Integrand size = 30, antiderivative size = 86 \[ \int \frac {\text {arcsinh}(a+b x)^2}{\left (1+a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\frac {\text {arcsinh}(a+b x)^2}{b}+\frac {(a+b x) \text {arcsinh}(a+b x)^2}{b \sqrt {1+(a+b x)^2}}-\frac {2 \text {arcsinh}(a+b x) \log \left (1+e^{2 \text {arcsinh}(a+b x)}\right )}{b}-\frac {\operatorname {PolyLog}\left (2,-e^{2 \text {arcsinh}(a+b x)}\right )}{b} \]
arcsinh(b*x+a)^2/b-2*arcsinh(b*x+a)*ln(1+(b*x+a+(1+(b*x+a)^2)^(1/2))^2)/b- polylog(2,-(b*x+a+(1+(b*x+a)^2)^(1/2))^2)/b+(b*x+a)*arcsinh(b*x+a)^2/b/(1+ (b*x+a)^2)^(1/2)
Time = 0.30 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.14 \[ \int \frac {\text {arcsinh}(a+b x)^2}{\left (1+a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\frac {\text {arcsinh}(a+b x) \left (\frac {\left (a+b x-\sqrt {1+a^2+2 a b x+b^2 x^2}\right ) \text {arcsinh}(a+b x)}{\sqrt {1+a^2+2 a b x+b^2 x^2}}-2 \log \left (1+e^{-2 \text {arcsinh}(a+b x)}\right )\right )+\operatorname {PolyLog}\left (2,-e^{-2 \text {arcsinh}(a+b x)}\right )}{b} \]
(ArcSinh[a + b*x]*(((a + b*x - Sqrt[1 + a^2 + 2*a*b*x + b^2*x^2])*ArcSinh[ a + b*x])/Sqrt[1 + a^2 + 2*a*b*x + b^2*x^2] - 2*Log[1 + E^(-2*ArcSinh[a + b*x])]) + PolyLog[2, -E^(-2*ArcSinh[a + b*x])])/b
Result contains complex when optimal does not.
Time = 0.59 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.08, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {6275, 6202, 6212, 3042, 26, 4201, 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.
\(\displaystyle \int \frac {\text {arcsinh}(a+b x)^2}{\left (a^2+2 a b x+b^2 x^2+1\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 6275 |
\(\displaystyle \frac {\int \frac {\text {arcsinh}(a+b x)^2}{\left ((a+b x)^2+1\right )^{3/2}}d(a+b x)}{b}\) |
\(\Big \downarrow \) 6202 |
\(\displaystyle \frac {\frac {(a+b x) \text {arcsinh}(a+b x)^2}{\sqrt {(a+b x)^2+1}}-2 \int \frac {(a+b x) \text {arcsinh}(a+b x)}{(a+b x)^2+1}d(a+b x)}{b}\) |
\(\Big \downarrow \) 6212 |
\(\displaystyle \frac {\frac {(a+b x) \text {arcsinh}(a+b x)^2}{\sqrt {(a+b x)^2+1}}-2 \int \frac {(a+b x) \text {arcsinh}(a+b x)}{\sqrt {(a+b x)^2+1}}d\text {arcsinh}(a+b x)}{b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {(a+b x) \text {arcsinh}(a+b x)^2}{\sqrt {(a+b x)^2+1}}-2 \int -i \text {arcsinh}(a+b x) \tan (i \text {arcsinh}(a+b x))d\text {arcsinh}(a+b x)}{b}\) |
\(\Big \downarrow \) 26 |
\(\displaystyle \frac {\frac {(a+b x) \text {arcsinh}(a+b x)^2}{\sqrt {(a+b x)^2+1}}+2 i \int \text {arcsinh}(a+b x) \tan (i \text {arcsinh}(a+b x))d\text {arcsinh}(a+b x)}{b}\) |
\(\Big \downarrow \) 4201 |
\(\displaystyle \frac {\frac {(a+b x) \text {arcsinh}(a+b x)^2}{\sqrt {(a+b x)^2+1}}+2 i \left (2 i \int \frac {e^{2 \text {arcsinh}(a+b x)} \text {arcsinh}(a+b x)}{1+e^{2 \text {arcsinh}(a+b x)}}d\text {arcsinh}(a+b x)-\frac {1}{2} i \text {arcsinh}(a+b x)^2\right )}{b}\) |
\(\Big \downarrow \) 2620 |
\(\displaystyle \frac {\frac {(a+b x) \text {arcsinh}(a+b x)^2}{\sqrt {(a+b x)^2+1}}+2 i \left (2 i \left (\frac {1}{2} \text {arcsinh}(a+b x) \log \left (e^{2 \text {arcsinh}(a+b x)}+1\right )-\frac {1}{2} \int \log \left (1+e^{2 \text {arcsinh}(a+b x)}\right )d\text {arcsinh}(a+b x)\right )-\frac {1}{2} i \text {arcsinh}(a+b x)^2\right )}{b}\) |
\(\Big \downarrow \) 2715 |
\(\displaystyle \frac {\frac {(a+b x) \text {arcsinh}(a+b x)^2}{\sqrt {(a+b x)^2+1}}+2 i \left (2 i \left (\frac {1}{2} \text {arcsinh}(a+b x) \log \left (e^{2 \text {arcsinh}(a+b x)}+1\right )-\frac {1}{4} \int e^{-2 \text {arcsinh}(a+b x)} \log \left (1+e^{2 \text {arcsinh}(a+b x)}\right )de^{2 \text {arcsinh}(a+b x)}\right )-\frac {1}{2} i \text {arcsinh}(a+b x)^2\right )}{b}\) |
\(\Big \downarrow \) 2838 |
\(\displaystyle \frac {\frac {(a+b x) \text {arcsinh}(a+b x)^2}{\sqrt {(a+b x)^2+1}}+2 i \left (2 i \left (\frac {1}{2} \text {arcsinh}(a+b x) \log \left (e^{2 \text {arcsinh}(a+b x)}+1\right )+\frac {1}{4} \operatorname {PolyLog}(2,-a-b x)\right )-\frac {1}{2} i \text {arcsinh}(a+b x)^2\right )}{b}\) |
