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

Optimal. Leaf size=133 \[ \frac{a^3 \sin ^5(c+d x)}{5 d}+\frac{3 a^3 \sin ^4(c+d x)}{4 d}+\frac{a^3 \sin ^3(c+d x)}{3 d}-\frac{5 a^3 \sin ^2(c+d x)}{2 d}-\frac{5 a^3 \sin (c+d x)}{d}-\frac{a^3 \csc ^2(c+d x)}{2 d}-\frac{3 a^3 \csc (c+d x)}{d}+\frac{a^3 \log (\sin (c+d x))}{d} \]

[Out]

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

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Rubi [A]  time = 0.127574, antiderivative size = 133, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.103, Rules used = {2836, 12, 88} \[ \frac{a^3 \sin ^5(c+d x)}{5 d}+\frac{3 a^3 \sin ^4(c+d x)}{4 d}+\frac{a^3 \sin ^3(c+d x)}{3 d}-\frac{5 a^3 \sin ^2(c+d x)}{2 d}-\frac{5 a^3 \sin (c+d x)}{d}-\frac{a^3 \csc ^2(c+d x)}{2 d}-\frac{3 a^3 \csc (c+d x)}{d}+\frac{a^3 \log (\sin (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.147773, size = 86, normalized size = 0.65 \[ -\frac{a^3 \left (-12 \sin ^5(c+d x)-45 \sin ^4(c+d x)-20 \sin ^3(c+d x)+150 \sin ^2(c+d x)+300 \sin (c+d x)+30 \csc ^2(c+d x)+180 \csc (c+d x)-60 \log (\sin (c+d x))\right )}{60 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a^3*(180*Csc[c + d*x] + 30*Csc[c + d*x]^2 - 60*Log[Sin[c + d*x]] + 300*Sin[c + d*x] + 150*Sin[c + d*x]^2 - 2
0*Sin[c + d*x]^3 - 45*Sin[c + d*x]^4 - 12*Sin[c + d*x]^5))/(60*d)

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Maple [A]  time = 0.092, size = 154, normalized size = 1.2 \begin{align*} -{\frac{112\,{a}^{3}\sin \left ( dx+c \right ) }{15\,d}}-{\frac{14\,{a}^{3} \left ( \cos \left ( dx+c \right ) \right ) ^{4}\sin \left ( dx+c \right ) }{5\,d}}-{\frac{56\,{a}^{3} \left ( \cos \left ( dx+c \right ) \right ) ^{2}\sin \left ( dx+c \right ) }{15\,d}}+{\frac{{a}^{3} \left ( \cos \left ( dx+c \right ) \right ) ^{4}}{4\,d}}+{\frac{{a}^{3} \left ( \cos \left ( dx+c \right ) \right ) ^{2}}{2\,d}}+{\frac{{a}^{3}\ln \left ( \sin \left ( dx+c \right ) \right ) }{d}}-3\,{\frac{{a}^{3} \left ( \cos \left ( dx+c \right ) \right ) ^{6}}{d\sin \left ( dx+c \right ) }}-{\frac{{a}^{3} \left ( \cos \left ( dx+c \right ) \right ) ^{6}}{2\,d \left ( \sin \left ( dx+c \right ) \right ) ^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-112/15*a^3*sin(d*x+c)/d-14/5/d*a^3*cos(d*x+c)^4*sin(d*x+c)-56/15/d*a^3*cos(d*x+c)^2*sin(d*x+c)+1/4/d*cos(d*x+
c)^4*a^3+1/2/d*a^3*cos(d*x+c)^2+a^3*ln(sin(d*x+c))/d-3/d*a^3/sin(d*x+c)*cos(d*x+c)^6-1/2/d*a^3/sin(d*x+c)^2*co
s(d*x+c)^6

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Maxima [A]  time = 1.17822, size = 143, normalized size = 1.08 \begin{align*} \frac{12 \, a^{3} \sin \left (d x + c\right )^{5} + 45 \, a^{3} \sin \left (d x + c\right )^{4} + 20 \, a^{3} \sin \left (d x + c\right )^{3} - 150 \, a^{3} \sin \left (d x + c\right )^{2} + 60 \, a^{3} \log \left (\sin \left (d x + c\right )\right ) - 300 \, a^{3} \sin \left (d x + c\right ) - \frac{30 \,{\left (6 \, a^{3} \sin \left (d x + c\right ) + a^{3}\right )}}{\sin \left (d x + c\right )^{2}}}{60 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/60*(12*a^3*sin(d*x + c)^5 + 45*a^3*sin(d*x + c)^4 + 20*a^3*sin(d*x + c)^3 - 150*a^3*sin(d*x + c)^2 + 60*a^3*
log(sin(d*x + c)) - 300*a^3*sin(d*x + c) - 30*(6*a^3*sin(d*x + c) + a^3)/sin(d*x + c)^2)/d

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Fricas [A]  time = 1.57352, size = 363, normalized size = 2.73 \begin{align*} \frac{360 \, a^{3} \cos \left (d x + c\right )^{6} + 120 \, a^{3} \cos \left (d x + c\right )^{4} - 855 \, a^{3} \cos \left (d x + c\right )^{2} + 615 \, 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 ) + 32 \,{\left (3 \, a^{3} \cos \left (d x + c\right )^{6} - 14 \, a^{3} \cos \left (d x + c\right )^{4} - 56 \, a^{3} \cos \left (d x + c\right )^{2} + 112 \, a^{3}\right )} \sin \left (d x + c\right )}{480 \,{\left (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)^3*(a+a*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

1/480*(360*a^3*cos(d*x + c)^6 + 120*a^3*cos(d*x + c)^4 - 855*a^3*cos(d*x + c)^2 + 615*a^3 + 480*(a^3*cos(d*x +
c)^2 - a^3)*log(1/2*sin(d*x + c)) + 32*(3*a^3*cos(d*x + c)^6 - 14*a^3*cos(d*x + c)^4 - 56*a^3*cos(d*x + c)^2
+ 112*a^3)*sin(d*x + c))/(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)**3*(a+a*sin(d*x+c))**3,x)

[Out]

Timed out

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Giac [A]  time = 1.33728, size = 162, normalized size = 1.22 \begin{align*} \frac{12 \, a^{3} \sin \left (d x + c\right )^{5} + 45 \, a^{3} \sin \left (d x + c\right )^{4} + 20 \, a^{3} \sin \left (d x + c\right )^{3} - 150 \, a^{3} \sin \left (d x + c\right )^{2} + 60 \, a^{3} \log \left ({\left | \sin \left (d x + c\right ) \right |}\right ) - 300 \, a^{3} \sin \left (d x + c\right ) - \frac{30 \,{\left (3 \, a^{3} \sin \left (d x + c\right )^{2} + 6 \, a^{3} \sin \left (d x + c\right ) + a^{3}\right )}}{\sin \left (d x + c\right )^{2}}}{60 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/60*(12*a^3*sin(d*x + c)^5 + 45*a^3*sin(d*x + c)^4 + 20*a^3*sin(d*x + c)^3 - 150*a^3*sin(d*x + c)^2 + 60*a^3*
log(abs(sin(d*x + c))) - 300*a^3*sin(d*x + c) - 30*(3*a^3*sin(d*x + c)^2 + 6*a^3*sin(d*x + c) + a^3)/sin(d*x +
c)^2)/d