3.1.84 \(\int \frac {\csc ^3(c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx\) [84]

Optimal. Leaf size=224 \[ -\frac {39 \tanh ^{-1}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {a+a \sin (c+d x)}}\right )}{4 a^{5/2} d}+\frac {219 \tanh ^{-1}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {2} \sqrt {a+a \sin (c+d x)}}\right )}{16 \sqrt {2} a^{5/2} d}+\frac {\cot (c+d x) \csc (c+d x)}{4 d (a+a \sin (c+d x))^{5/2}}+\frac {19 \cot (c+d x) \csc (c+d x)}{16 a d (a+a \sin (c+d x))^{3/2}}+\frac {63 \cot (c+d x)}{16 a^2 d \sqrt {a+a \sin (c+d x)}}-\frac {31 \cot (c+d x) \csc (c+d x)}{16 a^2 d \sqrt {a+a \sin (c+d x)}} \]

[Out]

-39/4*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)*csc(d*x+c)/d/(a+a*sin(d*x+c)
)^(5/2)+19/16*cot(d*x+c)*csc(d*x+c)/a/d/(a+a*sin(d*x+c))^(3/2)+219/32*arctanh(1/2*cos(d*x+c)*a^(1/2)*2^(1/2)/(
a+a*sin(d*x+c))^(1/2))*2^(1/2)/a^(5/2)/d+63/16*cot(d*x+c)/a^2/d/(a+a*sin(d*x+c))^(1/2)-31/16*cot(d*x+c)*csc(d*
x+c)/a^2/d/(a+a*sin(d*x+c))^(1/2)

