3.633 \(\int \frac{\cot ^6(c+d x)}{a+a \sin (c+d x)} \, dx\)

Optimal. Leaf size=82 \[ -\frac{\cot ^5(c+d x)}{5 a d}+\frac{3 \tanh ^{-1}(\cos (c+d x))}{8 a d}+\frac{\cot ^3(c+d x) \csc (c+d x)}{4 a d}-\frac{3 \cot (c+d x) \csc (c+d x)}{8 a d} \]

[Out]

(3*ArcTanh[Cos[c + d*x]])/(8*a*d) - Cot[c + d*x]^5/(5*a*d) - (3*Cot[c + d*x]*Csc[c + d*x])/(8*a*d) + (Cot[c +
d*x]^3*Csc[c + d*x])/(4*a*d)

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Rubi [A]  time = 0.110803, antiderivative size = 82, normalized size of antiderivative = 1., number of steps used = 6, 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} \[ -\frac{\cot ^5(c+d x)}{5 a d}+\frac{3 \tanh ^{-1}(\cos (c+d x))}{8 a d}+\frac{\cot ^3(c+d x) \csc (c+d x)}{4 a d}-\frac{3 \cot (c+d x) \csc (c+d x)}{8 a d} \]

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]^6/(a + a*Sin[c + d*x]),x]

[Out]

(3*ArcTanh[Cos[c + d*x]])/(8*a*d) - Cot[c + d*x]^5/(5*a*d) - (3*Cot[c + d*x]*Csc[c + d*x])/(8*a*d) + (Cot[c +
d*x]^3*Csc[c + d*x])/(4*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 ^6(c+d x)}{a+a \sin (c+d x)} \, dx &=-\frac{\int \cot ^4(c+d x) \csc (c+d x) \, dx}{a}+\frac{\int \cot ^4(c+d x) \csc ^2(c+d x) \, dx}{a}\\ &=\frac{\cot ^3(c+d x) \csc (c+d x)}{4 a d}+\frac{3 \int \cot ^2(c+d x) \csc (c+d x) \, dx}{4 a}+\frac{\operatorname{Subst}\left (\int x^4 \, dx,x,-\cot (c+d x)\right )}{a d}\\ &=-\frac{\cot ^5(c+d x)}{5 a d}-\frac{3 \cot (c+d x) \csc (c+d x)}{8 a d}+\frac{\cot ^3(c+d x) \csc (c+d x)}{4 a d}-\frac{3 \int \csc (c+d x) \, dx}{8 a}\\ &=\frac{3 \tanh ^{-1}(\cos (c+d x))}{8 a d}-\frac{\cot ^5(c+d x)}{5 a d}-\frac{3 \cot (c+d x) \csc (c+d x)}{8 a d}+\frac{\cot ^3(c+d x) \csc (c+d x)}{4 a d}\\ \end{align*}

Mathematica [B]  time = 0.757315, size = 189, normalized size = 2.3 \[ -\frac{\csc ^5(c+d x) \left (20 \sin (2 (c+d x))-50 \sin (4 (c+d x))+80 \cos (c+d x)+40 \cos (3 (c+d x))+8 \cos (5 (c+d x))+150 \sin (c+d x) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )-75 \sin (3 (c+d x)) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )+15 \sin (5 (c+d x)) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )-150 \sin (c+d x) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )+75 \sin (3 (c+d x)) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )-15 \sin (5 (c+d x)) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )}{640 a d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[c + d*x]^6/(a + a*Sin[c + d*x]),x]

[Out]

-(Csc[c + d*x]^5*(80*Cos[c + d*x] + 40*Cos[3*(c + d*x)] + 8*Cos[5*(c + d*x)] - 150*Log[Cos[(c + d*x)/2]]*Sin[c
+ d*x] + 150*Log[Sin[(c + d*x)/2]]*Sin[c + d*x] + 20*Sin[2*(c + d*x)] + 75*Log[Cos[(c + d*x)/2]]*Sin[3*(c + d
*x)] - 75*Log[Sin[(c + d*x)/2]]*Sin[3*(c + d*x)] - 50*Sin[4*(c + d*x)] - 15*Log[Cos[(c + d*x)/2]]*Sin[5*(c + d
*x)] + 15*Log[Sin[(c + d*x)/2]]*Sin[5*(c + d*x)]))/(640*a*d)

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Maple [B]  time = 0.145, size = 208, normalized size = 2.5 \begin{align*}{\frac{1}{160\,da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{5}}-{\frac{1}{64\,da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{4}}-{\frac{1}{32\,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}{16\,da}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }-{\frac{1}{16\,da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1}}-{\frac{1}{160\,da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-5}}+{\frac{1}{64\,da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-4}}-{\frac{3}{8\,da}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) }+{\frac{1}{32\,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(cos(d*x+c)^6*csc(d*x+c)^6/(a+a*sin(d*x+c)),x)

