\(\int \frac {\cos (c+d x) \cot (c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx\) [339]

Optimal result

Integrand size = 27, antiderivative size = 63 \[ \int \frac {\cos (c+d x) \cot (c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx=-\frac {2 \text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {a+a \sin (c+d x)}}\right )}{\sqrt {a} d}+\frac {2 \cos (c+d x)}{d \sqrt {a+a \sin (c+d x)}} \]

[Out]

-2*arctanh(cos(d*x+c)*a^(1/2)/(a+a*sin(d*x+c))^(1/2))/d/a^(1/2)+2*cos(d*x+c)/d/(a+a*sin(d*x+c))^(1/2)

Rubi [A] (verified)

Time = 0.18 (sec) , antiderivative size = 63, 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 = {2953, 3060, 2852, 212} \[ \int \frac {\cos (c+d x) \cot (c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx=\frac {2 \cos (c+d x)}{d \sqrt {a \sin (c+d x)+a}}-\frac {2 \text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {a \sin (c+d x)+a}}\right )}{\sqrt {a} d} \]

[In]

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

[Out]

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

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 2852

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

Rule 2953

Int[cos[(e_.) + (f_.)*(x_)]^2*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)
, x_Symbol] :> Dist[1/b^2, Int[(d*Sin[e + f*x])^n*(a + b*Sin[e + f*x])^(m + 1)*(a - b*Sin[e + f*x]), x], x] /;
FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 - b^2, 0] && (ILtQ[m, 0] ||  !IGtQ[n, 0])

Rule 3060

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.
) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[-2*b*B*Cos[e + f*x]*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(2*n + 3)*Sqrt
[a + b*Sin[e + f*x]])), x] + Dist[(A*b*d*(2*n + 3) - B*(b*c - 2*a*d*(n + 1)))/(b*d*(2*n + 3)), Int[Sqrt[a + b*
Sin[e + f*x]]*(c + d*Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] &&
EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] &&  !LtQ[n, -1]

Rubi steps \begin{align*} \text {integral}& = \frac {\int \csc (c+d x) (a-a \sin (c+d x)) \sqrt {a+a \sin (c+d x)} \, dx}{a^2} \\ & = \frac {2 \cos (c+d x)}{d \sqrt {a+a \sin (c+d x)}}+\frac {\int \csc (c+d x) \sqrt {a+a \sin (c+d x)} \, dx}{a} \\ & = \frac {2 \cos (c+d x)}{d \sqrt {a+a \sin (c+d x)}}-\frac {2 \text {Subst}\left (\int \frac {1}{a-x^2} \, dx,x,\frac {a \cos (c+d x)}{\sqrt {a+a \sin (c+d x)}}\right )}{d} \\ & = -\frac {2 \text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {a+a \sin (c+d x)}}\right )}{\sqrt {a} d}+\frac {2 \cos (c+d x)}{d \sqrt {a+a \sin (c+d x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.38 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.84 \[ \int \frac {\cos (c+d x) \cot (c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx=\frac {\left (2 \cos \left (\frac {1}{2} (c+d x)\right )-\log \left (1+\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+\log \left (1-\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )-2 \sin \left (\frac {1}{2} (c+d x)\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}{d \sqrt {a (1+\sin (c+d x))}} \]

[In]

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

[Out]

((2*Cos[(c + d*x)/2] - Log[1 + Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + Log[1 - Cos[(c + d*x)/2] + Sin[(c + d*x)
/2]] - 2*Sin[(c + d*x)/2])*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]))/(d*Sqrt[a*(1 + Sin[c + d*x])])

Maple [A] (verified)

Time = 0.09 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.38

method	result	size
default	\(\frac {2 \left (1+\sin \left (d x +c \right )\right ) \sqrt {-a \left (\sin \left (d x +c \right )-1\right )}\, \left (-\sqrt {a}\, \operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (d x +c \right )}}{\sqrt {a}}\right )+\sqrt {a -a \sin \left (d x +c \right )}\right )}{a \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d}\)	\(87\)

[In]

int(cos(d*x+c)^2*csc(d*x+c)/(a+a*sin(d*x+c))^(1/2),x,method=_RETURNVERBOSE)

[Out]

2*(1+sin(d*x+c))*(-a*(sin(d*x+c)-1))^(1/2)*(-a^(1/2)*arctanh((a-a*sin(d*x+c))^(1/2)/a^(1/2))+(a-a*sin(d*x+c))^
(1/2))/a/cos(d*x+c)/(a+a*sin(d*x+c))^(1/2)/d

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 236 vs. \(2 (55) = 110\).

