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

Optimal. Leaf size=93 \[ \frac{\cot ^3(c+d x)}{a^3 d}+\frac{4 \cot (c+d x)}{a^3 d}-\frac{15 \tanh ^{-1}(\cos (c+d x))}{8 a^3 d}-\frac{\cot (c+d x) \csc ^3(c+d x)}{4 a^3 d}-\frac{15 \cot (c+d x) \csc (c+d x)}{8 a^3 d} \]

[Out]

(-15*ArcTanh[Cos[c + d*x]])/(8*a^3*d) + (4*Cot[c + d*x])/(a^3*d) + Cot[c + d*x]^3/(a^3*d) - (15*Cot[c + d*x]*C
sc[c + d*x])/(8*a^3*d) - (Cot[c + d*x]*Csc[c + d*x]^3)/(4*a^3*d)

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Rubi [A]  time = 0.204519, antiderivative size = 93, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 6, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.222, Rules used = {2869, 2757, 3767, 8, 3768, 3770} \[ \frac{\cot ^3(c+d x)}{a^3 d}+\frac{4 \cot (c+d x)}{a^3 d}-\frac{15 \tanh ^{-1}(\cos (c+d x))}{8 a^3 d}-\frac{\cot (c+d x) \csc ^3(c+d x)}{4 a^3 d}-\frac{15 \cot (c+d x) \csc (c+d x)}{8 a^3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-15*ArcTanh[Cos[c + d*x]])/(8*a^3*d) + (4*Cot[c + d*x])/(a^3*d) + Cot[c + d*x]^3/(a^3*d) - (15*Cot[c + d*x]*C
sc[c + d*x])/(8*a^3*d) - (Cot[c + d*x]*Csc[c + d*x]^3)/(4*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 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{\cos (c+d x) \cot ^5(c+d x)}{(a+a \sin (c+d x))^3} \, dx &=\frac{\int \csc ^5(c+d x) (a-a \sin (c+d x))^3 \, dx}{a^6}\\ &=\frac{\int \left (-a^3 \csc ^2(c+d x)+3 a^3 \csc ^3(c+d x)-3 a^3 \csc ^4(c+d x)+a^3 \csc ^5(c+d x)\right ) \, dx}{a^6}\\ &=-\frac{\int \csc ^2(c+d x) \, dx}{a^3}+\frac{\int \csc ^5(c+d x) \, dx}{a^3}+\frac{3 \int \csc ^3(c+d x) \, dx}{a^3}-\frac{3 \int \csc ^4(c+d x) \, dx}{a^3}\\ &=-\frac{3 \cot (c+d x) \csc (c+d x)}{2 a^3 d}-\frac{\cot (c+d x) \csc ^3(c+d x)}{4 a^3 d}+\frac{3 \int \csc ^3(c+d x) \, dx}{4 a^3}+\frac{3 \int \csc (c+d x) \, dx}{2 a^3}+\frac{\operatorname{Subst}(\int 1 \, dx,x,\cot (c+d x))}{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{3 \tanh ^{-1}(\cos (c+d x))}{2 a^3 d}+\frac{4 \cot (c+d x)}{a^3 d}+\frac{\cot ^3(c+d x)}{a^3 d}-\frac{15 \cot (c+d x) \csc (c+d x)}{8 a^3 d}-\frac{\cot (c+d x) \csc ^3(c+d x)}{4 a^3 d}+\frac{3 \int \csc (c+d x) \, dx}{8 a^3}\\ &=-\frac{15 \tanh ^{-1}(\cos (c+d x))}{8 a^3 d}+\frac{4 \cot (c+d x)}{a^3 d}+\frac{\cot ^3(c+d x)}{a^3 d}-\frac{15 \cot (c+d x) \csc (c+d x)}{8 a^3 d}-\frac{\cot (c+d x) \csc ^3(c+d x)}{4 a^3 d}\\ \end{align*}

Mathematica [A]  time = 2.15347, size = 125, normalized size = 1.34 \[ -\frac{\csc ^4(c+d x) \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^6 \left (-56 \sin (2 (c+d x))+46 \cos (c+d x)+6 (8 \sin (c+d x)-5) \cos (3 (c+d x))+120 \sin ^4(c+d x) \left (\log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )\right )\right )}{64 a^3 d (\sin (c+d x)+1)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Csc[c + d*x]^4*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^6*(46*Cos[c + d*x] + 120*(Log[Cos[(c + d*x)/2]] - Log[S
in[(c + d*x)/2]])*Sin[c + d*x]^4 + 6*Cos[3*(c + d*x)]*(-5 + 8*Sin[c + d*x]) - 56*Sin[2*(c + d*x)]))/(64*a^3*d*
(1 + Sin[c + d*x])^3)

