Integrand size = 38, antiderivative size = 98 \[ \int \frac {a+b \arctan \left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )}{1-c^2 x^2} \, dx=-\frac {a \log \left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )}{c}-\frac {i b \operatorname {PolyLog}\left (2,-\frac {i \sqrt {1-c x}}{\sqrt {1+c x}}\right )}{2 c}+\frac {i b \operatorname {PolyLog}\left (2,\frac {i \sqrt {1-c x}}{\sqrt {1+c x}}\right )}{2 c} \]
-a*ln((-c*x+1)^(1/2)/(c*x+1)^(1/2))/c-1/2*I*b*polylog(2,-I*(-c*x+1)^(1/2)/ (c*x+1)^(1/2))/c+1/2*I*b*polylog(2,I*(-c*x+1)^(1/2)/(c*x+1)^(1/2))/c
Time = 0.03 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.95 \[ \int \frac {a+b \arctan \left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )}{1-c^2 x^2} \, dx=-\frac {a \log \left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )+\frac {1}{2} i b \operatorname {PolyLog}\left (2,-\frac {i \sqrt {1-c x}}{\sqrt {1+c x}}\right )-\frac {1}{2} i b \operatorname {PolyLog}\left (2,\frac {i \sqrt {1-c x}}{\sqrt {1+c x}}\right )}{c} \]
-((a*Log[Sqrt[1 - c*x]/Sqrt[1 + c*x]] + (I/2)*b*PolyLog[2, ((-I)*Sqrt[1 - c*x])/Sqrt[1 + c*x]] - (I/2)*b*PolyLog[2, (I*Sqrt[1 - c*x])/Sqrt[1 + c*x]] )/c)
Time = 0.33 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.95, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.079, Rules used = {7232, 5355, 2838}
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.
\(\displaystyle \int \frac {a+b \arctan \left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )}{1-c^2 x^2} \, dx\) |
\(\Big \downarrow \) 7232 |
\(\displaystyle -\frac {\int \frac {\sqrt {c x+1} \left (a+b \arctan \left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )\right )}{\sqrt {1-c x}}d\frac {\sqrt {1-c x}}{\sqrt {c x+1}}}{c}\) |
\(\Big \downarrow \) 5355 |
\(\displaystyle -\frac {\frac {1}{2} i b \int \frac {\sqrt {c x+1} \log \left (1-\frac {i \sqrt {1-c x}}{\sqrt {c x+1}}\right )}{\sqrt {1-c x}}d\frac {\sqrt {1-c x}}{\sqrt {c x+1}}-\frac {1}{2} i b \int \frac {\sqrt {c x+1} \log \left (\frac {i \sqrt {1-c x}}{\sqrt {c x+1}}+1\right )}{\sqrt {1-c x}}d\frac {\sqrt {1-c x}}{\sqrt {c x+1}}+a \log \left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )}{c}\) |
\(\Big \downarrow \) 2838 |
\(\displaystyle -\frac {a \log \left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )+\frac {1}{2} i b \operatorname {PolyLog}\left (2,-\frac {i \sqrt {1-c x}}{\sqrt {c x+1}}\right )-\frac {1}{2} i b \operatorname {PolyLog}\left (2,\frac {i \sqrt {1-c x}}{\sqrt {c x+1}}\right )}{c}\) |
-((a*Log[Sqrt[1 - c*x]/Sqrt[1 + c*x]] + (I/2)*b*PolyLog[2, ((-I)*Sqrt[1 - c*x])/Sqrt[1 + c*x]] - (I/2)*b*PolyLog[2, (I*Sqrt[1 - c*x])/Sqrt[1 + c*x]] )/c)
3.1.34.3.1 Defintions of rubi rules used
Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> Simp[-PolyLog[2 , (-c)*e*x^n]/n, x] /; FreeQ[{c, d, e, n}, x] && EqQ[c*d, 1]
Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))/(x_), x_Symbol] :> Simp[a*Log[x], x] + (Simp[I*(b/2) Int[Log[1 - I*c*x]/x, x], x] - Simp[I*(b/2) Int[Log[1 + I*c*x]/x, x], x]) /; FreeQ[{a, b, c}, x]
Int[((a_.) + (b_.)*(F_)[((c_.)*Sqrt[(d_.) + (e_.)*(x_)])/Sqrt[(f_.) + (g_.) *(x_)]])^(n_.)/((A_.) + (C_.)*(x_)^2), x_Symbol] :> Simp[2*e*(g/(C*(e*f - d *g))) Subst[Int[(a + b*F[c*x])^n/x, x], x, Sqrt[d + e*x]/Sqrt[f + g*x]], x] /; FreeQ[{a, b, c, d, e, f, g, A, C, F}, x] && EqQ[C*d*f - A*e*g, 0] && EqQ[e*f + d*g, 0] && IGtQ[n, 0]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 367 vs. \(2 (78 ) = 156\).
