\(\int \cos ^2(c+d x) \cot ^2(c+d x) \sqrt {a+a \sin (c+d x)} \, dx\) [446]

Optimal result

Integrand size = 31, antiderivative size = 148 \[ \int \cos ^2(c+d x) \cot ^2(c+d x) \sqrt {a+a \sin (c+d x)} \, dx=-\frac {\sqrt {a} \text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {a+a \sin (c+d x)}}\right )}{d}+\frac {61 a \cos (c+d x)}{15 d \sqrt {a+a \sin (c+d x)}}+\frac {4 \cos (c+d x) \sqrt {a+a \sin (c+d x)}}{15 d}-\frac {\cot (c+d x) \sqrt {a+a \sin (c+d x)}}{d}-\frac {2 \cos (c+d x) (a+a \sin (c+d x))^{3/2}}{5 a d} \]

[Out]

-2/5*cos(d*x+c)*(a+a*sin(d*x+c))^(3/2)/a/d-arctanh(cos(d*x+c)*a^(1/2)/(a+a*sin(d*x+c))^(1/2))*a^(1/2)/d+61/15*
a*cos(d*x+c)/d/(a+a*sin(d*x+c))^(1/2)+4/15*cos(d*x+c)*(a+a*sin(d*x+c))^(1/2)/d-cot(d*x+c)*(a+a*sin(d*x+c))^(1/
2)/d

Rubi [A] (verified)

Time = 0.36 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.258, Rules used = {2960, 2838, 2830, 2725, 3123, 3060, 2852, 212} \[ \int \cos ^2(c+d x) \cot ^2(c+d x) \sqrt {a+a \sin (c+d x)} \, dx=-\frac {\sqrt {a} \text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {a \sin (c+d x)+a}}\right )}{d}-\frac {2 \cos (c+d x) (a \sin (c+d x)+a)^{3/2}}{5 a d}+\frac {4 \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{15 d}+\frac {61 a \cos (c+d x)}{15 d \sqrt {a \sin (c+d x)+a}}-\frac {\cot (c+d x) \sqrt {a \sin (c+d x)+a}}{d} \]

[In]

Int[Cos[c + d*x]^2*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) + (61*a*Cos[c + d*x])/(15*d*Sqrt[a + a
*Sin[c + d*x]]) + (4*Cos[c + d*x]*Sqrt[a + a*Sin[c + d*x]])/(15*d) - (Cot[c + d*x]*Sqrt[a + a*Sin[c + d*x]])/d
- (2*Cos[c + d*x]*(a + a*Sin[c + d*x])^(3/2))/(5*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 2725

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[-2*b*(Cos[c + d*x]/(d*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 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]

