\(\int \sqrt {x} \tan (\sqrt {x}) \, dx\) [62]

Optimal result

Integrand size = 12, antiderivative size = 70 \[ \int \sqrt {x} \tan \left (\sqrt {x}\right ) \, dx=\frac {2}{3} i x^{3/2}-2 x \log \left (1+e^{2 i \sqrt {x}}\right )+2 i \sqrt {x} \operatorname {PolyLog}\left (2,-e^{2 i \sqrt {x}}\right )-\operatorname {PolyLog}\left (3,-e^{2 i \sqrt {x}}\right ) \] Output:

2/3*I*x^(3/2)-2*x*ln(1+exp(2*I*x^(1/2)))+2*I*x^(1/2)*polylog(2,-exp(2*I*x^ 
(1/2)))-polylog(3,-exp(2*I*x^(1/2)))
 

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.00 \[ \int \sqrt {x} \tan \left (\sqrt {x}\right ) \, dx=\frac {2}{3} i x^{3/2}-2 x \log \left (1+e^{2 i \sqrt {x}}\right )+2 i \sqrt {x} \operatorname {PolyLog}\left (2,-e^{2 i \sqrt {x}}\right )-\operatorname {PolyLog}\left (3,-e^{2 i \sqrt {x}}\right ) \] Input:

Integrate[Sqrt[x]*Tan[Sqrt[x]],x]
 

Output:

((2*I)/3)*x^(3/2) - 2*x*Log[1 + E^((2*I)*Sqrt[x])] + (2*I)*Sqrt[x]*PolyLog 
[2, -E^((2*I)*Sqrt[x])] - PolyLog[3, -E^((2*I)*Sqrt[x])]
 

Rubi [A] (verified)

Time = 0.43 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.29, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.583, Rules used = {4234, 3042, 4202, 2620, 3011, 2720, 7143}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \sqrt {x} \tan \left (\sqrt {x}\right ) \, dx\)

\(\Big \downarrow \) 4234

\(\displaystyle 2 \int x \tan \left (\sqrt {x}\right )d\sqrt {x}\)

\(\Big \downarrow \) 3042

\(\displaystyle 2 \int x \tan \left (\sqrt {x}\right )d\sqrt {x}\)

\(\Big \downarrow \) 4202

\(\displaystyle 2 \left (\frac {1}{3} i x^{3/2}-2 i \int \frac {e^{2 i \sqrt {x}} x}{1+e^{2 i \sqrt {x}}}d\sqrt {x}\right )\)

\(\Big \downarrow \) 2620

\(\displaystyle 2 \left (\frac {1}{3} i x^{3/2}-2 i \left (i \int \sqrt {x} \log \left (1+e^{2 i \sqrt {x}}\right )d\sqrt {x}-\frac {1}{2} i x \log \left (1+e^{2 i \sqrt {x}}\right )\right )\right )\)

\(\Big \downarrow \) 3011

\(\displaystyle 2 \left (\frac {1}{3} i x^{3/2}-2 i \left (i \left (\frac {1}{2} i \sqrt {x} \operatorname {PolyLog}\left (2,-e^{2 i \sqrt {x}}\right )-\frac {1}{2} i \int \operatorname {PolyLog}\left (2,-e^{2 i \sqrt {x}}\right )d\sqrt {x}\right )-\frac {1}{2} i x \log \left (1+e^{2 i \sqrt {x}}\right )\right )\right )\)

\(\Big \downarrow \) 2720

\(\displaystyle 2 \left (\frac {1}{3} i x^{3/2}-2 i \left (i \left (\frac {1}{2} i \sqrt {x} \operatorname {PolyLog}\left (2,-e^{2 i \sqrt {x}}\right )-\frac {1}{4} \int \frac {\operatorname {PolyLog}\left (2,-e^{2 i \sqrt {x}}\right )}{\sqrt {x}}de^{2 i \sqrt {x}}\right )-\frac {1}{2} i x \log \left (1+e^{2 i \sqrt {x}}\right )\right )\right )\)

\(\Big \downarrow \) 7143

\(\displaystyle 2 \left (\frac {1}{3} i x^{3/2}-2 i \left (i \left (\frac {1}{2} i \sqrt {x} \operatorname {PolyLog}\left (2,-e^{2 i \sqrt {x}}\right )-\frac {1}{4} \operatorname {PolyLog}\left (3,-e^{2 i \sqrt {x}}\right )\right )-\frac {1}{2} i x \log \left (1+e^{2 i \sqrt {x}}\right )\right )\right )\)

Input:

Int[Sqrt[x]*Tan[Sqrt[x]],x]
 

Output:

2*((I/3)*x^(3/2) - (2*I)*((-1/2*I)*x*Log[1 + E^((2*I)*Sqrt[x])] + I*((I/2) 
*Sqrt[x]*PolyLog[2, -E^((2*I)*Sqrt[x])] - PolyLog[3, -E^((2*I)*Sqrt[x])]/4 
)))
 

