3.400 \(\int \cot ^4(c+d x) \csc (c+d x) (a+a \sin (c+d x))^3 \, dx\)

Optimal. Leaf size=138 \[ -\frac{3 a^3 \cos (c+d x)}{d}-\frac{a^3 \cot ^3(c+d x)}{d}+\frac{2 a^3 \cot (c+d x)}{d}-\frac{a^3 \sin (c+d x) \cos (c+d x)}{2 d}+\frac{33 a^3 \tanh ^{-1}(\cos (c+d x))}{8 d}-\frac{a^3 \cot (c+d x) \csc ^3(c+d x)}{4 d}-\frac{7 a^3 \cot (c+d x) \csc (c+d x)}{8 d}+\frac{3 a^3 x}{2} \]

[Out]

(3*a^3*x)/2 + (33*a^3*ArcTanh[Cos[c + d*x]])/(8*d) - (3*a^3*Cos[c + d*x])/d + (2*a^3*Cot[c + d*x])/d - (a^3*Co
t[c + d*x]^3)/d - (7*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) - (a^3*Cos
[c + d*x]*Sin[c + d*x])/(2*d)

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Rubi [A]  time = 0.192219, antiderivative size = 138, normalized size of antiderivative = 1., number of steps used = 15, 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} \[ -\frac{3 a^3 \cos (c+d x)}{d}-\frac{a^3 \cot ^3(c+d x)}{d}+\frac{2 a^3 \cot (c+d x)}{d}-\frac{a^3 \sin (c+d x) \cos (c+d x)}{2 d}+\frac{33 a^3 \tanh ^{-1}(\cos (c+d x))}{8 d}-\frac{a^3 \cot (c+d x) \csc ^3(c+d x)}{4 d}-\frac{7 a^3 \cot (c+d x) \csc (c+d x)}{8 d}+\frac{3 a^3 x}{2} \]

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]^4*Csc[c + d*x]*(a + a*Sin[c + d*x])^3,x]

[Out]

(3*a^3*x)/2 + (33*a^3*ArcTanh[Cos[c + d*x]])/(8*d) - (3*a^3*Cos[c + d*x])/d + (2*a^3*Cot[c + d*x])/d - (a^3*Co
t[c + d*x]^3)/d - (7*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) - (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 ^4(c+d x) \csc (c+d x) (a+a \sin (c+d x))^3 \, dx &=\frac{\int \left (a^7-5 a^7 \csc (c+d x)-5 a^7 \csc ^2(c+d x)+a^7 \csc ^3(c+d x)+3 a^7 \csc ^4(c+d x)+a^7 \csc ^5(c+d x)+3 a^7 \sin (c+d x)+a^7 \sin ^2(c+d x)\right ) \, dx}{a^4}\\ &=a^3 x+a^3 \int \csc ^3(c+d x) \, dx+a^3 \int \csc ^5(c+d x) \, dx+a^3 \int \sin ^2(c+d x) \, dx+\left (3 a^3\right ) \int \csc ^4(c+d x) \, dx+\left (3 a^3\right ) \int \sin (c+d x) \, dx-\left (5 a^3\right ) \int \csc (c+d x) \, dx-\left (5 a^3\right ) \int \csc ^2(c+d x) \, dx\\ &=a^3 x+\frac{5 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 \cot (c+d x) \csc ^3(c+d x)}{4 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{1}{4} \left (3 a^3\right ) \int \csc ^3(c+d x) \, dx-\frac{\left (3 a^3\right ) \operatorname{Subst}\left (\int \left (1+x^2\right ) \, dx,x,\cot (c+d x)\right )}{d}+\frac{\left (5 a^3\right ) \operatorname{Subst}(\int 1 \, dx,x,\cot (c+d x))}{d}\\ &=\frac{3 a^3 x}{2}+\frac{9 a^3 \tanh ^{-1}(\cos (c+d x))}{2 d}-\frac{3 a^3 \cos (c+d x)}{d}+\frac{2 a^3 \cot (c+d x)}{d}-\frac{a^3 \cot ^3(c+d x)}{d}-\frac{7 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{a^3 \cos (c+d x) \sin (c+d x)}{2 d}+\frac{1}{8} \left (3 a^3\right ) \int \csc (c+d x) \, dx\\ &=\frac{3 a^3 x}{2}+\frac{33 a^3 \tanh ^{-1}(\cos (c+d x))}{8 d}-\frac{3 a^3 \cos (c+d x)}{d}+\frac{2 a^3 \cot (c+d x)}{d}-\frac{a^3 \cot ^3(c+d x)}{d}-\frac{7 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{a^3 \cos (c+d x) \sin (c+d x)}{2 d}\\ \end{align*}

