\(\int \frac {\cos (c+d x) \cot ^3(c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx\) [487]

Optimal result

Integrand size = 29, antiderivative size = 153 \[ \int \frac {\cos (c+d x) \cot ^3(c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx=-\frac {23 \text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {a+a \sin (c+d x)}}\right )}{4 a^{5/2} d}+\frac {4 \sqrt {2} \text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {2} \sqrt {a+a \sin (c+d x)}}\right )}{a^{5/2} d}+\frac {9 \cot (c+d x)}{4 a^2 d \sqrt {a+a \sin (c+d x)}}-\frac {\cot (c+d x) \csc (c+d x)}{2 a^2 d \sqrt {a+a \sin (c+d x)}} \]

[Out]

-23/4*arctanh(cos(d*x+c)*a^(1/2)/(a+a*sin(d*x+c))^(1/2))/a^(5/2)/d+4*arctanh(1/2*cos(d*x+c)*a^(1/2)*2^(1/2)/(a
+a*sin(d*x+c))^(1/2))/a^(5/2)/d*2^(1/2)+9/4*cot(d*x+c)/a^2/d/(a+a*sin(d*x+c))^(1/2)-1/2*cot(d*x+c)*csc(d*x+c)/
a^2/d/(a+a*sin(d*x+c))^(1/2)

Rubi [A] (verified)

Time = 0.54 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.276, Rules used = {2959, 2858, 3064, 2728, 212, 2852, 3123, 3063} \[ \int \frac {\cos (c+d x) \cot ^3(c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx=-\frac {23 \text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {a \sin (c+d x)+a}}\right )}{4 a^{5/2} d}+\frac {4 \sqrt {2} \text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {2} \sqrt {a \sin (c+d x)+a}}\right )}{a^{5/2} d}+\frac {9 \cot (c+d x)}{4 a^2 d \sqrt {a \sin (c+d x)+a}}-\frac {\cot (c+d x) \csc (c+d x)}{2 a^2 d \sqrt {a \sin (c+d x)+a}} \]

[In]

Int[(Cos[c + d*x]*Cot[c + d*x]^3)/(a + a*Sin[c + d*x])^(5/2),x]

[Out]

(-23*ArcTanh[(Sqrt[a]*Cos[c + d*x])/Sqrt[a + a*Sin[c + d*x]]])/(4*a^(5/2)*d) + (4*Sqrt[2]*ArcTanh[(Sqrt[a]*Cos
[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sin[c + d*x]])])/(a^(5/2)*d) + (9*Cot[c + d*x])/(4*a^2*d*Sqrt[a + a*Sin[c + d*x
]]) - (Cot[c + d*x]*Csc[c + d*x])/(2*a^2*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 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 2858

