3.184 \(\int \frac{z^2 \sqrt{1+\cos (z)}}{\sqrt{1-\cos (z)}} \, dz\)

Optimal. Leaf size=300 \[ \frac{2 i z \sin (z) \text{PolyLog}\left (2,-e^{i z}\right )}{\sqrt{1-\cos (z)} \sqrt{\cos (z)+1}}-\frac{2 i z \sin (z) \text{PolyLog}\left (2,e^{i z}\right )}{\sqrt{1-\cos (z)} \sqrt{\cos (z)+1}}-\frac{i z \sin (z) \text{PolyLog}\left (2,e^{2 i z}\right )}{\sqrt{1-\cos (z)} \sqrt{\cos (z)+1}}-\frac{2 \sin (z) \text{PolyLog}\left (3,-e^{i z}\right )}{\sqrt{1-\cos (z)} \sqrt{\cos (z)+1}}+\frac{2 \sin (z) \text{PolyLog}\left (3,e^{i z}\right )}{\sqrt{1-\cos (z)} \sqrt{\cos (z)+1}}+\frac{\sin (z) \text{PolyLog}\left (3,e^{2 i z}\right )}{2 \sqrt{1-\cos (z)} \sqrt{\cos (z)+1}}-\frac{i z^3 \sin (z)}{3 \sqrt{1-\cos (z)} \sqrt{\cos (z)+1}}+\frac{z^2 \log \left (1-e^{2 i z}\right ) \sin (z)}{\sqrt{1-\cos (z)} \sqrt{\cos (z)+1}}-\frac{2 z^2 \sin (z) \tanh ^{-1}\left (e^{i z}\right )}{\sqrt{1-\cos (z)} \sqrt{\cos (z)+1}} \]

[Out]

((-I/3)*z^3*Sin[z])/(Sqrt[1 - Cos[z]]*Sqrt[1 + Cos[z]]) - (2*z^2*ArcTanh[E^(I*z)]*Sin[z])/(Sqrt[1 - Cos[z]]*Sq
rt[1 + Cos[z]]) + (z^2*Log[1 - E^((2*I)*z)]*Sin[z])/(Sqrt[1 - Cos[z]]*Sqrt[1 + Cos[z]]) + ((2*I)*z*PolyLog[2,
-E^(I*z)]*Sin[z])/(Sqrt[1 - Cos[z]]*Sqrt[1 + Cos[z]]) - ((2*I)*z*PolyLog[2, E^(I*z)]*Sin[z])/(Sqrt[1 - Cos[z]]
*Sqrt[1 + Cos[z]]) - (I*z*PolyLog[2, E^((2*I)*z)]*Sin[z])/(Sqrt[1 - Cos[z]]*Sqrt[1 + Cos[z]]) - (2*PolyLog[3,
-E^(I*z)]*Sin[z])/(Sqrt[1 - Cos[z]]*Sqrt[1 + Cos[z]]) + (2*PolyLog[3, E^(I*z)]*Sin[z])/(Sqrt[1 - Cos[z]]*Sqrt[
1 + Cos[z]]) + (PolyLog[3, E^((2*I)*z)]*Sin[z])/(2*Sqrt[1 - Cos[z]]*Sqrt[1 + Cos[z]])