Mathematica [A]  time = 1.20033, size = 215, normalized size = 1.56 \[ \frac{a^3 (\sin (c+d x)+1)^3 \left (96 (c+d x)-16 \sin (2 (c+d x))-192 \cos (c+d x)-96 \tan \left (\frac{1}{2} (c+d x)\right )+96 \cot \left (\frac{1}{2} (c+d x)\right )-\csc ^4\left (\frac{1}{2} (c+d x)\right )-14 \csc ^2\left (\frac{1}{2} (c+d x)\right )+\sec ^4\left (\frac{1}{2} (c+d x)\right )+14 \sec ^2\left (\frac{1}{2} (c+d x)\right )-264 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )+264 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )+64 \sin ^4\left (\frac{1}{2} (c+d x)\right ) \csc ^3(c+d x)-4 \sin (c+d x) \csc ^4\left (\frac{1}{2} (c+d x)\right )\right )}{64 d \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^6} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[c + d*x]^4*Csc[c + d*x]*(a + a*Sin[c + d*x])^3,x]

[Out]

(a^3*(1 + Sin[c + d*x])^3*(96*(c + d*x) - 192*Cos[c + d*x] + 96*Cot[(c + d*x)/2] - 14*Csc[(c + d*x)/2]^2 - Csc
[(c + d*x)/2]^4 + 264*Log[Cos[(c + d*x)/2]] - 264*Log[Sin[(c + d*x)/2]] + 14*Sec[(c + d*x)/2]^2 + Sec[(c + d*x
)/2]^4 + 64*Csc[c + d*x]^3*Sin[(c + d*x)/2]^4 - 4*Csc[(c + d*x)/2]^4*Sin[c + d*x] - 16*Sin[2*(c + d*x)] - 96*T
an[(c + d*x)/2]))/(64*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^6)

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Maple [A]  time = 0.088, size = 215, normalized size = 1.6 \begin{align*} -{\frac{{a}^{3} \left ( \cos \left ( dx+c \right ) \right ) ^{5}}{d\sin \left ( dx+c \right ) }}-{\frac{{a}^{3} \left ( \cos \left ( dx+c \right ) \right ) ^{3}\sin \left ( dx+c \right ) }{d}}-{\frac{3\,{a}^{3}\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{2\,d}}+{\frac{3\,{a}^{3}x}{2}}+{\frac{3\,{a}^{3}c}{2\,d}}-{\frac{11\,{a}^{3} \left ( \cos \left ( dx+c \right ) \right ) ^{5}}{8\,d \left ( \sin \left ( dx+c \right ) \right ) ^{2}}}-{\frac{11\,{a}^{3} \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{8\,d}}-{\frac{33\,{a}^{3}\cos \left ( dx+c \right ) }{8\,d}}-{\frac{33\,{a}^{3}\ln \left ( \csc \left ( dx+c \right ) -\cot \left ( dx+c \right ) \right ) }{8\,d}}-{\frac{{a}^{3} \left ( \cot \left ( dx+c \right ) \right ) ^{3}}{d}}+3\,{\frac{{a}^{3}\cot \left ( dx+c \right ) }{d}}-{\frac{{a}^{3} \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))^3,x)

[Out]

