Integrand size = 22, antiderivative size = 229 \[ \int \frac {\sqrt {c+a^2 c x^2} \arctan (a x)}{x} \, dx=\sqrt {c+a^2 c x^2} \arctan (a x)-\frac {2 c \sqrt {1+a^2 x^2} \arctan (a x) \text {arctanh}\left (\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{\sqrt {c+a^2 c x^2}}-\sqrt {c} \text {arctanh}\left (\frac {a \sqrt {c} x}{\sqrt {c+a^2 c x^2}}\right )+\frac {i c \sqrt {1+a^2 x^2} \operatorname {PolyLog}\left (2,-\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{\sqrt {c+a^2 c x^2}}-\frac {i c \sqrt {1+a^2 x^2} \operatorname {PolyLog}\left (2,\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{\sqrt {c+a^2 c x^2}} \]
-arctanh(a*x*c^(1/2)/(a^2*c*x^2+c)^(1/2))*c^(1/2)-2*c*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)+I*c *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)-I*c*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)+arctan(a*x)*(a^2*c*x^2+c)^(1/2)
Time = 0.25 (sec) , antiderivative size = 164, normalized size of antiderivative = 0.72 \[ \int \frac {\sqrt {c+a^2 c x^2} \arctan (a x)}{x} \, dx=\frac {\sqrt {c+a^2 c x^2} \left (\sqrt {1+a^2 x^2} \arctan (a x)+\arctan (a x) \log \left (1-e^{i \arctan (a x)}\right )-\arctan (a x) \log \left (1+e^{i \arctan (a x)}\right )+\log \left (\cos \left (\frac {1}{2} \arctan (a x)\right )-\sin \left (\frac {1}{2} \arctan (a x)\right )\right )-\log \left (\cos \left (\frac {1}{2} \arctan (a x)\right )+\sin \left (\frac {1}{2} \arctan (a x)\right )\right )+i \operatorname {PolyLog}\left (2,-e^{i \arctan (a x)}\right )-i \operatorname {PolyLog}\left (2,e^{i \arctan (a x)}\right )\right )}{\sqrt {1+a^2 x^2}} \]
(Sqrt[c + a^2*c*x^2]*(Sqrt[1 + a^2*x^2]*ArcTan[a*x] + ArcTan[a*x]*Log[1 - E^(I*ArcTan[a*x])] - ArcTan[a*x]*Log[1 + E^(I*ArcTan[a*x])] + Log[Cos[ArcT an[a*x]/2] - Sin[ArcTan[a*x]/2]] - Log[Cos[ArcTan[a*x]/2] + Sin[ArcTan[a*x ]/2]] + I*PolyLog[2, -E^(I*ArcTan[a*x])] - I*PolyLog[2, E^(I*ArcTan[a*x])] ))/Sqrt[1 + a^2*x^2]
Time = 0.53 (sec) , antiderivative size = 175, normalized size of antiderivative = 0.76, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {5481, 224, 219, 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.
\(\displaystyle \int \frac {\arctan (a x) \sqrt {a^2 c x^2+c}}{x} \, dx\) |
\(\Big \downarrow \) 5481 |
\(\displaystyle c \int \frac {\arctan (a x)}{x \sqrt {a^2 c x^2+c}}dx-a c \int \frac {1}{\sqrt {a^2 c x^2+c}}dx+\arctan (a x) \sqrt {a^2 c x^2+c}\) |
\(\Big \downarrow \) 224 |
\(\displaystyle c \int \frac {\arctan (a x)}{x \sqrt {a^2 c x^2+c}}dx-a c \int \frac {1}{1-\frac {a^2 c x^2}{a^2 c x^2+c}}d\frac {x}{\sqrt {a^2 c x^2+c}}+\arctan (a x) \sqrt {a^2 c x^2+c}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle c \int \frac {\arctan (a x)}{x \sqrt {a^2 c x^2+c}}dx+\arctan (a x) \sqrt {a^2 c x^2+c}-\sqrt {c} \text {arctanh}\left (\frac {a \sqrt {c} x}{\sqrt {a^2 c x^2+c}}\right )\) |
\(\Big \downarrow \) 5493 |
\(\displaystyle \frac {c \sqrt {a^2 x^2+1} \int \frac {\arctan (a x)}{x \sqrt {a^2 x^2+1}}dx}{\sqrt {a^2 c x^2+c}}+\arctan (a x) \sqrt {a^2 c x^2+c}-\sqrt {c} \text {arctanh}\left (\frac {a \sqrt {c} x}{\sqrt {a^2 c x^2+c}}\right )\) |
