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

Optimal. Leaf size=74 \[ -\frac{2 \cos (c+d x)}{a^2 d}-\frac{\cot (c+d x)}{a^2 d}+\frac{\sin (c+d x) \cos (c+d x)}{2 a^2 d}+\frac{2 \tanh ^{-1}(\cos (c+d x))}{a^2 d}-\frac{x}{2 a^2} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.203037, antiderivative size = 74, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.241, Rules used = {2875, 2709, 3770, 3767, 8, 2638, 2635} \[ -\frac{2 \cos (c+d x)}{a^2 d}-\frac{\cot (c+d x)}{a^2 d}+\frac{\sin (c+d x) \cos (c+d x)}{2 a^2 d}+\frac{2 \tanh ^{-1}(\cos (c+d x))}{a^2 d}-\frac{x}{2 a^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-x/(2*a^2) + (2*ArcTanh[Cos[c + d*x]])/(a^2*d) - (2*Cos[c + d*x])/(a^2*d) - Cot[c + d*x]/(a^2*d) + (Cos[c + d*
x]*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 2709

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*tan[(e_.) + (f_.)*(x_)]^(p_), x_Symbol] :> Dist[a^p, Int[Expan
dIntegrand[(Sin[e + f*x]^p*(a + b*Sin[e + f*x])^(m - p/2))/(a - b*Sin[e + f*x])^(p/2), x], x], x] /; FreeQ[{a,
b, e, f}, x] && EqQ[a^2 - b^2, 0] && IntegersQ[m, p/2] && (LtQ[p, 0] || GtQ[m - p/2, 0])

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]

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]

Rubi steps

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

Mathematica [A]  time = 0.523539, size = 116, normalized size = 1.57 \[ \frac{\left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^4 \left (-2 (c+d x)+\sin (2 (c+d x))-8 \cos (c+d x)+2 \tan \left (\frac{1}{2} (c+d x)\right )-2 \cot \left (\frac{1}{2} (c+d x)\right )-8 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )+8 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )}{4 d (a \sin (c+d x)+a)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^4*(-2*(c + d*x) - 8*Cos[c + d*x] - 2*Cot[(c + d*x)/2] + 8*Log[Cos[(c +
d*x)/2]] - 8*Log[Sin[(c + d*x)/2]] + Sin[2*(c + d*x)] + 2*Tan[(c + d*x)/2]))/(4*d*(a + a*Sin[c + d*x])^2)

________________________________________________________________________________________

Maple [B]  time = 0.152, size = 196, normalized size = 2.7 \begin{align*}{\frac{1}{2\,d{a}^{2}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }-{\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 ) ^{-2}}-4\,{\frac{ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}{d{a}^{2} \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{2}}}+{\frac{1}{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 ) ^{-2}}-4\,{\frac{1}{d{a}^{2} \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{2}}}-{\frac{1}{d{a}^{2}}\arctan \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) }-{\frac{1}{2\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1}}-2\,{\frac{\ln \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) }{d{a}^{2}}} \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))^2,x)

[Out]

1/2/d/a^2*tan(1/2*d*x+1/2*c)-1/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^2*tan(1/2*d*x+1/2*c)^3-4/d/a^2/(1+tan(1/2*d*x+1/
2*c)^2)^2*tan(1/2*d*x+1/2*c)^2+1/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^2*tan(1/2*d*x+1/2*c)-4/d/a^2/(1+tan(1/2*d*x+1/
2*c)^2)^2-1/d/a^2*arctan(tan(1/2*d*x+1/2*c))-1/2/d/a^2/tan(1/2*d*x+1/2*c)-2/d/a^2*ln(tan(1/2*d*x+1/2*c))

________________________________________________________________________________________

Maxima [B]  time = 1.50557, size = 273, normalized size = 3.69 \begin{align*} -\frac{\frac{\frac{8 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{8 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac{3 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + 1}{\frac{a^{2} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{2 \, a^{2} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac{a^{2} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}} + \frac{2 \, \arctan \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}} + \frac{4 \, \log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}} - \frac{\sin \left (d x + c\right )}{a^{2}{\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))^2,x, algorithm="maxima")

[Out]

-1/2*((8*sin(d*x + c)/(cos(d*x + c) + 1) + 8*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 3*sin(d*x + c)^4/(cos(d*x +
c) + 1)^4 + 1)/(a^2*sin(d*x + c)/(cos(d*x + c) + 1) + 2*a^2*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + a^2*sin(d*x
+ c)^5/(cos(d*x + c) + 1)^5) + 2*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a^2 + 4*log(sin(d*x + c)/(cos(d*x +
c) + 1))/a^2 - sin(d*x + c)/(a^2*(cos(d*x + c) + 1)))/d

________________________________________________________________________________________

Fricas [A]  time = 1.20094, size = 254, normalized size = 3.43 \begin{align*} -\frac{\cos \left (d x + c\right )^{3} +{\left (d x + 4 \, \cos \left (d x + c\right )\right )} \sin \left (d x + c\right ) - 2 \, \log \left (\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) \sin \left (d x + c\right ) + 2 \, \log \left (-\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) \sin \left (d x + c\right ) + \cos \left (d x + c\right )}{2 \, a^{2} 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))^2,x, algorithm="fricas")

[Out]

-1/2*(cos(d*x + c)^3 + (d*x + 4*cos(d*x + c))*sin(d*x + c) - 2*log(1/2*cos(d*x + c) + 1/2)*sin(d*x + c) + 2*lo
g(-1/2*cos(d*x + c) + 1/2)*sin(d*x + c) + cos(d*x + c))/(a^2*d*sin(d*x + c))

________________________________________________________________________________________

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))**2,x)

[Out]

Timed out

________________________________________________________________________________________

Giac [A]  time = 1.33712, size = 177, normalized size = 2.39 \begin{align*} -\frac{\frac{d x + c}{a^{2}} + \frac{4 \, \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right )}{a^{2}} - \frac{\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{a^{2}} - \frac{4 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1}{a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )} + \frac{2 \,{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 4 \, \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 ) + 4\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1\right )}^{2} a^{2}}}{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))^2,x, algorithm="giac")

[Out]

-1/2*((d*x + c)/a^2 + 4*log(abs(tan(1/2*d*x + 1/2*c)))/a^2 - tan(1/2*d*x + 1/2*c)/a^2 - (4*tan(1/2*d*x + 1/2*c
) - 1)/(a^2*tan(1/2*d*x + 1/2*c)) + 2*(tan(1/2*d*x + 1/2*c)^3 + 4*tan(1/2*d*x + 1/2*c)^2 - tan(1/2*d*x + 1/2*c
) + 4)/((tan(1/2*d*x + 1/2*c)^2 + 1)^2*a^2))/d