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

Optimal. Leaf size=108 \[ \frac {3}{d \left (a^3 \sin (c+d x)+a^3\right )}-\frac {\csc ^2(c+d x)}{2 a^3 d}+\frac {3 \csc (c+d x)}{a^3 d}+\frac {6 \log (\sin (c+d x))}{a^3 d}-\frac {6 \log (\sin (c+d x)+1)}{a^3 d}+\frac {1}{2 a d (a \sin (c+d x)+a)^2} \]

[Out]

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

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Rubi [A]  time = 0.10, antiderivative size = 108, normalized size of antiderivative = 1.00, 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} \[ \frac {3}{d \left (a^3 \sin (c+d x)+a^3\right )}-\frac {\csc ^2(c+d x)}{2 a^3 d}+\frac {3 \csc (c+d x)}{a^3 d}+\frac {6 \log (\sin (c+d x))}{a^3 d}-\frac {6 \log (\sin (c+d x)+1)}{a^3 d}+\frac {1}{2 a d (a \sin (c+d x)+a)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

Rubi steps

\begin {align*} \int \frac {\cot (c+d x) \csc ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {a^3}{x^3 (a+x)^3} \, dx,x,a \sin (c+d x)\right )}{a d}\\ &=\frac {a^2 \operatorname {Subst}\left (\int \frac {1}{x^3 (a+x)^3} \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac {a^2 \operatorname {Subst}\left (\int \left (\frac {1}{a^3 x^3}-\frac {3}{a^4 x^2}+\frac {6}{a^5 x}-\frac {1}{a^3 (a+x)^3}-\frac {3}{a^4 (a+x)^2}-\frac {6}{a^5 (a+x)}\right ) \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac {3 \csc (c+d x)}{a^3 d}-\frac {\csc ^2(c+d x)}{2 a^3 d}+\frac {6 \log (\sin (c+d x))}{a^3 d}-\frac {6 \log (1+\sin (c+d x))}{a^3 d}+\frac {1}{2 a d (a+a \sin (c+d x))^2}+\frac {3}{d \left (a^3+a^3 \sin (c+d x)\right )}\\ \end {align*}

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Mathematica [A]  time = 0.57, size = 71, normalized size = 0.66 \[ \frac {\frac {6}{\sin (c+d x)+1}+\frac {1}{(\sin (c+d x)+1)^2}-\csc ^2(c+d x)+6 \csc (c+d x)+12 \log (\sin (c+d x))-12 \log (\sin (c+d x)+1)}{2 a^3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(6*Csc[c + d*x] - Csc[c + d*x]^2 + 12*Log[Sin[c + d*x]] - 12*Log[1 + Sin[c + d*x]] + (1 + Sin[c + d*x])^(-2) +
6/(1 + Sin[c + d*x]))/(2*a^3*d)

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fricas [A]  time = 0.51, size = 196, normalized size = 1.81 \[ -\frac {18 \, \cos \left (d x + c\right )^{2} - 12 \, {\left (\cos \left (d x + c\right )^{4} - 3 \, \cos \left (d x + c\right )^{2} - 2 \, {\left (\cos \left (d x + c\right )^{2} - 1\right )} \sin \left (d x + c\right ) + 2\right )} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) + 12 \, {\left (\cos \left (d x + c\right )^{4} - 3 \, \cos \left (d x + c\right )^{2} - 2 \, {\left (\cos \left (d x + c\right )^{2} - 1\right )} \sin \left (d x + c\right ) + 2\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + 4 \, {\left (3 \, \cos \left (d x + c\right )^{2} - 4\right )} \sin \left (d x + c\right ) - 17}{2 \, {\left (a^{3} d \cos \left (d x + c\right )^{4} - 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 2 \, a^{3} d - 2 \, {\left (a^{3} d \cos \left (d x + c\right )^{2} - a^{3} d\right )} \sin \left (d x + c\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(18*cos(d*x + c)^2 - 12*(cos(d*x + c)^4 - 3*cos(d*x + c)^2 - 2*(cos(d*x + c)^2 - 1)*sin(d*x + c) + 2)*log
(1/2*sin(d*x + c)) + 12*(cos(d*x + c)^4 - 3*cos(d*x + c)^2 - 2*(cos(d*x + c)^2 - 1)*sin(d*x + c) + 2)*log(sin(
d*x + c) + 1) + 4*(3*cos(d*x + c)^2 - 4)*sin(d*x + c) - 17)/(a^3*d*cos(d*x + c)^4 - 3*a^3*d*cos(d*x + c)^2 + 2
*a^3*d - 2*(a^3*d*cos(d*x + c)^2 - a^3*d)*sin(d*x + c))

