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

Optimal. Leaf size=91 \[ \frac{a^3 \cos (c+d x)}{d}-\frac{a^3 \cot ^3(c+d x)}{3 d}-\frac{3 a^3 \cot (c+d x)}{d}+\frac{a^3 \tanh ^{-1}(\cos (c+d x))}{2 d}-\frac{3 a^3 \cot (c+d x) \csc (c+d x)}{2 d}-3 a^3 x \]

[Out]

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

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Rubi [A]  time = 0.170831, antiderivative size = 91, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 6, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.207, Rules used = {2872, 3770, 3767, 8, 3768, 2638} \[ \frac{a^3 \cos (c+d x)}{d}-\frac{a^3 \cot ^3(c+d x)}{3 d}-\frac{3 a^3 \cot (c+d x)}{d}+\frac{a^3 \tanh ^{-1}(\cos (c+d x))}{2 d}-\frac{3 a^3 \cot (c+d x) \csc (c+d x)}{2 d}-3 a^3 x \]

Antiderivative was successfully verified.

[In]

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

[Out]

-3*a^3*x + (a^3*ArcTanh[Cos[c + d*x]])/(2*d) + (a^3*Cos[c + d*x])/d - (3*a^3*Cot[c + d*x])/d - (a^3*Cot[c + d*
x]^3)/(3*d) - (3*a^3*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 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 ^2(c+d x) \csc ^2(c+d x) (a+a \sin (c+d x))^3 \, dx &=\frac{\int \left (-3 a^5-2 a^5 \csc (c+d x)+2 a^5 \csc ^2(c+d x)+3 a^5 \csc ^3(c+d x)+a^5 \csc ^4(c+d x)-a^5 \sin (c+d x)\right ) \, dx}{a^2}\\ &=-3 a^3 x+a^3 \int \csc ^4(c+d x) \, dx-a^3 \int \sin (c+d x) \, dx-\left (2 a^3\right ) \int \csc (c+d x) \, dx+\left (2 a^3\right ) \int \csc ^2(c+d x) \, dx+\left (3 a^3\right ) \int \csc ^3(c+d x) \, dx\\ &=-3 a^3 x+\frac{2 a^3 \tanh ^{-1}(\cos (c+d x))}{d}+\frac{a^3 \cos (c+d x)}{d}-\frac{3 a^3 \cot (c+d x) \csc (c+d x)}{2 d}+\frac{1}{2} \left (3 a^3\right ) \int \csc (c+d x) \, dx-\frac{a^3 \operatorname{Subst}\left (\int \left (1+x^2\right ) \, dx,x,\cot (c+d x)\right )}{d}-\frac{\left (2 a^3\right ) \operatorname{Subst}(\int 1 \, dx,x,\cot (c+d x))}{d}\\ &=-3 a^3 x+\frac{a^3 \tanh ^{-1}(\cos (c+d x))}{2 d}+\frac{a^3 \cos (c+d x)}{d}-\frac{3 a^3 \cot (c+d x)}{d}-\frac{a^3 \cot ^3(c+d x)}{3 d}-\frac{3 a^3 \cot (c+d x) \csc (c+d x)}{2 d}\\ \end{align*}

Mathematica [A]  time = 0.396047, size = 148, normalized size = 1.63 \[ \frac{a^3 \left (24 \cos (c+d x)+32 \tan \left (\frac{1}{2} (c+d x)\right )-32 \cot \left (\frac{1}{2} (c+d x)\right )-9 \csc ^2\left (\frac{1}{2} (c+d x)\right )+9 \sec ^2\left (\frac{1}{2} (c+d x)\right )-12 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )+12 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )+8 \sin ^4\left (\frac{1}{2} (c+d x)\right ) \csc ^3(c+d x)-\frac{1}{2} \sin (c+d x) \csc ^4\left (\frac{1}{2} (c+d x)\right )-72 c-72 d x\right )}{24 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*(-72*c - 72*d*x + 24*Cos[c + d*x] - 32*Cot[(c + d*x)/2] - 9*Csc[(c + d*x)/2]^2 + 12*Log[Cos[(c + d*x)/2]]
- 12*Log[Sin[(c + d*x)/2]] + 9*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 + 32*Tan[(c + d*x)/2]))/(24*d)