________________________________________________________________________________________

Rubi [A]
time = 0.44, antiderivative size = 224, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {2845, 3057, 3063, 3064, 2728, 212, 2852} \begin {gather*} -\frac {39 \tanh ^{-1}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {a \sin (c+d x)+a}}\right )}{4 a^{5/2} d}+\frac {219 \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 {63 \cot (c+d x)}{16 a^2 d \sqrt {a \sin (c+d x)+a}}-\frac {31 \cot (c+d x) \csc (c+d x)}{16 a^2 d \sqrt {a \sin (c+d x)+a}}+\frac {19 \cot (c+d x) \csc (c+d x)}{16 a d (a \sin (c+d x)+a)^{3/2}}+\frac {\cot (c+d x) \csc (c+d x)}{4 d (a \sin (c+d x)+a)^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-39*ArcTanh[(Sqrt[a]*Cos[c + d*x])/Sqrt[a + a*Sin[c + d*x]]])/(4*a^(5/2)*d) + (219*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]*Csc[c + d*x])/(4*d*(a + a*Sin
[c + d*x])^(5/2)) + (19*Cot[c + d*x]*Csc[c + d*x])/(16*a*d*(a + a*Sin[c + d*x])^(3/2)) + (63*Cot[c + d*x])/(16
*a^2*d*Sqrt[a + a*Sin[c + d*x]]) - (31*Cot[c + d*x]*Csc[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*} \int \frac {\csc ^3(c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx &=\frac {\cot (c+d x) \csc (c+d x)}{4 d (a+a \sin (c+d x))^{5/2}}+\frac {\int \frac {\csc ^3(c+d x) \left (6 a-\frac {7}{2} a \sin (c+d x)\right )}{(a+a \sin (c+d x))^{3/2}} \, dx}{4 a^2}\\ &=\frac {\cot (c+d x) \csc (c+d x)}{4 d (a+a \sin (c+d x))^{5/2}}+\frac {19 \cot (c+d x) \csc (c+d x)}{16 a d (a+a \sin (c+d x))^{3/2}}+\frac {\int \frac {\csc ^3(c+d x) \left (31 a^2-\frac {95}{4} a^2 \sin (c+d x)\right )}{\sqrt {a+a \sin (c+d x)}} \, dx}{8 a^4}\\ &=\frac {\cot (c+d x) \csc (c+d x)}{4 d (a+a \sin (c+d x))^{5/2}}+\frac {19 \cot (c+d x) \csc (c+d x)}{16 a d (a+a \sin (c+d x))^{3/2}}-\frac {31 \cot (c+d x) \csc (c+d x)}{16 a^2 d \sqrt {a+a \sin (c+d x)}}+\frac {\int \frac {\csc ^2(c+d x) \left (-63 a^3+\frac {93}{2} a^3 \sin (c+d x)\right )}{\sqrt {a+a \sin (c+d x)}} \, dx}{16 a^5}\\ &=\frac {\cot (c+d x) \csc (c+d x)}{4 d (a+a \sin (c+d x))^{5/2}}+\frac {19 \cot (c+d x) \csc (c+d x)}{16 a d (a+a \sin (c+d x))^{3/2}}+\frac {63 \cot (c+d x)}{16 a^2 d \sqrt {a+a \sin (c+d x)}}-\frac {31 \cot (c+d x) \csc (c+d x)}{16 a^2 d \sqrt {a+a \sin (c+d x)}}+\frac {\int \frac {\csc (c+d x) \left (78 a^4-\frac {63}{2} a^4 \sin (c+d x)\right )}{\sqrt {a+a \sin (c+d x)}} \, dx}{16 a^6}\\ &=\frac {\cot (c+d x) \csc (c+d x)}{4 d (a+a \sin (c+d x))^{5/2}}+\frac {19 \cot (c+d x) \csc (c+d x)}{16 a d (a+a \sin (c+d x))^{3/2}}+\frac {63 \cot (c+d x)}{16 a^2 d \sqrt {a+a \sin (c+d x)}}-\frac {31 \cot (c+d x) \csc (c+d x)}{16 a^2 d \sqrt {a+a \sin (c+d x)}}+\frac {39 \int \csc (c+d x) \sqrt {a+a \sin (c+d x)} \, dx}{8 a^3}-\frac {219 \int \frac {1}{\sqrt {a+a \sin (c+d x)}} \, dx}{32 a^2}\\ &=\frac {\cot (c+d x) \csc (c+d x)}{4 d (a+a \sin (c+d x))^{5/2}}+\frac {19 \cot (c+d x) \csc (c+d x)}{16 a d (a+a \sin (c+d x))^{3/2}}+\frac {63 \cot (c+d x)}{16 a^2 d \sqrt {a+a \sin (c+d x)}}-\frac {31 \cot (c+d x) \csc (c+d x)}{16 a^2 d \sqrt {a+a \sin (c+d x)}}-\frac {39 \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 {219 \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 {39 \tanh ^{-1}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {a+a \sin (c+d x)}}\right )}{4 a^{5/2} d}+\frac {219 \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) \csc (c+d x)}{4 d (a+a \sin (c+d x))^{5/2}}+\frac {19 \cot (c+d x) \csc (c+d x)}{16 a d (a+a \sin (c+d x))^{3/2}}+\frac {63 \cot (c+d x)}{16 a^2 d \sqrt {a+a \sin (c+d x)}}-\frac {31 \cot (c+d x) \csc (c+d x)}{16 a^2 d \sqrt {a+a \sin (c+d x)}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C] Result contains complex when optimal does not.
time = 0.78, size = 680, normalized size = 3.04 \begin {gather*} \frac {\left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right ) \left (-16 \sin \left (\frac {1}{2} (c+d x)\right )+8 \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )-108 \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+54 \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^3-40 \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^4-(438+438 i) (-1)^{3/4} \tanh ^{-1}\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+20 \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-\csc ^2\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 )^4-156 \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+156 \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+\sec ^2\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+\frac {2 \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^4}{\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 ) \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 {2 \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^4}{\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 ) \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 )}+20 \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 )}{32 d (a (1+\sin (c+d x)))^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((Cos[(c + d*x)/2] + Sin[(c + d*x)/2])*(-16*Sin[(c + d*x)/2] + 8*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]) - 108*S
in[(c + d*x)/2]*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2 + 54*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^3 - 40*(Cos
[(c + d*x)/2] + Sin[(c + d*x)/2])^4 - (438 + 438*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 + 20*Cot[(c + d*x)/4]*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^4
 - Csc[(c + d*x)/4]^2*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^4 - 156*Log[1 + Cos[(c + d*x)/2] - Sin[(c + d*x)/2
]]*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^4 + 156*Log[1 - Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]*(Cos[(c + d*x)/2
] + Sin[(c + d*x)/2])^4 + Sec[(c + d*x)/4]^2*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^4 + (2*(Cos[(c + d*x)/2] +
Sin[(c + d*x)/2])^4)/(Cos[(c + d*x)/4] - Sin[(c + d*x)/4])^2 - (40*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]) - (2*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^4)/(Cos[(c + d
*x)/4] + Sin[(c + d*x)/4])^2 + (40*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]) + 20*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^4*Tan[(c + d*x)/4]))/(32*d*(a*(1 + Sin[c + d*x
]))^(5/2))

