3.24 \(\int \cot ^4(c+d x) (a+a \sin (c+d x))^2 \, dx\)

Optimal. Leaf size=98 \[ -\frac{2 a^2 \cos (c+d x)}{d}-\frac{a^2 \cot ^3(c+d x)}{3 d}-\frac{a^2 \sin (c+d x) \cos (c+d x)}{2 d}+\frac{3 a^2 \tanh ^{-1}(\cos (c+d x))}{d}-\frac{a^2 \cot (c+d x) \csc (c+d x)}{d}-\frac{a^2 x}{2} \]

[Out]

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

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Rubi [A]  time = 0.161405, antiderivative size = 98, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 7, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {2709, 3770, 3767, 8, 3768, 2638, 2635} \[ -\frac{2 a^2 \cos (c+d x)}{d}-\frac{a^2 \cot ^3(c+d x)}{3 d}-\frac{a^2 \sin (c+d x) \cos (c+d x)}{2 d}+\frac{3 a^2 \tanh ^{-1}(\cos (c+d x))}{d}-\frac{a^2 \cot (c+d x) \csc (c+d x)}{d}-\frac{a^2 x}{2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 5.58143, size = 191, normalized size = 1.95 \[ \frac{a^2 (\sin (c+d x)+1)^2 \left (-12 (c+d x)-6 \sin (2 (c+d x))-48 \cos (c+d x)-4 \tan \left (\frac{1}{2} (c+d x)\right )+4 \cot \left (\frac{1}{2} (c+d x)\right )-6 \csc ^2\left (\frac{1}{2} (c+d x)\right )+6 \sec ^2\left (\frac{1}{2} (c+d x)\right )-72 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )+72 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )+8 \sin ^4\left (\frac{1}{2} (c+d x)\right ) \csc ^3(c+d x)-\frac{1}{2} \sin (c+d x) \csc ^4\left (\frac{1}{2} (c+d x)\right )\right )}{24 d \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(1 + Sin[c + d*x])^2*(-12*(c + d*x) - 48*Cos[c + d*x] + 4*Cot[(c + d*x)/2] - 6*Csc[(c + d*x)/2]^2 + 72*Lo
g[Cos[(c + d*x)/2]] - 72*Log[Sin[(c + d*x)/2]] + 6*Sec[(c + d*x)/2]^2 + 8*Csc[c + d*x]^3*Sin[(c + d*x)/2]^4 -
(Csc[(c + d*x)/2]^4*Sin[c + d*x])/2 - 6*Sin[2*(c + d*x)] - 4*Tan[(c + d*x)/2]))/(24*d*(Cos[(c + d*x)/2] + Sin[
(c + d*x)/2])^4)

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Maple [B]  time = 0.044, size = 190, normalized size = 1.9 \begin{align*} -{\frac{{a}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{5}}{d\sin \left ( dx+c \right ) }}-{\frac{{a}^{2}\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{d}}-{\frac{3\,\cos \left ( dx+c \right ){a}^{2}\sin \left ( dx+c \right ) }{2\,d}}-{\frac{{a}^{2}x}{2}}-{\frac{{a}^{2}c}{2\,d}}-{\frac{{a}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{5}}{d \left ( \sin \left ( dx+c \right ) \right ) ^{2}}}-{\frac{{a}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{d}}-3\,{\frac{\cos \left ( dx+c \right ){a}^{2}}{d}}-3\,{\frac{{a}^{2}\ln \left ( \csc \left ( dx+c \right ) -\cot \left ( dx+c \right ) \right ) }{d}}-{\frac{{a}^{2} \left ( \cot \left ( dx+c \right ) \right ) ^{3}}{3\,d}}+{\frac{{a}^{2}\cot \left ( dx+c \right ) }{d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(d*x+c)^4*(a+a*sin(d*x+c))^2,x)

[Out]

-1/d*a^2/sin(d*x+c)*cos(d*x+c)^5-1/d*a^2*sin(d*x+c)*cos(d*x+c)^3-3/2*a^2*cos(d*x+c)*sin(d*x+c)/d-1/2*a^2*x-1/2
/d*a^2*c-1/d*a^2/sin(d*x+c)^2*cos(d*x+c)^5-1/d*a^2*cos(d*x+c)^3-3*a^2*cos(d*x+c)/d-3/d*a^2*ln(csc(d*x+c)-cot(d
*x+c))-1/3*a^2*cot(d*x+c)^3/d+a^2*cot(d*x+c)/d

