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

Optimal. Leaf size=119 \[ -\frac{a^2 \cos ^5(c+d x)}{5 d}+\frac{a^2 \cos ^3(c+d x)}{3 d}+\frac{a^2 \cos (c+d x)}{d}+\frac{a^2 \sin (c+d x) \cos ^3(c+d x)}{2 d}+\frac{3 a^2 \sin (c+d x) \cos (c+d x)}{4 d}-\frac{a^2 \tanh ^{-1}(\cos (c+d x))}{d}+\frac{3 a^2 x}{4} \]

[Out]

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

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Rubi [A]  time = 0.142898, antiderivative size = 119, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 8, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.296, Rules used = {2873, 2635, 8, 2592, 302, 206, 2565, 30} \[ -\frac{a^2 \cos ^5(c+d x)}{5 d}+\frac{a^2 \cos ^3(c+d x)}{3 d}+\frac{a^2 \cos (c+d x)}{d}+\frac{a^2 \sin (c+d x) \cos ^3(c+d x)}{2 d}+\frac{3 a^2 \sin (c+d x) \cos (c+d x)}{4 d}-\frac{a^2 \tanh ^{-1}(\cos (c+d x))}{d}+\frac{3 a^2 x}{4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rubi steps

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

Mathematica [A]  time = 0.82845, size = 96, normalized size = 0.81 \[ \frac{a^2 \left (270 \cos (c+d x)+5 \cos (3 (c+d x))-3 \cos (5 (c+d x))+15 \left (8 \sin (2 (c+d x))+\sin (4 (c+d x))+4 \left (4 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )-4 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )+3 c+3 d x\right )\right )\right )}{240 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(270*Cos[c + d*x] + 5*Cos[3*(c + d*x)] - 3*Cos[5*(c + d*x)] + 15*(4*(3*c + 3*d*x - 4*Log[Cos[(c + d*x)/2]
] + 4*Log[Sin[(c + d*x)/2]]) + 8*Sin[2*(c + d*x)] + Sin[4*(c + d*x)])))/(240*d)

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Maple [A]  time = 0.07, size = 127, normalized size = 1.1 \begin{align*} -{\frac{{a}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{5}}{5\,d}}+{\frac{{a}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{3}\sin \left ( dx+c \right ) }{2\,d}}+{\frac{3\,{a}^{2}\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{4\,d}}+{\frac{3\,{a}^{2}x}{4}}+{\frac{3\,c{a}^{2}}{4\,d}}+{\frac{{a}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{3\,d}}+{\frac{{a}^{2}\cos \left ( dx+c \right ) }{d}}+{\frac{{a}^{2}\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))^2,x)

[Out]

-1/5*a^2*cos(d*x+c)^5/d+1/2*a^2*cos(d*x+c)^3*sin(d*x+c)/d+3/4*a^2*cos(d*x+c)*sin(d*x+c)/d+3/4*a^2*x+3/4/d*c*a^
2+1/3*a^2*cos(d*x+c)^3/d+a^2*cos(d*x+c)/d+1/d*a^2*ln(csc(d*x+c)-cot(d*x+c))

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Maxima [A]  time = 1.10069, size = 132, normalized size = 1.11 \begin{align*} -\frac{48 \, a^{2} \cos \left (d x + c\right )^{5} - 40 \,{\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} - 15 \,{\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^{2}}{240 \, 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))^2,x, algorithm="maxima")

[Out]

-1/240*(48*a^2*cos(d*x + c)^5 - 40*(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 - 15*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*a^2)/d

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Fricas [A]  time = 1.52003, size = 309, normalized size = 2.6 \begin{align*} -\frac{12 \, a^{2} \cos \left (d x + c\right )^{5} - 20 \, a^{2} \cos \left (d x + c\right )^{3} - 45 \, a^{2} d x - 60 \, a^{2} \cos \left (d x + c\right ) + 30 \, a^{2} \log \left (\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) - 30 \, a^{2} \log \left (-\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) - 15 \,{\left (2 \, 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 )}{60 \, 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))^2,x, algorithm="fricas")

[Out]

-1/60*(12*a^2*cos(d*x + c)^5 - 20*a^2*cos(d*x + c)^3 - 45*a^2*d*x - 60*a^2*cos(d*x + c) + 30*a^2*log(1/2*cos(d
*x + c) + 1/2) - 30*a^2*log(-1/2*cos(d*x + c) + 1/2) - 15*(2*a^2*cos(d*x + c)^3 + 3*a^2*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))**2,x)

[Out]

Timed out

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Giac [A]  time = 1.32986, size = 244, normalized size = 2.05 \begin{align*} \frac{45 \,{\left (d x + c\right )} a^{2} + 60 \, a^{2} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right ) - \frac{2 \,{\left (75 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{9} - 60 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{8} + 30 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 360 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{6} - 320 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 30 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 280 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 75 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 68 \, a^{2}\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1\right )}^{5}}}{60 \, 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))^2,x, algorithm="giac")

[Out]

1/60*(45*(d*x + c)*a^2 + 60*a^2*log(abs(tan(1/2*d*x + 1/2*c))) - 2*(75*a^2*tan(1/2*d*x + 1/2*c)^9 - 60*a^2*tan
(1/2*d*x + 1/2*c)^8 + 30*a^2*tan(1/2*d*x + 1/2*c)^7 - 360*a^2*tan(1/2*d*x + 1/2*c)^6 - 320*a^2*tan(1/2*d*x + 1
/2*c)^4 - 30*a^2*tan(1/2*d*x + 1/2*c)^3 - 280*a^2*tan(1/2*d*x + 1/2*c)^2 - 75*a^2*tan(1/2*d*x + 1/2*c) - 68*a^
2)/(tan(1/2*d*x + 1/2*c)^2 + 1)^5)/d