Integrand size = 22, antiderivative size = 242 \[ \int \frac {\arctan (a x)}{x^3 \sqrt {c+a^2 c x^2}} \, dx=-\frac {a \sqrt {c+a^2 c x^2}}{2 c x}-\frac {\sqrt {c+a^2 c x^2} \arctan (a x)}{2 c x^2}+\frac {a^2 \sqrt {1+a^2 x^2} \arctan (a x) \text {arctanh}\left (\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{\sqrt {c+a^2 c x^2}}-\frac {i a^2 \sqrt {1+a^2 x^2} \operatorname {PolyLog}\left (2,-\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{2 \sqrt {c+a^2 c x^2}}+\frac {i a^2 \sqrt {1+a^2 x^2} \operatorname {PolyLog}\left (2,\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{2 \sqrt {c+a^2 c x^2}} \]
a^2*arctan(a*x)*arctanh((1+I*a*x)^(1/2)/(1-I*a*x)^(1/2))*(a^2*x^2+1)^(1/2) /(a^2*c*x^2+c)^(1/2)-1/2*I*a^2*polylog(2,-(1+I*a*x)^(1/2)/(1-I*a*x)^(1/2)) *(a^2*x^2+1)^(1/2)/(a^2*c*x^2+c)^(1/2)+1/2*I*a^2*polylog(2,(1+I*a*x)^(1/2) /(1-I*a*x)^(1/2))*(a^2*x^2+1)^(1/2)/(a^2*c*x^2+c)^(1/2)-1/2*a*(a^2*c*x^2+c )^(1/2)/c/x-1/2*arctan(a*x)*(a^2*c*x^2+c)^(1/2)/c/x^2
Time = 0.76 (sec) , antiderivative size = 165, normalized size of antiderivative = 0.68 \[ \int \frac {\arctan (a x)}{x^3 \sqrt {c+a^2 c x^2}} \, dx=\frac {a^2 \sqrt {1+a^2 x^2} \left (-2 \cot \left (\frac {1}{2} \arctan (a x)\right )-\arctan (a x) \csc ^2\left (\frac {1}{2} \arctan (a x)\right )-4 \arctan (a x) \log \left (1-e^{i \arctan (a x)}\right )+4 \arctan (a x) \log \left (1+e^{i \arctan (a x)}\right )-4 i \operatorname {PolyLog}\left (2,-e^{i \arctan (a x)}\right )+4 i \operatorname {PolyLog}\left (2,e^{i \arctan (a x)}\right )+\arctan (a x) \sec ^2\left (\frac {1}{2} \arctan (a x)\right )-2 \tan \left (\frac {1}{2} \arctan (a x)\right )\right )}{8 \sqrt {c \left (1+a^2 x^2\right )}} \]
(a^2*Sqrt[1 + a^2*x^2]*(-2*Cot[ArcTan[a*x]/2] - ArcTan[a*x]*Csc[ArcTan[a*x ]/2]^2 - 4*ArcTan[a*x]*Log[1 - E^(I*ArcTan[a*x])] + 4*ArcTan[a*x]*Log[1 + E^(I*ArcTan[a*x])] - (4*I)*PolyLog[2, -E^(I*ArcTan[a*x])] + (4*I)*PolyLog[ 2, E^(I*ArcTan[a*x])] + ArcTan[a*x]*Sec[ArcTan[a*x]/2]^2 - 2*Tan[ArcTan[a* x]/2]))/(8*Sqrt[c*(1 + a^2*x^2)])
Time = 0.52 (sec) , antiderivative size = 184, normalized size of antiderivative = 0.76, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {5497, 242, 5493, 5489}
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.
| \(\displaystyle \int \frac {\arctan (a x)}{x^3 \sqrt {a^2 c x^2+c}} \, dx\) |
| \(\Big \downarrow \) 5497 |
| \(\displaystyle -\frac {1}{2} a^2 \int \frac {\arctan (a x)}{x \sqrt {a^2 c x^2+c}}dx+\frac {1}{2} a \int \frac {1}{x^2 \sqrt {a^2 c x^2+c}}dx-\frac {\arctan (a x) \sqrt {a^2 c x^2+c}}{2 c x^2}\) |
| \(\Big \downarrow \) 242 |
| \(\displaystyle -\frac {1}{2} a^2 \int \frac {\arctan (a x)}{x \sqrt {a^2 c x^2+c}}dx-\frac {\arctan (a x) \sqrt {a^2 c x^2+c}}{2 c x^2}-\frac {a \sqrt {a^2 c x^2+c}}{2 c x}\) |
| \(\Big \downarrow \) 5493 |
| \(\displaystyle -\frac {a^2 \sqrt {a^2 x^2+1} \int \frac {\arctan (a x)}{x \sqrt {a^2 x^2+1}}dx}{2 \sqrt {a^2 c x^2+c}}-\frac {\arctan (a x) \sqrt {a^2 c x^2+c}}{2 c x^2}-\frac {a \sqrt {a^2 c x^2+c}}{2 c x}\) |
| \(\Big \downarrow \) 5489 |
| \(\displaystyle -\frac {a^2 \sqrt {a^2 x^2+1} \left (-2 \arctan (a x) \text {arctanh}\left (\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right )+i \operatorname {PolyLog}\left (2,-\frac {\sqrt {i a x+1}}{\sqrt {1-i a x}}\right )-i \operatorname {PolyLog}\left (2,\frac {\sqrt {i a x+1}}{\sqrt {1-i a x}}\right )\right )}{2 \sqrt {a^2 c x^2+c}}-\frac {\arctan (a x) \sqrt {a^2 c x^2+c}}{2 c x^2}-\frac {a \sqrt {a^2 c x^2+c}}{2 c x}\) |
