3.612 \(\int \cos ^2(c+d x) \cot ^4(c+d x) (a+a \sin (c+d x))^3 \, dx\)

Optimal. Leaf size=176 \[ \frac{a^3 \cos ^5(c+d x)}{5 d}-\frac{2 a^3 \cos ^3(c+d x)}{3 d}-\frac{5 a^3 \cos (c+d x)}{d}-\frac{a^3 \cot ^3(c+d x)}{3 d}-\frac{a^3 \cot (c+d x)}{d}+\frac{3 a^3 \sin ^3(c+d x) \cos (c+d x)}{4 d}-\frac{23 a^3 \sin (c+d x) \cos (c+d x)}{8 d}+\frac{13 a^3 \tanh ^{-1}(\cos (c+d x))}{2 d}-\frac{3 a^3 \cot (c+d x) \csc (c+d x)}{2 d}-\frac{25 a^3 x}{8} \]

[Out]

(-25*a^3*x)/8 + (13*a^3*ArcTanh[Cos[c + d*x]])/(2*d) - (5*a^3*Cos[c + d*x])/d - (2*a^3*Cos[c + d*x]^3)/(3*d) +
(a^3*Cos[c + d*x]^5)/(5*d) - (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) - (23*a^3*Cos[c + d*x]*Sin[c + d*x])/(8*d) + (3*a^3*Cos[c + d*x]*Sin[c + d*x]^3)/(4*d)

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Rubi [A]  time = 0.211157, antiderivative size = 176, normalized size of antiderivative = 1., number of steps used = 15, number of rules used = 8, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.276, Rules used = {2872, 3770, 3768, 3767, 2638, 2635, 8, 2633} \[ \frac{a^3 \cos ^5(c+d x)}{5 d}-\frac{2 a^3 \cos ^3(c+d x)}{3 d}-\frac{5 a^3 \cos (c+d x)}{d}-\frac{a^3 \cot ^3(c+d x)}{3 d}-\frac{a^3 \cot (c+d x)}{d}+\frac{3 a^3 \sin ^3(c+d x) \cos (c+d x)}{4 d}-\frac{23 a^3 \sin (c+d x) \cos (c+d x)}{8 d}+\frac{13 a^3 \tanh ^{-1}(\cos (c+d x))}{2 d}-\frac{3 a^3 \cot (c+d x) \csc (c+d x)}{2 d}-\frac{25 a^3 x}{8} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^2*Cot[c + d*x]^4*(a + a*Sin[c + d*x])^3,x]

[Out]

(-25*a^3*x)/8 + (13*a^3*ArcTanh[Cos[c + d*x]])/(2*d) - (5*a^3*Cos[c + d*x])/d - (2*a^3*Cos[c + d*x]^3)/(3*d) +
(a^3*Cos[c + d*x]^5)/(5*d) - (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) - (23*a^3*Cos[c + d*x]*Sin[c + d*x])/(8*d) + (3*a^3*Cos[c + d*x]*Sin[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 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 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 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]

Rule 8

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

Rule 2633

Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[Expand[(1 - x^2)^((n - 1)/2), x], x], x
, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 0]

