3.85 \(\int \frac {\cot ^3(c+d x)}{(a+a \sin (c+d x))^4} \, dx\)

Optimal. Leaf size=106 \[ \frac {5}{d \left (a^4 \sin (c+d x)+a^4\right )}-\frac {\csc ^2(c+d x)}{2 a^4 d}+\frac {4 \csc (c+d x)}{a^4 d}+\frac {9 \log (\sin (c+d x))}{a^4 d}-\frac {9 \log (\sin (c+d x)+1)}{a^4 d}+\frac {1}{d \left (a^2 \sin (c+d x)+a^2\right )^2} \]

[Out]

4*csc(d*x+c)/a^4/d-1/2*csc(d*x+c)^2/a^4/d+9*ln(sin(d*x+c))/a^4/d-9*ln(1+sin(d*x+c))/a^4/d+1/d/(a^2+a^2*sin(d*x
+c))^2+5/d/(a^4+a^4*sin(d*x+c))

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Rubi [A]  time = 0.08, antiderivative size = 106, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {2707, 77} \[ \frac {5}{d \left (a^4 \sin (c+d x)+a^4\right )}+\frac {1}{d \left (a^2 \sin (c+d x)+a^2\right )^2}-\frac {\csc ^2(c+d x)}{2 a^4 d}+\frac {4 \csc (c+d x)}{a^4 d}+\frac {9 \log (\sin (c+d x))}{a^4 d}-\frac {9 \log (\sin (c+d x)+1)}{a^4 d} \]

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]^3/(a + a*Sin[c + d*x])^4,x]

[Out]

(4*Csc[c + d*x])/(a^4*d) - Csc[c + d*x]^2/(2*a^4*d) + (9*Log[Sin[c + d*x]])/(a^4*d) - (9*Log[1 + Sin[c + d*x]]
)/(a^4*d) + 1/(d*(a^2 + a^2*Sin[c + d*x])^2) + 5/(d*(a^4 + a^4*Sin[c + d*x]))

Rule 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rule 2707

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(p_.), x_Symbol] :> Dist[1/f, Subst[I
nt[(x^p*(a + x)^(m - (p + 1)/2))/(a - x)^((p + 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x]
&& EqQ[a^2 - b^2, 0] && IntegerQ[(p + 1)/2]

Rubi steps

\begin {align*} \int \frac {\cot ^3(c+d x)}{(a+a \sin (c+d x))^4} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {a-x}{x^3 (a+x)^3} \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac {\operatorname {Subst}\left (\int \left (\frac {1}{a^2 x^3}-\frac {4}{a^3 x^2}+\frac {9}{a^4 x}-\frac {2}{a^2 (a+x)^3}-\frac {5}{a^3 (a+x)^2}-\frac {9}{a^4 (a+x)}\right ) \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac {4 \csc (c+d x)}{a^4 d}-\frac {\csc ^2(c+d x)}{2 a^4 d}+\frac {9 \log (\sin (c+d x))}{a^4 d}-\frac {9 \log (1+\sin (c+d x))}{a^4 d}+\frac {1}{d \left (a^2+a^2 \sin (c+d x)\right )^2}+\frac {5}{d \left (a^4+a^4 \sin (c+d x)\right )}\\ \end {align*}

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Mathematica [A]  time = 0.80, size = 73, normalized size = 0.69 \[ \frac {\frac {10}{\sin (c+d x)+1}+\frac {2}{(\sin (c+d x)+1)^2}-\csc ^2(c+d x)+8 \csc (c+d x)+18 \log (\sin (c+d x))-18 \log (\sin (c+d x)+1)}{2 a^4 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[c + d*x]^3/(a + a*Sin[c + d*x])^4,x]

[Out]

(8*Csc[c + d*x] - Csc[c + d*x]^2 + 18*Log[Sin[c + d*x]] - 18*Log[1 + Sin[c + d*x]] + 2/(1 + Sin[c + d*x])^2 +
10/(1 + Sin[c + d*x]))/(2*a^4*d)

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fricas [A]  time = 0.45, size = 196, normalized size = 1.85 \[ -\frac {27 \, \cos \left (d x + c\right )^{2} - 18 \, {\left (\cos \left (d x + c\right )^{4} - 3 \, \cos \left (d x + c\right )^{2} - 2 \, {\left (\cos \left (d x + c\right )^{2} - 1\right )} \sin \left (d x + c\right ) + 2\right )} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) + 18 \, {\left (\cos \left (d x + c\right )^{4} - 3 \, \cos \left (d x + c\right )^{2} - 2 \, {\left (\cos \left (d x + c\right )^{2} - 1\right )} \sin \left (d x + c\right ) + 2\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + 6 \, {\left (3 \, \cos \left (d x + c\right )^{2} - 4\right )} \sin \left (d x + c\right ) - 26}{2 \, {\left (a^{4} d \cos \left (d x + c\right )^{4} - 3 \, a^{4} d \cos \left (d x + c\right )^{2} + 2 \, a^{4} d - 2 \, {\left (a^{4} d \cos \left (d x + c\right )^{2} - a^{4} d\right )} \sin \left (d x + c\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^3/(a+a*sin(d*x+c))^4,x, algorithm="fricas")

[Out]

-1/2*(27*cos(d*x + c)^2 - 18*(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)) + 18*(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) + 6*(3*cos(d*x + c)^2 - 4)*sin(d*x + c) - 26)/(a^4*d*cos(d*x + c)^4 - 3*a^4*d*cos(d*x + c)^2 + 2
*a^4*d - 2*(a^4*d*cos(d*x + c)^2 - a^4*d)*sin(d*x + c))

