Integrand size = 19, antiderivative size = 69 \[ \int \frac {\csc ^{-1}(a+b x)}{\frac {a d}{b}+d x} \, dx=\frac {i \csc ^{-1}(a+b x)^2}{2 d}-\frac {\csc ^{-1}(a+b x) \log \left (1-e^{2 i \csc ^{-1}(a+b x)}\right )}{d}+\frac {i \operatorname {PolyLog}\left (2,e^{2 i \csc ^{-1}(a+b x)}\right )}{2 d} \]
1/2*I*arccsc(b*x+a)^2/d-arccsc(b*x+a)*ln(1-(I/(b*x+a)+(1-1/(b*x+a)^2)^(1/2 ))^2)/d+1/2*I*polylog(2,(I/(b*x+a)+(1-1/(b*x+a)^2)^(1/2))^2)/d
Time = 0.06 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.86 \[ \int \frac {\csc ^{-1}(a+b x)}{\frac {a d}{b}+d x} \, dx=\frac {-\csc ^{-1}(a+b x) \log \left (1-e^{2 i \csc ^{-1}(a+b x)}\right )+\frac {1}{2} i \left (\csc ^{-1}(a+b x)^2+\operatorname {PolyLog}\left (2,e^{2 i \csc ^{-1}(a+b x)}\right )\right )}{d} \]
(-(ArcCsc[a + b*x]*Log[1 - E^((2*I)*ArcCsc[a + b*x])]) + (I/2)*(ArcCsc[a + b*x]^2 + PolyLog[2, E^((2*I)*ArcCsc[a + b*x])]))/d
Time = 0.49 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.12, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.579, Rules used = {5780, 27, 5742, 5136, 3042, 25, 4200, 25, 2620, 2715, 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.
| \(\displaystyle \int \frac {\csc ^{-1}(a+b x)}{\frac {a d}{b}+d x} \, dx\) |
| \(\Big \downarrow \) 5780 |
| \(\displaystyle \frac {\int \frac {b \csc ^{-1}(a+b x)}{d (a+b x)}d(a+b x)}{b}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {\csc ^{-1}(a+b x)}{a+b x}d(a+b x)}{d}\) |
| \(\Big \downarrow \) 5742 |
| \(\displaystyle -\frac {\int (a+b x) \arcsin \left (\frac {1}{a+b x}\right )d\frac {1}{a+b x}}{d}\) |
| \(\Big \downarrow \) 5136 |
| \(\displaystyle -\frac {\int (a+b x) \sqrt {1-\frac {1}{(a+b x)^2}} \arcsin \left (\frac {1}{a+b x}\right )d\arcsin \left (\frac {1}{a+b x}\right )}{d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {\int -\arcsin \left (\frac {1}{a+b x}\right ) \tan \left (\arcsin \left (\frac {1}{a+b x}\right )+\frac {\pi }{2}\right )d\arcsin \left (\frac {1}{a+b x}\right )}{d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\int \arcsin \left (\frac {1}{a+b x}\right ) \tan \left (\arcsin \left (\frac {1}{a+b x}\right )+\frac {\pi }{2}\right )d\arcsin \left (\frac {1}{a+b x}\right )}{d}\) |
| \(\Big \downarrow \) 4200 |
| \(\displaystyle -\frac {2 i \int -\frac {e^{2 i \arcsin \left (\frac {1}{a+b x}\right )} \arcsin \left (\frac {1}{a+b x}\right )}{1-e^{2 i \arcsin \left (\frac {1}{a+b x}\right )}}d\arcsin \left (\frac {1}{a+b x}\right )-\frac {i}{2 (a+b x)^2}}{d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {-2 i \int \frac {e^{2 i \arcsin \left (\frac {1}{a+b x}\right )} \arcsin \left (\frac {1}{a+b x}\right )}{1-e^{2 i \arcsin \left (\frac {1}{a+b x}\right )}}d\arcsin \left (\frac {1}{a+b x}\right )-\frac {i}{2 (a+b x)^2}}{d}\) |
| \(\Big \downarrow \) 2620 |
