3.727 \(\int \frac{\cos ^7(c+d x) \cot (c+d x)}{(a+a \sin (c+d x))^2} \, dx\)

Optimal. Leaf size=119 \[ -\frac{\cos ^5(c+d x)}{5 a^2 d}+\frac{\cos ^3(c+d x)}{3 a^2 d}+\frac{\cos (c+d x)}{a^2 d}-\frac{\sin (c+d x) \cos ^3(c+d x)}{2 a^2 d}-\frac{3 \sin (c+d x) \cos (c+d x)}{4 a^2 d}-\frac{\tanh ^{-1}(\cos (c+d x))}{a^2 d}-\frac{3 x}{4 a^2} \]

[Out]

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

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Rubi [A]  time = 0.236758, antiderivative size = 119, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 9, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {2875, 2873, 2635, 8, 2592, 302, 206, 2565, 30} \[ -\frac{\cos ^5(c+d x)}{5 a^2 d}+\frac{\cos ^3(c+d x)}{3 a^2 d}+\frac{\cos (c+d x)}{a^2 d}-\frac{\sin (c+d x) \cos ^3(c+d x)}{2 a^2 d}-\frac{3 \sin (c+d x) \cos (c+d x)}{4 a^2 d}-\frac{\tanh ^{-1}(\cos (c+d x))}{a^2 d}-\frac{3 x}{4 a^2} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[c + d*x]^7*Cot[c + d*x])/(a + a*Sin[c + d*x])^2,x]

[Out]

(-3*x)/(4*a^2) - ArcTanh[Cos[c + d*x]]/(a^2*d) + Cos[c + d*x]/(a^2*d) + Cos[c + d*x]^3/(3*a^2*d) - Cos[c + d*x
]^5/(5*a^2*d) - (3*Cos[c + d*x]*Sin[c + d*x])/(4*a^2*d) - (Cos[c + d*x]^3*Sin[c + d*x])/(2*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 2635

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 2592

Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> With[{ff = FreeFactors[S
in[e + f*x], x]}, Dist[ff/f, Subst[Int[(ff*x)^(m + n)/(a^2 - ff^2*x^2)^((n + 1)/2), x], x, (a*Sin[e + f*x])/ff
], x]] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n + 1)/2]

Rule 302

Int[(x_)^(m_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^n, x], x] /; FreeQ[{a,
b}, x] && IGtQ[m, 0] && IGtQ[n, 0] && GtQ[m, 2*n - 1]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 2565

Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_.)*sin[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> -Dist[(a*f)^(-1), Subst[
Int[x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a*Cos[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2]
&&  !(IntegerQ[(m - 1)/2] && GtQ[m, 0] && LeQ[m, n])

