3.255 \(\int \frac{\cot (c+d x) \csc (c+d x)}{(a+a \sin (c+d x))^4} \, dx\)

Optimal. Leaf size=111 \[ -\frac{3}{d \left (a^4 \sin (c+d x)+a^4\right )}-\frac{1}{d \left (a^2 \sin (c+d x)+a^2\right )^2}-\frac{\csc (c+d x)}{a^4 d}-\frac{4 \log (\sin (c+d x))}{a^4 d}+\frac{4 \log (\sin (c+d x)+1)}{a^4 d}-\frac{1}{3 a d (a \sin (c+d x)+a)^3} \]

[Out]

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

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Rubi [A]  time = 0.0936986, antiderivative size = 111, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.12, Rules used = {2833, 12, 44} \[ -\frac{3}{d \left (a^4 \sin (c+d x)+a^4\right )}-\frac{1}{d \left (a^2 \sin (c+d x)+a^2\right )^2}-\frac{\csc (c+d x)}{a^4 d}-\frac{4 \log (\sin (c+d x))}{a^4 d}+\frac{4 \log (\sin (c+d x)+1)}{a^4 d}-\frac{1}{3 a d (a \sin (c+d x)+a)^3} \]

Antiderivative was successfully verified.

[In]

Int[(Cot[c + d*x]*Csc[c + d*x])/(a + a*Sin[c + d*x])^4,x]

[Out]

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

Mathematica [A]  time = 0.985034, size = 73, normalized size = 0.66 \[ -\frac{\frac{9}{\sin (c+d x)+1}+\frac{3}{(\sin (c+d x)+1)^2}+\frac{1}{(\sin (c+d x)+1)^3}+3 \csc (c+d x)+12 \log (\sin (c+d x))-12 \log (\sin (c+d x)+1)}{3 a^4 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cot[c + d*x]*Csc[c + d*x])/(a + a*Sin[c + d*x])^4,x]

[Out]

-(3*Csc[c + d*x] + 12*Log[Sin[c + d*x]] - 12*Log[1 + Sin[c + d*x]] + (1 + Sin[c + d*x])^(-3) + 3/(1 + Sin[c +
d*x])^2 + 9/(1 + Sin[c + d*x]))/(3*a^4*d)

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Maple [A]  time = 0.05, size = 104, normalized size = 0.9 \begin{align*} -{\frac{1}{3\,d{a}^{4} \left ( 1+\sin \left ( dx+c \right ) \right ) ^{3}}}-{\frac{1}{d{a}^{4} \left ( 1+\sin \left ( dx+c \right ) \right ) ^{2}}}-3\,{\frac{1}{d{a}^{4} \left ( 1+\sin \left ( dx+c \right ) \right ) }}+4\,{\frac{\ln \left ( 1+\sin \left ( dx+c \right ) \right ) }{d{a}^{4}}}-{\frac{1}{d{a}^{4}\sin \left ( dx+c \right ) }}-4\,{\frac{\ln \left ( \sin \left ( dx+c \right ) \right ) }{d{a}^{4}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.05201, size = 154, normalized size = 1.39 \begin{align*} -\frac{\frac{12 \, \sin \left (d x + c\right )^{3} + 30 \, \sin \left (d x + c\right )^{2} + 22 \, \sin \left (d x + c\right ) + 3}{a^{4} \sin \left (d x + c\right )^{4} + 3 \, a^{4} \sin \left (d x + c\right )^{3} + 3 \, a^{4} \sin \left (d x + c\right )^{2} + a^{4} \sin \left (d x + c\right )} - \frac{12 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{4}} + \frac{12 \, \log \left (\sin \left (d x + c\right )\right )}{a^{4}}}{3 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*((12*sin(d*x + c)^3 + 30*sin(d*x + c)^2 + 22*sin(d*x + c) + 3)/(a^4*sin(d*x + c)^4 + 3*a^4*sin(d*x + c)^3
+ 3*a^4*sin(d*x + c)^2 + a^4*sin(d*x + c)) - 12*log(sin(d*x + c) + 1)/a^4 + 12*log(sin(d*x + c))/a^4)/d

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Fricas [A]  time = 1.44199, size = 525, normalized size = 4.73 \begin{align*} \frac{30 \, \cos \left (d x + c\right )^{2} - 12 \,{\left (\cos \left (d x + c\right )^{4} - 5 \, \cos \left (d x + c\right )^{2} -{\left (3 \, \cos \left (d x + c\right )^{2} - 4\right )} \sin \left (d x + c\right ) + 4\right )} \log \left (\frac{1}{2} \, \sin \left (d x + c\right )\right ) + 12 \,{\left (\cos \left (d x + c\right )^{4} - 5 \, \cos \left (d x + c\right )^{2} -{\left (3 \, \cos \left (d x + c\right )^{2} - 4\right )} \sin \left (d x + c\right ) + 4\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (6 \, \cos \left (d x + c\right )^{2} - 17\right )} \sin \left (d x + c\right ) - 33}{3 \,{\left (a^{4} d \cos \left (d x + c\right )^{4} - 5 \, a^{4} d \cos \left (d x + c\right )^{2} + 4 \, a^{4} d -{\left (3 \, a^{4} d \cos \left (d x + c\right )^{2} - 4 \, a^{4} 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)^2/(a+a*sin(d*x+c))^4,x, algorithm="fricas")

[Out]

1/3*(30*cos(d*x + c)^2 - 12*(cos(d*x + c)^4 - 5*cos(d*x + c)^2 - (3*cos(d*x + c)^2 - 4)*sin(d*x + c) + 4)*log(
1/2*sin(d*x + c)) + 12*(cos(d*x + c)^4 - 5*cos(d*x + c)^2 - (3*cos(d*x + c)^2 - 4)*sin(d*x + c) + 4)*log(sin(d
*x + c) + 1) + 2*(6*cos(d*x + c)^2 - 17)*sin(d*x + c) - 33)/(a^4*d*cos(d*x + c)^4 - 5*a^4*d*cos(d*x + c)^2 + 4
*a^4*d - (3*a^4*d*cos(d*x + c)^2 - 4*a^4*d)*sin(d*x + c))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{\cos{\left (c + d x \right )} \csc ^{2}{\left (c + d x \right )}}{\sin ^{4}{\left (c + d x \right )} + 4 \sin ^{3}{\left (c + d x \right )} + 6 \sin ^{2}{\left (c + d x \right )} + 4 \sin{\left (c + d x \right )} + 1}\, dx}{a^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(cos(c + d*x)*csc(c + d*x)**2/(sin(c + d*x)**4 + 4*sin(c + d*x)**3 + 6*sin(c + d*x)**2 + 4*sin(c + d*x
) + 1), x)/a**4

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Giac [A]  time = 1.2458, size = 117, normalized size = 1.05 \begin{align*} \frac{\frac{12 \, \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a^{4}} - \frac{12 \, \log \left ({\left | \sin \left (d x + c\right ) \right |}\right )}{a^{4}} - \frac{12 \, \sin \left (d x + c\right )^{3} + 30 \, \sin \left (d x + c\right )^{2} + 22 \, \sin \left (d x + c\right ) + 3}{a^{4}{\left (\sin \left (d x + c\right ) + 1\right )}^{3} \sin \left (d x + c\right )}}{3 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(12*log(abs(sin(d*x + c) + 1))/a^4 - 12*log(abs(sin(d*x + c)))/a^4 - (12*sin(d*x + c)^3 + 30*sin(d*x + c)^
2 + 22*sin(d*x + c) + 3)/(a^4*(sin(d*x + c) + 1)^3*sin(d*x + c)))/d