\(\int \frac {\cos ^3(c+d x) \cot (c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx\) [466]

Optimal result

Integrand size = 29, antiderivative size = 130 \[ \int \frac {\cos ^3(c+d x) \cot (c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx=-\frac {2 \text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {a+a \sin (c+d x)}}\right )}{\sqrt {a} d}+\frac {32 \cos (c+d x)}{15 d \sqrt {a+a \sin (c+d x)}}-\frac {2 \cos (c+d x) \sin ^2(c+d x)}{5 d \sqrt {a+a \sin (c+d x)}}+\frac {2 \cos (c+d x) \sqrt {a+a \sin (c+d x)}}{15 a d} \]

[Out]

-2*arctanh(cos(d*x+c)*a^(1/2)/(a+a*sin(d*x+c))^(1/2))/d/a^(1/2)+32/15*cos(d*x+c)/d/(a+a*sin(d*x+c))^(1/2)-2/5*
cos(d*x+c)*sin(d*x+c)^2/d/(a+a*sin(d*x+c))^(1/2)+2/15*cos(d*x+c)*(a+a*sin(d*x+c))^(1/2)/a/d

Rubi [A] (verified)

Time = 0.43 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.345, Rules used = {2960, 2857, 3047, 3102, 2830, 2728, 212, 3125, 3064, 2852} \[ \int \frac {\cos ^3(c+d x) \cot (c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx=-\frac {2 \text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {a \sin (c+d x)+a}}\right )}{\sqrt {a} d}-\frac {2 \sin ^2(c+d x) \cos (c+d x)}{5 d \sqrt {a \sin (c+d x)+a}}+\frac {2 \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{15 a d}+\frac {32 \cos (c+d x)}{15 d \sqrt {a \sin (c+d x)+a}} \]

[In]

Int[(Cos[c + d*x]^3*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) + (32*Cos[c + d*x])/(15*d*Sqrt[a + a
*Sin[c + d*x]]) - (2*Cos[c + d*x]*Sin[c + d*x]^2)/(5*d*Sqrt[a + a*Sin[c + d*x]]) + (2*Cos[c + d*x]*Sqrt[a + a*
Sin[c + d*x]])/(15*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 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 2857

Int[((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)/Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> Simp
[-2*d*Cos[e + f*x]*((c + d*Sin[e + f*x])^(n - 1)/(f*(2*n - 1)*Sqrt[a + b*Sin[e + f*x]])), x] - Dist[1/(b*(2*n
- 1)), Int[((c + d*Sin[e + f*x])^(n - 2)/Sqrt[a + b*Sin[e + f*x]])*Simp[a*c*d - b*(2*d^2*(n - 1) + c^2*(2*n -
1)) + d*(a*d - b*c*(4*n - 3))*Sin[e + f*x], x], 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] && GtQ[n, 1] && IntegerQ[2*n]

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 3047

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(
e_.) + (f_.)*(x_)]), x_Symbol] :> Int[(a + b*Sin[e + f*x])^m*(A*c + (B*c + A*d)*Sin[e + f*x] + B*d*Sin[e + f*x
]^2), x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 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 3102

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (
f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*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*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Sin[e + f*x], x],
x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] &&  !LtQ[m, -1]