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rubi steps

\begin{align*} \int \frac{\cos ^7(c+d x) \cot (c+d x)}{(a+a \sin (c+d x))^2} \, dx &=\frac{\int \cos ^3(c+d x) \cot (c+d x) (a-a \sin (c+d x))^2 \, dx}{a^4}\\ &=\frac{\int \left (-2 a^2 \cos ^4(c+d x)+a^2 \cos ^3(c+d x) \cot (c+d x)+a^2 \cos ^4(c+d x) \sin (c+d x)\right ) \, dx}{a^4}\\ &=\frac{\int \cos ^3(c+d x) \cot (c+d x) \, dx}{a^2}+\frac{\int \cos ^4(c+d x) \sin (c+d x) \, dx}{a^2}-\frac{2 \int \cos ^4(c+d x) \, dx}{a^2}\\ &=-\frac{\cos ^3(c+d x) \sin (c+d x)}{2 a^2 d}-\frac{3 \int \cos ^2(c+d x) \, dx}{2 a^2}-\frac{\operatorname{Subst}\left (\int x^4 \, dx,x,\cos (c+d x)\right )}{a^2 d}-\frac{\operatorname{Subst}\left (\int \frac{x^4}{1-x^2} \, dx,x,\cos (c+d x)\right )}{a^2 d}\\ &=-\frac{\cos ^5(c+d x)}{5 a^2 d}-\frac{3 \cos (c+d x) \sin (c+d x)}{4 a^2 d}-\frac{\cos ^3(c+d x) \sin (c+d x)}{2 a^2 d}-\frac{3 \int 1 \, dx}{4 a^2}-\frac{\operatorname{Subst}\left (\int \left (-1-x^2+\frac{1}{1-x^2}\right ) \, dx,x,\cos (c+d x)\right )}{a^2 d}\\ &=-\frac{3 x}{4 a^2}+\frac{\cos (c+d x)}{a^2 d}+\frac{\cos ^3(c+d x)}{3 a^2 d}-\frac{\cos ^5(c+d x)}{5 a^2 d}-\frac{3 \cos (c+d x) \sin (c+d x)}{4 a^2 d}-\frac{\cos ^3(c+d x) \sin (c+d x)}{2 a^2 d}-\frac{\operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,\cos (c+d x)\right )}{a^2 d}\\ &=-\frac{3 x}{4 a^2}-\frac{\tanh ^{-1}(\cos (c+d x))}{a^2 d}+\frac{\cos (c+d x)}{a^2 d}+\frac{\cos ^3(c+d x)}{3 a^2 d}-\frac{\cos ^5(c+d x)}{5 a^2 d}-\frac{3 \cos (c+d x) \sin (c+d x)}{4 a^2 d}-\frac{\cos ^3(c+d x) \sin (c+d x)}{2 a^2 d}\\ \end{align*}

Mathematica [A]  time = 0.701307, size = 93, normalized size = 0.78 \[ \frac{270 \cos (c+d x)+5 \cos (3 (c+d x))-3 \left (40 \sin (2 (c+d x))+5 \sin (4 (c+d x))+\cos (5 (c+d x))-80 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )+80 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )+60 c+60 d x\right )}{240 a^2 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cos[c + d*x]^7*Cot[c + d*x])/(a + a*Sin[c + d*x])^2,x]

[Out]

(270*Cos[c + d*x] + 5*Cos[3*(c + d*x)] - 3*(60*c + 60*d*x + Cos[5*(c + d*x)] + 80*Log[Cos[(c + d*x)/2]] - 80*L
og[Sin[(c + d*x)/2]] + 40*Sin[2*(c + d*x)] + 5*Sin[4*(c + d*x)]))/(240*a^2*d)

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Maple [B]  time = 0.132, size = 329, normalized size = 2.8 \begin{align*}{\frac{5}{2\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{9} \left ( 1+ \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2} \right ) ^{-5}}+2\,{\frac{ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{8}}{d{a}^{2} \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{5}}}+{\frac{1}{d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{7} \left ( 1+ \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2} \right ) ^{-5}}+12\,{\frac{ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{6}}{d{a}^{2} \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{5}}}+{\frac{32}{3\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{4} \left ( 1+ \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2} \right ) ^{-5}}-{\frac{1}{d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3} \left ( 1+ \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2} \right ) ^{-5}}+{\frac{28}{3\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2} \left ( 1+ \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2} \right ) ^{-5}}-{\frac{5}{2\,d{a}^{2}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \left ( 1+ \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2} \right ) ^{-5}}+{\frac{34}{15\,d{a}^{2}} \left ( 1+ \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2} \right ) ^{-5}}-{\frac{3}{2\,d{a}^{2}}\arctan \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) }+{\frac{1}{d{a}^{2}}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^8*csc(d*x+c)/(a+a*sin(d*x+c))^2,x)

[Out]

5/2/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^5*tan(1/2*d*x+1/2*c)^9+2/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^5*tan(1/2*d*x+1/2*c
)^8+1/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^5*tan(1/2*d*x+1/2*c)^7+12/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^5*tan(1/2*d*x+1/
2*c)^6+32/3/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^5*tan(1/2*d*x+1/2*c)^4-1/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^5*tan(1/2*d
*x+1/2*c)^3+28/3/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^5*tan(1/2*d*x+1/2*c)^2-5/2/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^5*ta
n(1/2*d*x+1/2*c)+34/15/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^5-3/2/d/a^2*arctan(tan(1/2*d*x+1/2*c))+1/d/a^2*ln(tan(1/
2*d*x+1/2*c))

