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

Optimal. Leaf size=100 \[ -\frac{a^3 \cot ^3(c+d x)}{d}-\frac{a^3 \cot (c+d x)}{d}+\frac{13 a^3 \tanh ^{-1}(\cos (c+d x))}{8 d}-\frac{a^3 \cot (c+d x) \csc ^3(c+d x)}{4 d}-\frac{11 a^3 \cot (c+d x) \csc (c+d x)}{8 d}-a^3 x \]

[Out]

-(a^3*x) + (13*a^3*ArcTanh[Cos[c + d*x]])/(8*d) - (a^3*Cot[c + d*x])/d - (a^3*Cot[c + d*x]^3)/d - (11*a^3*Cot[
c + d*x]*Csc[c + d*x])/(8*d) - (a^3*Cot[c + d*x]*Csc[c + d*x]^3)/(4*d)

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Rubi [A]  time = 0.22124, antiderivative size = 100, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 8, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.276, Rules used = {2873, 3473, 8, 2611, 3770, 2607, 30, 3768} \[ -\frac{a^3 \cot ^3(c+d x)}{d}-\frac{a^3 \cot (c+d x)}{d}+\frac{13 a^3 \tanh ^{-1}(\cos (c+d x))}{8 d}-\frac{a^3 \cot (c+d x) \csc ^3(c+d x)}{4 d}-\frac{11 a^3 \cot (c+d x) \csc (c+d x)}{8 d}-a^3 x \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a^3*x) + (13*a^3*ArcTanh[Cos[c + d*x]])/(8*d) - (a^3*Cot[c + d*x])/d - (a^3*Cot[c + d*x]^3)/d - (11*a^3*Cot[
c + d*x]*Csc[c + d*x])/(8*d) - (a^3*Cot[c + d*x]*Csc[c + d*x]^3)/(4*d)

Rule 2873

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

Rule 3473

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(b*Tan[c + d*x])^(n - 1))/(d*(n - 1)), x] - Dis
t[b^2, Int[(b*Tan[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 2611

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(a*Sec[e
+ f*x])^m*(b*Tan[e + f*x])^(n - 1))/(f*(m + n - 1)), x] - Dist[(b^2*(n - 1))/(m + n - 1), Int[(a*Sec[e + f*x])
^m*(b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && GtQ[n, 1] && NeQ[m + n - 1, 0] && Integers
Q[2*m, 2*n]

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rule 2607

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[1/f, Subst[Int[(b*x)
^n*(1 + x^2)^(m/2 - 1), x], x, Tan[e + f*x]], x] /; FreeQ[{b, e, f, n}, x] && IntegerQ[m/2] &&  !(IntegerQ[(n
- 1)/2] && LtQ[0, n, m - 1])

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

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]

Rubi steps

\begin{align*} \int \cot ^2(c+d x) \csc ^3(c+d x) (a+a \sin (c+d x))^3 \, dx &=\int \left (a^3 \cot ^2(c+d x)+3 a^3 \cot ^2(c+d x) \csc (c+d x)+3 a^3 \cot ^2(c+d x) \csc ^2(c+d x)+a^3 \cot ^2(c+d x) \csc ^3(c+d x)\right ) \, dx\\ &=a^3 \int \cot ^2(c+d x) \, dx+a^3 \int \cot ^2(c+d x) \csc ^3(c+d x) \, dx+\left (3 a^3\right ) \int \cot ^2(c+d x) \csc (c+d x) \, dx+\left (3 a^3\right ) \int \cot ^2(c+d x) \csc ^2(c+d x) \, dx\\ &=-\frac{a^3 \cot (c+d x)}{d}-\frac{3 a^3 \cot (c+d x) \csc (c+d x)}{2 d}-\frac{a^3 \cot (c+d x) \csc ^3(c+d x)}{4 d}-\frac{1}{4} a^3 \int \csc ^3(c+d x) \, dx-a^3 \int 1 \, dx-\frac{1}{2} \left (3 a^3\right ) \int \csc (c+d x) \, dx+\frac{\left (3 a^3\right ) \operatorname{Subst}\left (\int x^2 \, dx,x,-\cot (c+d x)\right )}{d}\\ &=-a^3 x+\frac{3 a^3 \tanh ^{-1}(\cos (c+d x))}{2 d}-\frac{a^3 \cot (c+d x)}{d}-\frac{a^3 \cot ^3(c+d x)}{d}-\frac{11 a^3 \cot (c+d x) \csc (c+d x)}{8 d}-\frac{a^3 \cot (c+d x) \csc ^3(c+d x)}{4 d}-\frac{1}{8} a^3 \int \csc (c+d x) \, dx\\ &=-a^3 x+\frac{13 a^3 \tanh ^{-1}(\cos (c+d x))}{8 d}-\frac{a^3 \cot (c+d x)}{d}-\frac{a^3 \cot ^3(c+d x)}{d}-\frac{11 a^3 \cot (c+d x) \csc (c+d x)}{8 d}-\frac{a^3 \cot (c+d x) \csc ^3(c+d x)}{4 d}\\ \end{align*}

Mathematica [A]  time = 0.530498, size = 133, normalized size = 1.33 \[ \frac{a^3 \left (-22 \csc ^2\left (\frac{1}{2} (c+d x)\right )+\sec ^4\left (\frac{1}{2} (c+d x)\right )+22 \sec ^2\left (\frac{1}{2} (c+d x)\right )-(4 \sin (c+d x)+1) \csc ^4\left (\frac{1}{2} (c+d x)\right )-8 \left (13 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )-13 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )-8 \sin ^4\left (\frac{1}{2} (c+d x)\right ) \csc ^3(c+d x)+8 c+8 d x\right )\right )}{64 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*(-22*Csc[(c + d*x)/2]^2 + 22*Sec[(c + d*x)/2]^2 + Sec[(c + d*x)/2]^4 - 8*(8*c + 8*d*x - 13*Log[Cos[(c + d
*x)/2]] + 13*Log[Sin[(c + d*x)/2]] - 8*Csc[c + d*x]^3*Sin[(c + d*x)/2]^4) - Csc[(c + d*x)/2]^4*(1 + 4*Sin[c +
d*x])))/(64*d)

