3.279 \(\int \cot ^2(c+d x) (a+a \sin (c+d x))^2 \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.104235, antiderivative size = 74, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {2709, 3770, 3767, 8, 2638, 2635} \[ \frac{2 a^2 \cos (c+d x)}{d}-\frac{a^2 \cot (c+d x)}{d}+\frac{a^2 \sin (c+d x) \cos (c+d x)}{2 d}-\frac{2 a^2 \tanh ^{-1}(\cos (c+d x))}{d}-\frac{a^2 x}{2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [A]  time = 0.556689, size = 94, normalized size = 1.27 \[ -\frac{a^2 \csc \left (\frac{1}{2} (c+d x)\right ) \sec \left (\frac{1}{2} (c+d x)\right ) \left (7 \cos (c+d x)+\cos (3 (c+d x))+4 \sin (c+d x) \left (-4 \cos (c+d x)-4 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )+4 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )+c+d x\right )\right )}{16 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a^2*Csc[(c + d*x)/2]*Sec[(c + d*x)/2]*(7*Cos[c + d*x] + Cos[3*(c + d*x)] + 4*(c + d*x - 4*Cos[c + d*x] + 4*L
og[Cos[(c + d*x)/2]] - 4*Log[Sin[(c + d*x)/2]])*Sin[c + d*x]))/(16*d)

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Maple [A]  time = 0.064, size = 89, normalized size = 1.2 \begin{align*}{\frac{{a}^{2}\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{2\,d}}-{\frac{{a}^{2}x}{2}}-{\frac{c{a}^{2}}{2\,d}}+2\,{\frac{{a}^{2}\cos \left ( dx+c \right ) }{d}}+2\,{\frac{{a}^{2}\ln \left ( \csc \left ( dx+c \right ) -\cot \left ( dx+c \right ) \right ) }{d}}-{\frac{{a}^{2}\cot \left ( dx+c \right ) }{d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*a^2*cos(d*x+c)*sin(d*x+c)/d-1/2*a^2*x-1/2/d*c*a^2+2*a^2*cos(d*x+c)/d+2/d*a^2*ln(csc(d*x+c)-cot(d*x+c))-a^2
*cot(d*x+c)/d

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Maxima [A]  time = 1.59901, size = 107, normalized size = 1.45 \begin{align*} \frac{{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} a^{2} - 4 \,{\left (d x + c + \frac{1}{\tan \left (d x + c\right )}\right )} a^{2} + 4 \, a^{2}{\left (2 \, \cos \left (d x + c\right ) - \log \left (\cos \left (d x + c\right ) + 1\right ) + \log \left (\cos \left (d x + c\right ) - 1\right )\right )}}{4 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*csc(d*x+c)^2*(a+a*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.79059, size = 281, normalized size = 3.8 \begin{align*} -\frac{a^{2} \cos \left (d x + c\right )^{3} + 2 \, a^{2} \log \left (\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) \sin \left (d x + c\right ) - 2 \, a^{2} \log \left (-\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) \sin \left (d x + c\right ) + a^{2} \cos \left (d x + c\right ) +{\left (a^{2} d x - 4 \, a^{2} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{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)^2*csc(d*x+c)^2*(a+a*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

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

[Out]

Timed out

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Giac [B]  time = 1.33705, size = 193, normalized size = 2.61 \begin{align*} -\frac{{\left (d x + c\right )} a^{2} - 4 \, a^{2} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right ) - a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + \frac{4 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + a^{2}}{\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )} + \frac{2 \,{\left (a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 4 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 4 \, a^{2}\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1\right )}^{2}}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*csc(d*x+c)^2*(a+a*sin(d*x+c))^2,x, algorithm="giac")

[Out]

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