{1}{8\,da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(d*x+c)^4/(a+a*sin(d*x+c)),x)

[Out]

1/24/d/a*tan(1/2*d*x+1/2*c)^3-1/8/d/a*tan(1/2*d*x+1/2*c)^2-1/8/d/a*tan(1/2*d*x+1/2*c)+1/8/d/a/tan(1/2*d*x+1/2*
c)+1/2/d/a*ln(tan(1/2*d*x+1/2*c))-1/24/d/a/tan(1/2*d*x+1/2*c)^3+1/8/d/a/tan(1/2*d*x+1/2*c)^2

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Maxima [B]  time = 1.02688, size = 209, normalized size = 3.6 \begin{align*} -\frac{\frac{\frac{3 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{3 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac{\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a} - \frac{12 \, \log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a} - \frac{{\left (\frac{3 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{3 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - 1\right )}{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}{a \sin \left (d x + c\right )^{3}}}{24 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/24*((3*sin(d*x + c)/(cos(d*x + c) + 1) + 3*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - sin(d*x + c)^3/(cos(d*x +
c) + 1)^3)/a - 12*log(sin(d*x + c)/(cos(d*x + c) + 1))/a - (3*sin(d*x + c)/(cos(d*x + c) + 1) + 3*sin(d*x + c)
^2/(cos(d*x + c) + 1)^2 - 1)*(cos(d*x + c) + 1)^3/(a*sin(d*x + c)^3))/d

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Fricas [B]  time = 1.41589, size = 311, normalized size = 5.36 \begin{align*} \frac{4 \, \cos \left (d x + c\right )^{3} - 3 \,{\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 ) \sin \left (d x + c\right ) + 3 \,{\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 ) \sin \left (d x + c\right ) - 6 \, \cos \left (d x + c\right ) \sin \left (d x + c\right )}{12 \,{\left (a d \cos \left (d x + c\right )^{2} - a d\right )} \sin \left (d x + c\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(4*cos(d*x + c)^3 - 3*(cos(d*x + c)^2 - 1)*log(1/2*cos(d*x + c) + 1/2)*sin(d*x + c) + 3*(cos(d*x + c)^2 -
1)*log(-1/2*cos(d*x + c) + 1/2)*sin(d*x + c) - 6*cos(d*x + c)*sin(d*x + c))/((a*d*cos(d*x + c)^2 - a*d)*sin(d
*x + c))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{\cot ^{4}{\left (c + d x \right )}}{\sin{\left (c + d x \right )} + 1}\, dx}{a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)**4/(a+a*sin(d*x+c)),x)

[Out]

Integral(cot(c + d*x)**4/(sin(c + d*x) + 1), x)/a

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Giac [B]  time = 1.41595, size = 171, normalized size = 2.95 \begin{align*} \frac{\frac{12 \, \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right )}{a} + \frac{a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 3 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 3 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{a^{3}} - \frac{22 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 3 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 3 \, \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 )^{3}}}{24 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(12*log(abs(tan(1/2*d*x + 1/2*c)))/a + (a^2*tan(1/2*d*x + 1/2*c)^3 - 3*a^2*tan(1/2*d*x + 1/2*c)^2 - 3*a^2
*tan(1/2*d*x + 1/2*c))/a^3 - (22*tan(1/2*d*x + 1/2*c)^3 - 3*tan(1/2*d*x + 1/2*c)^2 - 3*tan(1/2*d*x + 1/2*c) +
1)/(a*tan(1/2*d*x + 1/2*c)^3))/d