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Maple [A]  time = 0.083, size = 141, normalized size = 1.4 \begin{align*} -{a}^{3}x-{\frac{{a}^{3}\cot \left ( dx+c \right ) }{d}}-{\frac{{a}^{3}c}{d}}-{\frac{13\,{a}^{3} \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{8\,d \left ( \sin \left ( dx+c \right ) \right ) ^{2}}}-{\frac{13\,{a}^{3}\cos \left ( dx+c \right ) }{8\,d}}-{\frac{13\,{a}^{3}\ln \left ( \csc \left ( dx+c \right ) -\cot \left ( dx+c \right ) \right ) }{8\,d}}-{\frac{{a}^{3} \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{d \left ( \sin \left ( dx+c \right ) \right ) ^{3}}}-{\frac{{a}^{3} \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{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)^2*csc(d*x+c)^5*(a+a*sin(d*x+c))^3,x)

[Out]

-a^3*x-a^3*cot(d*x+c)/d-1/d*a^3*c-13/8/d*a^3/sin(d*x+c)^2*cos(d*x+c)^3-13/8*a^3*cos(d*x+c)/d-13/8/d*a^3*ln(csc
(d*x+c)-cot(d*x+c))-1/d*a^3/sin(d*x+c)^3*cos(d*x+c)^3-1/4/d*a^3/sin(d*x+c)^4*cos(d*x+c)^3

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Maxima [A]  time = 1.73384, size = 198, normalized size = 1.98 \begin{align*} -\frac{16 \,{\left (d x + c + \frac{1}{\tan \left (d x + c\right )}\right )} a^{3} + a^{3}{\left (\frac{2 \,{\left (\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )\right )}}{\cos \left (d x + c\right )^{4} - 2 \, \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 )} - 12 \, 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 )} + \frac{16 \, a^{3}}{\tan \left (d x + c\right )^{3}}}{16 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*csc(d*x+c)^5*(a+a*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

-1/16*(16*(d*x + c + 1/tan(d*x + c))*a^3 + a^3*(2*(cos(d*x + c)^3 + cos(d*x + c))/(cos(d*x + c)^4 - 2*cos(d*x
+ c)^2 + 1) - log(cos(d*x + c) + 1) + log(cos(d*x + c) - 1)) - 12*a^3*(2*cos(d*x + c)/(cos(d*x + c)^2 - 1) + l
og(cos(d*x + c) + 1) - log(cos(d*x + c) - 1)) + 16*a^3/tan(d*x + c)^3)/d

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Fricas [B]  time = 1.69957, size = 497, normalized size = 4.97 \begin{align*} -\frac{16 \, a^{3} d x \cos \left (d x + c\right )^{4} - 32 \, a^{3} d x \cos \left (d x + c\right )^{2} - 22 \, a^{3} \cos \left (d x + c\right )^{3} + 16 \, a^{3} d x + 16 \, a^{3} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + 26 \, a^{3} \cos \left (d x + c\right ) - 13 \,{\left (a^{3} \cos \left (d x + c\right )^{4} - 2 \, 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 ) + 13 \,{\left (a^{3} \cos \left (d x + c\right )^{4} - 2 \, 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 )}{16 \,{\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)^2*csc(d*x+c)^5*(a+a*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

-1/16*(16*a^3*d*x*cos(d*x + c)^4 - 32*a^3*d*x*cos(d*x + c)^2 - 22*a^3*cos(d*x + c)^3 + 16*a^3*d*x + 16*a^3*cos
(d*x + c)*sin(d*x + c) + 26*a^3*cos(d*x + c) - 13*(a^3*cos(d*x + c)^4 - 2*a^3*cos(d*x + c)^2 + a^3)*log(1/2*co
s(d*x + c) + 1/2) + 13*(a^3*cos(d*x + c)^4 - 2*a^3*cos(d*x + c)^2 + a^3)*log(-1/2*cos(d*x + c) + 1/2))/(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)**2*csc(d*x+c)**5*(a+a*sin(d*x+c))**3,x)

[Out]

Timed out

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Giac [A]  time = 1.39205, size = 235, normalized size = 2.35 \begin{align*} \frac{3 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 24 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 72 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 192 \,{\left (d x + c\right )} a^{3} - 312 \, a^{3} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right ) + 24 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + \frac{650 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 24 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 72 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 24 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 3 \, a^{3}}{\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)^2*csc(d*x+c)^5*(a+a*sin(d*x+c))^3,x, algorithm="giac")

[Out]

1/192*(3*a^3*tan(1/2*d*x + 1/2*c)^4 + 24*a^3*tan(1/2*d*x + 1/2*c)^3 + 72*a^3*tan(1/2*d*x + 1/2*c)^2 - 192*(d*x
+ c)*a^3 - 312*a^3*log(abs(tan(1/2*d*x + 1/2*c))) + 24*a^3*tan(1/2*d*x + 1/2*c) + (650*a^3*tan(1/2*d*x + 1/2*
c)^4 - 24*a^3*tan(1/2*d*x + 1/2*c)^3 - 72*a^3*tan(1/2*d*x + 1/2*c)^2 - 24*a^3*tan(1/2*d*x + 1/2*c) - 3*a^3)/ta
n(1/2*d*x + 1/2*c)^4)/d