\(\int \frac {-2 \cot ^2(x)+\sin (x)}{(1+5 \tan ^2(x))^{3/2}} \, dx\) [436]

Optimal result

Integrand size = 22, antiderivative size = 94 \[ \int \frac {-2 \cot ^2(x)+\sin (x)}{\left (1+5 \tan ^2(x)\right )^{3/2}} \, dx=-\frac {1}{4} \text {arctanh}\left (\frac {2 \tan (x)}{\sqrt {1+5 \tan ^2(x)}}\right )-\frac {\cos (x)}{4 \sqrt {1+5 \tan ^2(x)}}-\frac {5 \cot (x)}{2 \sqrt {1+5 \tan ^2(x)}}-\frac {1}{8} \cos (x) \sqrt {1+5 \tan ^2(x)}+\frac {9}{2} \cot (x) \sqrt {1+5 \tan ^2(x)} \]

[Out]

-1/4*arctanh(2*tan(x)/(1+5*tan(x)^2)^(1/2))-1/4*cos(x)/(1+5*tan(x)^2)^(1/2)-5/2*cot(x)/(1+5*tan(x)^2)^(1/2)-1/
8*cos(x)*(1+5*tan(x)^2)^(1/2)+9/2*cot(x)*(1+5*tan(x)^2)^(1/2)

Rubi [A] (verified)

Time = 0.14 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.455, Rules used = {4462, 12, 3751, 483, 597, 385, 212, 3745, 277, 197} \[ \int \frac {-2 \cot ^2(x)+\sin (x)}{\left (1+5 \tan ^2(x)\right )^{3/2}} \, dx=-\frac {1}{4} \text {arctanh}\left (\frac {2 \tan (x)}{\sqrt {5 \tan ^2(x)+1}}\right )-\frac {5 \sec (x)}{8 \sqrt {5 \sec ^2(x)-4}}+\frac {\cos (x)}{4 \sqrt {5 \sec ^2(x)-4}}+\frac {9}{2} \sqrt {5 \tan ^2(x)+1} \cot (x)-\frac {5 \cot (x)}{2 \sqrt {5 \tan ^2(x)+1}} \]

[In]

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

[Out]

-1/4*ArcTanh[(2*Tan[x])/Sqrt[1 + 5*Tan[x]^2]] + Cos[x]/(4*Sqrt[-4 + 5*Sec[x]^2]) - (5*Sec[x])/(8*Sqrt[-4 + 5*S
ec[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 197

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 212

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

Rule 277

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 385

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 483

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 597

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 3745

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 3751

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 4462

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*} \text {integral}& = \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 )+\text {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 \text {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} \text {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} \text {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} \text {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} \text {Subst}\left (\int \frac {1}{1-4 x^2} \, dx,x,\frac {\tan (x)}{\sqrt {1+5 \tan ^2(x)}}\right ) \\ & = -\frac {1}{4} \text {arctanh}\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*}

Mathematica [A] (verified)

Time = 0.60 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.39 \[ \int \frac {-2 \cot ^2(x)+\sin (x)}{\left (1+5 \tan ^2(x)\right )^{3/2}} \, dx=-\frac {(-3+2 \cos (2 x))^{3/2} \left (-1+2 \cot ^2(x) \csc (x)\right ) \sin ^2(x) \left (-2 \text {arcsinh}(2 \sin (x)) \left (4+\csc ^2(x)\right )+\left (-2+164 \csc (x)-3 \csc ^2(x)+16 \csc ^3(x)\right ) \sqrt {1+4 \sin ^2(x)}\right ) \tan (x)}{2 \sqrt {-(3-2 \cos (2 x))^2} \left (5+\cot ^2(x)\right ) (4+4 \cos (2 x)-3 \sin (x)+\sin (3 x)) \sqrt {1+5 \tan ^2(x)}} \]

[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])

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(150\) vs. \(2(74)=148\).

