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

Optimal. Leaf size=126 \[ -\frac{4}{d \left (a^3 \sin (c+d x)+a^3\right )}-\frac{\csc ^3(c+d x)}{3 a^3 d}+\frac{3 \csc ^2(c+d x)}{2 a^3 d}-\frac{6 \csc (c+d x)}{a^3 d}-\frac{10 \log (\sin (c+d x))}{a^3 d}+\frac{10 \log (\sin (c+d x)+1)}{a^3 d}-\frac{1}{2 a d (a \sin (c+d x)+a)^2} \]

[Out]

(-6*Csc[c + d*x])/(a^3*d) + (3*Csc[c + d*x]^2)/(2*a^3*d) - Csc[c + d*x]^3/(3*a^3*d) - (10*Log[Sin[c + d*x]])/(
a^3*d) + (10*Log[1 + Sin[c + d*x]])/(a^3*d) - 1/(2*a*d*(a + a*Sin[c + d*x])^2) - 4/(d*(a^3 + a^3*Sin[c + d*x])
)

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Rubi [A]  time = 0.112726, antiderivative size = 126, 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} \[ -\frac{4}{d \left (a^3 \sin (c+d x)+a^3\right )}-\frac{\csc ^3(c+d x)}{3 a^3 d}+\frac{3 \csc ^2(c+d x)}{2 a^3 d}-\frac{6 \csc (c+d x)}{a^3 d}-\frac{10 \log (\sin (c+d x))}{a^3 d}+\frac{10 \log (\sin (c+d x)+1)}{a^3 d}-\frac{1}{2 a d (a \sin (c+d x)+a)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-6*Csc[c + d*x])/(a^3*d) + (3*Csc[c + d*x]^2)/(2*a^3*d) - Csc[c + d*x]^3/(3*a^3*d) - (10*Log[Sin[c + d*x]])/(
a^3*d) + (10*Log[1 + Sin[c + d*x]])/(a^3*d) - 1/(2*a*d*(a + a*Sin[c + d*x])^2) - 4/(d*(a^3 + a^3*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))^3} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{a^4}{x^4 (a+x)^3} \, dx,x,a \sin (c+d x)\right )}{a d}\\ &=\frac{a^3 \operatorname{Subst}\left (\int \frac{1}{x^4 (a+x)^3} \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac{a^3 \operatorname{Subst}\left (\int \left (\frac{1}{a^3 x^4}-\frac{3}{a^4 x^3}+\frac{6}{a^5 x^2}-\frac{10}{a^6 x}+\frac{1}{a^4 (a+x)^3}+\frac{4}{a^5 (a+x)^2}+\frac{10}{a^6 (a+x)}\right ) \, dx,x,a \sin (c+d x)\right )}{d}\\ &=-\frac{6 \csc (c+d x)}{a^3 d}+\frac{3 \csc ^2(c+d x)}{2 a^3 d}-\frac{\csc ^3(c+d x)}{3 a^3 d}-\frac{10 \log (\sin (c+d x))}{a^3 d}+\frac{10 \log (1+\sin (c+d x))}{a^3 d}-\frac{1}{2 a d (a+a \sin (c+d x))^2}-\frac{4}{d \left (a^3+a^3 \sin (c+d x)\right )}\\ \end{align*}

Mathematica [A]  time = 5.7232, size = 81, normalized size = 0.64 \[ -\frac{\frac{3 (8 \sin (c+d x)+9)}{(\sin (c+d x)+1)^2}+2 \csc ^3(c+d x)-9 \csc ^2(c+d x)+36 \csc (c+d x)+60 \log (\sin (c+d x))-60 \log (\sin (c+d x)+1)}{6 a^3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(36*Csc[c + d*x] - 9*Csc[c + d*x]^2 + 2*Csc[c + d*x]^3 + 60*Log[Sin[c + d*x]] - 60*Log[1 + Sin[c + d*x]] + (3
*(9 + 8*Sin[c + d*x]))/(1 + Sin[c + d*x])^2)/(6*a^3*d)

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Maple [A]  time = 0.062, size = 118, normalized size = 0.9 \begin{align*} -{\frac{1}{2\,{a}^{3}d \left ( 1+\sin \left ( dx+c \right ) \right ) ^{2}}}-4\,{\frac{1}{{a}^{3}d \left ( 1+\sin \left ( dx+c \right ) \right ) }}+10\,{\frac{\ln \left ( 1+\sin \left ( dx+c \right ) \right ) }{{a}^{3}d}}-{\frac{1}{3\,{a}^{3}d \left ( \sin \left ( dx+c \right ) \right ) ^{3}}}+{\frac{3}{2\,{a}^{3}d \left ( \sin \left ( dx+c \right ) \right ) ^{2}}}-6\,{\frac{1}{{a}^{3}d\sin \left ( dx+c \right ) }}-10\,{\frac{\ln \left ( \sin \left ( dx+c \right ) \right ) }{{a}^{3}d}} \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))^3,x)

[Out]

