\(\int \frac {\tan ^2(\sqrt {x})}{\sqrt {x}} \, dx\) [59]

Optimal result

Integrand size = 14, antiderivative size = 16 \[ \int \frac {\tan ^2\left (\sqrt {x}\right )}{\sqrt {x}} \, dx=-2 \sqrt {x}+2 \tan \left (\sqrt {x}\right ) \]

[Out]

-2*x^(1/2)+2*tan(x^(1/2))

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {3832, 3554, 8} \[ \int \frac {\tan ^2\left (\sqrt {x}\right )}{\sqrt {x}} \, dx=2 \tan \left (\sqrt {x}\right )-2 \sqrt {x} \]

[In]

Int[Tan[Sqrt[x]]^2/Sqrt[x],x]

[Out]

-2*Sqrt[x] + 2*Tan[Sqrt[x]]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 3554

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b*((b*Tan[c + d*x])^(n - 1)/(d*(n - 1))), x] - Dis
t[b^2, Int[(b*Tan[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1]

Rule 3832

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

Rubi steps \begin{align*} \text {integral}& = 2 \text {Subst}\left (\int \tan ^2(x) \, dx,x,\sqrt {x}\right ) \\ & = 2 \tan \left (\sqrt {x}\right )-2 \text {Subst}\left (\int 1 \, dx,x,\sqrt {x}\right ) \\ & = -2 \sqrt {x}+2 \tan \left (\sqrt {x}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.12 \[ \int \frac {\tan ^2\left (\sqrt {x}\right )}{\sqrt {x}} \, dx=-2 \arctan \left (\tan \left (\sqrt {x}\right )\right )+2 \tan \left (\sqrt {x}\right ) \]

[In]

Integrate[Tan[Sqrt[x]]^2/Sqrt[x],x]

[Out]

-2*ArcTan[Tan[Sqrt[x]]] + 2*Tan[Sqrt[x]]

Maple [A] (verified)

Time = 0.36 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.94

method	result	size
derivativedivides	\(2 \tan \left (\sqrt {x}\right )-2 \arctan \left (\tan \left (\sqrt {x}\right )\right )\)	\(15\)
default	\(2 \tan \left (\sqrt {x}\right )-2 \arctan \left (\tan \left (\sqrt {x}\right )\right )\)	\(15\)

[In]

int(tan(x^(1/2))^2/x^(1/2),x,method=_RETURNVERBOSE)

[Out]

2*tan(x^(1/2))-2*arctan(tan(x^(1/2)))

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.75 \[ \int \frac {\tan ^2\left (\sqrt {x}\right )}{\sqrt {x}} \, dx=-2 \, \sqrt {x} + 2 \, \tan \left (\sqrt {x}\right ) \]

[In]

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

[Out]

-2*sqrt(x) + 2*tan(sqrt(x))

Sympy [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.88 \[ \int \frac {\tan ^2\left (\sqrt {x}\right )}{\sqrt {x}} \, dx=- 2 \sqrt {x} + 2 \tan {\left (\sqrt {x} \right )} \]

[In]

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

[Out]

-2*sqrt(x) + 2*tan(sqrt(x))

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.75 \[ \int \frac {\tan ^2\left (\sqrt {x}\right )}{\sqrt {x}} \, dx=-2 \, \sqrt {x} + 2 \, \tan \left (\sqrt {x}\right ) \]

[In]

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

[Out]

-2*sqrt(x) + 2*tan(sqrt(x))

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.75 \[ \int \frac {\tan ^2\left (\sqrt {x}\right )}{\sqrt {x}} \, dx=-2 \, \sqrt {x} + 2 \, \tan \left (\sqrt {x}\right ) \]

[In]

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

[Out]

-2*sqrt(x) + 2*tan(sqrt(x))

Mupad [B] (verification not implemented)

Time = 26.94 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.25 \[ \int \frac {\tan ^2\left (\sqrt {x}\right )}{\sqrt {x}} \, dx=-2\,\sqrt {x}+\frac {4{}\mathrm {i}}{{\mathrm {e}}^{\sqrt {x}\,2{}\mathrm {i}}+1} \]

[In]

int(tan(x^(1/2))^2/x^(1/2),x)

[Out]

4i/(exp(x^(1/2)*2i) + 1) - 2*x^(1/2)