\(\int \frac {\cot ^2(c+d x) \csc ^2(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx\) [342]

Optimal result

Integrand size = 31, antiderivative size = 135 \[ \int \frac {\cot ^2(c+d x) \csc ^2(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {a+a \sin (c+d x)}}\right )}{8 \sqrt {a} d}+\frac {\cot (c+d x)}{8 d \sqrt {a+a \sin (c+d x)}}+\frac {\cot (c+d x) \csc (c+d x)}{12 d \sqrt {a+a \sin (c+d x)}}-\frac {\cot (c+d x) \csc ^2(c+d x)}{3 d \sqrt {a+a \sin (c+d x)}} \]

[Out]

1/8*arctanh(cos(d*x+c)*a^(1/2)/(a+a*sin(d*x+c))^(1/2))/d/a^(1/2)+1/8*cot(d*x+c)/d/(a+a*sin(d*x+c))^(1/2)+1/12*
cot(d*x+c)*csc(d*x+c)/d/(a+a*sin(d*x+c))^(1/2)-1/3*cot(d*x+c)*csc(d*x+c)^2/d/(a+a*sin(d*x+c))^(1/2)

Rubi [A] (verified)

Time = 0.30 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.161, Rules used = {2953, 3059, 2851, 2852, 212} \[ \int \frac {\cot ^2(c+d x) \csc ^2(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {a \sin (c+d x)+a}}\right )}{8 \sqrt {a} d}+\frac {\cot (c+d x)}{8 d \sqrt {a \sin (c+d x)+a}}-\frac {\cot (c+d x) \csc ^2(c+d x)}{3 d \sqrt {a \sin (c+d x)+a}}+\frac {\cot (c+d x) \csc (c+d x)}{12 d \sqrt {a \sin (c+d x)+a}} \]

[In]

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

[Out]

ArcTanh[(Sqrt[a]*Cos[c + d*x])/Sqrt[a + a*Sin[c + d*x]]]/(8*Sqrt[a]*d) + Cot[c + d*x]/(8*d*Sqrt[a + a*Sin[c +
d*x]]) + (Cot[c + d*x]*Csc[c + d*x])/(12*d*Sqrt[a + a*Sin[c + d*x]]) - (Cot[c + d*x]*Csc[c + d*x]^2)/(3*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 2851

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

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 3059

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.
) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b^2)*(B*c - A*d)*Cos[e + f*x]*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(n
+ 1)*(b*c + a*d)*Sqrt[a + b*Sin[e + f*x]])), x] + Dist[(A*b*d*(2*n + 3) - B*(b*c - 2*a*d*(n + 1)))/(2*d*(n +
1)*(b*c + a*d)), Int[Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d, e, f,
A, B}, 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 ^4(c+d x) (a-a \sin (c+d x)) \sqrt {a+a \sin (c+d x)} \, dx}{a^2} \\ & = -\frac {\cot (c+d x) \csc ^2(c+d x)}{3 d \sqrt {a+a \sin (c+d x)}}-\frac {\int \csc ^3(c+d x) \sqrt {a+a \sin (c+d x)} \, dx}{6 a} \\ & = \frac {\cot (c+d x) \csc (c+d x)}{12 d \sqrt {a+a \sin (c+d x)}}-\frac {\cot (c+d x) \csc ^2(c+d x)}{3 d \sqrt {a+a \sin (c+d x)}}-\frac {\int \csc ^2(c+d x) \sqrt {a+a \sin (c+d x)} \, dx}{8 a} \\ & = \frac {\cot (c+d x)}{8 d \sqrt {a+a \sin (c+d x)}}+\frac {\cot (c+d x) \csc (c+d x)}{12 d \sqrt {a+a \sin (c+d x)}}-\frac {\cot (c+d x) \csc ^2(c+d x)}{3 d \sqrt {a+a \sin (c+d x)}}-\frac {\int \csc (c+d x) \sqrt {a+a \sin (c+d x)} \, dx}{16 a} \\ & = \frac {\cot (c+d x)}{8 d \sqrt {a+a \sin (c+d x)}}+\frac {\cot (c+d x) \csc (c+d x)}{12 d \sqrt {a+a \sin (c+d x)}}-\frac {\cot (c+d x) \csc ^2(c+d x)}{3 d \sqrt {a+a \sin (c+d x)}}+\frac {\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 )}{8 d} \\ & = \frac {\text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {a+a \sin (c+d x)}}\right )}{8 \sqrt {a} d}+\frac {\cot (c+d x)}{8 d \sqrt {a+a \sin (c+d x)}}+\frac {\cot (c+d x) \csc (c+d x)}{12 d \sqrt {a+a \sin (c+d x)}}-\frac {\cot (c+d x) \csc ^2(c+d x)}{3 d \sqrt {a+a \sin (c+d x)}} \\ \end{align*}

