3.652 \(\int \frac{\cot ^6(c+d x)}{(a+a \sin (c+d x))^3} \, dx\)

Optimal. Leaf size=114 \[ -\frac{\cot ^5(c+d x)}{5 a^3 d}-\frac{5 \cot ^3(c+d x)}{3 a^3 d}-\frac{4 \cot (c+d x)}{a^3 d}+\frac{13 \tanh ^{-1}(\cos (c+d x))}{8 a^3 d}+\frac{3 \cot (c+d x) \csc ^3(c+d x)}{4 a^3 d}+\frac{13 \cot (c+d x) \csc (c+d x)}{8 a^3 d} \]

[Out]

(13*ArcTanh[Cos[c + d*x]])/(8*a^3*d) - (4*Cot[c + d*x])/(a^3*d) - (5*Cot[c + d*x]^3)/(3*a^3*d) - Cot[c + d*x]^
5/(5*a^3*d) + (13*Cot[c + d*x]*Csc[c + d*x])/(8*a^3*d) + (3*Cot[c + d*x]*Csc[c + d*x]^3)/(4*a^3*d)

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Rubi [A]  time = 0.17691, antiderivative size = 114, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 5, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.238, Rules used = {2708, 2757, 3768, 3770, 3767} \[ -\frac{\cot ^5(c+d x)}{5 a^3 d}-\frac{5 \cot ^3(c+d x)}{3 a^3 d}-\frac{4 \cot (c+d x)}{a^3 d}+\frac{13 \tanh ^{-1}(\cos (c+d x))}{8 a^3 d}+\frac{3 \cot (c+d x) \csc ^3(c+d x)}{4 a^3 d}+\frac{13 \cot (c+d x) \csc (c+d x)}{8 a^3 d} \]

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]^6/(a + a*Sin[c + d*x])^3,x]

[Out]

(13*ArcTanh[Cos[c + d*x]])/(8*a^3*d) - (4*Cot[c + d*x])/(a^3*d) - (5*Cot[c + d*x]^3)/(3*a^3*d) - Cot[c + d*x]^
5/(5*a^3*d) + (13*Cot[c + d*x]*Csc[c + d*x])/(8*a^3*d) + (3*Cot[c + d*x]*Csc[c + d*x]^3)/(4*a^3*d)

Rule 2708

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

Rule 2757

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

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]

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]

Rubi steps

\begin{align*} \int \frac{\cot ^6(c+d x)}{(a+a \sin (c+d x))^3} \, dx &=\frac{\int \csc ^6(c+d x) (a-a \sin (c+d x))^3 \, dx}{a^6}\\ &=\frac{\int \left (-a^3 \csc ^3(c+d x)+3 a^3 \csc ^4(c+d x)-3 a^3 \csc ^5(c+d x)+a^3 \csc ^6(c+d x)\right ) \, dx}{a^6}\\ &=-\frac{\int \csc ^3(c+d x) \, dx}{a^3}+\frac{\int \csc ^6(c+d x) \, dx}{a^3}+\frac{3 \int \csc ^4(c+d x) \, dx}{a^3}-\frac{3 \int \csc ^5(c+d x) \, dx}{a^3}\\ &=\frac{\cot (c+d x) \csc (c+d x)}{2 a^3 d}+\frac{3 \cot (c+d x) \csc ^3(c+d x)}{4 a^3 d}-\frac{\int \csc (c+d x) \, dx}{2 a^3}-\frac{9 \int \csc ^3(c+d x) \, dx}{4 a^3}-\frac{\operatorname{Subst}\left (\int \left (1+2 x^2+x^4\right ) \, dx,x,\cot (c+d x)\right )}{a^3 d}-\frac{3 \operatorname{Subst}\left (\int \left (1+x^2\right ) \, dx,x,\cot (c+d x)\right )}{a^3 d}\\ &=\frac{\tanh ^{-1}(\cos (c+d x))}{2 a^3 d}-\frac{4 \cot (c+d x)}{a^3 d}-\frac{5 \cot ^3(c+d x)}{3 a^3 d}-\frac{\cot ^5(c+d x)}{5 a^3 d}+\frac{13 \cot (c+d x) \csc (c+d x)}{8 a^3 d}+\frac{3 \cot (c+d x) \csc ^3(c+d x)}{4 a^3 d}-\frac{9 \int \csc (c+d x) \, dx}{8 a^3}\\ &=\frac{13 \tanh ^{-1}(\cos (c+d x))}{8 a^3 d}-\frac{4 \cot (c+d x)}{a^3 d}-\frac{5 \cot ^3(c+d x)}{3 a^3 d}-\frac{\cot ^5(c+d x)}{5 a^3 d}+\frac{13 \cot (c+d x) \csc (c+d x)}{8 a^3 d}+\frac{3 \cot (c+d x) \csc ^3(c+d x)}{4 a^3 d}\\ \end{align*}

Mathematica [A]  time = 1.81144, size = 189, normalized size = 1.66 \[ \frac{\csc ^5(c+d x) \left (1500 \sin (2 (c+d x))-390 \sin (4 (c+d x))-1600 \cos (c+d x)+1520 \cos (3 (c+d x))-304 \cos (5 (c+d x))-1950 \sin (c+d x) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )+975 \sin (3 (c+d x)) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )-195 \sin (5 (c+d x)) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )+1950 \sin (c+d x) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )-975 \sin (3 (c+d x)) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )+195 \sin (5 (c+d x)) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )}{1920 a^3 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[c + d*x]^6/(a + a*Sin[c + d*x])^3,x]

[Out]

