3.426 \(\int \frac{\csc ^2(x) (-2 \cos ^3(x) (-1+\sin (x))+\cos (2 x) \sin (x))}{\sqrt{-5+\sin ^2(x)}} \, dx\)

Optimal. Leaf size=111 \[ 2 \sqrt{\sin ^2(x)-5}-\frac{2 \tan ^{-1}\left (\frac{\sqrt{\sin ^2(x)-5}}{\sqrt{5}}\right )}{\sqrt{5}}-2 \tanh ^{-1}\left (\frac{\sin (x)}{\sqrt{\sin ^2(x)-5}}\right )+\frac{2}{5} \sqrt{\sin ^2(x)-5} \csc (x)+2 \tan ^{-1}\left (\frac{\cos (x)}{\sqrt{\sin ^2(x)-5}}\right )-\frac{\tan ^{-1}\left (\frac{\sqrt{5} \cos (x)}{\sqrt{\sin ^2(x)-5}}\right )}{\sqrt{5}} \]

[Out]

2*ArcTan[Cos[x]/Sqrt[-5 + Sin[x]^2]] - ArcTan[(Sqrt[5]*Cos[x])/Sqrt[-5 + Sin[x]^2]]/Sqrt[5] - (2*ArcTan[Sqrt[-
5 + Sin[x]^2]/Sqrt[5]])/Sqrt[5] - 2*ArcTanh[Sin[x]/Sqrt[-5 + Sin[x]^2]] + 2*Sqrt[-5 + Sin[x]^2] + (2*Csc[x]*Sq
rt[-5 + Sin[x]^2])/5

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Rubi [A]  time = 0.574979, antiderivative size = 119, normalized size of antiderivative = 1.07, number of steps used = 18, number of rules used = 13, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.394, Rules used = {4401, 4356, 451, 217, 206, 4366, 6725, 203, 261, 1010, 377, 444, 63} \[ 2 \sqrt{-\cos ^2(x)-4}-2 \tanh ^{-1}\left (\frac{\sin (x)}{\sqrt{\sin ^2(x)-5}}\right )+2 \tan ^{-1}\left (\frac{\cos (x)}{\sqrt{-\cos ^2(x)-4}}\right )-\frac{\tan ^{-1}\left (\frac{\sqrt{5} \cos (x)}{\sqrt{-\cos ^2(x)-4}}\right )}{\sqrt{5}}-\frac{2 \tan ^{-1}\left (\frac{\sqrt{-\cos ^2(x)-4}}{\sqrt{5}}\right )}{\sqrt{5}}+\frac{2}{5} \sqrt{\sin ^2(x)-5} \csc (x) \]

Antiderivative was successfully verified.

[In]

Int[(Csc[x]^2*(-2*Cos[x]^3*(-1 + Sin[x]) + Cos[2*x]*Sin[x]))/Sqrt[-5 + Sin[x]^2],x]

[Out]

2*ArcTan[Cos[x]/Sqrt[-4 - Cos[x]^2]] - ArcTan[(Sqrt[5]*Cos[x])/Sqrt[-4 - Cos[x]^2]]/Sqrt[5] - (2*ArcTan[Sqrt[-
4 - Cos[x]^2]/Sqrt[5]])/Sqrt[5] - 2*ArcTanh[Sin[x]/Sqrt[-5 + Sin[x]^2]] + 2*Sqrt[-4 - Cos[x]^2] + (2*Csc[x]*Sq
rt[-5 + Sin[x]^2])/5

Rule 4401

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

Rule 4356

Int[(u_)*(F_)[(c_.)*((a_.) + (b_.)*(x_))], x_Symbol] :> With[{d = FreeFactors[Sin[c*(a + b*x)], x]}, Dist[d/(b
*c), Subst[Int[SubstFor[1, Sin[c*(a + b*x)]/d, u, x], x], x, Sin[c*(a + b*x)]/d], x] /; FunctionOfQ[Sin[c*(a +
b*x)]/d, u, x]] /; FreeQ[{a, b, c}, x] && (EqQ[F, Cos] || EqQ[F, cos])

