3.436 \(\int \frac {-2 \cot ^2(x)+\sin (x)}{(1+5 \tan ^2(x))^{3/2}} \, dx\)

Optimal. Leaf size=94 \[ -\frac {1}{8} \cos (x) \sqrt {5 \tan ^2(x)+1}-\frac {\cos (x)}{4 \sqrt {5 \tan ^2(x)+1}}-\frac {1}{4} \tanh ^{-1}\left (\frac {2 \tan (x)}{\sqrt {5 \tan ^2(x)+1}}\right )+\frac {9}{2} \sqrt {5 \tan ^2(x)+1} \cot (x)-\frac {5 \cot (x)}{2 \sqrt {5 \tan ^2(x)+1}} \]

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Rubi [A]  time = 0.19, antiderivative size = 94, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 10, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.454, Rules used = {4377, 12, 3670, 472, 583, 377, 206, 3664, 271, 191} \[ -\frac {5 \sec (x)}{8 \sqrt {5 \sec ^2(x)-4}}+\frac {\cos (x)}{4 \sqrt {5 \sec ^2(x)-4}}-\frac {1}{4} \tanh ^{-1}\left (\frac {2 \tan (x)}{\sqrt {5 \tan ^2(x)+1}}\right )+\frac {9}{2} \sqrt {5 \tan ^2(x)+1} \cot (x)-\frac {5 \cot (x)}{2 \sqrt {5 \tan ^2(x)+1}} \]

Antiderivative was successfully verified.

[In]

Int[(-2*Cot[x]^2 + Sin[x])/(1 + 5*Tan[x]^2)^(3/2),x]

[Out]

-ArcTanh[(2*Tan[x])/Sqrt[1 + 5*Tan[x]^2]]/4 + Cos[x]/(4*Sqrt[-4 + 5*Sec[x]^2]) - (5*Sec[x])/(8*Sqrt[-4 + 5*Sec
[x]^2]) - (5*Cot[x])/(2*Sqrt[1 + 5*Tan[x]^2]) + (9*Cot[x]*Sqrt[1 + 5*Tan[x]^2])/2

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 191

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^(p + 1))/a, x] /; FreeQ[{a, b, n, p}, x] &
& EqQ[1/n + p + 1, 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 271

Int[(x_)^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x^(m + 1)*(a + b*x^n)^(p + 1))/(a*(m + 1)), x]
- Dist[(b*(m + n*(p + 1) + 1))/(a*(m + 1)), Int[x^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, m, n, p}, x]
&& ILtQ[Simplify[(m + 1)/n + p + 1], 0] && NeQ[m, -1]

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 472

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> -Simp[(b*(e*x
)^(m + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(a*e*n*(b*c - a*d)*(p + 1)), x] + Dist[1/(a*n*(b*c - a*d)*(
p + 1)), Int[(e*x)^m*(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[c*b*(m + 1) + n*(b*c - a*d)*(p + 1) + d*b*(m + n*(
p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, m, q}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[p
, -1] && IntBinomialQ[a, b, c, d, e, m, n, p, q, x]

Rule 583

Int[((g_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)),
x_Symbol] :> Simp[(e*(g*x)^(m + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(a*c*g*(m + 1)), x] + Dist[1/(a*c*
g^n*(m + 1)), Int[(g*x)^(m + n)*(a + b*x^n)^p*(c + d*x^n)^q*Simp[a*f*c*(m + 1) - e*(b*c + a*d)*(m + n + 1) - e
*n*(b*c*p + a*d*q) - b*e*d*(m + n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p, q}, x] &&
IGtQ[n, 0] && LtQ[m, -1]

Rule 3664

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]^2)^(p_.), x_Symbol] :> With[{ff = Free
Factors[Sec[e + f*x], x]}, Dist[1/(f*ff^m), Subst[Int[((-1 + ff^2*x^2)^((m - 1)/2)*(a - b + b*ff^2*x^2)^p)/x^(
m + 1), x], x, Sec[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[(m - 1)/2]

Rule 3670

Int[((d_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_.), x_Symbol]
:> With[{ff = FreeFactors[Tan[e + f*x], x]}, Dist[(c*ff)/f, Subst[Int[(((d*ff*x)/c)^m*(a + b*(ff*x)^n)^p)/(c^
2 + ff^2*x^2), x], x, (c*Tan[e + f*x])/ff], x]] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && (IGtQ[p, 0] || EqQ
[n, 2] || EqQ[n, 4] || (IntegerQ[p] && RationalQ[n]))

