3.239 \(\int \frac{\cot (c+d x) \csc ^3(c+d x)}{(a+a \sin (c+d x))^2} \, dx\)

Optimal. Leaf size=101 \[ -\frac{1}{d \left (a^2 \sin (c+d x)+a^2\right )}-\frac{\csc ^3(c+d x)}{3 a^2 d}+\frac{\csc ^2(c+d x)}{a^2 d}-\frac{3 \csc (c+d x)}{a^2 d}-\frac{4 \log (\sin (c+d x))}{a^2 d}+\frac{4 \log (\sin (c+d x)+1)}{a^2 d} \]

[Out]

(-3*Csc[c + d*x])/(a^2*d) + Csc[c + d*x]^2/(a^2*d) - Csc[c + d*x]^3/(3*a^2*d) - (4*Log[Sin[c + d*x]])/(a^2*d)
+ (4*Log[1 + Sin[c + d*x]])/(a^2*d) - 1/(d*(a^2 + a^2*Sin[c + d*x]))

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Rubi [A]  time = 0.0986431, antiderivative size = 101, 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 = {2833, 12, 44} \[ -\frac{1}{d \left (a^2 \sin (c+d x)+a^2\right )}-\frac{\csc ^3(c+d x)}{3 a^2 d}+\frac{\csc ^2(c+d x)}{a^2 d}-\frac{3 \csc (c+d x)}{a^2 d}-\frac{4 \log (\sin (c+d x))}{a^2 d}+\frac{4 \log (\sin (c+d x)+1)}{a^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*Csc[c + d*x])/(a^2*d) + Csc[c + d*x]^2/(a^2*d) - Csc[c + d*x]^3/(3*a^2*d) - (4*Log[Sin[c + d*x]])/(a^2*d)
+ (4*Log[1 + Sin[c + d*x]])/(a^2*d) - 1/(d*(a^2 + a^2*Sin[c + d*x]))

Rule 2833

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

Rule 12

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

Rule 44

Int[((a_) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*
x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int \frac{\cot (c+d x) \csc ^3(c+d x)}{(a+a \sin (c+d x))^2} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{a^4}{x^4 (a+x)^2} \, dx,x,a \sin (c+d x)\right )}{a d}\\ &=\frac{a^3 \operatorname{Subst}\left (\int \frac{1}{x^4 (a+x)^2} \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac{a^3 \operatorname{Subst}\left (\int \left (\frac{1}{a^2 x^4}-\frac{2}{a^3 x^3}+\frac{3}{a^4 x^2}-\frac{4}{a^5 x}+\frac{1}{a^4 (a+x)^2}+\frac{4}{a^5 (a+x)}\right ) \, dx,x,a \sin (c+d x)\right )}{d}\\ &=-\frac{3 \csc (c+d x)}{a^2 d}+\frac{\csc ^2(c+d x)}{a^2 d}-\frac{\csc ^3(c+d x)}{3 a^2 d}-\frac{4 \log (\sin (c+d x))}{a^2 d}+\frac{4 \log (1+\sin (c+d x))}{a^2 d}-\frac{1}{d \left (a^2+a^2 \sin (c+d x)\right )}\\ \end{align*}

Mathematica [A]  time = 2.21165, size = 98, normalized size = 0.97 \[ -\frac{1}{a^2 d (\sin (c+d x)+1)}-\frac{\csc ^3(c+d x)}{3 a^2 d}+\frac{\csc ^2(c+d x)}{a^2 d}-\frac{3 \csc (c+d x)}{a^2 d}-\frac{4 \log (\sin (c+d x))}{a^2 d}+\frac{4 \log (\sin (c+d x)+1)}{a^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*Csc[c + d*x])/(a^2*d) + Csc[c + d*x]^2/(a^2*d) - Csc[c + d*x]^3/(3*a^2*d) - (4*Log[Sin[c + d*x]])/(a^2*d)
+ (4*Log[1 + Sin[c + d*x]])/(a^2*d) - 1/(a^2*d*(1 + Sin[c + d*x]))

