3.526 \(\int \cos (c+d x) \cot ^4(c+d x) (a+a \sin (c+d x))^3 \, dx\)

Optimal. Leaf size=131 \[ \frac{a^3 \sin ^4(c+d x)}{4 d}+\frac{a^3 \sin ^3(c+d x)}{d}+\frac{a^3 \sin ^2(c+d x)}{2 d}-\frac{5 a^3 \sin (c+d x)}{d}-\frac{a^3 \csc ^3(c+d x)}{3 d}-\frac{3 a^3 \csc ^2(c+d x)}{2 d}-\frac{a^3 \csc (c+d x)}{d}-\frac{5 a^3 \log (\sin (c+d x))}{d} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.109789, antiderivative size = 131, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.111, Rules used = {2836, 12, 88} \[ \frac{a^3 \sin ^4(c+d x)}{4 d}+\frac{a^3 \sin ^3(c+d x)}{d}+\frac{a^3 \sin ^2(c+d x)}{2 d}-\frac{5 a^3 \sin (c+d x)}{d}-\frac{a^3 \csc ^3(c+d x)}{3 d}-\frac{3 a^3 \csc ^2(c+d x)}{2 d}-\frac{a^3 \csc (c+d x)}{d}-\frac{5 a^3 \log (\sin (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]*Cot[c + d*x]^4*(a + a*Sin[c + d*x])^3,x]

[Out]

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

Rule 2836

Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)
*(x_)])^(n_.), x_Symbol] :> Dist[1/(b^p*f), Subst[Int[(a + x)^(m + (p - 1)/2)*(a - x)^((p - 1)/2)*(c + (d*x)/b
)^n, x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, c, d, m, n}, x] && IntegerQ[(p - 1)/2] && EqQ[a^2 - b^2,
0]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rubi steps

\begin{align*} \int \cos (c+d x) \cot ^4(c+d x) (a+a \sin (c+d x))^3 \, dx &=\frac{\operatorname{Subst}\left (\int \frac{a^4 (a-x)^2 (a+x)^5}{x^4} \, dx,x,a \sin (c+d x)\right )}{a^5 d}\\ &=\frac{\operatorname{Subst}\left (\int \frac{(a-x)^2 (a+x)^5}{x^4} \, dx,x,a \sin (c+d x)\right )}{a d}\\ &=\frac{\operatorname{Subst}\left (\int \left (-5 a^3+\frac{a^7}{x^4}+\frac{3 a^6}{x^3}+\frac{a^5}{x^2}-\frac{5 a^4}{x}+a^2 x+3 a x^2+x^3\right ) \, dx,x,a \sin (c+d x)\right )}{a d}\\ &=-\frac{a^3 \csc (c+d x)}{d}-\frac{3 a^3 \csc ^2(c+d x)}{2 d}-\frac{a^3 \csc ^3(c+d x)}{3 d}-\frac{5 a^3 \log (\sin (c+d x))}{d}-\frac{5 a^3 \sin (c+d x)}{d}+\frac{a^3 \sin ^2(c+d x)}{2 d}+\frac{a^3 \sin ^3(c+d x)}{d}+\frac{a^3 \sin ^4(c+d x)}{4 d}\\ \end{align*}

Mathematica [A]  time = 0.25712, size = 86, normalized size = 0.66 \[ -\frac{a^3 \left (-3 \sin ^4(c+d x)-12 \sin ^3(c+d x)-6 \sin ^2(c+d x)+60 \sin (c+d x)+4 \csc ^3(c+d x)+18 \csc ^2(c+d x)+12 \csc (c+d x)+60 \log (\sin (c+d x))\right )}{12 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]*Cot[c + d*x]^4*(a + a*Sin[c + d*x])^3,x]

[Out]

-(a^3*(12*Csc[c + d*x] + 18*Csc[c + d*x]^2 + 4*Csc[c + d*x]^3 + 60*Log[Sin[c + d*x]] + 60*Sin[c + d*x] - 6*Sin
[c + d*x]^2 - 12*Sin[c + d*x]^3 - 3*Sin[c + d*x]^4))/(12*d)

________________________________________________________________________________________

Maple [A]  time = 0.086, size = 179, normalized size = 1.4 \begin{align*} -{\frac{5\, \left ( \cos \left ( dx+c \right ) \right ) ^{4}{a}^{3}}{4\,d}}-{\frac{5\,{a}^{3} \left ( \cos \left ( dx+c \right ) \right ) ^{2}}{2\,d}}-5\,{\frac{{a}^{3}\ln \left ( \sin \left ( dx+c \right ) \right ) }{d}}-2\,{\frac{{a}^{3} \left ( \cos \left ( dx+c \right ) \right ) ^{6}}{d\sin \left ( dx+c \right ) }}-{\frac{16\,{a}^{3}\sin \left ( dx+c \right ) }{3\,d}}-2\,{\frac{ \left ( \cos \left ( dx+c \right ) \right ) ^{4}{a}^{3}\sin \left ( dx+c \right ) }{d}}-{\frac{8\,{a}^{3} \left ( \cos \left ( dx+c \right ) \right ) ^{2}\sin \left ( dx+c \right ) }{3\,d}}-{\frac{3\,{a}^{3} \left ( \cos \left ( dx+c \right ) \right ) ^{6}}{2\,d \left ( \sin \left ( dx+c \right ) \right ) ^{2}}}-{\frac{{a}^{3} \left ( \cos \left ( dx+c \right ) \right ) ^{6}}{3\,d \left ( \sin \left ( dx+c \right ) \right ) ^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^5*csc(d*x+c)^4*(a+a*sin(d*x+c))^3,x)

