3.12 \(\int \cot ^4(c+d x) (a+a \sin (c+d x)) \, dx\)

Optimal. Leaf size=82 \[ -\frac{3 a \cos (c+d x)}{2 d}-\frac{a \cot ^3(c+d x)}{3 d}+\frac{a \cot (c+d x)}{d}-\frac{a \cos (c+d x) \cot ^2(c+d x)}{2 d}+\frac{3 a \tanh ^{-1}(\cos (c+d x))}{2 d}+a x \]

[Out]

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

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Rubi [A]  time = 0.0797018, antiderivative size = 82, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.368, Rules used = {2710, 2592, 288, 321, 206, 3473, 8} \[ -\frac{3 a \cos (c+d x)}{2 d}-\frac{a \cot ^3(c+d x)}{3 d}+\frac{a \cot (c+d x)}{d}-\frac{a \cos (c+d x) \cot ^2(c+d x)}{2 d}+\frac{3 a \tanh ^{-1}(\cos (c+d x))}{2 d}+a x \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2710

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((g_.)*tan[(e_.) + (f_.)*(x_)])^(p_.), x_Symbol] :> Int[Expan
dIntegrand[(g*Tan[e + f*x])^p, (a + b*Sin[e + f*x])^m, x], x] /; FreeQ[{a, b, e, f, g, p}, x] && EqQ[a^2 - b^2
, 0] && IGtQ[m, 0]

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 288

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*n*(p + 1)), x] - Dist[(c^n*(m - n + 1))/(b*n*(p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

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 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]

Rubi steps

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

Mathematica [C]  time = 0.0531241, size = 125, normalized size = 1.52 \[ -\frac{a \cot ^3(c+d x) \, _2F_1\left (-\frac{3}{2},1;-\frac{1}{2};-\tan ^2(c+d x)\right )}{3 d}-\frac{a \cos (c+d x)}{d}-\frac{a \csc ^2\left (\frac{1}{2} (c+d x)\right )}{8 d}+\frac{a \sec ^2\left (\frac{1}{2} (c+d x)\right )}{8 d}-\frac{3 a \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )}{2 d}+\frac{3 a \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a*Cos[c + d*x])/d) - (a*Csc[(c + d*x)/2]^2)/(8*d) - (a*Cot[c + d*x]^3*Hypergeometric2F1[-3/2, 1, -1/2, -Tan
[c + d*x]^2])/(3*d) + (3*a*Log[Cos[(c + d*x)/2]])/(2*d) - (3*a*Log[Sin[(c + d*x)/2]])/(2*d) + (a*Sec[(c + d*x)
/2]^2)/(8*d)

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Maple [A]  time = 0.034, size = 106, normalized size = 1.3 \begin{align*} -{\frac{a \left ( \cos \left ( dx+c \right ) \right ) ^{5}}{2\,d \left ( \sin \left ( dx+c \right ) \right ) ^{2}}}-{\frac{a \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{2\,d}}-{\frac{3\,\cos \left ( dx+c \right ) a}{2\,d}}-{\frac{3\,a\ln \left ( \csc \left ( dx+c \right ) -\cot \left ( dx+c \right ) \right ) }{2\,d}}-{\frac{a \left ( \cot \left ( dx+c \right ) \right ) ^{3}}{3\,d}}+{\frac{a\cot \left ( dx+c \right ) }{d}}+ax+{\frac{ca}{d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(d*x+c)^4*(a+a*sin(d*x+c)),x)

[Out]

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

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Maxima [A]  time = 1.69055, size = 124, normalized size = 1.51 \begin{align*} \frac{4 \,{\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{\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(cot(d*x+c)^4*(a+a*sin(d*x+c)),x, algorithm="maxima")

[Out]

1/12*(4*(3*d*x + 3*c + (3*tan(d*x + c)^2 - 1)/tan(d*x + c)^3)*a + 3*a*(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 [B]  time = 1.50448, size = 425, normalized size = 5.18 \begin{align*} \frac{16 \, a \cos \left (d x + c\right )^{3} + 9 \,{\left (a \cos \left (d x + c\right )^{2} - a\right )} \log \left (\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) \sin \left (d x + c\right ) - 9 \,{\left (a \cos \left (d x + c\right )^{2} - a\right )} \log \left (-\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) \sin \left (d x + c\right ) - 12 \, a \cos \left (d x + c\right ) + 6 \,{\left (2 \, a d x \cos \left (d x + c\right )^{2} - 2 \, a \cos \left (d x + c\right )^{3} - 2 \, a d x + 3 \, a \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 )} \sin \left (d x + c\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^4*(a+a*sin(d*x+c)),x, algorithm="fricas")

[Out]

1/12*(16*a*cos(d*x + c)^3 + 9*(a*cos(d*x + c)^2 - a)*log(1/2*cos(d*x + c) + 1/2)*sin(d*x + c) - 9*(a*cos(d*x +
c)^2 - a)*log(-1/2*cos(d*x + c) + 1/2)*sin(d*x + c) - 12*a*cos(d*x + c) + 6*(2*a*d*x*cos(d*x + c)^2 - 2*a*cos
(d*x + c)^3 - 2*a*d*x + 3*a*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]  time = 0., size = 0, normalized size = 0. \begin{align*} a \left (\int \sin{\left (c + d x \right )} \cot ^{4}{\left (c + d x \right )}\, dx + \int \cot ^{4}{\left (c + d x \right )}\, dx\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)**4*(a+a*sin(d*x+c)),x)

[Out]

a*(Integral(sin(c + d*x)*cot(c + d*x)**4, x) + Integral(cot(c + d*x)**4, x))

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Giac [A]  time = 1.27813, size = 190, normalized size = 2.32 \begin{align*} \frac{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 3 \, a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 24 \,{\left (d x + c\right )} a - 36 \, a \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right ) - 15 \, a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - \frac{48 \, a}{\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1} + \frac{66 \, a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 15 \, a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 3 \, a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - a}{\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3}}}{24 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^4*(a+a*sin(d*x+c)),x, algorithm="giac")

[Out]

1/24*(a*tan(1/2*d*x + 1/2*c)^3 + 3*a*tan(1/2*d*x + 1/2*c)^2 + 24*(d*x + c)*a - 36*a*log(abs(tan(1/2*d*x + 1/2*
c))) - 15*a*tan(1/2*d*x + 1/2*c) - 48*a/(tan(1/2*d*x + 1/2*c)^2 + 1) + (66*a*tan(1/2*d*x + 1/2*c)^3 + 15*a*tan
(1/2*d*x + 1/2*c)^2 - 3*a*tan(1/2*d*x + 1/2*c) - a)/tan(1/2*d*x + 1/2*c)^3)/d