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Maple [A]  time = 0.055, size = 99, normalized size = 1. \begin{align*} -{\frac{1}{d{a}^{2} \left ( 1+\sin \left ( dx+c \right ) \right ) }}+4\,{\frac{\ln \left ( 1+\sin \left ( dx+c \right ) \right ) }{d{a}^{2}}}-{\frac{1}{3\,d{a}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{3}}}+{\frac{1}{d{a}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{2}}}-3\,{\frac{1}{d{a}^{2}\sin \left ( dx+c \right ) }}-4\,{\frac{\ln \left ( \sin \left ( dx+c \right ) \right ) }{d{a}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)*csc(d*x+c)^4/(a+a*sin(d*x+c))^2,x)

[Out]

-1/d/a^2/(1+sin(d*x+c))+4*ln(1+sin(d*x+c))/a^2/d-1/3/d/a^2/sin(d*x+c)^3+1/d/a^2/sin(d*x+c)^2-3/d/a^2/sin(d*x+c
)-4*ln(sin(d*x+c))/a^2/d

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Maxima [A]  time = 1.11096, size = 122, normalized size = 1.21 \begin{align*} -\frac{\frac{12 \, \sin \left (d x + c\right )^{3} + 6 \, \sin \left (d x + c\right )^{2} - 2 \, \sin \left (d x + c\right ) + 1}{a^{2} \sin \left (d x + c\right )^{4} + a^{2} \sin \left (d x + c\right )^{3}} - \frac{12 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{2}} + \frac{12 \, \log \left (\sin \left (d x + c\right )\right )}{a^{2}}}{3 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)*csc(d*x+c)^4/(a+a*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.53725, size = 508, normalized size = 5.03 \begin{align*} \frac{6 \, \cos \left (d x + c\right )^{2} - 12 \,{\left (\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} -{\left (\cos \left (d x + c\right )^{2} - 1\right )} \sin \left (d x + c\right ) + 1\right )} \log \left (\frac{1}{2} \, \sin \left (d x + c\right )\right ) + 12 \,{\left (\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} -{\left (\cos \left (d x + c\right )^{2} - 1\right )} \sin \left (d x + c\right ) + 1\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (6 \, \cos \left (d x + c\right )^{2} - 5\right )} \sin \left (d x + c\right ) - 7}{3 \,{\left (a^{2} d \cos \left (d x + c\right )^{4} - 2 \, a^{2} d \cos \left (d x + c\right )^{2} + a^{2} d -{\left (a^{2} d \cos \left (d x + c\right )^{2} - a^{2} d\right )} \sin \left (d x + c\right )\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)*csc(d*x+c)^4/(a+a*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

1/3*(6*cos(d*x + c)^2 - 12*(cos(d*x + c)^4 - 2*cos(d*x + c)^2 - (cos(d*x + c)^2 - 1)*sin(d*x + c) + 1)*log(1/2
*sin(d*x + c)) + 12*(cos(d*x + c)^4 - 2*cos(d*x + c)^2 - (cos(d*x + c)^2 - 1)*sin(d*x + c) + 1)*log(sin(d*x +
c) + 1) + 2*(6*cos(d*x + c)^2 - 5)*sin(d*x + c) - 7)/(a^2*d*cos(d*x + c)^4 - 2*a^2*d*cos(d*x + c)^2 + a^2*d -
(a^2*d*cos(d*x + c)^2 - a^2*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)*csc(d*x+c)**4/(a+a*sin(d*x+c))**2,x)

[Out]

Timed out

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Giac [A]  time = 1.24538, size = 139, normalized size = 1.38 \begin{align*} -\frac{\frac{12 \, \log \left ({\left | -\frac{a}{a \sin \left (d x + c\right ) + a} + 1 \right |}\right )}{a^{2}} + \frac{3}{{\left (a \sin \left (d x + c\right ) + a\right )} a} + \frac{\frac{30 \, a}{a \sin \left (d x + c\right ) + a} - \frac{18 \, a^{2}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{2}} - 13}{a^{2}{\left (\frac{a}{a \sin \left (d x + c\right ) + a} - 1\right )}^{3}}}{3 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)*csc(d*x+c)^4/(a+a*sin(d*x+c))^2,x, algorithm="giac")

[Out]

-1/3*(12*log(abs(-a/(a*sin(d*x + c) + a) + 1))/a^2 + 3/((a*sin(d*x + c) + a)*a) + (30*a/(a*sin(d*x + c) + a) -
18*a^2/(a*sin(d*x + c) + a)^2 - 13)/(a^2*(a/(a*sin(d*x + c) + a) - 1)^3))/d