3.289 \(\int \cot ^2(c+d x) \csc (c+d x) (a+a \sin (c+d x))^3 \, dx\)

Optimal. Leaf size=98 \[ \frac{3 a^3 \cos (c+d x)}{d}-\frac{3 a^3 \cot (c+d x)}{d}+\frac{a^3 \sin (c+d x) \cos (c+d x)}{2 d}-\frac{5 a^3 \tanh ^{-1}(\cos (c+d x))}{2 d}-\frac{a^3 \cot (c+d x) \csc (c+d x)}{2 d}-\frac{5 a^3 x}{2} \]

[Out]

(-5*a^3*x)/2 - (5*a^3*ArcTanh[Cos[c + d*x]])/(2*d) + (3*a^3*Cos[c + d*x])/d - (3*a^3*Cot[c + d*x])/d - (a^3*Co
t[c + d*x]*Csc[c + d*x])/(2*d) + (a^3*Cos[c + d*x]*Sin[c + d*x])/(2*d)

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Rubi [A]  time = 0.149492, antiderivative size = 98, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 7, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.259, Rules used = {2872, 3770, 3767, 8, 3768, 2638, 2635} \[ \frac{3 a^3 \cos (c+d x)}{d}-\frac{3 a^3 \cot (c+d x)}{d}+\frac{a^3 \sin (c+d x) \cos (c+d x)}{2 d}-\frac{5 a^3 \tanh ^{-1}(\cos (c+d x))}{2 d}-\frac{a^3 \cot (c+d x) \csc (c+d x)}{2 d}-\frac{5 a^3 x}{2} \]

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]^2*Csc[c + d*x]*(a + a*Sin[c + d*x])^3,x]

[Out]

(-5*a^3*x)/2 - (5*a^3*ArcTanh[Cos[c + d*x]])/(2*d) + (3*a^3*Cos[c + d*x])/d - (3*a^3*Cot[c + d*x])/d - (a^3*Co
t[c + d*x]*Csc[c + d*x])/(2*d) + (a^3*Cos[c + d*x]*Sin[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 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 ^2(c+d x) \csc (c+d x) (a+a \sin (c+d x))^3 \, dx &=\frac{\int \left (-2 a^5+2 a^5 \csc (c+d x)+3 a^5 \csc ^2(c+d x)+a^5 \csc ^3(c+d x)-3 a^5 \sin (c+d x)-a^5 \sin ^2(c+d x)\right ) \, dx}{a^2}\\ &=-2 a^3 x+a^3 \int \csc ^3(c+d x) \, dx-a^3 \int \sin ^2(c+d x) \, dx+\left (2 a^3\right ) \int \csc (c+d x) \, dx+\left (3 a^3\right ) \int \csc ^2(c+d x) \, dx-\left (3 a^3\right ) \int \sin (c+d x) \, dx\\ &=-2 a^3 x-\frac{2 a^3 \tanh ^{-1}(\cos (c+d x))}{d}+\frac{3 a^3 \cos (c+d x)}{d}-\frac{a^3 \cot (c+d x) \csc (c+d x)}{2 d}+\frac{a^3 \cos (c+d x) \sin (c+d x)}{2 d}-\frac{1}{2} a^3 \int 1 \, dx+\frac{1}{2} a^3 \int \csc (c+d x) \, dx-\frac{\left (3 a^3\right ) \operatorname{Subst}(\int 1 \, dx,x,\cot (c+d x))}{d}\\ &=-\frac{5 a^3 x}{2}-\frac{5 a^3 \tanh ^{-1}(\cos (c+d x))}{2 d}+\frac{3 a^3 \cos (c+d x)}{d}-\frac{3 a^3 \cot (c+d x)}{d}-\frac{a^3 \cot (c+d x) \csc (c+d x)}{2 d}+\frac{a^3 \cos (c+d x) \sin (c+d x)}{2 d}\\ \end{align*}

Mathematica [A]  time = 1.03916, size = 112, normalized size = 1.14 \[ \frac{a^3 \left (2 \sin (2 (c+d x))+24 \cos (c+d x)+12 \tan \left (\frac{1}{2} (c+d x)\right )-12 \cot \left (\frac{1}{2} (c+d x)\right )-\csc ^2\left (\frac{1}{2} (c+d x)\right )+\sec ^2\left (\frac{1}{2} (c+d x)\right )+20 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )-20 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )-20 c-20 d x\right )}{8 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[c + d*x]^2*Csc[c + d*x]*(a + a*Sin[c + d*x])^3,x]

