Integrand size = 27, antiderivative size = 156 \[ \int \frac {\arctan \left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right )}{x^{7/2}} \, dx=-\frac {4 \sqrt {-e} \sqrt {d+e x^2}}{15 d x^{3/2}}-\frac {2 \arctan \left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right )}{5 x^{5/2}}-\frac {2 \sqrt {-e} e^{3/4} \left (\sqrt {d}+\sqrt {e} x\right ) \sqrt {\frac {d+e x^2}{\left (\sqrt {d}+\sqrt {e} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{e} \sqrt {x}}{\sqrt [4]{d}}\right ),\frac {1}{2}\right )}{15 d^{5/4} \sqrt {d+e x^2}} \]
-2/5*arctan(x*(-e)^(1/2)/(e*x^2+d)^(1/2))/x^(5/2)-4/15*(-e)^(1/2)*(e*x^2+d )^(1/2)/d/x^(3/2)-2/15*e^(3/4)*(cos(2*arctan(e^(1/4)*x^(1/2)/d^(1/4)))^2)^ (1/2)/cos(2*arctan(e^(1/4)*x^(1/2)/d^(1/4)))*EllipticF(sin(2*arctan(e^(1/4 )*x^(1/2)/d^(1/4))),1/2*2^(1/2))*(-e)^(1/2)*(d^(1/2)+x*e^(1/2))*((e*x^2+d) /(d^(1/2)+x*e^(1/2))^2)^(1/2)/d^(5/4)/(e*x^2+d)^(1/2)
Result contains complex when optimal does not.
Time = 0.20 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.96 \[ \int \frac {\arctan \left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right )}{x^{7/2}} \, dx=-\frac {2 \left (2 \sqrt {-e} x \sqrt {d+e x^2}+3 d \arctan \left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right )\right )}{15 d x^{5/2}}+\frac {4 i (-e)^{3/2} \sqrt {1+\frac {d}{e x^2}} x \operatorname {EllipticF}\left (i \text {arcsinh}\left (\frac {\sqrt {\frac {i \sqrt {d}}{\sqrt {e}}}}{\sqrt {x}}\right ),-1\right )}{15 d \sqrt {\frac {i \sqrt {d}}{\sqrt {e}}} \sqrt {d+e x^2}} \]
(-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])
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 |
\(\displaystyle \frac {2}{5} \sqrt {-e} \left (-\frac {e^{3/4} \left (\sqrt {d}+\sqrt {e} x\right ) \sqrt {\frac {d+e x^2}{\left (\sqrt {d}+\sqrt {e} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{e} \sqrt {x}}{\sqrt [4]{d}}\right ),\frac {1}{2}\right )}{3 d^{5/4} \sqrt {d+e x^2}}-\frac {2 \sqrt {d+e x^2}}{3 d x^{3/2}}\right )-\frac {2 \arctan \left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right )}{5 x^{5/2}}\) |
(-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
3.1.22.3.1 Defintions of rubi rules used
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]
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]
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]
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]
\[\int \frac {\arctan \left (\frac {x \sqrt {-e}}{\sqrt {e \,x^{2}+d}}\right )}{x^{\frac {7}{2}}}d x\]
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.10 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.47 \[ \int \frac {\arctan \left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right )}{x^{7/2}} \, dx=-\frac {2 \, {\left (2 \, \sqrt {-e} \sqrt {e} x^{3} {\rm weierstrassPInverse}\left (-\frac {4 \, d}{e}, 0, x\right ) + 2 \, \sqrt {e x^{2} + d} \sqrt {-e} x^{\frac {3}{2}} + 3 \, d \sqrt {x} \arctan \left (\frac {\sqrt {-e} x}{\sqrt {e x^{2} + d}}\right )\right )}}{15 \, d x^{3}} \]
-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)
Result contains complex when optimal does not.
Time = 44.90 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.50 \[ \int \frac {\arctan \left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right )}{x^{7/2}} \, dx=- \frac {2 \operatorname {atan}{\left (\frac {x \sqrt {- e}}{\sqrt {d + e x^{2}}} \right )}}{5 x^{\frac {5}{2}}} + \frac {\sqrt {- e} \Gamma \left (- \frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{4}, \frac {1}{2} \\ \frac {1}{4} \end {matrix}\middle | {\frac {e x^{2} e^{i \pi }}{d}} \right )}}{5 \sqrt {d} x^{\frac {3}{2}} \Gamma \left (\frac {1}{4}\right )} \]
-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))
\[ \int \frac {\arctan \left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right )}{x^{7/2}} \, dx=\int { \frac {\arctan \left (\frac {\sqrt {-e} x}{\sqrt {e x^{2} + d}}\right )}{x^{\frac {7}{2}}} \,d x } \]
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)
\[ \int \frac {\arctan \left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right )}{x^{7/2}} \, dx=\int { \frac {\arctan \left (\frac {\sqrt {-e} x}{\sqrt {e x^{2} + d}}\right )}{x^{\frac {7}{2}}} \,d x } \]
Timed out. \[ \int \frac {\arctan \left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right )}{x^{7/2}} \, dx=\int \frac {\mathrm {atan}\left (\frac {\sqrt {-e}\,x}{\sqrt {e\,x^2+d}}\right )}{x^{7/2}} \,d x \]