Integrand size = 25, antiderivative size = 119 \[ \int \frac {\arctan \left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right )}{x^6} \, dx=-\frac {\sqrt {-e} \sqrt {d+e x^2}}{20 d x^4}-\frac {3 (-e)^{3/2} \sqrt {d+e x^2}}{40 d^2 x^2}-\frac {\arctan \left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right )}{5 x^5}-\frac {3 (-e)^{5/2} \text {arctanh}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{40 d^{5/2}} \]
-1/5*arctan(x*(-e)^(1/2)/(e*x^2+d)^(1/2))/x^5-3/40*(-e)^(5/2)*arctanh((e*x ^2+d)^(1/2)/d^(1/2))/d^(5/2)-3/40*(-e)^(3/2)*(e*x^2+d)^(1/2)/d^2/x^2-1/20* (-e)^(1/2)*(e*x^2+d)^(1/2)/d/x^4
Time = 0.08 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.96 \[ \int \frac {\arctan \left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right )}{x^6} \, dx=\sqrt {-e} \left (-\frac {1}{20 d x^4}+\frac {3 e}{40 d^2 x^2}\right ) \sqrt {d+e x^2}-\frac {3 e^{5/2} \arctan \left (\frac {\sqrt {d} \sqrt {-e}}{\sqrt {e} \sqrt {d+e x^2}}\right )}{40 d^{5/2}}-\frac {\arctan \left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right )}{5 x^5} \]
Sqrt[-e]*(-1/20*1/(d*x^4) + (3*e)/(40*d^2*x^2))*Sqrt[d + e*x^2] - (3*e^(5/ 2)*ArcTan[(Sqrt[d]*Sqrt[-e])/(Sqrt[e]*Sqrt[d + e*x^2])])/(40*d^(5/2)) - Ar cTan[(Sqrt[-e]*x)/Sqrt[d + e*x^2]]/(5*x^5)
Time = 0.26 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.97, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {5674, 243, 52, 52, 73, 221}
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.
\(\displaystyle \int \frac {\arctan \left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right )}{x^6} \, dx\) |
\(\Big \downarrow \) 5674 |
\(\displaystyle \frac {1}{5} \sqrt {-e} \int \frac {1}{x^5 \sqrt {e x^2+d}}dx-\frac {\arctan \left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right )}{5 x^5}\) |
\(\Big \downarrow \) 243 |
\(\displaystyle \frac {1}{10} \sqrt {-e} \int \frac {1}{x^6 \sqrt {e x^2+d}}dx^2-\frac {\arctan \left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right )}{5 x^5}\) |
\(\Big \downarrow \) 52 |
\(\displaystyle \frac {1}{10} \sqrt {-e} \left (-\frac {3 e \int \frac {1}{x^4 \sqrt {e x^2+d}}dx^2}{4 d}-\frac {\sqrt {d+e x^2}}{2 d x^4}\right )-\frac {\arctan \left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right )}{5 x^5}\) |
\(\Big \downarrow \) 52 |
\(\displaystyle \frac {1}{10} \sqrt {-e} \left (-\frac {3 e \left (-\frac {e \int \frac {1}{x^2 \sqrt {e x^2+d}}dx^2}{2 d}-\frac {\sqrt {d+e x^2}}{d x^2}\right )}{4 d}-\frac {\sqrt {d+e x^2}}{2 d x^4}\right )-\frac {\arctan \left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right )}{5 x^5}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {1}{10} \sqrt {-e} \left (-\frac {3 e \left (-\frac {\int \frac {1}{\frac {x^4}{e}-\frac {d}{e}}d\sqrt {e x^2+d}}{d}-\frac {\sqrt {d+e x^2}}{d x^2}\right )}{4 d}-\frac {\sqrt {d+e x^2}}{2 d x^4}\right )-\frac {\arctan \left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right )}{5 x^5}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {1}{10} \sqrt {-e} \left (-\frac {3 e \left (\frac {e \text {arctanh}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{d^{3/2}}-\frac {\sqrt {d+e x^2}}{d x^2}\right )}{4 d}-\frac {\sqrt {d+e x^2}}{2 d x^4}\right )-\frac {\arctan \left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right )}{5 x^5}\) |