________________________________________________________________________________________

Rubi [A]  time = 0.438342, antiderivative size = 300, normalized size of antiderivative = 1., number of steps used = 15, number of rules used = 8, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.364, Rules used = {4605, 6742, 3717, 2190, 2531, 2282, 6589, 4183} \[ \frac{2 i z \sin (z) \text{PolyLog}\left (2,-e^{i z}\right )}{\sqrt{1-\cos (z)} \sqrt{\cos (z)+1}}-\frac{2 i z \sin (z) \text{PolyLog}\left (2,e^{i z}\right )}{\sqrt{1-\cos (z)} \sqrt{\cos (z)+1}}-\frac{i z \sin (z) \text{PolyLog}\left (2,e^{2 i z}\right )}{\sqrt{1-\cos (z)} \sqrt{\cos (z)+1}}-\frac{2 \sin (z) \text{PolyLog}\left (3,-e^{i z}\right )}{\sqrt{1-\cos (z)} \sqrt{\cos (z)+1}}+\frac{2 \sin (z) \text{PolyLog}\left (3,e^{i z}\right )}{\sqrt{1-\cos (z)} \sqrt{\cos (z)+1}}+\frac{\sin (z) \text{PolyLog}\left (3,e^{2 i z}\right )}{2 \sqrt{1-\cos (z)} \sqrt{\cos (z)+1}}-\frac{i z^3 \sin (z)}{3 \sqrt{1-\cos (z)} \sqrt{\cos (z)+1}}+\frac{z^2 \log \left (1-e^{2 i z}\right ) \sin (z)}{\sqrt{1-\cos (z)} \sqrt{\cos (z)+1}}-\frac{2 z^2 \sin (z) \tanh ^{-1}\left (e^{i z}\right )}{\sqrt{1-\cos (z)} \sqrt{\cos (z)+1}} \]

Antiderivative was successfully verified.

[In]

Int[(z^2*Sqrt[1 + Cos[z]])/Sqrt[1 - Cos[z]],z]

[Out]

((-I/3)*z^3*Sin[z])/(Sqrt[1 - Cos[z]]*Sqrt[1 + Cos[z]]) - (2*z^2*ArcTanh[E^(I*z)]*Sin[z])/(Sqrt[1 - Cos[z]]*Sq
rt[1 + Cos[z]]) + (z^2*Log[1 - E^((2*I)*z)]*Sin[z])/(Sqrt[1 - Cos[z]]*Sqrt[1 + Cos[z]]) + ((2*I)*z*PolyLog[2,
-E^(I*z)]*Sin[z])/(Sqrt[1 - Cos[z]]*Sqrt[1 + Cos[z]]) - ((2*I)*z*PolyLog[2, E^(I*z)]*Sin[z])/(Sqrt[1 - Cos[z]]
*Sqrt[1 + Cos[z]]) - (I*z*PolyLog[2, E^((2*I)*z)]*Sin[z])/(Sqrt[1 - Cos[z]]*Sqrt[1 + Cos[z]]) - (2*PolyLog[3,
-E^(I*z)]*Sin[z])/(Sqrt[1 - Cos[z]]*Sqrt[1 + Cos[z]]) + (2*PolyLog[3, E^(I*z)]*Sin[z])/(Sqrt[1 - Cos[z]]*Sqrt[
1 + Cos[z]]) + (PolyLog[3, E^((2*I)*z)]*Sin[z])/(2*Sqrt[1 - Cos[z]]*Sqrt[1 + Cos[z]])

Rule 4605

Int[(Cos[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(Cos[(e_.) + (f_.)*(x_)]*(d_.) + (c_))^(n_)*((g_.) + (h_.)*(x_
))^(p_.), x_Symbol] :> Dist[(a^IntPart[m]*c^IntPart[m]*(a + b*Cos[e + f*x])^FracPart[m]*(c + d*Cos[e + f*x])^F
racPart[m])/Sin[e + f*x]^(2*FracPart[m]), Int[(g + h*x)^p*Sin[e + f*x]^(2*m)*(c + d*Cos[e + f*x])^(n - m), x],
x] /; FreeQ[{a, b, c, d, e, f, g, h}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && IntegerQ[p] && IntegerQ
[2*m] && IGeQ[n - m, 0]

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rule 3717

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + Pi*(k_.) + (f_.)*(x_)], x_Symbol] :> Simp[(I*(c + d*x)^(m + 1))/(d*
(m + 1)), x] - Dist[2*I, Int[((c + d*x)^m*E^(2*I*k*Pi)*E^(2*I*(e + f*x)))/(1 + E^(2*I*k*Pi)*E^(2*I*(e + f*x)))
, x], x] /; FreeQ[{c, d, e, f}, x] && IntegerQ[4*k] && IGtQ[m, 0]

