\(\int \frac {\csc ^2(c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx\) [83]

Optimal result

Integrand size = 23, antiderivative size = 174 \[ \int \frac {\csc ^2(c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx=\frac {5 \text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {a+a \sin (c+d x)}}\right )}{a^{5/2} d}-\frac {115 \text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {2} \sqrt {a+a \sin (c+d x)}}\right )}{16 \sqrt {2} a^{5/2} d}+\frac {\cot (c+d x)}{4 d (a+a \sin (c+d x))^{5/2}}+\frac {15 \cot (c+d x)}{16 a d (a+a \sin (c+d x))^{3/2}}-\frac {35 \cot (c+d x)}{16 a^2 d \sqrt {a+a \sin (c+d x)}} \]

[Out]

5*arctanh(cos(d*x+c)*a^(1/2)/(a+a*sin(d*x+c))^(1/2))/a^(5/2)/d+1/4*cot(d*x+c)/d/(a+a*sin(d*x+c))^(5/2)+15/16*c
ot(d*x+c)/a/d/(a+a*sin(d*x+c))^(3/2)-115/32*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)-35/16*cot(d*x+c)/a^2/d/(a+a*sin(d*x+c))^(1/2)

Rubi [A] (verified)

Time = 0.38 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {2845, 3057, 3063, 3064, 2728, 212, 2852} \[ \int \frac {\csc ^2(c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx=\frac {5 \text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {a \sin (c+d x)+a}}\right )}{a^{5/2} d}-\frac {115 \text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {2} \sqrt {a \sin (c+d x)+a}}\right )}{16 \sqrt {2} a^{5/2} d}-\frac {35 \cot (c+d x)}{16 a^2 d \sqrt {a \sin (c+d x)+a}}+\frac {15 \cot (c+d x)}{16 a d (a \sin (c+d x)+a)^{3/2}}+\frac {\cot (c+d x)}{4 d (a \sin (c+d x)+a)^{5/2}} \]

[In]

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

[Out]

(5*ArcTanh[(Sqrt[a]*Cos[c + d*x])/Sqrt[a + a*Sin[c + d*x]]])/(a^(5/2)*d) - (115*ArcTanh[(Sqrt[a]*Cos[c + d*x])
/(Sqrt[2]*Sqrt[a + a*Sin[c + d*x]])])/(16*Sqrt[2]*a^(5/2)*d) + Cot[c + d*x]/(4*d*(a + a*Sin[c + d*x])^(5/2)) +
(15*Cot[c + d*x])/(16*a*d*(a + a*Sin[c + d*x])^(3/2)) - (35*Cot[c + d*x])/(16*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 2845

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[b^2*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^(n + 1)/(a*f*(2*m + 1)*(b*c - a*d))), x] + Dis
t[1/(a*(2*m + 1)*(b*c - a*d)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[b*c*(m + 1) - a*d*
(2*m + n + 2) + b*d*(m + n + 2)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d,
0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -1] &&  !GtQ[n, 0] && (IntegersQ[2*m, 2*n] || (IntegerQ
[m] && EqQ[c, 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 3057

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*(A*b - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*
x])^(n + 1)/(a*f*(2*m + 1)*(b*c - a*d))), x] + Dist[1/(a*(2*m + 1)*(b*c - a*d)), Int[(a + b*Sin[e + f*x])^(m +
1)*(c + d*Sin[e + f*x])^n*Simp[B*(a*c*m + b*d*(n + 1)) + A*(b*c*(m + 1) - a*d*(2*m + n + 2)) + d*(A*b - a*B)*
(m + n + 2)*Sin[e + f*x], x], 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[m, -2^(-1)] &&  !GtQ[n, 0] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c,
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]