-1/5*ArcTan[(Sqrt[-e]*x)/Sqrt[d + e*x^2]]/x^5 + (Sqrt[-e]*(-1/2*Sqrt[d + e *x^2]/(d*x^4) - (3*e*(-(Sqrt[d + e*x^2]/(d*x^2)) + (e*ArcTanh[Sqrt[d + e*x ^2]/Sqrt[d]])/d^(3/2)))/(4*d)))/10
3.1.17.3.1 Defintions of rubi rules used
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(( m + n + 2)/((b*c - a*d)*(m + 1))) Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && ILtQ[m, -1] && FractionQ[n] && LtQ[n, 0]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/2 Subst[In t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I ntegerQ[(m - 1)/2]
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]
Time = 0.03 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.50
method | result | size |
default | \(-\frac {\arctan \left (\frac {x \sqrt {-e}}{\sqrt {e \,x^{2}+d}}\right )}{5 x^{5}}-\frac {\sqrt {-e}\, e \left (-\frac {\sqrt {e \,x^{2}+d}}{2 d \,x^{2}}+\frac {e \ln \left (\frac {2 d +2 \sqrt {d}\, \sqrt {e \,x^{2}+d}}{x}\right )}{2 d^{\frac {3}{2}}}\right )}{5 d}+\frac {\sqrt {-e}\, \left (-\frac {\left (e \,x^{2}+d \right )^{\frac {3}{2}}}{4 d \,x^{4}}-\frac {e \left (-\frac {\left (e \,x^{2}+d \right )^{\frac {3}{2}}}{2 d \,x^{2}}+\frac {e \left (\sqrt {e \,x^{2}+d}-\sqrt {d}\, \ln \left (\frac {2 d +2 \sqrt {d}\, \sqrt {e \,x^{2}+d}}{x}\right )\right )}{2 d}\right )}{4 d}\right )}{5 d}\) | \(178\) |
parts | \(-\frac {\arctan \left (\frac {x \sqrt {-e}}{\sqrt {e \,x^{2}+d}}\right )}{5 x^{5}}-\frac {\sqrt {-e}\, e \left (-\frac {\sqrt {e \,x^{2}+d}}{2 d \,x^{2}}+\frac {e \ln \left (\frac {2 d +2 \sqrt {d}\, \sqrt {e \,x^{2}+d}}{x}\right )}{2 d^{\frac {3}{2}}}\right )}{5 d}+\frac {\sqrt {-e}\, \left (-\frac {\left (e \,x^{2}+d \right )^{\frac {3}{2}}}{4 d \,x^{4}}-\frac {e \left (-\frac {\left (e \,x^{2}+d \right )^{\frac {3}{2}}}{2 d \,x^{2}}+\frac {e \left (\sqrt {e \,x^{2}+d}-\sqrt {d}\, \ln \left (\frac {2 d +2 \sqrt {d}\, \sqrt {e \,x^{2}+d}}{x}\right )\right )}{2 d}\right )}{4 d}\right )}{5 d}\) | \(178\) |
-1/5*arctan(x*(-e)^(1/2)/(e*x^2+d)^(1/2))/x^5-1/5*(-e)^(1/2)*e/d*(-1/2/d/x ^2*(e*x^2+d)^(1/2)+1/2*e/d^(3/2)*ln((2*d+2*d^(1/2)*(e*x^2+d)^(1/2))/x))+1/ 5*(-e)^(1/2)/d*(-1/4/d/x^4*(e*x^2+d)^(3/2)-1/4*e/d*(-1/2/d/x^2*(e*x^2+d)^( 3/2)+1/2*e/d*((e*x^2+d)^(1/2)-d^(1/2)*ln((2*d+2*d^(1/2)*(e*x^2+d)^(1/2))/x ))))
Time = 0.31 (sec) , antiderivative size = 228, normalized size of antiderivative = 1.92 \[ \int \frac {\arctan \left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right )}{x^6} \, dx=\left [\frac {3 \, e^{2} x^{5} \sqrt {-\frac {e}{d}} \log \left (-\frac {e^{2} x^{2} + 2 \, \sqrt {e x^{2} + d} d \sqrt {-e} \sqrt {-\frac {e}{d}} + 2 \, d e}{x^{2}}\right ) - 16 \, d^{2} \arctan \left (\frac {\sqrt {-e} x}{\sqrt {e x^{2} + d}}\right ) + 2 \, {\left (3 \, e x^{3} - 2 \, d x\right )} \sqrt {e x^{2} + d} \sqrt {-e}}{80 \, d^{2} x^{5}}, -\frac {3 \, e^{2} x^{5} \sqrt {\frac {e}{d}} \arctan \left (\frac {\sqrt {e x^{2} + d} d \sqrt {-e} \sqrt {\frac {e}{d}}}{e^{2} x^{2} + d e}\right ) + 8 \, d^{2} \arctan \left (\frac {\sqrt {-e} x}{\sqrt {e x^{2} + d}}\right ) - {\left (3 \, e x^{3} - 2 \, d x\right )} \sqrt {e x^{2} + d} \sqrt {-e}}{40 \, d^{2} x^{5}}\right ] \]