Time = 0.32 (sec) , antiderivative size = 368, normalized size of antiderivative = 3.76
method | result | size |
default | \(-\frac {a \ln \left (c x -1\right )}{2 c}+\frac {a \ln \left (c x +1\right )}{2 c}-b \left (\frac {\arctan \left (\frac {\sqrt {-c x +1}}{\sqrt {c x +1}}\right ) \ln \left (1-\frac {1+\frac {i \sqrt {-c x +1}}{\sqrt {c x +1}}}{\sqrt {\frac {-c x +1}{c x +1}+1}}\right )}{c}-\frac {i \operatorname {polylog}\left (2, \frac {1+\frac {i \sqrt {-c x +1}}{\sqrt {c x +1}}}{\sqrt {\frac {-c x +1}{c x +1}+1}}\right )}{c}-\frac {\arctan \left (\frac {\sqrt {-c x +1}}{\sqrt {c x +1}}\right ) \ln \left (1+\frac {\left (1+\frac {i \sqrt {-c x +1}}{\sqrt {c x +1}}\right )^{2}}{\frac {-c x +1}{c x +1}+1}\right )}{c}+\frac {i \operatorname {polylog}\left (2, -\frac {\left (1+\frac {i \sqrt {-c x +1}}{\sqrt {c x +1}}\right )^{2}}{\frac {-c x +1}{c x +1}+1}\right )}{2 c}+\frac {\arctan \left (\frac {\sqrt {-c x +1}}{\sqrt {c x +1}}\right ) \ln \left (1+\frac {1+\frac {i \sqrt {-c x +1}}{\sqrt {c x +1}}}{\sqrt {\frac {-c x +1}{c x +1}+1}}\right )}{c}-\frac {i \operatorname {polylog}\left (2, -\frac {1+\frac {i \sqrt {-c x +1}}{\sqrt {c x +1}}}{\sqrt {\frac {-c x +1}{c x +1}+1}}\right )}{c}\right )\) | \(368\) |
parts | \(-\frac {a \ln \left (c x -1\right )}{2 c}+\frac {a \ln \left (c x +1\right )}{2 c}-b \left (\frac {\arctan \left (\frac {\sqrt {-c x +1}}{\sqrt {c x +1}}\right ) \ln \left (1-\frac {1+\frac {i \sqrt {-c x +1}}{\sqrt {c x +1}}}{\sqrt {\frac {-c x +1}{c x +1}+1}}\right )}{c}-\frac {i \operatorname {polylog}\left (2, \frac {1+\frac {i \sqrt {-c x +1}}{\sqrt {c x +1}}}{\sqrt {\frac {-c x +1}{c x +1}+1}}\right )}{c}-\frac {\arctan \left (\frac {\sqrt {-c x +1}}{\sqrt {c x +1}}\right ) \ln \left (1+\frac {\left (1+\frac {i \sqrt {-c x +1}}{\sqrt {c x +1}}\right )^{2}}{\frac {-c x +1}{c x +1}+1}\right )}{c}+\frac {i \operatorname {polylog}\left (2, -\frac {\left (1+\frac {i \sqrt {-c x +1}}{\sqrt {c x +1}}\right )^{2}}{\frac {-c x +1}{c x +1}+1}\right )}{2 c}+\frac {\arctan \left (\frac {\sqrt {-c x +1}}{\sqrt {c x +1}}\right ) \ln \left (1+\frac {1+\frac {i \sqrt {-c x +1}}{\sqrt {c x +1}}}{\sqrt {\frac {-c x +1}{c x +1}+1}}\right )}{c}-\frac {i \operatorname {polylog}\left (2, -\frac {1+\frac {i \sqrt {-c x +1}}{\sqrt {c x +1}}}{\sqrt {\frac {-c x +1}{c x +1}+1}}\right )}{c}\right )\) | \(368\) |