(Csc[c + d*x]^5*(-1600*Cos[c + d*x] + 1520*Cos[3*(c + d*x)] - 304*Cos[5*(c + d*x)] + 1950*Log[Cos[(c + d*x)/2]
]*Sin[c + d*x] - 1950*Log[Sin[(c + d*x)/2]]*Sin[c + d*x] + 1500*Sin[2*(c + d*x)] - 975*Log[Cos[(c + d*x)/2]]*S
in[3*(c + d*x)] + 975*Log[Sin[(c + d*x)/2]]*Sin[3*(c + d*x)] - 390*Sin[4*(c + d*x)] + 195*Log[Cos[(c + d*x)/2]
]*Sin[5*(c + d*x)] - 195*Log[Sin[(c + d*x)/2]]*Sin[5*(c + d*x)]))/(1920*a^3*d)

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Maple [A]  time = 0.2, size = 208, normalized size = 1.8 \begin{align*}{\frac{1}{160\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{5}}-{\frac{3}{64\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{4}}+{\frac{17}{96\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}-{\frac{1}{2\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}}+{\frac{23}{16\,d{a}^{3}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }-{\frac{23}{16\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1}}-{\frac{1}{160\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-5}}+{\frac{3}{64\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-4}}-{\frac{13}{8\,d{a}^{3}}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) }-{\frac{17}{96\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-3}}+{\frac{1}{2\,d{a}^{3}} \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))^3,x)

[Out]

1/160/d/a^3*tan(1/2*d*x+1/2*c)^5-3/64/d/a^3*tan(1/2*d*x+1/2*c)^4+17/96/d/a^3*tan(1/2*d*x+1/2*c)^3-1/2/d/a^3*ta
n(1/2*d*x+1/2*c)^2+23/16/d/a^3*tan(1/2*d*x+1/2*c)-23/16/d/a^3/tan(1/2*d*x+1/2*c)-1/160/d/a^3/tan(1/2*d*x+1/2*c
)^5+3/64/d/a^3/tan(1/2*d*x+1/2*c)^4-13/8/d/a^3*ln(tan(1/2*d*x+1/2*c))-17/96/d/a^3/tan(1/2*d*x+1/2*c)^3+1/2/d/a
^3/tan(1/2*d*x+1/2*c)^2

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Maxima [B]  time = 1.0308, size = 316, normalized size = 2.77 \begin{align*} \frac{\frac{\frac{1380 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac{480 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{170 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac{45 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac{6 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a^{3}} - \frac{1560 \, \log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}} + \frac{{\left (\frac{45 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac{170 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{480 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac{1380 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - 6\right )}{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}{a^{3} \sin \left (d x + c\right )^{5}}}{960 \, 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))^3,x, algorithm="maxima")

[Out]

1/960*((1380*sin(d*x + c)/(cos(d*x + c) + 1) - 480*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 170*sin(d*x + c)^3/(c
os(d*x + c) + 1)^3 - 45*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 6*sin(d*x + c)^5/(cos(d*x + c) + 1)^5)/a^3 - 156
0*log(sin(d*x + c)/(cos(d*x + c) + 1))/a^3 + (45*sin(d*x + c)/(cos(d*x + c) + 1) - 170*sin(d*x + c)^2/(cos(d*x
+ c) + 1)^2 + 480*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 - 1380*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - 6)*(cos(d*
x + c) + 1)^5/(a^3*sin(d*x + c)^5))/d

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Fricas [A]  time = 1.14746, size = 502, normalized size = 4.4 \begin{align*} -\frac{608 \, \cos \left (d x + c\right )^{5} - 1520 \, \cos \left (d x + c\right )^{3} - 195 \,{\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 ) + 195 \,{\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 (13 \, \cos \left (d x + c\right )^{3} - 19 \, \cos \left (d x + c\right )\right )} \sin \left (d x + c\right ) + 960 \, \cos \left (d x + c\right )}{240 \,{\left (a^{3} d \cos \left (d x + c\right )^{4} - 2 \, a^{3} d \cos \left (d x + c\right )^{2} + a^{3} 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))^3,x, algorithm="fricas")

[Out]

-1/240*(608*cos(d*x + c)^5 - 1520*cos(d*x + c)^3 - 195*(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) + 195*(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*(13*cos(d*x + c)^3 - 19*cos(d*x + c))*sin(d*x + c) + 960*cos(d*x + c))/((a^3*d*cos(d*x + c)^4 - 2*a^3
*d*cos(d*x + c)^2 + a^3*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))**3,x)

[Out]

Timed out

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Giac [A]  time = 1.32235, size = 252, normalized size = 2.21 \begin{align*} -\frac{\frac{1560 \, \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right )}{a^{3}} - \frac{3562 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 1380 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 480 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 170 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 45 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 6}{a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5}} - \frac{6 \, a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 45 \, a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 170 \, a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 480 \, a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1380 \, a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{a^{15}}}{960 \, 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))^3,x, algorithm="giac")

[Out]

-1/960*(1560*log(abs(tan(1/2*d*x + 1/2*c)))/a^3 - (3562*tan(1/2*d*x + 1/2*c)^5 - 1380*tan(1/2*d*x + 1/2*c)^4 +
480*tan(1/2*d*x + 1/2*c)^3 - 170*tan(1/2*d*x + 1/2*c)^2 + 45*tan(1/2*d*x + 1/2*c) - 6)/(a^3*tan(1/2*d*x + 1/2
*c)^5) - (6*a^12*tan(1/2*d*x + 1/2*c)^5 - 45*a^12*tan(1/2*d*x + 1/2*c)^4 + 170*a^12*tan(1/2*d*x + 1/2*c)^3 - 4
80*a^12*tan(1/2*d*x + 1/2*c)^2 + 1380*a^12*tan(1/2*d*x + 1/2*c))/a^15)/d