Rule 4377

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

Rubi steps

\begin {align*} \int \frac {-2 \cot ^2(x)+\sin (x)}{\left (1+5 \tan ^2(x)\right )^{3/2}} \, dx &=\int -\frac {2 \cot ^2(x)}{\left (1+5 \tan ^2(x)\right )^{3/2}} \, dx+\int \frac {\sin (x)}{\left (1+5 \tan ^2(x)\right )^{3/2}} \, dx\\ &=-\left (2 \int \frac {\cot ^2(x)}{\left (1+5 \tan ^2(x)\right )^{3/2}} \, dx\right )+\operatorname {Subst}\left (\int \frac {1}{x^2 \left (-4+5 x^2\right )^{3/2}} \, dx,x,\sec (x)\right )\\ &=\frac {\cos (x)}{4 \sqrt {-4+5 \sec ^2(x)}}-2 \operatorname {Subst}\left (\int \frac {1}{x^2 \left (1+x^2\right ) \left (1+5 x^2\right )^{3/2}} \, dx,x,\tan (x)\right )+\frac {5}{2} \operatorname {Subst}\left (\int \frac {1}{\left (-4+5 x^2\right )^{3/2}} \, dx,x,\sec (x)\right )\\ &=\frac {\cos (x)}{4 \sqrt {-4+5 \sec ^2(x)}}-\frac {5 \sec (x)}{8 \sqrt {-4+5 \sec ^2(x)}}-\frac {5 \cot (x)}{2 \sqrt {1+5 \tan ^2(x)}}+\frac {1}{2} \operatorname {Subst}\left (\int \frac {-9-10 x^2}{x^2 \left (1+x^2\right ) \sqrt {1+5 x^2}} \, dx,x,\tan (x)\right )\\ &=\frac {\cos (x)}{4 \sqrt {-4+5 \sec ^2(x)}}-\frac {5 \sec (x)}{8 \sqrt {-4+5 \sec ^2(x)}}-\frac {5 \cot (x)}{2 \sqrt {1+5 \tan ^2(x)}}+\frac {9}{2} \cot (x) \sqrt {1+5 \tan ^2(x)}-\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{\left (1+x^2\right ) \sqrt {1+5 x^2}} \, dx,x,\tan (x)\right )\\ &=\frac {\cos (x)}{4 \sqrt {-4+5 \sec ^2(x)}}-\frac {5 \sec (x)}{8 \sqrt {-4+5 \sec ^2(x)}}-\frac {5 \cot (x)}{2 \sqrt {1+5 \tan ^2(x)}}+\frac {9}{2} \cot (x) \sqrt {1+5 \tan ^2(x)}-\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{1-4 x^2} \, dx,x,\frac {\tan (x)}{\sqrt {1+5 \tan ^2(x)}}\right )\\ &=-\frac {1}{4} \tanh ^{-1}\left (\frac {2 \tan (x)}{\sqrt {1+5 \tan ^2(x)}}\right )+\frac {\cos (x)}{4 \sqrt {-4+5 \sec ^2(x)}}-\frac {5 \sec (x)}{8 \sqrt {-4+5 \sec ^2(x)}}-\frac {5 \cot (x)}{2 \sqrt {1+5 \tan ^2(x)}}+\frac {9}{2} \cot (x) \sqrt {1+5 \tan ^2(x)}\\ \end {align*}

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Mathematica [A]  time = 0.52, size = 131, normalized size = 1.39 \[ -\frac {\sin ^2(x) (2 \cos (2 x)-3)^{3/2} \tan (x) \left (2 \cot ^2(x) \csc (x)-1\right ) \left (\sqrt {4 \sin ^2(x)+1} \left (16 \csc ^3(x)-3 \csc ^2(x)+164 \csc (x)-2\right )-2 \left (\csc ^2(x)+4\right ) \sinh ^{-1}(2 \sin (x))\right )}{2 \sqrt {-(3-2 \cos (2 x))^2} \sqrt {5 \tan ^2(x)+1} \left (\cot ^2(x)+5\right ) (-3 \sin (x)+\sin (3 x)+4 \cos (2 x)+4)} \]

Antiderivative was successfully verified.

