3.67 \(\int \frac {\csc ^3(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx\)

Optimal. Leaf size=146 \[ \frac {\cot (c+d x)}{4 d \sqrt {a \sin (c+d x)+a}}-\frac {7 \tanh ^{-1}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {a \sin (c+d x)+a}}\right )}{4 \sqrt {a} d}+\frac {\sqrt {2} \tanh ^{-1}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {2} \sqrt {a \sin (c+d x)+a}}\right )}{\sqrt {a} d}-\frac {\cot (c+d x) \csc (c+d x)}{2 d \sqrt {a \sin (c+d x)+a}} \]

[Out]

-7/4*arctanh(cos(d*x+c)*a^(1/2)/(a+a*sin(d*x+c))^(1/2))/d/a^(1/2)+arctanh(1/2*cos(d*x+c)*a^(1/2)*2^(1/2)/(a+a*
sin(d*x+c))^(1/2))*2^(1/2)/d/a^(1/2)+1/4*cot(d*x+c)/d/(a+a*sin(d*x+c))^(1/2)-1/2*cot(d*x+c)*csc(d*x+c)/d/(a+a*
sin(d*x+c))^(1/2)

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Rubi [A]  time = 0.34, antiderivative size = 146, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {2779, 2984, 2985, 2649, 206, 2773} \[ \frac {\cot (c+d x)}{4 d \sqrt {a \sin (c+d x)+a}}-\frac {7 \tanh ^{-1}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {a \sin (c+d x)+a}}\right )}{4 \sqrt {a} d}+\frac {\sqrt {2} \tanh ^{-1}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {2} \sqrt {a \sin (c+d x)+a}}\right )}{\sqrt {a} d}-\frac {\cot (c+d x) \csc (c+d x)}{2 d \sqrt {a \sin (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

Int[Csc[c + d*x]^3/Sqrt[a + a*Sin[c + d*x]],x]

[Out]

(-7*ArcTanh[(Sqrt[a]*Cos[c + d*x])/Sqrt[a + a*Sin[c + d*x]]])/(4*Sqrt[a]*d) + (Sqrt[2]*ArcTanh[(Sqrt[a]*Cos[c
+ d*x])/(Sqrt[2]*Sqrt[a + a*Sin[c + d*x]])])/(Sqrt[a]*d) + Cot[c + d*x]/(4*d*Sqrt[a + a*Sin[c + d*x]]) - (Cot[
c + d*x]*Csc[c + d*x])/(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 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 2779

Int[((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)/Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> -Sim
p[(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 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 ^3(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx &=-\frac {\cot (c+d x) \csc (c+d x)}{2 d \sqrt {a+a \sin (c+d x)}}-\frac {\int \frac {\csc ^2(c+d x) (a-3 a \sin (c+d x))}{\sqrt {a+a \sin (c+d x)}} \, dx}{4 a}\\ &=\frac {\cot (c+d x)}{4 d \sqrt {a+a \sin (c+d x)}}-\frac {\cot (c+d x) \csc (c+d x)}{2 d \sqrt {a+a \sin (c+d x)}}-\frac {\int \frac {\csc (c+d x) \left (-\frac {7 a^2}{2}+\frac {1}{2} a^2 \sin (c+d x)\right )}{\sqrt {a+a \sin (c+d x)}} \, dx}{4 a^2}\\ &=\frac {\cot (c+d x)}{4 d \sqrt {a+a \sin (c+d x)}}-\frac {\cot (c+d x) \csc (c+d x)}{2 d \sqrt {a+a \sin (c+d x)}}+\frac {7 \int \csc (c+d x) \sqrt {a+a \sin (c+d x)} \, dx}{8 a}-\int \frac {1}{\sqrt {a+a \sin (c+d x)}} \, dx\\ &=\frac {\cot (c+d x)}{4 d \sqrt {a+a \sin (c+d x)}}-\frac {\cot (c+d x) \csc (c+d x)}{2 d \sqrt {a+a \sin (c+d x)}}-\frac {7 \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 d}+\frac {2 \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 )}{d}\\ &=-\frac {7 \tanh ^{-1}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {a+a \sin (c+d x)}}\right )}{4 \sqrt {a} d}+\frac {\sqrt {2} \tanh ^{-1}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {2} \sqrt {a+a \sin (c+d x)}}\right )}{\sqrt {a} d}+\frac {\cot (c+d x)}{4 d \sqrt {a+a \sin (c+d x)}}-\frac {\cot (c+d x) \csc (c+d x)}{2 d \sqrt {a+a \sin (c+d x)}}\\ \end {align*}

