\(\int \frac {\cot ^4(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx\) [478]

Optimal result

Integrand size = 23, antiderivative size = 144 \[ \int \frac {\cot ^4(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx=-\frac {\text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {a+a \sin (c+d x)}}\right )}{8 a^{3/2} d}-\frac {\cot (c+d x)}{8 a d \sqrt {a+a \sin (c+d x)}}+\frac {11 \cot (c+d x) \csc (c+d x)}{12 a d \sqrt {a+a \sin (c+d x)}}-\frac {\cot (c+d x) \csc ^2(c+d x) \sqrt {a+a \sin (c+d x)}}{3 a^2 d} \]

[Out]

-1/8*arctanh(cos(d*x+c)*a^(1/2)/(a+a*sin(d*x+c))^(1/2))/a^(3/2)/d-1/8*cot(d*x+c)/a/d/(a+a*sin(d*x+c))^(1/2)+11
/12*cot(d*x+c)*csc(d*x+c)/a/d/(a+a*sin(d*x+c))^(1/2)-1/3*cot(d*x+c)*csc(d*x+c)^2*(a+a*sin(d*x+c))^(1/2)/a^2/d

Rubi [A] (verified)

Time = 0.40 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {2796, 2851, 2852, 212, 3123, 3059} \[ \int \frac {\cot ^4(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx=-\frac {\text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {a \sin (c+d x)+a}}\right )}{8 a^{3/2} d}-\frac {\cot (c+d x) \csc ^2(c+d x) \sqrt {a \sin (c+d x)+a}}{3 a^2 d}-\frac {\cot (c+d x)}{8 a d \sqrt {a \sin (c+d x)+a}}+\frac {11 \cot (c+d x) \csc (c+d x)}{12 a d \sqrt {a \sin (c+d x)+a}} \]

[In]

Int[Cot[c + d*x]^4/(a + a*Sin[c + d*x])^(3/2),x]

[Out]

-1/8*ArcTanh[(Sqrt[a]*Cos[c + d*x])/Sqrt[a + a*Sin[c + d*x]]]/(a^(3/2)*d) - Cot[c + d*x]/(8*a*d*Sqrt[a + a*Sin
[c + d*x]]) + (11*Cot[c + d*x]*Csc[c + d*x])/(12*a*d*Sqrt[a + a*Sin[c + d*x]]) - (Cot[c + d*x]*Csc[c + d*x]^2*
Sqrt[a + a*Sin[c + d*x]])/(3*a^2*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 2796

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)/tan[(e_.) + (f_.)*(x_)]^4, x_Symbol] :> Dist[-2/(a*b), Int[(a
+ b*Sin[e + f*x])^(m + 2)/Sin[e + f*x]^3, x], x] + Dist[1/a^2, Int[(a + b*Sin[e + f*x])^(m + 2)*((1 + Sin[e +
f*x]^2)/Sin[e + f*x]^4), x], x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0] && IntegerQ[m - 1/2] && LtQ[m,
-1]

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 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]

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}& = \frac {\int \csc ^4(c+d x) \sqrt {a+a \sin (c+d x)} \left (1+\sin ^2(c+d x)\right ) \, dx}{a^2}-\frac {2 \int \csc ^3(c+d x) \sqrt {a+a \sin (c+d x)} \, dx}{a^2} \\ & = \frac {\cot (c+d x) \csc (c+d x)}{a d \sqrt {a+a \sin (c+d x)}}-\frac {\cot (c+d x) \csc ^2(c+d x) \sqrt {a+a \sin (c+d x)}}{3 a^2 d}+\frac {\int \csc ^3(c+d x) \sqrt {a+a \sin (c+d x)} \left (\frac {a}{2}+\frac {9}{2} a \sin (c+d x)\right ) \, dx}{3 a^3}-\frac {3 \int \csc ^2(c+d x) \sqrt {a+a \sin (c+d x)} \, dx}{2 a^2} \\ & = \frac {3 \cot (c+d x)}{2 a d \sqrt {a+a \sin (c+d x)}}+\frac {11 \cot (c+d x) \csc (c+d x)}{12 a d \sqrt {a+a \sin (c+d x)}}-\frac {\cot (c+d x) \csc ^2(c+d x) \sqrt {a+a \sin (c+d x)}}{3 a^2 d}-\frac {3 \int \csc (c+d x) \sqrt {a+a \sin (c+d x)} \, dx}{4 a^2}+\frac {13 \int \csc ^2(c+d x) \sqrt {a+a \sin (c+d x)} \, dx}{8 a^2} \\ & = -\frac {\cot (c+d x)}{8 a d \sqrt {a+a \sin (c+d x)}}+\frac {11 \cot (c+d x) \csc (c+d x)}{12 a d \sqrt {a+a \sin (c+d x)}}-\frac {\cot (c+d x) \csc ^2(c+d x) \sqrt {a+a \sin (c+d x)}}{3 a^2 d}+\frac {13 \int \csc (c+d x) \sqrt {a+a \sin (c+d x)} \, dx}{16 a^2}+\frac {3 \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 )}{2 a d} \\ & = \frac {3 \text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {a+a \sin (c+d x)}}\right )}{2 a^{3/2} d}-\frac {\cot (c+d x)}{8 a d \sqrt {a+a \sin (c+d x)}}+\frac {11 \cot (c+d x) \csc (c+d x)}{12 a d \sqrt {a+a \sin (c+d x)}}-\frac {\cot (c+d x) \csc ^2(c+d x) \sqrt {a+a \sin (c+d x)}}{3 a^2 d}-\frac {13 \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 a d} \\ & = -\frac {\text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {a+a \sin (c+d x)}}\right )}{8 a^{3/2} d}-\frac {\cot (c+d x)}{8 a d \sqrt {a+a \sin (c+d x)}}+\frac {11 \cot (c+d x) \csc (c+d x)}{12 a d \sqrt {a+a \sin (c+d x)}}-\frac {\cot (c+d x) \csc ^2(c+d x) \sqrt {a+a \sin (c+d x)}}{3 a^2 d} \\ \end{align*}

