3.348 \(\int \frac {\cot ^2(c+d x) \csc (c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx\)

Optimal. Leaf size=153 \[ -\frac {11 \tanh ^{-1}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {a \sin (c+d x)+a}}\right )}{4 a^{3/2} d}+\frac {2 \sqrt {2} \tanh ^{-1}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {2} \sqrt {a \sin (c+d x)+a}}\right )}{a^{3/2} d}+\frac {5 \cot (c+d x)}{4 a d \sqrt {a \sin (c+d x)+a}}-\frac {\cot (c+d x) \csc (c+d x)}{2 a d \sqrt {a \sin (c+d x)+a}} \]

[Out]

-11/4*arctanh(cos(d*x+c)*a^(1/2)/(a+a*sin(d*x+c))^(1/2))/a^(3/2)/d+2*arctanh(1/2*cos(d*x+c)*a^(1/2)*2^(1/2)/(a
+a*sin(d*x+c))^(1/2))/a^(3/2)/d*2^(1/2)+5/4*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/
d/(a+a*sin(d*x+c))^(1/2)

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Rubi [A]  time = 0.54, antiderivative size = 153, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.207, Rules used = {2874, 2984, 2985, 2649, 206, 2773} \[ -\frac {11 \tanh ^{-1}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {a \sin (c+d x)+a}}\right )}{4 a^{3/2} d}+\frac {2 \sqrt {2} \tanh ^{-1}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {2} \sqrt {a \sin (c+d x)+a}}\right )}{a^{3/2} d}+\frac {5 \cot (c+d x)}{4 a d \sqrt {a \sin (c+d x)+a}}-\frac {\cot (c+d x) \csc (c+d x)}{2 a d \sqrt {a \sin (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-11*ArcTanh[(Sqrt[a]*Cos[c + d*x])/Sqrt[a + a*Sin[c + d*x]]])/(4*a^(3/2)*d) + (2*Sqrt[2]*ArcTanh[(Sqrt[a]*Cos
[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sin[c + d*x]])])/(a^(3/2)*d) + (5*Cot[c + d*x])/(4*a*d*Sqrt[a + a*Sin[c + d*x]]
) - (Cot[c + d*x]*Csc[c + d*x])/(2*a*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 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 2874

Int[cos[(e_.) + (f_.)*(x_)]^2*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)
, x_Symbol] :> Dist[1/b^2, Int[(d*Sin[e + f*x])^n*(a + b*Sin[e + f*x])^(m + 1)*(a - b*Sin[e + f*x]), x], x] /;
FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 - b^2, 0] && (ILtQ[m, 0] ||  !IGtQ[n, 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 {\cot ^2(c+d x) \csc (c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx &=\frac {\int \frac {\csc ^3(c+d x) (a-a \sin (c+d x))}{\sqrt {a+a \sin (c+d x)}} \, dx}{a^2}\\ &=-\frac {\cot (c+d x) \csc (c+d x)}{2 a d \sqrt {a+a \sin (c+d x)}}+\frac {\int \frac {\csc ^2(c+d x) \left (-\frac {5 a^2}{2}+\frac {3}{2} a^2 \sin (c+d x)\right )}{\sqrt {a+a \sin (c+d x)}} \, dx}{2 a^3}\\ &=\frac {5 \cot (c+d x)}{4 a d \sqrt {a+a \sin (c+d x)}}-\frac {\cot (c+d x) \csc (c+d x)}{2 a d \sqrt {a+a \sin (c+d x)}}+\frac {\int \frac {\csc (c+d x) \left (\frac {11 a^3}{4}-\frac {5}{4} a^3 \sin (c+d x)\right )}{\sqrt {a+a \sin (c+d x)}} \, dx}{2 a^4}\\ &=\frac {5 \cot (c+d x)}{4 a d \sqrt {a+a \sin (c+d x)}}-\frac {\cot (c+d x) \csc (c+d x)}{2 a d \sqrt {a+a \sin (c+d x)}}+\frac {11 \int \csc (c+d x) \sqrt {a+a \sin (c+d x)} \, dx}{8 a^2}-\frac {2 \int \frac {1}{\sqrt {a+a \sin (c+d x)}} \, dx}{a}\\ &=\frac {5 \cot (c+d x)}{4 a d \sqrt {a+a \sin (c+d x)}}-\frac {\cot (c+d x) \csc (c+d x)}{2 a d \sqrt {a+a \sin (c+d x)}}-\frac {11 \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 )}{4 a d}+\frac {4 \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 )}{a d}\\ &=-\frac {11 \tanh ^{-1}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {a+a \sin (c+d x)}}\right )}{4 a^{3/2} d}+\frac {2 \sqrt {2} \tanh ^{-1}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {2} \sqrt {a+a \sin (c+d x)}}\right )}{a^{3/2} d}+\frac {5 \cot (c+d x)}{4 a d \sqrt {a+a \sin (c+d x)}}-\frac {\cot (c+d x) \csc (c+d x)}{2 a d \sqrt {a+a \sin (c+d x)}}\\ \end {align*}

