\(\int \frac {\arctan (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}})}{\sqrt {x}} \, dx\) [27]

Optimal result

Integrand size = 27, antiderivative size = 260 \[ \int \frac {\arctan \left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right )}{\sqrt {x}} \, dx=-\frac {4 \sqrt {-e} \sqrt {x} \sqrt {d+e x^2}}{\sqrt {e} \left (\sqrt {d}+\sqrt {e} x\right )}+2 \sqrt {x} \arctan \left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right )+\frac {4 \sqrt [4]{d} \sqrt {-e} \left (\sqrt {d}+\sqrt {e} x\right ) \sqrt {\frac {d+e x^2}{\left (\sqrt {d}+\sqrt {e} x\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{e} \sqrt {x}}{\sqrt [4]{d}}\right )|\frac {1}{2}\right )}{e^{3/4} \sqrt {d+e x^2}}-\frac {2 \sqrt [4]{d} \sqrt {-e} \left (\sqrt {d}+\sqrt {e} x\right ) \sqrt {\frac {d+e x^2}{\left (\sqrt {d}+\sqrt {e} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{e} \sqrt {x}}{\sqrt [4]{d}}\right ),\frac {1}{2}\right )}{e^{3/4} \sqrt {d+e x^2}} \] Output:

-4*(-e)^(1/2)*x^(1/2)*(e*x^2+d)^(1/2)/e^(1/2)/(d^(1/2)+e^(1/2)*x)+2*x^(1/2 
)*arctan((-e)^(1/2)*x/(e*x^2+d)^(1/2))+4*d^(1/4)*(-e)^(1/2)*(d^(1/2)+e^(1/ 
2)*x)*((e*x^2+d)/(d^(1/2)+e^(1/2)*x)^2)^(1/2)*EllipticE(sin(2*arctan(e^(1/ 
4)*x^(1/2)/d^(1/4))),1/2*2^(1/2))/e^(3/4)/(e*x^2+d)^(1/2)-2*d^(1/4)*(-e)^( 
1/2)*(d^(1/2)+e^(1/2)*x)*((e*x^2+d)/(d^(1/2)+e^(1/2)*x)^2)^(1/2)*InverseJa 
cobiAM(2*arctan(e^(1/4)*x^(1/2)/d^(1/4)),1/2*2^(1/2))/e^(3/4)/(e*x^2+d)^(1 
/2)
                                                                                    
                                                                                    
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 0.07 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.34 \[ \int \frac {\arctan \left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right )}{\sqrt {x}} \, dx=2 \sqrt {x} \arctan \left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right )-\frac {4 \sqrt {-e} x^{3/2} \sqrt {1+\frac {e x^2}{d}} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-\frac {e x^2}{d}\right )}{3 \sqrt {d+e x^2}} \] Input:

Integrate[ArcTan[(Sqrt[-e]*x)/Sqrt[d + e*x^2]]/Sqrt[x],x]
 

Output:

2*Sqrt[x]*ArcTan[(Sqrt[-e]*x)/Sqrt[d + e*x^2]] - (4*Sqrt[-e]*x^(3/2)*Sqrt[ 
1 + (e*x^2)/d]*Hypergeometric2F1[1/2, 3/4, 7/4, -((e*x^2)/d)])/(3*Sqrt[d + 
 e*x^2])
 

Rubi [A] (verified)

Time = 0.35 (sec) , antiderivative size = 253, normalized size of antiderivative = 0.97, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {5674, 266, 834, 27, 761, 1510}

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.

\(\displaystyle \int \frac {\arctan \left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right )}{\sqrt {x}} \, dx\)

\(\Big \downarrow \) 5674

\(\displaystyle 2 \sqrt {x} \arctan \left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right )-2 \sqrt {-e} \int \frac {\sqrt {x}}{\sqrt {e x^2+d}}dx\)

\(\Big \downarrow \) 266

\(\displaystyle 2 \sqrt {x} \arctan \left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right )-4 \sqrt {-e} \int \frac {x}{\sqrt {e x^2+d}}d\sqrt {x}\)

\(\Big \downarrow \) 834

\(\displaystyle 2 \sqrt {x} \arctan \left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right )-4 \sqrt {-e} \left (\frac {\sqrt {d} \int \frac {1}{\sqrt {e x^2+d}}d\sqrt {x}}{\sqrt {e}}-\frac {\sqrt {d} \int \frac {\sqrt {d}-\sqrt {e} x}{\sqrt {d} \sqrt {e x^2+d}}d\sqrt {x}}{\sqrt {e}}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle 2 \sqrt {x} \arctan \left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right )-4 \sqrt {-e} \left (\frac {\sqrt {d} \int \frac {1}{\sqrt {e x^2+d}}d\sqrt {x}}{\sqrt {e}}-\frac {\int \frac {\sqrt {d}-\sqrt {e} x}{\sqrt {e x^2+d}}d\sqrt {x}}{\sqrt {e}}\right )\)