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(292\) vs. \(2(135)=270\).

Time = 1.52 (sec) , antiderivative size = 292, normalized size of antiderivative = 2.16 \[ \int \frac {\cot ^2(c+d x) \csc ^2(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx=\frac {\csc ^9\left (\frac {1}{2} (c+d x)\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right ) \left (-60 \cos \left (\frac {1}{2} (c+d x)\right )+2 \cos \left (\frac {3}{2} (c+d x)\right )-6 \cos \left (\frac {5}{2} (c+d x)\right )+60 \sin \left (\frac {1}{2} (c+d x)\right )+9 \log \left (1+\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right ) \sin (c+d x)-9 \log \left (1-\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right ) \sin (c+d x)+2 \sin \left (\frac {3}{2} (c+d x)\right )+6 \sin \left (\frac {5}{2} (c+d x)\right )-3 \log \left (1+\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right ) \sin (3 (c+d x))+3 \log \left (1-\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right ) \sin (3 (c+d x))\right )}{24 d \left (\csc ^2\left (\frac {1}{4} (c+d x)\right )-\sec ^2\left (\frac {1}{4} (c+d x)\right )\right )^3 \sqrt {a (1+\sin (c+d x))}} \]

[In]

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

[Out]

(Csc[(c + d*x)/2]^9*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])*(-60*Cos[(c + d*x)/2] + 2*Cos[(3*(c + d*x))/2] - 6*C
os[(5*(c + d*x))/2] + 60*Sin[(c + d*x)/2] + 9*Log[1 + Cos[(c + d*x)/2] - Sin[(c + d*x)/2]]*Sin[c + d*x] - 9*Lo
g[1 - Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]*Sin[c + d*x] + 2*Sin[(3*(c + d*x))/2] + 6*Sin[(5*(c + d*x))/2] - 3*
Log[1 + Cos[(c + d*x)/2] - Sin[(c + d*x)/2]]*Sin[3*(c + d*x)] + 3*Log[1 - Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]
*Sin[3*(c + d*x)]))/(24*d*(Csc[(c + d*x)/4]^2 - Sec[(c + d*x)/4]^2)^3*Sqrt[a*(1 + Sin[c + d*x])])

Maple [A] (verified)

Time = 0.11 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.07

method	result	size
default	\(\frac {\left (1+\sin \left (d x +c \right )\right ) \sqrt {-a \left (\sin \left (d x +c \right )-1\right )}\, \left (3 \left (-a \left (\sin \left (d x +c \right )-1\right )\right )^{\frac {5}{2}} a^{\frac {5}{2}}-8 \left (-a \left (\sin \left (d x +c \right )-1\right )\right )^{\frac {3}{2}} a^{\frac {7}{2}}+3 \,\operatorname {arctanh}\left (\frac {\sqrt {-a \left (\sin \left (d x +c \right )-1\right )}}{\sqrt {a}}\right ) a^{5} \left (\sin ^{3}\left (d x +c \right )\right )-3 \sqrt {-a \left (\sin \left (d x +c \right )-1\right )}\, a^{\frac {9}{2}}\right )}{24 a^{\frac {11}{2}} \sin \left (d x +c \right )^{3} \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d}\)	\(144\)

[In]

int(cos(d*x+c)^2*csc(d*x+c)^4/(a+a*sin(d*x+c))^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/24*(1+sin(d*x+c))*(-a*(sin(d*x+c)-1))^(1/2)/a^(11/2)*(3*(-a*(sin(d*x+c)-1))^(5/2)*a^(5/2)-8*(-a*(sin(d*x+c)-
1))^(3/2)*a^(7/2)+3*arctanh((-a*(sin(d*x+c)-1))^(1/2)/a^(1/2))*a^5*sin(d*x+c)^3-3*(-a*(sin(d*x+c)-1))^(1/2)*a^
(9/2))/sin(d*x+c)^3/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. 367 vs. \(2 (115) = 230\).