[In]

Integrate[(-2*Cot[x]^2 + Sin[x])/(1 + 5*Tan[x]^2)^(3/2),x]

[Out]

-1/2*((-3 + 2*Cos[2*x])^(3/2)*(-1 + 2*Cot[x]^2*Csc[x])*Sin[x]^2*(-2*ArcSinh[2*Sin[x]]*(4 + Csc[x]^2) + (-2 + 1
64*Csc[x] - 3*Csc[x]^2 + 16*Csc[x]^3)*Sqrt[1 + 4*Sin[x]^2])*Tan[x])/(Sqrt[-(3 - 2*Cos[2*x])^2]*(5 + Cot[x]^2)*
(4 + 4*Cos[2*x] - 3*Sin[x] + Sin[3*x])*Sqrt[1 + 5*Tan[x]^2])

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {-2 \cot ^2(x)+\sin (x)}{\left (1+5 \tan ^2(x)\right )^{3/2}} \, dx \]

Verification is Not applicable to the result.

[In]

IntegrateAlgebraic[(-2*Cot[x]^2 + Sin[x])/(1 + 5*Tan[x]^2)^(3/2),x]

[Out]

Could not integrate

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fricas [A]  time = 1.12, size = 97, normalized size = 1.03 \[ \frac {2 \, {\left (4 \, \cos \relax (x)^{2} - 5\right )} \log \left (\sqrt {-\frac {4 \, \cos \relax (x)^{2} - 5}{\cos \relax (x)^{2}}} \cos \relax (x) - 2 \, \sin \relax (x)\right ) \sin \relax (x) + {\left (164 \, \cos \relax (x)^{3} - {\left (2 \, \cos \relax (x)^{3} - 5 \, \cos \relax (x)\right )} \sin \relax (x) - 180 \, \cos \relax (x)\right )} \sqrt {-\frac {4 \, \cos \relax (x)^{2} - 5}{\cos \relax (x)^{2}}}}{8 \, {\left (4 \, \cos \relax (x)^{2} - 5\right )} \sin \relax (x)} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(2*(4*cos(x)^2 - 5)*log(sqrt(-(4*cos(x)^2 - 5)/cos(x)^2)*cos(x) - 2*sin(x))*sin(x) + (164*cos(x)^3 - (2*co
s(x)^3 - 5*cos(x))*sin(x) - 180*cos(x))*sqrt(-(4*cos(x)^2 - 5)/cos(x)^2))/((4*cos(x)^2 - 5)*sin(x))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int -\frac {2 \, \cot \relax (x)^{2} - \sin \relax (x)}{{\left (5 \, \tan \relax (x)^{2} + 1\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(-(2*cot(x)^2 - sin(x))/(5*tan(x)^2 + 1)^(3/2), x)

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maple [C]  time = 1.14, size = 975, normalized size = 10.37