[1/80*(3*e^2*x^5*sqrt(-e/d)*log(-(e^2*x^2 + 2*sqrt(e*x^2 + d)*d*sqrt(-e)*s qrt(-e/d) + 2*d*e)/x^2) - 16*d^2*arctan(sqrt(-e)*x/sqrt(e*x^2 + d)) + 2*(3 *e*x^3 - 2*d*x)*sqrt(e*x^2 + d)*sqrt(-e))/(d^2*x^5), -1/40*(3*e^2*x^5*sqrt (e/d)*arctan(sqrt(e*x^2 + d)*d*sqrt(-e)*sqrt(e/d)/(e^2*x^2 + d*e)) + 8*d^2 *arctan(sqrt(-e)*x/sqrt(e*x^2 + d)) - (3*e*x^3 - 2*d*x)*sqrt(e*x^2 + d)*sq rt(-e))/(d^2*x^5)]
Time = 6.19 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.24 \[ \int \frac {\arctan \left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right )}{x^6} \, dx=- \frac {\operatorname {atan}{\left (\frac {x \sqrt {- e}}{\sqrt {d + e x^{2}}} \right )}}{5 x^{5}} - \frac {\sqrt {- e}}{20 \sqrt {e} x^{5} \sqrt {\frac {d}{e x^{2}} + 1}} + \frac {\sqrt {e} \sqrt {- e}}{40 d x^{3} \sqrt {\frac {d}{e x^{2}} + 1}} + \frac {3 e^{\frac {3}{2}} \sqrt {- e}}{40 d^{2} x \sqrt {\frac {d}{e x^{2}} + 1}} - \frac {3 e^{2} \sqrt {- e} \operatorname {asinh}{\left (\frac {\sqrt {d}}{\sqrt {e} x} \right )}}{40 d^{\frac {5}{2}}} \]
-atan(x*sqrt(-e)/sqrt(d + e*x**2))/(5*x**5) - sqrt(-e)/(20*sqrt(e)*x**5*sq rt(d/(e*x**2) + 1)) + sqrt(e)*sqrt(-e)/(40*d*x**3*sqrt(d/(e*x**2) + 1)) + 3*e**(3/2)*sqrt(-e)/(40*d**2*x*sqrt(d/(e*x**2) + 1)) - 3*e**2*sqrt(-e)*asi nh(sqrt(d)/(sqrt(e)*x))/(40*d**(5/2))
\[ \int \frac {\arctan \left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right )}{x^6} \, dx=\int { \frac {\arctan \left (\frac {\sqrt {-e} x}{\sqrt {e x^{2} + d}}\right )}{x^{6}} \,d x } \]
1/5*(5*d*sqrt(-e)*x^5*integrate(-1/5*sqrt(e*x^2 + d)/(e^2*x^9 + d*e*x^7 - (e*x^7 + d*x^5)*(e*x^2 + d)), x) - arctan2(sqrt(-e)*x, sqrt(e*x^2 + d)))/x ^5
Time = 0.34 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.97 \[ \int \frac {\arctan \left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right )}{x^6} \, dx=-\frac {\frac {3 \, e^{4} \arctan \left (\frac {\sqrt {-e^{2} x^{2} - d e}}{\sqrt {d e}}\right )}{\sqrt {d e} d^{2}} + \frac {5 \, \sqrt {-e^{2} x^{2} - d e} d e^{5} + 3 \, {\left (-e^{2} x^{2} - d e\right )}^{\frac {3}{2}} e^{4}}{d^{2} e^{4} x^{4}}}{40 \, e} - \frac {\arctan \left (\frac {\sqrt {-e} x}{\sqrt {e x^{2} + d}}\right )}{5 \, x^{5}} \]
-1/40*(3*e^4*arctan(sqrt(-e^2*x^2 - d*e)/sqrt(d*e))/(sqrt(d*e)*d^2) + (5*s qrt(-e^2*x^2 - d*e)*d*e^5 + 3*(-e^2*x^2 - d*e)^(3/2)*e^4)/(d^2*e^4*x^4))/e - 1/5*arctan(sqrt(-e)*x/sqrt(e*x^2 + d))/x^5
Timed out. \[ \int \frac {\arctan \left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right )}{x^6} \, dx=\int \frac {\mathrm {atan}\left (\frac {\sqrt {-e}\,x}{\sqrt {e\,x^2+d}}\right )}{x^6} \,d x \]