Rule 451

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(c*(e*x)^(m
+ 1)*(a + b*x^n)^(p + 1))/(a*e*(m + 1)), x] + Dist[d/e^n, Int[(e*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a,
b, c, d, e, m, n, p}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n*(p + 1) + 1, 0] && (IntegerQ[n] || GtQ[e, 0]) && (
(GtQ[n, 0] && LtQ[m, -1]) || (LtQ[n, 0] && GtQ[m + n, -1]))

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

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 4366

Int[(u_)*(F_)[(c_.)*((a_.) + (b_.)*(x_))]^(n_), x_Symbol] :> With[{d = FreeFactors[Cos[c*(a + b*x)], x]}, -Dis
t[d/(b*c), Subst[Int[SubstFor[(1 - d^2*x^2)^((n - 1)/2), Cos[c*(a + b*x)]/d, u, x], x], x, Cos[c*(a + b*x)]/d]
, x] /; FunctionOfQ[Cos[c*(a + b*x)]/d, u, x]] /; FreeQ[{a, b, c}, x] && IntegerQ[(n - 1)/2] && NonsumQ[u] &&
(EqQ[F, Sin] || EqQ[F, sin])

Rule 6725

Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a + b*x^n), x]}, Int[v, x]
/; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ[n, 0]

Rule 203

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

Rule 261

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a + b*x^n)^(p + 1)/(b*n*(p + 1)), x] /; FreeQ
[{a, b, m, n, p}, x] && EqQ[m, n - 1] && NeQ[p, -1]

Rule 1010

Int[((g_) + (h_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_)*((d_) + (f_.)*(x_)^2)^(q_), x_Symbol] :> Dist[g, Int[(a + c
*x^2)^p*(d + f*x^2)^q, x], x] + Dist[h, Int[x*(a + c*x^2)^p*(d + f*x^2)^q, x], x] /; FreeQ[{a, c, d, f, g, h,
p, q}, x]

Rule 377

Int[((a_) + (b_.)*(x_)^(n_))^(p_)/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Subst[Int[1/(c - (b*c - a*d)*x^n), x]
, x, x/(a + b*x^n)^(1/n)] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && EqQ[n*p + 1, 0] && IntegerQ[n]

Rule 444

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && EqQ[m
- n + 1, 0]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rubi steps

