3.647 \(\int \frac{\cos ^5(c+d x) \cot (c+d x)}{(a+a \sin (c+d x))^3} \, dx\)

Optimal. Leaf size=60 \[ -\frac{3 \cos (c+d x)}{a^3 d}+\frac{\sin (c+d x) \cos (c+d x)}{2 a^3 d}-\frac{\tanh ^{-1}(\cos (c+d x))}{a^3 d}-\frac{7 x}{2 a^3} \]

[Out]

(-7*x)/(2*a^3) - ArcTanh[Cos[c + d*x]]/(a^3*d) - (3*Cos[c + d*x])/(a^3*d) + (Cos[c + d*x]*Sin[c + d*x])/(2*a^3
*d)

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Rubi [A]  time = 0.147155, antiderivative size = 60, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.222, Rules used = {2869, 2757, 3770, 2638, 2635, 8} \[ -\frac{3 \cos (c+d x)}{a^3 d}+\frac{\sin (c+d x) \cos (c+d x)}{2 a^3 d}-\frac{\tanh ^{-1}(\cos (c+d x))}{a^3 d}-\frac{7 x}{2 a^3} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[c + d*x]^5*Cot[c + d*x])/(a + a*Sin[c + d*x])^3,x]

[Out]

(-7*x)/(2*a^3) - ArcTanh[Cos[c + d*x]]/(a^3*d) - (3*Cos[c + d*x])/(a^3*d) + (Cos[c + d*x]*Sin[c + d*x])/(2*a^3
*d)

Rule 2869

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

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 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rule 2638

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Cos[c + d*x]/d, x] /; FreeQ[{c, d}, x]

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]

Rubi steps

\begin{align*} \int \frac{\cos ^5(c+d x) \cot (c+d x)}{(a+a \sin (c+d x))^3} \, dx &=\frac{\int \csc (c+d x) (a-a \sin (c+d x))^3 \, dx}{a^6}\\ &=\frac{\int \left (-3 a^3+a^3 \csc (c+d x)+3 a^3 \sin (c+d x)-a^3 \sin ^2(c+d x)\right ) \, dx}{a^6}\\ &=-\frac{3 x}{a^3}+\frac{\int \csc (c+d x) \, dx}{a^3}-\frac{\int \sin ^2(c+d x) \, dx}{a^3}+\frac{3 \int \sin (c+d x) \, dx}{a^3}\\ &=-\frac{3 x}{a^3}-\frac{\tanh ^{-1}(\cos (c+d x))}{a^3 d}-\frac{3 \cos (c+d x)}{a^3 d}+\frac{\cos (c+d x) \sin (c+d x)}{2 a^3 d}-\frac{\int 1 \, dx}{2 a^3}\\ &=-\frac{7 x}{2 a^3}-\frac{\tanh ^{-1}(\cos (c+d x))}{a^3 d}-\frac{3 \cos (c+d x)}{a^3 d}+\frac{\cos (c+d x) \sin (c+d x)}{2 a^3 d}\\ \end{align*}

Mathematica [A]  time = 0.217078, size = 63, normalized size = 1.05 \[ \frac{\sin (2 (c+d x))-12 \cos (c+d x)-2 \left (-2 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )+2 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )+7 c+7 d x\right )}{4 a^3 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cos[c + d*x]^5*Cot[c + d*x])/(a + a*Sin[c + d*x])^3,x]

[Out]

(-12*Cos[c + d*x] - 2*(7*c + 7*d*x + 2*Log[Cos[(c + d*x)/2]] - 2*Log[Sin[(c + d*x)/2]]) + Sin[2*(c + d*x)])/(4
*a^3*d)

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Maple [B]  time = 0.151, size = 159, normalized size = 2.7 \begin{align*} -{\frac{1}{d{a}^{3}} \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 ) ^{-2}}-6\,{\frac{ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}{d{a}^{3} \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{2}}}+{\frac{1}{d{a}^{3}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \left ( 1+ \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2} \right ) ^{-2}}-6\,{\frac{1}{d{a}^{3} \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{2}}}-7\,{\frac{\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) }{d{a}^{3}}}+{\frac{1}{d{a}^{3}}\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)^6*csc(d*x+c)/(a+a*sin(d*x+c))^3,x)

[Out]

-1/d/a^3/(1+tan(1/2*d*x+1/2*c)^2)^2*tan(1/2*d*x+1/2*c)^3-6/d/a^3/(1+tan(1/2*d*x+1/2*c)^2)^2*tan(1/2*d*x+1/2*c)
^2+1/d/a^3/(1+tan(1/2*d*x+1/2*c)^2)^2*tan(1/2*d*x+1/2*c)-6/d/a^3/(1+tan(1/2*d*x+1/2*c)^2)^2-7/d/a^3*arctan(tan
(1/2*d*x+1/2*c))+1/d/a^3*ln(tan(1/2*d*x+1/2*c))

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Maxima [B]  time = 1.52224, size = 217, normalized size = 3.62 \begin{align*} \frac{\frac{\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac{6 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac{\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - 6}{a^{3} + \frac{2 \, a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{a^{3} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}} - \frac{7 \, \arctan \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}} + \frac{\log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*csc(d*x+c)/(a+a*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

((sin(d*x + c)/(cos(d*x + c) + 1) - 6*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - sin(d*x + c)^3/(cos(d*x + c) + 1)^
3 - 6)/(a^3 + 2*a^3*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + a^3*sin(d*x + c)^4/(cos(d*x + c) + 1)^4) - 7*arctan(
sin(d*x + c)/(cos(d*x + c) + 1))/a^3 + log(sin(d*x + c)/(cos(d*x + c) + 1))/a^3)/d

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Fricas [A]  time = 1.11585, size = 173, normalized size = 2.88 \begin{align*} -\frac{7 \, d x - \cos \left (d x + c\right ) \sin \left (d x + c\right ) + 6 \, \cos \left (d x + c\right ) + \log \left (\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) - \log \left (-\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right )}{2 \, a^{3} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*csc(d*x+c)/(a+a*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

-1/2*(7*d*x - cos(d*x + c)*sin(d*x + c) + 6*cos(d*x + c) + log(1/2*cos(d*x + c) + 1/2) - log(-1/2*cos(d*x + c)
+ 1/2))/(a^3*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)/(a+a*sin(d*x+c))**3,x)

[Out]

Timed out

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Giac [A]  time = 1.28735, size = 120, normalized size = 2. \begin{align*} -\frac{\frac{7 \,{\left (d x + c\right )}}{a^{3}} - \frac{2 \, \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right )}{a^{3}} + \frac{2 \,{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 6 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 6\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1\right )}^{2} a^{3}}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*csc(d*x+c)/(a+a*sin(d*x+c))^3,x, algorithm="giac")

[Out]

-1/2*(7*(d*x + c)/a^3 - 2*log(abs(tan(1/2*d*x + 1/2*c)))/a^3 + 2*(tan(1/2*d*x + 1/2*c)^3 + 6*tan(1/2*d*x + 1/2
*c)^2 - tan(1/2*d*x + 1/2*c) + 6)/((tan(1/2*d*x + 1/2*c)^2 + 1)^2*a^3))/d