Rubi steps \begin{align*} \text {integral}& = \frac {\cot (c+d x)}{4 d (a+a \sin (c+d x))^{5/2}}+\frac {\int \frac {\csc ^2(c+d x) \left (5 a-\frac {5}{2} a \sin (c+d x)\right )}{(a+a \sin (c+d x))^{3/2}} \, dx}{4 a^2} \\ & = \frac {\cot (c+d x)}{4 d (a+a \sin (c+d x))^{5/2}}+\frac {15 \cot (c+d x)}{16 a d (a+a \sin (c+d x))^{3/2}}+\frac {\int \frac {\csc ^2(c+d x) \left (\frac {35 a^2}{2}-\frac {45}{4} a^2 \sin (c+d x)\right )}{\sqrt {a+a \sin (c+d x)}} \, dx}{8 a^4} \\ & = \frac {\cot (c+d x)}{4 d (a+a \sin (c+d x))^{5/2}}+\frac {15 \cot (c+d x)}{16 a d (a+a \sin (c+d x))^{3/2}}-\frac {35 \cot (c+d x)}{16 a^2 d \sqrt {a+a \sin (c+d x)}}+\frac {\int \frac {\csc (c+d x) \left (-20 a^3+\frac {35}{4} a^3 \sin (c+d x)\right )}{\sqrt {a+a \sin (c+d x)}} \, dx}{8 a^5} \\ & = \frac {\cot (c+d x)}{4 d (a+a \sin (c+d x))^{5/2}}+\frac {15 \cot (c+d x)}{16 a d (a+a \sin (c+d x))^{3/2}}-\frac {35 \cot (c+d x)}{16 a^2 d \sqrt {a+a \sin (c+d x)}}-\frac {5 \int \csc (c+d x) \sqrt {a+a \sin (c+d x)} \, dx}{2 a^3}+\frac {115 \int \frac {1}{\sqrt {a+a \sin (c+d x)}} \, dx}{32 a^2} \\ & = \frac {\cot (c+d x)}{4 d (a+a \sin (c+d x))^{5/2}}+\frac {15 \cot (c+d x)}{16 a d (a+a \sin (c+d x))^{3/2}}-\frac {35 \cot (c+d x)}{16 a^2 d \sqrt {a+a \sin (c+d x)}}+\frac {5 \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 {115 \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 )}{16 a^2 d} \\ & = \frac {5 \text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {a+a \sin (c+d x)}}\right )}{a^{5/2} d}-\frac {115 \text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {2} \sqrt {a+a \sin (c+d x)}}\right )}{16 \sqrt {2} a^{5/2} d}+\frac {\cot (c+d x)}{4 d (a+a \sin (c+d x))^{5/2}}+\frac {15 \cot (c+d x)}{16 a d (a+a \sin (c+d x))^{3/2}}-\frac {35 \cot (c+d x)}{16 a^2 d \sqrt {a+a \sin (c+d x)}} \\ \end{align*}

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 3.37 (sec) , antiderivative size = 509, normalized size of antiderivative = 2.93 \[ \int \frac {\csc ^2(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 ) \left (8 \sin \left (\frac {1}{2} (c+d x)\right )-4 \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+38 \sin \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 )^2-19 \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^3+8 \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^4+(115+115 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 ) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^4-4 \cot \left (\frac {1}{4} (c+d x)\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^4+40 \log \left (1+\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^4-40 \log \left (1-\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^4+\frac {8 \sin \left (\frac {1}{4} (c+d x)\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^4}{\cos \left (\frac {1}{4} (c+d x)\right )-\sin \left (\frac {1}{4} (c+d x)\right )}-\frac {8 \sin \left (\frac {1}{4} (c+d x)\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^4}{\cos \left (\frac {1}{4} (c+d x)\right )+\sin \left (\frac {1}{4} (c+d x)\right )}-4 \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^4 \tan \left (\frac {1}{4} (c+d x)\right )\right )}{16 d (a (1+\sin (c+d x)))^{5/2}} \]

[In]

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

[Out]

