\(\int \frac {\arctan (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}})}{x^{15/2}} \, dx\) [24]

Optimal result

Integrand size = 27, antiderivative size = 216 \[ \int \frac {\arctan \left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right )}{x^{15/2}} \, dx=-\frac {4 \sqrt {-e} \sqrt {d+e x^2}}{143 d x^{11/2}}-\frac {36 (-e)^{3/2} \sqrt {d+e x^2}}{1001 d^2 x^{7/2}}-\frac {60 (-e)^{5/2} \sqrt {d+e x^2}}{1001 d^3 x^{3/2}}-\frac {2 \arctan \left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right )}{13 x^{13/2}}-\frac {30 \sqrt {-e} e^{11/4} \left (\sqrt {d}+\sqrt {e} x\right ) \sqrt {\frac {d+e x^2}{\left (\sqrt {d}+\sqrt {e} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{e} \sqrt {x}}{\sqrt [4]{d}}\right ),\frac {1}{2}\right )}{1001 d^{13/4} \sqrt {d+e x^2}} \] Output:

-4/143*(-e)^(1/2)*(e*x^2+d)^(1/2)/d/x^(11/2)-36/1001*(-e)^(3/2)*(e*x^2+d)^ 
(1/2)/d^2/x^(7/2)-60/1001*(-e)^(5/2)*(e*x^2+d)^(1/2)/d^3/x^(3/2)-2/13*arct 
an((-e)^(1/2)*x/(e*x^2+d)^(1/2))/x^(13/2)-30/1001*(-e)^(1/2)*e^(11/4)*(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^(13/4)/(e*x^2+d)^(1/2)
                                                                                    
                                                                                    
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.40 (sec) , antiderivative size = 171, normalized size of antiderivative = 0.79 \[ \int \frac {\arctan \left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right )}{x^{15/2}} \, dx=\frac {2 \left (-\frac {2 \sqrt {-e} \sqrt {d+e x^2} \left (7 d^2 x-9 d e x^3+15 e^2 x^5\right )}{d^3}-77 \arctan \left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right )+\frac {30 i (-e)^{7/2} \sqrt {1+\frac {d}{e x^2}} x^{15/2} \operatorname {EllipticF}\left (i \text {arcsinh}\left (\frac {\sqrt {\frac {i \sqrt {d}}{\sqrt {e}}}}{\sqrt {x}}\right ),-1\right )}{d^3 \sqrt {\frac {i \sqrt {d}}{\sqrt {e}}} \sqrt {d+e x^2}}\right )}{1001 x^{13/2}} \] Input:

Integrate[ArcTan[(Sqrt[-e]*x)/Sqrt[d + e*x^2]]/x^(15/2),x]
 

Output:

(2*((-2*Sqrt[-e]*Sqrt[d + e*x^2]*(7*d^2*x - 9*d*e*x^3 + 15*e^2*x^5))/d^3 - 
 77*ArcTan[(Sqrt[-e]*x)/Sqrt[d + e*x^2]] + ((30*I)*(-e)^(7/2)*Sqrt[1 + d/( 
e*x^2)]*x^(15/2)*EllipticF[I*ArcSinh[Sqrt[(I*Sqrt[d])/Sqrt[e]]/Sqrt[x]], - 
1])/(d^3*Sqrt[(I*Sqrt[d])/Sqrt[e]]*Sqrt[d + e*x^2])))/(1001*x^(13/2))
 

Rubi [A] (verified)

Time = 0.33 (sec) , antiderivative size = 218, normalized size of antiderivative = 1.01, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {5674, 264, 264, 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.