method	result	size
default	\(\frac {i \left (6 i \cos \relax (x ) \sin \relax (x ) \sqrt {-\frac {4 \left (\cos ^{2}\relax (x )\right )-5}{\left (1+\cos \relax (x )\right )^{2}}}\, \arctanh \left (\frac {\sqrt {-16}\, \cos \relax (x ) \left (-1+\cos \relax (x )\right )}{2 \sin \relax (x )^{2} \sqrt {-\frac {4 \left (\cos ^{2}\relax (x )\right )-5}{\left (1+\cos \relax (x )\right )^{2}}}}\right )+4 i \cos \relax (x ) \sin \relax (x ) \sqrt {2}\, \sqrt {\frac {2 \cos \relax (x ) \sqrt {5}-4 \cos \relax (x )-2 \sqrt {5}+5}{1+\cos \relax (x )}}\, \sqrt {-\frac {2 \left (2 \cos \relax (x ) \sqrt {5}+4 \cos \relax (x )-2 \sqrt {5}-5\right )}{1+\cos \relax (x )}}\, \EllipticF \left (\frac {i \left (-1+\cos \relax (x )\right ) \left (-2+\sqrt {5}\right )}{\sin \relax (x )}, 9+4 \sqrt {5}\right )-3 i \sin \relax (x ) \arctanh \left (\frac {\sqrt {-16}\, \cos \relax (x ) \left (-1+\cos \relax (x )\right )}{2 \sin \relax (x )^{2} \sqrt {-\frac {4 \left (\cos ^{2}\relax (x )\right )-5}{\left (1+\cos \relax (x )\right )^{2}}}}\right ) \sqrt {-\frac {4 \left (\cos ^{2}\relax (x )\right )-5}{\left (1+\cos \relax (x )\right )^{2}}}\, \sqrt {5}-3 i \cos \relax (x ) \sqrt {5}\, \sqrt {-\frac {4 \left (\cos ^{2}\relax (x )\right )-5}{\left (1+\cos \relax (x )\right )^{2}}}\, \sin \relax (x ) \arctanh \left (\frac {\sqrt {-16}\, \cos \relax (x ) \left (-1+\cos \relax (x )\right )}{2 \sin \relax (x )^{2} \sqrt {-\frac {4 \left (\cos ^{2}\relax (x )\right )-5}{\left (1+\cos \relax (x )\right )^{2}}}}\right )-8 i \cos \relax (x ) \sqrt {2}\, \sqrt {\frac {2 \cos \relax (x ) \sqrt {5}-4 \cos \relax (x )-2 \sqrt {5}+5}{1+\cos \relax (x )}}\, \sqrt {-\frac {2 \left (2 \cos \relax (x ) \sqrt {5}+4 \cos \relax (x )-2 \sqrt {5}-5\right )}{1+\cos \relax (x )}}\, \sin \relax (x ) \EllipticPi \left (\frac {\sqrt {-9+4 \sqrt {5}}\, \left (-1+\cos \relax (x )\right )}{\sin \relax (x )}, -\frac {1}{-9+4 \sqrt {5}}, \frac {\sqrt {-9-4 \sqrt {5}}}{\sqrt {-9+4 \sqrt {5}}}\right )-3 \cos \relax (x ) \sqrt {5}\, \sqrt {-\frac {4 \left (\cos ^{2}\relax (x )\right )-5}{\left (1+\cos \relax (x )\right )^{2}}}\, \sin \relax (x ) \arctan \left (\frac {2 \cos \relax (x ) \left (-1+\cos \relax (x )\right )}{\sin \relax (x )^{2} \sqrt {-\frac {4 \left (\cos ^{2}\relax (x )\right )-5}{\left (1+\cos \relax (x )\right )^{2}}}}\right )-8 i \sqrt {2}\, \sqrt {\frac {2 \cos \relax (x ) \sqrt {5}-4 \cos \relax (x )-2 \sqrt {5}+5}{1+\cos \relax (x )}}\, \sqrt {-\frac {2 \left (2 \cos \relax (x ) \sqrt {5}+4 \cos \relax (x )-2 \sqrt {5}-5\right )}{1+\cos \relax (x )}}\, \sin \relax (x ) \EllipticPi \left (\frac {\sqrt {-9+4 \sqrt {5}}\, \left (-1+\cos \relax (x )\right )}{\sin \relax (x )}, -\frac {1}{-9+4 \sqrt {5}}, \frac {\sqrt {-9-4 \sqrt {5}}}{\sqrt {-9+4 \sqrt {5}}}\right )+4 i \sqrt {2}\, \sqrt {\frac {2 \cos \relax (x ) \sqrt {5}-4 \cos \relax (x )-2 \sqrt {5}+5}{1+\cos \relax (x )}}\, \sqrt {-\frac {2 \left (2 \cos \relax (x ) \sqrt {5}+4 \cos \relax (x )-2 \sqrt {5}-5\right )}{1+\cos \relax (x )}}\, \sin \relax (x ) \EllipticF \left (\frac {i \left (-1+\cos \relax (x )\right ) \left (-2+\sqrt {5}\right )}{\sin \relax (x )}, 9+4 \sqrt {5}\right )+2 \left (\cos ^{2}\relax (x )\right ) \sin \relax (x ) \sqrt {5}+6 \cos \relax (x ) \sin \relax (x ) \sqrt {-\frac {4 \left (\cos ^{2}\relax (x )\right )-5}{\left (1+\cos \relax (x )\right )^{2}}}\, \arctan \left (\frac {2 \cos \relax (x ) \left (-1+\cos \relax (x )\right )}{\sin \relax (x )^{2} \sqrt {-\frac {4 \left (\cos ^{2}\relax (x )\right )-5}{\left (1+\cos \relax (x )\right )^{2}}}}\right )-3 \sin \relax (x ) \arctan \left (\frac {2 \cos \relax (x ) \left (-1+\cos \relax (x )\right )}{\sin \relax (x )^{2} \sqrt {-\frac {4 \left (\cos ^{2}\relax (x )\right )-5}{\left (1+\cos \relax (x )\right )^{2}}}}\right ) \sqrt {-\frac {4 \left (\cos ^{2}\relax (x )\right )-5}{\left (1+\cos \relax (x )\right )^{2}}}\, \sqrt {5}+6 i \sqrt {-\frac {4 \left (\cos ^{2}\relax (x )\right )-5}{\left (1+\cos \relax (x )\right )^{2}}}\, \sin \relax (x ) \arctanh \left (\frac {\sqrt {-16}\, \cos \relax (x ) \left (-1+\cos \relax (x )\right )}{2 \sin \relax (x )^{2} \sqrt {-\frac {4 \left (\cos ^{2}\relax (x )\right )-5}{\left (1+\cos \relax (x )\right )^{2}}}}\right )-4 \left (\cos ^{2}\relax (x )\right ) \sin \relax (x )-164 \left (\cos ^{2}\relax (x )\right ) \sqrt {5}+6 \arctan \left (\frac {2 \cos \relax (x ) \left (-1+\cos \relax (x )\right )}{\sin \relax (x )^{2} \sqrt {-\frac {4 \left (\cos ^{2}\relax (x )\right )-5}{\left (1+\cos \relax (x )\right )^{2}}}}\right ) \sqrt {-\frac {4 \left (\cos ^{2}\relax (x )\right )-5}{\left (1+\cos \relax (x )\right )^{2}}}\, \sin \relax (x )+328 \left (\cos ^{2}\relax (x )\right )-5 \sin \relax (x ) \sqrt {5}+10 \sin \relax (x )+180 \sqrt {5}-360\right ) \left (\cos ^{3}\relax (x )\right ) \left (-\frac {4 \left (\cos ^{2}\relax (x )\right )-5}{\cos \relax (x )^{2}}\right )^{\frac {3}{2}}}{8 \sqrt {-9+4 \sqrt {5}}\, \left (\sqrt {5}+2\right )^{2} \left (-2+\sqrt {5}\right )^{2} \left (4 \left (\cos ^{2}\relax (x )\right )-5\right )^{2} \sin \relax (x )}\)	\(975\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-2*cot(x)^2+sin(x))/(1+5*tan(x)^2)^(3/2),x,method=_RETURNVERBOSE)

