\(\int \cot ^2(c+d x) \sqrt {a+a \sin (c+d x)} \, dx\) [326]

Optimal result

Integrand size = 23, antiderivative size = 89 \[ \int \cot ^2(c+d x) \sqrt {a+a \sin (c+d x)} \, dx=-\frac {\sqrt {a} \text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {a+a \sin (c+d x)}}\right )}{d}+\frac {3 a \cos (c+d x)}{d \sqrt {a+a \sin (c+d x)}}-\frac {\cot (c+d x) \sqrt {a+a \sin (c+d x)}}{d} \]

[Out]

-arctanh(cos(d*x+c)*a^(1/2)/(a+a*sin(d*x+c))^(1/2))*a^(1/2)/d+3*a*cos(d*x+c)/d/(a+a*sin(d*x+c))^(1/2)-cot(d*x+
c)*(a+a*sin(d*x+c))^(1/2)/d

Rubi [A] (verified)

Time = 0.15 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {2795, 3060, 2852, 212} \[ \int \cot ^2(c+d x) \sqrt {a+a \sin (c+d x)} \, dx=-\frac {\sqrt {a} \text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {a \sin (c+d x)+a}}\right )}{d}+\frac {3 a \cos (c+d x)}{d \sqrt {a \sin (c+d x)+a}}-\frac {\cot (c+d x) \sqrt {a \sin (c+d x)+a}}{d} \]

[In]

Int[Cot[c + d*x]^2*Sqrt[a + a*Sin[c + d*x]],x]

[Out]

-((Sqrt[a]*ArcTanh[(Sqrt[a]*Cos[c + d*x])/Sqrt[a + a*Sin[c + d*x]]])/d) + (3*a*Cos[c + d*x])/(d*Sqrt[a + a*Sin
[c + d*x]]) - (Cot[c + d*x]*Sqrt[a + a*Sin[c + d*x]])/d

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 2795

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

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 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 {\cot (c+d x) \sqrt {a+a \sin (c+d x)}}{d}+\frac {\int \csc (c+d x) \left (\frac {a}{2}-\frac {3}{2} a \sin (c+d x)\right ) \sqrt {a+a \sin (c+d x)} \, dx}{a} \\ & = \frac {3 a \cos (c+d x)}{d \sqrt {a+a \sin (c+d x)}}-\frac {\cot (c+d x) \sqrt {a+a \sin (c+d x)}}{d}+\frac {1}{2} \int \csc (c+d x) \sqrt {a+a \sin (c+d x)} \, dx \\ & = \frac {3 a \cos (c+d x)}{d \sqrt {a+a \sin (c+d x)}}-\frac {\cot (c+d x) \sqrt {a+a \sin (c+d x)}}{d}-\frac {a \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 {\sqrt {a} \text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {a+a \sin (c+d x)}}\right )}{d}+\frac {3 a \cos (c+d x)}{d \sqrt {a+a \sin (c+d x)}}-\frac {\cot (c+d x) \sqrt {a+a \sin (c+d x)}}{d} \\ \end{align*}

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(206\) vs. \(2(89)=178\).

Time = 0.95 (sec) , antiderivative size = 206, normalized size of antiderivative = 2.31 \[ \int \cot ^2(c+d x) \sqrt {a+a \sin (c+d x)} \, dx=\frac {\csc ^4\left (\frac {1}{2} (c+d x)\right ) \sqrt {a (1+\sin (c+d x))} \left (-4 \cos \left (\frac {1}{2} (c+d x)\right )+2 \cos \left (\frac {3}{2} (c+d x)\right )+4 \sin \left (\frac {1}{2} (c+d x)\right )-\log \left (1+\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right ) \sin (c+d x)+\log \left (1-\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right ) \sin (c+d x)+2 \sin \left (\frac {3}{2} (c+d x)\right )\right )}{d \left (1+\cot \left (\frac {1}{2} (c+d x)\right )\right ) \left (\csc \left (\frac {1}{4} (c+d x)\right )-\sec \left (\frac {1}{4} (c+d x)\right )\right ) \left (\csc \left (\frac {1}{4} (c+d x)\right )+\sec \left (\frac {1}{4} (c+d x)\right )\right )} \]

[In]

Integrate[Cot[c + d*x]^2*Sqrt[a + a*Sin[c + d*x]],x]

