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

Optimal. Leaf size=89 \[ \frac{a \cos ^3(c+d x)}{3 d}+\frac{a \cos (c+d x)}{d}+\frac{a \sin (c+d x) \cos ^3(c+d x)}{4 d}+\frac{3 a \sin (c+d x) \cos (c+d x)}{8 d}-\frac{a \tanh ^{-1}(\cos (c+d x))}{d}+\frac{3 a x}{8} \]

[Out]

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

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Rubi [A]  time = 0.086916, antiderivative size = 89, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.24, Rules used = {2838, 2592, 302, 206, 2635, 8} \[ \frac{a \cos ^3(c+d x)}{3 d}+\frac{a \cos (c+d x)}{d}+\frac{a \sin (c+d x) \cos ^3(c+d x)}{4 d}+\frac{3 a \sin (c+d x) \cos (c+d x)}{8 d}-\frac{a \tanh ^{-1}(\cos (c+d x))}{d}+\frac{3 a x}{8} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2838

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((a_) + (b_.)*sin[(e_.) + (f_.)
*(x_)]), x_Symbol] :> Dist[a, Int[(g*Cos[e + f*x])^p*(d*Sin[e + f*x])^n, x], x] + Dist[b/d, Int[(g*Cos[e + f*x
])^p*(d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, g, n, p}, 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 302

Int[(x_)^(m_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^n, x], x] /; FreeQ[{a,
b}, x] && IGtQ[m, 0] && IGtQ[n, 0] && GtQ[m, 2*n - 1]

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 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]

Rubi steps

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

Mathematica [A]  time = 0.146551, size = 81, normalized size = 0.91 \[ \frac{a \left (120 \cos (c+d x)+8 \cos (3 (c+d x))+3 \left (8 \sin (2 (c+d x))+\sin (4 (c+d x))+32 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )-32 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )+12 c+12 d x\right )\right )}{96 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*(120*Cos[c + d*x] + 8*Cos[3*(c + d*x)] + 3*(12*c + 12*d*x - 32*Log[Cos[(c + d*x)/2]] + 32*Log[Sin[(c + d*x)
/2]] + 8*Sin[2*(c + d*x)] + Sin[4*(c + d*x)])))/(96*d)

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Maple [A]  time = 0.05, size = 97, normalized size = 1.1 \begin{align*}{\frac{a \left ( \cos \left ( dx+c \right ) \right ) ^{3}\sin \left ( dx+c \right ) }{4\,d}}+{\frac{3\,\cos \left ( dx+c \right ) a\sin \left ( dx+c \right ) }{8\,d}}+{\frac{3\,ax}{8}}+{\frac{3\,ca}{8\,d}}+{\frac{a \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{3\,d}}+{\frac{\cos \left ( dx+c \right ) a}{d}}+{\frac{a\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)^4*csc(d*x+c)*(a+a*sin(d*x+c)),x)

[Out]

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

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Maxima [A]  time = 1.14849, size = 109, normalized size = 1.22 \begin{align*} \frac{16 \,{\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 + 3 \,{\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} a}{96 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/96*(16*(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 + 3*(12*d*x
+ 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*a)/d

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Fricas [A]  time = 1.54472, size = 252, normalized size = 2.83 \begin{align*} \frac{8 \, a \cos \left (d x + c\right )^{3} + 9 \, a d x + 24 \, a \cos \left (d x + c\right ) - 12 \, a \log \left (\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) + 12 \, a \log \left (-\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) + 3 \,{\left (2 \, a \cos \left (d x + c\right )^{3} + 3 \, a \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{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)*(a+a*sin(d*x+c)),x, algorithm="fricas")

[Out]

1/24*(8*a*cos(d*x + c)^3 + 9*a*d*x + 24*a*cos(d*x + c) - 12*a*log(1/2*cos(d*x + c) + 1/2) + 12*a*log(-1/2*cos(
d*x + c) + 1/2) + 3*(2*a*cos(d*x + c)^3 + 3*a*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)**4*csc(d*x+c)*(a+a*sin(d*x+c)),x)

[Out]

Timed out

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Giac [A]  time = 1.46771, size = 196, normalized size = 2.2 \begin{align*} \frac{9 \,{\left (d x + c\right )} a + 24 \, a \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right ) - \frac{2 \,{\left (15 \, a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 48 \, a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{6} - 9 \, a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 96 \, a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 9 \, a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 80 \, a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 15 \, a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 32 \, a\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1\right )}^{4}}}{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)*(a+a*sin(d*x+c)),x, algorithm="giac")

[Out]

1/24*(9*(d*x + c)*a + 24*a*log(abs(tan(1/2*d*x + 1/2*c))) - 2*(15*a*tan(1/2*d*x + 1/2*c)^7 - 48*a*tan(1/2*d*x
+ 1/2*c)^6 - 9*a*tan(1/2*d*x + 1/2*c)^5 - 96*a*tan(1/2*d*x + 1/2*c)^4 + 9*a*tan(1/2*d*x + 1/2*c)^3 - 80*a*tan(
1/2*d*x + 1/2*c)^2 - 15*a*tan(1/2*d*x + 1/2*c) - 32*a)/(tan(1/2*d*x + 1/2*c)^2 + 1)^4)/d