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giac [A]  time = 1.18, size = 185, normalized size = 1.75 \[ -\frac {\frac {144 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{4}} - \frac {72 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a^{4}} + \frac {108 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 16 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1}{a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}} + \frac {a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 16 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{8}} - \frac {4 \, {\left (75 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 272 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 402 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 272 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 75\right )}}{a^{4} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}^{4}}}{8 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^3/(a+a*sin(d*x+c))^4,x, algorithm="giac")

[Out]

-1/8*(144*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^4 - 72*log(abs(tan(1/2*d*x + 1/2*c)))/a^4 + (108*tan(1/2*d*x +
1/2*c)^2 - 16*tan(1/2*d*x + 1/2*c) + 1)/(a^4*tan(1/2*d*x + 1/2*c)^2) + (a^4*tan(1/2*d*x + 1/2*c)^2 - 16*a^4*ta
n(1/2*d*x + 1/2*c))/a^8 - 4*(75*tan(1/2*d*x + 1/2*c)^4 + 272*tan(1/2*d*x + 1/2*c)^3 + 402*tan(1/2*d*x + 1/2*c)
^2 + 272*tan(1/2*d*x + 1/2*c) + 75)/(a^4*(tan(1/2*d*x + 1/2*c) + 1)^4))/d

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maple [A]  time = 0.34, size = 101, normalized size = 0.95 \[ -\frac {1}{2 a^{4} d \sin \left (d x +c \right )^{2}}+\frac {4}{a^{4} d \sin \left (d x +c \right )}+\frac {9 \ln \left (\sin \left (d x +c \right )\right )}{a^{4} d}+\frac {1}{a^{4} d \left (1+\sin \left (d x +c \right )\right )^{2}}+\frac {5}{a^{4} d \left (1+\sin \left (d x +c \right )\right )}-\frac {9 \ln \left (1+\sin \left (d x +c \right )\right )}{a^{4} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(d*x+c)^3/(a+a*sin(d*x+c))^4,x)

[Out]

-1/2/a^4/d/sin(d*x+c)^2+4/a^4/d/sin(d*x+c)+9*ln(sin(d*x+c))/a^4/d+1/a^4/d/(1+sin(d*x+c))^2+5/a^4/d/(1+sin(d*x+
c))-9*ln(1+sin(d*x+c))/a^4/d

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maxima [A]  time = 0.33, size = 103, normalized size = 0.97 \[ \frac {\frac {18 \, \sin \left (d x + c\right )^{3} + 27 \, \sin \left (d x + c\right )^{2} + 6 \, \sin \left (d x + c\right ) - 1}{a^{4} \sin \left (d x + c\right )^{4} + 2 \, a^{4} \sin \left (d x + c\right )^{3} + a^{4} \sin \left (d x + c\right )^{2}} - \frac {18 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{4}} + \frac {18 \, \log \left (\sin \left (d x + c\right )\right )}{a^{4}}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^3/(a+a*sin(d*x+c))^4,x, algorithm="maxima")

[Out]

1/2*((18*sin(d*x + c)^3 + 27*sin(d*x + c)^2 + 6*sin(d*x + c) - 1)/(a^4*sin(d*x + c)^4 + 2*a^4*sin(d*x + c)^3 +
a^4*sin(d*x + c)^2) - 18*log(sin(d*x + c) + 1)/a^4 + 18*log(sin(d*x + c))/a^4)/d

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mupad [B]  time = 6.61, size = 228, normalized size = 2.15 \[ \frac {9\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a^4\,d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{8\,a^4\,d}-\frac {48\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\frac {129\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{2}+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-29\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-6\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+\frac {1}{2}}{d\,\left (4\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+16\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+24\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+16\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+4\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\right )}-\frac {18\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )}{a^4\,d}+\frac {2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{a^4\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(c + d*x)^3/(a + a*sin(c + d*x))^4,x)

[Out]

(9*log(tan(c/2 + (d*x)/2)))/(a^4*d) - tan(c/2 + (d*x)/2)^2/(8*a^4*d) - (10*tan(c/2 + (d*x)/2)^3 - 29*tan(c/2 +
(d*x)/2)^2 - 6*tan(c/2 + (d*x)/2) + (129*tan(c/2 + (d*x)/2)^4)/2 + 48*tan(c/2 + (d*x)/2)^5 + 1/2)/(d*(4*a^4*t
an(c/2 + (d*x)/2)^2 + 16*a^4*tan(c/2 + (d*x)/2)^3 + 24*a^4*tan(c/2 + (d*x)/2)^4 + 16*a^4*tan(c/2 + (d*x)/2)^5
+ 4*a^4*tan(c/2 + (d*x)/2)^6)) - (18*log(tan(c/2 + (d*x)/2) + 1))/(a^4*d) + (2*tan(c/2 + (d*x)/2))/(a^4*d)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {\cot ^{3}{\left (c + d x \right )}}{\sin ^{4}{\left (c + d x \right )} + 4 \sin ^{3}{\left (c + d x \right )} + 6 \sin ^{2}{\left (c + d x \right )} + 4 \sin {\left (c + d x \right )} + 1}\, dx}{a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)**3/(a+a*sin(d*x+c))**4,x)

[Out]

Integral(cot(c + d*x)**3/(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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