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

Optimal. Leaf size=133 \[ \frac{a^3 \sin ^6(c+d x)}{6 d}+\frac{3 a^3 \sin ^5(c+d x)}{5 d}+\frac{a^3 \sin ^4(c+d x)}{4 d}-\frac{5 a^3 \sin ^3(c+d x)}{3 d}-\frac{5 a^3 \sin ^2(c+d x)}{2 d}+\frac{a^3 \sin (c+d x)}{d}-\frac{a^3 \csc (c+d x)}{d}+\frac{3 a^3 \log (\sin (c+d x))}{d} \]

[Out]

-((a^3*Csc[c + d*x])/d) + (3*a^3*Log[Sin[c + d*x]])/d + (a^3*Sin[c + d*x])/d - (5*a^3*Sin[c + d*x]^2)/(2*d) -
(5*a^3*Sin[c + d*x]^3)/(3*d) + (a^3*Sin[c + d*x]^4)/(4*d) + (3*a^3*Sin[c + d*x]^5)/(5*d) + (a^3*Sin[c + d*x]^6
)/(6*d)

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Rubi [A]  time = 0.123447, 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} \[ \frac{a^3 \sin ^6(c+d x)}{6 d}+\frac{3 a^3 \sin ^5(c+d x)}{5 d}+\frac{a^3 \sin ^4(c+d x)}{4 d}-\frac{5 a^3 \sin ^3(c+d x)}{3 d}-\frac{5 a^3 \sin ^2(c+d x)}{2 d}+\frac{a^3 \sin (c+d x)}{d}-\frac{a^3 \csc (c+d x)}{d}+\frac{3 a^3 \log (\sin (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.164678, size = 86, normalized size = 0.65 \[ -\frac{a^3 \left (-10 \sin ^6(c+d x)-36 \sin ^5(c+d x)-15 \sin ^4(c+d x)+100 \sin ^3(c+d x)+150 \sin ^2(c+d x)-60 \sin (c+d x)+60 \csc (c+d x)-180 \log (\sin (c+d x))\right )}{60 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a^3*(60*Csc[c + d*x] - 180*Log[Sin[c + d*x]] - 60*Sin[c + d*x] + 150*Sin[c + d*x]^2 + 100*Sin[c + d*x]^3 - 1
5*Sin[c + d*x]^4 - 36*Sin[c + d*x]^5 - 10*Sin[c + d*x]^6))/(60*d)

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Maple [A]  time = 0.082, size = 147, normalized size = 1.1 \begin{align*} -{\frac{{a}^{3} \left ( \cos \left ( dx+c \right ) \right ) ^{6}}{6\,d}}-{\frac{16\,{a}^{3}\sin \left ( dx+c \right ) }{15\,d}}-{\frac{2\,{a}^{3} \left ( \cos \left ( dx+c \right ) \right ) ^{4}\sin \left ( dx+c \right ) }{5\,d}}-{\frac{8\,{a}^{3} \left ( \cos \left ( dx+c \right ) \right ) ^{2}\sin \left ( dx+c \right ) }{15\,d}}+{\frac{3\,{a}^{3} \left ( \cos \left ( dx+c \right ) \right ) ^{4}}{4\,d}}+{\frac{3\,{a}^{3} \left ( \cos \left ( dx+c \right ) \right ) ^{2}}{2\,d}}+3\,{\frac{{a}^{3}\ln \left ( \sin \left ( dx+c \right ) \right ) }{d}}-{\frac{{a}^{3} \left ( \cos \left ( dx+c \right ) \right ) ^{6}}{d\sin \left ( dx+c \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^5*csc(d*x+c)^2*(a+a*sin(d*x+c))^3,x)

[Out]

-1/6/d*a^3*cos(d*x+c)^6-16/15*a^3*sin(d*x+c)/d-2/5/d*a^3*cos(d*x+c)^4*sin(d*x+c)-8/15/d*a^3*cos(d*x+c)^2*sin(d
*x+c)+3/4/d*cos(d*x+c)^4*a^3+3/2/d*a^3*cos(d*x+c)^2+3*a^3*ln(sin(d*x+c))/d-1/d*a^3/sin(d*x+c)*cos(d*x+c)^6