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

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 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}& = \frac {\int \frac {\csc ^3(c+d x) \left (1+\sin ^2(c+d x)\right )}{\sqrt {a+a \sin (c+d x)}} \, dx}{a^2}-\frac {2 \int \frac {\csc ^2(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx}{a^2} \\ & = \frac {2 \cot (c+d x)}{a^2 d \sqrt {a+a \sin (c+d x)}}-\frac {\cot (c+d x) \csc (c+d x)}{2 a^2 d \sqrt {a+a \sin (c+d x)}}+\frac {\int \frac {\csc ^2(c+d x) \left (-\frac {a}{2}+\frac {7}{2} a \sin (c+d x)\right )}{\sqrt {a+a \sin (c+d x)}} \, dx}{2 a^3}+\frac {\int \frac {\csc (c+d x) (a-a \sin (c+d x))}{\sqrt {a+a \sin (c+d x)}} \, dx}{a^3} \\ & = \frac {9 \cot (c+d x)}{4 a^2 d \sqrt {a+a \sin (c+d x)}}-\frac {\cot (c+d x) \csc (c+d x)}{2 a^2 d \sqrt {a+a \sin (c+d x)}}+\frac {\int \frac {\csc (c+d x) \left (\frac {15 a^2}{4}-\frac {1}{4} a^2 \sin (c+d x)\right )}{\sqrt {a+a \sin (c+d x)}} \, dx}{2 a^4}+\frac {\int \csc (c+d x) \sqrt {a+a \sin (c+d x)} \, dx}{a^3}-\frac {2 \int \frac {1}{\sqrt {a+a \sin (c+d x)}} \, dx}{a^2} \\ & = \frac {9 \cot (c+d x)}{4 a^2 d \sqrt {a+a \sin (c+d x)}}-\frac {\cot (c+d x) \csc (c+d x)}{2 a^2 d \sqrt {a+a \sin (c+d x)}}+\frac {15 \int \csc (c+d x) \sqrt {a+a \sin (c+d x)} \, dx}{8 a^3}-\frac {2 \int \frac {1}{\sqrt {a+a \sin (c+d x)}} \, dx}{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^2 d}+\frac {4 \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 )}{a^2 d} \\ & = -\frac {2 \text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {a+a \sin (c+d x)}}\right )}{a^{5/2} d}+\frac {2 \sqrt {2} \text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {2} \sqrt {a+a \sin (c+d x)}}\right )}{a^{5/2} d}+\frac {9 \cot (c+d x)}{4 a^2 d \sqrt {a+a \sin (c+d x)}}-\frac {\cot (c+d x) \csc (c+d x)}{2 a^2 d \sqrt {a+a \sin (c+d x)}}-\frac {15 \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^2 d}+\frac {4 \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 )}{a^2 d} \\ & = -\frac {23 \text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {a+a \sin (c+d x)}}\right )}{4 a^{5/2} d}+\frac {4 \sqrt {2} \text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {2} \sqrt {a+a \sin (c+d x)}}\right )}{a^{5/2} d}+\frac {9 \cot (c+d x)}{4 a^2 d \sqrt {a+a \sin (c+d x)}}-\frac {\cot (c+d x) \csc (c+d x)}{2 a^2 d \sqrt {a+a \sin (c+d x)}} \\ \end{align*}

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 2.82 (sec) , antiderivative size = 309, normalized size of antiderivative = 2.02 \[ \int \frac {\cos (c+d x) \cot ^3(c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx=\frac {\left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^5 \left (-40-(256+256 i) (-1)^{3/4} \text {arctanh}\left (\left (\frac {1}{2}+\frac {i}{2}\right ) (-1)^{3/4} \left (-1+\tan \left (\frac {1}{4} (c+d x)\right )\right )\right )+20 \cot \left (\frac {1}{4} (c+d x)\right )-\csc ^2\left (\frac {1}{4} (c+d x)\right )-92 \log \left (1+\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+92 \log \left (1-\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+\sec ^2\left (\frac {1}{4} (c+d x)\right )+\frac {2}{\left (\cos \left (\frac {1}{4} (c+d x)\right )-\sin \left (\frac {1}{4} (c+d x)\right )\right )^2}-\frac {40 \sin \left (\frac {1}{4} (c+d x)\right )}{\cos \left (\frac {1}{4} (c+d x)\right )-\sin \left (\frac {1}{4} (c+d x)\right )}-\frac {2}{\left (\cos \left (\frac {1}{4} (c+d x)\right )+\sin \left (\frac {1}{4} (c+d x)\right )\right )^2}+\frac {40 \sin \left (\frac {1}{4} (c+d x)\right )}{\cos \left (\frac {1}{4} (c+d x)\right )+\sin \left (\frac {1}{4} (c+d x)\right )}+20 \tan \left (\frac {1}{4} (c+d x)\right )\right )}{32 d (a (1+\sin (c+d x)))^{5/2}} \]

[In]

Integrate[(Cos[c + d*x]*Cot[c + d*x]^3)/(a + a*Sin[c + d*x])^(5/2),x]

[Out]

((Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^5*(-40 - (256 + 256*I)*(-1)^(3/4)*ArcTanh[(1/2 + I/2)*(-1)^(3/4)*(-1 +
Tan[(c + d*x)/4])] + 20*Cot[(c + d*x)/4] - Csc[(c + d*x)/4]^2 - 92*Log[1 + Cos[(c + d*x)/2] - Sin[(c + d*x)/2]
] + 92*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 - (40*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 + (40*Sin[(c + d*x)/4])/(Cos[(c + d*x)/4] + Sin[(c + d*x)/4]) + 20*Tan[(c + d*x)/4]))/(32*d*(a*(1 + Sin[c +
d*x]))^(5/2))

Maple [A] (verified)

Time = 0.14 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.07

method	result	size
default	\(-\frac {\left (1+\sin \left (d x +c \right )\right ) \sqrt {-a \left (\sin \left (d x +c \right )-1\right )}\, \left (23 \,\operatorname {arctanh}\left (\frac {\sqrt {-a \left (\sin \left (d x +c \right )-1\right )}}{\sqrt {a}}\right ) a^{3} \left (\sin ^{2}\left (d x +c \right )\right )-16 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-a \left (\sin \left (d x +c \right )-1\right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{3} \left (\sin ^{2}\left (d x +c \right )\right )+9 \left (-a \left (\sin \left (d x +c \right )-1\right )\right )^{\frac {3}{2}} a^{\frac {3}{2}}-7 \sqrt {-a \left (\sin \left (d x +c \right )-1\right )}\, a^{\frac {5}{2}}\right )}{4 a^{\frac {11}{2}} \sin \left (d x +c \right )^{2} \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d}\)	\(164\)

[In]

int(cos(d*x+c)^4*csc(d*x+c)^3/(a+a*sin(d*x+c))^(5/2),x,method=_RETURNVERBOSE)

[Out]

-1/4/a^(11/2)*(1+sin(d*x+c))*(-a*(sin(d*x+c)-1))^(1/2)*(23*arctanh((-a*(sin(d*x+c)-1))^(1/2)/a^(1/2))*a^3*sin(
d*x+c)^2-16*2^(1/2)*arctanh(1/2*(-a*(sin(d*x+c)-1))^(1/2)*2^(1/2)/a^(1/2))*a^3*sin(d*x+c)^2+9*(-a*(sin(d*x+c)-
1))^(3/2)*a^(3/2)-7*(-a*(sin(d*x+c)-1))^(1/2)*a^(5/2))/sin(d*x+c)^2/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. 508 vs. \(2 (128) = 256\).

Time = 0.31 (sec) , antiderivative size = 508, normalized size of antiderivative = 3.32 \[ \int \frac {\cos (c+d x) \cot ^3(c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx=\frac {23 \, {\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 ) + \frac {32 \, \sqrt {2} {\left (a \cos \left (d x + c\right )^{3} + a \cos \left (d x + c\right )^{2} - a \cos \left (d x + c\right ) + {\left (a \cos \left (d x + c\right )^{2} - a\right )} \sin \left (d x + c\right ) - a\right )} \log \left (-\frac {\cos \left (d x + c\right )^{2} - {\left (\cos \left (d x + c\right ) - 2\right )} \sin \left (d x + c\right ) + \frac {2 \, \sqrt {2} \sqrt {a \sin \left (d x + c\right ) + a} {\left (\cos \left (d x + c\right ) - \sin \left (d x + c\right ) + 1\right )}}{\sqrt {a}} + 3 \, \cos \left (d x + c\right ) + 2}{\cos \left (d x + c\right )^{2} - {\left (\cos \left (d x + c\right ) + 2\right )} \sin \left (d x + c\right ) - \cos \left (d x + c\right ) - 2}\right )}{\sqrt {a}} - 4 \, {\left (9 \, \cos \left (d x + c\right )^{2} + {\left (9 \, \cos \left (d x + c\right ) + 11\right )} \sin \left (d x + c\right ) - 2 \, \cos \left (d x + c\right ) - 11\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{16 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + a^{3} d \cos \left (d x + c\right )^{2} - a^{3} d \cos \left (d x + c\right ) - a^{3} d + {\left (a^{3} d \cos \left (d x + c\right )^{2} - a^{3} 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))^(5/2),x, algorithm="fricas")