[Out]

-5/4/d*cos(d*x+c)^4*a^3-5/2/d*a^3*cos(d*x+c)^2-5*a^3*ln(sin(d*x+c))/d-2/d*a^3/sin(d*x+c)*cos(d*x+c)^6-16/3*a^3
*sin(d*x+c)/d-2/d*a^3*cos(d*x+c)^4*sin(d*x+c)-8/3/d*a^3*cos(d*x+c)^2*sin(d*x+c)-3/2/d*a^3/sin(d*x+c)^2*cos(d*x
+c)^6-1/3/d*a^3/sin(d*x+c)^3*cos(d*x+c)^6

________________________________________________________________________________________

Maxima [A]  time = 1.03676, size = 146, normalized size = 1.11 \begin{align*} \frac{3 \, a^{3} \sin \left (d x + c\right )^{4} + 12 \, a^{3} \sin \left (d x + c\right )^{3} + 6 \, a^{3} \sin \left (d x + c\right )^{2} - 60 \, a^{3} \log \left (\sin \left (d x + c\right )\right ) - 60 \, a^{3} \sin \left (d x + c\right ) - \frac{2 \,{\left (6 \, a^{3} \sin \left (d x + c\right )^{2} + 9 \, a^{3} \sin \left (d x + c\right ) + 2 \, a^{3}\right )}}{\sin \left (d x + c\right )^{3}}}{12 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*csc(d*x+c)^4*(a+a*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

1/12*(3*a^3*sin(d*x + c)^4 + 12*a^3*sin(d*x + c)^3 + 6*a^3*sin(d*x + c)^2 - 60*a^3*log(sin(d*x + c)) - 60*a^3*
sin(d*x + c) - 2*(6*a^3*sin(d*x + c)^2 + 9*a^3*sin(d*x + c) + 2*a^3)/sin(d*x + c)^3)/d

________________________________________________________________________________________

Fricas [A]  time = 1.53772, size = 396, normalized size = 3.02 \begin{align*} \frac{96 \, a^{3} \cos \left (d x + c\right )^{6} + 192 \, a^{3} \cos \left (d x + c\right )^{4} - 768 \, a^{3} \cos \left (d x + c\right )^{2} + 512 \, a^{3} - 480 \,{\left (a^{3} \cos \left (d x + c\right )^{2} - a^{3}\right )} \log \left (\frac{1}{2} \, \sin \left (d x + c\right )\right ) \sin \left (d x + c\right ) + 3 \,{\left (8 \, a^{3} \cos \left (d x + c\right )^{6} - 40 \, a^{3} \cos \left (d x + c\right )^{4} + 45 \, a^{3} \cos \left (d x + c\right )^{2} + 35 \, a^{3}\right )} \sin \left (d x + c\right )}{96 \,{\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(cos(d*x+c)^5*csc(d*x+c)^4*(a+a*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

1/96*(96*a^3*cos(d*x + c)^6 + 192*a^3*cos(d*x + c)^4 - 768*a^3*cos(d*x + c)^2 + 512*a^3 - 480*(a^3*cos(d*x + c
)^2 - a^3)*log(1/2*sin(d*x + c))*sin(d*x + c) + 3*(8*a^3*cos(d*x + c)^6 - 40*a^3*cos(d*x + c)^4 + 45*a^3*cos(d
*x + c)^2 + 35*a^3)*sin(d*x + c))/((d*cos(d*x + c)^2 - d)*sin(d*x + c))

________________________________________________________________________________________

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)**5*csc(d*x+c)**4*(a+a*sin(d*x+c))**3,x)

[Out]

Timed out

________________________________________________________________________________________

Giac [A]  time = 1.36269, size = 165, normalized size = 1.26 \begin{align*} \frac{3 \, a^{3} \sin \left (d x + c\right )^{4} + 12 \, a^{3} \sin \left (d x + c\right )^{3} + 6 \, a^{3} \sin \left (d x + c\right )^{2} - 60 \, a^{3} \log \left ({\left | \sin \left (d x + c\right ) \right |}\right ) - 60 \, a^{3} \sin \left (d x + c\right ) + \frac{2 \,{\left (55 \, a^{3} \sin \left (d x + c\right )^{3} - 6 \, a^{3} \sin \left (d x + c\right )^{2} - 9 \, a^{3} \sin \left (d x + c\right ) - 2 \, a^{3}\right )}}{\sin \left (d x + c\right )^{3}}}{12 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*csc(d*x+c)^4*(a+a*sin(d*x+c))^3,x, algorithm="giac")

[Out]

1/12*(3*a^3*sin(d*x + c)^4 + 12*a^3*sin(d*x + c)^3 + 6*a^3*sin(d*x + c)^2 - 60*a^3*log(abs(sin(d*x + c))) - 60
*a^3*sin(d*x + c) + 2*(55*a^3*sin(d*x + c)^3 - 6*a^3*sin(d*x + c)^2 - 9*a^3*sin(d*x + c) - 2*a^3)/sin(d*x + c)
^3)/d