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giac [A]  time = 0.19, size = 86, normalized size = 0.80 \[ -\frac {\frac {12 \, \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a^{3}} - \frac {12 \, \log \left ({\left | \sin \left (d x + c\right ) \right |}\right )}{a^{3}} - \frac {12 \, \sin \left (d x + c\right )^{3} + 18 \, \sin \left (d x + c\right )^{2} + 4 \, \sin \left (d x + c\right ) - 1}{{\left (\sin \left (d x + c\right )^{2} + \sin \left (d x + c\right )\right )}^{2} a^{3}}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(12*log(abs(sin(d*x + c) + 1))/a^3 - 12*log(abs(sin(d*x + c)))/a^3 - (12*sin(d*x + c)^3 + 18*sin(d*x + c)
^2 + 4*sin(d*x + c) - 1)/((sin(d*x + c)^2 + sin(d*x + c))^2*a^3))/d

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maple [A]  time = 0.31, size = 102, normalized size = 0.94 \[ -\frac {1}{2 a^{3} d \sin \left (d x +c \right )^{2}}+\frac {3}{a^{3} d \sin \left (d x +c \right )}+\frac {6 \ln \left (\sin \left (d x +c \right )\right )}{a^{3} d}+\frac {1}{2 d \,a^{3} \left (1+\sin \left (d x +c \right )\right )^{2}}+\frac {3}{d \,a^{3} \left (1+\sin \left (d x +c \right )\right )}-\frac {6 \ln \left (1+\sin \left (d x +c \right )\right )}{a^{3} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.31, size = 103, normalized size = 0.95 \[ \frac {\frac {12 \, \sin \left (d x + c\right )^{3} + 18 \, \sin \left (d x + c\right )^{2} + 4 \, \sin \left (d x + c\right ) - 1}{a^{3} \sin \left (d x + c\right )^{4} + 2 \, a^{3} \sin \left (d x + c\right )^{3} + a^{3} \sin \left (d x + c\right )^{2}} - \frac {12 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{3}} + \frac {12 \, \log \left (\sin \left (d x + c\right )\right )}{a^{3}}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*((12*sin(d*x + c)^3 + 18*sin(d*x + c)^2 + 4*sin(d*x + c) - 1)/(a^3*sin(d*x + c)^4 + 2*a^3*sin(d*x + c)^3 +
a^3*sin(d*x + c)^2) - 12*log(sin(d*x + c) + 1)/a^3 + 12*log(sin(d*x + c))/a^3)/d

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mupad [B]  time = 8.68, size = 227, normalized size = 2.10 \[ \frac {6\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a^3\,d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{8\,a^3\,d}+\frac {-26\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5-\frac {65\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{2}+2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+21\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-\frac {1}{2}}{d\,\left (4\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+16\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+24\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+16\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+4\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\right )}-\frac {12\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )}{a^3\,d}+\frac {3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,a^3\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(6*log(tan(c/2 + (d*x)/2)))/(a^3*d) - tan(c/2 + (d*x)/2)^2/(8*a^3*d) + (4*tan(c/2 + (d*x)/2) + 21*tan(c/2 + (d
*x)/2)^2 + 2*tan(c/2 + (d*x)/2)^3 - (65*tan(c/2 + (d*x)/2)^4)/2 - 26*tan(c/2 + (d*x)/2)^5 - 1/2)/(d*(4*a^3*tan
(c/2 + (d*x)/2)^2 + 16*a^3*tan(c/2 + (d*x)/2)^3 + 24*a^3*tan(c/2 + (d*x)/2)^4 + 16*a^3*tan(c/2 + (d*x)/2)^5 +
4*a^3*tan(c/2 + (d*x)/2)^6)) - (12*log(tan(c/2 + (d*x)/2) + 1))/(a^3*d) + (3*tan(c/2 + (d*x)/2))/(2*a^3*d)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {\cos {\left (c + d x \right )} \csc ^{3}{\left (c + d x \right )}}{\sin ^{3}{\left (c + d x \right )} + 3 \sin ^{2}{\left (c + d x \right )} + 3 \sin {\left (c + d x \right )} + 1}\, dx}{a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)*csc(d*x+c)**3/(a+a*sin(d*x+c))**3,x)

[Out]

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

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