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Maple [A]  time = 0.079, size = 117, normalized size = 1.3 \begin{align*} -{\frac{{a}^{3}\cos \left ( dx+c \right ) }{2\,d}}-{\frac{{a}^{3}\ln \left ( \csc \left ( dx+c \right ) -\cot \left ( dx+c \right ) \right ) }{2\,d}}-3\,{a}^{3}x-3\,{\frac{{a}^{3}\cot \left ( dx+c \right ) }{d}}-3\,{\frac{{a}^{3}c}{d}}-{\frac{3\,{a}^{3} \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{2\,d \left ( \sin \left ( dx+c \right ) \right ) ^{2}}}-{\frac{{a}^{3} \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{3\,d \left ( \sin \left ( dx+c \right ) \right ) ^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^2*csc(d*x+c)^4*(a+a*sin(d*x+c))^3,x)

[Out]

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

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Maxima [A]  time = 1.75237, size = 158, normalized size = 1.74 \begin{align*} -\frac{36 \,{\left (d x + c + \frac{1}{\tan \left (d x + c\right )}\right )} a^{3} - 9 \, 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 )} + \frac{4 \, a^{3}}{\tan \left (d x + c\right )^{3}}}{12 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*csc(d*x+c)^4*(a+a*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

-1/12*(36*(d*x + c + 1/tan(d*x + c))*a^3 - 9*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)) + 4*a^3/tan(
d*x + c)^3)/d

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Fricas [B]  time = 1.78067, size = 451, normalized size = 4.96 \begin{align*} -\frac{32 \, a^{3} \cos \left (d x + c\right )^{3} - 36 \, a^{3} \cos \left (d x + c\right ) - 3 \,{\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 ) \sin \left (d x + c\right ) + 3 \,{\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 ) \sin \left (d x + c\right ) + 6 \,{\left (6 \, a^{3} d x \cos \left (d x + c\right )^{2} - 2 \, a^{3} \cos \left (d x + c\right )^{3} - 6 \, a^{3} d x - a^{3} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{12 \,{\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(cos(d*x+c)^2*csc(d*x+c)^4*(a+a*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

-1/12*(32*a^3*cos(d*x + c)^3 - 36*a^3*cos(d*x + c) - 3*(a^3*cos(d*x + c)^2 - a^3)*log(1/2*cos(d*x + c) + 1/2)*
sin(d*x + c) + 3*(a^3*cos(d*x + c)^2 - a^3)*log(-1/2*cos(d*x + c) + 1/2)*sin(d*x + c) + 6*(6*a^3*d*x*cos(d*x +
c)^2 - 2*a^3*cos(d*x + c)^3 - 6*a^3*d*x - a^3*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(-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)**4*(a+a*sin(d*x+c))**3,x)

[Out]

Timed out

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Giac [A]  time = 1.32951, size = 217, normalized size = 2.38 \begin{align*} \frac{a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 9 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 72 \,{\left (d x + c\right )} a^{3} - 12 \, a^{3} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right ) + 33 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + \frac{48 \, a^{3}}{\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1} + \frac{22 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 33 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 9 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - a^{3}}{\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(cos(d*x+c)^2*csc(d*x+c)^4*(a+a*sin(d*x+c))^3,x, algorithm="giac")

[Out]

1/24*(a^3*tan(1/2*d*x + 1/2*c)^3 + 9*a^3*tan(1/2*d*x + 1/2*c)^2 - 72*(d*x + c)*a^3 - 12*a^3*log(abs(tan(1/2*d*
x + 1/2*c))) + 33*a^3*tan(1/2*d*x + 1/2*c) + 48*a^3/(tan(1/2*d*x + 1/2*c)^2 + 1) + (22*a^3*tan(1/2*d*x + 1/2*c
)^3 - 33*a^3*tan(1/2*d*x + 1/2*c)^2 - 9*a^3*tan(1/2*d*x + 1/2*c) - a^3)/tan(1/2*d*x + 1/2*c)^3)/d