-1/2/d/a^3/(1+sin(d*x+c))^2-4/d/a^3/(1+sin(d*x+c))+10*ln(1+sin(d*x+c))/a^3/d-1/3/d/a^3/sin(d*x+c)^3+3/2/d/a^3/
sin(d*x+c)^2-6/d/a^3/sin(d*x+c)-10*ln(sin(d*x+c))/a^3/d

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Maxima [A]  time = 0.984024, size = 153, normalized size = 1.21 \begin{align*} -\frac{\frac{60 \, \sin \left (d x + c\right )^{4} + 90 \, \sin \left (d x + c\right )^{3} + 20 \, \sin \left (d x + c\right )^{2} - 5 \, \sin \left (d x + c\right ) + 2}{a^{3} \sin \left (d x + c\right )^{5} + 2 \, a^{3} \sin \left (d x + c\right )^{4} + a^{3} \sin \left (d x + c\right )^{3}} - \frac{60 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{3}} + \frac{60 \, \log \left (\sin \left (d x + c\right )\right )}{a^{3}}}{6 \, 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))^3,x, algorithm="maxima")

[Out]

-1/6*((60*sin(d*x + c)^4 + 90*sin(d*x + c)^3 + 20*sin(d*x + c)^2 - 5*sin(d*x + c) + 2)/(a^3*sin(d*x + c)^5 + 2
*a^3*sin(d*x + c)^4 + a^3*sin(d*x + c)^3) - 60*log(sin(d*x + c) + 1)/a^3 + 60*log(sin(d*x + c))/a^3)/d

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Fricas [B]  time = 1.54369, size = 641, normalized size = 5.09 \begin{align*} -\frac{60 \, \cos \left (d x + c\right )^{4} - 140 \, \cos \left (d x + c\right )^{2} + 60 \,{\left (2 \, \cos \left (d x + c\right )^{4} - 4 \, \cos \left (d x + c\right )^{2} +{\left (\cos \left (d x + c\right )^{4} - 3 \, \cos \left (d x + c\right )^{2} + 2\right )} \sin \left (d x + c\right ) + 2\right )} \log \left (\frac{1}{2} \, \sin \left (d x + c\right )\right ) - 60 \,{\left (2 \, \cos \left (d x + c\right )^{4} - 4 \, \cos \left (d x + c\right )^{2} +{\left (\cos \left (d x + c\right )^{4} - 3 \, \cos \left (d x + c\right )^{2} + 2\right )} \sin \left (d x + c\right ) + 2\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 5 \,{\left (18 \, \cos \left (d x + c\right )^{2} - 17\right )} \sin \left (d x + c\right ) + 82}{6 \,{\left (2 \, a^{3} d \cos \left (d x + c\right )^{4} - 4 \, a^{3} d \cos \left (d x + c\right )^{2} + 2 \, a^{3} d +{\left (a^{3} d \cos \left (d x + c\right )^{4} - 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 2 \, a^{3} 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))^3,x, algorithm="fricas")

[Out]

-1/6*(60*cos(d*x + c)^4 - 140*cos(d*x + c)^2 + 60*(2*cos(d*x + c)^4 - 4*cos(d*x + c)^2 + (cos(d*x + c)^4 - 3*c
os(d*x + c)^2 + 2)*sin(d*x + c) + 2)*log(1/2*sin(d*x + c)) - 60*(2*cos(d*x + c)^4 - 4*cos(d*x + c)^2 + (cos(d*
x + c)^4 - 3*cos(d*x + c)^2 + 2)*sin(d*x + c) + 2)*log(sin(d*x + c) + 1) - 5*(18*cos(d*x + c)^2 - 17)*sin(d*x
+ c) + 82)/(2*a^3*d*cos(d*x + c)^4 - 4*a^3*d*cos(d*x + c)^2 + 2*a^3*d + (a^3*d*cos(d*x + c)^4 - 3*a^3*d*cos(d*
x + c)^2 + 2*a^3*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))**3,x)

[Out]

Timed out

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Giac [A]  time = 1.2696, size = 131, normalized size = 1.04 \begin{align*} \frac{\frac{60 \, \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a^{3}} - \frac{60 \, \log \left ({\left | \sin \left (d x + c\right ) \right |}\right )}{a^{3}} - \frac{60 \, \sin \left (d x + c\right )^{4} + 90 \, \sin \left (d x + c\right )^{3} + 20 \, \sin \left (d x + c\right )^{2} - 5 \, \sin \left (d x + c\right ) + 2}{a^{3}{\left (\sin \left (d x + c\right ) + 1\right )}^{2} \sin \left (d x + c\right )^{3}}}{6 \, 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))^3,x, algorithm="giac")

[Out]

1/6*(60*log(abs(sin(d*x + c) + 1))/a^3 - 60*log(abs(sin(d*x + c)))/a^3 - (60*sin(d*x + c)^4 + 90*sin(d*x + c)^
3 + 20*sin(d*x + c)^2 - 5*sin(d*x + c) + 2)/(a^3*(sin(d*x + c) + 1)^2*sin(d*x + c)^3))/d