3.734 \(\int \frac{\cot ^8(c+d x)}{(a+a \sin (c+d x))^2} \, dx\)

Optimal. Leaf size=124 \[ -\frac{\cot ^7(c+d x)}{7 a^2 d}-\frac{2 \cot ^5(c+d x)}{5 a^2 d}+\frac{\tanh ^{-1}(\cos (c+d x))}{8 a^2 d}+\frac{\cot (c+d x) \csc ^5(c+d x)}{3 a^2 d}-\frac{7 \cot (c+d x) \csc ^3(c+d x)}{12 a^2 d}+\frac{\cot (c+d x) \csc (c+d x)}{8 a^2 d} \]

[Out]

ArcTanh[Cos[c + d*x]]/(8*a^2*d) - (2*Cot[c + d*x]^5)/(5*a^2*d) - Cot[c + d*x]^7/(7*a^2*d) + (Cot[c + d*x]*Csc[
c + d*x])/(8*a^2*d) - (7*Cot[c + d*x]*Csc[c + d*x]^3)/(12*a^2*d) + (Cot[c + d*x]*Csc[c + d*x]^5)/(3*a^2*d)

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Rubi [A]  time = 0.25998, antiderivative size = 124, normalized size of antiderivative = 1., number of steps used = 19, 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} \[ -\frac{\cot ^7(c+d x)}{7 a^2 d}-\frac{2 \cot ^5(c+d x)}{5 a^2 d}+\frac{\tanh ^{-1}(\cos (c+d x))}{8 a^2 d}+\frac{\cot (c+d x) \csc ^5(c+d x)}{3 a^2 d}-\frac{7 \cot (c+d x) \csc ^3(c+d x)}{12 a^2 d}+\frac{\cot (c+d x) \csc (c+d x)}{8 a^2 d} \]

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]^8/(a + a*Sin[c + d*x])^2,x]

[Out]

ArcTanh[Cos[c + d*x]]/(8*a^2*d) - (2*Cot[c + d*x]^5)/(5*a^2*d) - Cot[c + d*x]^7/(7*a^2*d) + (Cot[c + d*x]*Csc[
c + d*x])/(8*a^2*d) - (7*Cot[c + d*x]*Csc[c + d*x]^3)/(12*a^2*d) + (Cot[c + d*x]*Csc[c + d*x]^5)/(3*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 ^8(c+d x)}{(a+a \sin (c+d x))^2} \, dx &=\frac{\int \left (a^6 \csc ^2(c+d x)-2 a^6 \csc ^3(c+d x)-a^6 \csc ^4(c+d x)+4 a^6 \csc ^5(c+d x)-a^6 \csc ^6(c+d x)-2 a^6 \csc ^7(c+d x)+a^6 \csc ^8(c+d x)\right ) \, dx}{a^8}\\ &=\frac{\int \csc ^2(c+d x) \, dx}{a^2}-\frac{\int \csc ^4(c+d x) \, dx}{a^2}-\frac{\int \csc ^6(c+d x) \, dx}{a^2}+\frac{\int \csc ^8(c+d x) \, dx}{a^2}-\frac{2 \int \csc ^3(c+d x) \, dx}{a^2}-\frac{2 \int \csc ^7(c+d x) \, dx}{a^2}+\frac{4 \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)}{a^2 d}+\frac{\cot (c+d x) \csc ^5(c+d x)}{3 a^2 d}-\frac{\int \csc (c+d x) \, dx}{a^2}-\frac{5 \int \csc ^5(c+d x) \, dx}{3 a^2}+\frac{3 \int \csc ^3(c+d x) \, dx}{a^2}-\frac{\operatorname{Subst}(\int 1 \, dx,x,\cot (c+d x))}{a^2 d}+\frac{\operatorname{Subst}\left (\int \left (1+x^2\right ) \, dx,x,\cot (c+d x)\right )}{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{\operatorname{Subst}\left (\int \left (1+3 x^2+3 x^4+x^6\right ) \, dx,x,\cot (c+d x)\right )}{a^2 d}\\ &=\frac{\tanh ^{-1}(\cos (c+d x))}{a^2 d}-\frac{2 \cot ^5(c+d x)}{5 a^2 d}-\frac{\cot ^7(c+d x)}{7 a^2 d}-\frac{\cot (c+d x) \csc (c+d x)}{2 a^2 d}-\frac{7 \cot (c+d x) \csc ^3(c+d x)}{12 a^2 d}+\frac{\cot (c+d x) \csc ^5(c+d x)}{3 a^2 d}-\frac{5 \int \csc ^3(c+d x) \, dx}{4 a^2}+\frac{3 \int \csc (c+d x) \, dx}{2 a^2}\\ &=-\frac{\tanh ^{-1}(\cos (c+d x))}{2 a^2 d}-\frac{2 \cot ^5(c+d x)}{5 a^2 d}-\frac{\cot ^7(c+d x)}{7 a^2 d}+\frac{\cot (c+d x) \csc (c+d x)}{8 a^2 d}-\frac{7 \cot (c+d x) \csc ^3(c+d x)}{12 a^2 d}+\frac{\cot (c+d x) \csc ^5(c+d x)}{3 a^2 d}-\frac{5 \int \csc (c+d x) \, dx}{8 a^2}\\ &=\frac{\tanh ^{-1}(\cos (c+d x))}{8 a^2 d}-\frac{2 \cot ^5(c+d x)}{5 a^2 d}-\frac{\cot ^7(c+d x)}{7 a^2 d}+\frac{\cot (c+d x) \csc (c+d x)}{8 a^2 d}-\frac{7 \cot (c+d x) \csc ^3(c+d x)}{12 a^2 d}+\frac{\cot (c+d x) \csc ^5(c+d x)}{3 a^2 d}\\ \end{align*}