-1/d*a^3/sin(d*x+c)*cos(d*x+c)^5-a^3*cos(d*x+c)^3*sin(d*x+c)/d-3/2*a^3*cos(d*x+c)*sin(d*x+c)/d+3/2*a^3*x+3/2/d
*a^3*c-11/8/d*a^3/sin(d*x+c)^2*cos(d*x+c)^5-11/8*a^3*cos(d*x+c)^3/d-33/8*a^3*cos(d*x+c)/d-33/8/d*a^3*ln(csc(d*
x+c)-cot(d*x+c))-a^3*cot(d*x+c)^3/d+3*a^3*cot(d*x+c)/d-1/4/d*a^3/sin(d*x+c)^4*cos(d*x+c)^5

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Maxima [A]  time = 1.63978, size = 282, normalized size = 2.04 \begin{align*} -\frac{8 \,{\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^{3} - 16 \,{\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^{3} + a^{3}{\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^{3}{\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 )}}{16 \, 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))^3,x, algorithm="maxima")

[Out]

-1/16*(8*(3*d*x + 3*c + (3*tan(d*x + c)^2 + 2)/(tan(d*x + c)^3 + tan(d*x + c)))*a^3 - 16*(3*d*x + 3*c + (3*tan
(d*x + c)^2 - 1)/tan(d*x + c)^3)*a^3 + a^3*(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^3*(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 [A]  time = 1.26099, size = 593, normalized size = 4.3 \begin{align*} \frac{24 \, a^{3} d x \cos \left (d x + c\right )^{4} - 48 \, a^{3} \cos \left (d x + c\right )^{5} - 48 \, a^{3} d x \cos \left (d x + c\right )^{2} + 110 \, a^{3} \cos \left (d x + c\right )^{3} + 24 \, a^{3} d x - 66 \, a^{3} \cos \left (d x + c\right ) + 33 \,{\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 ) - 33 \,{\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 ) - 8 \,{\left (a^{3} \cos \left (d x + c\right )^{5} + 4 \, a^{3} \cos \left (d x + c\right )^{3} - 3 \, a^{3} \cos \left (d x + c\right )\right )} \sin \left (d x + c\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)^4*csc(d*x+c)^5*(a+a*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

1/16*(24*a^3*d*x*cos(d*x + c)^4 - 48*a^3*cos(d*x + c)^5 - 48*a^3*d*x*cos(d*x + c)^2 + 110*a^3*cos(d*x + c)^3 +
24*a^3*d*x - 66*a^3*cos(d*x + c) + 33*(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) - 33*(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) - 8*(a^3*cos(d*x +
c)^5 + 4*a^3*cos(d*x + c)^3 - 3*a^3*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))**3,x)

[Out]

Timed out

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Giac [A]  time = 1.48639, size = 325, normalized size = 2.36 \begin{align*} \frac{a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 8 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 16 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 96 \,{\left (d x + c\right )} a^{3} - 264 \, a^{3} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right ) - 88 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + \frac{64 \,{\left (a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 6 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 6 \, a^{3}\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1\right )}^{2}} + \frac{550 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 88 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 16 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 8 \, 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 )^{4}}}{64 \, 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))^3,x, algorithm="giac")

[Out]

1/64*(a^3*tan(1/2*d*x + 1/2*c)^4 + 8*a^3*tan(1/2*d*x + 1/2*c)^3 + 16*a^3*tan(1/2*d*x + 1/2*c)^2 + 96*(d*x + c)
*a^3 - 264*a^3*log(abs(tan(1/2*d*x + 1/2*c))) - 88*a^3*tan(1/2*d*x + 1/2*c) + 64*(a^3*tan(1/2*d*x + 1/2*c)^3 -
6*a^3*tan(1/2*d*x + 1/2*c)^2 - a^3*tan(1/2*d*x + 1/2*c) - 6*a^3)/(tan(1/2*d*x + 1/2*c)^2 + 1)^2 + (550*a^3*ta
n(1/2*d*x + 1/2*c)^4 + 88*a^3*tan(1/2*d*x + 1/2*c)^3 - 16*a^3*tan(1/2*d*x + 1/2*c)^2 - 8*a^3*tan(1/2*d*x + 1/2
*c) - a^3)/tan(1/2*d*x + 1/2*c)^4)/d