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

Optimal. Leaf size=124 \[ \frac{2 \cot ^5(c+d x)}{5 a^2 d}+\frac{2 \cot ^3(c+d x)}{3 a^2 d}+\frac{3 \tanh ^{-1}(\cos (c+d x))}{16 a^2 d}-\frac{\cot (c+d x) \csc ^5(c+d x)}{6 a^2 d}-\frac{5 \cot (c+d x) \csc ^3(c+d x)}{24 a^2 d}+\frac{3 \cot (c+d x) \csc (c+d x)}{16 a^2 d} \]

[Out]

(3*ArcTanh[Cos[c + d*x]])/(16*a^2*d) + (2*Cot[c + d*x]^3)/(3*a^2*d) + (2*Cot[c + d*x]^5)/(5*a^2*d) + (3*Cot[c
+ d*x]*Csc[c + d*x])/(16*a^2*d) - (5*Cot[c + d*x]*Csc[c + d*x]^3)/(24*a^2*d) - (Cot[c + d*x]*Csc[c + d*x]^5)/(
6*a^2*d)

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Rubi [A]  time = 0.33197, antiderivative size = 124, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 7, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.259, Rules used = {2875, 2873, 2611, 3768, 3770, 2607, 14} \[ \frac{2 \cot ^5(c+d x)}{5 a^2 d}+\frac{2 \cot ^3(c+d x)}{3 a^2 d}+\frac{3 \tanh ^{-1}(\cos (c+d x))}{16 a^2 d}-\frac{\cot (c+d x) \csc ^5(c+d x)}{6 a^2 d}-\frac{5 \cot (c+d x) \csc ^3(c+d x)}{24 a^2 d}+\frac{3 \cot (c+d x) \csc (c+d x)}{16 a^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*ArcTanh[Cos[c + d*x]])/(16*a^2*d) + (2*Cot[c + d*x]^3)/(3*a^2*d) + (2*Cot[c + d*x]^5)/(5*a^2*d) + (3*Cot[c
+ d*x]*Csc[c + d*x])/(16*a^2*d) - (5*Cot[c + d*x]*Csc[c + d*x]^3)/(24*a^2*d) - (Cot[c + d*x]*Csc[c + d*x]^5)/(
6*a^2*d)

Rule 2875

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

Rule 2873

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*
(x_)])^(m_), x_Symbol] :> Int[ExpandTrig[(g*cos[e + f*x])^p, (d*sin[e + f*x])^n*(a + b*sin[e + f*x])^m, x], x]
/; FreeQ[{a, b, d, e, f, g, n, p}, x] && EqQ[a^2 - b^2, 0] && IGtQ[m, 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 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 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 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rubi steps

\begin{align*} \int \frac{\cot ^6(c+d x) \csc (c+d x)}{(a+a \sin (c+d x))^2} \, dx &=\frac{\int \cot ^2(c+d x) \csc ^5(c+d x) (a-a \sin (c+d x))^2 \, dx}{a^4}\\ &=\frac{\int \left (a^2 \cot ^2(c+d x) \csc ^3(c+d x)-2 a^2 \cot ^2(c+d x) \csc ^4(c+d x)+a^2 \cot ^2(c+d x) \csc ^5(c+d x)\right ) \, dx}{a^4}\\ &=\frac{\int \cot ^2(c+d x) \csc ^3(c+d x) \, dx}{a^2}+\frac{\int \cot ^2(c+d x) \csc ^5(c+d x) \, dx}{a^2}-\frac{2 \int \cot ^2(c+d x) \csc ^4(c+d x) \, dx}{a^2}\\ &=-\frac{\cot (c+d x) \csc ^3(c+d x)}{4 a^2 d}-\frac{\cot (c+d x) \csc ^5(c+d x)}{6 a^2 d}-\frac{\int \csc ^5(c+d x) \, dx}{6 a^2}-\frac{\int \csc ^3(c+d x) \, dx}{4 a^2}-\frac{2 \operatorname{Subst}\left (\int x^2 \left (1+x^2\right ) \, dx,x,-\cot (c+d x)\right )}{a^2 d}\\ &=\frac{\cot (c+d x) \csc (c+d x)}{8 a^2 d}-\frac{5 \cot (c+d x) \csc ^3(c+d x)}{24 a^2 d}-\frac{\cot (c+d x) \csc ^5(c+d x)}{6 a^2 d}-\frac{\int \csc (c+d x) \, dx}{8 a^2}-\frac{\int \csc ^3(c+d x) \, dx}{8 a^2}-\frac{2 \operatorname{Subst}\left (\int \left (x^2+x^4\right ) \, dx,x,-\cot (c+d x)\right )}{a^2 d}\\ &=\frac{\tanh ^{-1}(\cos (c+d x))}{8 a^2 d}+\frac{2 \cot ^3(c+d x)}{3 a^2 d}+\frac{2 \cot ^5(c+d x)}{5 a^2 d}+\frac{3 \cot (c+d x) \csc (c+d x)}{16 a^2 d}-\frac{5 \cot (c+d x) \csc ^3(c+d x)}{24 a^2 d}-\frac{\cot (c+d x) \csc ^5(c+d x)}{6 a^2 d}-\frac{\int \csc (c+d x) \, dx}{16 a^2}\\ &=\frac{3 \tanh ^{-1}(\cos (c+d x))}{16 a^2 d}+\frac{2 \cot ^3(c+d x)}{3 a^2 d}+\frac{2 \cot ^5(c+d x)}{5 a^2 d}+\frac{3 \cot (c+d x) \csc (c+d x)}{16 a^2 d}-\frac{5 \cot (c+d x) \csc ^3(c+d x)}{24 a^2 d}-\frac{\cot (c+d x) \csc ^5(c+d x)}{6 a^2 d}\\ \end{align*}

