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

Optimal. Leaf size=143 \[ -\frac{3 a^3 \cos ^5(c+d x)}{5 d}+\frac{a^3 \cos ^3(c+d x)}{3 d}+\frac{a^3 \cos (c+d x)}{d}-\frac{a^3 \sin (c+d x) \cos ^5(c+d x)}{6 d}+\frac{19 a^3 \sin (c+d x) \cos ^3(c+d x)}{24 d}+\frac{19 a^3 \sin (c+d x) \cos (c+d x)}{16 d}-\frac{a^3 \tanh ^{-1}(\cos (c+d x))}{d}+\frac{19 a^3 x}{16} \]

[Out]

(19*a^3*x)/16 - (a^3*ArcTanh[Cos[c + d*x]])/d + (a^3*Cos[c + d*x])/d + (a^3*Cos[c + d*x]^3)/(3*d) - (3*a^3*Cos
[c + d*x]^5)/(5*d) + (19*a^3*Cos[c + d*x]*Sin[c + d*x])/(16*d) + (19*a^3*Cos[c + d*x]^3*Sin[c + d*x])/(24*d) -
(a^3*Cos[c + d*x]^5*Sin[c + d*x])/(6*d)

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Rubi [A]  time = 0.202189, antiderivative size = 143, normalized size of antiderivative = 1., number of steps used = 15, number of rules used = 9, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {2873, 2635, 8, 2592, 302, 206, 2565, 30, 2568} \[ -\frac{3 a^3 \cos ^5(c+d x)}{5 d}+\frac{a^3 \cos ^3(c+d x)}{3 d}+\frac{a^3 \cos (c+d x)}{d}-\frac{a^3 \sin (c+d x) \cos ^5(c+d x)}{6 d}+\frac{19 a^3 \sin (c+d x) \cos ^3(c+d x)}{24 d}+\frac{19 a^3 \sin (c+d x) \cos (c+d x)}{16 d}-\frac{a^3 \tanh ^{-1}(\cos (c+d x))}{d}+\frac{19 a^3 x}{16} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(19*a^3*x)/16 - (a^3*ArcTanh[Cos[c + d*x]])/d + (a^3*Cos[c + d*x])/d + (a^3*Cos[c + d*x]^3)/(3*d) - (3*a^3*Cos
[c + d*x]^5)/(5*d) + (19*a^3*Cos[c + d*x]*Sin[c + d*x])/(16*d) + (19*a^3*Cos[c + d*x]^3*Sin[c + d*x])/(24*d) -
(a^3*Cos[c + d*x]^5*Sin[c + d*x])/(6*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]

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 ^3(c+d x) \cot (c+d x) (a+a \sin (c+d x))^3 \, dx &=\int \left (3 a^3 \cos ^4(c+d x)+a^3 \cos ^3(c+d x) \cot (c+d x)+3 a^3 \cos ^4(c+d x) \sin (c+d x)+a^3 \cos ^4(c+d x) \sin ^2(c+d x)\right ) \, dx\\ &=a^3 \int \cos ^3(c+d x) \cot (c+d x) \, dx+a^3 \int \cos ^4(c+d x) \sin ^2(c+d x) \, dx+\left (3 a^3\right ) \int \cos ^4(c+d x) \, dx+\left (3 a^3\right ) \int \cos ^4(c+d x) \sin (c+d x) \, dx\\ &=\frac{3 a^3 \cos ^3(c+d x) \sin (c+d x)}{4 d}-\frac{a^3 \cos ^5(c+d x) \sin (c+d x)}{6 d}+\frac{1}{6} a^3 \int \cos ^4(c+d x) \, dx+\frac{1}{4} \left (9 a^3\right ) \int \cos ^2(c+d x) \, dx-\frac{a^3 \operatorname{Subst}\left (\int \frac{x^4}{1-x^2} \, dx,x,\cos (c+d x)\right )}{d}-\frac{\left (3 a^3\right ) \operatorname{Subst}\left (\int x^4 \, dx,x,\cos (c+d x)\right )}{d}\\ &=-\frac{3 a^3 \cos ^5(c+d x)}{5 d}+\frac{9 a^3 \cos (c+d x) \sin (c+d x)}{8 d}+\frac{19 a^3 \cos ^3(c+d x) \sin (c+d x)}{24 d}-\frac{a^3 \cos ^5(c+d x) \sin (c+d x)}{6 d}+\frac{1}{8} a^3 \int \cos ^2(c+d x) \, dx+\frac{1}{8} \left (9 a^3\right ) \int 1 \, dx-\frac{a^3 \operatorname{Subst}\left (\int \left (-1-x^2+\frac{1}{1-x^2}\right ) \, dx,x,\cos (c+d x)\right )}{d}\\ &=\frac{9 a^3 x}{8}+\frac{a^3 \cos (c+d x)}{d}+\frac{a^3 \cos ^3(c+d x)}{3 d}-\frac{3 a^3 \cos ^5(c+d x)}{5 d}+\frac{19 a^3 \cos (c+d x) \sin (c+d x)}{16 d}+\frac{19 a^3 \cos ^3(c+d x) \sin (c+d x)}{24 d}-\frac{a^3 \cos ^5(c+d x) \sin (c+d x)}{6 d}+\frac{1}{16} 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{19 a^3 x}{16}-\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)}{3 d}-\frac{3 a^3 \cos ^5(c+d x)}{5 d}+\frac{19 a^3 \cos (c+d x) \sin (c+d x)}{16 d}+\frac{19 a^3 \cos ^3(c+d x) \sin (c+d x)}{24 d}-\frac{a^3 \cos ^5(c+d x) \sin (c+d x)}{6 d}\\ \end{align*}