\(\displaystyle \int \frac {\arctan \left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right )}{x^{15/2}} \, dx\)

\(\Big \downarrow \) 5674

\(\displaystyle \frac {2}{13} \sqrt {-e} \int \frac {1}{x^{13/2} \sqrt {e x^2+d}}dx-\frac {2 \arctan \left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right )}{13 x^{13/2}}\)

\(\Big \downarrow \) 264

\(\displaystyle \frac {2}{13} \sqrt {-e} \left (-\frac {9 e \int \frac {1}{x^{9/2} \sqrt {e x^2+d}}dx}{11 d}-\frac {2 \sqrt {d+e x^2}}{11 d x^{11/2}}\right )-\frac {2 \arctan \left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right )}{13 x^{13/2}}\)

\(\Big \downarrow \) 264

\(\displaystyle \frac {2}{13} \sqrt {-e} \left (-\frac {9 e \left (-\frac {5 e \int \frac {1}{x^{5/2} \sqrt {e x^2+d}}dx}{7 d}-\frac {2 \sqrt {d+e x^2}}{7 d x^{7/2}}\right )}{11 d}-\frac {2 \sqrt {d+e x^2}}{11 d x^{11/2}}\right )-\frac {2 \arctan \left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right )}{13 x^{13/2}}\)

\(\Big \downarrow \) 264

\(\displaystyle \frac {2}{13} \sqrt {-e} \left (-\frac {9 e \left (-\frac {5 e \left (-\frac {e \int \frac {1}{\sqrt {x} \sqrt {e x^2+d}}dx}{3 d}-\frac {2 \sqrt {d+e x^2}}{3 d x^{3/2}}\right )}{7 d}-\frac {2 \sqrt {d+e x^2}}{7 d x^{7/2}}\right )}{11 d}-\frac {2 \sqrt {d+e x^2}}{11 d x^{11/2}}\right )-\frac {2 \arctan \left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right )}{13 x^{13/2}}\)

\(\Big \downarrow \) 266

\(\displaystyle \frac {2}{13} \sqrt {-e} \left (-\frac {9 e \left (-\frac {5 e \left (-\frac {2 e \int \frac {1}{\sqrt {e x^2+d}}d\sqrt {x}}{3 d}-\frac {2 \sqrt {d+e x^2}}{3 d x^{3/2}}\right )}{7 d}-\frac {2 \sqrt {d+e x^2}}{7 d x^{7/2}}\right )}{11 d}-\frac {2 \sqrt {d+e x^2}}{11 d x^{11/2}}\right )-\frac {2 \arctan \left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right )}{13 x^{13/2}}\)

\(\Big \downarrow \) 761

\(\displaystyle \frac {2}{13} \sqrt {-e} \left (-\frac {9 e \left (-\frac {5 e \left (-\frac {e^{3/4} \left (\sqrt {d}+\sqrt {e} x\right ) \sqrt {\frac {d+e x^2}{\left (\sqrt {d}+\sqrt {e} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{e} \sqrt {x}}{\sqrt [4]{d}}\right ),\frac {1}{2}\right )}{3 d^{5/4} \sqrt {d+e x^2}}-\frac {2 \sqrt {d+e x^2}}{3 d x^{3/2}}\right )}{7 d}-\frac {2 \sqrt {d+e x^2}}{7 d x^{7/2}}\right )}{11 d}-\frac {2 \sqrt {d+e x^2}}{11 d x^{11/2}}\right )-\frac {2 \arctan \left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right )}{13 x^{13/2}}\)

Input:

Int[ArcTan[(Sqrt[-e]*x)/Sqrt[d + e*x^2]]/x^(15/2),x]
 

Output:

(-2*ArcTan[(Sqrt[-e]*x)/Sqrt[d + e*x^2]])/(13*x^(13/2)) + (2*Sqrt[-e]*((-2 
*Sqrt[d + e*x^2])/(11*d*x^(11/2)) - (9*e*((-2*Sqrt[d + e*x^2])/(7*d*x^(7/2 
)) - (5*e*((-2*Sqrt[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])))/(7*d)))/(11*d)))/1 
3
 

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]

\[\int \frac {\arctan \left (\frac {\sqrt {-e}\, x}{\sqrt {e \,x^{2}+d}}\right )}{x^{\frac {15}{2}}}d x\]

Input:

int(arctan((-e)^(1/2)*x/(e*x^2+d)^(1/2))/x^(15/2),x)
 

Output:

int(arctan((-e)^(1/2)*x/(e*x^2+d)^(1/2))/x^(15/2),x)
 