((Cos[(c + d*x)/2] + Sin[(c + d*x)/2])*(8*Sin[(c + d*x)/2] - 4*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]) + 38*Sin[
(c + d*x)/2]*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2 - 19*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^3 + 8*(Cos[(c
+ d*x)/2] + Sin[(c + d*x)/2])^4 + (115 + 115*I)*(-1)^(3/4)*ArcTanh[(1/2 + I/2)*(-1)^(3/4)*(-1 + Tan[(c + d*x)/
4])]*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^4 - 4*Cot[(c + d*x)/4]*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^4 + 40
*Log[1 + Cos[(c + d*x)/2] - Sin[(c + d*x)/2]]*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^4 - 40*Log[1 - Cos[(c + d*
x)/2] + Sin[(c + d*x)/2]]*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^4 + (8*Sin[(c + d*x)/4]*(Cos[(c + d*x)/2] + Si
n[(c + d*x)/2])^4)/(Cos[(c + d*x)/4] - Sin[(c + d*x)/4]) - (8*Sin[(c + d*x)/4]*(Cos[(c + d*x)/2] + Sin[(c + d*
x)/2])^4)/(Cos[(c + d*x)/4] + Sin[(c + d*x)/4]) - 4*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^4*Tan[(c + d*x)/4]))
/(16*d*(a*(1 + Sin[c + d*x]))^(5/2))

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(355\) vs. \(2(145)=290\).

Time = 1.02 (sec) , antiderivative size = 356, normalized size of antiderivative = 2.05

method	result	size
default	\(-\frac {\left (115 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-a \left (\sin \left (d x +c \right )-1\right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) \left (\sin ^{3}\left (d x +c \right )\right ) a^{2}+230 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-a \left (\sin \left (d x +c \right )-1\right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) \left (\sin ^{2}\left (d x +c \right )\right ) a^{2}+32 \sqrt {-a \left (\sin \left (d x +c \right )-1\right )}\, a^{\frac {3}{2}} \left (\sin ^{2}\left (d x +c \right )\right )-160 \,\operatorname {arctanh}\left (\frac {\sqrt {-a \left (\sin \left (d x +c \right )-1\right )}}{\sqrt {a}}\right ) \left (\sin ^{3}\left (d x +c \right )\right ) a^{2}+115 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-a \left (\sin \left (d x +c \right )-1\right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{2} \sin \left (d x +c \right )+148 \sqrt {-a \left (\sin \left (d x +c \right )-1\right )}\, \sin \left (d x +c \right ) a^{\frac {3}{2}}-38 \left (-a \left (\sin \left (d x +c \right )-1\right )\right )^{\frac {3}{2}} \sqrt {a}\, \sin \left (d x +c \right )-320 \,\operatorname {arctanh}\left (\frac {\sqrt {-a \left (\sin \left (d x +c \right )-1\right )}}{\sqrt {a}}\right ) \left (\sin ^{2}\left (d x +c \right )\right ) a^{2}+32 \sqrt {-a \left (\sin \left (d x +c \right )-1\right )}\, a^{\frac {3}{2}}-160 \,\operatorname {arctanh}\left (\frac {\sqrt {-a \left (\sin \left (d x +c \right )-1\right )}}{\sqrt {a}}\right ) a^{2} \sin \left (d x +c \right )\right ) \sqrt {-a \left (\sin \left (d x +c \right )-1\right )}}{32 a^{\frac {9}{2}} \sin \left (d x +c \right ) \left (1+\sin \left (d x +c \right )\right ) \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d}\)	\(356\)

[In]

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

[Out]

