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

Optimal. Leaf size=116 \[ -\frac{a^2 \cos (c+d x)}{d}-\frac{2 a^2 \cot ^3(c+d x)}{3 d}+\frac{2 a^2 \cot (c+d x)}{d}+\frac{9 a^2 \tanh ^{-1}(\cos (c+d x))}{8 d}-\frac{a^2 \cot (c+d x) \csc ^3(c+d x)}{4 d}+\frac{a^2 \cot (c+d x) \csc (c+d x)}{8 d}+2 a^2 x \]

[Out]

2*a^2*x + (9*a^2*ArcTanh[Cos[c + d*x]])/(8*d) - (a^2*Cos[c + d*x])/d + (2*a^2*Cot[c + d*x])/d - (2*a^2*Cot[c +
d*x]^3)/(3*d) + (a^2*Cot[c + d*x]*Csc[c + d*x])/(8*d) - (a^2*Cot[c + d*x]*Csc[c + d*x]^3)/(4*d)

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Rubi [A]  time = 0.182647, antiderivative size = 116, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 6, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.222, Rules used = {2872, 3770, 3767, 8, 3768, 2638} \[ -\frac{a^2 \cos (c+d x)}{d}-\frac{2 a^2 \cot ^3(c+d x)}{3 d}+\frac{2 a^2 \cot (c+d x)}{d}+\frac{9 a^2 \tanh ^{-1}(\cos (c+d x))}{8 d}-\frac{a^2 \cot (c+d x) \csc ^3(c+d x)}{4 d}+\frac{a^2 \cot (c+d x) \csc (c+d x)}{8 d}+2 a^2 x \]

Antiderivative was successfully verified.

[In]

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

[Out]

2*a^2*x + (9*a^2*ArcTanh[Cos[c + d*x]])/(8*d) - (a^2*Cos[c + d*x])/d + (2*a^2*Cot[c + d*x])/d - (2*a^2*Cot[c +
d*x]^3)/(3*d) + (a^2*Cot[c + d*x]*Csc[c + d*x])/(8*d) - (a^2*Cot[c + d*x]*Csc[c + d*x]^3)/(4*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 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]

Rubi steps

\begin{align*} \int \cot ^4(c+d x) \csc (c+d x) (a+a \sin (c+d x))^2 \, dx &=\frac{\int \left (2 a^6-a^6 \csc (c+d x)-4 a^6 \csc ^2(c+d x)-a^6 \csc ^3(c+d x)+2 a^6 \csc ^4(c+d x)+a^6 \csc ^5(c+d x)+a^6 \sin (c+d x)\right ) \, dx}{a^4}\\ &=2 a^2 x-a^2 \int \csc (c+d x) \, dx-a^2 \int \csc ^3(c+d x) \, dx+a^2 \int \csc ^5(c+d x) \, dx+a^2 \int \sin (c+d x) \, dx+\left (2 a^2\right ) \int \csc ^4(c+d x) \, dx-\left (4 a^2\right ) \int \csc ^2(c+d x) \, dx\\ &=2 a^2 x+\frac{a^2 \tanh ^{-1}(\cos (c+d x))}{d}-\frac{a^2 \cos (c+d x)}{d}+\frac{a^2 \cot (c+d x) \csc (c+d x)}{2 d}-\frac{a^2 \cot (c+d x) \csc ^3(c+d x)}{4 d}-\frac{1}{2} a^2 \int \csc (c+d x) \, dx+\frac{1}{4} \left (3 a^2\right ) \int \csc ^3(c+d x) \, dx-\frac{\left (2 a^2\right ) \operatorname{Subst}\left (\int \left (1+x^2\right ) \, dx,x,\cot (c+d x)\right )}{d}+\frac{\left (4 a^2\right ) \operatorname{Subst}(\int 1 \, dx,x,\cot (c+d x))}{d}\\ &=2 a^2 x+\frac{3 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{2 a^2 \cot ^3(c+d x)}{3 d}+\frac{a^2 \cot (c+d x) \csc (c+d x)}{8 d}-\frac{a^2 \cot (c+d x) \csc ^3(c+d x)}{4 d}+\frac{1}{8} \left (3 a^2\right ) \int \csc (c+d x) \, dx\\ &=2 a^2 x+\frac{9 a^2 \tanh ^{-1}(\cos (c+d x))}{8 d}-\frac{a^2 \cos (c+d x)}{d}+\frac{2 a^2 \cot (c+d x)}{d}-\frac{2 a^2 \cot ^3(c+d x)}{3 d}+\frac{a^2 \cot (c+d x) \csc (c+d x)}{8 d}-\frac{a^2 \cot (c+d x) \csc ^3(c+d x)}{4 d}\\ \end{align*}