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Maxima [A]  time = 1.08016, size = 144, normalized size = 1.08 \begin{align*} \frac{10 \, a^{3} \sin \left (d x + c\right )^{6} + 36 \, a^{3} \sin \left (d x + c\right )^{5} + 15 \, a^{3} \sin \left (d x + c\right )^{4} - 100 \, a^{3} \sin \left (d x + c\right )^{3} - 150 \, a^{3} \sin \left (d x + c\right )^{2} + 180 \, a^{3} \log \left (\sin \left (d x + c\right )\right ) + 60 \, a^{3} \sin \left (d x + c\right ) - \frac{60 \, a^{3}}{\sin \left (d x + c\right )}}{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)^2*(a+a*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

1/60*(10*a^3*sin(d*x + c)^6 + 36*a^3*sin(d*x + c)^5 + 15*a^3*sin(d*x + c)^4 - 100*a^3*sin(d*x + c)^3 - 150*a^3
*sin(d*x + c)^2 + 180*a^3*log(sin(d*x + c)) + 60*a^3*sin(d*x + c) - 60*a^3/sin(d*x + c))/d

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Fricas [A]  time = 1.60819, size = 339, normalized size = 2.55 \begin{align*} -\frac{144 \, a^{3} \cos \left (d x + c\right )^{6} - 32 \, a^{3} \cos \left (d x + c\right )^{4} - 128 \, a^{3} \cos \left (d x + c\right )^{2} - 720 \, a^{3} \log \left (\frac{1}{2} \, \sin \left (d x + c\right )\right ) \sin \left (d x + c\right ) + 256 \, a^{3} + 5 \,{\left (8 \, a^{3} \cos \left (d x + c\right )^{6} - 36 \, a^{3} \cos \left (d x + c\right )^{4} - 72 \, a^{3} \cos \left (d x + c\right )^{2} + 47 \, a^{3}\right )} \sin \left (d x + c\right )}{240 \, d \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)^2*(a+a*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

-1/240*(144*a^3*cos(d*x + c)^6 - 32*a^3*cos(d*x + c)^4 - 128*a^3*cos(d*x + c)^2 - 720*a^3*log(1/2*sin(d*x + c)
)*sin(d*x + c) + 256*a^3 + 5*(8*a^3*cos(d*x + c)^6 - 36*a^3*cos(d*x + c)^4 - 72*a^3*cos(d*x + c)^2 + 47*a^3)*s
in(d*x + c))/(d*sin(d*x + c))

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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)**2*(a+a*sin(d*x+c))**3,x)

[Out]

Timed out

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Giac [A]  time = 1.34305, size = 162, normalized size = 1.22 \begin{align*} \frac{10 \, a^{3} \sin \left (d x + c\right )^{6} + 36 \, a^{3} \sin \left (d x + c\right )^{5} + 15 \, a^{3} \sin \left (d x + c\right )^{4} - 100 \, a^{3} \sin \left (d x + c\right )^{3} - 150 \, a^{3} \sin \left (d x + c\right )^{2} + 180 \, a^{3} \log \left ({\left | \sin \left (d x + c\right ) \right |}\right ) + 60 \, a^{3} \sin \left (d x + c\right ) - \frac{60 \,{\left (3 \, a^{3} \sin \left (d x + c\right ) + a^{3}\right )}}{\sin \left (d x + c\right )}}{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)^2*(a+a*sin(d*x+c))^3,x, algorithm="giac")

[Out]

1/60*(10*a^3*sin(d*x + c)^6 + 36*a^3*sin(d*x + c)^5 + 15*a^3*sin(d*x + c)^4 - 100*a^3*sin(d*x + c)^3 - 150*a^3
*sin(d*x + c)^2 + 180*a^3*log(abs(sin(d*x + c))) + 60*a^3*sin(d*x + c) - 60*(3*a^3*sin(d*x + c) + a^3)/sin(d*x
+ c))/d