3.592 \(\int \cos ^5(c+d x) \cot (c+d x) (a+a \sin (c+d x))^2 \, dx\)

Optimal. Leaf size=161 \[ -\frac{a^2 \cos ^7(c+d x)}{7 d}+\frac{a^2 \cos ^5(c+d x)}{5 d}+\frac{a^2 \cos ^3(c+d x)}{3 d}+\frac{a^2 \cos (c+d x)}{d}+\frac{a^2 \sin (c+d x) \cos ^5(c+d x)}{3 d}+\frac{5 a^2 \sin (c+d x) \cos ^3(c+d x)}{12 d}+\frac{5 a^2 \sin (c+d x) \cos (c+d x)}{8 d}-\frac{a^2 \tanh ^{-1}(\cos (c+d x))}{d}+\frac{5 a^2 x}{8} \]

[Out]

(5*a^2*x)/8 - (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) - (a^2*Cos[c + d*x]^7)/(7*d) + (5*a^2*Cos[c + d*x]*Sin[c + d*x])/(8*d) + (5*a^2*Cos[c + d*x]^3*
Sin[c + d*x])/(12*d) + (a^2*Cos[c + d*x]^5*Sin[c + d*x])/(3*d)

________________________________________________________________________________________

Rubi [A]  time = 0.164433, antiderivative size = 161, 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 = {2873, 2635, 8, 2592, 302, 206, 2565, 30} \[ -\frac{a^2 \cos ^7(c+d x)}{7 d}+\frac{a^2 \cos ^5(c+d x)}{5 d}+\frac{a^2 \cos ^3(c+d x)}{3 d}+\frac{a^2 \cos (c+d x)}{d}+\frac{a^2 \sin (c+d x) \cos ^5(c+d x)}{3 d}+\frac{5 a^2 \sin (c+d x) \cos ^3(c+d x)}{12 d}+\frac{5 a^2 \sin (c+d x) \cos (c+d x)}{8 d}-\frac{a^2 \tanh ^{-1}(\cos (c+d x))}{d}+\frac{5 a^2 x}{8} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^5*Cot[c + d*x]*(a + a*Sin[c + d*x])^2,x]

[Out]

(5*a^2*x)/8 - (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) - (a^2*Cos[c + d*x]^7)/(7*d) + (5*a^2*Cos[c + d*x]*Sin[c + d*x])/(8*d) + (5*a^2*Cos[c + d*x]^3*
Sin[c + d*x])/(12*d) + (a^2*Cos[c + d*x]^5*Sin[c + d*x])/(3*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 ^5(c+d x) \cot (c+d x) (a+a \sin (c+d x))^2 \, dx &=\int \left (2 a^2 \cos ^6(c+d x)+a^2 \cos ^5(c+d x) \cot (c+d x)+a^2 \cos ^6(c+d x) \sin (c+d x)\right ) \, dx\\ &=a^2 \int \cos ^5(c+d x) \cot (c+d x) \, dx+a^2 \int \cos ^6(c+d x) \sin (c+d x) \, dx+\left (2 a^2\right ) \int \cos ^6(c+d x) \, dx\\ &=\frac{a^2 \cos ^5(c+d x) \sin (c+d x)}{3 d}+\frac{1}{3} \left (5 a^2\right ) \int \cos ^4(c+d x) \, dx-\frac{a^2 \operatorname{Subst}\left (\int x^6 \, dx,x,\cos (c+d x)\right )}{d}-\frac{a^2 \operatorname{Subst}\left (\int \frac{x^6}{1-x^2} \, dx,x,\cos (c+d x)\right )}{d}\\ &=-\frac{a^2 \cos ^7(c+d x)}{7 d}+\frac{5 a^2 \cos ^3(c+d x) \sin (c+d x)}{12 d}+\frac{a^2 \cos ^5(c+d x) \sin (c+d x)}{3 d}+\frac{1}{4} \left (5 a^2\right ) \int \cos ^2(c+d x) \, dx-\frac{a^2 \operatorname{Subst}\left (\int \left (-1-x^2-x^4+\frac{1}{1-x^2}\right ) \, dx,x,\cos (c+d x)\right )}{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{a^2 \cos ^7(c+d x)}{7 d}+\frac{5 a^2 \cos (c+d x) \sin (c+d x)}{8 d}+\frac{5 a^2 \cos ^3(c+d x) \sin (c+d x)}{12 d}+\frac{a^2 \cos ^5(c+d x) \sin (c+d x)}{3 d}+\frac{1}{8} \left (5 a^2\right ) \int 1 \, dx-\frac{a^2 \operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,\cos (c+d x)\right )}{d}\\ &=\frac{5 a^2 x}{8}-\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{a^2 \cos ^7(c+d x)}{7 d}+\frac{5 a^2 \cos (c+d x) \sin (c+d x)}{8 d}+\frac{5 a^2 \cos ^3(c+d x) \sin (c+d x)}{12 d}+\frac{a^2 \cos ^5(c+d x) \sin (c+d x)}{3 d}\\ \end{align*}