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(294\) vs. \(2(144)=288\).

Time = 0.92 (sec) , antiderivative size = 294, normalized size of antiderivative = 2.04 \[ \int \frac {\cot ^4(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx=\frac {\csc ^9\left (\frac {1}{2} (c+d x)\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^3 \left (-132 \cos \left (\frac {1}{2} (c+d x)\right )+62 \cos \left (\frac {3}{2} (c+d x)\right )+6 \cos \left (\frac {5}{2} (c+d x)\right )+132 \sin \left (\frac {1}{2} (c+d x)\right )-9 \log \left (1+\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right ) \sin (c+d x)+9 \log \left (1-\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right ) \sin (c+d x)+62 \sin \left (\frac {3}{2} (c+d x)\right )-6 \sin \left (\frac {5}{2} (c+d x)\right )+3 \log \left (1+\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right ) \sin (3 (c+d x))-3 \log \left (1-\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right ) \sin (3 (c+d x))\right )}{24 d \left (\csc ^2\left (\frac {1}{4} (c+d x)\right )-\sec ^2\left (\frac {1}{4} (c+d x)\right )\right )^3 (a (1+\sin (c+d x)))^{3/2}} \]

[In]

Integrate[Cot[c + d*x]^4/(a + a*Sin[c + d*x])^(3/2),x]

[Out]

(Csc[(c + d*x)/2]^9*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^3*(-132*Cos[(c + d*x)/2] + 62*Cos[(3*(c + d*x))/2] +
6*Cos[(5*(c + d*x))/2] + 132*Sin[(c + d*x)/2] - 9*Log[1 + Cos[(c + d*x)/2] - Sin[(c + d*x)/2]]*Sin[c + d*x] +
9*Log[1 - Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]*Sin[c + d*x] + 62*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*(a*(1 + Sin[c + d*x]))^(3/2))

Maple [A] (verified)

Time = 0.12 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.00

method	result	size
default	\(-\frac {\left (1+\sin \left (d x +c \right )\right ) \sqrt {-a \left (\sin \left (d x +c \right )-1\right )}\, \left (3 \left (-a \left (\sin \left (d x +c \right )-1\right )\right )^{\frac {5}{2}} a^{\frac {3}{2}}+3 \,\operatorname {arctanh}\left (\frac {\sqrt {-a \left (\sin \left (d x +c \right )-1\right )}}{\sqrt {a}}\right ) a^{4} \left (\sin ^{3}\left (d x +c \right )\right )+8 \left (-a \left (\sin \left (d x +c \right )-1\right )\right )^{\frac {3}{2}} a^{\frac {5}{2}}-3 \sqrt {-a \left (\sin \left (d x +c \right )-1\right )}\, a^{\frac {7}{2}}\right )}{24 a^{\frac {11}{2}} \sin \left (d x +c \right )^{3} \cos \left (d x +c \right ) \sqrt {a \left (1+\sin \left (d x +c \right )\right )}\, d}\)	\(144\)

[In]

int(cos(d*x+c)^4*csc(d*x+c)^4/(a+a*sin(d*x+c))^(3/2),x,method=_RETURNVERBOSE)

[Out]

-1/24/a^(11/2)*(1+sin(d*x+c))*(-a*(sin(d*x+c)-1))^(1/2)*(3*(-a*(sin(d*x+c)-1))^(5/2)*a^(3/2)+3*arctanh((-a*(si
n(d*x+c)-1))^(1/2)/a^(1/2))*a^4*sin(d*x+c)^3+8*(-a*(sin(d*x+c)-1))^(3/2)*a^(5/2)-3*(-a*(sin(d*x+c)-1))^(1/2)*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. 383 vs. \(2 (124) = 248\).