[Out]

(a^3*(-20*c - 20*d*x + 24*Cos[c + d*x] - 12*Cot[(c + d*x)/2] - Csc[(c + d*x)/2]^2 - 20*Log[Cos[(c + d*x)/2]] +
20*Log[Sin[(c + d*x)/2]] + Sec[(c + d*x)/2]^2 + 2*Sin[2*(c + d*x)] + 12*Tan[(c + d*x)/2]))/(8*d)

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Maple [A]  time = 0.077, size = 113, normalized size = 1.2 \begin{align*}{\frac{{a}^{3}\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{2\,d}}-{\frac{5\,{a}^{3}x}{2}}-{\frac{5\,{a}^{3}c}{2\,d}}+{\frac{5\,{a}^{3}\cos \left ( dx+c \right ) }{2\,d}}+{\frac{5\,{a}^{3}\ln \left ( \csc \left ( dx+c \right ) -\cot \left ( dx+c \right ) \right ) }{2\,d}}-3\,{\frac{{a}^{3}\cot \left ( dx+c \right ) }{d}}-{\frac{{a}^{3} \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))^3,x)

[Out]

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

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Maxima [A]  time = 1.6246, size = 167, normalized size = 1.7 \begin{align*} \frac{{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} a^{3} - 12 \,{\left (d x + c + \frac{1}{\tan \left (d x + c\right )}\right )} a^{3} + a^{3}{\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 )} + 6 \, a^{3}{\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))^3,x, algorithm="maxima")

[Out]

1/4*((2*d*x + 2*c + sin(2*d*x + 2*c))*a^3 - 12*(d*x + c + 1/tan(d*x + c))*a^3 + a^3*(2*cos(d*x + c)/(cos(d*x +
c)^2 - 1) + log(cos(d*x + c) + 1) - log(cos(d*x + c) - 1)) + 6*a^3*(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 = 1.88082, size = 397, normalized size = 4.05 \begin{align*} -\frac{10 \, a^{3} d x \cos \left (d x + c\right )^{2} - 12 \, a^{3} \cos \left (d x + c\right )^{3} - 10 \, a^{3} d x + 10 \, a^{3} \cos \left (d x + c\right ) + 5 \,{\left (a^{3} \cos \left (d x + c\right )^{2} - a^{3}\right )} \log \left (\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) - 5 \,{\left (a^{3} \cos \left (d x + c\right )^{2} - a^{3}\right )} \log \left (-\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) - 2 \,{\left (a^{3} \cos \left (d x + c\right )^{3} + 5 \, a^{3} \cos \left (d x + c\right )\right )} \sin \left (d x + c\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))^3,x, algorithm="fricas")

[Out]

-1/4*(10*a^3*d*x*cos(d*x + c)^2 - 12*a^3*cos(d*x + c)^3 - 10*a^3*d*x + 10*a^3*cos(d*x + c) + 5*(a^3*cos(d*x +
c)^2 - a^3)*log(1/2*cos(d*x + c) + 1/2) - 5*(a^3*cos(d*x + c)^2 - a^3)*log(-1/2*cos(d*x + c) + 1/2) - 2*(a^3*c
os(d*x + c)^3 + 5*a^3*cos(d*x + c))*sin(d*x + c))/(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))**3,x)

[Out]

Timed out

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Giac [B]  time = 1.40233, size = 248, normalized size = 2.53 \begin{align*} \frac{a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 20 \,{\left (d x + c\right )} a^{3} + 20 \, a^{3} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right ) + 12 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - \frac{10 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{6} + 20 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 27 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 16 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 36 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 12 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + a^{3}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\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))^3,x, algorithm="giac")

[Out]

1/8*(a^3*tan(1/2*d*x + 1/2*c)^2 - 20*(d*x + c)*a^3 + 20*a^3*log(abs(tan(1/2*d*x + 1/2*c))) + 12*a^3*tan(1/2*d*
x + 1/2*c) - (10*a^3*tan(1/2*d*x + 1/2*c)^6 + 20*a^3*tan(1/2*d*x + 1/2*c)^5 - 27*a^3*tan(1/2*d*x + 1/2*c)^4 +
16*a^3*tan(1/2*d*x + 1/2*c)^3 - 36*a^3*tan(1/2*d*x + 1/2*c)^2 + 12*a^3*tan(1/2*d*x + 1/2*c) + a^3)/(tan(1/2*d*
x + 1/2*c)^3 + tan(1/2*d*x + 1/2*c))^2)/d