Integrand size = 14, antiderivative size = 195 \[ \int \sqrt {a+a x^2} \cot ^{-1}(x) \, dx=\frac {1}{2} \sqrt {a+a x^2}+\frac {1}{2} x \sqrt {a+a x^2} \cot ^{-1}(x)-\frac {i a \sqrt {1+x^2} \cot ^{-1}(x) \arctan \left (\frac {\sqrt {1+i x}}{\sqrt {1-i x}}\right )}{\sqrt {a+a x^2}}-\frac {i a \sqrt {1+x^2} \operatorname {PolyLog}\left (2,-\frac {i \sqrt {1+i x}}{\sqrt {1-i x}}\right )}{2 \sqrt {a+a x^2}}+\frac {i a \sqrt {1+x^2} \operatorname {PolyLog}\left (2,\frac {i \sqrt {1+i x}}{\sqrt {1-i x}}\right )}{2 \sqrt {a+a x^2}} \]
-I*a*arccot(x)*arctan((1+I*x)^(1/2)/(1-I*x)^(1/2))*(x^2+1)^(1/2)/(a*x^2+a) ^(1/2)-1/2*I*a*polylog(2,-I*(1+I*x)^(1/2)/(1-I*x)^(1/2))*(x^2+1)^(1/2)/(a* x^2+a)^(1/2)+1/2*I*a*polylog(2,I*(1+I*x)^(1/2)/(1-I*x)^(1/2))*(x^2+1)^(1/2 )/(a*x^2+a)^(1/2)+1/2*(a*x^2+a)^(1/2)+1/2*x*arccot(x)*(a*x^2+a)^(1/2)
Time = 1.00 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.70 \[ \int \sqrt {a+a x^2} \cot ^{-1}(x) \, dx=-\frac {\left (a \left (1+x^2\right )\right )^{3/2} \left (-2 \cot \left (\frac {1}{2} \cot ^{-1}(x)\right )-\cot ^{-1}(x) \csc ^2\left (\frac {1}{2} \cot ^{-1}(x)\right )+4 \cot ^{-1}(x) \log \left (1-e^{i \cot ^{-1}(x)}\right )-4 \cot ^{-1}(x) \log \left (1+e^{i \cot ^{-1}(x)}\right )+4 i \operatorname {PolyLog}\left (2,-e^{i \cot ^{-1}(x)}\right )-4 i \operatorname {PolyLog}\left (2,e^{i \cot ^{-1}(x)}\right )+\cot ^{-1}(x) \sec ^2\left (\frac {1}{2} \cot ^{-1}(x)\right )-2 \tan \left (\frac {1}{2} \cot ^{-1}(x)\right )\right )}{8 a \left (1+\frac {1}{x^2}\right )^{3/2} x^3} \]
-1/8*((a*(1 + x^2))^(3/2)*(-2*Cot[ArcCot[x]/2] - ArcCot[x]*Csc[ArcCot[x]/2 ]^2 + 4*ArcCot[x]*Log[1 - E^(I*ArcCot[x])] - 4*ArcCot[x]*Log[1 + E^(I*ArcC ot[x])] + (4*I)*PolyLog[2, -E^(I*ArcCot[x])] - (4*I)*PolyLog[2, E^(I*ArcCo t[x])] + ArcCot[x]*Sec[ArcCot[x]/2]^2 - 2*Tan[ArcCot[x]/2]))/(a*(1 + x^(-2 ))^(3/2)*x^3)
Time = 0.38 (sec) , antiderivative size = 154, normalized size of antiderivative = 0.79, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {5414, 5426, 5422}
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.
