Integrand size = 14, antiderivative size = 93 \[ \int \frac {\left (a+b \sec ^{-1}(c x)\right )^2}{x} \, dx=\frac {i \left (a+b \sec ^{-1}(c x)\right )^3}{3 b}-\left (a+b \sec ^{-1}(c x)\right )^2 \log \left (1+e^{2 i \sec ^{-1}(c x)}\right )+i b \left (a+b \sec ^{-1}(c x)\right ) \operatorname {PolyLog}\left (2,-e^{2 i \sec ^{-1}(c x)}\right )-\frac {1}{2} b^2 \operatorname {PolyLog}\left (3,-e^{2 i \sec ^{-1}(c x)}\right ) \]
1/3*I*(a+b*arcsec(c*x))^3/b-(a+b*arcsec(c*x))^2*ln(1+(1/c/x+I*(1-1/c^2/x^2 )^(1/2))^2)+I*b*(a+b*arcsec(c*x))*polylog(2,-(1/c/x+I*(1-1/c^2/x^2)^(1/2)) ^2)-1/2*b^2*polylog(3,-(1/c/x+I*(1-1/c^2/x^2)^(1/2))^2)
Time = 0.10 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.39 \[ \int \frac {\left (a+b \sec ^{-1}(c x)\right )^2}{x} \, dx=i a b \sec ^{-1}(c x)^2+\frac {1}{3} i b^2 \sec ^{-1}(c x)^3-2 a b \sec ^{-1}(c x) \log \left (1+e^{2 i \sec ^{-1}(c x)}\right )-b^2 \sec ^{-1}(c x)^2 \log \left (1+e^{2 i \sec ^{-1}(c x)}\right )+a^2 \log (c x)+i b \left (a+b \sec ^{-1}(c x)\right ) \operatorname {PolyLog}\left (2,-e^{2 i \sec ^{-1}(c x)}\right )-\frac {1}{2} b^2 \operatorname {PolyLog}\left (3,-e^{2 i \sec ^{-1}(c x)}\right ) \]
I*a*b*ArcSec[c*x]^2 + (I/3)*b^2*ArcSec[c*x]^3 - 2*a*b*ArcSec[c*x]*Log[1 + E^((2*I)*ArcSec[c*x])] - b^2*ArcSec[c*x]^2*Log[1 + E^((2*I)*ArcSec[c*x])] + a^2*Log[c*x] + I*b*(a + b*ArcSec[c*x])*PolyLog[2, -E^((2*I)*ArcSec[c*x]) ] - (b^2*PolyLog[3, -E^((2*I)*ArcSec[c*x])])/2
Time = 0.52 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.15, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {5745, 3042, 4202, 2620, 3011, 2720, 7143}
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.
| \(\displaystyle \int \frac {\left (a+b \sec ^{-1}(c x)\right )^2}{x} \, dx\) |
| \(\Big \downarrow \) 5745 |
| \(\displaystyle \int c x \sqrt {1-\frac {1}{c^2 x^2}} \left (a+b \sec ^{-1}(c x)\right )^2d\sec ^{-1}(c x)\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \tan \left (\sec ^{-1}(c x)\right ) \left (a+b \sec ^{-1}(c x)\right )^2d\sec ^{-1}(c x)\) |
| \(\Big \downarrow \) 4202 |
| \(\displaystyle \frac {i \left (a+b \sec ^{-1}(c x)\right )^3}{3 b}-2 i \int \frac {e^{2 i \sec ^{-1}(c x)} \left (a+b \sec ^{-1}(c x)\right )^2}{1+e^{2 i \sec ^{-1}(c x)}}d\sec ^{-1}(c x)\) |
| \(\Big \downarrow \) 2620 |
| \(\displaystyle \frac {i \left (a+b \sec ^{-1}(c x)\right )^3}{3 b}-2 i \left (i b \int \left (a+b \sec ^{-1}(c x)\right ) \log \left (1+e^{2 i \sec ^{-1}(c x)}\right )d\sec ^{-1}(c x)-\frac {1}{2} i \log \left (1+e^{2 i \sec ^{-1}(c x)}\right ) \left (a+b \sec ^{-1}(c x)\right )^2\right )\) |
| \(\Big \downarrow \) 3011 |