Time = 4.21 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.61

method	result	size
default	\(-\frac {\left (\sec ^{3}\left (x \right )\right ) \csc \left (x \right ) \left (4 \left (\cos ^{2}\left (x \right )\right )-5\right ) \left (-2 \cos \left (x \right ) \sin \left (x \right ) \operatorname {arctanh}\left (\frac {2 \sin \left (x \right )}{\left (\cos \left (x \right )+1\right ) \sqrt {-\frac {4 \left (\cos ^{2}\left (x \right )\right )-5}{\left (\cos \left (x \right )+1\right )^{2}}}}\right ) \sqrt {-\frac {4 \left (\cos ^{2}\left (x \right )\right )-5}{\left (\cos \left (x \right )+1\right )^{2}}}+2 \left (\cos ^{2}\left (x \right )\right ) \sin \left (x \right )-2 \sin \left (x \right ) \operatorname {arctanh}\left (\frac {2 \sin \left (x \right )}{\left (\cos \left (x \right )+1\right ) \sqrt {-\frac {4 \left (\cos ^{2}\left (x \right )\right )-5}{\left (\cos \left (x \right )+1\right )^{2}}}}\right ) \sqrt {-\frac {4 \left (\cos ^{2}\left (x \right )\right )-5}{\left (\cos \left (x \right )+1\right )^{2}}}-164 \left (\cos ^{2}\left (x \right )\right )-5 \sin \left (x \right )+180\right )}{8 {\left (5 \left (\sec ^{2}\left (x \right )\right )-4\right )}^{\frac {3}{2}}}\)	\(151\)
parts	\(-\frac {8 \cos \left (x \right )-30 \sec \left (x \right )+25 \left (\sec ^{3}\left (x \right )\right )}{8 {\left (5 \left (\sec ^{2}\left (x \right )\right )-4\right )}^{\frac {3}{2}} \left (-2+\sqrt {5}\right )^{2} \left (2+\sqrt {5}\right )^{2}}+\frac {\left (\sec ^{3}\left (x \right )\right ) \csc \left (x \right ) \left (4 \left (\cos ^{2}\left (x \right )\right )-5\right ) \left (\cos \left (x \right ) \sin \left (x \right ) \operatorname {arctanh}\left (\frac {2 \sin \left (x \right )}{\left (\cos \left (x \right )+1\right ) \sqrt {-\frac {4 \left (\cos ^{2}\left (x \right )\right )-5}{\left (\cos \left (x \right )+1\right )^{2}}}}\right ) \sqrt {-\frac {4 \left (\cos ^{2}\left (x \right )\right )-5}{\left (\cos \left (x \right )+1\right )^{2}}}+\sin \left (x \right ) \operatorname {arctanh}\left (\frac {2 \sin \left (x \right )}{\left (\cos \left (x \right )+1\right ) \sqrt {-\frac {4 \left (\cos ^{2}\left (x \right )\right )-5}{\left (\cos \left (x \right )+1\right )^{2}}}}\right ) \sqrt {-\frac {4 \left (\cos ^{2}\left (x \right )\right )-5}{\left (\cos \left (x \right )+1\right )^{2}}}+82 \left (\cos ^{2}\left (x \right )\right )-90\right )}{4 {\left (5 \left (\sec ^{2}\left (x \right )\right )-4\right )}^{\frac {3}{2}}}\)	\(179\)

[In]

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

[Out]

-1/8*sec(x)^3*csc(x)*(4*cos(x)^2-5)*(-2*cos(x)*sin(x)*arctanh(2*sin(x)/(cos(x)+1)/(-(4*cos(x)^2-5)/(cos(x)+1)^
2)^(1/2))*(-(4*cos(x)^2-5)/(cos(x)+1)^2)^(1/2)+2*cos(x)^2*sin(x)-2*sin(x)*arctanh(2*sin(x)/(cos(x)+1)/(-(4*cos
(x)^2-5)/(cos(x)+1)^2)^(1/2))*(-(4*cos(x)^2-5)/(cos(x)+1)^2)^(1/2)-164*cos(x)^2-5*sin(x)+180)/(5*sec(x)^2-4)^(
3/2)

Fricas [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.03 \[ \int \frac {-2 \cot ^2(x)+\sin (x)}{\left (1+5 \tan ^2(x)\right )^{3/2}} \, dx=\frac {2 \, {\left (4 \, \cos \left (x\right )^{2} - 5\right )} \log \left (\sqrt {-\frac {4 \, \cos \left (x\right )^{2} - 5}{\cos \left (x\right )^{2}}} \cos \left (x\right ) - 2 \, \sin \left (x\right )\right ) \sin \left (x\right ) + {\left (164 \, \cos \left (x\right )^{3} - {\left (2 \, \cos \left (x\right )^{3} - 5 \, \cos \left (x\right )\right )} \sin \left (x\right ) - 180 \, \cos \left (x\right )\right )} \sqrt {-\frac {4 \, \cos \left (x\right )^{2} - 5}{\cos \left (x\right )^{2}}}}{8 \, {\left (4 \, \cos \left (x\right )^{2} - 5\right )} \sin \left (x\right )} \]

[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))

Sympy [F]

\[ \int \frac {-2 \cot ^2(x)+\sin (x)}{\left (1+5 \tan ^2(x)\right )^{3/2}} \, dx=- \int \left (- \frac {\sin {\left (x \right )}}{5 \sqrt {5 \tan ^{2}{\left (x \right )} + 1} \tan ^{2}{\left (x \right )} + \sqrt {5 \tan ^{2}{\left (x \right )} + 1}}\right )\, dx - \int \frac {2 \cot ^{2}{\left (x \right )}}{5 \sqrt {5 \tan ^{2}{\left (x \right )} + 1} \tan ^{2}{\left (x \right )} + \sqrt {5 \tan ^{2}{\left (x \right )} + 1}}\, dx \]

[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)

Maxima [F(-1)]

Timed out. \[ \int \frac {-2 \cot ^2(x)+\sin (x)}{\left (1+5 \tan ^2(x)\right )^{3/2}} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Giac [F]

\[ \int \frac {-2 \cot ^2(x)+\sin (x)}{\left (1+5 \tan ^2(x)\right )^{3/2}} \, dx=\int { -\frac {2 \, \cot \left (x\right )^{2} - \sin \left (x\right )}{{\left (5 \, \tan \left (x\right )^{2} + 1\right )}^{\frac {3}{2}}} \,d x } \]

[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)

Mupad [F(-1)]

Timed out. \[ \int \frac {-2 \cot ^2(x)+\sin (x)}{\left (1+5 \tan ^2(x)\right )^{3/2}} \, dx=\int \frac {\sin \left (x\right )-2\,{\mathrm {cot}\left (x\right )}^2}{{\left (5\,{\mathrm {tan}\left (x\right )}^2+1\right )}^{3/2}} \,d x \]

[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)