\(\int \frac {\cot ^4(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx\) [469]

Optimal result

Integrand size = 23, antiderivative size = 135 \[ \int \frac {\cot ^4(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx=-\frac {7 \text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {a+a \sin (c+d x)}}\right )}{8 \sqrt {a} d}+\frac {9 \cot (c+d x)}{8 d \sqrt {a+a \sin (c+d x)}}+\frac {\cot (c+d x) \csc (c+d x)}{12 d \sqrt {a+a \sin (c+d x)}}-\frac {\cot (c+d x) \csc ^2(c+d x)}{3 d \sqrt {a+a \sin (c+d x)}} \]

[Out]

-7/8*arctanh(cos(d*x+c)*a^(1/2)/(a+a*sin(d*x+c))^(1/2))/d/a^(1/2)+9/8*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 = 0.46 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {2797, 2728, 212, 3123, 3063, 3064, 2852} \[ \int \frac {\cot ^4(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx=-\frac {7 \text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {a \sin (c+d x)+a}}\right )}{8 \sqrt {a} d}+\frac {9 \cot (c+d x)}{8 d \sqrt {a \sin (c+d x)+a}}-\frac {\cot (c+d x) \csc ^2(c+d x)}{3 d \sqrt {a \sin (c+d x)+a}}+\frac {\cot (c+d x) \csc (c+d x)}{12 d \sqrt {a \sin (c+d x)+a}} \]

[In]

Int[Cot[c + d*x]^4/Sqrt[a + a*Sin[c + d*x]],x]

[Out]

(-7*ArcTanh[(Sqrt[a]*Cos[c + d*x])/Sqrt[a + a*Sin[c + d*x]]])/(8*Sqrt[a]*d) + (9*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 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 2797

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)/tan[(e_.) + (f_.)*(x_)]^4, x_Symbol] :> Int[(a + b*Sin[e + f*x
])^m, x] + Int[(a + b*Sin[e + f*x])^m*((1 - 2*Sin[e + f*x]^2)/Sin[e + f*x]^4), 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 3063

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(B*c - A*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x]
)^(n + 1)/(f*(n + 1)*(c^2 - d^2))), x] + Dist[1/(b*(n + 1)*(c^2 - d^2)), Int[(a + b*Sin[e + f*x])^m*(c + d*Sin
[e + f*x])^(n + 1)*Simp[A*(a*d*m + b*c*(n + 1)) - B*(a*c*m + b*d*(n + 1)) + b*(B*c - A*d)*(m + n + 2)*Sin[e +
f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2
- d^2, 0] && LtQ[n, -1] && (IntegerQ[n] || EqQ[m + 1/2, 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 {1}{\sqrt {a+a \sin (c+d x)}} \, dx+\int \frac {\csc ^4(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) \csc ^2(c+d x)}{3 d \sqrt {a+a \sin (c+d x)}}+\frac {\int \frac {\csc ^3(c+d x) \left (-\frac {a}{2}-\frac {7}{2} a \sin (c+d x)\right )}{\sqrt {a+a \sin (c+d x)}} \, dx}{3 a}-\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 {\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 {\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 \frac {\csc ^2(c+d x) \left (-\frac {27 a^2}{4}-\frac {3}{4} a^2 \sin (c+d x)\right )}{\sqrt {a+a \sin (c+d x)}} \, dx}{6 a^2} \\ & = -\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 {9 \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 \frac {\csc (c+d x) \left (\frac {21 a^3}{8}-\frac {27}{8} a^3 \sin (c+d x)\right )}{\sqrt {a+a \sin (c+d x)}} \, dx}{6 a^3} \\ & = -\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 {9 \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 {7 \int \csc (c+d x) \sqrt {a+a \sin (c+d x)} \, dx}{16 a}-\int \frac {1}{\sqrt {a+a \sin (c+d x)}} \, dx \\ & = -\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 {9 \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 {7 \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 {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 {7 \text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {a+a \sin (c+d x)}}\right )}{8 \sqrt {a} d}+\frac {9 \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 = 0.75 (sec) , antiderivative size = 292, normalized size of antiderivative = 2.16 \[ \int \frac {\cot ^4(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx=\frac {\csc ^9\left (\frac {1}{2} (c+d x)\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right ) \left (36 \cos \left (\frac {1}{2} (c+d x)\right )-46 \cos \left (\frac {3}{2} (c+d x)\right )-54 \cos \left (\frac {5}{2} (c+d x)\right )-36 \sin \left (\frac {1}{2} (c+d x)\right )-63 \log \left (1+\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right ) \sin (c+d x)+63 \log \left (1-\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right ) \sin (c+d x)-46 \sin \left (\frac {3}{2} (c+d x)\right )+54 \sin \left (\frac {5}{2} (c+d x)\right )+21 \log \left (1+\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right ) \sin (3 (c+d x))-21 \log \left (1-\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right ) \sin (3 (c+d x))\right )}{24 d \left (\csc ^2\left (\frac {1}{4} (c+d x)\right )-\sec ^2\left (\frac {1}{4} (c+d x)\right )\right )^3 \sqrt {a (1+\sin (c+d x))}} \]

[In]

