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

Optimal. Leaf size=99 \[ -\frac{a^3 \cos ^3(c+d x)}{d}+\frac{a^3 \cos (c+d x)}{d}-\frac{a^3 \sin (c+d x) \cos ^3(c+d x)}{4 d}+\frac{13 a^3 \sin (c+d x) \cos (c+d x)}{8 d}-\frac{a^3 \tanh ^{-1}(\cos (c+d x))}{d}+\frac{13 a^3 x}{8} \]

[Out]

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

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Rubi [A]  time = 0.169719, antiderivative size = 99, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 9, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.36, Rules used = {2873, 2635, 8, 2592, 321, 206, 2565, 30, 2568} \[ -\frac{a^3 \cos ^3(c+d x)}{d}+\frac{a^3 \cos (c+d x)}{d}-\frac{a^3 \sin (c+d x) \cos ^3(c+d x)}{4 d}+\frac{13 a^3 \sin (c+d x) \cos (c+d x)}{8 d}-\frac{a^3 \tanh ^{-1}(\cos (c+d x))}{d}+\frac{13 a^3 x}{8} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(13*a^3*x)/8 - (a^3*ArcTanh[Cos[c + d*x]])/d + (a^3*Cos[c + d*x])/d - (a^3*Cos[c + d*x]^3)/d + (13*a^3*Cos[c +
d*x]*Sin[c + d*x])/(8*d) - (a^3*Cos[c + d*x]^3*Sin[c + d*x])/(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 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 2592

Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> With[{ff = FreeFactors[S
in[e + f*x], x]}, Dist[ff/f, Subst[Int[(ff*x)^(m + n)/(a^2 - ff^2*x^2)^((n + 1)/2), x], x, (a*Sin[e + f*x])/ff
], x]] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n + 1)/2]

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 2565

Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_.)*sin[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> -Dist[(a*f)^(-1), Subst[
Int[x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a*Cos[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2]
&&  !(IntegerQ[(m - 1)/2] && GtQ[m, 0] && LeQ[m, n])

Rule 30

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

Rule 2568

Int[(cos[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> -Simp[(a*(b*Cos[e
+ f*x])^(n + 1)*(a*Sin[e + f*x])^(m - 1))/(b*f*(m + n)), x] + Dist[(a^2*(m - 1))/(m + n), Int[(b*Cos[e + f*x])
^n*(a*Sin[e + f*x])^(m - 2), x], x] /; FreeQ[{a, b, e, f, n}, x] && GtQ[m, 1] && NeQ[m + n, 0] && IntegersQ[2*
m, 2*n]

Rubi steps

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

Mathematica [A]  time = 0.619033, size = 82, normalized size = 0.83 \[ \frac{a^3 \left (24 \sin (2 (c+d x))-\sin (4 (c+d x))+8 \cos (c+d x)-8 \cos (3 (c+d x))+32 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )-32 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )+52 c+52 d x\right )}{32 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*(52*c + 52*d*x + 8*Cos[c + d*x] - 8*Cos[3*(c + d*x)] - 32*Log[Cos[(c + d*x)/2]] + 32*Log[Sin[(c + d*x)/2]
] + 24*Sin[2*(c + d*x)] - Sin[4*(c + d*x)]))/(32*d)

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Maple [A]  time = 0.072, size = 111, normalized size = 1.1 \begin{align*} -{\frac{{a}^{3} \left ( \cos \left ( dx+c \right ) \right ) ^{3}\sin \left ( dx+c \right ) }{4\,d}}+{\frac{13\,{a}^{3}\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{8\,d}}+{\frac{13\,{a}^{3}x}{8}}+{\frac{13\,{a}^{3}c}{8\,d}}-{\frac{{a}^{3} \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{d}}+{\frac{{a}^{3}\cos \left ( dx+c \right ) }{d}}+{\frac{{a}^{3}\ln \left ( \csc \left ( dx+c \right ) -\cot \left ( dx+c \right ) \right ) }{d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.10688, size = 134, normalized size = 1.35 \begin{align*} -\frac{32 \, a^{3} \cos \left (d x + c\right )^{3} -{\left (4 \, d x + 4 \, c - \sin \left (4 \, d x + 4 \, c\right )\right )} a^{3} - 24 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} a^{3} - 16 \, 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 )}}{32 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/32*(32*a^3*cos(d*x + c)^3 - (4*d*x + 4*c - sin(4*d*x + 4*c))*a^3 - 24*(2*d*x + 2*c + sin(2*d*x + 2*c))*a^3
- 16*a^3*(2*cos(d*x + c) - log(cos(d*x + c) + 1) + log(cos(d*x + c) - 1)))/d

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Fricas [A]  time = 1.76301, size = 267, normalized size = 2.7 \begin{align*} -\frac{8 \, a^{3} \cos \left (d x + c\right )^{3} - 13 \, a^{3} d x - 8 \, a^{3} \cos \left (d x + c\right ) + 4 \, a^{3} \log \left (\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) - 4 \, a^{3} \log \left (-\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) +{\left (2 \, a^{3} \cos \left (d x + c\right )^{3} - 13 \, a^{3} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{8 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(8*a^3*cos(d*x + c)^3 - 13*a^3*d*x - 8*a^3*cos(d*x + c) + 4*a^3*log(1/2*cos(d*x + c) + 1/2) - 4*a^3*log(-
1/2*cos(d*x + c) + 1/2) + (2*a^3*cos(d*x + c)^3 - 13*a^3*cos(d*x + c))*sin(d*x + c))/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)*(a+a*sin(d*x+c))**3,x)

[Out]

Timed out

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Giac [A]  time = 1.3275, size = 194, normalized size = 1.96 \begin{align*} \frac{13 \,{\left (d x + c\right )} a^{3} + 8 \, a^{3} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right ) - \frac{2 \,{\left (11 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 16 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{6} + 19 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 19 \, 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} - 11 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1\right )}^{4}}}{8 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(13*(d*x + c)*a^3 + 8*a^3*log(abs(tan(1/2*d*x + 1/2*c))) - 2*(11*a^3*tan(1/2*d*x + 1/2*c)^7 + 16*a^3*tan(1
/2*d*x + 1/2*c)^6 + 19*a^3*tan(1/2*d*x + 1/2*c)^5 - 19*a^3*tan(1/2*d*x + 1/2*c)^3 - 16*a^3*tan(1/2*d*x + 1/2*c
)^2 - 11*a^3*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 + 1)^4)/d