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

Optimal. Leaf size=66 \[ \frac{1}{5} \left (5 \tan ^2(x)+1\right )^{5/2}-\frac{4}{3} \left (5 \tan ^2(x)+1\right )^{3/2}+16 \sqrt{5 \tan ^2(x)+1}-32 \tan ^{-1}\left (\frac{1}{2} \sqrt{5 \tan ^2(x)+1}\right ) \]

[Out]

-32*ArcTan[Sqrt[1 + 5*Tan[x]^2]/2] + 16*Sqrt[1 + 5*Tan[x]^2] - (4*(1 + 5*Tan[x]^2)^(3/2))/3 + (1 + 5*Tan[x]^2)
^(5/2)/5

________________________________________________________________________________________

Rubi [A]  time = 0.0706004, antiderivative size = 66, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {3670, 444, 50, 63, 203} \[ \frac{1}{5} \left (5 \tan ^2(x)+1\right )^{5/2}-\frac{4}{3} \left (5 \tan ^2(x)+1\right )^{3/2}+16 \sqrt{5 \tan ^2(x)+1}-32 \tan ^{-1}\left (\frac{1}{2} \sqrt{5 \tan ^2(x)+1}\right ) \]

Antiderivative was successfully verified.

[In]

Int[Tan[x]*(1 + 5*Tan[x]^2)^(5/2),x]

[Out]

-32*ArcTan[Sqrt[1 + 5*Tan[x]^2]/2] + 16*Sqrt[1 + 5*Tan[x]^2] - (4*(1 + 5*Tan[x]^2)^(3/2))/3 + (1 + 5*Tan[x]^2)
^(5/2)/5

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 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && 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 \tan (x) \left (1+5 \tan ^2(x)\right )^{5/2} \, dx &=\operatorname{Subst}\left (\int \frac{x \left (1+5 x^2\right )^{5/2}}{1+x^2} \, dx,x,\tan (x)\right )\\ &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{(1+5 x)^{5/2}}{1+x} \, dx,x,\tan ^2(x)\right )\\ &=\frac{1}{5} \left (1+5 \tan ^2(x)\right )^{5/2}-2 \operatorname{Subst}\left (\int \frac{(1+5 x)^{3/2}}{1+x} \, dx,x,\tan ^2(x)\right )\\ &=-\frac{4}{3} \left (1+5 \tan ^2(x)\right )^{3/2}+\frac{1}{5} \left (1+5 \tan ^2(x)\right )^{5/2}+8 \operatorname{Subst}\left (\int \frac{\sqrt{1+5 x}}{1+x} \, dx,x,\tan ^2(x)\right )\\ &=16 \sqrt{1+5 \tan ^2(x)}-\frac{4}{3} \left (1+5 \tan ^2(x)\right )^{3/2}+\frac{1}{5} \left (1+5 \tan ^2(x)\right )^{5/2}-32 \operatorname{Subst}\left (\int \frac{1}{(1+x) \sqrt{1+5 x}} \, dx,x,\tan ^2(x)\right )\\ &=16 \sqrt{1+5 \tan ^2(x)}-\frac{4}{3} \left (1+5 \tan ^2(x)\right )^{3/2}+\frac{1}{5} \left (1+5 \tan ^2(x)\right )^{5/2}-\frac{64}{5} \operatorname{Subst}\left (\int \frac{1}{\frac{4}{5}+\frac{x^2}{5}} \, dx,x,\sqrt{1+5 \tan ^2(x)}\right )\\ &=-32 \tan ^{-1}\left (\frac{1}{2} \sqrt{1+5 \tan ^2(x)}\right )+16 \sqrt{1+5 \tan ^2(x)}-\frac{4}{3} \left (1+5 \tan ^2(x)\right )^{3/2}+\frac{1}{5} \left (1+5 \tan ^2(x)\right )^{5/2}\\ \end{align*}

Mathematica [C]  time = 0.225238, size = 49, normalized size = 0.74 \[ \frac{5 \sqrt{5} \left (5 \tan ^2(x)+1\right )^{5/2} \, _2F_1\left (-\frac{5}{2},-\frac{5}{2};-\frac{3}{2};\frac{4 \cos ^2(x)}{5}\right )}{(3-2 \cos (2 x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(5*Sqrt[5]*Hypergeometric2F1[-5/2, -5/2, -3/2, (4*Cos[x]^2)/5]*(1 + 5*Tan[x]^2)^(5/2))/(3 - 2*Cos[2*x])^(5/2)

________________________________________________________________________________________

Maple [A]  time = 0.025, size = 61, normalized size = 0.9 \begin{align*} 5\, \left ( \tan \left ( x \right ) \right ) ^{4}\sqrt{1+5\, \left ( \tan \left ( x \right ) \right ) ^{2}}-{\frac{14\, \left ( \tan \left ( x \right ) \right ) ^{2}}{3}\sqrt{1+5\, \left ( \tan \left ( x \right ) \right ) ^{2}}}+{\frac{223}{15}\sqrt{1+5\, \left ( \tan \left ( x \right ) \right ) ^{2}}}-32\,\arctan \left ( 1/2\,\sqrt{1+5\, \left ( \tan \left ( x \right ) \right ) ^{2}} \right ) \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]

5*tan(x)^4*(1+5*tan(x)^2)^(1/2)-14/3*tan(x)^2*(1+5*tan(x)^2)^(1/2)+223/15*(1+5*tan(x)^2)^(1/2)-32*arctan(1/2*(
1+5*tan(x)^2)^(1/2))

________________________________________________________________________________________

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

________________________________________________________________________________________

Fricas [A]  time = 3.06739, size = 157, normalized size = 2.38 \begin{align*} \frac{1}{15} \,{\left (75 \, \tan \left (x\right )^{4} - 70 \, \tan \left (x\right )^{2} + 223\right )} \sqrt{5 \, \tan \left (x\right )^{2} + 1} - 16 \, \arctan \left (\frac{5 \, \tan \left (x\right )^{2} - 3}{4 \, \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="fricas")

[Out]

1/15*(75*tan(x)^4 - 70*tan(x)^2 + 223)*sqrt(5*tan(x)^2 + 1) - 16*arctan(1/4*(5*tan(x)^2 - 3)/sqrt(5*tan(x)^2 +
1))

________________________________________________________________________________________

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

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

________________________________________________________________________________________

Giac [A]  time = 1.08715, size = 70, normalized size = 1.06 \begin{align*} \frac{1}{5} \,{\left (5 \, \tan \left (x\right )^{2} + 1\right )}^{\frac{5}{2}} - \frac{4}{3} \,{\left (5 \, \tan \left (x\right )^{2} + 1\right )}^{\frac{3}{2}} + 16 \, \sqrt{5 \, \tan \left (x\right )^{2} + 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]

1/5*(5*tan(x)^2 + 1)^(5/2) - 4/3*(5*tan(x)^2 + 1)^(3/2) + 16*sqrt(5*tan(x)^2 + 1) - 32*arctan(1/2*sqrt(5*tan(x
)^2 + 1))