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Mathematica [C]  time = 3.56, size = 309, normalized size = 2.02 \[ \frac {\left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^3 \left (12 \tan \left (\frac {1}{4} (c+d x)\right )+12 \cot \left (\frac {1}{4} (c+d x)\right )-\csc ^2\left (\frac {1}{4} (c+d x)\right )+\sec ^2\left (\frac {1}{4} (c+d x)\right )-\frac {24 \sin \left (\frac {1}{4} (c+d x)\right )}{\cos \left (\frac {1}{4} (c+d x)\right )-\sin \left (\frac {1}{4} (c+d x)\right )}+\frac {24 \sin \left (\frac {1}{4} (c+d x)\right )}{\sin \left (\frac {1}{4} (c+d x)\right )+\cos \left (\frac {1}{4} (c+d x)\right )}+\frac {2}{\left (\cos \left (\frac {1}{4} (c+d x)\right )-\sin \left (\frac {1}{4} (c+d x)\right )\right )^2}-\frac {2}{\left (\sin \left (\frac {1}{4} (c+d x)\right )+\cos \left (\frac {1}{4} (c+d x)\right )\right )^2}-(128+128 i) (-1)^{3/4} \tanh ^{-1}\left (\left (\frac {1}{2}+\frac {i}{2}\right ) (-1)^{3/4} \left (\tan \left (\frac {1}{4} (c+d x)\right )-1\right )\right )-44 \log \left (-\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )+1\right )+44 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )-\cos \left (\frac {1}{2} (c+d x)\right )+1\right )-24\right )}{32 d (a (\sin (c+d x)+1))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^3*(-24 - (128 + 128*I)*(-1)^(3/4)*ArcTanh[(1/2 + I/2)*(-1)^(3/4)*(-1 +
Tan[(c + d*x)/4])] + 12*Cot[(c + d*x)/4] - Csc[(c + d*x)/4]^2 - 44*Log[1 + Cos[(c + d*x)/2] - Sin[(c + d*x)/2]
] + 44*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))

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fricas [B]  time = 0.50, size = 508, normalized size = 3.32 \[ \frac {11 \, {\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 {16 \, \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 (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 )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(11*(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)) + 16*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*(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))