\begin{align*} \int \frac{\csc ^2(x) \left (-2 \cos ^3(x) (-1+\sin (x))+\cos (2 x) \sin (x)\right )}{\sqrt{-5+\sin ^2(x)}} \, dx &=\int \left (\frac{2 \cos (x) \cot ^2(x)}{\sqrt{-5+\sin ^2(x)}}+\frac{\left (-2 \cos ^3(x)+\cos (2 x)\right ) \csc (x)}{\sqrt{-5+\sin ^2(x)}}\right ) \, dx\\ &=2 \int \frac{\cos (x) \cot ^2(x)}{\sqrt{-5+\sin ^2(x)}} \, dx+\int \frac{\left (-2 \cos ^3(x)+\cos (2 x)\right ) \csc (x)}{\sqrt{-5+\sin ^2(x)}} \, dx\\ &=2 \operatorname{Subst}\left (\int \frac{1-x^2}{x^2 \sqrt{-5+x^2}} \, dx,x,\sin (x)\right )-\operatorname{Subst}\left (\int \frac{-1+2 x^2-2 x^3}{\sqrt{-4-x^2} \left (1-x^2\right )} \, dx,x,\cos (x)\right )\\ &=\frac{2}{5} \csc (x) \sqrt{-5+\sin ^2(x)}-2 \operatorname{Subst}\left (\int \frac{1}{\sqrt{-5+x^2}} \, dx,x,\sin (x)\right )-\operatorname{Subst}\left (\int \left (-\frac{2}{\sqrt{-4-x^2}}+\frac{2 x}{\sqrt{-4-x^2}}+\frac{1-2 x}{\sqrt{-4-x^2} \left (1-x^2\right )}\right ) \, dx,x,\cos (x)\right )\\ &=\frac{2}{5} \csc (x) \sqrt{-5+\sin ^2(x)}+2 \operatorname{Subst}\left (\int \frac{1}{\sqrt{-4-x^2}} \, dx,x,\cos (x)\right )-2 \operatorname{Subst}\left (\int \frac{x}{\sqrt{-4-x^2}} \, dx,x,\cos (x)\right )-2 \operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,\frac{\sin (x)}{\sqrt{-5+\sin ^2(x)}}\right )-\operatorname{Subst}\left (\int \frac{1-2 x}{\sqrt{-4-x^2} \left (1-x^2\right )} \, dx,x,\cos (x)\right )\\ &=-2 \tanh ^{-1}\left (\frac{\sin (x)}{\sqrt{-5+\sin ^2(x)}}\right )+2 \sqrt{-4-\cos ^2(x)}+\frac{2}{5} \csc (x) \sqrt{-5+\sin ^2(x)}+2 \operatorname{Subst}\left (\int \frac{x}{\sqrt{-4-x^2} \left (1-x^2\right )} \, dx,x,\cos (x)\right )+2 \operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,\frac{\cos (x)}{\sqrt{-4-\cos ^2(x)}}\right )-\operatorname{Subst}\left (\int \frac{1}{\sqrt{-4-x^2} \left (1-x^2\right )} \, dx,x,\cos (x)\right )\\ &=2 \tan ^{-1}\left (\frac{\cos (x)}{\sqrt{-4-\cos ^2(x)}}\right )-2 \tanh ^{-1}\left (\frac{\sin (x)}{\sqrt{-5+\sin ^2(x)}}\right )+2 \sqrt{-4-\cos ^2(x)}+\frac{2}{5} \csc (x) \sqrt{-5+\sin ^2(x)}+\operatorname{Subst}\left (\int \frac{1}{\sqrt{-4-x} (1-x)} \, dx,x,\cos ^2(x)\right )-\operatorname{Subst}\left (\int \frac{1}{1+5 x^2} \, dx,x,\frac{\cos (x)}{\sqrt{-4-\cos ^2(x)}}\right )\\ &=2 \tan ^{-1}\left (\frac{\cos (x)}{\sqrt{-4-\cos ^2(x)}}\right )-\frac{\tan ^{-1}\left (\frac{\sqrt{5} \cos (x)}{\sqrt{-4-\cos ^2(x)}}\right )}{\sqrt{5}}-2 \tanh ^{-1}\left (\frac{\sin (x)}{\sqrt{-5+\sin ^2(x)}}\right )+2 \sqrt{-4-\cos ^2(x)}+\frac{2}{5} \csc (x) \sqrt{-5+\sin ^2(x)}-2 \operatorname{Subst}\left (\int \frac{1}{5+x^2} \, dx,x,\sqrt{-4-\cos ^2(x)}\right )\\ &=2 \tan ^{-1}\left (\frac{\cos (x)}{\sqrt{-4-\cos ^2(x)}}\right )-\frac{\tan ^{-1}\left (\frac{\sqrt{5} \cos (x)}{\sqrt{-4-\cos ^2(x)}}\right )}{\sqrt{5}}-\frac{2 \tan ^{-1}\left (\frac{\sqrt{-4-\cos ^2(x)}}{\sqrt{5}}\right )}{\sqrt{5}}-2 \tanh ^{-1}\left (\frac{\sin (x)}{\sqrt{-5+\sin ^2(x)}}\right )+2 \sqrt{-4-\cos ^2(x)}+\frac{2}{5} \csc (x) \sqrt{-5+\sin ^2(x)}\\ \end{align*}