(((a + b*x)*ArcSinh[a + b*x]^2)/Sqrt[1 + (a + b*x)^2] + (2*I)*((-1/2*I)*Ar cSinh[a + b*x]^2 + (2*I)*((ArcSinh[a + b*x]*Log[1 + E^(2*ArcSinh[a + b*x]) ])/2 + PolyLog[2, -a - b*x]/4)))/b
3.3.79.3.1 Defintions of rubi rules used
Int[(Complex[0, a_])*(Fx_), x_Symbol] :> Simp[(Complex[Identity[0], a]) I nt[Fx, x], x] /; FreeQ[a, x] && EqQ[a^2, 1]
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_.) + (Complex[0, fz_])*(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*fz*x))/(1 + E^(2*((-I)*e + f*fz*x)))), x], x] /; FreeQ[{c, d, e, f, fz}, x] && IGtQ[m, 0]
Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)/((d_) + (e_.)*(x_)^2)^(3/2), x_Symbol] :> Simp[x*((a + b*ArcSinh[c*x])^n/(d*Sqrt[d + e*x^2])), x] - Simp [b*c*(n/d)*Simp[Sqrt[1 + c^2*x^2]/Sqrt[d + e*x^2]] Int[x*((a + b*ArcSinh[ c*x])^(n - 1)/(1 + c^2*x^2)), x], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*d] && GtQ[n, 0]
Int[(((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*(x_))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Simp[1/e Subst[Int[(a + b*x)^n*Tanh[x], x], x, ArcSinh[c*x] ], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*d] && IGtQ[n, 0]
Int[((a_.) + ArcSinh[(c_) + (d_.)*(x_)]*(b_.))^(n_.)*((A_.) + (B_.)*(x_) + (C_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[1/d Subst[Int[(C/d^2 + (C/d^2)*x^2 )^p*(a + b*ArcSinh[x])^n, x], x, c + d*x], x] /; FreeQ[{a, b, c, d, A, B, C , n, p}, x] && EqQ[B*(1 + c^2) - 2*A*c*d, 0] && EqQ[2*c*C - B*d, 0]
Time = 0.63 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.95
method | result | size |
default | \(-\frac {\left (b^{2} x^{2}-\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\, b x +2 a b x -a \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}+a^{2}+1\right ) \operatorname {arcsinh}\left (b x +a \right )^{2}}{b \left (b^{2} x^{2}+2 a b x +a^{2}+1\right )}+\frac {2 \operatorname {arcsinh}\left (b x +a \right )^{2}}{b}-\frac {2 \,\operatorname {arcsinh}\left (b x +a \right ) \ln \left (1+\left (b x +a +\sqrt {1+\left (b x +a \right )^{2}}\right )^{2}\right )}{b}-\frac {\operatorname {polylog}\left (2, -\left (b x +a +\sqrt {1+\left (b x +a \right )^{2}}\right )^{2}\right )}{b}\) | \(168\) |
-(b^2*x^2-(b^2*x^2+2*a*b*x+a^2+1)^(1/2)*b*x+2*a*b*x-a*(b^2*x^2+2*a*b*x+a^2 +1)^(1/2)+a^2+1)/b/(b^2*x^2+2*a*b*x+a^2+1)*arcsinh(b*x+a)^2+2*arcsinh(b*x+ a)^2/b-2*arcsinh(b*x+a)*ln(1+(b*x+a+(1+(b*x+a)^2)^(1/2))^2)/b-polylog(2,-( b*x+a+(1+(b*x+a)^2)^(1/2))^2)/b
\[ \int \frac {\text {arcsinh}(a+b x)^2}{\left (1+a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\int { \frac {\operatorname {arsinh}\left (b x + a\right )^{2}}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )}^{\frac {3}{2}}} \,d x } \]
integral(sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)*arcsinh(b*x + a)^2/(b^4*x^4 + 4 *a*b^3*x^3 + 2*(3*a^2 + 1)*b^2*x^2 + a^4 + 4*(a^3 + a)*b*x + 2*a^2 + 1), x )
\[ \int \frac {\text {arcsinh}(a+b x)^2}{\left (1+a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\int \frac {\operatorname {asinh}^{2}{\left (a + b x \right )}}{\left (a^{2} + 2 a b x + b^{2} x^{2} + 1\right )^{\frac {3}{2}}}\, dx \]
\[ \int \frac {\text {arcsinh}(a+b x)^2}{\left (1+a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\int { \frac {\operatorname {arsinh}\left (b x + a\right )^{2}}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )}^{\frac {3}{2}}} \,d x } \]
\[ \int \frac {\text {arcsinh}(a+b x)^2}{\left (1+a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\int { \frac {\operatorname {arsinh}\left (b x + a\right )^{2}}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )}^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {\text {arcsinh}(a+b x)^2}{\left (1+a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\int \frac {{\mathrm {asinh}\left (a+b\,x\right )}^2}{{\left (a^2+2\,a\,b\,x+b^2\,x^2+1\right )}^{3/2}} \,d x \]