3.642 \(\int \frac{\cot ^6(c+d x)}{(a+a \sin (c+d x))^2} \, dx\)

Optimal. Leaf size=100 \[ -\frac{\cot ^5(c+d x)}{5 a^2 d}-\frac{2 \cot ^3(c+d x)}{3 a^2 d}-\frac{\tanh ^{-1}(\cos (c+d x))}{4 a^2 d}+\frac{\cot (c+d x) \csc ^3(c+d x)}{2 a^2 d}-\frac{\cot (c+d x) \csc (c+d x)}{4 a^2 d} \]

[Out]

-ArcTanh[Cos[c + d*x]]/(4*a^2*d) - (2*Cot[c + d*x]^3)/(3*a^2*d) - Cot[c + d*x]^5/(5*a^2*d) - (Cot[c + d*x]*Csc
[c + d*x])/(4*a^2*d) + (Cot[c + d*x]*Csc[c + d*x]^3)/(2*a^2*d)

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Rubi [A]  time = 0.14719, antiderivative size = 100, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 5, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.238, Rules used = {2709, 3767, 8, 3768, 3770} \[ -\frac{\cot ^5(c+d x)}{5 a^2 d}-\frac{2 \cot ^3(c+d x)}{3 a^2 d}-\frac{\tanh ^{-1}(\cos (c+d x))}{4 a^2 d}+\frac{\cot (c+d x) \csc ^3(c+d x)}{2 a^2 d}-\frac{\cot (c+d x) \csc (c+d x)}{4 a^2 d} \]

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]^6/(a + a*Sin[c + d*x])^2,x]

[Out]

-ArcTanh[Cos[c + d*x]]/(4*a^2*d) - (2*Cot[c + d*x]^3)/(3*a^2*d) - Cot[c + d*x]^5/(5*a^2*d) - (Cot[c + d*x]*Csc
[c + d*x])/(4*a^2*d) + (Cot[c + d*x]*Csc[c + d*x]^3)/(2*a^2*d)

Rule 2709

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*tan[(e_.) + (f_.)*(x_)]^(p_), x_Symbol] :> Dist[a^p, Int[Expan
dIntegrand[(Sin[e + f*x]^p*(a + b*Sin[e + f*x])^(m - p/2))/(a - b*Sin[e + f*x])^(p/2), x], x], x] /; FreeQ[{a,
b, e, f}, x] && EqQ[a^2 - b^2, 0] && IntegersQ[m, p/2] && (LtQ[p, 0] || GtQ[m - p/2, 0])

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 3768

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Csc[c + d*x])^(n - 1))/(d*(n -
1)), x] + Dist[(b^2*(n - 2))/(n - 1), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1
] && IntegerQ[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))^2} \, dx &=\frac{\int \left (-a^4 \csc ^2(c+d x)+2 a^4 \csc ^3(c+d x)-2 a^4 \csc ^5(c+d x)+a^4 \csc ^6(c+d x)\right ) \, dx}{a^6}\\ &=-\frac{\int \csc ^2(c+d x) \, dx}{a^2}+\frac{\int \csc ^6(c+d x) \, dx}{a^2}+\frac{2 \int \csc ^3(c+d x) \, dx}{a^2}-\frac{2 \int \csc ^5(c+d x) \, dx}{a^2}\\ &=-\frac{\cot (c+d x) \csc (c+d x)}{a^2 d}+\frac{\cot (c+d x) \csc ^3(c+d x)}{2 a^2 d}+\frac{\int \csc (c+d x) \, dx}{a^2}-\frac{3 \int \csc ^3(c+d x) \, dx}{2 a^2}+\frac{\operatorname{Subst}(\int 1 \, dx,x,\cot (c+d x))}{a^2 d}-\frac{\operatorname{Subst}\left (\int \left (1+2 x^2+x^4\right ) \, dx,x,\cot (c+d x)\right )}{a^2 d}\\ &=-\frac{\tanh ^{-1}(\cos (c+d x))}{a^2 d}-\frac{2 \cot ^3(c+d x)}{3 a^2 d}-\frac{\cot ^5(c+d x)}{5 a^2 d}-\frac{\cot (c+d x) \csc (c+d x)}{4 a^2 d}+\frac{\cot (c+d x) \csc ^3(c+d x)}{2 a^2 d}-\frac{3 \int \csc (c+d x) \, dx}{4 a^2}\\ &=-\frac{\tanh ^{-1}(\cos (c+d x))}{4 a^2 d}-\frac{2 \cot ^3(c+d x)}{3 a^2 d}-\frac{\cot ^5(c+d x)}{5 a^2 d}-\frac{\cot (c+d x) \csc (c+d x)}{4 a^2 d}+\frac{\cot (c+d x) \csc ^3(c+d x)}{2 a^2 d}\\ \end{align*}