Mathematica [B]  time = 1.08262, size = 251, normalized size = 2.02 \[ -\frac{\csc ^7(c+d x) \left (-2170 \sin (2 (c+d x))-3080 \sin (4 (c+d x))-210 \sin (6 (c+d x))+5880 \cos (c+d x)+2184 \cos (3 (c+d x))-168 \cos (5 (c+d x))-216 \cos (7 (c+d x))+3675 \sin (c+d x) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )-2205 \sin (3 (c+d x)) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )+735 \sin (5 (c+d x)) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )-105 \sin (7 (c+d x)) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )-3675 \sin (c+d x) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )+2205 \sin (3 (c+d x)) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )-735 \sin (5 (c+d x)) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )+105 \sin (7 (c+d x)) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )}{53760 a^2 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[c + d*x]^8/(a + a*Sin[c + d*x])^2,x]

[Out]

-(Csc[c + d*x]^7*(5880*Cos[c + d*x] + 2184*Cos[3*(c + d*x)] - 168*Cos[5*(c + d*x)] - 216*Cos[7*(c + d*x)] - 36
75*Log[Cos[(c + d*x)/2]]*Sin[c + d*x] + 3675*Log[Sin[(c + d*x)/2]]*Sin[c + d*x] - 2170*Sin[2*(c + d*x)] + 2205
*Log[Cos[(c + d*x)/2]]*Sin[3*(c + d*x)] - 2205*Log[Sin[(c + d*x)/2]]*Sin[3*(c + d*x)] - 3080*Sin[4*(c + d*x)]
- 735*Log[Cos[(c + d*x)/2]]*Sin[5*(c + d*x)] + 735*Log[Sin[(c + d*x)/2]]*Sin[5*(c + d*x)] - 210*Sin[6*(c + d*x
)] + 105*Log[Cos[(c + d*x)/2]]*Sin[7*(c + d*x)] - 105*Log[Sin[(c + d*x)/2]]*Sin[7*(c + d*x)]))/(53760*a^2*d)

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Maple [B]  time = 0.183, size = 284, normalized size = 2.3 \begin{align*}{\frac{1}{896\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{7}}-{\frac{1}{192\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{6}}+{\frac{3}{640\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{5}}+{\frac{1}{64\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{4}}-{\frac{5}{128\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}+{\frac{1}{64\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}}+{\frac{11}{128\,d{a}^{2}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }-{\frac{1}{896\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-7}}-{\frac{11}{128\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1}}-{\frac{3}{640\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-5}}-{\frac{1}{64\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-4}}-{\frac{1}{8\,d{a}^{2}}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) }+{\frac{1}{192\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-6}}+{\frac{5}{128\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-3}}-{\frac{1}{64\,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)^8*csc(d*x+c)^8/(a+a*sin(d*x+c))^2,x)

[Out]

1/896/d/a^2*tan(1/2*d*x+1/2*c)^7-1/192/d/a^2*tan(1/2*d*x+1/2*c)^6+3/640/d/a^2*tan(1/2*d*x+1/2*c)^5+1/64/d/a^2*
tan(1/2*d*x+1/2*c)^4-5/128/d/a^2*tan(1/2*d*x+1/2*c)^3+1/64/d/a^2*tan(1/2*d*x+1/2*c)^2+11/128/d/a^2*tan(1/2*d*x
+1/2*c)-1/896/d/a^2/tan(1/2*d*x+1/2*c)^7-11/128/d/a^2/tan(1/2*d*x+1/2*c)-3/640/d/a^2/tan(1/2*d*x+1/2*c)^5-1/64
/d/a^2/tan(1/2*d*x+1/2*c)^4-1/8/d/a^2*ln(tan(1/2*d*x+1/2*c))+1/192/d/a^2/tan(1/2*d*x+1/2*c)^6+5/128/d/a^2/tan(
1/2*d*x+1/2*c)^3-1/64/d/a^2/tan(1/2*d*x+1/2*c)^2