Rule 3125

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

Rubi steps \begin{align*} \text {integral}& = \int \frac {\sin ^3(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx+\int \frac {\csc (c+d x) \left (1-2 \sin ^2(c+d x)\right )}{\sqrt {a+a \sin (c+d x)}} \, dx \\ & = \frac {4 \cos (c+d x)}{d \sqrt {a+a \sin (c+d x)}}-\frac {2 \cos (c+d x) \sin ^2(c+d x)}{5 d \sqrt {a+a \sin (c+d x)}}-\frac {\int \frac {\sin (c+d x) (-4 a+a \sin (c+d x))}{\sqrt {a+a \sin (c+d x)}} \, dx}{5 a}+\frac {2 \int \frac {\csc (c+d x) \left (\frac {a}{2}+a \sin (c+d x)\right )}{\sqrt {a+a \sin (c+d x)}} \, dx}{a} \\ & = \frac {4 \cos (c+d x)}{d \sqrt {a+a \sin (c+d x)}}-\frac {2 \cos (c+d x) \sin ^2(c+d x)}{5 d \sqrt {a+a \sin (c+d x)}}-\frac {\int \frac {-4 a \sin (c+d x)+a \sin ^2(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx}{5 a}+\frac {\int \csc (c+d x) \sqrt {a+a \sin (c+d x)} \, dx}{a}+\int \frac {1}{\sqrt {a+a \sin (c+d x)}} \, dx \\ & = \frac {4 \cos (c+d x)}{d \sqrt {a+a \sin (c+d x)}}-\frac {2 \cos (c+d x) \sin ^2(c+d x)}{5 d \sqrt {a+a \sin (c+d x)}}+\frac {2 \cos (c+d x) \sqrt {a+a \sin (c+d x)}}{15 a d}-\frac {2 \int \frac {\frac {a^2}{2}-7 a^2 \sin (c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx}{15 a^2}-\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 {Subst}\left (\int \frac {1}{2 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 {\sqrt {2} \text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {2} \sqrt {a+a \sin (c+d x)}}\right )}{\sqrt {a} d}+\frac {32 \cos (c+d x)}{15 d \sqrt {a+a \sin (c+d x)}}-\frac {2 \cos (c+d x) \sin ^2(c+d x)}{5 d \sqrt {a+a \sin (c+d x)}}+\frac {2 \cos (c+d x) \sqrt {a+a \sin (c+d x)}}{15 a d}-\int \frac {1}{\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 {\sqrt {2} \text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {2} \sqrt {a+a \sin (c+d x)}}\right )}{\sqrt {a} d}+\frac {32 \cos (c+d x)}{15 d \sqrt {a+a \sin (c+d x)}}-\frac {2 \cos (c+d x) \sin ^2(c+d x)}{5 d \sqrt {a+a \sin (c+d x)}}+\frac {2 \cos (c+d x) \sqrt {a+a \sin (c+d x)}}{15 a d}+\frac {2 \text {Subst}\left (\int \frac {1}{2 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 {32 \cos (c+d x)}{15 d \sqrt {a+a \sin (c+d x)}}-\frac {2 \cos (c+d x) \sin ^2(c+d x)}{5 d \sqrt {a+a \sin (c+d x)}}+\frac {2 \cos (c+d x) \sqrt {a+a \sin (c+d x)}}{15 a d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.63 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.30 \[ \int \frac {\cos ^3(c+d x) \cot (c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx=\frac {\left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right ) \left (60 \cos \left (\frac {1}{2} (c+d x)\right )+5 \cos \left (\frac {3}{2} (c+d x)\right )+3 \cos \left (\frac {5}{2} (c+d x)\right )-30 \log \left (1+\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+30 \log \left (1-\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )-60 \sin \left (\frac {1}{2} (c+d x)\right )+5 \sin \left (\frac {3}{2} (c+d x)\right )-3 \sin \left (\frac {5}{2} (c+d x)\right )\right )}{30 d \sqrt {a (1+\sin (c+d x))}} \]

[In]

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

[Out]

((Cos[(c + d*x)/2] + Sin[(c + d*x)/2])*(60*Cos[(c + d*x)/2] + 5*Cos[(3*(c + d*x))/2] + 3*Cos[(5*(c + d*x))/2]
- 30*Log[1 + Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 30*Log[1 - Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] - 60*Sin[(
c + d*x)/2] + 5*Sin[(3*(c + d*x))/2] - 3*Sin[(5*(c + d*x))/2]))/(30*d*Sqrt[a*(1 + Sin[c + d*x])])

Maple [A] (verified)

Time = 0.11 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.95

method	result	size
default	\(-\frac {2 \left (1+\sin \left (d x +c \right )\right ) \sqrt {-a \left (\sin \left (d x +c \right )-1\right )}\, \left (15 a^{\frac {5}{2}} \operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (d x +c \right )}}{\sqrt {a}}\right )+3 \left (a -a \sin \left (d x +c \right )\right )^{\frac {5}{2}}-5 a \left (a -a \sin \left (d x +c \right )\right )^{\frac {3}{2}}-15 a^{2} \sqrt {a -a \sin \left (d x +c \right )}\right )}{15 a^{3} \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d}\)	\(123\)