Time = 0.30 (sec) , antiderivative size = 367, normalized size of antiderivative = 2.72 \[ \int \frac {\cot ^2(c+d x) \csc ^2(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx=\frac {3 \, {\left (\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} - {\left (\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2} - \cos \left (d x + c\right ) - 1\right )} \sin \left (d x + c\right ) + 1\right )} \sqrt {a} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 7 \, a \cos \left (d x + c\right )^{2} + 4 \, {\left (\cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right ) + 3\right )} \sin \left (d x + c\right ) - 2 \, \cos \left (d x + c\right ) - 3\right )} \sqrt {a \sin \left (d x + c\right ) + a} \sqrt {a} - 9 \, a \cos \left (d x + c\right ) + {\left (a \cos \left (d x + c\right )^{2} + 8 \, a \cos \left (d x + c\right ) - a\right )} \sin \left (d x + c\right ) - a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right )^{2} - 1\right )} \sin \left (d x + c\right ) - \cos \left (d x + c\right ) - 1}\right ) - 4 \, {\left (3 \, \cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2} - {\left (3 \, \cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 7\right )} \sin \left (d x + c\right ) + 5 \, \cos \left (d x + c\right ) + 7\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{96 \, {\left (a d \cos \left (d x + c\right )^{4} - 2 \, a d \cos \left (d x + c\right )^{2} + a d - {\left (a d \cos \left (d x + c\right )^{3} + a d \cos \left (d x + c\right )^{2} - a d \cos \left (d x + c\right ) - a d\right )} \sin \left (d x + c\right )\right )}} \]

[In]

integrate(cos(d*x+c)^2*csc(d*x+c)^4/(a+a*sin(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

1/96*(3*(cos(d*x + c)^4 - 2*cos(d*x + c)^2 - (cos(d*x + c)^3 + 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*(3*cos(d*x + c)^3 + cos(d*x + c)^2 - (3*cos(d*x + c)^2 + 2*cos(d*x + c) + 7)*sin(d*x + c) + 5*cos(d
*x + c) + 7)*sqrt(a*sin(d*x + c) + a))/(a*d*cos(d*x + c)^4 - 2*a*d*cos(d*x + c)^2 + a*d - (a*d*cos(d*x + c)^3
+ a*d*cos(d*x + c)^2 - a*d*cos(d*x + c) - a*d)*sin(d*x + c))

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

integrate(cos(d*x+c)^2*csc(d*x+c)^4/(a+a*sin(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.24 \[ \int \frac {\cot ^2(c+d x) \csc ^2(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx=\frac {\sqrt {2} \sqrt {a} {\left (\frac {3 \, \sqrt {2} \log \left (\frac {{\left | -2 \, \sqrt {2} + 4 \, \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}}{{\left | 2 \, \sqrt {2} + 4 \, \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}}\right )}{a \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} + \frac {4 \, {\left (12 \, \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 16 \, \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 3 \, \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (2 \, \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{3} a \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}\right )}}{96 \, d} \]

[In]

integrate(cos(d*x+c)^2*csc(d*x+c)^4/(a+a*sin(d*x+c))^(1/2),x, algorithm="giac")

[Out]

1/96*sqrt(2)*sqrt(a)*(3*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*(12*sin(-1/4*pi + 1/2*d*x + 1/2*c)^5 -
16*sin(-1/4*pi + 1/2*d*x + 1/2*c)^3 - 3*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)^3*a*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))))/d

Mupad [F(-1)]

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

[In]

int(cos(c + d*x)^2/(sin(c + d*x)^4*(a + a*sin(c + d*x))^(1/2)),x)

[Out]

int(cos(c + d*x)^2/(sin(c + d*x)^4*(a + a*sin(c + d*x))^(1/2)), x)