[Out]

1/16*(23*(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)) + 32*sqrt(2
)*(a*cos(d*x + c)^3 + a*cos(d*x + c)^2 - a*cos(d*x + c) + (a*cos(d*x + c)^2 - a)*sin(d*x + c) - a)*log(-(cos(d
*x + c)^2 - (cos(d*x + c) - 2)*sin(d*x + c) + 2*sqrt(2)*sqrt(a*sin(d*x + c) + a)*(cos(d*x + c) - sin(d*x + c)
+ 1)/sqrt(a) + 3*cos(d*x + c) + 2)/(cos(d*x + c)^2 - (cos(d*x + c) + 2)*sin(d*x + c) - cos(d*x + c) - 2))/sqrt
(a) - 4*(9*cos(d*x + c)^2 + (9*cos(d*x + c) + 11)*sin(d*x + c) - 2*cos(d*x + c) - 11)*sqrt(a*sin(d*x + c) + a)
)/(a^3*d*cos(d*x + c)^3 + a^3*d*cos(d*x + c)^2 - a^3*d*cos(d*x + c) - a^3*d + (a^3*d*cos(d*x + c)^2 - a^3*d)*s
in(d*x + c))

Sympy [F(-1)]

Timed out. \[ \int \frac {\cos (c+d x) \cot ^3(c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx=\text {Timed out} \]

[In]

integrate(cos(d*x+c)**4*csc(d*x+c)**3/(a+a*sin(d*x+c))**(5/2),x)

[Out]

Timed out

Maxima [F(-1)]

Timed out. \[ \int \frac {\cos (c+d x) \cot ^3(c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx=\text {Timed out} \]

[In]

integrate(cos(d*x+c)^4*csc(d*x+c)^3/(a+a*sin(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

Timed out

Giac [A] (verification not implemented)

none

Time = 0.43 (sec) , antiderivative size = 224, normalized size of antiderivative = 1.46 \[ \int \frac {\cos (c+d x) \cot ^3(c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx=-\frac {\sqrt {2} \sqrt {a} {\left (\frac {23 \, \sqrt {2} \log \left (\frac {{\left | -2 \, \sqrt {2} + 4 \, \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}}{{\left | 2 \, \sqrt {2} + 4 \, \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}}\right )}{a^{3} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} + \frac {32 \, \log \left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}{a^{3} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} - \frac {32 \, \log \left (-\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}{a^{3} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} - \frac {4 \, {\left (18 \, \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 7 \, \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (2 \, \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2} a^{3} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}\right )}}{16 \, d} \]

[In]

integrate(cos(d*x+c)^4*csc(d*x+c)^3/(a+a*sin(d*x+c))^(5/2),x, algorithm="giac")

[Out]

-1/16*sqrt(2)*sqrt(a)*(23*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^3*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))) + 32*log(sin(-1/4*pi + 1/2*d*x + 1/2*c
) + 1)/(a^3*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))) - 32*log(-sin(-1/4*pi + 1/2*d*x + 1/2*c) + 1)/(a^3*sgn(cos(-1
/4*pi + 1/2*d*x + 1/2*c))) - 4*(18*sin(-1/4*pi + 1/2*d*x + 1/2*c)^3 - 7*sin(-1/4*pi + 1/2*d*x + 1/2*c))/((2*si
n(-1/4*pi + 1/2*d*x + 1/2*c)^2 - 1)^2*a^3*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))))/d

Mupad [F(-1)]

Timed out. \[ \int \frac {\cos (c+d x) \cot ^3(c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^4}{{\sin \left (c+d\,x\right )}^3\,{\left (a+a\,\sin \left (c+d\,x\right )\right )}^{5/2}} \,d x \]

[In]

int(cos(c + d*x)^4/(sin(c + d*x)^3*(a + a*sin(c + d*x))^(5/2)),x)

[Out]

int(cos(c + d*x)^4/(sin(c + d*x)^3*(a + a*sin(c + d*x))^(5/2)), x)