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giac [B]  time = 0.89, size = 620, normalized size = 4.05 \[ \frac {\sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} {\left (\frac {\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{2} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )} - \frac {6}{a^{2} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}\right )} - \frac {{\left (44 \, \sqrt {2} \sqrt {a} \arctan \left (\frac {\sqrt {2} \sqrt {a} + \sqrt {a}}{\sqrt {-a}}\right ) - 96 \, \sqrt {2} \sqrt {a} \arctan \left (\frac {\sqrt {a}}{\sqrt {-a}}\right ) - 22 \, \sqrt {2} \sqrt {-a} \log \left (\sqrt {2} \sqrt {a} + \sqrt {a}\right ) + 66 \, \sqrt {a} \arctan \left (\frac {\sqrt {2} \sqrt {a} + \sqrt {a}}{\sqrt {-a}}\right ) - 128 \, \sqrt {a} \arctan \left (\frac {\sqrt {a}}{\sqrt {-a}}\right ) - 33 \, \sqrt {-a} \log \left (\sqrt {2} \sqrt {a} + \sqrt {a}\right ) - 30 \, \sqrt {2} \sqrt {-a} - 38 \, \sqrt {-a}\right )} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}{2 \, \sqrt {2} \sqrt {-a} a^{\frac {3}{2}} + 3 \, \sqrt {-a} a^{\frac {3}{2}}} - \frac {32 \, \sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} + \sqrt {a}\right )}}{2 \, \sqrt {-a}}\right )}{\sqrt {-a} a \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )} + \frac {22 \, \arctan \left (-\frac {\sqrt {a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )} - \frac {11 \, \log \left ({\left | -\sqrt {a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} \right |}\right )}{a^{\frac {3}{2}} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )} + \frac {2 \, {\left ({\left (\sqrt {a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{3} - 6 \, {\left (\sqrt {a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{2} \sqrt {a} + {\left (\sqrt {a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )} a + 6 \, a^{\frac {3}{2}}\right )}}{{\left ({\left (\sqrt {a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{2} - a\right )}^{2} a \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}}{8 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a)*(tan(1/2*d*x + 1/2*c)/(a^2*sgn(tan(1/2*d*x + 1/2*c) + 1)) - 6/(a^2*sgn
(tan(1/2*d*x + 1/2*c) + 1))) - (44*sqrt(2)*sqrt(a)*arctan((sqrt(2)*sqrt(a) + sqrt(a))/sqrt(-a)) - 96*sqrt(2)*s
qrt(a)*arctan(sqrt(a)/sqrt(-a)) - 22*sqrt(2)*sqrt(-a)*log(sqrt(2)*sqrt(a) + sqrt(a)) + 66*sqrt(a)*arctan((sqrt
(2)*sqrt(a) + sqrt(a))/sqrt(-a)) - 128*sqrt(a)*arctan(sqrt(a)/sqrt(-a)) - 33*sqrt(-a)*log(sqrt(2)*sqrt(a) + sq
rt(a)) - 30*sqrt(2)*sqrt(-a) - 38*sqrt(-a))*sgn(tan(1/2*d*x + 1/2*c) + 1)/(2*sqrt(2)*sqrt(-a)*a^(3/2) + 3*sqrt
(-a)*a^(3/2)) - 32*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) + sqrt(a))/sqrt(-a))/(sqrt(-a)*a*sgn(tan(1/2*d*x + 1/2*c) + 1)) + 22*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*sgn(tan(1/2*d*x + 1/2*c) + 1)) - 11*log(abs(-sqrt
(a)*tan(1/2*d*x + 1/2*c) + sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a)))/(a^(3/2)*sgn(tan(1/2*d*x + 1/2*c) + 1)) + 2*((
sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^3 - 6*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(
a*tan(1/2*d*x + 1/2*c)^2 + a))^2*sqrt(a) + (sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))
*a + 6*a^(3/2))/(((sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^2 - a)^2*a*sgn(tan(1/2*d
*x + 1/2*c) + 1)))/d

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maple [A]  time = 1.37, size = 164, normalized size = 1.07 \[ -\frac {\left (1+\sin \left (d x +c \right )\right ) \sqrt {-a \left (\sin \left (d x +c \right )-1\right )}\, \left (11 a^{4} \arctanh \left (\frac {\sqrt {-a \left (\sin \left (d x +c \right )-1\right )}}{\sqrt {a}}\right ) \left (\sin ^{2}\left (d x +c \right )\right )+5 \left (-a \left (\sin \left (d x +c \right )-1\right )\right )^{\frac {3}{2}} a^{\frac {5}{2}}-8 \sqrt {2}\, \arctanh \left (\frac {\sqrt {-a \left (\sin \left (d x +c \right )-1\right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{4} \left (\sin ^{2}\left (d x +c \right )\right )-3 \sqrt {-a \left (\sin \left (d x +c \right )-1\right )}\, a^{\frac {7}{2}}\right )}{4 a^{\frac {11}{2}} \sin \left (d x +c \right )^{2} \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4/a^(11/2)*(1+sin(d*x+c))*(-a*(sin(d*x+c)-1))^(1/2)*(11*a^4*arctanh((-a*(sin(d*x+c)-1))^(1/2)/a^(1/2))*sin(
d*x+c)^2+5*(-a*(sin(d*x+c)-1))^(3/2)*a^(5/2)-8*2^(1/2)*arctanh(1/2*(-a*(sin(d*x+c)-1))^(1/2)*2^(1/2)/a^(1/2))*
a^4*sin(d*x+c)^2-3*(-a*(sin(d*x+c)-1))^(1/2)*a^(7/2))/sin(d*x+c)^2/cos(d*x+c)/(a+a*sin(d*x+c))^(1/2)/d

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maxima [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\cos \left (c+d\,x\right )}^2}{{\sin \left (c+d\,x\right )}^3\,{\left (a+a\,\sin \left (c+d\,x\right )\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cos ^{2}{\left (c + d x \right )} \csc ^{3}{\left (c + d x \right )}}{\left (a \left (\sin {\left (c + d x \right )} + 1\right )\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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