Mathematica [A]  time = 0.671966, size = 229, normalized size = 1.85 \[ -\frac{\csc ^6(c+d x) \left (-960 \sin (2 (c+d x))-384 \sin (4 (c+d x))+64 \sin (6 (c+d x))+1500 \cos (c+d x)-130 \cos (3 (c+d x))-90 \cos (5 (c+d x))+450 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )+675 \cos (2 (c+d x)) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )-270 \cos (4 (c+d x)) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )+45 \cos (6 (c+d x)) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )-450 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )-675 \cos (2 (c+d x)) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )+270 \cos (4 (c+d x)) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )-45 \cos (6 (c+d x)) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )\right )}{7680 a^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Csc[c + d*x]^6*(1500*Cos[c + d*x] - 130*Cos[3*(c + d*x)] - 90*Cos[5*(c + d*x)] - 450*Log[Cos[(c + d*x)/2]] +
675*Cos[2*(c + d*x)]*Log[Cos[(c + d*x)/2]] - 270*Cos[4*(c + d*x)]*Log[Cos[(c + d*x)/2]] + 45*Cos[6*(c + d*x)]
*Log[Cos[(c + d*x)/2]] + 450*Log[Sin[(c + d*x)/2]] - 675*Cos[2*(c + d*x)]*Log[Sin[(c + d*x)/2]] + 270*Cos[4*(c
+ d*x)]*Log[Sin[(c + d*x)/2]] - 45*Cos[6*(c + d*x)]*Log[Sin[(c + d*x)/2]] - 960*Sin[2*(c + d*x)] - 384*Sin[4*
(c + d*x)] + 64*Sin[6*(c + d*x)]))/(7680*a^2*d)

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Maple [B]  time = 0.176, size = 246, normalized size = 2. \begin{align*}{\frac{1}{384\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{6}}-{\frac{1}{80\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{5}}+{\frac{3}{128\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{4}}-{\frac{1}{48\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}-{\frac{1}{128\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}}+{\frac{1}{8\,d{a}^{2}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }-{\frac{1}{8\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1}}+{\frac{1}{80\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-5}}-{\frac{3}{128\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-4}}-{\frac{3}{16\,d{a}^{2}}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) }-{\frac{1}{384\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-6}}+{\frac{1}{48\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-3}}+{\frac{1}{128\,d{a}^{2}} \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)^7/(a+a*sin(d*x+c))^2,x)

[Out]

1/384/d/a^2*tan(1/2*d*x+1/2*c)^6-1/80/d/a^2*tan(1/2*d*x+1/2*c)^5+3/128/d/a^2*tan(1/2*d*x+1/2*c)^4-1/48/d/a^2*t
an(1/2*d*x+1/2*c)^3-1/128/d/a^2*tan(1/2*d*x+1/2*c)^2+1/8/d/a^2*tan(1/2*d*x+1/2*c)-1/8/d/a^2/tan(1/2*d*x+1/2*c)
+1/80/d/a^2/tan(1/2*d*x+1/2*c)^5-3/128/d/a^2/tan(1/2*d*x+1/2*c)^4-3/16/d/a^2*ln(tan(1/2*d*x+1/2*c))-1/384/d/a^
2/tan(1/2*d*x+1/2*c)^6+1/48/d/a^2/tan(1/2*d*x+1/2*c)^3+1/128/d/a^2/tan(1/2*d*x+1/2*c)^2