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Maxima [B]  time = 1.56378, size = 450, normalized size = 3.78 \begin{align*} -\frac{\frac{\frac{75 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac{280 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{30 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac{320 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac{360 \, \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} - \frac{30 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} - \frac{60 \, \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}} - \frac{75 \, \sin \left (d x + c\right )^{9}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{9}} - 68}{a^{2} + \frac{5 \, a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{10 \, a^{2} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac{10 \, a^{2} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac{5 \, a^{2} \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}} + \frac{a^{2} \sin \left (d x + c\right )^{10}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{10}}} + \frac{45 \, \arctan \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}} - \frac{30 \, \log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}}}{30 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/30*((75*sin(d*x + c)/(cos(d*x + c) + 1) - 280*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 30*sin(d*x + c)^3/(cos(
d*x + c) + 1)^3 - 320*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - 360*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 - 30*sin(d
*x + c)^7/(cos(d*x + c) + 1)^7 - 60*sin(d*x + c)^8/(cos(d*x + c) + 1)^8 - 75*sin(d*x + c)^9/(cos(d*x + c) + 1)
^9 - 68)/(a^2 + 5*a^2*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 10*a^2*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 10*a^
2*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + 5*a^2*sin(d*x + c)^8/(cos(d*x + c) + 1)^8 + a^2*sin(d*x + c)^10/(cos(d
*x + c) + 1)^10) + 45*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a^2 - 30*log(sin(d*x + c)/(cos(d*x + c) + 1))/a^
2)/d

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Fricas [A]  time = 1.15343, size = 274, normalized size = 2.3 \begin{align*} -\frac{12 \, \cos \left (d x + c\right )^{5} - 20 \, \cos \left (d x + c\right )^{3} + 45 \, d x + 15 \,{\left (2 \, \cos \left (d x + c\right )^{3} + 3 \, \cos \left (d x + c\right )\right )} \sin \left (d x + c\right ) - 60 \, \cos \left (d x + c\right ) + 30 \, \log \left (\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) - 30 \, \log \left (-\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right )}{60 \, a^{2} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/60*(12*cos(d*x + c)^5 - 20*cos(d*x + c)^3 + 45*d*x + 15*(2*cos(d*x + c)^3 + 3*cos(d*x + c))*sin(d*x + c) -
60*cos(d*x + c) + 30*log(1/2*cos(d*x + c) + 1/2) - 30*log(-1/2*cos(d*x + c) + 1/2))/(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)**8*csc(d*x+c)/(a+a*sin(d*x+c))**2,x)

[Out]

Timed out

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Giac [A]  time = 1.312, size = 211, normalized size = 1.77 \begin{align*} -\frac{\frac{45 \,{\left (d x + c\right )}}{a^{2}} - \frac{60 \, \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right )}{a^{2}} - \frac{2 \,{\left (75 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{9} + 60 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{8} + 30 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 360 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{6} + 320 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 30 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 280 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 75 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 68\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1\right )}^{5} a^{2}}}{60 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/60*(45*(d*x + c)/a^2 - 60*log(abs(tan(1/2*d*x + 1/2*c)))/a^2 - 2*(75*tan(1/2*d*x + 1/2*c)^9 + 60*tan(1/2*d*
x + 1/2*c)^8 + 30*tan(1/2*d*x + 1/2*c)^7 + 360*tan(1/2*d*x + 1/2*c)^6 + 320*tan(1/2*d*x + 1/2*c)^4 - 30*tan(1/
2*d*x + 1/2*c)^3 + 280*tan(1/2*d*x + 1/2*c)^2 - 75*tan(1/2*d*x + 1/2*c) + 68)/((tan(1/2*d*x + 1/2*c)^2 + 1)^5*
a^2))/d