3.441 \(\int \frac{\tan (x)}{(1+5 \tan ^2(x))^{5/2}} \, dx\)

Optimal. Leaf size=54 \[ \frac{1}{32} \tan ^{-1}\left (\frac{1}{2} \sqrt{5 \tan ^2(x)+1}\right )+\frac{1}{16 \sqrt{5 \tan ^2(x)+1}}-\frac{1}{12 \left (5 \tan ^2(x)+1\right )^{3/2}} \]

[Out]

ArcTan[Sqrt[1 + 5*Tan[x]^2]/2]/32 - 1/(12*(1 + 5*Tan[x]^2)^(3/2)) + 1/(16*Sqrt[1 + 5*Tan[x]^2])

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Rubi [A]  time = 0.0601066, antiderivative size = 54, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {3670, 444, 51, 63, 203} \[ \frac{1}{32} \tan ^{-1}\left (\frac{1}{2} \sqrt{5 \tan ^2(x)+1}\right )+\frac{1}{16 \sqrt{5 \tan ^2(x)+1}}-\frac{1}{12 \left (5 \tan ^2(x)+1\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[Tan[x]/(1 + 5*Tan[x]^2)^(5/2),x]

[Out]

ArcTan[Sqrt[1 + 5*Tan[x]^2]/2]/32 - 1/(12*(1 + 5*Tan[x]^2)^(3/2)) + 1/(16*Sqrt[1 + 5*Tan[x]^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 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 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]

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]

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])

Rubi steps

\begin{align*} \int \frac{\tan (x)}{\left (1+5 \tan ^2(x)\right )^{5/2}} \, dx &=\operatorname{Subst}\left (\int \frac{x}{\left (1+x^2\right ) \left (1+5 x^2\right )^{5/2}} \, dx,x,\tan (x)\right )\\ &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{1}{(1+x) (1+5 x)^{5/2}} \, dx,x,\tan ^2(x)\right )\\ &=-\frac{1}{12 \left (1+5 \tan ^2(x)\right )^{3/2}}-\frac{1}{8} \operatorname{Subst}\left (\int \frac{1}{(1+x) (1+5 x)^{3/2}} \, dx,x,\tan ^2(x)\right )\\ &=-\frac{1}{12 \left (1+5 \tan ^2(x)\right )^{3/2}}+\frac{1}{16 \sqrt{1+5 \tan ^2(x)}}+\frac{1}{32} \operatorname{Subst}\left (\int \frac{1}{(1+x) \sqrt{1+5 x}} \, dx,x,\tan ^2(x)\right )\\ &=-\frac{1}{12 \left (1+5 \tan ^2(x)\right )^{3/2}}+\frac{1}{16 \sqrt{1+5 \tan ^2(x)}}+\frac{1}{80} \operatorname{Subst}\left (\int \frac{1}{\frac{4}{5}+\frac{x^2}{5}} \, dx,x,\sqrt{1+5 \tan ^2(x)}\right )\\ &=\frac{1}{32} \tan ^{-1}\left (\frac{1}{2} \sqrt{1+5 \tan ^2(x)}\right )-\frac{1}{12 \left (1+5 \tan ^2(x)\right )^{3/2}}+\frac{1}{16 \sqrt{1+5 \tan ^2(x)}}\\ \end{align*}

Mathematica [A]  time = 0.681154, size = 71, normalized size = 1.31 \[ \frac{(2 \cos (2 x)-3) \sec ^5(x) \left (-6 \cos (x)+8 \cos (3 x)-3 (2 \cos (2 x)-3)^{3/2} \log \left (2 \cos (x)+\sqrt{2 \cos (2 x)-3}\right )\right )}{96 \left (5 \tan ^2(x)+1\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[x]/(1 + 5*Tan[x]^2)^(5/2),x]

[Out]

((-3 + 2*Cos[2*x])*(-6*Cos[x] + 8*Cos[3*x] - 3*(-3 + 2*Cos[2*x])^(3/2)*Log[2*Cos[x] + Sqrt[-3 + 2*Cos[2*x]]])*
Sec[x]^5)/(96*(1 + 5*Tan[x]^2)^(5/2))

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Maple [A]  time = 0.02, size = 41, normalized size = 0.8 \begin{align*}{\frac{1}{32}\arctan \left ({\frac{1}{2}\sqrt{1+5\, \left ( \tan \left ( x \right ) \right ) ^{2}}} \right ) }+{\frac{1}{16}{\frac{1}{\sqrt{1+5\, \left ( \tan \left ( x \right ) \right ) ^{2}}}}}-{\frac{1}{12} \left ( 1+5\, \left ( \tan \left ( x \right ) \right ) ^{2} \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(x)/(1+5*tan(x)^2)^(5/2),x)

[Out]

1/32*arctan(1/2*(1+5*tan(x)^2)^(1/2))+1/16/(1+5*tan(x)^2)^(1/2)-1/12/(1+5*tan(x)^2)^(3/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\tan \left (x\right )}{{\left (5 \, \tan \left (x\right )^{2} + 1\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(tan(x)/(5*tan(x)^2 + 1)^(5/2), x)

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Fricas [A]  time = 3.07106, size = 227, normalized size = 4.2 \begin{align*} \frac{3 \,{\left (25 \, \tan \left (x\right )^{4} + 10 \, \tan \left (x\right )^{2} + 1\right )} \arctan \left (\frac{5 \, \tan \left (x\right )^{2} - 3}{4 \, \sqrt{5 \, \tan \left (x\right )^{2} + 1}}\right ) + 4 \,{\left (15 \, \tan \left (x\right )^{2} - 1\right )} \sqrt{5 \, \tan \left (x\right )^{2} + 1}}{192 \,{\left (25 \, \tan \left (x\right )^{4} + 10 \, \tan \left (x\right )^{2} + 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/192*(3*(25*tan(x)^4 + 10*tan(x)^2 + 1)*arctan(1/4*(5*tan(x)^2 - 3)/sqrt(5*tan(x)^2 + 1)) + 4*(15*tan(x)^2 -
1)*sqrt(5*tan(x)^2 + 1))/(25*tan(x)^4 + 10*tan(x)^2 + 1)

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Sympy [A]  time = 5.40261, size = 46, normalized size = 0.85 \begin{align*} \frac{\operatorname{atan}{\left (\frac{\sqrt{5 \tan ^{2}{\left (x \right )} + 1}}{2} \right )}}{32} + \frac{1}{16 \sqrt{5 \tan ^{2}{\left (x \right )} + 1}} - \frac{1}{12 \left (5 \tan ^{2}{\left (x \right )} + 1\right )^{\frac{3}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(x)/(1+5*tan(x)**2)**(5/2),x)

[Out]

atan(sqrt(5*tan(x)**2 + 1)/2)/32 + 1/(16*sqrt(5*tan(x)**2 + 1)) - 1/(12*(5*tan(x)**2 + 1)**(3/2))

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Giac [A]  time = 1.07443, size = 49, normalized size = 0.91 \begin{align*} \frac{15 \, \tan \left (x\right )^{2} - 1}{48 \,{\left (5 \, \tan \left (x\right )^{2} + 1\right )}^{\frac{3}{2}}} + \frac{1}{32} \, \arctan \left (\frac{1}{2} \, \sqrt{5 \, \tan \left (x\right )^{2} + 1}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/48*(15*tan(x)^2 - 1)/(5*tan(x)^2 + 1)^(3/2) + 1/32*arctan(1/2*sqrt(5*tan(x)^2 + 1))