\(\int \frac {\arctan (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}})}{x^{7/2}} \, dx\) [22]

Optimal result

Integrand size = 27, antiderivative size = 156 \[ \int \frac {\arctan \left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right )}{x^{7/2}} \, dx=-\frac {4 \sqrt {-e} \sqrt {d+e x^2}}{15 d x^{3/2}}-\frac {2 \arctan \left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right )}{5 x^{5/2}}-\frac {2 \sqrt {-e} e^{3/4} \left (\sqrt {d}+\sqrt {e} x\right ) \sqrt {\frac {d+e x^2}{\left (\sqrt {d}+\sqrt {e} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{e} \sqrt {x}}{\sqrt [4]{d}}\right ),\frac {1}{2}\right )}{15 d^{5/4} \sqrt {d+e x^2}} \] Output:

-4/15*(-e)^(1/2)*(e*x^2+d)^(1/2)/d/x^(3/2)-2/5*arctan((-e)^(1/2)*x/(e*x^2+ 
d)^(1/2))/x^(5/2)-2/15*(-e)^(1/2)*e^(3/4)*(d^(1/2)+e^(1/2)*x)*((e*x^2+d)/( 
d^(1/2)+e^(1/2)*x)^2)^(1/2)*InverseJacobiAM(2*arctan(e^(1/4)*x^(1/2)/d^(1/ 
4)),1/2*2^(1/2))/d^(5/4)/(e*x^2+d)^(1/2)
                                                                                    
                                                                                    
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.19 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.96 \[ \int \frac {\arctan \left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right )}{x^{7/2}} \, dx=-\frac {2 \left (2 \sqrt {-e} x \sqrt {d+e x^2}+3 d \arctan \left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right )\right )}{15 d x^{5/2}}+\frac {4 i (-e)^{3/2} \sqrt {1+\frac {d}{e x^2}} x \operatorname {EllipticF}\left (i \text {arcsinh}\left (\frac {\sqrt {\frac {i \sqrt {d}}{\sqrt {e}}}}{\sqrt {x}}\right ),-1\right )}{15 d \sqrt {\frac {i \sqrt {d}}{\sqrt {e}}} \sqrt {d+e x^2}} \] Input:

Integrate[ArcTan[(Sqrt[-e]*x)/Sqrt[d + e*x^2]]/x^(7/2),x]
 

Output:

(-2*(2*Sqrt[-e]*x*Sqrt[d + e*x^2] + 3*d*ArcTan[(Sqrt[-e]*x)/Sqrt[d + e*x^2 
]]))/(15*d*x^(5/2)) + (((4*I)/15)*(-e)^(3/2)*Sqrt[1 + d/(e*x^2)]*x*Ellipti 
cF[I*ArcSinh[Sqrt[(I*Sqrt[d])/Sqrt[e]]/Sqrt[x]], -1])/(d*Sqrt[(I*Sqrt[d])/ 
Sqrt[e]]*Sqrt[d + e*x^2])
 

Rubi [A] (verified)

Time = 0.28 (sec) , antiderivative size = 154, normalized size of antiderivative = 0.99, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {5674, 264, 266, 761}

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.

\(\displaystyle \int \frac {\arctan \left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right )}{x^{7/2}} \, dx\)

\(\Big \downarrow \) 5674

\(\displaystyle \frac {2}{5} \sqrt {-e} \int \frac {1}{x^{5/2} \sqrt {e x^2+d}}dx-\frac {2 \arctan \left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right )}{5 x^{5/2}}\)

\(\Big \downarrow \) 264

\(\displaystyle \frac {2}{5} \sqrt {-e} \left (-\frac {e \int \frac {1}{\sqrt {x} \sqrt {e x^2+d}}dx}{3 d}-\frac {2 \sqrt {d+e x^2}}{3 d x^{3/2}}\right )-\frac {2 \arctan \left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right )}{5 x^{5/2}}\)

\(\Big \downarrow \) 266

\(\displaystyle \frac {2}{5} \sqrt {-e} \left (-\frac {2 e \int \frac {1}{\sqrt {e x^2+d}}d\sqrt {x}}{3 d}-\frac {2 \sqrt {d+e x^2}}{3 d x^{3/2}}\right )-\frac {2 \arctan \left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right )}{5 x^{5/2}}\)

\(\Big \downarrow \) 761

\(\displaystyle \frac {2}{5} \sqrt {-e} \left (-\frac {e^{3/4} \left (\sqrt {d}+\sqrt {e} x\right ) \sqrt {\frac {d+e x^2}{\left (\sqrt {d}+\sqrt {e} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{e} \sqrt {x}}{\sqrt [4]{d}}\right ),\frac {1}{2}\right )}{3 d^{5/4} \sqrt {d+e x^2}}-\frac {2 \sqrt {d+e x^2}}{3 d x^{3/2}}\right )-\frac {2 \arctan \left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right )}{5 x^{5/2}}\)

Input:

Int[ArcTan[(Sqrt[-e]*x)/Sqrt[d + e*x^2]]/x^(7/2),x]
 

Output:

(-2*ArcTan[(Sqrt[-e]*x)/Sqrt[d + e*x^2]])/(5*x^(5/2)) + (2*Sqrt[-e]*((-2*S 
qrt[d + e*x^2])/(3*d*x^(3/2)) - (e^(3/4)*(Sqrt[d] + Sqrt[e]*x)*Sqrt[(d + e 
*x^2)/(Sqrt[d] + Sqrt[e]*x)^2]*EllipticF[2*ArcTan[(e^(1/4)*Sqrt[x])/d^(1/4 
)], 1/2])/(3*d^(5/4)*Sqrt[d + e*x^2])))/5
 