Rule 2190

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/((a_) + (b_.)*((F_)^((g_.)*((e_.) +
(f_.)*(x_))))^(n_.)), x_Symbol] :> Simp[((c + d*x)^m*Log[1 + (b*(F^(g*(e + f*x)))^n)/a])/(b*f*g*n*Log[F]), x]
- Dist[(d*m)/(b*f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*Log[1 + (b*(F^(g*(e + f*x)))^n)/a], x], x] /; FreeQ[{F,
a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]

Rule 2531

Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.)*(x_))^(m_.), x_Symbol] :> -Simp[((
f + g*x)^m*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)])/(b*c*n*Log[F]), x] + Dist[(g*m)/(b*c*n*Log[F]), Int[(f + g*x)
^(m - 1)*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)], x], x] /; FreeQ[{F, a, b, c, e, f, g, n}, x] && GtQ[m, 0]

Rule 2282

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 6589

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[PolyLog[n + 1, c*(a
+ b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d, e, n, p}, x] && EqQ[b*d, a*e]

Rule 4183

Int[csc[(e_.) + (f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(-2*(c + d*x)^m*ArcTanh[E^(I*(e + f*
x))])/f, x] + (-Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Log[1 - E^(I*(e + f*x))], x], x] + Dist[(d*m)/f, Int[(c +
d*x)^(m - 1)*Log[1 + E^(I*(e + f*x))], x], x]) /; FreeQ[{c, d, e, f}, x] && IGtQ[m, 0]