| \(\displaystyle \frac {i \left (a+b \sec ^{-1}(c x)\right )^3}{3 b}-2 i \left (i b \left (\frac {1}{2} i \operatorname {PolyLog}\left (2,-e^{2 i \sec ^{-1}(c x)}\right ) \left (a+b \sec ^{-1}(c x)\right )-\frac {1}{2} i b \int \operatorname {PolyLog}\left (2,-e^{2 i \sec ^{-1}(c x)}\right )d\sec ^{-1}(c x)\right )-\frac {1}{2} i \log \left (1+e^{2 i \sec ^{-1}(c x)}\right ) \left (a+b \sec ^{-1}(c x)\right )^2\right )\) |
| \(\Big \downarrow \) 2720 |
| \(\displaystyle \frac {i \left (a+b \sec ^{-1}(c x)\right )^3}{3 b}-2 i \left (i b \left (\frac {1}{2} i \operatorname {PolyLog}\left (2,-e^{2 i \sec ^{-1}(c x)}\right ) \left (a+b \sec ^{-1}(c x)\right )-\frac {1}{4} b \int e^{-2 i \sec ^{-1}(c x)} \operatorname {PolyLog}\left (2,-e^{2 i \sec ^{-1}(c x)}\right )de^{2 i \sec ^{-1}(c x)}\right )-\frac {1}{2} i \log \left (1+e^{2 i \sec ^{-1}(c x)}\right ) \left (a+b \sec ^{-1}(c x)\right )^2\right )\) |
| \(\Big \downarrow \) 7143 |
| \(\displaystyle \frac {i \left (a+b \sec ^{-1}(c x)\right )^3}{3 b}-2 i \left (i b \left (\frac {1}{2} i \operatorname {PolyLog}\left (2,-e^{2 i \sec ^{-1}(c x)}\right ) \left (a+b \sec ^{-1}(c x)\right )-\frac {1}{4} b \operatorname {PolyLog}\left (3,-e^{2 i \sec ^{-1}(c x)}\right )\right )-\frac {1}{2} i \log \left (1+e^{2 i \sec ^{-1}(c x)}\right ) \left (a+b \sec ^{-1}(c x)\right )^2\right )\) |
((I/3)*(a + b*ArcSec[c*x])^3)/b - (2*I)*((-1/2*I)*(a + b*ArcSec[c*x])^2*Lo g[1 + E^((2*I)*ArcSec[c*x])] + I*b*((I/2)*(a + b*ArcSec[c*x])*PolyLog[2, - E^((2*I)*ArcSec[c*x])] - (b*PolyLog[3, -E^((2*I)*ArcSec[c*x])])/4))
3.1.19.3.1 Defintions of rubi rules used
Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/ ((a_) + (b_.)*((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)), x_Symbol] :> Simp [((c + d*x)^m/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], x] - Si mp[d*(m/(b*f*g*n*Log[F])) Int[(c + d*x)^(m - 1)*Log[1 + b*((F^(g*(e + f*x )))^n/a)], x], x] /; FreeQ[{F, a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]
Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Simp[v/D[v, x] Subst[Int[FunctionOfExponentialFunction[u, x]/x, x], x, v], x]] /; Funct ionOfExponentialQ[u, x] && !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; FreeQ [{a, m, n}, x] && IntegerQ[m*n]] && !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x)) *(F_)[v_] /; FreeQ[{a, b, c}, x] && InverseFunctionQ[F[x]]]
Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.) *(x_))^(m_.), x_Symbol] :> Simp[(-(f + g*x)^m)*(PolyLog[2, (-e)*(F^(c*(a + b*x)))^n]/(b*c*n*Log[F])), x] + Simp[g*(m/(b*c*n*Log[F])) Int[(f + g*x)^( m - 1)*PolyLog[2, (-e)*(F^(c*(a + b*x)))^n], x], x] /; FreeQ[{F, a, b, c, e , f, g, n}, x] && GtQ[m, 0]
Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[I *((c + d*x)^(m + 1)/(d*(m + 1))), x] - Simp[2*I Int[(c + d*x)^m*(E^(2*I*( e + f*x))/(1 + E^(2*I*(e + f*x)))), x], x] /; FreeQ[{c, d, e, f}, x] && IGt Q[m, 0]