Time = 0.28 (sec) , antiderivative size = 383, normalized size of antiderivative = 2.66 \[ \int \frac {\cot ^4(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx=\frac {3 \, {\left (\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} - {\left (\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2} - \cos \left (d x + c\right ) - 1\right )} \sin \left (d x + c\right ) + 1\right )} \sqrt {a} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 7 \, a \cos \left (d x + c\right )^{2} - 4 \, {\left (\cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right ) + 3\right )} \sin \left (d x + c\right ) - 2 \, \cos \left (d x + c\right ) - 3\right )} \sqrt {a \sin \left (d x + c\right ) + a} \sqrt {a} - 9 \, a \cos \left (d x + c\right ) + {\left (a \cos \left (d x + c\right )^{2} + 8 \, a \cos \left (d x + c\right ) - a\right )} \sin \left (d x + c\right ) - a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right )^{2} - 1\right )} \sin \left (d x + c\right ) - \cos \left (d x + c\right ) - 1}\right ) + 4 \, {\left (3 \, \cos \left (d x + c\right )^{3} + 17 \, \cos \left (d x + c\right )^{2} - {\left (3 \, \cos \left (d x + c\right )^{2} - 14 \, \cos \left (d x + c\right ) - 25\right )} \sin \left (d x + c\right ) - 11 \, \cos \left (d x + c\right ) - 25\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{96 \, {\left (a^{2} d \cos \left (d x + c\right )^{4} - 2 \, a^{2} d \cos \left (d x + c\right )^{2} + a^{2} d - {\left (a^{2} d \cos \left (d x + c\right )^{3} + a^{2} d \cos \left (d x + c\right )^{2} - a^{2} d \cos \left (d x + c\right ) - a^{2} d\right )} \sin \left (d x + c\right )\right )}} \]

[In]

integrate(cos(d*x+c)^4*csc(d*x+c)^4/(a+a*sin(d*x+c))^(3/2),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 + 17*cos(d*x + c)^2 - (3*cos(d*x + c)^2 - 14*cos(d*x + c) - 25)*sin(d*x + c) - 11
*cos(d*x + c) - 25)*sqrt(a*sin(d*x + c) + a))/(a^2*d*cos(d*x + c)^4 - 2*a^2*d*cos(d*x + c)^2 + a^2*d - (a^2*d*
cos(d*x + c)^3 + a^2*d*cos(d*x + c)^2 - a^2*d*cos(d*x + c) - a^2*d)*sin(d*x + c))

Sympy [F(-1)]

Timed out. \[ \int \frac {\cot ^4(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx=\text {Timed out} \]

[In]

integrate(cos(d*x+c)**4*csc(d*x+c)**4/(a+a*sin(d*x+c))**(3/2),x)

[Out]

Timed out

Maxima [F(-1)]

Timed out. \[ \int \frac {\cot ^4(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx=\text {Timed out} \]

[In]

integrate(cos(d*x+c)^4*csc(d*x+c)^4/(a+a*sin(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

Timed out

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.17 \[ \int \frac {\cot ^4(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx=-\frac {\sqrt {2} \sqrt {a} {\left (\frac {3 \, \sqrt {2} \log \left (\frac {{\left | -2 \, \sqrt {2} + 4 \, \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}}{{\left | 2 \, \sqrt {2} + 4 \, \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}}\right )}{a^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} + \frac {4 \, {\left (12 \, \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 16 \, \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 3 \, \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (2 \, \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{3} a^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}\right )}}{96 \, d} \]

[In]

integrate(cos(d*x+c)^4*csc(d*x+c)^4/(a+a*sin(d*x+c))^(3/2),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^2*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^2*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))))/d

Mupad [F(-1)]

Timed out. \[ \int \frac {\cot ^4(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^4}{{\sin \left (c+d\,x\right )}^4\,{\left (a+a\,\sin \left (c+d\,x\right )\right )}^{3/2}} \,d x \]

[In]

int(cos(c + d*x)^4/(sin(c + d*x)^4*(a + a*sin(c + d*x))^(3/2)),x)

[Out]

int(cos(c + d*x)^4/(sin(c + d*x)^4*(a + a*sin(c + d*x))^(3/2)), x)