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

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

[Out]

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

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Rubi [A]  time = 0.175151, antiderivative size = 60, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.207, Rules used = {2869, 2757, 3770, 3767, 8, 3768} \[ \frac{3 \cot (c+d x)}{a^3 d}-\frac{7 \tanh ^{-1}(\cos (c+d x))}{2 a^3 d}-\frac{\cot (c+d x) \csc (c+d x)}{2 a^3 d}-\frac{x}{a^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rubi steps

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

Mathematica [B]  time = 0.47218, size = 126, normalized size = 2.1 \[ \frac{\left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^6 \left (-8 (c+d x)-12 \tan \left (\frac{1}{2} (c+d x)\right )+12 \cot \left (\frac{1}{2} (c+d x)\right )-\csc ^2\left (\frac{1}{2} (c+d x)\right )+\sec ^2\left (\frac{1}{2} (c+d x)\right )+28 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )-28 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )}{8 d (a \sin (c+d x)+a)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^6*(-8*(c + d*x) + 12*Cot[(c + d*x)/2] - Csc[(c + d*x)/2]^2 - 28*Log[Cos
[(c + d*x)/2]] + 28*Log[Sin[(c + d*x)/2]] + Sec[(c + d*x)/2]^2 - 12*Tan[(c + d*x)/2]))/(8*d*(a + a*Sin[c + d*x
])^3)

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Maple [A]  time = 0.174, size = 112, normalized size = 1.9 \begin{align*}{\frac{1}{8\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}}-{\frac{3}{2\,d{a}^{3}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }-2\,{\frac{\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) }{d{a}^{3}}}-{\frac{1}{8\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-2}}+{\frac{3}{2\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1}}+{\frac{7}{2\,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)^3/(a+a*sin(d*x+c))^3,x)

[Out]

1/8/d/a^3*tan(1/2*d*x+1/2*c)^2-3/2/d/a^3*tan(1/2*d*x+1/2*c)-2/d/a^3*arctan(tan(1/2*d*x+1/2*c))-1/8/d/a^3/tan(1
/2*d*x+1/2*c)^2+3/2/d/a^3/tan(1/2*d*x+1/2*c)+7/2/d/a^3*ln(tan(1/2*d*x+1/2*c))

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Maxima [B]  time = 1.50917, size = 186, normalized size = 3.1 \begin{align*} -\frac{\frac{\frac{12 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac{\sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}}{a^{3}} + \frac{16 \, \arctan \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}} - \frac{28 \, \log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}} - \frac{{\left (\frac{12 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}{a^{3} \sin \left (d x + c\right )^{2}}}{8 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*((12*sin(d*x + c)/(cos(d*x + c) + 1) - sin(d*x + c)^2/(cos(d*x + c) + 1)^2)/a^3 + 16*arctan(sin(d*x + c)/
(cos(d*x + c) + 1))/a^3 - 28*log(sin(d*x + c)/(cos(d*x + c) + 1))/a^3 - (12*sin(d*x + c)/(cos(d*x + c) + 1) -
1)*(cos(d*x + c) + 1)^2/(a^3*sin(d*x + c)^2))/d

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Fricas [A]  time = 1.10356, size = 301, normalized size = 5.02 \begin{align*} -\frac{4 \, d x \cos \left (d x + c\right )^{2} - 4 \, d x + 7 \,{\left (\cos \left (d x + c\right )^{2} - 1\right )} \log \left (\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) - 7 \,{\left (\cos \left (d x + c\right )^{2} - 1\right )} \log \left (-\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) + 12 \, \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 2 \, \cos \left (d x + c\right )}{4 \,{\left (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)^3/(a+a*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

-1/4*(4*d*x*cos(d*x + c)^2 - 4*d*x + 7*(cos(d*x + c)^2 - 1)*log(1/2*cos(d*x + c) + 1/2) - 7*(cos(d*x + c)^2 -
1)*log(-1/2*cos(d*x + c) + 1/2) + 12*cos(d*x + c)*sin(d*x + c) - 2*cos(d*x + c))/(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)**3/(a+a*sin(d*x+c))**3,x)

[Out]

Timed out

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Giac [A]  time = 1.36977, size = 146, normalized size = 2.43 \begin{align*} -\frac{\frac{8 \,{\left (d x + c\right )}}{a^{3}} - \frac{28 \, \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right )}{a^{3}} + \frac{42 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 12 \, \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 )^{2}} - \frac{a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 12 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{a^{6}}}{8 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(8*(d*x + c)/a^3 - 28*log(abs(tan(1/2*d*x + 1/2*c)))/a^3 + (42*tan(1/2*d*x + 1/2*c)^2 - 12*tan(1/2*d*x +
1/2*c) + 1)/(a^3*tan(1/2*d*x + 1/2*c)^2) - (a^3*tan(1/2*d*x + 1/2*c)^2 - 12*a^3*tan(1/2*d*x + 1/2*c))/a^6)/d