________________________________________________________________________________________

Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(403\) vs. \(2(189)=378\).
time = 3.31, size = 404, normalized size = 1.80

method result size
default \(\frac {\left (219 \sqrt {2}\, \arctanh \left (\frac {\sqrt {-a \left (\sin \left (d x +c \right )-1\right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) \left (\sin ^{4}\left (d x +c \right )\right ) a^{2}+438 \sqrt {2}\, \arctanh \left (\frac {\sqrt {-a \left (\sin \left (d x +c \right )-1\right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) \left (\sin ^{3}\left (d x +c \right )\right ) a^{2}-312 \arctanh \left (\frac {\sqrt {-a \left (\sin \left (d x +c \right )-1\right )}}{\sqrt {a}}\right ) \left (\sin ^{4}\left (d x +c \right )\right ) a^{2}+219 \sqrt {2}\, \arctanh \left (\frac {\sqrt {-a \left (\sin \left (d x +c \right )-1\right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) \left (\sin ^{2}\left (d x +c \right )\right ) a^{2}-624 \arctanh \left (\frac {\sqrt {-a \left (\sin \left (d x +c \right )-1\right )}}{\sqrt {a}}\right ) \left (\sin ^{3}\left (d x +c \right )\right ) a^{2}-126 \left (-a \left (\sin \left (d x +c \right )-1\right )\right )^{\frac {3}{2}} \sqrt {a}\, \left (\sin ^{2}\left (d x +c \right )\right )+172 \sqrt {-a \left (\sin \left (d x +c \right )-1\right )}\, a^{\frac {3}{2}} \left (\sin ^{2}\left (d x +c \right )\right )-312 \arctanh \left (\frac {\sqrt {-a \left (\sin \left (d x +c \right )-1\right )}}{\sqrt {a}}\right ) \left (\sin ^{2}\left (d x +c \right )\right ) a^{2}-144 \left (-a \left (\sin \left (d x +c \right )-1\right )\right )^{\frac {3}{2}} \sqrt {a}\, \sin \left (d x +c \right )+112 \sqrt {-a \left (\sin \left (d x +c \right )-1\right )}\, a^{\frac {3}{2}} \sin \left (d x +c \right )-72 \left (-a \left (\sin \left (d x +c \right )-1\right )\right )^{\frac {3}{2}} \sqrt {a}+56 \sqrt {-a \left (\sin \left (d x +c \right )-1\right )}\, a^{\frac {3}{2}}\right ) \sqrt {-a \left (\sin \left (d x +c \right )-1\right )}}{32 a^{\frac {9}{2}} \sin \left (d x +c \right )^{2} \left (1+\sin \left (d x +c \right )\right ) \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d}\) \(404\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/32*(219*2^(1/2)*arctanh(1/2*(-a*(sin(d*x+c)-1))^(1/2)*2^(1/2)/a^(1/2))*sin(d*x+c)^4*a^2+438*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^2-312*arctanh((-a*(sin(d*x+c)-1))^(1/2)/a^(1/2))
*sin(d*x+c)^4*a^2+219*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^2-624*arct
anh((-a*(sin(d*x+c)-1))^(1/2)/a^(1/2))*sin(d*x+c)^3*a^2-126*(-a*(sin(d*x+c)-1))^(3/2)*a^(1/2)*sin(d*x+c)^2+172
*(-a*(sin(d*x+c)-1))^(1/2)*a^(3/2)*sin(d*x+c)^2-312*arctanh((-a*(sin(d*x+c)-1))^(1/2)/a^(1/2))*sin(d*x+c)^2*a^
2-144*(-a*(sin(d*x+c)-1))^(3/2)*a^(1/2)*sin(d*x+c)+112*(-a*(sin(d*x+c)-1))^(1/2)*a^(3/2)*sin(d*x+c)-72*(-a*(si
n(d*x+c)-1))^(3/2)*a^(1/2)+56*(-a*(sin(d*x+c)-1))^(1/2)*a^(3/2))*(-a*(sin(d*x+c)-1))^(1/2)/a^(9/2)/sin(d*x+c)^
2/(1+sin(d*x+c))/cos(d*x+c)/(a+a*sin(d*x+c))^(1/2)/d