-1/2*(a*Sqrt[c + a^2*c*x^2])/(c*x) - (Sqrt[c + a^2*c*x^2]*ArcTan[a*x])/(2* c*x^2) - (a^2*Sqrt[1 + a^2*x^2]*(-2*ArcTan[a*x]*ArcTanh[Sqrt[1 + I*a*x]/Sq rt[1 - I*a*x]] + I*PolyLog[2, -(Sqrt[1 + I*a*x]/Sqrt[1 - I*a*x])] - I*Poly Log[2, Sqrt[1 + I*a*x]/Sqrt[1 - I*a*x]]))/(2*Sqrt[c + a^2*c*x^2])
3.3.30.3.1 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] /; FreeQ[{a, b, c, m, p}, x ] && EqQ[m + 2*p + 3, 0] && NeQ[m, -1]
Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))/((x_)*Sqrt[(d_) + (e_.)*(x_)^2]), x_ Symbol] :> Simp[(-2/Sqrt[d])*(a + b*ArcTan[c*x])*ArcTanh[Sqrt[1 + I*c*x]/Sq rt[1 - I*c*x]], x] + (Simp[I*(b/Sqrt[d])*PolyLog[2, -Sqrt[1 + I*c*x]/Sqrt[1 - I*c*x]], x] - Simp[I*(b/Sqrt[d])*PolyLog[2, Sqrt[1 + I*c*x]/Sqrt[1 - I*c *x]], x]) /; FreeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*d] && GtQ[d, 0]
Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)/((x_)*Sqrt[(d_) + (e_.)*(x_)^2 ]), x_Symbol] :> Simp[Sqrt[1 + c^2*x^2]/Sqrt[d + e*x^2] Int[(a + b*ArcTan [c*x])^p/(x*Sqrt[1 + c^2*x^2]), x], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[ e, c^2*d] && IGtQ[p, 0] && !GtQ[d, 0]
Int[(((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_))/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp[(f*x)^(m + 1)*Sqrt[d + e*x^2]*((a + b*Ar cTan[c*x])^p/(d*f*(m + 1))), x] + (-Simp[b*c*(p/(f*(m + 1))) Int[(f*x)^(m + 1)*((a + b*ArcTan[c*x])^(p - 1)/Sqrt[d + e*x^2]), x], x] - Simp[c^2*((m + 2)/(f^2*(m + 1))) Int[(f*x)^(m + 2)*((a + b*ArcTan[c*x])^p/Sqrt[d + e*x ^2]), x], x]) /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[e, c^2*d] && GtQ[p, 0] && LtQ[m, -1] && NeQ[m, -2]
Time = 0.45 (sec) , antiderivative size = 175, normalized size of antiderivative = 0.72
| method | result | size |
| default | \(-\frac {\left (a x +\arctan \left (a x \right )\right ) \sqrt {c \left (a x -i\right ) \left (a x +i\right )}}{2 c \,x^{2}}-\frac {i a^{2} \left (i \arctan \left (a x \right ) \ln \left (\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}+1\right )-i \arctan \left (a x \right ) \ln \left (1-\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right )+\operatorname {polylog}\left (2, -\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right )-\operatorname {polylog}\left (2, \frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right )\right ) \sqrt {c \left (a x -i\right ) \left (a x +i\right )}}{2 \sqrt {a^{2} x^{2}+1}\, c}\) | \(175\) |
-1/2*(a*x+arctan(a*x))*(c*(a*x-I)*(I+a*x))^(1/2)/c/x^2-1/2*I*a^2*(I*arctan (a*x)*ln((1+I*a*x)/(a^2*x^2+1)^(1/2)+1)-I*arctan(a*x)*ln(1-(1+I*a*x)/(a^2* x^2+1)^(1/2))+polylog(2,-(1+I*a*x)/(a^2*x^2+1)^(1/2))-polylog(2,(1+I*a*x)/ (a^2*x^2+1)^(1/2)))*(c*(a*x-I)*(I+a*x))^(1/2)/(a^2*x^2+1)^(1/2)/c
\[ \int \frac {\arctan (a x)}{x^3 \sqrt {c+a^2 c x^2}} \, dx=\int { \frac {\arctan \left (a x\right )}{\sqrt {a^{2} c x^{2} + c} x^{3}} \,d x } \]
\[ \int \frac {\arctan (a x)}{x^3 \sqrt {c+a^2 c x^2}} \, dx=\int \frac {\operatorname {atan}{\left (a x \right )}}{x^{3} \sqrt {c \left (a^{2} x^{2} + 1\right )}}\, dx \]
\[ \int \frac {\arctan (a x)}{x^3 \sqrt {c+a^2 c x^2}} \, dx=\int { \frac {\arctan \left (a x\right )}{\sqrt {a^{2} c x^{2} + c} x^{3}} \,d x } \]
\[ \int \frac {\arctan (a x)}{x^3 \sqrt {c+a^2 c x^2}} \, dx=\int { \frac {\arctan \left (a x\right )}{\sqrt {a^{2} c x^{2} + c} x^{3}} \,d x } \]
Timed out. \[ \int \frac {\arctan (a x)}{x^3 \sqrt {c+a^2 c x^2}} \, dx=\int \frac {\mathrm {atan}\left (a\,x\right )}{x^3\,\sqrt {c\,a^2\,x^2+c}} \,d x \]