Defintions of rubi rules used

rule 264

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c*x)^( 
m + 1)*((a + b*x^2)^(p + 1)/(a*c*(m + 1))), x] - Simp[b*((m + 2*p + 3)/(a*c 
^2*(m + 1)))   Int[(c*x)^(m + 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, p 
}, x] && LtQ[m, -1] && IntBinomialQ[a, b, c, 2, m, p, 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 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]

\[\int \frac {\arctan \left (\frac {\sqrt {-e}\, x}{\sqrt {e \,x^{2}+d}}\right )}{x^{\frac {7}{2}}}d x\]

Input:

int(arctan((-e)^(1/2)*x/(e*x^2+d)^(1/2))/x^(7/2),x)
 

Output:

int(arctan((-e)^(1/2)*x/(e*x^2+d)^(1/2))/x^(7/2),x)
 

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.47 \[ \int \frac {\arctan \left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right )}{x^{7/2}} \, dx=-\frac {2 \, {\left (2 \, \sqrt {-e} \sqrt {e} x^{3} {\rm weierstrassPInverse}\left (-\frac {4 \, d}{e}, 0, x\right ) + 2 \, \sqrt {e x^{2} + d} \sqrt {-e} x^{\frac {3}{2}} + 3 \, d \sqrt {x} \arctan \left (\frac {\sqrt {-e} x}{\sqrt {e x^{2} + d}}\right )\right )}}{15 \, d x^{3}} \] Input:

integrate(arctan((-e)^(1/2)*x/(e*x^2+d)^(1/2))/x^(7/2),x, algorithm="frica 
s")
 

Output:

-2/15*(2*sqrt(-e)*sqrt(e)*x^3*weierstrassPInverse(-4*d/e, 0, x) + 2*sqrt(e 
*x^2 + d)*sqrt(-e)*x^(3/2) + 3*d*sqrt(x)*arctan(sqrt(-e)*x/sqrt(e*x^2 + d) 
))/(d*x^3)
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 44.00 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.50 \[ \int \frac {\arctan \left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right )}{x^{7/2}} \, dx=- \frac {2 \operatorname {atan}{\left (\frac {x \sqrt {- e}}{\sqrt {d + e x^{2}}} \right )}}{5 x^{\frac {5}{2}}} + \frac {\sqrt {- e} \Gamma \left (- \frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{4}, \frac {1}{2} \\ \frac {1}{4} \end {matrix}\middle | {\frac {e x^{2} e^{i \pi }}{d}} \right )}}{5 \sqrt {d} x^{\frac {3}{2}} \Gamma \left (\frac {1}{4}\right )} \] Input:

integrate(atan((-e)**(1/2)*x/(e*x**2+d)**(1/2))/x**(7/2),x)
                                                                                    
                                                                                    
 

Output:

-2*atan(x*sqrt(-e)/sqrt(d + e*x**2))/(5*x**(5/2)) + sqrt(-e)*gamma(-3/4)*h 
yper((-3/4, 1/2), (1/4,), e*x**2*exp_polar(I*pi)/d)/(5*sqrt(d)*x**(3/2)*ga 
mma(1/4))
 

Maxima [F]

\[ \int \frac {\arctan \left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right )}{x^{7/2}} \, dx=\int { \frac {\arctan \left (\frac {\sqrt {-e} x}{\sqrt {e x^{2} + d}}\right )}{x^{\frac {7}{2}}} \,d x } \] Input:

integrate(arctan((-e)^(1/2)*x/(e*x^2+d)^(1/2))/x^(7/2),x, algorithm="maxim 
a")
 

Output:

2/5*(5*d*sqrt(-e)*x^(5/2)*integrate(-1/5*sqrt(e*x^2 + d)*x/((e^2*x^4 + d*e 
*x^2)*x^(7/2) - (e*x^2 + d)*e^(log(e*x^2 + d) + 7/2*log(x))), x) - arctan2 
(sqrt(-e)*x, sqrt(e*x^2 + d)))/x^(5/2)
 

Giac [F]

\[ \int \frac {\arctan \left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right )}{x^{7/2}} \, dx=\int { \frac {\arctan \left (\frac {\sqrt {-e} x}{\sqrt {e x^{2} + d}}\right )}{x^{\frac {7}{2}}} \,d x } \] Input:

integrate(arctan((-e)^(1/2)*x/(e*x^2+d)^(1/2))/x^(7/2),x, algorithm="giac" 
)
 

Output:

integrate(arctan(sqrt(-e)*x/sqrt(e*x^2 + d))/x^(7/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\arctan \left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right )}{x^{7/2}} \, dx=\int \frac {\mathrm {atan}\left (\frac {\sqrt {-e}\,x}{\sqrt {e\,x^2+d}}\right )}{x^{7/2}} \,d x \] Input:

int(atan(((-e)^(1/2)*x)/(d + e*x^2)^(1/2))/x^(7/2),x)
 

Output:

int(atan(((-e)^(1/2)*x)/(d + e*x^2)^(1/2))/x^(7/2), x)
 

Reduce [F]

\[ \int \frac {\arctan \left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right )}{x^{7/2}} \, dx=\int \frac {\sqrt {x}\, \mathit {atan} \left (\frac {\sqrt {e}\, i x}{\sqrt {e \,x^{2}+d}}\right )}{x^{4}}d x \] Input:

int(atan((-e)^(1/2)*x/(e*x^2+d)^(1/2))/x^(7/2),x)
 

Output:

int((sqrt(x)*atan((sqrt(e)*i*x)/sqrt(d + e*x**2)))/x**4,x)