-1/32*(115*2^(1/2)*arctanh(1/2*(-a*(sin(d*x+c)-1))^(1/2)*2^(1/2)/a^(1/2))*sin(d*x+c)^3*a^2+230*2^(1/2)*arctanh
(1/2*(-a*(sin(d*x+c)-1))^(1/2)*2^(1/2)/a^(1/2))*sin(d*x+c)^2*a^2+32*(-a*(sin(d*x+c)-1))^(1/2)*a^(3/2)*sin(d*x+
c)^2-160*arctanh((-a*(sin(d*x+c)-1))^(1/2)/a^(1/2))*sin(d*x+c)^3*a^2+115*2^(1/2)*arctanh(1/2*(-a*(sin(d*x+c)-1
))^(1/2)*2^(1/2)/a^(1/2))*a^2*sin(d*x+c)+148*(-a*(sin(d*x+c)-1))^(1/2)*sin(d*x+c)*a^(3/2)-38*(-a*(sin(d*x+c)-1
))^(3/2)*a^(1/2)*sin(d*x+c)-320*arctanh((-a*(sin(d*x+c)-1))^(1/2)/a^(1/2))*sin(d*x+c)^2*a^2+32*(-a*(sin(d*x+c)
-1))^(1/2)*a^(3/2)-160*arctanh((-a*(sin(d*x+c)-1))^(1/2)/a^(1/2))*a^2*sin(d*x+c))*(-a*(sin(d*x+c)-1))^(1/2)/a^
(9/2)/sin(d*x+c)/(1+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. 631 vs. \(2 (145) = 290\).

Time = 0.28 (sec) , antiderivative size = 631, normalized size of antiderivative = 3.63 \[ \int \frac {\csc ^2(c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx=\frac {115 \, \sqrt {2} {\left (\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{3} - 5 \, \cos \left (d x + c\right )^{2} - {\left (\cos \left (d x + c\right )^{3} + 3 \, \cos \left (d x + c\right )^{2} - 2 \, \cos \left (d x + c\right ) - 4\right )} \sin \left (d x + c\right ) + 2 \, \cos \left (d x + c\right ) + 4\right )} \sqrt {a} \log \left (-\frac {a \cos \left (d x + c\right )^{2} - 2 \, \sqrt {2} \sqrt {a \sin \left (d x + c\right ) + a} \sqrt {a} {\left (\cos \left (d x + c\right ) - \sin \left (d x + c\right ) + 1\right )} + 3 \, a \cos \left (d x + c\right ) - {\left (a \cos \left (d x + c\right ) - 2 \, a\right )} \sin \left (d x + c\right ) + 2 \, a}{\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 ) + 80 \, {\left (\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{3} - 5 \, \cos \left (d x + c\right )^{2} - {\left (\cos \left (d x + c\right )^{3} + 3 \, \cos \left (d x + c\right )^{2} - 2 \, \cos \left (d x + c\right ) - 4\right )} \sin \left (d x + c\right ) + 2 \, \cos \left (d x + c\right ) + 4\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 (35 \, \cos \left (d x + c\right )^{3} - 20 \, \cos \left (d x + c\right )^{2} - {\left (35 \, \cos \left (d x + c\right )^{2} + 55 \, \cos \left (d x + c\right ) + 4\right )} \sin \left (d x + c\right ) - 51 \, \cos \left (d x + c\right ) + 4\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{64 \, {\left (a^{3} d \cos \left (d x + c\right )^{4} - 2 \, a^{3} d \cos \left (d x + c\right )^{3} - 5 \, a^{3} d \cos \left (d x + c\right )^{2} + 2 \, a^{3} d \cos \left (d x + c\right ) + 4 \, a^{3} d - {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} - 2 \, a^{3} d \cos \left (d x + c\right ) - 4 \, a^{3} d\right )} \sin \left (d x + c\right )\right )}} \]

[In]

integrate(csc(d*x+c)^2/(a+a*sin(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