Int[((a_.) + ArcSec[(c_.)*(x_)]*(b_.))^(n_)*(x_)^(m_.), x_Symbol] :> Simp[1 /c^(m + 1) Subst[Int[(a + b*x)^n*Sec[x]^(m + 1)*Tan[x], x], x, ArcSec[c*x ]], x] /; FreeQ[{a, b, c}, x] && IntegerQ[n] && IntegerQ[m] && (GtQ[n, 0] | | LtQ[m, -1])
Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_S ymbol] :> Simp[PolyLog[n + 1, c*(a + b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d , e, n, p}, x] && EqQ[b*d, a*e]
Time = 0.77 (sec) , antiderivative size = 206, normalized size of antiderivative = 2.22
| method | result | size |
| parts | \(a^{2} \ln \left (x \right )+b^{2} \left (\frac {i \operatorname {arcsec}\left (c x \right )^{3}}{3}-\operatorname {arcsec}\left (c x \right )^{2} \ln \left (1+{\left (\frac {1}{c x}+i \sqrt {1-\frac {1}{c^{2} x^{2}}}\right )}^{2}\right )+i \operatorname {arcsec}\left (c x \right ) \operatorname {polylog}\left (2, -{\left (\frac {1}{c x}+i \sqrt {1-\frac {1}{c^{2} x^{2}}}\right )}^{2}\right )-\frac {\operatorname {polylog}\left (3, -{\left (\frac {1}{c x}+i \sqrt {1-\frac {1}{c^{2} x^{2}}}\right )}^{2}\right )}{2}\right )+i a b \operatorname {arcsec}\left (c x \right )^{2}-2 a b \,\operatorname {arcsec}\left (c x \right ) \ln \left (1+{\left (\frac {1}{c x}+i \sqrt {1-\frac {1}{c^{2} x^{2}}}\right )}^{2}\right )+i a b \operatorname {polylog}\left (2, -{\left (\frac {1}{c x}+i \sqrt {1-\frac {1}{c^{2} x^{2}}}\right )}^{2}\right )\) | \(206\) |
| derivativedivides | \(a^{2} \ln \left (c x \right )+b^{2} \left (\frac {i \operatorname {arcsec}\left (c x \right )^{3}}{3}-\operatorname {arcsec}\left (c x \right )^{2} \ln \left (1+{\left (\frac {1}{c x}+i \sqrt {1-\frac {1}{c^{2} x^{2}}}\right )}^{2}\right )+i \operatorname {arcsec}\left (c x \right ) \operatorname {polylog}\left (2, -{\left (\frac {1}{c x}+i \sqrt {1-\frac {1}{c^{2} x^{2}}}\right )}^{2}\right )-\frac {\operatorname {polylog}\left (3, -{\left (\frac {1}{c x}+i \sqrt {1-\frac {1}{c^{2} x^{2}}}\right )}^{2}\right )}{2}\right )+i a b \operatorname {arcsec}\left (c x \right )^{2}-2 a b \,\operatorname {arcsec}\left (c x \right ) \ln \left (1+{\left (\frac {1}{c x}+i \sqrt {1-\frac {1}{c^{2} x^{2}}}\right )}^{2}\right )+i a b \operatorname {polylog}\left (2, -{\left (\frac {1}{c x}+i \sqrt {1-\frac {1}{c^{2} x^{2}}}\right )}^{2}\right )\) | \(208\) |
