Integrand size = 25, antiderivative size = 91 \[ \int \frac {\arctan \left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right )}{x^4} \, dx=-\frac {\sqrt {-e} \sqrt {d+e x^2}}{6 d x^2}-\frac {\arctan \left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right )}{3 x^3}-\frac {(-e)^{3/2} \text {arctanh}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{6 d^{3/2}} \]
-1/3*arctan(x*(-e)^(1/2)/(e*x^2+d)^(1/2))/x^3-1/6*(-e)^(3/2)*arctanh((e*x^ 2+d)^(1/2)/d^(1/2))/d^(3/2)-1/6*(-e)^(1/2)*(e*x^2+d)^(1/2)/d/x^2
Time = 0.06 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.11 \[ \int \frac {\arctan \left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right )}{x^4} \, dx=-\frac {\sqrt {-e} \sqrt {d+e x^2}}{6 d x^2}+\frac {e^{3/2} \arctan \left (\frac {\sqrt {d} \sqrt {-e}}{\sqrt {e} \sqrt {d+e x^2}}\right )}{6 d^{3/2}}-\frac {\arctan \left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right )}{3 x^3} \]
-1/6*(Sqrt[-e]*Sqrt[d + e*x^2])/(d*x^2) + (e^(3/2)*ArcTan[(Sqrt[d]*Sqrt[-e ])/(Sqrt[e]*Sqrt[d + e*x^2])])/(6*d^(3/2)) - ArcTan[(Sqrt[-e]*x)/Sqrt[d + e*x^2]]/(3*x^3)
Time = 0.25 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.93, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {5674, 243, 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.
\(\displaystyle \int \frac {\arctan \left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right )}{x^4} \, dx\) |
\(\Big \downarrow \) 5674 |
\(\displaystyle \frac {1}{3} \sqrt {-e} \int \frac {1}{x^3 \sqrt {e x^2+d}}dx-\frac {\arctan \left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right )}{3 x^3}\) |
\(\Big \downarrow \) 243 |
\(\displaystyle \frac {1}{6} \sqrt {-e} \int \frac {1}{x^4 \sqrt {e x^2+d}}dx^2-\frac {\arctan \left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right )}{3 x^3}\) |
\(\Big \downarrow \) 52 |
\(\displaystyle \frac {1}{6} \sqrt {-e} \left (-\frac {e \int \frac {1}{x^2 \sqrt {e x^2+d}}dx^2}{2 d}-\frac {\sqrt {d+e x^2}}{d x^2}\right )-\frac {\arctan \left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right )}{3 x^3}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {1}{6} \sqrt {-e} \left (-\frac {\int \frac {1}{\frac {x^4}{e}-\frac {d}{e}}d\sqrt {e x^2+d}}{d}-\frac {\sqrt {d+e x^2}}{d x^2}\right )-\frac {\arctan \left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right )}{3 x^3}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {1}{6} \sqrt {-e} \left (\frac {e \text {arctanh}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{d^{3/2}}-\frac {\sqrt {d+e x^2}}{d x^2}\right )-\frac {\arctan \left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right )}{3 x^3}\) |
-1/3*ArcTan[(Sqrt[-e]*x)/Sqrt[d + e*x^2]]/x^3 + (Sqrt[-e]*(-(Sqrt[d + e*x^ 2]/(d*x^2)) + (e*ArcTanh[Sqrt[d + e*x^2]/Sqrt[d]])/d^(3/2)))/6
3.1.16.3.1 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 = 0.02 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.43
method | result | size |
default | \(-\frac {\arctan \left (\frac {x \sqrt {-e}}{\sqrt {e \,x^{2}+d}}\right )}{3 x^{3}}+\frac {\sqrt {-e}\, e \ln \left (\frac {2 d +2 \sqrt {d}\, \sqrt {e \,x^{2}+d}}{x}\right )}{3 d^{\frac {3}{2}}}+\frac {\sqrt {-e}\, \left (-\frac {\left (e \,x^{2}+d \right )^{\frac {3}{2}}}{2 d \,x^{2}}+\frac {e \left (\sqrt {e \,x^{2}+d}-\sqrt {d}\, \ln \left (\frac {2 d +2 \sqrt {d}\, \sqrt {e \,x^{2}+d}}{x}\right )\right )}{2 d}\right )}{3 d}\) | \(130\) |