[Out]

1/8*I/(-9+4*5^(1/2))^(1/2)/(5^(1/2)+2)^2/(-2+5^(1/2))^2/(4*cos(x)^2-5)^2*(-8*I*2^(1/2)*((2*cos(x)*5^(1/2)-4*co
s(x)-2*5^(1/2)+5)/(1+cos(x)))^(1/2)*(-2*(2*cos(x)*5^(1/2)+4*cos(x)-2*5^(1/2)-5)/(1+cos(x)))^(1/2)*sin(x)*Ellip
ticPi((-9+4*5^(1/2))^(1/2)*(-1+cos(x))/sin(x),-1/(-9+4*5^(1/2)),(-9-4*5^(1/2))^(1/2)/(-9+4*5^(1/2))^(1/2))-3*I
*sin(x)*arctanh(1/2*(-16)^(1/2)*cos(x)*(-1+cos(x))/sin(x)^2/(-(4*cos(x)^2-5)/(1+cos(x))^2)^(1/2))*(-(4*cos(x)^
2-5)/(1+cos(x))^2)^(1/2)*5^(1/2)+4*I*cos(x)*sin(x)*2^(1/2)*((2*cos(x)*5^(1/2)-4*cos(x)-2*5^(1/2)+5)/(1+cos(x))
)^(1/2)*(-2*(2*cos(x)*5^(1/2)+4*cos(x)-2*5^(1/2)-5)/(1+cos(x)))^(1/2)*EllipticF(I*(-1+cos(x))*(-2+5^(1/2))/sin
(x),9+4*5^(1/2))-8*I*cos(x)*2^(1/2)*((2*cos(x)*5^(1/2)-4*cos(x)-2*5^(1/2)+5)/(1+cos(x)))^(1/2)*(-2*(2*cos(x)*5
^(1/2)+4*cos(x)-2*5^(1/2)-5)/(1+cos(x)))^(1/2)*sin(x)*EllipticPi((-9+4*5^(1/2))^(1/2)*(-1+cos(x))/sin(x),-1/(-
9+4*5^(1/2)),(-9-4*5^(1/2))^(1/2)/(-9+4*5^(1/2))^(1/2))+6*I*(-(4*cos(x)^2-5)/(1+cos(x))^2)^(1/2)*sin(x)*arctan
h(1/2*(-16)^(1/2)*cos(x)*(-1+cos(x))/sin(x)^2/(-(4*cos(x)^2-5)/(1+cos(x))^2)^(1/2))-3*cos(x)*5^(1/2)*(-(4*cos(
x)^2-5)/(1+cos(x))^2)^(1/2)*sin(x)*arctan(2*cos(x)*(-1+cos(x))/sin(x)^2/(-(4*cos(x)^2-5)/(1+cos(x))^2)^(1/2))-
3*I*cos(x)*5^(1/2)*(-(4*cos(x)^2-5)/(1+cos(x))^2)^(1/2)*sin(x)*arctanh(1/2*(-16)^(1/2)*cos(x)*(-1+cos(x))/sin(
x)^2/(-(4*cos(x)^2-5)/(1+cos(x))^2)^(1/2))+6*I*cos(x)*sin(x)*(-(4*cos(x)^2-5)/(1+cos(x))^2)^(1/2)*arctanh(1/2*
(-16)^(1/2)*cos(x)*(-1+cos(x))/sin(x)^2/(-(4*cos(x)^2-5)/(1+cos(x))^2)^(1/2))+2*cos(x)^2*sin(x)*5^(1/2)+6*cos(
x)*sin(x)*(-(4*cos(x)^2-5)/(1+cos(x))^2)^(1/2)*arctan(2*cos(x)*(-1+cos(x))/sin(x)^2/(-(4*cos(x)^2-5)/(1+cos(x)
)^2)^(1/2))-3*sin(x)*arctan(2*cos(x)*(-1+cos(x))/sin(x)^2/(-(4*cos(x)^2-5)/(1+cos(x))^2)^(1/2))*(-(4*cos(x)^2-
5)/(1+cos(x))^2)^(1/2)*5^(1/2)+4*I*2^(1/2)*((2*cos(x)*5^(1/2)-4*cos(x)-2*5^(1/2)+5)/(1+cos(x)))^(1/2)*(-2*(2*c
os(x)*5^(1/2)+4*cos(x)-2*5^(1/2)-5)/(1+cos(x)))^(1/2)*sin(x)*EllipticF(I*(-1+cos(x))*(-2+5^(1/2))/sin(x),9+4*5
^(1/2))-4*cos(x)^2*sin(x)-164*cos(x)^2*5^(1/2)+6*arctan(2*cos(x)*(-1+cos(x))/sin(x)^2/(-(4*cos(x)^2-5)/(1+cos(
x))^2)^(1/2))*(-(4*cos(x)^2-5)/(1+cos(x))^2)^(1/2)*sin(x)+328*cos(x)^2-5*sin(x)*5^(1/2)+10*sin(x)+180*5^(1/2)-
360)*cos(x)^3*(-(4*cos(x)^2-5)/cos(x)^2)^(3/2)/sin(x)

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

[Out]

Timed out

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\sin \relax (x)-2\,{\mathrm {cot}\relax (x)}^2}{{\left (5\,{\mathrm {tan}\relax (x)}^2+1\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - \int \left (- \frac {\sin {\relax (x )}}{5 \sqrt {5 \tan ^{2}{\relax (x )} + 1} \tan ^{2}{\relax (x )} + \sqrt {5 \tan ^{2}{\relax (x )} + 1}}\right )\, dx - \int \frac {2 \cot ^{2}{\relax (x )}}{5 \sqrt {5 \tan ^{2}{\relax (x )} + 1} \tan ^{2}{\relax (x )} + \sqrt {5 \tan ^{2}{\relax (x )} + 1}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-Integral(-sin(x)/(5*sqrt(5*tan(x)**2 + 1)*tan(x)**2 + sqrt(5*tan(x)**2 + 1)), x) - Integral(2*cot(x)**2/(5*sq
rt(5*tan(x)**2 + 1)*tan(x)**2 + sqrt(5*tan(x)**2 + 1)), x)

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