-1/2*a/c*ln(c*x-1)+1/2*a/c*ln(c*x+1)-b*(1/c*arctan((-c*x+1)^(1/2)/(c*x+1)^ (1/2))*ln(1-(1+I*(-c*x+1)^(1/2)/(c*x+1)^(1/2))/((-c*x+1)/(c*x+1)+1)^(1/2)) -I/c*polylog(2,(1+I*(-c*x+1)^(1/2)/(c*x+1)^(1/2))/((-c*x+1)/(c*x+1)+1)^(1/ 2))-1/c*arctan((-c*x+1)^(1/2)/(c*x+1)^(1/2))*ln(1+(1+I*(-c*x+1)^(1/2)/(c*x +1)^(1/2))^2/((-c*x+1)/(c*x+1)+1))+1/2*I/c*polylog(2,-(1+I*(-c*x+1)^(1/2)/ (c*x+1)^(1/2))^2/((-c*x+1)/(c*x+1)+1))+1/c*arctan((-c*x+1)^(1/2)/(c*x+1)^( 1/2))*ln(1+(1+I*(-c*x+1)^(1/2)/(c*x+1)^(1/2))/((-c*x+1)/(c*x+1)+1)^(1/2))- I/c*polylog(2,-(1+I*(-c*x+1)^(1/2)/(c*x+1)^(1/2))/((-c*x+1)/(c*x+1)+1)^(1/ 2)))
\[ \int \frac {a+b \arctan \left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )}{1-c^2 x^2} \, dx=\int { -\frac {b \arctan \left (\frac {\sqrt {-c x + 1}}{\sqrt {c x + 1}}\right ) + a}{c^{2} x^{2} - 1} \,d x } \]
\[ \int \frac {a+b \arctan \left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )}{1-c^2 x^2} \, dx=- \int \frac {a}{c^{2} x^{2} - 1}\, dx - \int \frac {b \operatorname {atan}{\left (\frac {\sqrt {- c x + 1}}{\sqrt {c x + 1}} \right )}}{c^{2} x^{2} - 1}\, dx \]
-Integral(a/(c**2*x**2 - 1), x) - Integral(b*atan(sqrt(-c*x + 1)/sqrt(c*x + 1))/(c**2*x**2 - 1), x)
\[ \int \frac {a+b \arctan \left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )}{1-c^2 x^2} \, dx=\int { -\frac {b \arctan \left (\frac {\sqrt {-c x + 1}}{\sqrt {c x + 1}}\right ) + a}{c^{2} x^{2} - 1} \,d x } \]
1/2*a*(log(c*x + 1)/c - log(c*x - 1)/c) + 1/2*((log(c*x + 1) - log(-c*x + 1))*arctan2(sqrt(-c*x + 1), sqrt(c*x + 1)) - 2*c*integrate(1/2*(e^(1/2*log (c*x + 1) + 1/2*log(-c*x + 1))*log(c*x + 1) - e^(1/2*log(c*x + 1) + 1/2*lo g(-c*x + 1))*log(-c*x + 1))/((c^2*x^2 - 1)*(c*x + 1) - (c^2*x^2 - 1)*(c*x - 1)), x))*b/c
\[ \int \frac {a+b \arctan \left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )}{1-c^2 x^2} \, dx=\int { -\frac {b \arctan \left (\frac {\sqrt {-c x + 1}}{\sqrt {c x + 1}}\right ) + a}{c^{2} x^{2} - 1} \,d x } \]
Timed out. \[ \int \frac {a+b \arctan \left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )}{1-c^2 x^2} \, dx=\int -\frac {a+b\,\mathrm {atan}\left (\frac {\sqrt {1-c\,x}}{\sqrt {c\,x+1}}\right )}{c^2\,x^2-1} \,d x \]