\(\int \frac {\cos (c+d x) \cot ^3(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx\) [477]

Optimal result

Integrand size = 29, antiderivative size = 106 \[ \int \frac {\cos (c+d x) \cot ^3(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx=-\frac {3 \text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {a+a \sin (c+d x)}}\right )}{4 a^{3/2} d}+\frac {7 \cot (c+d x)}{4 a d \sqrt {a+a \sin (c+d x)}}-\frac {\cot (c+d x) \csc (c+d x) \sqrt {a+a \sin (c+d x)}}{2 a^2 d} \]

[Out]

-3/4*arctanh(cos(d*x+c)*a^(1/2)/(a+a*sin(d*x+c))^(1/2))/a^(3/2)/d+7/4*cot(d*x+c)/a/d/(a+a*sin(d*x+c))^(1/2)-1/
2*cot(d*x+c)*csc(d*x+c)*(a+a*sin(d*x+c))^(1/2)/a^2/d

Rubi [A] (verified)

Time = 0.38 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.207, Rules used = {2959, 2851, 2852, 212, 3123, 3059} \[ \int \frac {\cos (c+d x) \cot ^3(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx=-\frac {3 \text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {a \sin (c+d x)+a}}\right )}{4 a^{3/2} d}-\frac {\cot (c+d x) \csc (c+d x) \sqrt {a \sin (c+d x)+a}}{2 a^2 d}+\frac {7 \cot (c+d x)}{4 a d \sqrt {a \sin (c+d x)+a}} \]

[In]

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

[Out]

(-3*ArcTanh[(Sqrt[a]*Cos[c + d*x])/Sqrt[a + a*Sin[c + d*x]]])/(4*a^(3/2)*d) + (7*Cot[c + d*x])/(4*a*d*Sqrt[a +
a*Sin[c + d*x]]) - (Cot[c + d*x]*Csc[c + d*x]*Sqrt[a + a*Sin[c + d*x]])/(2*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 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 2959

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

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 ^3(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 ^2(c+d x) \sqrt {a+a \sin (c+d x)} \, dx}{a^2} \\ & = \frac {2 \cot (c+d x)}{a d \sqrt {a+a \sin (c+d x)}}-\frac {\cot (c+d x) \csc (c+d x) \sqrt {a+a \sin (c+d x)}}{2 a^2 d}+\frac {\int \csc ^2(c+d x) \sqrt {a+a \sin (c+d x)} \left (\frac {a}{2}+\frac {5}{2} a \sin (c+d x)\right ) \, dx}{2 a^3}-\frac {\int \csc (c+d x) \sqrt {a+a \sin (c+d x)} \, dx}{a^2} \\ & = \frac {7 \cot (c+d x)}{4 a d \sqrt {a+a \sin (c+d x)}}-\frac {\cot (c+d x) \csc (c+d x) \sqrt {a+a \sin (c+d x)}}{2 a^2 d}+\frac {11 \int \csc (c+d x) \sqrt {a+a \sin (c+d x)} \, dx}{8 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 )}{a d} \\ & = \frac {2 \text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {a+a \sin (c+d x)}}\right )}{a^{3/2} d}+\frac {7 \cot (c+d x)}{4 a d \sqrt {a+a \sin (c+d x)}}-\frac {\cot (c+d x) \csc (c+d x) \sqrt {a+a \sin (c+d x)}}{2 a^2 d}-\frac {11 \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 )}{4 a d} \\ & = -\frac {3 \text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {a+a \sin (c+d x)}}\right )}{4 a^{3/2} d}+\frac {7 \cot (c+d x)}{4 a d \sqrt {a+a \sin (c+d x)}}-\frac {\cot (c+d x) \csc (c+d x) \sqrt {a+a \sin (c+d x)}}{2 a^2 d} \\ \end{align*}

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(274\) vs. \(2(106)=212\).

Time = 1.68 (sec) , antiderivative size = 274, normalized size of antiderivative = 2.58 \[ \int \frac {\cos (c+d x) \cot ^3(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx=\frac {\left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^3 \left (-24+12 \cot \left (\frac {1}{4} (c+d x)\right )-\csc ^2\left (\frac {1}{4} (c+d x)\right )-12 \log \left (1+\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+12 \log \left (1-\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+\sec ^2\left (\frac {1}{4} (c+d x)\right )+\frac {2}{\left (\cos \left (\frac {1}{4} (c+d x)\right )-\sin \left (\frac {1}{4} (c+d x)\right )\right )^2}-\frac {24 \sin \left (\frac {1}{4} (c+d x)\right )}{\cos \left (\frac {1}{4} (c+d x)\right )-\sin \left (\frac {1}{4} (c+d x)\right )}-\frac {2}{\left (\cos \left (\frac {1}{4} (c+d x)\right )+\sin \left (\frac {1}{4} (c+d x)\right )\right )^2}+\frac {24 \sin \left (\frac {1}{4} (c+d x)\right )}{\cos \left (\frac {1}{4} (c+d x)\right )+\sin \left (\frac {1}{4} (c+d x)\right )}+12 \tan \left (\frac {1}{4} (c+d x)\right )\right )}{32 d (a (1+\sin (c+d x)))^{3/2}} \]

