\(\int \frac {\cos ^2(c+d x) \cot ^2(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx\) [467]

Optimal result

Integrand size = 31, antiderivative size = 119 \[ \int \frac {\cos ^2(c+d x) \cot ^2(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {a+a \sin (c+d x)}}\right )}{\sqrt {a} d}+\frac {4 \cos (c+d x)}{3 d \sqrt {a+a \sin (c+d x)}}-\frac {\cot (c+d x)}{d \sqrt {a+a \sin (c+d x)}}-\frac {2 \cos (c+d x) \sqrt {a+a \sin (c+d x)}}{3 a d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.39 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.258, Rules used = {2960, 2838, 2830, 2728, 212, 3123, 3064, 2852} \[ \int \frac {\cos ^2(c+d x) \cot ^2(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {a \sin (c+d x)+a}}\right )}{\sqrt {a} d}-\frac {2 \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{3 a d}+\frac {4 \cos (c+d x)}{3 d \sqrt {a \sin (c+d x)+a}}-\frac {\cot (c+d x)}{d \sqrt {a \sin (c+d x)+a}} \]

[In]

Int[(Cos[c + d*x]^2*Cot[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]]]/(Sqrt[a]*d) + (4*Cos[c + d*x])/(3*d*Sqrt[a + a*Sin[c
+ d*x]]) - Cot[c + d*x]/(d*Sqrt[a + a*Sin[c + d*x]]) - (2*Cos[c + d*x]*Sqrt[a + a*Sin[c + d*x]])/(3*a*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 2728

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

Rule 2830

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

Rule 2838

Int[sin[(e_.) + (f_.)*(x_)]^2*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(-Cos[e + f*x])*(
(a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Dist[1/(b*(m + 2)), Int[(a + b*Sin[e + f*x])^m*(b*(m + 1) -
a*Sin[e + f*x]), x], x] /; FreeQ[{a, b, e, f, m}, x] && EqQ[a^2 - b^2, 0] &&  !LtQ[m, -2^(-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 2960

Int[cos[(e_.) + (f_.)*(x_)]^4*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)
, x_Symbol] :> Dist[1/d^4, Int[(d*Sin[e + f*x])^(n + 4)*(a + b*Sin[e + f*x])^m, x], x] + Int[(d*Sin[e + f*x])^
n*(a + b*Sin[e + f*x])^m*(1 - 2*Sin[e + f*x]^2), x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 - b^2, 0] &&
!IGtQ[m, 0]

Rule 3064

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_
.) + (f_.)*(x_)])), x_Symbol] :> Dist[(A*b - a*B)/(b*c - a*d), Int[1/Sqrt[a + b*Sin[e + f*x]], x], x] + Dist[(
B*c - A*d)/(b*c - a*d), Int[Sqrt[a + b*Sin[e + f*x]]/(c + d*Sin[e + f*x]), 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]

Rule 3123

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

Rubi steps \begin{align*} \text {integral}& = \int \frac {\sin ^2(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx+\int \frac {\csc ^2(c+d x) \left (1-2 \sin ^2(c+d x)\right )}{\sqrt {a+a \sin (c+d x)}} \, dx \\ & = -\frac {\cot (c+d x)}{d \sqrt {a+a \sin (c+d x)}}-\frac {2 \cos (c+d x) \sqrt {a+a \sin (c+d x)}}{3 a d}+\frac {2 \int \frac {\frac {a}{2}-a \sin (c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx}{3 a}+\frac {\int \frac {\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 {4 \cos (c+d x)}{3 d \sqrt {a+a \sin (c+d x)}}-\frac {\cot (c+d x)}{d \sqrt {a+a \sin (c+d x)}}-\frac {2 \cos (c+d x) \sqrt {a+a \sin (c+d x)}}{3 a d}-\frac {\int \csc (c+d x) \sqrt {a+a \sin (c+d x)} \, dx}{2 a} \\ & = \frac {4 \cos (c+d x)}{3 d \sqrt {a+a \sin (c+d x)}}-\frac {\cot (c+d x)}{d \sqrt {a+a \sin (c+d x)}}-\frac {2 \cos (c+d x) \sqrt {a+a \sin (c+d x)}}{3 a d}+\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 )}{d} \\ & = \frac {\text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {a+a \sin (c+d x)}}\right )}{\sqrt {a} d}+\frac {4 \cos (c+d x)}{3 d \sqrt {a+a \sin (c+d x)}}-\frac {\cot (c+d x)}{d \sqrt {a+a \sin (c+d x)}}-\frac {2 \cos (c+d x) \sqrt {a+a \sin (c+d x)}}{3 a d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.93 (sec) , antiderivative size = 190, normalized size of antiderivative = 1.60 \[ \int \frac {\cos ^2(c+d x) \cot ^2(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx=\frac {\csc \left (\frac {1}{4} (c+d x)\right ) \sec \left (\frac {1}{4} (c+d x)\right ) \left (-10 \cos \left (\frac {1}{2} (c+d x)\right )+3 \cos \left (\frac {3}{2} (c+d x)\right )+\cos \left (\frac {5}{2} (c+d x)\right )+10 \sin \left (\frac {1}{2} (c+d x)\right )+3 \log \left (1+\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right ) \sin (c+d x)-3 \log \left (1-\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right ) \sin (c+d x)+3 \sin \left (\frac {3}{2} (c+d x)\right )-\sin \left (\frac {5}{2} (c+d x)\right )\right ) \left (1+\tan \left (\frac {1}{2} (c+d x)\right )\right )}{24 d \sqrt {a (1+\sin (c+d x))}} \]