Rubi steps

\begin{align*} \int \frac{z^2 \sqrt{1+\cos (z)}}{\sqrt{1-\cos (z)}} \, dz &=\frac{\sin (z) \int z^2 (1+\cos (z)) \csc (z) \, dz}{\sqrt{1-\cos (z)} \sqrt{1+\cos (z)}}\\ &=\frac{\sin (z) \int \left (z^2 \cot (z)+z^2 \csc (z)\right ) \, dz}{\sqrt{1-\cos (z)} \sqrt{1+\cos (z)}}\\ &=\frac{\sin (z) \int z^2 \cot (z) \, dz}{\sqrt{1-\cos (z)} \sqrt{1+\cos (z)}}+\frac{\sin (z) \int z^2 \csc (z) \, dz}{\sqrt{1-\cos (z)} \sqrt{1+\cos (z)}}\\ &=-\frac{i z^3 \sin (z)}{3 \sqrt{1-\cos (z)} \sqrt{1+\cos (z)}}-\frac{2 z^2 \tanh ^{-1}\left (e^{i z}\right ) \sin (z)}{\sqrt{1-\cos (z)} \sqrt{1+\cos (z)}}-\frac{(2 i \sin (z)) \int \frac{e^{2 i z} z^2}{1-e^{2 i z}} \, dz}{\sqrt{1-\cos (z)} \sqrt{1+\cos (z)}}-\frac{(2 \sin (z)) \int z \log \left (1-e^{i z}\right ) \, dz}{\sqrt{1-\cos (z)} \sqrt{1+\cos (z)}}+\frac{(2 \sin (z)) \int z \log \left (1+e^{i z}\right ) \, dz}{\sqrt{1-\cos (z)} \sqrt{1+\cos (z)}}\\ &=-\frac{i z^3 \sin (z)}{3 \sqrt{1-\cos (z)} \sqrt{1+\cos (z)}}-\frac{2 z^2 \tanh ^{-1}\left (e^{i z}\right ) \sin (z)}{\sqrt{1-\cos (z)} \sqrt{1+\cos (z)}}+\frac{z^2 \log \left (1-e^{2 i z}\right ) \sin (z)}{\sqrt{1-\cos (z)} \sqrt{1+\cos (z)}}+\frac{2 i z \text{Li}_2\left (-e^{i z}\right ) \sin (z)}{\sqrt{1-\cos (z)} \sqrt{1+\cos (z)}}-\frac{2 i z \text{Li}_2\left (e^{i z}\right ) \sin (z)}{\sqrt{1-\cos (z)} \sqrt{1+\cos (z)}}-\frac{(2 i \sin (z)) \int \text{Li}_2\left (-e^{i z}\right ) \, dz}{\sqrt{1-\cos (z)} \sqrt{1+\cos (z)}}+\frac{(2 i \sin (z)) \int \text{Li}_2\left (e^{i z}\right ) \, dz}{\sqrt{1-\cos (z)} \sqrt{1+\cos (z)}}-\frac{(2 \sin (z)) \int z \log \left (1-e^{2 i z}\right ) \, dz}{\sqrt{1-\cos (z)} \sqrt{1+\cos (z)}}\\ &=-\frac{i z^3 \sin (z)}{3 \sqrt{1-\cos (z)} \sqrt{1+\cos (z)}}-\frac{2 z^2 \tanh ^{-1}\left (e^{i z}\right ) \sin (z)}{\sqrt{1-\cos (z)} \sqrt{1+\cos (z)}}+\frac{z^2 \log \left (1-e^{2 i z}\right ) \sin (z)}{\sqrt{1-\cos (z)} \sqrt{1+\cos (z)}}+\frac{2 i z \text{Li}_2\left (-e^{i z}\right ) \sin (z)}{\sqrt{1-\cos (z)} \sqrt{1+\cos (z)}}-\frac{2 i z \text{Li}_2\left (e^{i z}\right ) \sin (z)}{\sqrt{1-\cos (z)} \sqrt{1+\cos (z)}}-\frac{i z \text{Li}_2\left (e^{2 i z}\right ) \sin (z)}{\sqrt{1-\cos (z)} \sqrt{1+\cos (z)}}+\frac{(i \sin (z)) \int \text{Li}_2\left (e^{2 i z}\right ) \, dz}{\sqrt{1-\cos (z)} \sqrt{1+\cos (z)}}-\frac{(2 \sin (z)) \operatorname{Subst}\left (\int \frac{\text{Li}_2(-z)}{z} \, dz,z,e^{i z}\right )}{\sqrt{1-\cos (z)} \sqrt{1+\cos (z)}}+\frac{(2 \sin (z)) \operatorname{Subst}\left (\int \frac{\text{Li}_2(z)}{z} \, dz,z,e^{i z}\right )}{\sqrt{1-\cos (z)} \sqrt{1+\cos (z)}}\\ &=-\frac{i z^3 \sin (z)}{3 \sqrt{1-\cos (z)} \sqrt{1+\cos (z)}}-\frac{2 z^2 \tanh ^{-1}\left (e^{i z}\right ) \sin (z)}{\sqrt{1-\cos (z)} \sqrt{1+\cos (z)}}+\frac{z^2 \log \left (1-e^{2 i z}\right ) \sin (z)}{\sqrt{1-\cos (z)} \sqrt{1+\cos (z)}}+\frac{2 i z \text{Li}_2\left (-e^{i z}\right ) \sin (z)}{\sqrt{1-\cos (z)} \sqrt{1+\cos (z)}}-\frac{2 i z \text{Li}_2\left (e^{i z}\right ) \sin (z)}{\sqrt{1-\cos (z)} \sqrt{1+\cos (z)}}-\frac{i z \text{Li}_2\left (e^{2 i z}\right ) \sin (z)}{\sqrt{1-\cos (z)} \sqrt{1+\cos (z)}}-\frac{2 \text{Li}_3\left (-e^{i z}\right ) \sin (z)}{\sqrt{1-\cos (z)} \sqrt{1+\cos (z)}}+\frac{2 \text{Li}_3\left (e^{i z}\right ) \sin (z)}{\sqrt{1-\cos (z)} \sqrt{1+\cos (z)}}+\frac{\sin (z) \operatorname{Subst}\left (\int \frac{\text{Li}_2(z)}{z} \, dz,z,e^{2 i z}\right )}{2 \sqrt{1-\cos (z)} \sqrt{1+\cos (z)}}\\ &=-\frac{i z^3 \sin (z)}{3 \sqrt{1-\cos (z)} \sqrt{1+\cos (z)}}-\frac{2 z^2 \tanh ^{-1}\left (e^{i z}\right ) \sin (z)}{\sqrt{1-\cos (z)} \sqrt{1+\cos (z)}}+\frac{z^2 \log \left (1-e^{2 i z}\right ) \sin (z)}{\sqrt{1-\cos (z)} \sqrt{1+\cos (z)}}+\frac{2 i z \text{Li}_2\left (-e^{i z}\right ) \sin (z)}{\sqrt{1-\cos (z)} \sqrt{1+\cos (z)}}-\frac{2 i z \text{Li}_2\left (e^{i z}\right ) \sin (z)}{\sqrt{1-\cos (z)} \sqrt{1+\cos (z)}}-\frac{i z \text{Li}_2\left (e^{2 i z}\right ) \sin (z)}{\sqrt{1-\cos (z)} \sqrt{1+\cos (z)}}-\frac{2 \text{Li}_3\left (-e^{i z}\right ) \sin (z)}{\sqrt{1-\cos (z)} \sqrt{1+\cos (z)}}+\frac{2 \text{Li}_3\left (e^{i z}\right ) \sin (z)}{\sqrt{1-\cos (z)} \sqrt{1+\cos (z)}}+\frac{\text{Li}_3\left (e^{2 i z}\right ) \sin (z)}{2 \sqrt{1-\cos (z)} \sqrt{1+\cos (z)}}\\ \end{align*}