Defintions of rubi rules used

rule 2620

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/ 
((a_) + (b_.)*((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)), x_Symbol] :> Simp 
[((c + d*x)^m/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], x] - Si 
mp[d*(m/(b*f*g*n*Log[F]))   Int[(c + d*x)^(m - 1)*Log[1 + b*((F^(g*(e + f*x 
)))^n/a)], x], x] /; FreeQ[{F, a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]
 

rule 2720

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Simp[v/D[v, x] 
   Subst[Int[FunctionOfExponentialFunction[u, x]/x, x], x, v], x]] /; Funct 
ionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; FreeQ 
[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x)) 
*(F_)[v_] /; FreeQ[{a, b, c}, x] && InverseFunctionQ[F[x]]]
 

rule 3011

Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.) 
*(x_))^(m_.), x_Symbol] :> Simp[(-(f + g*x)^m)*(PolyLog[2, (-e)*(F^(c*(a + 
b*x)))^n]/(b*c*n*Log[F])), x] + Simp[g*(m/(b*c*n*Log[F]))   Int[(f + g*x)^( 
m - 1)*PolyLog[2, (-e)*(F^(c*(a + b*x)))^n], x], x] /; FreeQ[{F, a, b, c, e 
, f, g, n}, x] && GtQ[m, 0]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4202

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[I 
*((c + d*x)^(m + 1)/(d*(m + 1))), x] - Simp[2*I   Int[(c + d*x)^m*(E^(2*I*( 
e + f*x))/(1 + E^(2*I*(e + f*x)))), x], x] /; FreeQ[{c, d, e, f}, x] && IGt 
Q[m, 0]
 

rule 4234

Int[(x_)^(m_.)*((a_.) + (b_.)*Tan[(c_.) + (d_.)*(x_)^(n_)])^(p_.), x_Symbol 
] :> Simp[1/n   Subst[Int[x^(Simplify[(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]
 

rule 7143

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_S 
ymbol] :> Simp[PolyLog[n + 1, c*(a + b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d 
, e, n, p}, x] && EqQ[b*d, a*e]
 

Maple [F]

\[\int \sqrt {x}\, \tan \left (\sqrt {x}\right )d x\]

Input:

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

Output:

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

Fricas [F]

\[ \int \sqrt {x} \tan \left (\sqrt {x}\right ) \, dx=\int { \sqrt {x} \tan \left (\sqrt {x}\right ) \,d x } \] Input:

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

Output:

integral(sqrt(x)*tan(sqrt(x)), x)
 

Sympy [F]

\[ \int \sqrt {x} \tan \left (\sqrt {x}\right ) \, dx=\int \sqrt {x} \tan {\left (\sqrt {x} \right )}\, dx \] Input:

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

Output:

Integral(sqrt(x)*tan(sqrt(x)), x)
 

Maxima [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.14 \[ \int \sqrt {x} \tan \left (\sqrt {x}\right ) \, dx=-2 i \, x \arctan \left (\sin \left (2 \, \sqrt {x}\right ), \cos \left (2 \, \sqrt {x}\right ) + 1\right ) - x \log \left (\cos \left (2 \, \sqrt {x}\right )^{2} + \sin \left (2 \, \sqrt {x}\right )^{2} + 2 \, \cos \left (2 \, \sqrt {x}\right ) + 1\right ) + \frac {2}{3} i \, x^{\frac {3}{2}} + 2 i \, \sqrt {x} {\rm Li}_2\left (-e^{\left (2 i \, \sqrt {x}\right )}\right ) - {\rm Li}_{3}(-e^{\left (2 i \, \sqrt {x}\right )}) \] Input:

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

Output:

-2*I*x*arctan2(sin(2*sqrt(x)), cos(2*sqrt(x)) + 1) - x*log(cos(2*sqrt(x))^ 
2 + sin(2*sqrt(x))^2 + 2*cos(2*sqrt(x)) + 1) + 2/3*I*x^(3/2) + 2*I*sqrt(x) 
*dilog(-e^(2*I*sqrt(x))) - polylog(3, -e^(2*I*sqrt(x)))
 

Giac [F]

\[ \int \sqrt {x} \tan \left (\sqrt {x}\right ) \, dx=\int { \sqrt {x} \tan \left (\sqrt {x}\right ) \,d x } \] Input:

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

Output:

integrate(sqrt(x)*tan(sqrt(x)), x)
 

Mupad [F(-1)]

Timed out. \[ \int \sqrt {x} \tan \left (\sqrt {x}\right ) \, dx=\int \sqrt {x}\,\mathrm {tan}\left (\sqrt {x}\right ) \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int \sqrt {x} \tan \left (\sqrt {x}\right ) \, dx=\int \sqrt {x}\, \tan \left (\sqrt {x}\right )d x \] Input:

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

Output:

int(sqrt(x)*tan(sqrt(x)),x)