Mathematica [C]  time = 2.32738, size = 338, normalized size = 3.05 \[ \frac{(16-32 i) \sqrt{5} \sqrt{\frac{(1+2 i) (\cos (x)-2 i)}{\cos (x)+1}} \sqrt{\frac{(1-2 i) (\cos (x)+2 i)}{\cos (x)+1}} \cos ^2\left (\frac{x}{2}\right ) \text{EllipticF}\left (\sin ^{-1}\left (\frac{(1+2 i) \tan \left (\frac{x}{2}\right )}{\sqrt{5}}\right ),-\frac{7}{25}+\frac{24 i}{25}\right )-(32-64 i) \sqrt{5} \sqrt{\frac{(1+2 i) (\cos (x)-2 i)}{\cos (x)+1}} \sqrt{\frac{(1-2 i) (\cos (x)+2 i)}{\cos (x)+1}} \cos ^2\left (\frac{x}{2}\right ) \Pi \left (\frac{3}{5}+\frac{4 i}{5};\sin ^{-1}\left (\frac{(1+2 i) \tan \left (\frac{x}{2}\right )}{\sqrt{5}}\right )|-\frac{7}{25}+\frac{24 i}{25}\right )-5 \left (18 \csc (x)+10 i \sqrt{2} \sqrt{-\cos (2 x)-9} \log \left (\sqrt{-\cos (2 x)-9}+i \sqrt{2} \cos (x)\right )+\sqrt{10} \sqrt{-\cos (2 x)-9} \tan ^{-1}\left (\frac{\sqrt{10} \cos (x)}{\sqrt{-\cos (2 x)-9}}\right )+2 \sqrt{10} \sqrt{-\cos (2 x)-9} \tan ^{-1}\left (\frac{\sqrt{-\cos (2 x)-9}}{\sqrt{10}}\right )+2 \cos (2 x) \csc (x)+5 \sin (3 x) \csc (x)+85\right )}{25 \sqrt{2} \sqrt{-\cos (2 x)-9}} \]

Antiderivative was successfully verified.

[In]

Integrate[(Csc[x]^2*(-2*Cos[x]^3*(-1 + Sin[x]) + Cos[2*x]*Sin[x]))/Sqrt[-5 + Sin[x]^2],x]

[Out]

((16 - 32*I)*Sqrt[5]*Cos[x/2]^2*Sqrt[((1 + 2*I)*(-2*I + Cos[x]))/(1 + Cos[x])]*Sqrt[((1 - 2*I)*(2*I + Cos[x]))
/(1 + Cos[x])]*EllipticF[ArcSin[((1 + 2*I)*Tan[x/2])/Sqrt[5]], -7/25 + (24*I)/25] - (32 - 64*I)*Sqrt[5]*Cos[x/
2]^2*Sqrt[((1 + 2*I)*(-2*I + Cos[x]))/(1 + Cos[x])]*Sqrt[((1 - 2*I)*(2*I + Cos[x]))/(1 + Cos[x])]*EllipticPi[3
/5 + (4*I)/5, ArcSin[((1 + 2*I)*Tan[x/2])/Sqrt[5]], -7/25 + (24*I)/25] - 5*(85 + Sqrt[10]*ArcTan[(Sqrt[10]*Cos
[x])/Sqrt[-9 - Cos[2*x]]]*Sqrt[-9 - Cos[2*x]] + 2*Sqrt[10]*ArcTan[Sqrt[-9 - Cos[2*x]]/Sqrt[10]]*Sqrt[-9 - Cos[
2*x]] + 18*Csc[x] + 2*Cos[2*x]*Csc[x] + (10*I)*Sqrt[2]*Sqrt[-9 - Cos[2*x]]*Log[I*Sqrt[2]*Cos[x] + Sqrt[-9 - Co
s[2*x]]] + 5*Csc[x]*Sin[3*x]))/(25*Sqrt[2]*Sqrt[-9 - Cos[2*x]])

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Maple [A]  time = 0.184, size = 130, normalized size = 1.2 \begin{align*} -2\,\ln \left ( \sin \left ( x \right ) +\sqrt{-5+ \left ( \sin \left ( x \right ) \right ) ^{2}} \right ) +2\,\sqrt{-5+ \left ( \sin \left ( x \right ) \right ) ^{2}}+{\frac{2\,\sqrt{5}}{5}\arctan \left ({\sqrt{5}{\frac{1}{\sqrt{-5+ \left ( \sin \left ( x \right ) \right ) ^{2}}}}} \right ) }+{\frac{2}{5\,\sin \left ( x \right ) }\sqrt{-5+ \left ( \sin \left ( x \right ) \right ) ^{2}}}+{\frac{1}{10\,\cos \left ( x \right ) }\sqrt{ \left ( -5+ \left ( \sin \left ( x \right ) \right ) ^{2} \right ) \left ( \cos \left ( x \right ) \right ) ^{2}} \left ( \sqrt{5}\arctan \left ({\frac{ \left ( 3\, \left ( \sin \left ( x \right ) \right ) ^{2}-5 \right ) \sqrt{5}}{5}{\frac{1}{\sqrt{- \left ( \cos \left ( x \right ) \right ) ^{4}-4\, \left ( \cos \left ( x \right ) \right ) ^{2}}}}} \right ) +10\,\arcsin \left ( 1+1/2\, \left ( \cos \left ( x \right ) \right ) ^{2} \right ) \right ){\frac{1}{\sqrt{-5+ \left ( \sin \left ( x \right ) \right ) ^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-2*cos(x)^3*(-1+sin(x))+cos(2*x)*sin(x))/sin(x)^2/(-5+sin(x)^2)^(1/2),x)