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Mathematica [C]  time = 3.40, size = 307, normalized size = 2.10 \[ \frac {\left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right ) \left (4 \tan \left (\frac {1}{4} (c+d x)\right )+4 \cot \left (\frac {1}{4} (c+d x)\right )-\csc ^2\left (\frac {1}{4} (c+d x)\right )+\sec ^2\left (\frac {1}{4} (c+d x)\right )-\frac {8 \sin \left (\frac {1}{4} (c+d x)\right )}{\cos \left (\frac {1}{4} (c+d x)\right )-\sin \left (\frac {1}{4} (c+d x)\right )}+\frac {8 \sin \left (\frac {1}{4} (c+d x)\right )}{\sin \left (\frac {1}{4} (c+d x)\right )+\cos \left (\frac {1}{4} (c+d x)\right )}+\frac {2}{\left (\cos \left (\frac {1}{4} (c+d x)\right )-\sin \left (\frac {1}{4} (c+d x)\right )\right )^2}-\frac {2}{\left (\sin \left (\frac {1}{4} (c+d x)\right )+\cos \left (\frac {1}{4} (c+d x)\right )\right )^2}-(64+64 i) (-1)^{3/4} \tanh ^{-1}\left (\left (\frac {1}{2}+\frac {i}{2}\right ) (-1)^{3/4} \left (\tan \left (\frac {1}{4} (c+d x)\right )-1\right )\right )-28 \log \left (-\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )+1\right )+28 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )-\cos \left (\frac {1}{2} (c+d x)\right )+1\right )-8\right )}{32 d \sqrt {a (\sin (c+d x)+1)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[c + d*x]^3/Sqrt[a + a*Sin[c + d*x]],x]

[Out]

((Cos[(c + d*x)/2] + Sin[(c + d*x)/2])*(-8 - (64 + 64*I)*(-1)^(3/4)*ArcTanh[(1/2 + I/2)*(-1)^(3/4)*(-1 + Tan[(
c + d*x)/4])] + 4*Cot[(c + d*x)/4] - Csc[(c + d*x)/4]^2 - 28*Log[1 + Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 28
*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
- (8*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 + (8
*Sin[(c + d*x)/4])/(Cos[(c + d*x)/4] + Sin[(c + d*x)/4]) + 4*Tan[(c + d*x)/4]))/(32*d*Sqrt[a*(1 + Sin[c + d*x]
)])

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fricas [B]  time = 0.48, size = 492, normalized size = 3.37 \[ \frac {7 \, {\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 {8 \, \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 (\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}}{16 \, {\left (a d \cos \left (d x + c\right )^{3} + a d \cos \left (d x + c\right )^{2} - a d \cos \left (d x + c\right ) - a d + {\left (a d \cos \left (d x + c\right )^{2} - a 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)^3/(a+a*sin(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

1/16*(7*(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)) + 8*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*(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))/(a*d*c
os(d*x + c)^3 + a*d*cos(d*x + c)^2 - a*d*cos(d*x + c) - a*d + (a*d*cos(d*x + c)^2 - a*d)*sin(d*x + c))