Mathematica [A]  time = 0.0838063, size = 85, normalized size = 0.28 \[ \frac{\sqrt{\cos (z)+1} \tan \left (\frac{z}{2}\right ) \left (12 i z \text{PolyLog}\left (2,e^{-i z}\right )+12 \text{PolyLog}\left (3,e^{-i z}\right )+i z^3+6 z^2 \log \left (1-e^{-i z}\right )-i \pi ^3\right )}{3 \sqrt{1-\cos (z)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(z^2*Sqrt[1 + Cos[z]])/Sqrt[1 - Cos[z]],z]

[Out]

(Sqrt[1 + Cos[z]]*((-I)*Pi^3 + I*z^3 + 6*z^2*Log[1 - E^((-I)*z)] + (12*I)*z*PolyLog[2, E^((-I)*z)] + 12*PolyLo
g[3, E^((-I)*z)])*Tan[z/2])/(3*Sqrt[1 - Cos[z]])

________________________________________________________________________________________

Maple [A]  time = 0.067, size = 154, normalized size = 0.5 \begin{align*}{\frac{ \left ({{\rm e}^{iz}}-1 \right ){z}^{3}}{3\,{{\rm e}^{iz}}+3}\sqrt{ \left ({{\rm e}^{iz}}+1 \right ) ^{2}{{\rm e}^{-iz}}}{\frac{1}{\sqrt{- \left ({{\rm e}^{iz}}-1 \right ) ^{2}{{\rm e}^{-iz}}}}}}+{\frac{2\,i \left ({{\rm e}^{iz}}-1 \right ) \left ({\frac{i}{3}}{z}^{3}-{z}^{2}\ln \left ( 1-{{\rm e}^{iz}} \right ) +2\,iz{\it polylog} \left ( 2,{{\rm e}^{iz}} \right ) -2\,{\it polylog} \left ( 3,{{\rm e}^{iz}} \right ) \right ) }{{{\rm e}^{iz}}+1}\sqrt{ \left ({{\rm e}^{iz}}+1 \right ) ^{2}{{\rm e}^{-iz}}}{\frac{1}{\sqrt{- \left ({{\rm e}^{iz}}-1 \right ) ^{2}{{\rm e}^{-iz}}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(z^2*(1+cos(z))^(1/2)/(1-cos(z))^(1/2),z)