| \(\displaystyle -\frac {-2 i \left (\frac {1}{2} i \arcsin \left (\frac {1}{a+b x}\right ) \log \left (1-e^{2 i \arcsin \left (\frac {1}{a+b x}\right )}\right )-\frac {1}{2} i \int \log \left (1-e^{2 i \arcsin \left (\frac {1}{a+b x}\right )}\right )d\arcsin \left (\frac {1}{a+b x}\right )\right )-\frac {i}{2 (a+b x)^2}}{d}\) |
| \(\Big \downarrow \) 2715 |
| \(\displaystyle -\frac {-2 i \left (\frac {1}{2} i \arcsin \left (\frac {1}{a+b x}\right ) \log \left (1-e^{2 i \arcsin \left (\frac {1}{a+b x}\right )}\right )-\frac {1}{4} \int (a+b x) \log (-a-b x+1)de^{2 i \arcsin \left (\frac {1}{a+b x}\right )}\right )-\frac {i}{2 (a+b x)^2}}{d}\) |
| \(\Big \downarrow \) 2838 |
| \(\displaystyle -\frac {-2 i \left (\frac {1}{4} \operatorname {PolyLog}\left (2,e^{2 i \arcsin \left (\frac {1}{a+b x}\right )}\right )+\frac {1}{2} i \arcsin \left (\frac {1}{a+b x}\right ) \log \left (1-e^{2 i \arcsin \left (\frac {1}{a+b x}\right )}\right )\right )-\frac {i}{2 (a+b x)^2}}{d}\) |
-(((-1/2*I)/(a + b*x)^2 - (2*I)*((I/2)*ArcSin[(a + b*x)^(-1)]*Log[1 - E^(( 2*I)*ArcSin[(a + b*x)^(-1)])] + PolyLog[2, E^((2*I)*ArcSin[(a + b*x)^(-1)] )]/4))/d)
3.1.49.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Simp[1/(d*e*n*Log[F]) Subst[Int[Log[a + b*x]/x, x], x, (F^(e*(c + d*x) ))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]
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[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + Pi*(k_.) + (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*k*Pi)*(E^(2*I*(e + f*x))/(1 + E^(2*I*k*Pi)*E^(2*I*(e + f*x)))), x] , x] /; FreeQ[{c, d, e, f}, x] && IntegerQ[4*k] && IGtQ[m, 0]
Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)/(x_), x_Symbol] :> Subst[Int[( a + b*x)^n*Cot[x], x], x, ArcSin[c*x]] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0]
Int[((a_.) + ArcCsc[(c_.)*(x_)]*(b_.))/(x_), x_Symbol] :> -Subst[Int[(a + b *ArcSin[x/c])/x, x], x, 1/x] /; FreeQ[{a, b, c}, x]
Int[((a_.) + ArcCsc[(c_) + (d_.)*(x_)]*(b_.))^(p_.)*((e_.) + (f_.)*(x_))^(m _.), x_Symbol] :> Simp[1/d Subst[Int[(f*(x/d))^m*(a + b*ArcCsc[x])^p, x], x, c + d*x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[d*e - c*f, 0] && IGtQ[p, 0]
Time = 1.20 (sec) , antiderivative size = 167, normalized size of antiderivative = 2.42
| method | result | size |
| derivativedivides | \(\frac {\frac {i b \operatorname {arccsc}\left (b x +a \right )^{2}}{2 d}-\frac {b \,\operatorname {arccsc}\left (b x +a \right ) \ln \left (1-\frac {i}{b x +a}-\sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )}{d}+\frac {i b \operatorname {polylog}\left (2, \frac {i}{b x +a}+\sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )}{d}-\frac {b \,\operatorname {arccsc}\left (b x +a \right ) \ln \left (1+\frac {i}{b x +a}+\sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )}{d}+\frac {i b \operatorname {polylog}\left (2, -\frac {i}{b x +a}-\sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )}{d}}{b}\) | \(167\) |
