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

Optimal. Leaf size=174 \[ \frac {5 \tanh ^{-1}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {a \sin (c+d x)+a}}\right )}{a^{5/2} d}-\frac {115 \tanh ^{-1}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {2} \sqrt {a \sin (c+d x)+a}}\right )}{16 \sqrt {2} a^{5/2} d}-\frac {35 \cot (c+d x)}{16 a^2 d \sqrt {a \sin (c+d x)+a}}+\frac {15 \cot (c+d x)}{16 a d (a \sin (c+d x)+a)^{3/2}}+\frac {\cot (c+d x)}{4 d (a \sin (c+d x)+a)^{5/2}} \]

[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)

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Rubi [A]  time = 0.51, antiderivative size = 174, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {2766, 2978, 2984, 2985, 2649, 206, 2773} \[ -\frac {35 \cot (c+d x)}{16 a^2 d \sqrt {a \sin (c+d x)+a}}+\frac {5 \tanh ^{-1}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {a \sin (c+d x)+a}}\right )}{a^{5/2} d}-\frac {115 \tanh ^{-1}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {2} \sqrt {a \sin (c+d x)+a}}\right )}{16 \sqrt {2} a^{5/2} d}+\frac {15 \cot (c+d x)}{16 a d (a \sin (c+d x)+a)^{3/2}}+\frac {\cot (c+d x)}{4 d (a \sin (c+d x)+a)^{5/2}} \]

Antiderivative was successfully verified.

[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 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 2649

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 2766

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 2773

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 2978

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 2984

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 2985

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*} \int \frac {\csc ^2(c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx &=\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 \operatorname {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 \operatorname {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 \tanh ^{-1}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {a+a \sin (c+d x)}}\right )}{a^{5/2} d}-\frac {115 \tanh ^{-1}\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*}

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Mathematica [C]  time = 0.63, size = 509, normalized size = 2.93 \[ \frac {\left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right ) \left (8 \sin \left (\frac {1}{2} (c+d x)\right )+\frac {8 \sin \left (\frac {1}{4} (c+d x)\right ) \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \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 (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^4}{\sin \left (\frac {1}{4} (c+d x)\right )+\cos \left (\frac {1}{4} (c+d x)\right )}+8 \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^4-19 \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^3+38 \sin \left (\frac {1}{2} (c+d x)\right ) \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^2-4 \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )+40 \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^4 \log \left (-\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )+1\right )-40 \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^4 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )-\cos \left (\frac {1}{2} (c+d x)\right )+1\right )-4 \tan \left (\frac {1}{4} (c+d x)\right ) \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^4-4 \cot \left (\frac {1}{4} (c+d x)\right ) \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^4+(115+115 i) (-1)^{3/4} \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^4 \tanh ^{-1}\left (\left (\frac {1}{2}+\frac {i}{2}\right ) (-1)^{3/4} \left (\tan \left (\frac {1}{4} (c+d x)\right )-1\right )\right )\right )}{16 d (a (\sin (c+d x)+1))^{5/2}} \]

Antiderivative was successfully verified.

[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))

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fricas [B]  time = 0.54, size = 631, normalized size = 3.63 \[ \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 )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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))