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Maxima [B]  time = 1.04902, size = 424, normalized size = 3.42 \begin{align*} \frac{\frac{\frac{1155 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{210 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac{525 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac{210 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac{63 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac{70 \, \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac{15 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}}{a^{2}} - \frac{1680 \, \log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}} + \frac{{\left (\frac{70 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac{63 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac{210 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac{525 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac{210 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac{1155 \, \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} - 15\right )}{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}{a^{2} \sin \left (d x + c\right )^{7}}}{13440 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^8*csc(d*x+c)^8/(a+a*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

1/13440*((1155*sin(d*x + c)/(cos(d*x + c) + 1) + 210*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - 525*sin(d*x + c)^3/
(cos(d*x + c) + 1)^3 + 210*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 63*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 70*s
in(d*x + c)^6/(cos(d*x + c) + 1)^6 + 15*sin(d*x + c)^7/(cos(d*x + c) + 1)^7)/a^2 - 1680*log(sin(d*x + c)/(cos(
d*x + c) + 1))/a^2 + (70*sin(d*x + c)/(cos(d*x + c) + 1) - 63*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - 210*sin(d*
x + c)^3/(cos(d*x + c) + 1)^3 + 525*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - 210*sin(d*x + c)^5/(cos(d*x + c) + 1
)^5 - 1155*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 - 15)*(cos(d*x + c) + 1)^7/(a^2*sin(d*x + c)^7))/d

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Fricas [A]  time = 1.13392, size = 585, normalized size = 4.72 \begin{align*} -\frac{432 \, \cos \left (d x + c\right )^{7} - 672 \, \cos \left (d x + c\right )^{5} - 105 \,{\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 ) \sin \left (d x + c\right ) + 105 \,{\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 ) \sin \left (d x + c\right ) + 70 \,{\left (3 \, \cos \left (d x + c\right )^{5} + 8 \, \cos \left (d x + c\right )^{3} - 3 \, \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{1680 \,{\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 )} \sin \left (d x + c\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^8*csc(d*x+c)^8/(a+a*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

-1/1680*(432*cos(d*x + c)^7 - 672*cos(d*x + c)^5 - 105*(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)*sin(d*x + c) + 105*(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)*sin(d*x + c) + 70*(3*cos(d*x + c)^5 + 8*cos(d*x + c)^3 - 3*cos(d*x + c))*sin(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)*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)**8*csc(d*x+c)**8/(a+a*sin(d*x+c))**2,x)

[Out]

Timed out

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Giac [B]  time = 1.32531, size = 331, normalized size = 2.67 \begin{align*} -\frac{\frac{1680 \, \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right )}{a^{2}} - \frac{4356 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 1155 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{6} - 210 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 525 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 210 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 63 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 70 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 15}{a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7}} - \frac{15 \, a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 70 \, a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{6} + 63 \, a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 210 \, a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 525 \, a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 210 \, a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1155 \, a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{a^{14}}}{13440 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^8*csc(d*x+c)^8/(a+a*sin(d*x+c))^2,x, algorithm="giac")

[Out]

-1/13440*(1680*log(abs(tan(1/2*d*x + 1/2*c)))/a^2 - (4356*tan(1/2*d*x + 1/2*c)^7 - 1155*tan(1/2*d*x + 1/2*c)^6
- 210*tan(1/2*d*x + 1/2*c)^5 + 525*tan(1/2*d*x + 1/2*c)^4 - 210*tan(1/2*d*x + 1/2*c)^3 - 63*tan(1/2*d*x + 1/2
*c)^2 + 70*tan(1/2*d*x + 1/2*c) - 15)/(a^2*tan(1/2*d*x + 1/2*c)^7) - (15*a^12*tan(1/2*d*x + 1/2*c)^7 - 70*a^12
*tan(1/2*d*x + 1/2*c)^6 + 63*a^12*tan(1/2*d*x + 1/2*c)^5 + 210*a^12*tan(1/2*d*x + 1/2*c)^4 - 525*a^12*tan(1/2*
d*x + 1/2*c)^3 + 210*a^12*tan(1/2*d*x + 1/2*c)^2 + 1155*a^12*tan(1/2*d*x + 1/2*c))/a^14)/d