| default | \(a^{2} \ln \left (c x \right )+b^{2} \left (\frac {i \operatorname {arcsec}\left (c x \right )^{3}}{3}-\operatorname {arcsec}\left (c x \right )^{2} \ln \left (1+{\left (\frac {1}{c x}+i \sqrt {1-\frac {1}{c^{2} x^{2}}}\right )}^{2}\right )+i \operatorname {arcsec}\left (c x \right ) \operatorname {polylog}\left (2, -{\left (\frac {1}{c x}+i \sqrt {1-\frac {1}{c^{2} x^{2}}}\right )}^{2}\right )-\frac {\operatorname {polylog}\left (3, -{\left (\frac {1}{c x}+i \sqrt {1-\frac {1}{c^{2} x^{2}}}\right )}^{2}\right )}{2}\right )+i a b \operatorname {arcsec}\left (c x \right )^{2}-2 a b \,\operatorname {arcsec}\left (c x \right ) \ln \left (1+{\left (\frac {1}{c x}+i \sqrt {1-\frac {1}{c^{2} x^{2}}}\right )}^{2}\right )+i a b \operatorname {polylog}\left (2, -{\left (\frac {1}{c x}+i \sqrt {1-\frac {1}{c^{2} x^{2}}}\right )}^{2}\right )\) | \(208\) |
a^2*ln(x)+b^2*(1/3*I*arcsec(c*x)^3-arcsec(c*x)^2*ln(1+(1/c/x+I*(1-1/c^2/x^ 2)^(1/2))^2)+I*arcsec(c*x)*polylog(2,-(1/c/x+I*(1-1/c^2/x^2)^(1/2))^2)-1/2 *polylog(3,-(1/c/x+I*(1-1/c^2/x^2)^(1/2))^2))+I*a*b*arcsec(c*x)^2-2*a*b*ar csec(c*x)*ln(1+(1/c/x+I*(1-1/c^2/x^2)^(1/2))^2)+I*a*b*polylog(2,-(1/c/x+I* (1-1/c^2/x^2)^(1/2))^2)
\[ \int \frac {\left (a+b \sec ^{-1}(c x)\right )^2}{x} \, dx=\int { \frac {{\left (b \operatorname {arcsec}\left (c x\right ) + a\right )}^{2}}{x} \,d x } \]
\[ \int \frac {\left (a+b \sec ^{-1}(c x)\right )^2}{x} \, dx=\int \frac {\left (a + b \operatorname {asec}{\left (c x \right )}\right )^{2}}{x}\, dx \]
\[ \int \frac {\left (a+b \sec ^{-1}(c x)\right )^2}{x} \, dx=\int { \frac {{\left (b \operatorname {arcsec}\left (c x\right ) + a\right )}^{2}}{x} \,d x } \]
-1/2*b^2*c^2*(log(c*x + 1)/c^2 + log(c*x - 1)/c^2)*log(c)^2 + b^2*c^2*inte grate(x^2*log(c^2*x^2)/(c^2*x^3 - x), x)*log(c) - 2*b^2*c^2*integrate(x^2* log(x)/(c^2*x^3 - x), x)*log(c) + 2*b^2*c^2*integrate(x^2*log(c^2*x^2)*log (x)/(c^2*x^3 - x), x) - b^2*c^2*integrate(x^2*log(x)^2/(c^2*x^3 - x), x) + 2*a*b*c^2*integrate(x^2*arctan(sqrt(c*x + 1)*sqrt(c*x - 1))/(c^2*x^3 - x) , x) + 1/2*b^2*(log(c*x + 1) + log(c*x - 1) - 2*log(x))*log(c)^2 + b^2*arc tan(sqrt(c*x + 1)*sqrt(c*x - 1))^2*log(x) - 1/4*b^2*log(c^2*x^2)^2*log(x) - b^2*integrate(log(c^2*x^2)/(c^2*x^3 - x), x)*log(c) + 2*b^2*integrate(lo g(x)/(c^2*x^3 - x), x)*log(c) - 2*b^2*integrate(sqrt(c*x + 1)*sqrt(c*x - 1 )*arctan(sqrt(c*x + 1)*sqrt(c*x - 1))*log(x)/(c^2*x^3 - x), x) - 2*b^2*int egrate(log(c^2*x^2)*log(x)/(c^2*x^3 - x), x) + b^2*integrate(log(x)^2/(c^2 *x^3 - x), x) - 2*a*b*integrate(arctan(sqrt(c*x + 1)*sqrt(c*x - 1))/(c^2*x ^3 - x), x) + a^2*log(x)
Exception generated. \[ \int \frac {\left (a+b \sec ^{-1}(c x)\right )^2}{x} \, dx=\text {Exception raised: RuntimeError} \]
Exception raised: RuntimeError >> an error occurred running a Giac command :INPUT:sage2OUTPUT:ln of unsigned or minus infinity Error: Bad Argument Va lue
Timed out. \[ \int \frac {\left (a+b \sec ^{-1}(c x)\right )^2}{x} \, dx=\int \frac {{\left (a+b\,\mathrm {acos}\left (\frac {1}{c\,x}\right )\right )}^2}{x} \,d x \]