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

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

[Out]

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

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Rubi [A]  time = 0.157049, antiderivative size = 98, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 8, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.296, Rules used = {2872, 3770, 3767, 8, 3768, 2638, 2635, 2633} \[ \frac{a^2 \cos ^3(c+d x)}{3 d}-\frac{2 a^2 \cot (c+d x)}{d}-\frac{a^2 \sin (c+d x) \cos (c+d x)}{d}+\frac{a^2 \tanh ^{-1}(\cos (c+d x))}{2 d}-\frac{a^2 \cot (c+d x) \csc (c+d x)}{2 d}-3 a^2 x \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [A]  time = 2.1272, size = 158, normalized size = 1.61 \[ \frac{a^2 (\sin (c+d x)+1)^2 \left (6 \cos (c+d x)+2 \cos (3 (c+d x))+3 \left (-4 \sin (2 (c+d x))+8 \tan \left (\frac{1}{2} (c+d x)\right )-8 \cot \left (\frac{1}{2} (c+d x)\right )-\csc ^2\left (\frac{1}{2} (c+d x)\right )+\sec ^2\left (\frac{1}{2} (c+d x)\right )-4 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )+4 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )-24 c-24 d x\right )\right )}{24 d \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(1 + Sin[c + d*x])^2*(6*Cos[c + d*x] + 2*Cos[3*(c + d*x)] + 3*(-24*c - 24*d*x - 8*Cot[(c + d*x)/2] - Csc[
(c + d*x)/2]^2 + 4*Log[Cos[(c + d*x)/2]] - 4*Log[Sin[(c + d*x)/2]] + Sec[(c + d*x)/2]^2 - 4*Sin[2*(c + d*x)] +
8*Tan[(c + d*x)/2])))/(24*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^4)

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Maple [A]  time = 0.081, size = 161, normalized size = 1.6 \begin{align*} -{\frac{{a}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{6\,d}}-{\frac{{a}^{2}\cos \left ( dx+c \right ) }{2\,d}}-{\frac{{a}^{2}\ln \left ( \csc \left ( dx+c \right ) -\cot \left ( dx+c \right ) \right ) }{2\,d}}-2\,{\frac{{a}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{5}}{d\sin \left ( dx+c \right ) }}-2\,{\frac{{a}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{3}\sin \left ( dx+c \right ) }{d}}-3\,{\frac{{a}^{2}\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{d}}-3\,{a}^{2}x-3\,{\frac{c{a}^{2}}{d}}-{\frac{{a}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{5}}{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)^4*csc(d*x+c)^3*(a+a*sin(d*x+c))^2,x)

[Out]

-1/6*a^2*cos(d*x+c)^3/d-1/2*a^2*cos(d*x+c)/d-1/2/d*a^2*ln(csc(d*x+c)-cot(d*x+c))-2/d*a^2/sin(d*x+c)*cos(d*x+c)
^5-2*a^2*cos(d*x+c)^3*sin(d*x+c)/d-3*a^2*cos(d*x+c)*sin(d*x+c)/d-3*a^2*x-3/d*c*a^2-1/2/d*a^2/sin(d*x+c)^2*cos(
d*x+c)^5

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Maxima [A]  time = 1.65571, size = 204, normalized size = 2.08 \begin{align*} \frac{2 \,{\left (2 \, \cos \left (d x + c\right )^{3} + 6 \, \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 )} a^{2} - 12 \,{\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^{2} + 3 \, a^{2}{\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 )}}{12 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(2*(2*cos(d*x + c)^3 + 6*cos(d*x + c) - 3*log(cos(d*x + c) + 1) + 3*log(cos(d*x + c) - 1))*a^2 - 12*(3*d*
x + 3*c + (3*tan(d*x + c)^2 + 2)/(tan(d*x + c)^3 + tan(d*x + c)))*a^2 + 3*a^2*(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 = 1.61137, size = 427, normalized size = 4.36 \begin{align*} \frac{4 \, a^{2} \cos \left (d x + c\right )^{5} - 36 \, a^{2} d x \cos \left (d x + c\right )^{2} - 4 \, a^{2} \cos \left (d x + c\right )^{3} + 36 \, a^{2} d x + 6 \, a^{2} \cos \left (d x + c\right ) + 3 \,{\left (a^{2} \cos \left (d x + c\right )^{2} - a^{2}\right )} \log \left (\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) - 3 \,{\left (a^{2} \cos \left (d x + c\right )^{2} - a^{2}\right )} \log \left (-\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) - 12 \,{\left (a^{2} \cos \left (d x + c\right )^{3} - 3 \, a^{2} \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 )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(4*a^2*cos(d*x + c)^5 - 36*a^2*d*x*cos(d*x + c)^2 - 4*a^2*cos(d*x + c)^3 + 36*a^2*d*x + 6*a^2*cos(d*x + c
) + 3*(a^2*cos(d*x + c)^2 - a^2)*log(1/2*cos(d*x + c) + 1/2) - 3*(a^2*cos(d*x + c)^2 - a^2)*log(-1/2*cos(d*x +
c) + 1/2) - 12*(a^2*cos(d*x + c)^3 - 3*a^2*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)**4*csc(d*x+c)**3*(a+a*sin(d*x+c))**2,x)

[Out]

Timed out

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Giac [A]  time = 1.38414, size = 240, normalized size = 2.45 \begin{align*} \frac{3 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 72 \,{\left (d x + c\right )} a^{2} - 12 \, a^{2} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right ) + 24 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + \frac{3 \,{\left (6 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 8 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - a^{2}\right )}}{\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2}} + \frac{16 \,{\left (3 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 3 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 3 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + a^{2}\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1\right )}^{3}}}{24 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(3*a^2*tan(1/2*d*x + 1/2*c)^2 - 72*(d*x + c)*a^2 - 12*a^2*log(abs(tan(1/2*d*x + 1/2*c))) + 24*a^2*tan(1/2
*d*x + 1/2*c) + 3*(6*a^2*tan(1/2*d*x + 1/2*c)^2 - 8*a^2*tan(1/2*d*x + 1/2*c) - a^2)/tan(1/2*d*x + 1/2*c)^2 + 1
6*(3*a^2*tan(1/2*d*x + 1/2*c)^5 + 3*a^2*tan(1/2*d*x + 1/2*c)^4 - 3*a^2*tan(1/2*d*x + 1/2*c) + a^2)/(tan(1/2*d*
x + 1/2*c)^2 + 1)^3)/d