[In]

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

[Out]

((Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^3*(-24 + 12*Cot[(c + d*x)/4] - Csc[(c + d*x)/4]^2 - 12*Log[1 + Cos[(c +
d*x)/2] - Sin[(c + d*x)/2]] + 12*Log[1 - Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + Sec[(c + d*x)/4]^2 + 2/(Cos[(
c + d*x)/4] - Sin[(c + d*x)/4])^2 - (24*Sin[(c + d*x)/4])/(Cos[(c + d*x)/4] - Sin[(c + d*x)/4]) - 2/(Cos[(c +
d*x)/4] + Sin[(c + d*x)/4])^2 + (24*Sin[(c + d*x)/4])/(Cos[(c + d*x)/4] + Sin[(c + d*x)/4]) + 12*Tan[(c + d*x)
/4]))/(32*d*(a*(1 + Sin[c + d*x]))^(3/2))

Maple [A] (verified)

Time = 0.13 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.19

method	result	size
default	\(-\frac {\left (1+\sin \left (d x +c \right )\right ) \sqrt {-a \left (\sin \left (d x +c \right )-1\right )}\, \left (3 \,\operatorname {arctanh}\left (\frac {\sqrt {-a \left (\sin \left (d x +c \right )-1\right )}}{\sqrt {a}}\right ) \left (\sin ^{2}\left (d x +c \right )\right ) a^{2}+5 \left (-a \left (\sin \left (d x +c \right )-1\right )\right )^{\frac {3}{2}} \sqrt {a}-3 \sqrt {-a \left (\sin \left (d x +c \right )-1\right )}\, a^{\frac {3}{2}}\right )}{4 a^{\frac {7}{2}} \sin \left (d x +c \right )^{2} \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d}\)	\(126\)

[In]

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

[Out]

-1/4/a^(7/2)*(1+sin(d*x+c))*(-a*(sin(d*x+c)-1))^(1/2)*(3*arctanh((-a*(sin(d*x+c)-1))^(1/2)/a^(1/2))*sin(d*x+c)
^2*a^2+5*(-a*(sin(d*x+c)-1))^(3/2)*a^(1/2)-3*(-a*(sin(d*x+c)-1))^(1/2)*a^(3/2))/sin(d*x+c)^2/cos(d*x+c)/(a+a*s
in(d*x+c))^(1/2)/d

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 337 vs. \(2 (90) = 180\).

Time = 0.28 (sec) , antiderivative size = 337, normalized size of antiderivative = 3.18 \[ \int \frac {\cos (c+d x) \cot ^3(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx=\frac {3 \, {\left (\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right )^{2} - 1\right )} \sin \left (d x + c\right ) - \cos \left (d x + c\right ) - 1\right )} \sqrt {a} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 7 \, a \cos \left (d x + c\right )^{2} - 4 \, {\left (\cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right ) + 3\right )} \sin \left (d x + c\right ) - 2 \, \cos \left (d x + c\right ) - 3\right )} \sqrt {a \sin \left (d x + c\right ) + a} \sqrt {a} - 9 \, a \cos \left (d x + c\right ) + {\left (a \cos \left (d x + c\right )^{2} + 8 \, a \cos \left (d x + c\right ) - a\right )} \sin \left (d x + c\right ) - a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right )^{2} - 1\right )} \sin \left (d x + c\right ) - \cos \left (d x + c\right ) - 1}\right ) - 4 \, {\left (5 \, \cos \left (d x + c\right )^{2} + {\left (5 \, \cos \left (d x + c\right ) + 7\right )} \sin \left (d x + c\right ) - 2 \, \cos \left (d x + c\right ) - 7\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{16 \, {\left (a^{2} d \cos \left (d x + c\right )^{3} + a^{2} d \cos \left (d x + c\right )^{2} - a^{2} d \cos \left (d x + c\right ) - a^{2} d + {\left (a^{2} d \cos \left (d x + c\right )^{2} - a^{2} d\right )} \sin \left (d x + c\right )\right )}} \]

[In]

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

[Out]

1/16*(3*(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)*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*(5*cos(d
*x + c)^2 + (5*cos(d*x + c) + 7)*sin(d*x + c) - 2*cos(d*x + c) - 7)*sqrt(a*sin(d*x + c) + a))/(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 + (a^2*d*cos(d*x + c)^2 - a^2*d)*sin(d*x + c))

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F(-1)]

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

[In]

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

[Out]

Timed out

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.43 \[ \int \frac {\cos (c+d x) \cot ^3(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx=-\frac {\sqrt {2} \sqrt {a} {\left (\frac {3 \, \sqrt {2} \log \left (\frac {{\left | -2 \, \sqrt {2} + 4 \, \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}}{{\left | 2 \, \sqrt {2} + 4 \, \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}}\right )}{a^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} - \frac {4 \, {\left (10 \, \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 3 \, \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (2 \, \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2} a^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}\right )}}{16 \, d} \]

[In]

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

[Out]

-1/16*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*(10*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)^2*a^2*sgn(cos(-1/4*pi + 1/2*d*
x + 1/2*c))))/d

Mupad [F(-1)]

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

[In]

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

[Out]

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