Mathematica [A]  time = 1.18024, size = 215, normalized size = 1.85 \[ -\frac{a^2 \sin (c+d x) (\sin (c+d x)+1)^2 \left (192 \cot (c+d x)+\csc ^4\left (\frac{1}{2} (c+d x)\right ) (3 \csc (c+d x)+8)-2 \csc ^2\left (\frac{1}{2} (c+d x)\right ) (3 \csc (c+d x)+64)+8 (8 \cos (c+d x)+7) \sec ^4\left (\frac{1}{2} (c+d x)\right )-48 \sin ^4\left (\frac{1}{2} (c+d x)\right ) \csc ^5(c+d x)+24 \sin ^2\left (\frac{1}{2} (c+d x)\right ) \csc ^3(c+d x)-24 \csc (c+d x) \left (16 (c+d x)-9 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )+9 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )\right )}{192 d \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a^2*(192*Cot[c + d*x] + Csc[(c + d*x)/2]^4*(8 + 3*Csc[c + d*x]) - 2*Csc[(c + d*x)/2]^2*(64 + 3*Csc[c + d*x])
- 24*Csc[c + d*x]*(16*(c + d*x) + 9*Log[Cos[(c + d*x)/2]] - 9*Log[Sin[(c + d*x)/2]]) + 8*(7 + 8*Cos[c + d*x])
*Sec[(c + d*x)/2]^4 + 24*Csc[c + d*x]^3*Sin[(c + d*x)/2]^2 - 48*Csc[c + d*x]^5*Sin[(c + d*x)/2]^4)*Sin[c + d*x
]*(1 + Sin[c + d*x])^2)/(192*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^4)

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Maple [A]  time = 0.08, size = 149, normalized size = 1.3 \begin{align*} -{\frac{3\,{a}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{5}}{8\,d \left ( \sin \left ( dx+c \right ) \right ) ^{2}}}-{\frac{3\,{a}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{8\,d}}-{\frac{9\,{a}^{2}\cos \left ( dx+c \right ) }{8\,d}}-{\frac{9\,{a}^{2}\ln \left ( \csc \left ( dx+c \right ) -\cot \left ( dx+c \right ) \right ) }{8\,d}}-{\frac{2\,{a}^{2} \left ( \cot \left ( dx+c \right ) \right ) ^{3}}{3\,d}}+2\,{\frac{{a}^{2}\cot \left ( dx+c \right ) }{d}}+2\,{a}^{2}x+2\,{\frac{c{a}^{2}}{d}}-{\frac{{a}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{5}}{4\,d \left ( \sin \left ( dx+c \right ) \right ) ^{4}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^4*csc(d*x+c)^5*(a+a*sin(d*x+c))^2,x)

[Out]

-3/8/d*a^2/sin(d*x+c)^2*cos(d*x+c)^5-3/8*a^2*cos(d*x+c)^3/d-9/8*a^2*cos(d*x+c)/d-9/8/d*a^2*ln(csc(d*x+c)-cot(d
*x+c))-2/3*a^2*cot(d*x+c)^3/d+2*a^2*cot(d*x+c)/d+2*a^2*x+2/d*c*a^2-1/4/d*a^2/sin(d*x+c)^4*cos(d*x+c)^5