parts | \(-\frac {\arctan \left (\frac {x \sqrt {-e}}{\sqrt {e \,x^{2}+d}}\right )}{3 x^{3}}+\frac {\sqrt {-e}\, e \ln \left (\frac {2 d +2 \sqrt {d}\, \sqrt {e \,x^{2}+d}}{x}\right )}{3 d^{\frac {3}{2}}}+\frac {\sqrt {-e}\, \left (-\frac {\left (e \,x^{2}+d \right )^{\frac {3}{2}}}{2 d \,x^{2}}+\frac {e \left (\sqrt {e \,x^{2}+d}-\sqrt {d}\, \ln \left (\frac {2 d +2 \sqrt {d}\, \sqrt {e \,x^{2}+d}}{x}\right )\right )}{2 d}\right )}{3 d}\) | \(130\) |
-1/3*arctan(x*(-e)^(1/2)/(e*x^2+d)^(1/2))/x^3+1/3*(-e)^(1/2)*e/d^(3/2)*ln( (2*d+2*d^(1/2)*(e*x^2+d)^(1/2))/x)+1/3*(-e)^(1/2)/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 = 0.30 (sec) , antiderivative size = 198, normalized size of antiderivative = 2.18 \[ \int \frac {\arctan \left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right )}{x^4} \, dx=\left [\frac {e x^{3} \sqrt {-\frac {e}{d}} \log \left (-\frac {e^{2} x^{2} - 2 \, \sqrt {e x^{2} + d} d \sqrt {-e} \sqrt {-\frac {e}{d}} + 2 \, d e}{x^{2}}\right ) - 2 \, \sqrt {e x^{2} + d} \sqrt {-e} x - 4 \, d \arctan \left (\frac {\sqrt {-e} x}{\sqrt {e x^{2} + d}}\right )}{12 \, d x^{3}}, \frac {e x^{3} \sqrt {\frac {e}{d}} \arctan \left (\frac {\sqrt {e x^{2} + d} d \sqrt {-e} \sqrt {\frac {e}{d}}}{e^{2} x^{2} + d e}\right ) - \sqrt {e x^{2} + d} \sqrt {-e} x - 2 \, d \arctan \left (\frac {\sqrt {-e} x}{\sqrt {e x^{2} + d}}\right )}{6 \, d x^{3}}\right ] \]
[1/12*(e*x^3*sqrt(-e/d)*log(-(e^2*x^2 - 2*sqrt(e*x^2 + d)*d*sqrt(-e)*sqrt( -e/d) + 2*d*e)/x^2) - 2*sqrt(e*x^2 + d)*sqrt(-e)*x - 4*d*arctan(sqrt(-e)*x /sqrt(e*x^2 + d)))/(d*x^3), 1/6*(e*x^3*sqrt(e/d)*arctan(sqrt(e*x^2 + d)*d* sqrt(-e)*sqrt(e/d)/(e^2*x^2 + d*e)) - sqrt(e*x^2 + d)*sqrt(-e)*x - 2*d*arc tan(sqrt(-e)*x/sqrt(e*x^2 + d)))/(d*x^3)]
Time = 3.89 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.90 \[ \int \frac {\arctan \left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right )}{x^4} \, dx=- \frac {\operatorname {atan}{\left (\frac {x \sqrt {- e}}{\sqrt {d + e x^{2}}} \right )}}{3 x^{3}} - \frac {\sqrt {e} \sqrt {- e} \sqrt {\frac {d}{e x^{2}} + 1}}{6 d x} + \frac {e \sqrt {- e} \operatorname {asinh}{\left (\frac {\sqrt {d}}{\sqrt {e} x} \right )}}{6 d^{\frac {3}{2}}} \]
-atan(x*sqrt(-e)/sqrt(d + e*x**2))/(3*x**3) - sqrt(e)*sqrt(-e)*sqrt(d/(e*x **2) + 1)/(6*d*x) + e*sqrt(-e)*asinh(sqrt(d)/(sqrt(e)*x))/(6*d**(3/2))
\[ \int \frac {\arctan \left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right )}{x^4} \, dx=\int { \frac {\arctan \left (\frac {\sqrt {-e} x}{\sqrt {e x^{2} + d}}\right )}{x^{4}} \,d x } \]
1/3*(3*d*sqrt(-e)*x^3*integrate(-1/3*sqrt(e*x^2 + d)/(e^2*x^7 + d*e*x^5 - (e*x^5 + d*x^3)*(e*x^2 + d)), x) - arctan2(sqrt(-e)*x, sqrt(e*x^2 + d)))/x ^3
Time = 0.34 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.96 \[ \int \frac {\arctan \left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right )}{x^4} \, dx=\frac {\frac {e^{3} \arctan \left (\frac {\sqrt {-e^{2} x^{2} - d e}}{\sqrt {d e}}\right )}{\sqrt {d e} d} - \frac {\sqrt {-e^{2} x^{2} - d e} e}{d x^{2}}}{6 \, e} - \frac {\arctan \left (\frac {\sqrt {-e} x}{\sqrt {e x^{2} + d}}\right )}{3 \, x^{3}} \]
1/6*(e^3*arctan(sqrt(-e^2*x^2 - d*e)/sqrt(d*e))/(sqrt(d*e)*d) - sqrt(-e^2* x^2 - d*e)*e/(d*x^2))/e - 1/3*arctan(sqrt(-e)*x/sqrt(e*x^2 + d))/x^3
Timed out. \[ \int \frac {\arctan \left (\frac {\sqrt {-e} x}{\sqrt {d+e x^2}}\right )}{x^4} \, dx=\int \frac {\mathrm {atan}\left (\frac {\sqrt {-e}\,x}{\sqrt {e\,x^2+d}}\right )}{x^4} \,d x \]