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

Optimal. Leaf size=73 \[ \frac{a^2 \cos (c+d x)}{d}-\frac{2 a^2 \cot (c+d x)}{d}-\frac{a^2 \tanh ^{-1}(\cos (c+d x))}{2 d}-\frac{a^2 \cot (c+d x) \csc (c+d x)}{2 d}-2 a^2 x \]

[Out]

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

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Rubi [A]  time = 0.129619, antiderivative size = 73, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.222, Rules used = {2872, 3767, 8, 3768, 3770, 2638} \[ \frac{a^2 \cos (c+d x)}{d}-\frac{2 a^2 \cot (c+d x)}{d}-\frac{a^2 \tanh ^{-1}(\cos (c+d x))}{2 d}-\frac{a^2 \cot (c+d x) \csc (c+d x)}{2 d}-2 a^2 x \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2872

Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(
m_), x_Symbol] :> Dist[1/a^p, Int[ExpandTrig[(d*sin[e + f*x])^n*(a - b*sin[e + f*x])^(p/2)*(a + b*sin[e + f*x]
)^(m + p/2), x], x], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 - b^2, 0] && IntegersQ[m, n, p/2] && ((GtQ[m,
0] && GtQ[p, 0] && LtQ[-m - p, n, -1]) || (GtQ[m, 2] && LtQ[p, 0] && GtQ[m + p/2, 0]))

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]

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

Mathematica [A]  time = 0.65086, size = 102, normalized size = 1.4 \[ \frac{a^2 \left (8 \cos (c+d x)+8 \tan \left (\frac{1}{2} (c+d x)\right )-8 \cot \left (\frac{1}{2} (c+d x)\right )-\csc ^2\left (\frac{1}{2} (c+d x)\right )+\sec ^2\left (\frac{1}{2} (c+d x)\right )+4 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )-4 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )-16 c-16 d x\right )}{8 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(-16*c - 16*d*x + 8*Cos[c + d*x] - 8*Cot[(c + d*x)/2] - Csc[(c + d*x)/2]^2 - 4*Log[Cos[(c + d*x)/2]] + 4*
Log[Sin[(c + d*x)/2]] + Sec[(c + d*x)/2]^2 + 8*Tan[(c + d*x)/2]))/(8*d)

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Maple [A]  time = 0.073, size = 93, normalized size = 1.3 \begin{align*}{\frac{{a}^{2}\cos \left ( dx+c \right ) }{2\,d}}+{\frac{{a}^{2}\ln \left ( \csc \left ( dx+c \right ) -\cot \left ( dx+c \right ) \right ) }{2\,d}}-2\,{a}^{2}x-2\,{\frac{{a}^{2}\cot \left ( dx+c \right ) }{d}}-2\,{\frac{c{a}^{2}}{d}}-{\frac{{a}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{2\,d \left ( \sin \left ( dx+c \right ) \right ) ^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*a^2*cos(d*x+c)/d+1/2/d*a^2*ln(csc(d*x+c)-cot(d*x+c))-2*a^2*x-2*a^2*cot(d*x+c)/d-2/d*c*a^2-1/2/d*a^2/sin(d*
x+c)^2*cos(d*x+c)^3

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Maxima [A]  time = 1.72393, size = 140, normalized size = 1.92 \begin{align*} -\frac{8 \,{\left (d x + c + \frac{1}{\tan \left (d x + c\right )}\right )} a^{2} - a^{2}{\left (\frac{2 \, \cos \left (d x + c\right )}{\cos \left (d x + c\right )^{2} - 1} + \log \left (\cos \left (d x + c\right ) + 1\right ) - \log \left (\cos \left (d x + c\right ) - 1\right )\right )} - 2 \, 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)^3*(a+a*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

-1/4*(8*(d*x + c + 1/tan(d*x + c))*a^2 - a^2*(2*cos(d*x + c)/(cos(d*x + c)^2 - 1) + log(cos(d*x + c) + 1) - lo
g(cos(d*x + c) - 1)) - 2*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 [B]  time = 1.73496, size = 352, normalized size = 4.82 \begin{align*} -\frac{8 \, a^{2} d x \cos \left (d x + c\right )^{2} - 4 \, a^{2} \cos \left (d x + c\right )^{3} - 8 \, a^{2} d x - 8 \, a^{2} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + 2 \, a^{2} \cos \left (d x + c\right ) +{\left (a^{2} \cos \left (d x + c\right )^{2} - a^{2}\right )} \log \left (\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) -{\left (a^{2} \cos \left (d x + c\right )^{2} - a^{2}\right )} \log \left (-\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right )}{4 \,{\left (d \cos \left (d x + c\right )^{2} - d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(8*a^2*d*x*cos(d*x + c)^2 - 4*a^2*cos(d*x + c)^3 - 8*a^2*d*x - 8*a^2*cos(d*x + c)*sin(d*x + c) + 2*a^2*co
s(d*x + c) + (a^2*cos(d*x + c)^2 - a^2)*log(1/2*cos(d*x + c) + 1/2) - (a^2*cos(d*x + c)^2 - a^2)*log(-1/2*cos(
d*x + c) + 1/2))/(d*cos(d*x + c)^2 - 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)**2*csc(d*x+c)**3*(a+a*sin(d*x+c))**2,x)

[Out]

Timed out

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Giac [A]  time = 1.45154, size = 173, normalized size = 2.37 \begin{align*} \frac{a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 16 \,{\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 ) + 8 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + \frac{16 \, a^{2}}{\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1} - \frac{6 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 8 \, 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 )^{2}}}{8 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(a^2*tan(1/2*d*x + 1/2*c)^2 - 16*(d*x + c)*a^2 + 4*a^2*log(abs(tan(1/2*d*x + 1/2*c))) + 8*a^2*tan(1/2*d*x
+ 1/2*c) + 16*a^2/(tan(1/2*d*x + 1/2*c)^2 + 1) - (6*a^2*tan(1/2*d*x + 1/2*c)^2 + 8*a^2*tan(1/2*d*x + 1/2*c) +
a^2)/tan(1/2*d*x + 1/2*c)^2)/d