[Out]

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

Maple [A] (verified)

Time = 0.11 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.45

method	result	size
default	\(\frac {\left (1+\sin \left (d x +c \right )\right ) \sqrt {-a \left (\sin \left (d x +c \right )-1\right )}\, \left (2 \sqrt {a -a \sin \left (d x +c \right )}\, \sin \left (d x +c \right ) a^{\frac {3}{2}}-\sqrt {a -a \sin \left (d x +c \right )}\, a^{\frac {3}{2}}-\operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (d x +c \right )}}{\sqrt {a}}\right ) a^{2} \sin \left (d x +c \right )\right )}{\sin \left (d x +c \right ) a^{\frac {3}{2}} \cos \left (d x +c \right ) \sqrt {a \left (1+\sin \left (d x +c \right )\right )}\, d}\)	\(129\)

[In]

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

[Out]

(1+sin(d*x+c))*(-a*(sin(d*x+c)-1))^(1/2)*(2*(a-a*sin(d*x+c))^(1/2)*sin(d*x+c)*a^(3/2)-(a-a*sin(d*x+c))^(1/2)*a
^(3/2)-arctanh((a-a*sin(d*x+c))^(1/2)/a^(1/2))*a^2*sin(d*x+c))/sin(d*x+c)/a^(3/2)/cos(d*x+c)/(a*(1+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. 279 vs. \(2 (79) = 158\).

Time = 0.27 (sec) , antiderivative size = 279, normalized size of antiderivative = 3.13 \[ \int \cot ^2(c+d x) \sqrt {a+a \sin (c+d x)} \, dx=\frac {{\left (\cos \left (d x + c\right )^{2} - {\left (\cos \left (d x + c\right ) + 1\right )} \sin \left (d x + c\right ) - 1\right )} \sqrt {a} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 7 \, a \cos \left (d x + c\right )^{2} - 4 \, {\left (\cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right ) + 3\right )} \sin \left (d x + c\right ) - 2 \, \cos \left (d x + c\right ) - 3\right )} \sqrt {a \sin \left (d x + c\right ) + a} \sqrt {a} - 9 \, a \cos \left (d x + c\right ) + {\left (a \cos \left (d x + c\right )^{2} + 8 \, a \cos \left (d x + c\right ) - a\right )} \sin \left (d x + c\right ) - a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right )^{2} - 1\right )} \sin \left (d x + c\right ) - \cos \left (d x + c\right ) - 1}\right ) - 4 \, {\left (2 \, \cos \left (d x + c\right )^{2} + {\left (2 \, \cos \left (d x + c\right ) + 3\right )} \sin \left (d x + c\right ) - \cos \left (d x + c\right ) - 3\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{4 \, {\left (d \cos \left (d x + c\right )^{2} - {\left (d \cos \left (d x + c\right ) + d\right )} \sin \left (d x + c\right ) - d\right )}} \]

[In]

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

[Out]

1/4*((cos(d*x + c)^2 - (cos(d*x + c) + 1)*sin(d*x + c) - 1)*sqrt(a)*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*(2*cos(d*x + c)^2 + (2*cos(d*x + c) + 3)*sin(
d*x + c) - cos(d*x + c) - 3)*sqrt(a*sin(d*x + c) + a))/(d*cos(d*x + c)^2 - (d*cos(d*x + c) + d)*sin(d*x + c) -
d)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.38 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.67 \[ \int \cot ^2(c+d x) \sqrt {a+a \sin (c+d x)} \, dx=-\frac {\sqrt {2} {\left (\sqrt {2} \log \left (\frac {{\left | -2 \, \sqrt {2} + 4 \, \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}}{{\left | 2 \, \sqrt {2} + 4 \, \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}}\right ) \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + 8 \, \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \frac {4 \, \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{2 \, \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1}\right )} \sqrt {a}}{4 \, d} \]

[In]

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

[Out]

-1/4*sqrt(2)*(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)))*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c)) + 8*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(-1/4*pi + 1/
2*d*x + 1/2*c) + 4*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(-1/4*pi + 1/2*d*x + 1/2*c)/(2*sin(-1/4*pi + 1/2*d*x
+ 1/2*c)^2 - 1))*sqrt(a)/d

Mupad [F(-1)]

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

[In]

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

[Out]

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