\(\Big \downarrow \) 761

\(\displaystyle 2 \sqrt {x} \arctan \left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right )-4 \sqrt {-e} \left (\frac {\sqrt [4]{d} \left (\sqrt {d}+\sqrt {e} x\right ) \sqrt {\frac {d+e x^2}{\left (\sqrt {d}+\sqrt {e} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{e} \sqrt {x}}{\sqrt [4]{d}}\right ),\frac {1}{2}\right )}{2 e^{3/4} \sqrt {d+e x^2}}-\frac {\int \frac {\sqrt {d}-\sqrt {e} x}{\sqrt {e x^2+d}}d\sqrt {x}}{\sqrt {e}}\right )\)

\(\Big \downarrow \) 1510

\(\displaystyle 2 \sqrt {x} \arctan \left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right )-4 \sqrt {-e} \left (\frac {\sqrt [4]{d} \left (\sqrt {d}+\sqrt {e} x\right ) \sqrt {\frac {d+e x^2}{\left (\sqrt {d}+\sqrt {e} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{e} \sqrt {x}}{\sqrt [4]{d}}\right ),\frac {1}{2}\right )}{2 e^{3/4} \sqrt {d+e x^2}}-\frac {\frac {\sqrt [4]{d} \left (\sqrt {d}+\sqrt {e} x\right ) \sqrt {\frac {d+e x^2}{\left (\sqrt {d}+\sqrt {e} x\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{e} \sqrt {x}}{\sqrt [4]{d}}\right )|\frac {1}{2}\right )}{\sqrt [4]{e} \sqrt {d+e x^2}}-\frac {\sqrt {x} \sqrt {d+e x^2}}{\sqrt {d}+\sqrt {e} x}}{\sqrt {e}}\right )\)

Input:

Int[ArcTan[(Sqrt[-e]*x)/Sqrt[d + e*x^2]]/Sqrt[x],x]
 

Output:

2*Sqrt[x]*ArcTan[(Sqrt[-e]*x)/Sqrt[d + e*x^2]] - 4*Sqrt[-e]*(-((-((Sqrt[x] 
*Sqrt[d + e*x^2])/(Sqrt[d] + Sqrt[e]*x)) + (d^(1/4)*(Sqrt[d] + Sqrt[e]*x)* 
Sqrt[(d + e*x^2)/(Sqrt[d] + Sqrt[e]*x)^2]*EllipticE[2*ArcTan[(e^(1/4)*Sqrt 
[x])/d^(1/4)], 1/2])/(e^(1/4)*Sqrt[d + e*x^2]))/Sqrt[e]) + (d^(1/4)*(Sqrt[ 
d] + Sqrt[e]*x)*Sqrt[(d + e*x^2)/(Sqrt[d] + Sqrt[e]*x)^2]*EllipticF[2*ArcT 
an[(e^(1/4)*Sqrt[x])/d^(1/4)], 1/2])/(2*e^(3/4)*Sqrt[d + e*x^2]))
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 266

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{k = De 
nominator[m]}, Simp[k/c   Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(2*k)/c^2)) 
^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && FractionQ[m] && I 
ntBinomialQ[a, b, c, 2, m, p, x]
 

rule 761

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 4]}, Simp[( 
1 + q^2*x^2)*(Sqrt[(a + b*x^4)/(a*(1 + q^2*x^2)^2)]/(2*q*Sqrt[a + b*x^4]))* 
EllipticF[2*ArcTan[q*x], 1/2], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]
 

rule 834

Int[(x_)^2/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 2]}, S 
imp[1/q   Int[1/Sqrt[a + b*x^4], x], x] - Simp[1/q   Int[(1 - q*x^2)/Sqrt[a 
 + b*x^4], x], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]
 

rule 1510

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = 
 Rt[c/a, 4]}, Simp[(-d)*x*(Sqrt[a + c*x^4]/(a*(1 + q^2*x^2))), x] + Simp[d* 
(1 + q^2*x^2)*(Sqrt[(a + c*x^4)/(a*(1 + q^2*x^2)^2)]/(q*Sqrt[a + c*x^4]))*E 
llipticE[2*ArcTan[q*x], 1/2], x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e 
}, x] && PosQ[c/a]
 

rule 5674

Int[ArcTan[((c_.)*(x_))/Sqrt[(a_.) + (b_.)*(x_)^2]]*((d_.)*(x_))^(m_.), x_S 
ymbol] :> Simp[(d*x)^(m + 1)*(ArcTan[(c*x)/Sqrt[a + b*x^2]]/(d*(m + 1))), x 
] - Simp[c/(d*(m + 1))   Int[(d*x)^(m + 1)/Sqrt[a + b*x^2], x], x] /; FreeQ 
[{a, b, c, d, m}, x] && EqQ[b + c^2, 0] && NeQ[m, -1]
 