\(\displaystyle \int \sqrt {a x^2+a} \cot ^{-1}(x) \, dx\) |
\(\Big \downarrow \) 5414 |
\(\displaystyle \frac {1}{2} a \int \frac {\cot ^{-1}(x)}{\sqrt {a x^2+a}}dx+\frac {1}{2} \sqrt {a x^2+a}+\frac {1}{2} x \sqrt {a x^2+a} \cot ^{-1}(x)\) |
\(\Big \downarrow \) 5426 |
\(\displaystyle \frac {a \sqrt {x^2+1} \int \frac {\cot ^{-1}(x)}{\sqrt {x^2+1}}dx}{2 \sqrt {a x^2+a}}+\frac {1}{2} \sqrt {a x^2+a}+\frac {1}{2} x \sqrt {a x^2+a} \cot ^{-1}(x)\) |
\(\Big \downarrow \) 5422 |
\(\displaystyle \frac {a \sqrt {x^2+1} \left (-2 i \arctan \left (\frac {\sqrt {1+i x}}{\sqrt {1-i x}}\right ) \cot ^{-1}(x)-i \operatorname {PolyLog}\left (2,-\frac {i \sqrt {i x+1}}{\sqrt {1-i x}}\right )+i \operatorname {PolyLog}\left (2,\frac {i \sqrt {i x+1}}{\sqrt {1-i x}}\right )\right )}{2 \sqrt {a x^2+a}}+\frac {1}{2} \sqrt {a x^2+a}+\frac {1}{2} x \sqrt {a x^2+a} \cot ^{-1}(x)\) |
Sqrt[a + a*x^2]/2 + (x*Sqrt[a + a*x^2]*ArcCot[x])/2 + (a*Sqrt[1 + x^2]*((- 2*I)*ArcCot[x]*ArcTan[Sqrt[1 + I*x]/Sqrt[1 - I*x]] - I*PolyLog[2, ((-I)*Sq rt[1 + I*x])/Sqrt[1 - I*x]] + I*PolyLog[2, (I*Sqrt[1 + I*x])/Sqrt[1 - I*x] ]))/(2*Sqrt[a + a*x^2])
3.1.65.3.1 Defintions of rubi rules used
Int[((a_.) + ArcCot[(c_.)*(x_)]*(b_.))*((d_) + (e_.)*(x_)^2)^(q_.), x_Symbo l] :> Simp[b*((d + e*x^2)^q/(2*c*q*(2*q + 1))), x] + (Simp[x*(d + e*x^2)^q* ((a + b*ArcCot[c*x])/(2*q + 1)), x] + Simp[2*d*(q/(2*q + 1)) Int[(d + e*x ^2)^(q - 1)*(a + b*ArcCot[c*x]), x], x]) /; FreeQ[{a, b, c, d, e}, x] && Eq Q[e, c^2*d] && GtQ[q, 0]
Int[((a_.) + ArcCot[(c_.)*(x_)]*(b_.))/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp[-2*I*(a + b*ArcCot[c*x])*(ArcTan[Sqrt[1 + I*c*x]/Sqrt[1 - I*c*x]]/ (c*Sqrt[d])), x] + (-Simp[I*b*(PolyLog[2, (-I)*(Sqrt[1 + I*c*x]/Sqrt[1 - I* c*x])]/(c*Sqrt[d])), x] + Simp[I*b*(PolyLog[2, I*(Sqrt[1 + I*c*x]/Sqrt[1 - I*c*x])]/(c*Sqrt[d])), x]) /; FreeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*d] && GtQ[d, 0]
Int[((a_.) + ArcCot[(c_.)*(x_)]*(b_.))^(p_.)/Sqrt[(d_) + (e_.)*(x_)^2], x_S ymbol] :> Simp[Sqrt[1 + c^2*x^2]/Sqrt[d + e*x^2] Int[(a + b*ArcCot[c*x])^ p/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 = 0.87 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.60
method | result | size |
default | \(\frac {\sqrt {a \left (i+x \right ) \left (x -i\right )}\, \left (x \,\operatorname {arccot}\left (x \right )+1\right )}{2}-\frac {i \sqrt {a \left (i+x \right ) \left (x -i\right )}\, \left (i \operatorname {arccot}\left (x \right ) \ln \left (\frac {i+x}{\sqrt {x^{2}+1}}+1\right )-i \operatorname {arccot}\left (x \right ) \ln \left (1-\frac {i+x}{\sqrt {x^{2}+1}}\right )+\operatorname {polylog}\left (2, -\frac {i+x}{\sqrt {x^{2}+1}}\right )-\operatorname {polylog}\left (2, \frac {i+x}{\sqrt {x^{2}+1}}\right )\right )}{2 \sqrt {x^{2}+1}}\) | \(117\) |
1/2*(a*(I+x)*(x-I))^(1/2)*(x*arccot(x)+1)-1/2*I*(a*(I+x)*(x-I))^(1/2)*(I*a rccot(x)*ln((I+x)/(x^2+1)^(1/2)+1)-I*arccot(x)*ln(1-(I+x)/(x^2+1)^(1/2))+p olylog(2,-(I+x)/(x^2+1)^(1/2))-polylog(2,(I+x)/(x^2+1)^(1/2)))/(x^2+1)^(1/ 2)
\[ \int \sqrt {a+a x^2} \cot ^{-1}(x) \, dx=\int { \sqrt {a x^{2} + a} \operatorname {arccot}\left (x\right ) \,d x } \]
\[ \int \sqrt {a+a x^2} \cot ^{-1}(x) \, dx=\int \sqrt {a \left (x^{2} + 1\right )} \operatorname {acot}{\left (x \right )}\, dx \]
\[ \int \sqrt {a+a x^2} \cot ^{-1}(x) \, dx=\int { \sqrt {a x^{2} + a} \operatorname {arccot}\left (x\right ) \,d x } \]
\[ \int \sqrt {a+a x^2} \cot ^{-1}(x) \, dx=\int { \sqrt {a x^{2} + a} \operatorname {arccot}\left (x\right ) \,d x } \]
Timed out. \[ \int \sqrt {a+a x^2} \cot ^{-1}(x) \, dx=\int \mathrm {acot}\left (x\right )\,\sqrt {a\,x^2+a} \,d x \]