Rubi steps

\begin{align*} \int \cos ^2(c+d x) \cot ^4(c+d x) (a+a \sin (c+d x))^3 \, dx &=\frac{\int \left (-6 a^9-8 a^9 \csc (c+d x)+3 a^9 \csc ^3(c+d x)+a^9 \csc ^4(c+d x)+6 a^9 \sin (c+d x)+8 a^9 \sin ^2(c+d x)-3 a^9 \sin ^4(c+d x)-a^9 \sin ^5(c+d x)\right ) \, dx}{a^6}\\ &=-6 a^3 x+a^3 \int \csc ^4(c+d x) \, dx-a^3 \int \sin ^5(c+d x) \, dx+\left (3 a^3\right ) \int \csc ^3(c+d x) \, dx-\left (3 a^3\right ) \int \sin ^4(c+d x) \, dx+\left (6 a^3\right ) \int \sin (c+d x) \, dx-\left (8 a^3\right ) \int \csc (c+d x) \, dx+\left (8 a^3\right ) \int \sin ^2(c+d x) \, dx\\ &=-6 a^3 x+\frac{8 a^3 \tanh ^{-1}(\cos (c+d x))}{d}-\frac{6 a^3 \cos (c+d x)}{d}-\frac{3 a^3 \cot (c+d x) \csc (c+d x)}{2 d}-\frac{4 a^3 \cos (c+d x) \sin (c+d x)}{d}+\frac{3 a^3 \cos (c+d x) \sin ^3(c+d x)}{4 d}+\frac{1}{2} \left (3 a^3\right ) \int \csc (c+d x) \, dx-\frac{1}{4} \left (9 a^3\right ) \int \sin ^2(c+d x) \, dx+\left (4 a^3\right ) \int 1 \, dx-\frac{a^3 \operatorname{Subst}\left (\int \left (1+x^2\right ) \, dx,x,\cot (c+d x)\right )}{d}+\frac{a^3 \operatorname{Subst}\left (\int \left (1-2 x^2+x^4\right ) \, dx,x,\cos (c+d x)\right )}{d}\\ &=-2 a^3 x+\frac{13 a^3 \tanh ^{-1}(\cos (c+d x))}{2 d}-\frac{5 a^3 \cos (c+d x)}{d}-\frac{2 a^3 \cos ^3(c+d x)}{3 d}+\frac{a^3 \cos ^5(c+d x)}{5 d}-\frac{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}-\frac{23 a^3 \cos (c+d x) \sin (c+d x)}{8 d}+\frac{3 a^3 \cos (c+d x) \sin ^3(c+d x)}{4 d}-\frac{1}{8} \left (9 a^3\right ) \int 1 \, dx\\ &=-\frac{25 a^3 x}{8}+\frac{13 a^3 \tanh ^{-1}(\cos (c+d x))}{2 d}-\frac{5 a^3 \cos (c+d x)}{d}-\frac{2 a^3 \cos ^3(c+d x)}{3 d}+\frac{a^3 \cos ^5(c+d x)}{5 d}-\frac{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}-\frac{23 a^3 \cos (c+d x) \sin (c+d x)}{8 d}+\frac{3 a^3 \cos (c+d x) \sin ^3(c+d x)}{4 d}\\ \end{align*}

Mathematica [A]  time = 1.38142, size = 219, normalized size = 1.24 \[ \frac{a^3 (\sin (c+d x)+1)^3 \left (-1500 (c+d x)-600 \sin (2 (c+d x))-45 \sin (4 (c+d x))-2580 \cos (c+d x)-50 \cos (3 (c+d x))+6 \cos (5 (c+d x))+160 \tan \left (\frac{1}{2} (c+d x)\right )-160 \cot \left (\frac{1}{2} (c+d x)\right )-180 \csc ^2\left (\frac{1}{2} (c+d x)\right )+180 \sec ^2\left (\frac{1}{2} (c+d x)\right )-3120 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )+3120 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )+160 \sin ^4\left (\frac{1}{2} (c+d x)\right ) \csc ^3(c+d x)-10 \sin (c+d x) \csc ^4\left (\frac{1}{2} (c+d x)\right )\right )}{480 d \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^6} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^2*Cot[c + d*x]^4*(a + a*Sin[c + d*x])^3,x]

[Out]

(a^3*(1 + Sin[c + d*x])^3*(-1500*(c + d*x) - 2580*Cos[c + d*x] - 50*Cos[3*(c + d*x)] + 6*Cos[5*(c + d*x)] - 16
0*Cot[(c + d*x)/2] - 180*Csc[(c + d*x)/2]^2 + 3120*Log[Cos[(c + d*x)/2]] - 3120*Log[Sin[(c + d*x)/2]] + 180*Se
c[(c + d*x)/2]^2 + 160*Csc[c + d*x]^3*Sin[(c + d*x)/2]^4 - 10*Csc[(c + d*x)/2]^4*Sin[c + d*x] - 600*Sin[2*(c +
d*x)] - 45*Sin[4*(c + d*x)] + 160*Tan[(c + d*x)/2]))/(480*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^6)