[In]

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

[Out]

(Csc[(c + d*x)/4]*Sec[(c + d*x)/4]*(-10*Cos[(c + d*x)/2] + 3*Cos[(3*(c + d*x))/2] + Cos[(5*(c + d*x))/2] + 10*
Sin[(c + d*x)/2] + 3*Log[1 + Cos[(c + d*x)/2] - Sin[(c + d*x)/2]]*Sin[c + d*x] - 3*Log[1 - Cos[(c + d*x)/2] +
Sin[(c + d*x)/2]]*Sin[c + d*x] + 3*Sin[(3*(c + d*x))/2] - Sin[(5*(c + d*x))/2])*(1 + Tan[(c + d*x)/2]))/(24*d*
Sqrt[a*(1 + Sin[c + d*x])])

Maple [A] (verified)

Time = 0.11 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.09

method	result	size
default	\(\frac {\left (1+\sin \left (d x +c \right )\right ) \sqrt {-a \left (\sin \left (d x +c \right )-1\right )}\, \left (2 \left (a -a \sin \left (d x +c \right )\right )^{\frac {3}{2}} \sin \left (d x +c \right ) \sqrt {a}+3 \,\operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (d x +c \right )}}{\sqrt {a}}\right ) a^{2} \sin \left (d x +c \right )-3 \sqrt {a -a \sin \left (d x +c \right )}\, a^{\frac {3}{2}}\right )}{3 a^{\frac {5}{2}} \sin \left (d x +c \right ) \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d}\)	\(130\)

[In]

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

[Out]

1/3*(1+sin(d*x+c))*(-a*(sin(d*x+c)-1))^(1/2)/a^(5/2)*(2*(a-a*sin(d*x+c))^(3/2)*sin(d*x+c)*a^(1/2)+3*arctanh((a
-a*sin(d*x+c))^(1/2)/a^(1/2))*a^2*sin(d*x+c)-3*(a-a*sin(d*x+c))^(1/2)*a^(3/2))/sin(d*x+c)/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. 306 vs. \(2 (103) = 206\).

Time = 0.28 (sec) , antiderivative size = 306, normalized size of antiderivative = 2.57 \[ \int \frac {\cos ^2(c+d x) \cot ^2(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx=\frac {3 \, {\left (\cos \left (d x + c\right )^{2} - {\left (\cos \left (d x + c\right ) + 1\right )} \sin \left (d x + c\right ) - 1\right )} \sqrt {a} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 7 \, a \cos \left (d x + c\right )^{2} + 4 \, {\left (\cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right ) + 3\right )} \sin \left (d x + c\right ) - 2 \, \cos \left (d x + c\right ) - 3\right )} \sqrt {a \sin \left (d x + c\right ) + a} \sqrt {a} - 9 \, a \cos \left (d x + c\right ) + {\left (a \cos \left (d x + c\right )^{2} + 8 \, a \cos \left (d x + c\right ) - a\right )} \sin \left (d x + c\right ) - a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right )^{2} - 1\right )} \sin \left (d x + c\right ) - \cos \left (d x + c\right ) - 1}\right ) - 4 \, {\left (2 \, \cos \left (d x + c\right )^{3} + 4 \, \cos \left (d x + c\right )^{2} - {\left (2 \, \cos \left (d x + c\right )^{2} - 2 \, \cos \left (d x + c\right ) - 7\right )} \sin \left (d x + c\right ) - 5 \, \cos \left (d x + c\right ) - 7\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{12 \, {\left (a d \cos \left (d x + c\right )^{2} - a d - {\left (a d \cos \left (d x + c\right ) + a d\right )} \sin \left (d x + c\right )\right )}} \]

[In]

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

[Out]

1/12*(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*(2*cos(d*x + c)^3 + 4*cos(d*x + c)^2 - (2*
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)^2 - a*d - (a*d*cos(d*x + c) + a*d)*sin(d*x + c))

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.50 \[ \int \frac {\cos ^2(c+d x) \cot ^2(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx=-\frac {\frac {8 \, \sqrt {2} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3}}{\sqrt {a} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} + \frac {3 \, \log \left ({\left | \frac {1}{2} \, \sqrt {2} + \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{\sqrt {a} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} - \frac {3 \, \log \left ({\left | -\frac {1}{2} \, \sqrt {2} + \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{\sqrt {a} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} + \frac {6 \, \sqrt {2} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (2 \, \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )} \sqrt {a} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{6 \, d} \]

[In]

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

[Out]

-1/6*(8*sqrt(2)*sin(-1/4*pi + 1/2*d*x + 1/2*c)^3/(sqrt(a)*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))) + 3*log(abs(1/2
*sqrt(2) + sin(-1/4*pi + 1/2*d*x + 1/2*c)))/(sqrt(a)*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))) - 3*log(abs(-1/2*sqr
t(2) + sin(-1/4*pi + 1/2*d*x + 1/2*c)))/(sqrt(a)*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))) + 6*sqrt(2)*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)*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))))/d

Mupad [F(-1)]

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

[In]

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

[Out]

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