[Out]

1/160/d/a*tan(1/2*d*x+1/2*c)^5-1/64/d/a*tan(1/2*d*x+1/2*c)^4-1/32/d/a*tan(1/2*d*x+1/2*c)^3+1/8/d/a*tan(1/2*d*x
+1/2*c)^2+1/16/d/a*tan(1/2*d*x+1/2*c)-1/16/d/a/tan(1/2*d*x+1/2*c)-1/160/d/a/tan(1/2*d*x+1/2*c)^5+1/64/d/a/tan(
1/2*d*x+1/2*c)^4-3/8/d/a*ln(tan(1/2*d*x+1/2*c))+1/32/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 = 1.01695, size = 316, normalized size = 3.85 \begin{align*} \frac{\frac{\frac{20 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{40 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac{10 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac{5 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac{2 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a} - \frac{120 \, \log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a} + \frac{{\left (\frac{5 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{10 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac{40 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac{20 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - 2\right )}{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}{a \sin \left (d x + c\right )^{5}}}{320 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*csc(d*x+c)^6/(a+a*sin(d*x+c)),x, algorithm="maxima")

[Out]

1/320*((20*sin(d*x + c)/(cos(d*x + c) + 1) + 40*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - 10*sin(d*x + c)^3/(cos(d
*x + c) + 1)^3 - 5*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 2*sin(d*x + c)^5/(cos(d*x + c) + 1)^5)/a - 120*log(si
n(d*x + c)/(cos(d*x + c) + 1))/a + (5*sin(d*x + c)/(cos(d*x + c) + 1) + 10*sin(d*x + c)^2/(cos(d*x + c) + 1)^2
- 40*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 - 20*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - 2)*(cos(d*x + c) + 1)^5/(
a*sin(d*x + c)^5))/d

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Fricas [B]  time = 1.09888, size = 431, normalized size = 5.26 \begin{align*} -\frac{16 \, \cos \left (d x + c\right )^{5} - 15 \,{\left (\cos \left (d x + c\right )^{4} - 2 \, \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 ) + 15 \,{\left (\cos \left (d x + c\right )^{4} - 2 \, \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 ) - 10 \,{\left (5 \, \cos \left (d x + c\right )^{3} - 3 \, \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{80 \,{\left (a d \cos \left (d x + c\right )^{4} - 2 \, 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(cos(d*x+c)^6*csc(d*x+c)^6/(a+a*sin(d*x+c)),x, algorithm="fricas")

[Out]

-1/80*(16*cos(d*x + c)^5 - 15*(cos(d*x + c)^4 - 2*cos(d*x + c)^2 + 1)*log(1/2*cos(d*x + c) + 1/2)*sin(d*x + c)
+ 15*(cos(d*x + c)^4 - 2*cos(d*x + c)^2 + 1)*log(-1/2*cos(d*x + c) + 1/2)*sin(d*x + c) - 10*(5*cos(d*x + c)^3
- 3*cos(d*x + c))*sin(d*x + c))/((a*d*cos(d*x + c)^4 - 2*a*d*cos(d*x + c)^2 + a*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)**6*csc(d*x+c)**6/(a+a*sin(d*x+c)),x)

[Out]

Timed out

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Giac [B]  time = 1.33802, size = 252, normalized size = 3.07 \begin{align*} -\frac{\frac{120 \, \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right )}{a} - \frac{2 \, a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 5 \, a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 10 \, a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 40 \, a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 20 \, a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{a^{5}} - \frac{274 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 20 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 40 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 10 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 5 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 2}{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5}}}{320 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*csc(d*x+c)^6/(a+a*sin(d*x+c)),x, algorithm="giac")

[Out]

-1/320*(120*log(abs(tan(1/2*d*x + 1/2*c)))/a - (2*a^4*tan(1/2*d*x + 1/2*c)^5 - 5*a^4*tan(1/2*d*x + 1/2*c)^4 -
10*a^4*tan(1/2*d*x + 1/2*c)^3 + 40*a^4*tan(1/2*d*x + 1/2*c)^2 + 20*a^4*tan(1/2*d*x + 1/2*c))/a^5 - (274*tan(1/
2*d*x + 1/2*c)^5 - 20*tan(1/2*d*x + 1/2*c)^4 - 40*tan(1/2*d*x + 1/2*c)^3 + 10*tan(1/2*d*x + 1/2*c)^2 + 5*tan(1
/2*d*x + 1/2*c) - 2)/(a*tan(1/2*d*x + 1/2*c)^5))/d