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Maxima [B]  time = 1.00403, size = 370, normalized size = 2.98 \begin{align*} \frac{\frac{\frac{240 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac{15 \, \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{45 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac{24 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac{5 \, \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}}}{a^{2}} - \frac{360 \, \log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}} + \frac{{\left (\frac{24 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac{45 \, \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{15 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac{240 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - 5\right )}{\left (\cos \left (d x + c\right ) + 1\right )}^{6}}{a^{2} \sin \left (d x + c\right )^{6}}}{1920 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*csc(d*x+c)^7/(a+a*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

1/1920*((240*sin(d*x + c)/(cos(d*x + c) + 1) - 15*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - 40*sin(d*x + c)^3/(cos
(d*x + c) + 1)^3 + 45*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - 24*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 + 5*sin(d*x
+ c)^6/(cos(d*x + c) + 1)^6)/a^2 - 360*log(sin(d*x + c)/(cos(d*x + c) + 1))/a^2 + (24*sin(d*x + c)/(cos(d*x +
c) + 1) - 45*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 40*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 15*sin(d*x + c)^4
/(cos(d*x + c) + 1)^4 - 240*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 5)*(cos(d*x + c) + 1)^6/(a^2*sin(d*x + c)^6)
)/d

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Fricas [A]  time = 1.14985, size = 524, normalized size = 4.23 \begin{align*} -\frac{90 \, \cos \left (d x + c\right )^{5} - 80 \, \cos \left (d x + c\right )^{3} - 45 \,{\left (\cos \left (d x + c\right )^{6} - 3 \, \cos \left (d x + c\right )^{4} + 3 \, \cos \left (d x + c\right )^{2} - 1\right )} \log \left (\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) + 45 \,{\left (\cos \left (d x + c\right )^{6} - 3 \, \cos \left (d x + c\right )^{4} + 3 \, \cos \left (d x + c\right )^{2} - 1\right )} \log \left (-\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) - 64 \,{\left (2 \, \cos \left (d x + c\right )^{5} - 5 \, \cos \left (d x + c\right )^{3}\right )} \sin \left (d x + c\right ) - 90 \, \cos \left (d x + c\right )}{480 \,{\left (a^{2} d \cos \left (d x + c\right )^{6} - 3 \, a^{2} d \cos \left (d x + c\right )^{4} + 3 \, a^{2} d \cos \left (d x + c\right )^{2} - a^{2} d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*csc(d*x+c)^7/(a+a*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

-1/480*(90*cos(d*x + c)^5 - 80*cos(d*x + c)^3 - 45*(cos(d*x + c)^6 - 3*cos(d*x + c)^4 + 3*cos(d*x + c)^2 - 1)*
log(1/2*cos(d*x + c) + 1/2) + 45*(cos(d*x + c)^6 - 3*cos(d*x + c)^4 + 3*cos(d*x + c)^2 - 1)*log(-1/2*cos(d*x +
c) + 1/2) - 64*(2*cos(d*x + c)^5 - 5*cos(d*x + c)^3)*sin(d*x + c) - 90*cos(d*x + c))/(a^2*d*cos(d*x + c)^6 -
3*a^2*d*cos(d*x + c)^4 + 3*a^2*d*cos(d*x + c)^2 - a^2*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)**6*csc(d*x+c)**7/(a+a*sin(d*x+c))**2,x)

[Out]

Timed out

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Giac [A]  time = 1.34646, size = 292, normalized size = 2.35 \begin{align*} -\frac{\frac{360 \, \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right )}{a^{2}} - \frac{882 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{6} - 240 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 15 \, \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} - 45 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 24 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 5}{a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{6}} - \frac{5 \, a^{10} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{6} - 24 \, a^{10} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 45 \, a^{10} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 40 \, a^{10} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 15 \, a^{10} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 240 \, a^{10} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{a^{12}}}{1920 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*csc(d*x+c)^7/(a+a*sin(d*x+c))^2,x, algorithm="giac")

[Out]

-1/1920*(360*log(abs(tan(1/2*d*x + 1/2*c)))/a^2 - (882*tan(1/2*d*x + 1/2*c)^6 - 240*tan(1/2*d*x + 1/2*c)^5 + 1
5*tan(1/2*d*x + 1/2*c)^4 + 40*tan(1/2*d*x + 1/2*c)^3 - 45*tan(1/2*d*x + 1/2*c)^2 + 24*tan(1/2*d*x + 1/2*c) - 5
)/(a^2*tan(1/2*d*x + 1/2*c)^6) - (5*a^10*tan(1/2*d*x + 1/2*c)^6 - 24*a^10*tan(1/2*d*x + 1/2*c)^5 + 45*a^10*tan
(1/2*d*x + 1/2*c)^4 - 40*a^10*tan(1/2*d*x + 1/2*c)^3 - 15*a^10*tan(1/2*d*x + 1/2*c)^2 + 240*a^10*tan(1/2*d*x +
1/2*c))/a^12)/d