Mathematica [A]  time = 0.414069, size = 112, normalized size = 0.7 \[ \frac{a^2 \left (3150 \sin (2 (c+d x))+630 \sin (4 (c+d x))+70 \sin (6 (c+d x))+8715 \cos (c+d x)+665 \cos (3 (c+d x))-21 \cos (5 (c+d x))-15 \cos (7 (c+d x))+6720 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )-6720 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )+4200 c+4200 d x\right )}{6720 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^5*Cot[c + d*x]*(a + a*Sin[c + d*x])^2,x]

[Out]

(a^2*(4200*c + 4200*d*x + 8715*Cos[c + d*x] + 665*Cos[3*(c + d*x)] - 21*Cos[5*(c + d*x)] - 15*Cos[7*(c + d*x)]
- 6720*Log[Cos[(c + d*x)/2]] + 6720*Log[Sin[(c + d*x)/2]] + 3150*Sin[2*(c + d*x)] + 630*Sin[4*(c + d*x)] + 70
*Sin[6*(c + d*x)]))/(6720*d)

________________________________________________________________________________________

Maple [A]  time = 0.073, size = 165, normalized size = 1. \begin{align*} -{\frac{{a}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{7}}{7\,d}}+{\frac{{a}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{5}\sin \left ( dx+c \right ) }{3\,d}}+{\frac{5\,{a}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{3}\sin \left ( dx+c \right ) }{12\,d}}+{\frac{5\,{a}^{2}\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{8\,d}}+{\frac{5\,{a}^{2}x}{8}}+{\frac{5\,c{a}^{2}}{8\,d}}+{\frac{{a}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{5}}{5\,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)^6*csc(d*x+c)*(a+a*sin(d*x+c))^2,x)

[Out]

-1/7*a^2*cos(d*x+c)^7/d+1/3*a^2*cos(d*x+c)^5*sin(d*x+c)/d+5/12*a^2*cos(d*x+c)^3*sin(d*x+c)/d+5/8*a^2*cos(d*x+c
)*sin(d*x+c)/d+5/8*a^2*x+5/8/d*c*a^2+1/5*a^2*cos(d*x+c)^5/d+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))

________________________________________________________________________________________

Maxima [A]  time = 1.15604, size = 166, normalized size = 1.03 \begin{align*} -\frac{480 \, a^{2} \cos \left (d x + c\right )^{7} - 112 \,{\left (6 \, \cos \left (d x + c\right )^{5} + 10 \, \cos \left (d x + c\right )^{3} + 30 \, \cos \left (d x + c\right ) - 15 \, \log \left (\cos \left (d x + c\right ) + 1\right ) + 15 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )} a^{2} + 35 \,{\left (4 \, \sin \left (2 \, d x + 2 \, c\right )^{3} - 60 \, d x - 60 \, c - 9 \, \sin \left (4 \, d x + 4 \, c\right ) - 48 \, \sin \left (2 \, d x + 2 \, c\right )\right )} a^{2}}{3360 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*csc(d*x+c)*(a+a*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