[In]

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

[Out]

-2/15*(1+sin(d*x+c))*(-a*(sin(d*x+c)-1))^(1/2)*(15*a^(5/2)*arctanh((a-a*sin(d*x+c))^(1/2)/a^(1/2))+3*(a-a*sin(
d*x+c))^(5/2)-5*a*(a-a*sin(d*x+c))^(3/2)-15*a^2*(a-a*sin(d*x+c))^(1/2))/a^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. 279 vs. \(2 (112) = 224\).

Time = 0.27 (sec) , antiderivative size = 279, normalized size of antiderivative = 2.15 \[ \int \frac {\cos ^3(c+d x) \cot (c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx=\frac {15 \, \sqrt {a} {\left (\cos \left (d x + c\right ) + \sin \left (d x + c\right ) + 1\right )} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 7 \, a \cos \left (d x + c\right )^{2} - 4 \, {\left (\cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right ) + 3\right )} \sin \left (d x + c\right ) - 2 \, \cos \left (d x + c\right ) - 3\right )} \sqrt {a \sin \left (d x + c\right ) + a} \sqrt {a} - 9 \, a \cos \left (d x + c\right ) + {\left (a \cos \left (d x + c\right )^{2} + 8 \, a \cos \left (d x + c\right ) - a\right )} \sin \left (d x + c\right ) - a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right )^{2} - 1\right )} \sin \left (d x + c\right ) - \cos \left (d x + c\right ) - 1}\right ) + 4 \, {\left (3 \, \cos \left (d x + c\right )^{3} + 4 \, \cos \left (d x + c\right )^{2} - {\left (3 \, \cos \left (d x + c\right )^{2} - \cos \left (d x + c\right ) + 13\right )} \sin \left (d x + c\right ) + 14 \, \cos \left (d x + c\right ) + 13\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{30 \, {\left (a d \cos \left (d x + c\right ) + a d \sin \left (d x + c\right ) + a d\right )}} \]

[In]

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

[Out]

1/30*(15*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*(3*cos(d*x + c)^3 + 4*cos(d*x + c)^2 - (3*cos(d*x + c)^2 - cos
(d*x + c) + 13)*sin(d*x + c) + 14*cos(d*x + c) + 13)*sqrt(a*sin(d*x + c) + a))/(a*d*cos(d*x + c) + a*d*sin(d*x
+ c) + a*d)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.36 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.16 \[ \int \frac {\cos ^3(c+d x) \cot (c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx=-\frac {\frac {15 \, \log \left (\frac {{\left | -2 \, \sqrt {2} + 4 \, \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}}{{\left | 2 \, \sqrt {2} + 4 \, \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}}\right )}{\sqrt {a} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} - \frac {2 \, \sqrt {2} {\left (12 \, a^{\frac {9}{2}} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 10 \, a^{\frac {9}{2}} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 15 \, a^{\frac {9}{2}} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{a^{5} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{15 \, d} \]

[In]

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

[Out]

-1/15*(15*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)))/(sqrt(a)*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))) - 2*sqrt(2)*(12*a^(9/2)*sin(-1/4*pi + 1/2*d*x + 1/2*c)^5 -
10*a^(9/2)*sin(-1/4*pi + 1/2*d*x + 1/2*c)^3 - 15*a^(9/2)*sin(-1/4*pi + 1/2*d*x + 1/2*c))/(a^5*sgn(cos(-1/4*pi
+ 1/2*d*x + 1/2*c))))/d

Mupad [F(-1)]

Timed out. \[ \int \frac {\cos ^3(c+d x) \cot (c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^4}{\sin \left (c+d\,x\right )\,\sqrt {a+a\,\sin \left (c+d\,x\right )}} \,d x \]

[In]

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

[Out]

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