\(\int \frac {\cos ^3(c+d x) \cot (c+d x)}{(a+a \sin (c+d x))^2} \, dx\) [425]

Optimal result

Integrand size = 27, antiderivative size = 36 \[ \int \frac {\cos ^3(c+d x) \cot (c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {2 x}{a^2}-\frac {\text {arctanh}(\cos (c+d x))}{a^2 d}-\frac {\cos (c+d x)}{a^2 d} \]

[Out]

-2*x/a^2-arctanh(cos(d*x+c))/a^2/d-cos(d*x+c)/a^2/d

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {2948, 2825, 2814, 3855} \[ \int \frac {\cos ^3(c+d x) \cot (c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {\text {arctanh}(\cos (c+d x))}{a^2 d}-\frac {\cos (c+d x)}{a^2 d}-\frac {2 x}{a^2} \]

[In]

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

[Out]

(-2*x)/a^2 - ArcTanh[Cos[c + d*x]]/(a^2*d) - Cos[c + d*x]/(a^2*d)

Rule 2814

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[b*(x/d)
, x] - Dist[(b*c - a*d)/d, Int[1/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d
, 0]

Rule 2825

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^2/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-b^2
)*(Cos[e + f*x]/(d*f)), x] + Dist[1/d, Int[Simp[a^2*d - b*(b*c - 2*a*d)*Sin[e + f*x], x]/(c + d*Sin[e + f*x]),
x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0]

Rule 2948

Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(
m_), x_Symbol] :> Dist[a^(2*m), Int[(d*Sin[e + f*x])^n/(a - b*Sin[e + f*x])^m, x], x] /; FreeQ[{a, b, d, e, f,
n}, x] && EqQ[a^2 - b^2, 0] && IntegersQ[m, p] && EqQ[2*m + p, 0]

Rule 3855

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps \begin{align*} \text {integral}& = \frac {\int \csc (c+d x) (a-a \sin (c+d x))^2 \, dx}{a^4} \\ & = -\frac {\cos (c+d x)}{a^2 d}+\frac {\int \csc (c+d x) \left (a^2-2 a^2 \sin (c+d x)\right ) \, dx}{a^4} \\ & = -\frac {2 x}{a^2}-\frac {\cos (c+d x)}{a^2 d}+\frac {\int \csc (c+d x) \, dx}{a^2} \\ & = -\frac {2 x}{a^2}-\frac {\text {arctanh}(\cos (c+d x))}{a^2 d}-\frac {\cos (c+d x)}{a^2 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.63 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.28 \[ \int \frac {\cos ^3(c+d x) \cot (c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {2 c+2 d x+\cos (c+d x)+\log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )}{a^2 d} \]

[In]

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

[Out]

-((2*c + 2*d*x + Cos[c + d*x] + Log[Cos[(c + d*x)/2]] - Log[Sin[(c + d*x)/2]])/(a^2*d))

Maple [A] (verified)

Time = 0.28 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.89

method	result	size
parallelrisch	\(\frac {-2 d x +1+\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\cos \left (d x +c \right )}{d \,a^{2}}\)	\(32\)
derivativedivides	\(\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {2}{1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}-4 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,a^{2}}\)	\(48\)
default	\(\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {2}{1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}-4 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,a^{2}}\)	\(48\)
risch	\(-\frac {2 x}{a^{2}}-\frac {{\mathrm e}^{i \left (d x +c \right )}}{2 d \,a^{2}}-\frac {{\mathrm e}^{-i \left (d x +c \right )}}{2 d \,a^{2}}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{d \,a^{2}}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{d \,a^{2}}\)	\(81\)
norman	\(\frac {-\frac {2}{a d}-\frac {2 x}{a}-\frac {6 x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a}-\frac {12 x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}-\frac {20 x \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}-\frac {24 x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}-\frac {24 x \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}-\frac {20 x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}-\frac {12 x \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}-\frac {6 x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}-\frac {2 x \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}-\frac {6 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d a}-\frac {10 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}-\frac {2 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}-\frac {14 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}-\frac {6 \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}-\frac {14 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}-\frac {10 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3} a \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}+\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,a^{2}}\)	\(348\)

[In]

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

[Out]

1/d/a^2*(-2*d*x+1+ln(tan(1/2*d*x+1/2*c))-cos(d*x+c))

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.25 \[ \int \frac {\cos ^3(c+d x) \cot (c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {4 \, d x + 2 \, \cos \left (d x + c\right ) + \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right )}{2 \, a^{2} d} \]

[In]

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

[Out]

-1/2*(4*d*x + 2*cos(d*x + c) + log(1/2*cos(d*x + c) + 1/2) - log(-1/2*cos(d*x + c) + 1/2))/(a^2*d)

Sympy [F]

\[ \int \frac {\cos ^3(c+d x) \cot (c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {\int \frac {\cos ^{4}{\left (c + d x \right )} \csc {\left (c + d x \right )}}{\sin ^{2}{\left (c + d x \right )} + 2 \sin {\left (c + d x \right )} + 1}\, dx}{a^{2}} \]

[In]

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

[Out]

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 82 vs. \(2 (36) = 72\).

Time = 0.33 (sec) , antiderivative size = 82, normalized size of antiderivative = 2.28 \[ \int \frac {\cos ^3(c+d x) \cot (c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {\frac {2}{a^{2} + \frac {a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}} + \frac {4 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}} - \frac {\log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}}}{d} \]

[In]

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

[Out]

-(2/(a^2 + a^2*sin(d*x + c)^2/(cos(d*x + c) + 1)^2) + 4*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a^2 - log(sin(
d*x + c)/(cos(d*x + c) + 1))/a^2)/d

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.44 \[ \int \frac {\cos ^3(c+d x) \cot (c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {\frac {2 \, {\left (d x + c\right )}}{a^{2}} - \frac {\log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a^{2}} + \frac {2}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )} a^{2}}}{d} \]

[In]

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

[Out]

-(2*(d*x + c)/a^2 - log(abs(tan(1/2*d*x + 1/2*c)))/a^2 + 2/((tan(1/2*d*x + 1/2*c)^2 + 1)*a^2))/d

Mupad [B] (verification not implemented)

Time = 10.27 (sec) , antiderivative size = 97, normalized size of antiderivative = 2.69 \[ \int \frac {\cos ^3(c+d x) \cot (c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {4\,\mathrm {atan}\left (\frac {16}{16\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+8}-\frac {8\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{16\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+8}\right )}{a^2\,d}-\frac {2}{d\,\left (a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+a^2\right )}+\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a^2\,d} \]

[In]

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

[Out]

(4*atan(16/(16*tan(c/2 + (d*x)/2) + 8) - (8*tan(c/2 + (d*x)/2))/(16*tan(c/2 + (d*x)/2) + 8)))/(a^2*d) - 2/(d*(
a^2*tan(c/2 + (d*x)/2)^2 + a^2)) + log(tan(c/2 + (d*x)/2))/(a^2*d)