3.648 \(\int \frac{\cos ^4(c+d x) \cot ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx\)

Optimal. Leaf size=49 \[ \frac{\cos (c+d x)}{a^3 d}-\frac{\cot (c+d x)}{a^3 d}+\frac{3 \tanh ^{-1}(\cos (c+d x))}{a^3 d}+\frac{3 x}{a^3} \]

[Out]

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

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Rubi [A]  time = 0.163287, antiderivative size = 49, normalized size of antiderivative = 1., number of steps used = 7, 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, 2638} \[ \frac{\cos (c+d x)}{a^3 d}-\frac{\cot (c+d x)}{a^3 d}+\frac{3 \tanh ^{-1}(\cos (c+d x))}{a^3 d}+\frac{3 x}{a^3} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[c + d*x]^4*Cot[c + d*x]^2)/(a + a*Sin[c + d*x])^3,x]

[Out]

(3*x)/a^3 + (3*ArcTanh[Cos[c + d*x]])/(a^3*d) + Cos[c + d*x]/(a^3*d) - Cot[c + d*x]/(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 2638

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

Rubi steps

\begin{align*} \int \frac{\cos ^4(c+d x) \cot ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx &=\frac{\int \csc ^2(c+d x) (a-a \sin (c+d x))^3 \, dx}{a^6}\\ &=\frac{\int \left (3 a^3-3 a^3 \csc (c+d x)+a^3 \csc ^2(c+d x)-a^3 \sin (c+d x)\right ) \, dx}{a^6}\\ &=\frac{3 x}{a^3}+\frac{\int \csc ^2(c+d x) \, dx}{a^3}-\frac{\int \sin (c+d x) \, dx}{a^3}-\frac{3 \int \csc (c+d x) \, dx}{a^3}\\ &=\frac{3 x}{a^3}+\frac{3 \tanh ^{-1}(\cos (c+d x))}{a^3 d}+\frac{\cos (c+d x)}{a^3 d}-\frac{\operatorname{Subst}(\int 1 \, dx,x,\cot (c+d x))}{a^3 d}\\ &=\frac{3 x}{a^3}+\frac{3 \tanh ^{-1}(\cos (c+d x))}{a^3 d}+\frac{\cos (c+d x)}{a^3 d}-\frac{\cot (c+d x)}{a^3 d}\\ \end{align*}

Mathematica [B]  time = 0.482605, size = 106, normalized size = 2.16 \[ \frac{\left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^6 \left (6 (c+d x)+2 \cos (c+d x)+\tan \left (\frac{1}{2} (c+d x)\right )-\cot \left (\frac{1}{2} (c+d x)\right )-6 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )+6 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )}{2 d (a \sin (c+d x)+a)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cos[c + d*x]^4*Cot[c + d*x]^2)/(a + a*Sin[c + d*x])^3,x]

[Out]

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

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Maple [A]  time = 0.165, size = 97, normalized size = 2. \begin{align*}{\frac{1}{2\,d{a}^{3}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }+2\,{\frac{1}{d{a}^{3} \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) }}+6\,{\frac{\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) }{d{a}^{3}}}-{\frac{1}{2\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1}}-3\,{\frac{\ln \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) }{d{a}^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^6*csc(d*x+c)^2/(a+a*sin(d*x+c))^3,x)

[Out]

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

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Maxima [B]  time = 1.55993, size = 213, normalized size = 4.35 \begin{align*} \frac{\frac{\frac{4 \, \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}} - 1}{\frac{a^{3} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{a^{3} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}} + \frac{12 \, \arctan \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}} - \frac{6 \, \log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}} + \frac{\sin \left (d x + c\right )}{a^{3}{\left (\cos \left (d x + c\right ) + 1\right )}}}{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)^2/(a+a*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

1/2*((4*sin(d*x + c)/(cos(d*x + c) + 1) - sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - 1)/(a^3*sin(d*x + c)/(cos(d*x
+ c) + 1) + a^3*sin(d*x + c)^3/(cos(d*x + c) + 1)^3) + 12*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a^3 - 6*log(
sin(d*x + c)/(cos(d*x + c) + 1))/a^3 + sin(d*x + c)/(a^3*(cos(d*x + c) + 1)))/d

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Fricas [A]  time = 1.16906, size = 235, normalized size = 4.8 \begin{align*} \frac{2 \,{\left (3 \, d x + \cos \left (d x + c\right )\right )} \sin \left (d x + c\right ) + 3 \, \log \left (\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) \sin \left (d x + c\right ) - 3 \, \log \left (-\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) \sin \left (d x + c\right ) - 2 \, \cos \left (d x + c\right )}{2 \, a^{3} d \sin \left (d x + c\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*csc(d*x+c)^2/(a+a*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

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

[Out]

Timed out

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Giac [B]  time = 1.37522, size = 150, normalized size = 3.06 \begin{align*} \frac{\frac{6 \,{\left (d x + c\right )}}{a^{3}} - \frac{6 \, \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right )}{a^{3}} + \frac{\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{a^{3}} + \frac{2 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 6 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )} 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)^2/(a+a*sin(d*x+c))^3,x, algorithm="giac")

[Out]

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