3.527 \(\int \cot ^5(c+d x) (a+a \sin (c+d x))^3 \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.070055, antiderivative size = 131, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.095, Rules used = {2707, 88} \[ \frac{a^3 \sin ^3(c+d x)}{3 d}+\frac{3 a^3 \sin ^2(c+d x)}{2 d}+\frac{a^3 \sin (c+d x)}{d}-\frac{a^3 \csc ^4(c+d x)}{4 d}-\frac{a^3 \csc ^3(c+d x)}{d}-\frac{a^3 \csc ^2(c+d x)}{2 d}+\frac{5 a^3 \csc (c+d x)}{d}-\frac{5 a^3 \log (\sin (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]^5*(a + a*Sin[c + d*x])^3,x]

[Out]

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

Rule 2707

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

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

Mathematica [A]  time = 0.523199, size = 86, normalized size = 0.66 \[ \frac{a^3 \left (4 \sin ^3(c+d x)+18 \sin ^2(c+d x)+12 \sin (c+d x)-3 \csc ^4(c+d x)-12 \csc ^3(c+d x)-6 \csc ^2(c+d x)+60 \csc (c+d x)-60 \log (\sin (c+d x))\right )}{12 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[c + d*x]^5*(a + a*Sin[c + d*x])^3,x]

[Out]

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

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Maple [A]  time = 0.091, size = 211, normalized size = 1.6 \begin{align*} 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{{a}^{3} \left ( \cos \left ( dx+c \right ) \right ) ^{4}\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{3\,{a}^{3} \left ( \cos \left ( dx+c \right ) \right ) ^{4}}{2\,d}}-3\,{\frac{{a}^{3} \left ( \cos \left ( dx+c \right ) \right ) ^{2}}{d}}-5\,{\frac{{a}^{3}\ln \left ( \sin \left ( dx+c \right ) \right ) }{d}}-{\frac{{a}^{3} \left ( \cos \left ( dx+c \right ) \right ) ^{6}}{d \left ( \sin \left ( dx+c \right ) \right ) ^{3}}}-{\frac{{a}^{3} \left ( \cot \left ( dx+c \right ) \right ) ^{4}}{4\,d}}+{\frac{{a}^{3} \left ( \cot \left ( dx+c \right ) \right ) ^{2}}{2\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.1401, size = 146, normalized size = 1.11 \begin{align*} \frac{4 \, a^{3} \sin \left (d x + c\right )^{3} + 18 \, a^{3} \sin \left (d x + c\right )^{2} - 60 \, a^{3} \log \left (\sin \left (d x + c\right )\right ) + 12 \, a^{3} \sin \left (d x + c\right ) + \frac{3 \,{\left (20 \, a^{3} \sin \left (d x + c\right )^{3} - 2 \, a^{3} \sin \left (d x + c\right )^{2} - 4 \, a^{3} \sin \left (d x + c\right ) - a^{3}\right )}}{\sin \left (d x + c\right )^{4}}}{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)^5*(a+a*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

1/12*(4*a^3*sin(d*x + c)^3 + 18*a^3*sin(d*x + c)^2 - 60*a^3*log(sin(d*x + c)) + 12*a^3*sin(d*x + c) + 3*(20*a^
3*sin(d*x + c)^3 - 2*a^3*sin(d*x + c)^2 - 4*a^3*sin(d*x + c) - a^3)/sin(d*x + c)^4)/d

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Fricas [A]  time = 1.55544, size = 397, normalized size = 3.03 \begin{align*} -\frac{18 \, a^{3} \cos \left (d x + c\right )^{6} - 45 \, a^{3} \cos \left (d x + c\right )^{4} + 30 \, a^{3} \cos \left (d x + c\right )^{2} + 60 \,{\left (a^{3} \cos \left (d x + c\right )^{4} - 2 \, a^{3} \cos \left (d x + c\right )^{2} + a^{3}\right )} \log \left (\frac{1}{2} \, \sin \left (d x + c\right )\right ) + 4 \,{\left (a^{3} \cos \left (d x + c\right )^{6} - 6 \, a^{3} \cos \left (d x + c\right )^{4} + 24 \, a^{3} \cos \left (d x + c\right )^{2} - 16 \, a^{3}\right )} \sin \left (d x + c\right )}{12 \,{\left (d \cos \left (d x + c\right )^{4} - 2 \, d \cos \left (d x + c\right )^{2} + d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*(18*a^3*cos(d*x + c)^6 - 45*a^3*cos(d*x + c)^4 + 30*a^3*cos(d*x + c)^2 + 60*(a^3*cos(d*x + c)^4 - 2*a^3*
cos(d*x + c)^2 + a^3)*log(1/2*sin(d*x + c)) + 4*(a^3*cos(d*x + c)^6 - 6*a^3*cos(d*x + c)^4 + 24*a^3*cos(d*x +
c)^2 - 16*a^3)*sin(d*x + c))/(d*cos(d*x + c)^4 - 2*d*cos(d*x + c)^2 + 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)**5*csc(d*x+c)**5*(a+a*sin(d*x+c))**3,x)

[Out]

Timed out

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Giac [A]  time = 1.26071, size = 163, normalized size = 1.24 \begin{align*} \frac{4 \, a^{3} \sin \left (d x + c\right )^{3} + 18 \, a^{3} \sin \left (d x + c\right )^{2} - 60 \, a^{3} \log \left ({\left | \sin \left (d x + c\right ) \right |}\right ) + 12 \, a^{3} \sin \left (d x + c\right ) + \frac{125 \, a^{3} \sin \left (d x + c\right )^{4} + 60 \, a^{3} \sin \left (d x + c\right )^{3} - 6 \, a^{3} \sin \left (d x + c\right )^{2} - 12 \, a^{3} \sin \left (d x + c\right ) - 3 \, a^{3}}{\sin \left (d x + c\right )^{4}}}{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)^5*(a+a*sin(d*x+c))^3,x, algorithm="giac")

[Out]

1/12*(4*a^3*sin(d*x + c)^3 + 18*a^3*sin(d*x + c)^2 - 60*a^3*log(abs(sin(d*x + c))) + 12*a^3*sin(d*x + c) + (12
5*a^3*sin(d*x + c)^4 + 60*a^3*sin(d*x + c)^3 - 6*a^3*sin(d*x + c)^2 - 12*a^3*sin(d*x + c) - 3*a^3)/sin(d*x + c
)^4)/d