1/64*(115*sqrt(2)*(cos(d*x + c)^4 - 2*cos(d*x + c)^3 - 5*cos(d*x + c)^2 - (cos(d*x + c)^3 + 3*cos(d*x + c)^2 -
2*cos(d*x + c) - 4)*sin(d*x + c) + 2*cos(d*x + c) + 4)*sqrt(a)*log(-(a*cos(d*x + c)^2 - 2*sqrt(2)*sqrt(a*sin(
d*x + c) + a)*sqrt(a)*(cos(d*x + c) - sin(d*x + c) + 1) + 3*a*cos(d*x + c) - (a*cos(d*x + c) - 2*a)*sin(d*x +
c) + 2*a)/(cos(d*x + c)^2 - (cos(d*x + c) + 2)*sin(d*x + c) - cos(d*x + c) - 2)) + 80*(cos(d*x + c)^4 - 2*cos(
d*x + c)^3 - 5*cos(d*x + c)^2 - (cos(d*x + c)^3 + 3*cos(d*x + c)^2 - 2*cos(d*x + c) - 4)*sin(d*x + c) + 2*cos(
d*x + c) + 4)*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*c
os(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*(35*cos(d*x + c)^3 - 20*cos(d*x + c)^2 - (35*cos(d*x + c)^2 + 55*cos(d*x + c) + 4)*sin(d*x
+ c) - 51*cos(d*x + c) + 4)*sqrt(a*sin(d*x + c) + a))/(a^3*d*cos(d*x + c)^4 - 2*a^3*d*cos(d*x + c)^3 - 5*a^3*d
*cos(d*x + c)^2 + 2*a^3*d*cos(d*x + c) + 4*a^3*d - (a^3*d*cos(d*x + c)^3 + 3*a^3*d*cos(d*x + c)^2 - 2*a^3*d*co
s(d*x + c) - 4*a^3*d)*sin(d*x + c))

Sympy [F]

\[ \int \frac {\csc ^2(c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx=\int \frac {\csc ^{2}{\left (c + d x \right )}}{\left (a \left (\sin {\left (c + d x \right )} + 1\right )\right )^{\frac {5}{2}}}\, dx \]

[In]

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

[Out]

Integral(csc(c + d*x)**2/(a*(sin(c + d*x) + 1))**(5/2), x)

Maxima [F]

\[ \int \frac {\csc ^2(c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx=\int { \frac {\csc \left (d x + c\right )^{2}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {5}{2}}} \,d x } \]

[In]

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

[Out]

integrate(csc(d*x + c)^2/(a*sin(d*x + c) + a)^(5/2), x)

Giac [A] (verification not implemented)

none

Time = 0.61 (sec) , antiderivative size = 205, normalized size of antiderivative = 1.18 \[ \int \frac {\csc ^2(c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx=\frac {\frac {80 \, \log \left (\frac {{\left | -4 \, \sqrt {2} + 8 \, \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}}{{\left | 4 \, \sqrt {2} + 8 \, \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}}\right )}{a^{\frac {5}{2}} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} - \frac {32 \, \sqrt {2} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (2 \, \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )} a^{\frac {5}{2}} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} - \frac {\sqrt {2} {\left (19 \, \sqrt {a} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 21 \, \sqrt {a} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2} a^{3} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{32 \, d} \]

[In]

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

[Out]

1/32*(80*log(abs(-4*sqrt(2) + 8*sin(-1/4*pi + 1/2*d*x + 1/2*c))/abs(4*sqrt(2) + 8*sin(-1/4*pi + 1/2*d*x + 1/2*
c)))/(a^(5/2)*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))) - 32*sqrt(2)*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)*a^(5/2)*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))) - sqrt(2)*(19*sqrt(a)*sin(-1/4*pi + 1/
2*d*x + 1/2*c)^3 - 21*sqrt(a)*sin(-1/4*pi + 1/2*d*x + 1/2*c))/((sin(-1/4*pi + 1/2*d*x + 1/2*c)^2 - 1)^2*a^3*sg
n(cos(-1/4*pi + 1/2*d*x + 1/2*c))))/d

Mupad [F(-1)]

Timed out. \[ \int \frac {\csc ^2(c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx=\int \frac {1}{{\sin \left (c+d\,x\right )}^2\,{\left (a+a\,\sin \left (c+d\,x\right )\right )}^{5/2}} \,d x \]

[In]

int(1/(sin(c + d*x)^2*(a + a*sin(c + d*x))^(5/2)),x)

[Out]

int(1/(sin(c + d*x)^2*(a + a*sin(c + d*x))^(5/2)), x)