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Maxima [A]  time = 1.69817, size = 225, normalized size = 1.94 \begin{align*} \frac{32 \,{\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 \,{\left (5 \, \cos \left (d x + c\right )^{3} - 3 \, \cos \left (d x + c\right )\right )}}{\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} + 1} + 3 \, \log \left (\cos \left (d x + c\right ) + 1\right ) - 3 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )} + 12 \, 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 )}}{48 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*csc(d*x+c)^5*(a+a*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

1/48*(32*(3*d*x + 3*c + (3*tan(d*x + c)^2 - 1)/tan(d*x + c)^3)*a^2 - 3*a^2*(2*(5*cos(d*x + c)^3 - 3*cos(d*x +
c))/(cos(d*x + c)^4 - 2*cos(d*x + c)^2 + 1) + 3*log(cos(d*x + c) + 1) - 3*log(cos(d*x + c) - 1)) + 12*a^2*(2*c
os(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 = 1.67941, size = 566, normalized size = 4.88 \begin{align*} \frac{96 \, a^{2} d x \cos \left (d x + c\right )^{4} - 48 \, a^{2} \cos \left (d x + c\right )^{5} - 192 \, a^{2} d x \cos \left (d x + c\right )^{2} + 90 \, a^{2} \cos \left (d x + c\right )^{3} + 96 \, a^{2} d x - 54 \, a^{2} \cos \left (d x + c\right ) + 27 \,{\left (a^{2} \cos \left (d x + c\right )^{4} - 2 \, 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 ) - 27 \,{\left (a^{2} \cos \left (d x + c\right )^{4} - 2 \, 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 ) - 32 \,{\left (4 \, a^{2} \cos \left (d x + c\right )^{3} - 3 \, a^{2} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{48 \,{\left (d \cos \left (d x + c\right )^{4} - 2 \, 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)^4*csc(d*x+c)^5*(a+a*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

1/48*(96*a^2*d*x*cos(d*x + c)^4 - 48*a^2*cos(d*x + c)^5 - 192*a^2*d*x*cos(d*x + c)^2 + 90*a^2*cos(d*x + c)^3 +
96*a^2*d*x - 54*a^2*cos(d*x + c) + 27*(a^2*cos(d*x + c)^4 - 2*a^2*cos(d*x + c)^2 + a^2)*log(1/2*cos(d*x + c)
+ 1/2) - 27*(a^2*cos(d*x + c)^4 - 2*a^2*cos(d*x + c)^2 + a^2)*log(-1/2*cos(d*x + c) + 1/2) - 32*(4*a^2*cos(d*x
+ c)^3 - 3*a^2*cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)^4 - 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)**4*csc(d*x+c)**5*(a+a*sin(d*x+c))**2,x)

[Out]

Timed out

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Giac [A]  time = 1.32732, size = 219, normalized size = 1.89 \begin{align*} \frac{3 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 16 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 384 \,{\left (d x + c\right )} a^{2} - 216 \, a^{2} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right ) - 240 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - \frac{384 \, a^{2}}{\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1} + \frac{450 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 240 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 16 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 3 \, a^{2}}{\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4}}}{192 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*csc(d*x+c)^5*(a+a*sin(d*x+c))^2,x, algorithm="giac")

[Out]

1/192*(3*a^2*tan(1/2*d*x + 1/2*c)^4 + 16*a^2*tan(1/2*d*x + 1/2*c)^3 + 384*(d*x + c)*a^2 - 216*a^2*log(abs(tan(
1/2*d*x + 1/2*c))) - 240*a^2*tan(1/2*d*x + 1/2*c) - 384*a^2/(tan(1/2*d*x + 1/2*c)^2 + 1) + (450*a^2*tan(1/2*d*
x + 1/2*c)^4 + 240*a^2*tan(1/2*d*x + 1/2*c)^3 - 16*a^2*tan(1/2*d*x + 1/2*c) - 3*a^2)/tan(1/2*d*x + 1/2*c)^4)/d