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giac [B]  time = 3.05, size = 615, normalized size = 4.21 \[ \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 \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )} - \frac {2}{a \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}\right )} - \frac {{\left (42 \, \sqrt {2} a^{\frac {3}{2}} \arctan \left (\frac {\sqrt {2} \sqrt {a} + \sqrt {a}}{\sqrt {-a}}\right ) - 64 \, \sqrt {2} a^{\frac {3}{2}} \arctan \left (\frac {\sqrt {a}}{\sqrt {-a}}\right ) - 21 \, \sqrt {2} \sqrt {-a} a \log \left (\sqrt {2} \sqrt {a} + \sqrt {a}\right ) + 56 \, a^{\frac {3}{2}} \arctan \left (\frac {\sqrt {2} \sqrt {a} + \sqrt {a}}{\sqrt {-a}}\right ) - 96 \, a^{\frac {3}{2}} \arctan \left (\frac {\sqrt {a}}{\sqrt {-a}}\right ) - 28 \, \sqrt {-a} a \log \left (\sqrt {2} \sqrt {a} + \sqrt {a}\right ) - 18 \, \sqrt {2} \sqrt {-a} a - 28 \, \sqrt {-a} a\right )} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}{3 \, \sqrt {2} \sqrt {-a} a^{\frac {3}{2}} + 4 \, \sqrt {-a} a^{\frac {3}{2}}} - \frac {16 \, \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} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )} + \frac {14 \, \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} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )} - \frac {7 \, \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 )}{\sqrt {a} \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} - 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}\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 + 2 \, 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} \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(csc(d*x+c)^3/(a+a*sin(d*x+c))^(1/2),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*sgn(tan(1/2*d*x + 1/2*c) + 1)) - 2/(a*sgn(tan
(1/2*d*x + 1/2*c) + 1))) - (42*sqrt(2)*a^(3/2)*arctan((sqrt(2)*sqrt(a) + sqrt(a))/sqrt(-a)) - 64*sqrt(2)*a^(3/
2)*arctan(sqrt(a)/sqrt(-a)) - 21*sqrt(2)*sqrt(-a)*a*log(sqrt(2)*sqrt(a) + sqrt(a)) + 56*a^(3/2)*arctan((sqrt(2
)*sqrt(a) + sqrt(a))/sqrt(-a)) - 96*a^(3/2)*arctan(sqrt(a)/sqrt(-a)) - 28*sqrt(-a)*a*log(sqrt(2)*sqrt(a) + sqr
t(a)) - 18*sqrt(2)*sqrt(-a)*a - 28*sqrt(-a)*a)*sgn(tan(1/2*d*x + 1/2*c) + 1)/(3*sqrt(2)*sqrt(-a)*a^(3/2) + 4*s
qrt(-a)*a^(3/2)) - 16*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)*sgn(tan(1/2*d*x + 1/2*c) + 1)) + 14*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)*sgn(tan(1/2*d*x + 1/2*c) + 1)) - 7*log(abs(-sqrt(a
)*tan(1/2*d*x + 1/2*c) + sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a)))/(sqrt(a)*sgn(tan(1/2*d*x + 1/2*c) + 1)) + 2*((sq
rt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + 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*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
+ 2*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*sgn(tan(1/2*d*x +
1/2*c) + 1)))/d

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maple [A]  time = 1.00, size = 162, normalized size = 1.11 \[ -\frac {\left (1+\sin \left (d x +c \right )\right ) \sqrt {-a \left (\sin \left (d x +c \right )-1\right )}\, \left (7 a^{5} \arctanh \left (\frac {\sqrt {-a \left (\sin \left (d x +c \right )-1\right )}}{\sqrt {a}}\right ) \left (\sin ^{2}\left (d x +c \right )\right )+\left (-a \left (\sin \left (d x +c \right )-1\right )\right )^{\frac {3}{2}} a^{\frac {7}{2}}+\sqrt {-a \left (\sin \left (d x +c \right )-1\right )}\, a^{\frac {9}{2}}-4 \sqrt {2}\, \arctanh \left (\frac {\sqrt {-a \left (\sin \left (d x +c \right )-1\right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{5} \left (\sin ^{2}\left (d x +c \right )\right )\right )}{4 a^{\frac {11}{2}} \sin \left (d x +c \right )^{2} \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(1+sin(d*x+c))*(-a*(sin(d*x+c)-1))^(1/2)/a^(11/2)*(7*a^5*arctanh((-a*(sin(d*x+c)-1))^(1/2)/a^(1/2))*sin(d
*x+c)^2+(-a*(sin(d*x+c)-1))^(3/2)*a^(7/2)+(-a*(sin(d*x+c)-1))^(1/2)*a^(9/2)-4*2^(1/2)*arctanh(1/2*(-a*(sin(d*x
+c)-1))^(1/2)*2^(1/2)/a^(1/2))*a^5*sin(d*x+c)^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]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\csc \left (d x + c\right )^{3}}{\sqrt {a \sin \left (d x + c\right ) + a}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^3/(a+a*sin(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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