\(\Big \downarrow \) 5489 |
\(\displaystyle \frac {c \sqrt {a^2 x^2+1} \left (-2 \arctan (a x) \text {arctanh}\left (\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right )+i \operatorname {PolyLog}\left (2,-\frac {\sqrt {i a x+1}}{\sqrt {1-i a x}}\right )-i \operatorname {PolyLog}\left (2,\frac {\sqrt {i a x+1}}{\sqrt {1-i a x}}\right )\right )}{\sqrt {a^2 c x^2+c}}+\arctan (a x) \sqrt {a^2 c x^2+c}-\sqrt {c} \text {arctanh}\left (\frac {a \sqrt {c} x}{\sqrt {a^2 c x^2+c}}\right )\) |
Sqrt[c + a^2*c*x^2]*ArcTan[a*x] - Sqrt[c]*ArcTanh[(a*Sqrt[c]*x)/Sqrt[c + a ^2*c*x^2]] + (c*Sqrt[1 + a^2*x^2]*(-2*ArcTan[a*x]*ArcTanh[Sqrt[1 + I*a*x]/ Sqrt[1 - I*a*x]] + I*PolyLog[2, -(Sqrt[1 + I*a*x]/Sqrt[1 - I*a*x])] - I*Po lyLog[2, Sqrt[1 + I*a*x]/Sqrt[1 - I*a*x]]))/Sqrt[c + a^2*c*x^2]
3.3.4.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b}, x] && !GtQ[a, 0]
Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))*((f_.)*(x_))^(m_)*Sqrt[(d_) + (e_.)* (x_)^2], x_Symbol] :> Simp[(f*x)^(m + 1)*Sqrt[d + e*x^2]*((a + b*ArcTan[c*x ])/(f*(m + 2))), x] + (Simp[d/(m + 2) Int[(f*x)^m*((a + b*ArcTan[c*x])/Sq rt[d + e*x^2]), x], x] - Simp[b*c*(d/(f*(m + 2))) Int[(f*x)^(m + 1)/Sqrt[ d + e*x^2], x], x]) /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[e, c^2*d] && NeQ[m, -2]
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]
Time = 0.49 (sec) , antiderivative size = 151, normalized size of antiderivative = 0.66
method | result | size |
default | \(\sqrt {c \left (a x -i\right ) \left (a x +i\right )}\, \arctan \left (a x \right )-\frac {\sqrt {c \left (a x -i\right ) \left (a x +i\right )}\, \left (\arctan \left (a x \right ) \ln \left (\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}+1\right )-2 i \arctan \left (\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right )-i \operatorname {dilog}\left (\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right )-i \operatorname {dilog}\left (\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}+1\right )\right )}{\sqrt {a^{2} x^{2}+1}}\) | \(151\) |
(c*(a*x-I)*(I+a*x))^(1/2)*arctan(a*x)-(c*(a*x-I)*(I+a*x))^(1/2)*(arctan(a* x)*ln((1+I*a*x)/(a^2*x^2+1)^(1/2)+1)-2*I*arctan((1+I*a*x)/(a^2*x^2+1)^(1/2 ))-I*dilog((1+I*a*x)/(a^2*x^2+1)^(1/2))-I*dilog((1+I*a*x)/(a^2*x^2+1)^(1/2 )+1))/(a^2*x^2+1)^(1/2)
\[ \int \frac {\sqrt {c+a^2 c x^2} \arctan (a x)}{x} \, dx=\int { \frac {\sqrt {a^{2} c x^{2} + c} \arctan \left (a x\right )}{x} \,d x } \]
\[ \int \frac {\sqrt {c+a^2 c x^2} \arctan (a x)}{x} \, dx=\int \frac {\sqrt {c \left (a^{2} x^{2} + 1\right )} \operatorname {atan}{\left (a x \right )}}{x}\, dx \]
\[ \int \frac {\sqrt {c+a^2 c x^2} \arctan (a x)}{x} \, dx=\int { \frac {\sqrt {a^{2} c x^{2} + c} \arctan \left (a x\right )}{x} \,d x } \]
Exception generated. \[ \int \frac {\sqrt {c+a^2 c x^2} \arctan (a x)}{x} \, dx=\text {Exception raised: TypeError} \]
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value
Timed out. \[ \int \frac {\sqrt {c+a^2 c x^2} \arctan (a x)}{x} \, dx=\int \frac {\mathrm {atan}\left (a\,x\right )\,\sqrt {c\,a^2\,x^2+c}}{x} \,d x \]