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Maxima [A]  time = 1.64741, size = 188, normalized size = 1.92 \begin{align*} -\frac{3 \,{\left (3 \, d x + 3 \, c + \frac{3 \, \tan \left (d x + c\right )^{2} + 2}{\tan \left (d x + c\right )^{3} + \tan \left (d x + c\right )}\right )} a^{2} - 2 \,{\left (3 \, d x + 3 \, c + \frac{3 \, \tan \left (d x + c\right )^{2} - 1}{\tan \left (d x + c\right )^{3}}\right )} a^{2} - 3 \, a^{2}{\left (\frac{2 \, \cos \left (d x + c\right )}{\cos \left (d x + c\right )^{2} - 1} - 4 \, \cos \left (d x + c\right ) + 3 \, \log \left (\cos \left (d x + c\right ) + 1\right ) - 3 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )}}{6 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^4*(a+a*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 1.70258, size = 474, normalized size = 4.84 \begin{align*} \frac{3 \, a^{2} \cos \left (d x + c\right )^{5} - 4 \, a^{2} \cos \left (d x + c\right )^{3} + 3 \, a^{2} \cos \left (d x + c\right ) + 9 \,{\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 ) \sin \left (d x + c\right ) - 9 \,{\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 ) \sin \left (d x + c\right ) - 3 \,{\left (a^{2} d x \cos \left (d x + c\right )^{2} + 4 \, a^{2} \cos \left (d x + c\right )^{3} - a^{2} d x - 6 \, a^{2} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{6 \,{\left (d \cos \left (d x + c\right )^{2} - d\right )} \sin \left (d x + c\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^4*(a+a*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

1/6*(3*a^2*cos(d*x + c)^5 - 4*a^2*cos(d*x + c)^3 + 3*a^2*cos(d*x + c) + 9*(a^2*cos(d*x + c)^2 - a^2)*log(1/2*c
os(d*x + c) + 1/2)*sin(d*x + c) - 9*(a^2*cos(d*x + c)^2 - a^2)*log(-1/2*cos(d*x + c) + 1/2)*sin(d*x + c) - 3*(
a^2*d*x*cos(d*x + c)^2 + 4*a^2*cos(d*x + c)^3 - a^2*d*x - 6*a^2*cos(d*x + c))*sin(d*x + c))/((d*cos(d*x + c)^2
- d)*sin(d*x + c))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} a^{2} \left (\int 2 \sin{\left (c + d x \right )} \cot ^{4}{\left (c + d x \right )}\, dx + \int \sin ^{2}{\left (c + d x \right )} \cot ^{4}{\left (c + d x \right )}\, dx + \int \cot ^{4}{\left (c + d x \right )}\, dx\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)**4*(a+a*sin(d*x+c))**2,x)

[Out]

a**2*(Integral(2*sin(c + d*x)*cot(c + d*x)**4, x) + Integral(sin(c + d*x)**2*cot(c + d*x)**4, x) + Integral(co
t(c + d*x)**4, x))

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Giac [B]  time = 1.76195, size = 282, normalized size = 2.88 \begin{align*} \frac{a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 6 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 12 \,{\left (d x + c\right )} a^{2} - 72 \, a^{2} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right ) - 3 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + \frac{24 \,{\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}} + \frac{132 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 3 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 6 \, 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 )^{3}}}{24 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^4*(a+a*sin(d*x+c))^2,x, algorithm="giac")

[Out]

1/24*(a^2*tan(1/2*d*x + 1/2*c)^3 + 6*a^2*tan(1/2*d*x + 1/2*c)^2 - 12*(d*x + c)*a^2 - 72*a^2*log(abs(tan(1/2*d*
x + 1/2*c))) - 3*a^2*tan(1/2*d*x + 1/2*c) + 24*(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 + (132*a^2*tan(1/2*d*x + 1/2*c)^3 + 3*a^2*tan(1
/2*d*x + 1/2*c)^2 - 6*a^2*tan(1/2*d*x + 1/2*c) - a^2)/tan(1/2*d*x + 1/2*c)^3)/d