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 \sin ^2(c+d x) \sqrt {a+a \sin (c+d x)} \, dx+\int \csc ^2(c+d x) \sqrt {a+a \sin (c+d x)} \left (1-2 \sin ^2(c+d x)\right ) \, dx \\ & = -\frac {\cot (c+d x) \sqrt {a+a \sin (c+d x)}}{d}-\frac {2 \cos (c+d x) (a+a \sin (c+d x))^{3/2}}{5 a d}+\frac {2 \int \left (\frac {3 a}{2}-a \sin (c+d x)\right ) \sqrt {a+a \sin (c+d x)} \, dx}{5 a}+\frac {\int \csc (c+d x) \left (\frac {a}{2}-\frac {5}{2} a \sin (c+d x)\right ) \sqrt {a+a \sin (c+d x)} \, dx}{a} \\ & = \frac {5 a \cos (c+d x)}{d \sqrt {a+a \sin (c+d x)}}+\frac {4 \cos (c+d x) \sqrt {a+a \sin (c+d x)}}{15 d}-\frac {\cot (c+d x) \sqrt {a+a \sin (c+d x)}}{d}-\frac {2 \cos (c+d x) (a+a \sin (c+d x))^{3/2}}{5 a d}+\frac {7}{15} \int \sqrt {a+a \sin (c+d x)} \, dx+\frac {1}{2} \int \csc (c+d x) \sqrt {a+a \sin (c+d x)} \, dx \\ & = \frac {61 a \cos (c+d x)}{15 d \sqrt {a+a \sin (c+d x)}}+\frac {4 \cos (c+d x) \sqrt {a+a \sin (c+d x)}}{15 d}-\frac {\cot (c+d x) \sqrt {a+a \sin (c+d x)}}{d}-\frac {2 \cos (c+d x) (a+a \sin (c+d x))^{3/2}}{5 a 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 {61 a \cos (c+d x)}{15 d \sqrt {a+a \sin (c+d x)}}+\frac {4 \cos (c+d x) \sqrt {a+a \sin (c+d x)}}{15 d}-\frac {\cot (c+d x) \sqrt {a+a \sin (c+d x)}}{d}-\frac {2 \cos (c+d x) (a+a \sin (c+d x))^{3/2}}{5 a d} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.17 (sec) , antiderivative size = 258, normalized size of antiderivative = 1.74 \[ \int \cos ^2(c+d x) \cot ^2(c+d x) \sqrt {a+a \sin (c+d x)} \, dx=\frac {\csc ^4\left (\frac {1}{2} (c+d x)\right ) \sqrt {a (1+\sin (c+d x))} \left (-155 \cos \left (\frac {1}{2} (c+d x)\right )+87 \cos \left (\frac {3}{2} (c+d x)\right )+5 \cos \left (\frac {5}{2} (c+d x)\right )+3 \cos \left (\frac {7}{2} (c+d x)\right )+155 \sin \left (\frac {1}{2} (c+d x)\right )-30 \log \left (1+\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right ) \sin (c+d x)+30 \log \left (1-\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right ) \sin (c+d x)+87 \sin \left (\frac {3}{2} (c+d x)\right )-5 \sin \left (\frac {5}{2} (c+d x)\right )+3 \sin \left (\frac {7}{2} (c+d x)\right )\right )}{30 d \left (1+\cot \left (\frac {1}{2} (c+d x)\right )\right ) \left (\csc \left (\frac {1}{4} (c+d x)\right )-\sec \left (\frac {1}{4} (c+d x)\right )\right ) \left (\csc \left (\frac {1}{4} (c+d x)\right )+\sec \left (\frac {1}{4} (c+d x)\right )\right )} \]

[In]

Integrate[Cos[c + d*x]^2*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])]*(-155*Cos[(c + d*x)/2] + 87*Cos[(3*(c + d*x))/2] + 5*Cos[(5*(c
+ d*x))/2] + 3*Cos[(7*(c + d*x))/2] + 155*Sin[(c + d*x)/2] - 30*Log[1 + Cos[(c + d*x)/2] - Sin[(c + d*x)/2]]*S
in[c + d*x] + 30*Log[1 - Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]*Sin[c + d*x] + 87*Sin[(3*(c + d*x))/2] - 5*Sin[(
5*(c + d*x))/2] + 3*Sin[(7*(c + d*x))/2]))/(30*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 = 0.12 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.09

method	result	size
default	\(-\frac {\left (1+\sin \left (d x +c \right )\right ) \sqrt {-a \left (\sin \left (d x +c \right )-1\right )}\, \left (\sin \left (d x +c \right ) \left (6 \left (a -a \sin \left (d x +c \right )\right )^{\frac {5}{2}} a^{\frac {3}{2}}-20 \left (a -a \sin \left (d x +c \right )\right )^{\frac {3}{2}} a^{\frac {5}{2}}-30 \sqrt {a -a \sin \left (d x +c \right )}\, a^{\frac {7}{2}}+15 \,\operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (d x +c \right )}}{\sqrt {a}}\right ) a^{4}\right )+15 \sqrt {a -a \sin \left (d x +c \right )}\, a^{\frac {7}{2}}\right )}{15 a^{\frac {7}{2}} \sin \left (d x +c \right ) \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d}\)	\(162\)

[In]

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

[Out]

-1/15*(1+sin(d*x+c))*(-a*(sin(d*x+c)-1))^(1/2)/a^(7/2)*(sin(d*x+c)*(6*(a-a*sin(d*x+c))^(5/2)*a^(3/2)-20*(a-a*s
in(d*x+c))^(3/2)*a^(5/2)-30*(a-a*sin(d*x+c))^(1/2)*a^(7/2)+15*arctanh((a-a*sin(d*x+c))^(1/2)/a^(1/2))*a^4)+15*
(a-a*sin(d*x+c))^(1/2)*a^(7/2))/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. 320 vs. \(2 (128) = 256\).