________________________________________________________________________________________

Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 715 vs. \(2 (189) = 378\).
time = 0.39, size = 715, normalized size = 3.19 \begin {gather*} \frac {219 \, \sqrt {2} {\left (\cos \left (d x + c\right )^{5} + 3 \, \cos \left (d x + c\right )^{4} - 3 \, \cos \left (d x + c\right )^{3} - 7 \, \cos \left (d x + c\right )^{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} + 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 ) + 156 \, {\left (\cos \left (d x + c\right )^{5} + 3 \, \cos \left (d x + c\right )^{4} - 3 \, \cos \left (d x + c\right )^{3} - 7 \, \cos \left (d x + c\right )^{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} + 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 (63 \, \cos \left (d x + c\right )^{4} + 95 \, \cos \left (d x + c\right )^{3} - 51 \, \cos \left (d x + c\right )^{2} + {\left (63 \, \cos \left (d x + c\right )^{3} - 32 \, \cos \left (d x + c\right )^{2} - 83 \, \cos \left (d x + c\right ) + 4\right )} \sin \left (d x + c\right ) - 87 \, \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 )^{5} + 3 \, a^{3} d \cos \left (d x + c\right )^{4} - 3 \, a^{3} d \cos \left (d x + c\right )^{3} - 7 \, 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 )^{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\right )} \sin \left (d x + c\right )\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/64*(219*sqrt(2)*(cos(d*x + c)^5 + 3*cos(d*x + c)^4 - 3*cos(d*x + c)^3 - 7*cos(d*x + c)^2 + (cos(d*x + c)^4 -
 2*cos(d*x + c)^3 - 5*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)) + 156*(cos(d*x + c)^5 + 3*cos(d*x + c)^4 - 3*cos(d*x + c)^3 - 7*cos(d*x + c)^2 + (cos(d*x + c)^4 - 2*
cos(d*x + c)^3 - 5*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)*sq
rt(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*(63*cos(d*x +
c)^4 + 95*cos(d*x + c)^3 - 51*cos(d*x + c)^2 + (63*cos(d*x + c)^3 - 32*cos(d*x + c)^2 - 83*cos(d*x + c) + 4)*s
in(d*x + c) - 87*cos(d*x + c) - 4)*sqrt(a*sin(d*x + c) + a))/(a^3*d*cos(d*x + c)^5 + 3*a^3*d*cos(d*x + c)^4 -
3*a^3*d*cos(d*x + c)^3 - 7*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)^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)*sin(d*x + c))

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\csc ^{3}{\left (c + d x \right )}}{\left (a \left (\sin {\left (c + d x \right )} + 1\right )\right )^{\frac {5}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(csc(c + d*x)**3/(a*(sin(c + d*x) + 1))**(5/2), x)

________________________________________________________________________________________

Giac [A]
time = 0.58, size = 298, normalized size = 1.33 \begin {gather*} -\frac {\frac {219 \, \sqrt {2} \log \left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 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 {219 \, \sqrt {2} \log \left (-\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 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 {312 \, \log \left ({\left | \frac {1}{2} \, \sqrt {2} + \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 {312 \, \log \left ({\left | -\frac {1}{2} \, \sqrt {2} + \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 {2 \, \sqrt {2} {\left (252 \, \sqrt {a} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 568 \, \sqrt {a} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 399 \, \sqrt {a} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 85 \, \sqrt {a} \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 )^{4} - 3 \, \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 )}}{64 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/64*(219*sqrt(2)*log(sin(-1/4*pi + 1/2*d*x + 1/2*c) + 1)/(a^(5/2)*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))) - 219
*sqrt(2)*log(-sin(-1/4*pi + 1/2*d*x + 1/2*c) + 1)/(a^(5/2)*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))) - 312*log(abs(
1/2*sqrt(2) + 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))) + 312*log(abs(-1/
2*sqrt(2) + 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))) - 2*sqrt(2)*(252*sq
rt(a)*sin(-1/4*pi + 1/2*d*x + 1/2*c)^7 - 568*sqrt(a)*sin(-1/4*pi + 1/2*d*x + 1/2*c)^5 + 399*sqrt(a)*sin(-1/4*p
i + 1/2*d*x + 1/2*c)^3 - 85*sqrt(a)*sin(-1/4*pi + 1/2*d*x + 1/2*c))/((2*sin(-1/4*pi + 1/2*d*x + 1/2*c)^4 - 3*s
in(-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]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {1}{{\sin \left (c+d\,x\right )}^3\,{\left (a+a\,\sin \left (c+d\,x\right )\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(sin(c + d*x)^3*(a + a*sin(c + d*x))^(5/2)),x)

[Out]

int(1/(sin(c + d*x)^3*(a + a*sin(c + d*x))^(5/2)), x)

________________________________________________________________________________________