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Maple [A]  time = 0.089, size = 223, normalized size = 1.3 \begin{align*} -{\frac{13\,{a}^{3} \left ( \cos \left ( dx+c \right ) \right ) ^{5}}{10\,d}}-{\frac{13\,{a}^{3} \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{6\,d}}-{\frac{13\,{a}^{3}\cos \left ( dx+c \right ) }{2\,d}}-{\frac{13\,{a}^{3}\ln \left ( \csc \left ( dx+c \right ) -\cot \left ( dx+c \right ) \right ) }{2\,d}}-{\frac{5\,{a}^{3} \left ( \cos \left ( dx+c \right ) \right ) ^{7}}{3\,d\sin \left ( dx+c \right ) }}-{\frac{5\,{a}^{3} \left ( \cos \left ( dx+c \right ) \right ) ^{5}\sin \left ( dx+c \right ) }{3\,d}}-{\frac{25\,{a}^{3} \left ( \cos \left ( dx+c \right ) \right ) ^{3}\sin \left ( dx+c \right ) }{12\,d}}-{\frac{25\,{a}^{3}\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{8\,d}}-{\frac{25\,{a}^{3}x}{8}}-{\frac{25\,{a}^{3}c}{8\,d}}-{\frac{3\,{a}^{3} \left ( \cos \left ( dx+c \right ) \right ) ^{7}}{2\,d \left ( \sin \left ( dx+c \right ) \right ) ^{2}}}-{\frac{{a}^{3} \left ( \cos \left ( dx+c \right ) \right ) ^{7}}{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)^6*csc(d*x+c)^4*(a+a*sin(d*x+c))^3,x)

[Out]

-13/10*a^3*cos(d*x+c)^5/d-13/6*a^3*cos(d*x+c)^3/d-13/2*a^3*cos(d*x+c)/d-13/2/d*a^3*ln(csc(d*x+c)-cot(d*x+c))-5
/3/d*a^3/sin(d*x+c)*cos(d*x+c)^7-5/3*a^3*cos(d*x+c)^5*sin(d*x+c)/d-25/12*a^3*cos(d*x+c)^3*sin(d*x+c)/d-25/8*a^
3*cos(d*x+c)*sin(d*x+c)/d-25/8*a^3*x-25/8/d*a^3*c-3/2/d*a^3/sin(d*x+c)^2*cos(d*x+c)^7-1/3/d*a^3/sin(d*x+c)^3*c
os(d*x+c)^7

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Maxima [A]  time = 1.68217, size = 332, normalized size = 1.89 \begin{align*} \frac{4 \,{\left (6 \, \cos \left (d x + c\right )^{5} + 10 \, \cos \left (d x + c\right )^{3} + 30 \, \cos \left (d x + c\right ) - 15 \, \log \left (\cos \left (d x + c\right ) + 1\right ) + 15 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )} a^{3} - 30 \,{\left (4 \, \cos \left (d x + c\right )^{3} - \frac{6 \, \cos \left (d x + c\right )}{\cos \left (d x + c\right )^{2} - 1} + 24 \, \cos \left (d x + c\right ) - 15 \, \log \left (\cos \left (d x + c\right ) + 1\right ) + 15 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )} a^{3} - 45 \,{\left (15 \, d x + 15 \, c + \frac{15 \, \tan \left (d x + c\right )^{4} + 25 \, \tan \left (d x + c\right )^{2} + 8}{\tan \left (d x + c\right )^{5} + 2 \, \tan \left (d x + c\right )^{3} + \tan \left (d x + c\right )}\right )} a^{3} + 20 \,{\left (15 \, d x + 15 \, c + \frac{15 \, \tan \left (d x + c\right )^{4} + 10 \, \tan \left (d x + c\right )^{2} - 2}{\tan \left (d x + c\right )^{5} + \tan \left (d x + c\right )^{3}}\right )} a^{3}}{120 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*csc(d*x+c)^4*(a+a*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