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Maple [A]  time = 0.193, size = 170, normalized size = 1.8 \begin{align*}{\frac{1}{64\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{4}}-{\frac{1}{8\,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{13}{8\,d{a}^{3}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }+{\frac{13}{8\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1}}-{\frac{1}{64\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-4}}+{\frac{15}{8\,d{a}^{3}}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) }+{\frac{1}{8\,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)^5/(a+a*sin(d*x+c))^3,x)

[Out]

1/64/d/a^3*tan(1/2*d*x+1/2*c)^4-1/8/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)^2-13/8/d/a^3*tan(1
/2*d*x+1/2*c)+13/8/d/a^3/tan(1/2*d*x+1/2*c)-1/64/d/a^3/tan(1/2*d*x+1/2*c)^4+15/8/d/a^3*ln(tan(1/2*d*x+1/2*c))+
1/8/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 = 1.00906, size = 263, normalized size = 2.83 \begin{align*} -\frac{\frac{\frac{104 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac{32 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{8 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac{\sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}}{a^{3}} - \frac{120 \, \log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}} - \frac{{\left (\frac{8 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac{32 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{104 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - 1\right )}{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}{a^{3} \sin \left (d x + c\right )^{4}}}{64 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*csc(d*x+c)^5/(a+a*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

-1/64*((104*sin(d*x + c)/(cos(d*x + c) + 1) - 32*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 8*sin(d*x + c)^3/(cos(d
*x + c) + 1)^3 - sin(d*x + c)^4/(cos(d*x + c) + 1)^4)/a^3 - 120*log(sin(d*x + c)/(cos(d*x + c) + 1))/a^3 - (8*
sin(d*x + c)/(cos(d*x + c) + 1) - 32*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 104*sin(d*x + c)^3/(cos(d*x + c) +
1)^3 - 1)*(cos(d*x + c) + 1)^4/(a^3*sin(d*x + c)^4))/d

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Fricas [A]  time = 1.09012, size = 406, normalized size = 4.37 \begin{align*} \frac{30 \, \cos \left (d x + c\right )^{3} - 15 \,{\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 ) + 15 \,{\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 ) - 16 \,{\left (3 \, \cos \left (d x + c\right )^{3} - 4 \, \cos \left (d x + c\right )\right )} \sin \left (d x + c\right ) - 34 \, \cos \left (d x + c\right )}{16 \,{\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 )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*csc(d*x+c)^5/(a+a*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

1/16*(30*cos(d*x + c)^3 - 15*(cos(d*x + c)^4 - 2*cos(d*x + c)^2 + 1)*log(1/2*cos(d*x + c) + 1/2) + 15*(cos(d*x
+ c)^4 - 2*cos(d*x + c)^2 + 1)*log(-1/2*cos(d*x + c) + 1/2) - 16*(3*cos(d*x + c)^3 - 4*cos(d*x + c))*sin(d*x
+ c) - 34*cos(d*x + c))/(a^3*d*cos(d*x + c)^4 - 2*a^3*d*cos(d*x + c)^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)**5/(a+a*sin(d*x+c))**3,x)

[Out]

Timed out

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Giac [A]  time = 1.39788, size = 211, normalized size = 2.27 \begin{align*} \frac{\frac{120 \, \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right )}{a^{3}} - \frac{250 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 104 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 32 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 8 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1}{a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4}} + \frac{a^{9} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 8 \, a^{9} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 32 \, a^{9} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 104 \, a^{9} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{a^{12}}}{64 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*csc(d*x+c)^5/(a+a*sin(d*x+c))^3,x, algorithm="giac")

[Out]

1/64*(120*log(abs(tan(1/2*d*x + 1/2*c)))/a^3 - (250*tan(1/2*d*x + 1/2*c)^4 - 104*tan(1/2*d*x + 1/2*c)^3 + 32*t
an(1/2*d*x + 1/2*c)^2 - 8*tan(1/2*d*x + 1/2*c) + 1)/(a^3*tan(1/2*d*x + 1/2*c)^4) + (a^9*tan(1/2*d*x + 1/2*c)^4
- 8*a^9*tan(1/2*d*x + 1/2*c)^3 + 32*a^9*tan(1/2*d*x + 1/2*c)^2 - 104*a^9*tan(1/2*d*x + 1/2*c))/a^12)/d