Mathematica [A]  time = 0.980717, size = 102, normalized size = 0.71 \[ \frac{a^3 \left (735 \sin (2 (c+d x))+75 \sin (4 (c+d x))-5 \sin (6 (c+d x))+840 \cos (c+d x)-100 \cos (3 (c+d x))-36 \cos (5 (c+d x))+960 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )-960 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )+1140 c+1140 d x\right )}{960 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*(1140*c + 1140*d*x + 840*Cos[c + d*x] - 100*Cos[3*(c + d*x)] - 36*Cos[5*(c + d*x)] - 960*Log[Cos[(c + d*x
)/2]] + 960*Log[Sin[(c + d*x)/2]] + 735*Sin[2*(c + d*x)] + 75*Sin[4*(c + d*x)] - 5*Sin[6*(c + d*x)]))/(960*d)

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Maple [A]  time = 0.078, size = 149, normalized size = 1. \begin{align*} -{\frac{{a}^{3} \left ( \cos \left ( dx+c \right ) \right ) ^{5}\sin \left ( dx+c \right ) }{6\,d}}+{\frac{19\,{a}^{3} \left ( \cos \left ( dx+c \right ) \right ) ^{3}\sin \left ( dx+c \right ) }{24\,d}}+{\frac{19\,{a}^{3}\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{16\,d}}+{\frac{19\,{a}^{3}x}{16}}+{\frac{19\,{a}^{3}c}{16\,d}}-{\frac{3\,{a}^{3} \left ( \cos \left ( dx+c \right ) \right ) ^{5}}{5\,d}}+{\frac{{a}^{3} \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{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)^4*csc(d*x+c)*(a+a*sin(d*x+c))^3,x)

[Out]

-1/6*a^3*cos(d*x+c)^5*sin(d*x+c)/d+19/24*a^3*cos(d*x+c)^3*sin(d*x+c)/d+19/16*a^3*cos(d*x+c)*sin(d*x+c)/d+19/16
*a^3*x+19/16/d*a^3*c-3/5*a^3*cos(d*x+c)^5/d+1/3*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 = 1.07763, size = 182, normalized size = 1.27 \begin{align*} -\frac{576 \, a^{3} \cos \left (d x + c\right )^{5} - 160 \,{\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} - 5 \,{\left (4 \, \sin \left (2 \, d x + 2 \, c\right )^{3} + 12 \, d x + 12 \, c - 3 \, \sin \left (4 \, d x + 4 \, c\right )\right )} a^{3} - 90 \,{\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^{3}}{960 \, 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))^3,x, algorithm="maxima")

[Out]

-1/960*(576*a^3*cos(d*x + c)^5 - 160*(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 - 5*(4*sin(2*d*x + 2*c)^3 + 12*d*x + 12*c - 3*sin(4*d*x + 4*c))*a^3 - 90*(12*d*x + 12*c + s
in(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*a^3)/d

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Fricas [A]  time = 1.21633, size = 350, normalized size = 2.45 \begin{align*} -\frac{144 \, a^{3} \cos \left (d x + c\right )^{5} - 80 \, a^{3} \cos \left (d x + c\right )^{3} - 285 \, a^{3} d x - 240 \, a^{3} \cos \left (d x + c\right ) + 120 \, a^{3} \log \left (\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) - 120 \, a^{3} \log \left (-\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) + 5 \,{\left (8 \, a^{3} \cos \left (d x + c\right )^{5} - 38 \, a^{3} \cos \left (d x + c\right )^{3} - 57 \, a^{3} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{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))^3,x, algorithm="fricas")

[Out]

-1/240*(144*a^3*cos(d*x + c)^5 - 80*a^3*cos(d*x + c)^3 - 285*a^3*d*x - 240*a^3*cos(d*x + c) + 120*a^3*log(1/2*
cos(d*x + c) + 1/2) - 120*a^3*log(-1/2*cos(d*x + c) + 1/2) + 5*(8*a^3*cos(d*x + c)^5 - 38*a^3*cos(d*x + c)^3 -
57*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)**4*csc(d*x+c)*(a+a*sin(d*x+c))**3,x)

[Out]

Timed out

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Giac [A]  time = 1.34776, size = 309, normalized size = 2.16 \begin{align*} \frac{285 \,{\left (d x + c\right )} a^{3} + 240 \, a^{3} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right ) - \frac{2 \,{\left (435 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{11} + 240 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{10} + 865 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{9} - 1200 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{8} - 210 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 1760 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{6} + 210 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 1440 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 865 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 1296 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 435 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 176 \, a^{3}\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1\right )}^{6}}}{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))^3,x, algorithm="giac")

[Out]

1/240*(285*(d*x + c)*a^3 + 240*a^3*log(abs(tan(1/2*d*x + 1/2*c))) - 2*(435*a^3*tan(1/2*d*x + 1/2*c)^11 + 240*a
^3*tan(1/2*d*x + 1/2*c)^10 + 865*a^3*tan(1/2*d*x + 1/2*c)^9 - 1200*a^3*tan(1/2*d*x + 1/2*c)^8 - 210*a^3*tan(1/
2*d*x + 1/2*c)^7 - 1760*a^3*tan(1/2*d*x + 1/2*c)^6 + 210*a^3*tan(1/2*d*x + 1/2*c)^5 - 1440*a^3*tan(1/2*d*x + 1
/2*c)^4 - 865*a^3*tan(1/2*d*x + 1/2*c)^3 - 1296*a^3*tan(1/2*d*x + 1/2*c)^2 - 435*a^3*tan(1/2*d*x + 1/2*c) - 17
6*a^3)/(tan(1/2*d*x + 1/2*c)^2 + 1)^6)/d