| default | \(\frac {\frac {i b \operatorname {arccsc}\left (b x +a \right )^{2}}{2 d}-\frac {b \,\operatorname {arccsc}\left (b x +a \right ) \ln \left (1-\frac {i}{b x +a}-\sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )}{d}+\frac {i b \operatorname {polylog}\left (2, \frac {i}{b x +a}+\sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )}{d}-\frac {b \,\operatorname {arccsc}\left (b x +a \right ) \ln \left (1+\frac {i}{b x +a}+\sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )}{d}+\frac {i b \operatorname {polylog}\left (2, -\frac {i}{b x +a}-\sqrt {1-\frac {1}{\left (b x +a \right )^{2}}}\right )}{d}}{b}\) | \(167\) |
1/b*(1/2*I*b/d*arccsc(b*x+a)^2-b/d*arccsc(b*x+a)*ln(1-I/(b*x+a)-(1-1/(b*x+ a)^2)^(1/2))+I*b/d*polylog(2,I/(b*x+a)+(1-1/(b*x+a)^2)^(1/2))-b/d*arccsc(b *x+a)*ln(1+I/(b*x+a)+(1-1/(b*x+a)^2)^(1/2))+I*b/d*polylog(2,-I/(b*x+a)-(1- 1/(b*x+a)^2)^(1/2)))
\[ \int \frac {\csc ^{-1}(a+b x)}{\frac {a d}{b}+d x} \, dx=\int { \frac {\operatorname {arccsc}\left (b x + a\right )}{d x + \frac {a d}{b}} \,d x } \]
\[ \int \frac {\csc ^{-1}(a+b x)}{\frac {a d}{b}+d x} \, dx=\frac {b \int \frac {\operatorname {acsc}{\left (a + b x \right )}}{a + b x}\, dx}{d} \]
\[ \int \frac {\csc ^{-1}(a+b x)}{\frac {a d}{b}+d x} \, dx=\int { \frac {\operatorname {arccsc}\left (b x + a\right )}{d x + \frac {a d}{b}} \,d x } \]
1/2*(2*b*d*integrate(sqrt(b*x + a + 1)*sqrt(b*x + a - 1)*log(b*x + a)/(b^3 *d*x^3 + 3*a*b^2*d*x^2 + (3*a^2 - 1)*b*d*x + (a^3 - a)*d), x) - 2*I*b*d*in tegrate(log(b*x + a)/(b^3*d*x^3 + 3*a*b^2*d*x^2 + (3*a^2 - 1)*b*d*x + (a^3 - a)*d), x) + (2*arctan2(1, sqrt(b*x + a + 1)*sqrt(b*x + a - 1)) + I*log( -b*x - a + 1))*log(b*x + a) - I*log(b^2*x^2 + 2*a*b*x + a^2)*log(b*x + a) + I*log(b*x + a + 1)*log(b*x + a) + I*log(b*x + a)^2 + I*dilog(b*x + a) + I*dilog(-b*x - a))/d
Time = 0.50 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.65 \[ \int \frac {\csc ^{-1}(a+b x)}{\frac {a d}{b}+d x} \, dx=-\frac {1}{4} \, b^{2} {\left (\frac {2 \, {\left (b x + a\right )}^{2} \arcsin \left (\frac {1}{{\left ({\left (b x + a\right )} {\left (\frac {a}{b x + a} - 1\right )} - a\right )} {\left (\frac {a}{b x + a} - 1\right )} + a}\right )}{b^{3} d} + \frac {{\left (b x + a\right )} {\left (\sqrt {-\frac {1}{{\left (b x + a\right )}^{2}} + 1} - 1\right )} - \frac {1}{{\left (b x + a\right )} {\left (\sqrt {-\frac {1}{{\left (b x + a\right )}^{2}} + 1} - 1\right )}}}{b^{3} d}\right )} \]
-1/4*b^2*(2*(b*x + a)^2*arcsin(1/(((b*x + a)*(a/(b*x + a) - 1) - a)*(a/(b* x + a) - 1) + a))/(b^3*d) + ((b*x + a)*(sqrt(-1/(b*x + a)^2 + 1) - 1) - 1/ ((b*x + a)*(sqrt(-1/(b*x + a)^2 + 1) - 1)))/(b^3*d))
Timed out. \[ \int \frac {\csc ^{-1}(a+b x)}{\frac {a d}{b}+d x} \, dx=\int \frac {\mathrm {asin}\left (\frac {1}{a+b\,x}\right )}{d\,x+\frac {a\,d}{b}} \,d x \]