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giac [B]  time = 3.14, size = 671, normalized size = 3.86 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(115*sqrt(2)*arctan(-1/2*sqrt(2)*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a) + sqr
t(a))/sqrt(-a))/(sqrt(-a)*a^2*sgn(tan(1/2*d*x + 1/2*c) + 1)) - 80*arctan(-(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt
(a*tan(1/2*d*x + 1/2*c)^2 + a))/sqrt(-a))/(sqrt(-a)*a^2*sgn(tan(1/2*d*x + 1/2*c) + 1)) + 40*log(abs(-sqrt(a)*t
an(1/2*d*x + 1/2*c) + sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a)))/(a^(5/2)*sgn(tan(1/2*d*x + 1/2*c) + 1)) + 8*sqrt(a*
tan(1/2*d*x + 1/2*c)^2 + a)/(a^3*sgn(tan(1/2*d*x + 1/2*c) + 1)) + 16/(((sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*
tan(1/2*d*x + 1/2*c)^2 + a))^2 - a)*a^(3/2)*sgn(tan(1/2*d*x + 1/2*c) + 1)) + 2*(77*(sqrt(a)*tan(1/2*d*x + 1/2*
c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^7 + 283*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2
+ a))^6*sqrt(a) + 199*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^5*a - 299*(sqrt(a)*
tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^4*a^(3/2) + 15*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt
(a*tan(1/2*d*x + 1/2*c)^2 + a))^3*a^2 + 177*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a)
)^2*a^(5/2) - 107*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))*a^3 + 23*a^(7/2))/(((sqr
t(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^2 + 2*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*t
an(1/2*d*x + 1/2*c)^2 + a))*sqrt(a) - a)^4*a^2*sgn(tan(1/2*d*x + 1/2*c) + 1)))/d

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maple [B]  time = 1.12, size = 356, normalized size = 2.05 \[ -\frac {\left (115 \sqrt {2}\, \arctanh \left (\frac {\sqrt {-a \left (\sin \left (d x +c \right )-1\right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) \left (\sin ^{3}\left (d x +c \right )\right ) a^{2}+32 \sqrt {-a \left (\sin \left (d x +c \right )-1\right )}\, a^{\frac {3}{2}} \left (\sin ^{2}\left (d x +c \right )\right )+230 \sqrt {2}\, \arctanh \left (\frac {\sqrt {-a \left (\sin \left (d x +c \right )-1\right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) \left (\sin ^{2}\left (d x +c \right )\right ) a^{2}-160 \arctanh \left (\frac {\sqrt {-a \left (\sin \left (d x +c \right )-1\right )}}{\sqrt {a}}\right ) \left (\sin ^{3}\left (d x +c \right )\right ) a^{2}+148 \sqrt {-a \left (\sin \left (d x +c \right )-1\right )}\, \sin \left (d x +c \right ) a^{\frac {3}{2}}-38 \left (-a \left (\sin \left (d x +c \right )-1\right )\right )^{\frac {3}{2}} \sqrt {a}\, \sin \left (d x +c \right )+115 \sqrt {2}\, \arctanh \left (\frac {\sqrt {-a \left (\sin \left (d x +c \right )-1\right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{2} \sin \left (d x +c \right )-320 \arctanh \left (\frac {\sqrt {-a \left (\sin \left (d x +c \right )-1\right )}}{\sqrt {a}}\right ) \left (\sin ^{2}\left (d x +c \right )\right ) a^{2}+32 \sqrt {-a \left (\sin \left (d x +c \right )-1\right )}\, a^{\frac {3}{2}}-160 \arctanh \left (\frac {\sqrt {-a \left (\sin \left (d x +c \right )-1\right )}}{\sqrt {a}}\right ) a^{2} \sin \left (d x +c \right )\right ) \sqrt {-a \left (\sin \left (d x +c \right )-1\right )}}{32 a^{\frac {9}{2}} \left (1+\sin \left (d x +c \right )\right ) \sin \left (d x +c \right ) \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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+32*(-a*(sin(d*x+c)-
1))^(1/2)*a^(3/2)*sin(d*x+c)^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^2-160*arctanh((-a*(sin(d*x+c)-1))^(1/2)/a^(1/2))*sin(d*x+c)^3*a^2+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)+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)-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)/(1+sin(d*x+c))/sin(d*x+c)/cos(d*x+c)/(a+a*sin(d*x+c))^(1/2)/d

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\csc \left (d x + c\right )^{2}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{{\sin \left (c+d\,x\right )}^2\,{\left (a+a\,\sin \left (c+d\,x\right )\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\csc ^{2}{\left (c + d x \right )}}{\left (a \left (\sin {\left (c + d x \right )} + 1\right )\right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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