Integrate[Cot[c + d*x]^4/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])*(36*Cos[(c + d*x)/2] - 46*Cos[(3*(c + d*x))/2] - 54*
Cos[(5*(c + d*x))/2] - 36*Sin[(c + d*x)/2] - 63*Log[1 + Cos[(c + d*x)/2] - Sin[(c + d*x)/2]]*Sin[c + d*x] + 63
*Log[1 - Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]*Sin[c + d*x] - 46*Sin[(3*(c + d*x))/2] + 54*Sin[(5*(c + d*x))/2]
+ 21*Log[1 + Cos[(c + d*x)/2] - Sin[(c + d*x)/2]]*Sin[3*(c + d*x)] - 21*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 = 0.13 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.07

method	result	size
default	\(-\frac {\left (1+\sin \left (d x +c \right )\right ) \sqrt {-a \left (\sin \left (d x +c \right )-1\right )}\, \left (21 \,\operatorname {arctanh}\left (\frac {\sqrt {-a \left (\sin \left (d x +c \right )-1\right )}}{\sqrt {a}}\right ) a^{3} \left (\sin ^{3}\left (d x +c \right )\right )-27 \left (-a \left (\sin \left (d x +c \right )-1\right )\right )^{\frac {5}{2}} \sqrt {a}+56 \left (-a \left (\sin \left (d x +c \right )-1\right )\right )^{\frac {3}{2}} a^{\frac {3}{2}}-21 \sqrt {-a \left (\sin \left (d x +c \right )-1\right )}\, a^{\frac {5}{2}}\right )}{24 a^{\frac {7}{2}} \sin \left (d x +c \right )^{3} \cos \left (d x +c \right ) \sqrt {a \left (1+\sin \left (d x +c \right )\right )}\, d}\)	\(144\)

[In]

int(cos(d*x+c)^4*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)*(21*arctanh((-a*(sin(d*x+c)-1))^(1/2)/a^(1/2))*a^3*sin(d*x+c)^3
-27*(-a*(sin(d*x+c)-1))^(5/2)*a^(1/2)+56*(-a*(sin(d*x+c)-1))^(3/2)*a^(3/2)-21*(-a*(sin(d*x+c)-1))^(1/2)*a^(5/2
))/a^(7/2)/sin(d*x+c)^3/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. 369 vs. \(2 (115) = 230\).

Time = 0.30 (sec) , antiderivative size = 369, normalized size of antiderivative = 2.73 \[ \int \frac {\cot ^4(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx=\frac {21 \, {\left (\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} - {\left (\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2} - \cos \left (d x + c\right ) - 1\right )} \sin \left (d x + c\right ) + 1\right )} \sqrt {a} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 7 \, a \cos \left (d x + c\right )^{2} - 4 \, {\left (\cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right ) + 3\right )} \sin \left (d x + c\right ) - 2 \, \cos \left (d x + c\right ) - 3\right )} \sqrt {a \sin \left (d x + c\right ) + a} \sqrt {a} - 9 \, a \cos \left (d x + c\right ) + {\left (a \cos \left (d x + c\right )^{2} + 8 \, a \cos \left (d x + c\right ) - a\right )} \sin \left (d x + c\right ) - a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right )^{2} - 1\right )} \sin \left (d x + c\right ) - \cos \left (d x + c\right ) - 1}\right ) - 4 \, {\left (27 \, \cos \left (d x + c\right )^{3} + 25 \, \cos \left (d x + c\right )^{2} - {\left (27 \, \cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) - 17\right )} \sin \left (d x + c\right ) - 19 \, \cos \left (d x + c\right ) - 17\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{96 \, {\left (a d \cos \left (d x + c\right )^{4} - 2 \, a d \cos \left (d x + c\right )^{2} + a d - {\left (a d \cos \left (d x + c\right )^{3} + a d \cos \left (d x + c\right )^{2} - a d \cos \left (d x + c\right ) - a d\right )} \sin \left (d x + c\right )\right )}} \]

[In]

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

[Out]

1/96*(21*(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*(27*cos(d*x + c)^3 + 25*cos(d*x + c)^2 - (27*cos(d*x + c)^2 + 2*cos(d*x + c) - 17)*sin(d*x + c) -
19*cos(d*x + c) - 17)*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(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.48 (sec) , antiderivative size = 185, normalized size of antiderivative = 1.37 \[ \int \frac {\cot ^4(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx=\frac {\frac {21 \, \log \left ({\left | \frac {1}{2} \, \sqrt {2} + \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{\sqrt {a} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} - \frac {21 \, \log \left ({\left | -\frac {1}{2} \, \sqrt {2} + \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{\sqrt {a} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} + \frac {2 \, \sqrt {2} {\left (108 \, \sqrt {a} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 112 \, \sqrt {a} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 21 \, \sqrt {a} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (2 \, \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{3} a \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{48 \, d} \]

[In]

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

[Out]

1/48*(21*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)))
- 21*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))) + 2
*sqrt(2)*(108*sqrt(a)*sin(-1/4*pi + 1/2*d*x + 1/2*c)^5 - 112*sqrt(a)*sin(-1/4*pi + 1/2*d*x + 1/2*c)^3 + 21*sqr
t(a)*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. \[ \int \frac {\cot ^4(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^4}{{\sin \left (c+d\,x\right )}^4\,\sqrt {a+a\,\sin \left (c+d\,x\right )}} \,d x \]

[In]

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

[Out]

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