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.45 \[ \int \frac {\arctan \left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right )}{x^{15/2}} \, dx=-\frac {2 \, {\left (30 \, \sqrt {-e} e^{\frac {5}{2}} x^{7} {\rm weierstrassPInverse}\left (-\frac {4 \, d}{e}, 0, x\right ) + 77 \, d^{3} \sqrt {x} \arctan \left (\frac {\sqrt {-e} x}{\sqrt {e x^{2} + d}}\right ) + 2 \, {\left (15 \, e^{2} x^{5} - 9 \, d e x^{3} + 7 \, d^{2} x\right )} \sqrt {e x^{2} + d} \sqrt {-e} \sqrt {x}\right )}}{1001 \, d^{3} x^{7}} \] Input:

integrate(arctan((-e)^(1/2)*x/(e*x^2+d)^(1/2))/x^(15/2),x, algorithm="fric 
as")
 

Output:

-2/1001*(30*sqrt(-e)*e^(5/2)*x^7*weierstrassPInverse(-4*d/e, 0, x) + 77*d^ 
3*sqrt(x)*arctan(sqrt(-e)*x/sqrt(e*x^2 + d)) + 2*(15*e^2*x^5 - 9*d*e*x^3 + 
 7*d^2*x)*sqrt(e*x^2 + d)*sqrt(-e)*sqrt(x))/(d^3*x^7)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\arctan \left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right )}{x^{15/2}} \, dx=\text {Timed out} \] Input:

integrate(atan((-e)**(1/2)*x/(e*x**2+d)**(1/2))/x**(15/2),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {\arctan \left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right )}{x^{15/2}} \, dx=\int { \frac {\arctan \left (\frac {\sqrt {-e} x}{\sqrt {e x^{2} + d}}\right )}{x^{\frac {15}{2}}} \,d x } \] Input:

integrate(arctan((-e)^(1/2)*x/(e*x^2+d)^(1/2))/x^(15/2),x, algorithm="maxi 
ma")
 

Output:

2/13*(13*d*sqrt(-e)*x^(13/2)*integrate(-1/13*sqrt(e*x^2 + d)*x/((e^2*x^4 + 
 d*e*x^2)*x^(15/2) - (e*x^2 + d)*e^(log(e*x^2 + d) + 15/2*log(x))), x) - a 
rctan2(sqrt(-e)*x, sqrt(e*x^2 + d)))/x^(13/2)
 

Giac [F]

\[ \int \frac {\arctan \left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right )}{x^{15/2}} \, dx=\int { \frac {\arctan \left (\frac {\sqrt {-e} x}{\sqrt {e x^{2} + d}}\right )}{x^{\frac {15}{2}}} \,d x } \] Input:

integrate(arctan((-e)^(1/2)*x/(e*x^2+d)^(1/2))/x^(15/2),x, algorithm="giac 
")
 

Output:

integrate(arctan(sqrt(-e)*x/sqrt(e*x^2 + d))/x^(15/2), x)
                                                                                    
                                                                                    
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\arctan \left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right )}{x^{15/2}} \, dx=\int \frac {\mathrm {atan}\left (\frac {\sqrt {-e}\,x}{\sqrt {e\,x^2+d}}\right )}{x^{15/2}} \,d x \] Input:

int(atan(((-e)^(1/2)*x)/(d + e*x^2)^(1/2))/x^(15/2),x)
 

Output:

int(atan(((-e)^(1/2)*x)/(d + e*x^2)^(1/2))/x^(15/2), x)
 

Reduce [F]

\[ \int \frac {\arctan \left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right )}{x^{15/2}} \, dx=\int \frac {\sqrt {x}\, \mathit {atan} \left (\frac {\sqrt {e}\, i x}{\sqrt {e \,x^{2}+d}}\right )}{x^{8}}d x \] Input:

int(atan((-e)^(1/2)*x/(e*x^2+d)^(1/2))/x^(15/2),x)
 

Output:

int((sqrt(x)*atan((sqrt(e)*i*x)/sqrt(d + e*x**2)))/x**8,x)