[Out]

-2*ln(sin(x)+(-5+sin(x)^2)^(1/2))+2*(-5+sin(x)^2)^(1/2)+2/5*5^(1/2)*arctan(5^(1/2)/(-5+sin(x)^2)^(1/2))+2/5*(-
5+sin(x)^2)^(1/2)/sin(x)+1/10*((-5+sin(x)^2)*cos(x)^2)^(1/2)*(5^(1/2)*arctan(1/5*(3*sin(x)^2-5)*5^(1/2)/(-cos(
x)^4-4*cos(x)^2)^(1/2))+10*arcsin(1+1/2*cos(x)^2))/cos(x)/(-5+sin(x)^2)^(1/2)

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Maxima [C]  time = 1.51108, size = 155, normalized size = 1.4 \begin{align*} \frac{2}{5} \, \sqrt{5} \arcsin \left (\frac{\sqrt{5}}{{\left | \sin \left (x\right ) \right |}}\right ) - \frac{1}{10} i \, \sqrt{5} \operatorname{arsinh}\left (\frac{\cos \left (x\right )}{2 \,{\left (\cos \left (x\right ) + 1\right )}} - \frac{2}{\cos \left (x\right ) + 1}\right ) - \frac{1}{10} i \, \sqrt{5} \operatorname{arsinh}\left (-\frac{\cos \left (x\right )}{2 \,{\left (\cos \left (x\right ) - 1\right )}} - \frac{2}{\cos \left (x\right ) - 1}\right ) + 2 \, \sqrt{\sin \left (x\right )^{2} - 5} + \frac{2 \, \sqrt{\sin \left (x\right )^{2} - 5}}{5 \, \sin \left (x\right )} - 2 i \, \operatorname{arsinh}\left (\frac{1}{2} \, \cos \left (x\right )\right ) - 2 \, \log \left (2 \, \sqrt{\sin \left (x\right )^{2} - 5} + 2 \, \sin \left (x\right )\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*cos(x)^3*(-1+sin(x))+cos(2*x)*sin(x))/sin(x)^2/(-5+sin(x)^2)^(1/2),x, algorithm="maxima")

[Out]

2/5*sqrt(5)*arcsin(sqrt(5)/abs(sin(x))) - 1/10*I*sqrt(5)*arcsinh(1/2*cos(x)/(cos(x) + 1) - 2/(cos(x) + 1)) - 1
/10*I*sqrt(5)*arcsinh(-1/2*cos(x)/(cos(x) - 1) - 2/(cos(x) - 1)) + 2*sqrt(sin(x)^2 - 5) + 2/5*sqrt(sin(x)^2 -
5)/sin(x) - 2*I*arcsinh(1/2*cos(x)) - 2*log(2*sqrt(sin(x)^2 - 5) + 2*sin(x))

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Fricas [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: UnboundLocalError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*cos(x)^3*(-1+sin(x))+cos(2*x)*sin(x))/sin(x)^2/(-5+sin(x)^2)^(1/2),x, algorithm="fricas")

[Out]

Exception raised: UnboundLocalError

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*cos(x)**3*(-1+sin(x))+cos(2*x)*sin(x))/sin(x)**2/(-5+sin(x)**2)**(1/2),x)

[Out]