Mathematica [A]  time = 0.553905, size = 189, normalized size = 1.89 \[ -\frac{\csc ^5(c+d x) \left (-180 \sin (2 (c+d x))-30 \sin (4 (c+d x))+200 \cos (c+d x)+20 \cos (3 (c+d x))-28 \cos (5 (c+d x))-150 \sin (c+d x) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )+75 \sin (3 (c+d x)) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )-15 \sin (5 (c+d x)) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )+150 \sin (c+d x) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )-75 \sin (3 (c+d x)) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )+15 \sin (5 (c+d x)) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )}{960 a^2 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[c + d*x]^6/(a + a*Sin[c + d*x])^2,x]

[Out]

-(Csc[c + d*x]^5*(200*Cos[c + d*x] + 20*Cos[3*(c + d*x)] - 28*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] - 180*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)] - 30*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)]))/(960*a^2*d)

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Maple [A]  time = 0.161, size = 170, normalized size = 1.7 \begin{align*}{\frac{1}{160\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{5}}-{\frac{1}{32\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{4}}+{\frac{5}{96\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}-{\frac{3}{16\,d{a}^{2}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }+{\frac{3}{16\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1}}-{\frac{1}{160\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-5}}+{\frac{1}{32\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-4}}+{\frac{1}{4\,d{a}^{2}}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) }-{\frac{5}{96\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-3}} \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))^2,x)

[Out]

1/160/d/a^2*tan(1/2*d*x+1/2*c)^5-1/32/d/a^2*tan(1/2*d*x+1/2*c)^4+5/96/d/a^2*tan(1/2*d*x+1/2*c)^3-3/16/d/a^2*ta
n(1/2*d*x+1/2*c)+3/16/d/a^2/tan(1/2*d*x+1/2*c)-1/160/d/a^2/tan(1/2*d*x+1/2*c)^5+1/32/d/a^2/tan(1/2*d*x+1/2*c)^
4+1/4/d/a^2*ln(tan(1/2*d*x+1/2*c))-5/96/d/a^2/tan(1/2*d*x+1/2*c)^3

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Maxima [B]  time = 1.03594, size = 263, normalized size = 2.63 \begin{align*} -\frac{\frac{\frac{90 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac{25 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac{15 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac{3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a^{2}} - \frac{120 \, \log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}} - \frac{{\left (\frac{15 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac{25 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{90 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - 3\right )}{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}{a^{2} \sin \left (d x + c\right )^{5}}}{480 \, 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))^2,x, algorithm="maxima")

[Out]

-1/480*((90*sin(d*x + c)/(cos(d*x + c) + 1) - 25*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 15*sin(d*x + c)^4/(cos(
d*x + c) + 1)^4 - 3*sin(d*x + c)^5/(cos(d*x + c) + 1)^5)/a^2 - 120*log(sin(d*x + c)/(cos(d*x + c) + 1))/a^2 -
(15*sin(d*x + c)/(cos(d*x + c) + 1) - 25*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 90*sin(d*x + c)^4/(cos(d*x + c)
+ 1)^4 - 3)*(cos(d*x + c) + 1)^5/(a^2*sin(d*x + c)^5))/d

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Fricas [A]  time = 1.14992, size = 460, normalized size = 4.6 \begin{align*} \frac{56 \, \cos \left (d x + c\right )^{5} - 80 \, \cos \left (d x + c\right )^{3} - 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 ) + 30 \,{\left (\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{120 \,{\left (a^{2} d \cos \left (d x + c\right )^{4} - 2 \, a^{2} d \cos \left (d x + c\right )^{2} + a^{2} 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))^2,x, algorithm="fricas")

[Out]

1/120*(56*cos(d*x + c)^5 - 80*cos(d*x + c)^3 - 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) +
30*(cos(d*x + c)^3 + cos(d*x + c))*sin(d*x + c))/((a^2*d*cos(d*x + c)^4 - 2*a^2*d*cos(d*x + c)^2 + a^2*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))**2,x)

[Out]

Timed out

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Giac [A]  time = 1.38443, size = 212, normalized size = 2.12 \begin{align*} \frac{\frac{120 \, \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right )}{a^{2}} - \frac{274 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 90 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 25 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 15 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 3}{a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5}} + \frac{3 \, a^{8} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 15 \, a^{8} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 25 \, a^{8} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 90 \, a^{8} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{a^{10}}}{480 \, 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))^2,x, algorithm="giac")

[Out]

1/480*(120*log(abs(tan(1/2*d*x + 1/2*c)))/a^2 - (274*tan(1/2*d*x + 1/2*c)^5 - 90*tan(1/2*d*x + 1/2*c)^4 + 25*t
an(1/2*d*x + 1/2*c)^2 - 15*tan(1/2*d*x + 1/2*c) + 3)/(a^2*tan(1/2*d*x + 1/2*c)^5) + (3*a^8*tan(1/2*d*x + 1/2*c
)^5 - 15*a^8*tan(1/2*d*x + 1/2*c)^4 + 25*a^8*tan(1/2*d*x + 1/2*c)^3 - 90*a^8*tan(1/2*d*x + 1/2*c))/a^10)/d