Maple [F]

\[\int \frac {\arctan \left (\frac {\sqrt {-e}\, x}{\sqrt {e \,x^{2}+d}}\right )}{\sqrt {x}}d x\]

Input:

int(arctan((-e)^(1/2)*x/(e*x^2+d)^(1/2))/x^(1/2),x)
 

Output:

int(arctan((-e)^(1/2)*x/(e*x^2+d)^(1/2))/x^(1/2),x)
 

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.21 \[ \int \frac {\arctan \left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right )}{\sqrt {x}} \, dx=\frac {2 \, {\left (e \sqrt {x} \arctan \left (\frac {\sqrt {-e} x}{\sqrt {e x^{2} + d}}\right ) + 2 \, \sqrt {-e} \sqrt {e} {\rm weierstrassZeta}\left (-\frac {4 \, d}{e}, 0, {\rm weierstrassPInverse}\left (-\frac {4 \, d}{e}, 0, x\right )\right )\right )}}{e} \] Input:

integrate(arctan((-e)^(1/2)*x/(e*x^2+d)^(1/2))/x^(1/2),x, algorithm="frica 
s")
 

Output:

2*(e*sqrt(x)*arctan(sqrt(-e)*x/sqrt(e*x^2 + d)) + 2*sqrt(-e)*sqrt(e)*weier 
strassZeta(-4*d/e, 0, weierstrassPInverse(-4*d/e, 0, x)))/e
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 9.59 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.27 \[ \int \frac {\arctan \left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right )}{\sqrt {x}} \, dx=2 \sqrt {x} \operatorname {atan}{\left (\frac {x \sqrt {- e}}{\sqrt {d + e x^{2}}} \right )} - \frac {x^{\frac {3}{2}} \sqrt {- e} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {3}{4} \\ \frac {7}{4} \end {matrix}\middle | {\frac {e x^{2} e^{i \pi }}{d}} \right )}}{\sqrt {d} \Gamma \left (\frac {7}{4}\right )} \] Input:

integrate(atan((-e)**(1/2)*x/(e*x**2+d)**(1/2))/x**(1/2),x)
 

Output:

2*sqrt(x)*atan(x*sqrt(-e)/sqrt(d + e*x**2)) - x**(3/2)*sqrt(-e)*gamma(3/4) 
*hyper((1/2, 3/4), (7/4,), e*x**2*exp_polar(I*pi)/d)/(sqrt(d)*gamma(7/4))
 

Maxima [F]

\[ \int \frac {\arctan \left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right )}{\sqrt {x}} \, dx=\int { \frac {\arctan \left (\frac {\sqrt {-e} x}{\sqrt {e x^{2} + d}}\right )}{\sqrt {x}} \,d x } \] Input:

integrate(arctan((-e)^(1/2)*x/(e*x^2+d)^(1/2))/x^(1/2),x, algorithm="maxim 
a")
 

Output:

-2*d*sqrt(-e)*integrate(sqrt(e*x^2 + d)*x/((e*x^2 + d)*e^(log(e*x^2 + d) + 
 1/2*log(x)) - (e^2*x^4 + d*e*x^2)*sqrt(x)), x) + 2*sqrt(x)*arctan2(sqrt(- 
e)*x, sqrt(e*x^2 + d))
 

Giac [F]

\[ \int \frac {\arctan \left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right )}{\sqrt {x}} \, dx=\int { \frac {\arctan \left (\frac {\sqrt {-e} x}{\sqrt {e x^{2} + d}}\right )}{\sqrt {x}} \,d x } \] Input:

integrate(arctan((-e)^(1/2)*x/(e*x^2+d)^(1/2))/x^(1/2),x, algorithm="giac" 
)
 

Output:

integrate(arctan(sqrt(-e)*x/sqrt(e*x^2 + d))/sqrt(x), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\arctan \left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right )}{\sqrt {x}} \, dx=\int \frac {\mathrm {atan}\left (\frac {\sqrt {-e}\,x}{\sqrt {e\,x^2+d}}\right )}{\sqrt {x}} \,d x \] Input:

int(atan(((-e)^(1/2)*x)/(d + e*x^2)^(1/2))/x^(1/2),x)
 

Output:

int(atan(((-e)^(1/2)*x)/(d + e*x^2)^(1/2))/x^(1/2), x)
 

Reduce [F]

\[ \int \frac {\arctan \left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right )}{\sqrt {x}} \, dx=\int \frac {\sqrt {x}\, \mathit {atan} \left (\frac {\sqrt {e}\, i x}{\sqrt {e \,x^{2}+d}}\right )}{x}d x \] Input:

int(atan((-e)^(1/2)*x/(e*x^2+d)^(1/2))/x^(1/2),x)
 

Output:

int((sqrt(x)*atan((sqrt(e)*i*x)/sqrt(d + e*x**2)))/x,x)