Timed out

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Giac [C]  time = 1.78554, size = 356, normalized size = 3.21 \begin{align*} \pi \mathrm{sgn}\left (-2 i \, \sqrt{\cos \left (x\right )^{2} + 4} - 4 i\right ) \mathrm{sgn}\left (\cos \left (x\right )\right ) - \frac{1}{5} \, \sqrt{5}{\left (\pi \mathrm{sgn}\left (-2 i \, \sqrt{\cos \left (x\right )^{2} + 4} - 4 i\right ) \mathrm{sgn}\left (\cos \left (x\right )\right ) + 2 \, \arctan \left (\frac{\sqrt{5}{\left (\frac{{\left (i \, \sqrt{\cos \left (x\right )^{2} + 4} + 2 i\right )}^{2}}{\cos \left (x\right )^{2}} - 1\right )} \cos \left (x\right )}{5 \,{\left (-2 i \, \sqrt{\cos \left (x\right )^{2} + 4} - 4 i\right )}}\right )\right )} + \frac{1}{10} \, \sqrt{5}{\left (\pi \mathrm{sgn}\left (-2 i \, \sqrt{\cos \left (x\right )^{2} + 4} - 4 i\right ) \mathrm{sgn}\left (\cos \left (x\right )\right ) + 2 \, \arctan \left (\frac{\sqrt{5}{\left (\frac{{\left (-i \, \sqrt{\cos \left (x\right )^{2} + 4} - 2 i\right )}^{2}}{\cos \left (x\right )^{2}} - 1\right )} \cos \left (x\right )}{5 \,{\left (-2 i \, \sqrt{\cos \left (x\right )^{2} + 4} - 4 i\right )}}\right )\right )} - \frac{2}{5} \, \sqrt{5} \arctan \left (\frac{1}{5} \, \sqrt{5} \sqrt{\sin \left (x\right )^{2} - 5}\right ) + 2 \, \sqrt{\sin \left (x\right )^{2} - 5} + \frac{4}{{\left (\sqrt{\sin \left (x\right )^{2} - 5} - \sin \left (x\right )\right )}^{2} + 5} + 2 \, \arctan \left (\frac{{\left (\frac{{\left (i \, \sqrt{\cos \left (x\right )^{2} + 4} + 2 i\right )}^{2}}{\cos \left (x\right )^{2}} - 1\right )} \cos \left (x\right )}{-2 i \, \sqrt{\cos \left (x\right )^{2} + 4} - 4 i}\right ) + \log \left ({\left (\sqrt{\sin \left (x\right )^{2} - 5} - \sin \left (x\right )\right )}^{2}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*cos(x)^3*(-1+sin(x))+cos(2*x)*sin(x))/sin(x)^2/(-5+sin(x)^2)^(1/2),x, algorithm="giac")

[Out]

pi*sgn(-2*I*sqrt(cos(x)^2 + 4) - 4*I)*sgn(cos(x)) - 1/5*sqrt(5)*(pi*sgn(-2*I*sqrt(cos(x)^2 + 4) - 4*I)*sgn(cos
(x)) + 2*arctan(1/5*sqrt(5)*((I*sqrt(cos(x)^2 + 4) + 2*I)^2/cos(x)^2 - 1)*cos(x)/(-2*I*sqrt(cos(x)^2 + 4) - 4*
I))) + 1/10*sqrt(5)*(pi*sgn(-2*I*sqrt(cos(x)^2 + 4) - 4*I)*sgn(cos(x)) + 2*arctan(1/5*sqrt(5)*((-I*sqrt(cos(x)
^2 + 4) - 2*I)^2/cos(x)^2 - 1)*cos(x)/(-2*I*sqrt(cos(x)^2 + 4) - 4*I))) - 2/5*sqrt(5)*arctan(1/5*sqrt(5)*sqrt(
sin(x)^2 - 5)) + 2*sqrt(sin(x)^2 - 5) + 4/((sqrt(sin(x)^2 - 5) - sin(x))^2 + 5) + 2*arctan(((I*sqrt(cos(x)^2 +
4) + 2*I)^2/cos(x)^2 - 1)*cos(x)/(-2*I*sqrt(cos(x)^2 + 4) - 4*I)) + log((sqrt(sin(x)^2 - 5) - sin(x))^2)