1/120*(4*(6*cos(d*x + c)^5 + 10*cos(d*x + c)^3 + 30*cos(d*x + c) - 15*log(cos(d*x + c) + 1) + 15*log(cos(d*x +
c) - 1))*a^3 - 30*(4*cos(d*x + c)^3 - 6*cos(d*x + c)/(cos(d*x + c)^2 - 1) + 24*cos(d*x + c) - 15*log(cos(d*x
+ c) + 1) + 15*log(cos(d*x + c) - 1))*a^3 - 45*(15*d*x + 15*c + (15*tan(d*x + c)^4 + 25*tan(d*x + c)^2 + 8)/(t
an(d*x + c)^5 + 2*tan(d*x + c)^3 + tan(d*x + c)))*a^3 + 20*(15*d*x + 15*c + (15*tan(d*x + c)^4 + 10*tan(d*x +
c)^2 - 2)/(tan(d*x + c)^5 + tan(d*x + c)^3))*a^3)/d

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Fricas [A]  time = 1.26621, size = 601, normalized size = 3.41 \begin{align*} \frac{90 \, a^{3} \cos \left (d x + c\right )^{7} + 75 \, a^{3} \cos \left (d x + c\right )^{5} - 500 \, a^{3} \cos \left (d x + c\right )^{3} + 375 \, a^{3} \cos \left (d x + c\right ) + 390 \,{\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 ) - 390 \,{\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 ) +{\left (24 \, a^{3} \cos \left (d x + c\right )^{7} - 104 \, a^{3} \cos \left (d x + c\right )^{5} - 375 \, a^{3} d x \cos \left (d x + c\right )^{2} - 520 \, a^{3} \cos \left (d x + c\right )^{3} + 375 \, a^{3} d x + 780 \, a^{3} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{120 \,{\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)^6*csc(d*x+c)^4*(a+a*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

1/120*(90*a^3*cos(d*x + c)^7 + 75*a^3*cos(d*x + c)^5 - 500*a^3*cos(d*x + c)^3 + 375*a^3*cos(d*x + c) + 390*(a^
3*cos(d*x + c)^2 - a^3)*log(1/2*cos(d*x + c) + 1/2)*sin(d*x + c) - 390*(a^3*cos(d*x + c)^2 - a^3)*log(-1/2*cos
(d*x + c) + 1/2)*sin(d*x + c) + (24*a^3*cos(d*x + c)^7 - 104*a^3*cos(d*x + c)^5 - 375*a^3*d*x*cos(d*x + c)^2 -
520*a^3*cos(d*x + c)^3 + 375*a^3*d*x + 780*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)**6*csc(d*x+c)**4*(a+a*sin(d*x+c))**3,x)

[Out]

Timed out

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Giac [A]  time = 1.36642, size = 394, normalized size = 2.24 \begin{align*} \frac{5 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 45 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 375 \,{\left (d x + c\right )} a^{3} - 780 \, a^{3} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right ) + 45 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + \frac{5 \,{\left (286 \, 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} - 9 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - a^{3}\right )}}{\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3}} + \frac{2 \,{\left (345 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{9} - 720 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{8} + 330 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 2880 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{6} - 3680 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 330 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 2560 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 345 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 656 \, a^{3}\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1\right )}^{5}}}{120 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*csc(d*x+c)^4*(a+a*sin(d*x+c))^3,x, algorithm="giac")

[Out]

1/120*(5*a^3*tan(1/2*d*x + 1/2*c)^3 + 45*a^3*tan(1/2*d*x + 1/2*c)^2 - 375*(d*x + c)*a^3 - 780*a^3*log(abs(tan(
1/2*d*x + 1/2*c))) + 45*a^3*tan(1/2*d*x + 1/2*c) + 5*(286*a^3*tan(1/2*d*x + 1/2*c)^3 - 9*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 + 2*(345*a^3*tan(1/2*d*x + 1/2*c)^9 - 720*a^3
*tan(1/2*d*x + 1/2*c)^8 + 330*a^3*tan(1/2*d*x + 1/2*c)^7 - 2880*a^3*tan(1/2*d*x + 1/2*c)^6 - 3680*a^3*tan(1/2*
d*x + 1/2*c)^4 - 330*a^3*tan(1/2*d*x + 1/2*c)^3 - 2560*a^3*tan(1/2*d*x + 1/2*c)^2 - 345*a^3*tan(1/2*d*x + 1/2*
c) - 656*a^3)/(tan(1/2*d*x + 1/2*c)^2 + 1)^5)/d