[Out]

1/3*((exp(I*z)+1)^2*exp(-I*z))^(1/2)/(exp(I*z)+1)/(-(exp(I*z)-1)^2*exp(-I*z))^(1/2)*(exp(I*z)-1)*z^3+2*I*((exp
(I*z)+1)^2*exp(-I*z))^(1/2)/(exp(I*z)+1)/(-(exp(I*z)-1)^2*exp(-I*z))^(1/2)*(exp(I*z)-1)*(1/3*I*z^3-z^2*ln(1-ex
p(I*z))+2*I*z*polylog(2,exp(I*z))-2*polylog(3,exp(I*z)))

________________________________________________________________________________________

Maxima [A]  time = 1.56118, size = 76, normalized size = 0.25 \begin{align*} \frac{1}{3} i \, z^{3} + 2 i \, z^{2} \arctan \left (\sin \left (z\right ), -\cos \left (z\right ) + 1\right ) - z^{2} \log \left (\cos \left (z\right )^{2} + \sin \left (z\right )^{2} - 2 \, \cos \left (z\right ) + 1\right ) + 4 i \, z{\rm Li}_2\left (e^{\left (i \, z\right )}\right ) - 4 \,{\rm Li}_{3}(e^{\left (i \, z\right )}) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(z^2*(1+cos(z))^(1/2)/(1-cos(z))^(1/2),z, algorithm="maxima")

[Out]

1/3*I*z^3 + 2*I*z^2*arctan2(sin(z), -cos(z) + 1) - z^2*log(cos(z)^2 + sin(z)^2 - 2*cos(z) + 1) + 4*I*z*dilog(e
^(I*z)) - 4*polylog(3, e^(I*z))

________________________________________________________________________________________

Fricas [C]  time = 2.43017, size = 271, normalized size = 0.9 \begin{align*} z^{2} \log \left (-\cos \left (z\right ) + i \, \sin \left (z\right ) + 1\right ) + z^{2} \log \left (-\cos \left (z\right ) - i \, \sin \left (z\right ) + 1\right ) - 2 i \, z{\rm Li}_2\left (\cos \left (z\right ) + i \, \sin \left (z\right )\right ) + 2 i \, z{\rm Li}_2\left (\cos \left (z\right ) - i \, \sin \left (z\right )\right ) + 2 \,{\rm polylog}\left (3, \cos \left (z\right ) + i \, \sin \left (z\right )\right ) + 2 \,{\rm polylog}\left (3, \cos \left (z\right ) - i \, \sin \left (z\right )\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(z^2*(1+cos(z))^(1/2)/(1-cos(z))^(1/2),z, algorithm="fricas")

[Out]

z^2*log(-cos(z) + I*sin(z) + 1) + z^2*log(-cos(z) - I*sin(z) + 1) - 2*I*z*dilog(cos(z) + I*sin(z)) + 2*I*z*dil
og(cos(z) - I*sin(z)) + 2*polylog(3, cos(z) + I*sin(z)) + 2*polylog(3, cos(z) - I*sin(z))

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{z^{2} \sqrt{\cos{\left (z \right )} + 1}}{\sqrt{1 - \cos{\left (z \right )}}}\, dz \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(z**2*(1+cos(z))**(1/2)/(1-cos(z))**(1/2),z)

[Out]

Integral(z**2*sqrt(cos(z) + 1)/sqrt(1 - cos(z)), z)

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{z^{2} \sqrt{\cos \left (z\right ) + 1}}{\sqrt{-\cos \left (z\right ) + 1}}\,{d z} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(z^2*(1+cos(z))^(1/2)/(1-cos(z))^(1/2),z, algorithm="giac")

[Out]

integrate(z^2*sqrt(cos(z) + 1)/sqrt(-cos(z) + 1), z)