Time = 0.29 (sec) , antiderivative size = 320, normalized size of antiderivative = 2.16 \[ \int \cos ^2(c+d x) \cot ^2(c+d x) \sqrt {a+a \sin (c+d x)} \, dx=\frac {15 \, {\left (\cos \left (d x + c\right )^{2} - {\left (\cos \left (d x + c\right ) + 1\right )} \sin \left (d x + c\right ) - 1\right )} \sqrt {a} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 7 \, a \cos \left (d x + c\right )^{2} - 4 \, {\left (\cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right ) + 3\right )} \sin \left (d x + c\right ) - 2 \, \cos \left (d x + c\right ) - 3\right )} \sqrt {a \sin \left (d x + c\right ) + a} \sqrt {a} - 9 \, a \cos \left (d x + c\right ) + {\left (a \cos \left (d x + c\right )^{2} + 8 \, a \cos \left (d x + c\right ) - a\right )} \sin \left (d x + c\right ) - a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right )^{2} - 1\right )} \sin \left (d x + c\right ) - \cos \left (d x + c\right ) - 1}\right ) - 4 \, {\left (6 \, \cos \left (d x + c\right )^{4} + 8 \, \cos \left (d x + c\right )^{3} + 40 \, \cos \left (d x + c\right )^{2} + {\left (6 \, \cos \left (d x + c\right )^{3} - 2 \, \cos \left (d x + c\right )^{2} + 38 \, \cos \left (d x + c\right ) + 61\right )} \sin \left (d x + c\right ) - 23 \, \cos \left (d x + c\right ) - 61\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{60 \, {\left (d \cos \left (d x + c\right )^{2} - {\left (d \cos \left (d x + c\right ) + d\right )} \sin \left (d x + c\right ) - d\right )}} \]

[In]

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

[Out]

1/60*(15*(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*(6*cos(d*x + c)^4 + 8*cos(d*x + c)^3 + 40
*cos(d*x + c)^2 + (6*cos(d*x + c)^3 - 2*cos(d*x + c)^2 + 38*cos(d*x + c) + 61)*sin(d*x + c) - 23*cos(d*x + c)
- 61)*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(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F]

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

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

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 208, normalized size of antiderivative = 1.41 \[ \int \cos ^2(c+d x) \cot ^2(c+d x) \sqrt {a+a \sin (c+d x)} \, dx=\frac {\sqrt {2} {\left (96 \, \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 160 \, \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 15 \, \sqrt {2} \log \left (\frac {{\left | -2 \, \sqrt {2} + 4 \, \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}}{{\left | 2 \, \sqrt {2} + 4 \, \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}}\right ) \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) - 120 \, \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \frac {60 \, \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{2 \, \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1}\right )} \sqrt {a}}{60 \, d} \]

[In]

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

[Out]

1/60*sqrt(2)*(96*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(-1/4*pi + 1/2*d*x + 1/2*c)^5 - 160*sgn(cos(-1/4*pi +
1/2*d*x + 1/2*c))*sin(-1/4*pi + 1/2*d*x + 1/2*c)^3 - 15*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)) - 120*sgn(cos(
-1/4*pi + 1/2*d*x + 1/2*c))*sin(-1/4*pi + 1/2*d*x + 1/2*c) - 60*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(-1/4*p
i + 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. \[ \int \cos ^2(c+d x) \cot ^2(c+d x) \sqrt {a+a \sin (c+d x)} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^4\,\sqrt {a+a\,\sin \left (c+d\,x\right )}}{{\sin \left (c+d\,x\right )}^2} \,d x \]

[In]

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

[Out]

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