-1/3360*(480*a^2*cos(d*x + c)^7 - 112*(6*cos(d*x + c)^5 + 10*cos(d*x + c)^3 + 30*cos(d*x + c) - 15*log(cos(d*x
+ c) + 1) + 15*log(cos(d*x + c) - 1))*a^2 + 35*(4*sin(2*d*x + 2*c)^3 - 60*d*x - 60*c - 9*sin(4*d*x + 4*c) - 4
8*sin(2*d*x + 2*c))*a^2)/d

________________________________________________________________________________________

Fricas [A]  time = 1.24757, size = 386, normalized size = 2.4 \begin{align*} -\frac{120 \, a^{2} \cos \left (d x + c\right )^{7} - 168 \, a^{2} \cos \left (d x + c\right )^{5} - 280 \, a^{2} \cos \left (d x + c\right )^{3} - 525 \, a^{2} d x - 840 \, a^{2} \cos \left (d x + c\right ) + 420 \, a^{2} \log \left (\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) - 420 \, a^{2} \log \left (-\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) - 35 \,{\left (8 \, a^{2} \cos \left (d x + c\right )^{5} + 10 \, a^{2} \cos \left (d x + c\right )^{3} + 15 \, a^{2} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{840 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*csc(d*x+c)*(a+a*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

-1/840*(120*a^2*cos(d*x + c)^7 - 168*a^2*cos(d*x + c)^5 - 280*a^2*cos(d*x + c)^3 - 525*a^2*d*x - 840*a^2*cos(d
*x + c) + 420*a^2*log(1/2*cos(d*x + c) + 1/2) - 420*a^2*log(-1/2*cos(d*x + c) + 1/2) - 35*(8*a^2*cos(d*x + c)^
5 + 10*a^2*cos(d*x + c)^3 + 15*a^2*cos(d*x + c))*sin(d*x + c))/d

________________________________________________________________________________________

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)**6*csc(d*x+c)*(a+a*sin(d*x+c))**2,x)

[Out]

Timed out

________________________________________________________________________________________

Giac [A]  time = 1.30114, size = 331, normalized size = 2.06 \begin{align*} \frac{525 \,{\left (d x + c\right )} a^{2} + 840 \, a^{2} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right ) - \frac{2 \,{\left (1155 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{13} - 1680 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{12} + 980 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{11} - 10080 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{10} + 2975 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{9} - 16240 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{8} - 24640 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{6} - 2975 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 14448 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 980 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 6496 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1155 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1168 \, a^{2}\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1\right )}^{7}}}{840 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*csc(d*x+c)*(a+a*sin(d*x+c))^2,x, algorithm="giac")

[Out]

1/840*(525*(d*x + c)*a^2 + 840*a^2*log(abs(tan(1/2*d*x + 1/2*c))) - 2*(1155*a^2*tan(1/2*d*x + 1/2*c)^13 - 1680
*a^2*tan(1/2*d*x + 1/2*c)^12 + 980*a^2*tan(1/2*d*x + 1/2*c)^11 - 10080*a^2*tan(1/2*d*x + 1/2*c)^10 + 2975*a^2*
tan(1/2*d*x + 1/2*c)^9 - 16240*a^2*tan(1/2*d*x + 1/2*c)^8 - 24640*a^2*tan(1/2*d*x + 1/2*c)^6 - 2975*a^2*tan(1/
2*d*x + 1/2*c)^5 - 14448*a^2*tan(1/2*d*x + 1/2*c)^4 - 980*a^2*tan(1/2*d*x + 1/2*c)^3 - 6496*a^2*tan(1/2*d*x +
1/2*c)^2 - 1155*a^2*tan(1/2*d*x + 1/2*c) - 1168*a^2)/(tan(1/2*d*x + 1/2*c)^2 + 1)^7)/d