Time = 0.28 (sec) , antiderivative size = 236, normalized size of antiderivative = 3.75 \[ \int \frac {\cos (c+d x) \cot (c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx=\frac {\sqrt {a} {\left (\cos \left (d x + c\right ) + \sin \left (d x + c\right ) + 1\right )} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 7 \, a \cos \left (d x + c\right )^{2} - 4 \, {\left (\cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right ) + 3\right )} \sin \left (d x + c\right ) - 2 \, \cos \left (d x + c\right ) - 3\right )} \sqrt {a \sin \left (d x + c\right ) + a} \sqrt {a} - 9 \, a \cos \left (d x + c\right ) + {\left (a \cos \left (d x + c\right )^{2} + 8 \, a \cos \left (d x + c\right ) - a\right )} \sin \left (d x + c\right ) - a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right )^{2} - 1\right )} \sin \left (d x + c\right ) - \cos \left (d x + c\right ) - 1}\right ) + 4 \, \sqrt {a \sin \left (d x + c\right ) + a} {\left (\cos \left (d x + c\right ) - \sin \left (d x + c\right ) + 1\right )}}{2 \, {\left (a d \cos \left (d x + c\right ) + a d \sin \left (d x + c\right ) + a d\right )}} \]

[In]

integrate(cos(d*x+c)^2*csc(d*x+c)/(a+a*sin(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

1/2*(sqrt(a)*(cos(d*x + c) + sin(d*x + c) + 1)*log((a*cos(d*x + c)^3 - 7*a*cos(d*x + c)^2 - 4*(cos(d*x + c)^2
+ (cos(d*x + c) + 3)*sin(d*x + c) - 2*cos(d*x + c) - 3)*sqrt(a*sin(d*x + c) + a)*sqrt(a) - 9*a*cos(d*x + c) +
(a*cos(d*x + c)^2 + 8*a*cos(d*x + c) - a)*sin(d*x + c) - a)/(cos(d*x + c)^3 + cos(d*x + c)^2 + (cos(d*x + c)^2
- 1)*sin(d*x + c) - cos(d*x + c) - 1)) + 4*sqrt(a*sin(d*x + c) + a)*(cos(d*x + c) - sin(d*x + c) + 1))/(a*d*c
os(d*x + c) + a*d*sin(d*x + c) + a*d)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

integrate(cos(d*x+c)^2*csc(d*x+c)/(a+a*sin(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 112 vs. \(2 (55) = 110\).

Time = 0.38 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.78 \[ \int \frac {\cos (c+d x) \cot (c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx=-\frac {\sqrt {2} \sqrt {a} {\left (\frac {\sqrt {2} \log \left (\frac {{\left | -2 \, \sqrt {2} + 4 \, \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}}{{\left | 2 \, \sqrt {2} + 4 \, \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}}\right )}{a \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} + \frac {4 \, \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}\right )}}{2 \, d} \]

[In]

integrate(cos(d*x+c)^2*csc(d*x+c)/(a+a*sin(d*x+c))^(1/2),x, algorithm="giac")

[Out]

-1/2*sqrt(2)*sqrt(a)*(sqrt(2)*log(abs(-2*sqrt(2) + 4*sin(-1/4*pi + 1/2*d*x + 1/2*c))/abs(2*sqrt(2) + 4*sin(-1/
4*pi + 1/2*d*x + 1/2*c)))/(a*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))) + 4*sin(-1/4*pi + 1/2*d*x + 1/2*c)/(a*sgn(co
s(-1/4*pi + 1/2*d*x + 1/2*c))))/d

Mupad [F(-1)]

Timed out. \[ \int \frac {\cos (c+d x) \cot (c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^2}{\sin \left (c+d\,x\right )\,\sqrt {a+a\,\sin \left (c+d\,x\right )}} \,d x \]

[In]

int(cos(c + d*x)^2/(sin(c + d*x)*(a + a*sin(c + d*x))^(1/2)),x)

[Out]

int(cos(c + d*x)^2/(sin(c + d*x)*(a + a*sin(c + d*x))^(1/2)), x)