Integrand size = 30, antiderivative size = 36 \[ \int \frac {\sqrt {1+a^2+2 a b x+b^2 x^2}}{\text {arcsinh}(a+b x)^2} \, dx=-\frac {1+(a+b x)^2}{b \text {arcsinh}(a+b x)}+\frac {\text {Shi}(2 \text {arcsinh}(a+b x))}{b} \]
Time = 0.05 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.31 \[ \int \frac {\sqrt {1+a^2+2 a b x+b^2 x^2}}{\text {arcsinh}(a+b x)^2} \, dx=-\frac {1+a^2+2 a b x+b^2 x^2-\text {arcsinh}(a+b x) \text {Shi}(2 \text {arcsinh}(a+b x))}{b \text {arcsinh}(a+b x)} \]
-((1 + a^2 + 2*a*b*x + b^2*x^2 - ArcSinh[a + b*x]*SinhIntegral[2*ArcSinh[a + b*x]])/(b*ArcSinh[a + b*x]))
Time = 0.47 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.92, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {6275, 6205, 6195, 5971, 27, 3042, 26, 3779}
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.
\(\displaystyle \int \frac {\sqrt {a^2+2 a b x+b^2 x^2+1}}{\text {arcsinh}(a+b x)^2} \, dx\) |
\(\Big \downarrow \) 6275 |
\(\displaystyle \frac {\int \frac {\sqrt {(a+b x)^2+1}}{\text {arcsinh}(a+b x)^2}d(a+b x)}{b}\) |
\(\Big \downarrow \) 6205 |
\(\displaystyle \frac {2 \int \frac {a+b x}{\text {arcsinh}(a+b x)}d(a+b x)-\frac {(a+b x)^2+1}{\text {arcsinh}(a+b x)}}{b}\) |
\(\Big \downarrow \) 6195 |
\(\displaystyle \frac {2 \int \frac {(a+b x) \sqrt {(a+b x)^2+1}}{\text {arcsinh}(a+b x)}d\text {arcsinh}(a+b x)-\frac {(a+b x)^2+1}{\text {arcsinh}(a+b x)}}{b}\) |
\(\Big \downarrow \) 5971 |
\(\displaystyle \frac {2 \int \frac {\sinh (2 \text {arcsinh}(a+b x))}{2 \text {arcsinh}(a+b x)}d\text {arcsinh}(a+b x)-\frac {(a+b x)^2+1}{\text {arcsinh}(a+b x)}}{b}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {\sinh (2 \text {arcsinh}(a+b x))}{\text {arcsinh}(a+b x)}d\text {arcsinh}(a+b x)-\frac {(a+b x)^2+1}{\text {arcsinh}(a+b x)}}{b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {-\frac {(a+b x)^2+1}{\text {arcsinh}(a+b x)}+\int -\frac {i \sin (2 i \text {arcsinh}(a+b x))}{\text {arcsinh}(a+b x)}d\text {arcsinh}(a+b x)}{b}\) |
\(\Big \downarrow \) 26 |
\(\displaystyle \frac {-\frac {(a+b x)^2+1}{\text {arcsinh}(a+b x)}-i \int \frac {\sin (2 i \text {arcsinh}(a+b x))}{\text {arcsinh}(a+b x)}d\text {arcsinh}(a+b x)}{b}\) |
\(\Big \downarrow \) 3779 |
\(\displaystyle \frac {\text {Shi}(2 \text {arcsinh}(a+b x))-\frac {(a+b x)^2+1}{\text {arcsinh}(a+b x)}}{b}\) |
3.3.64.3.1 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[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[sin[(e_.) + (Complex[0, fz_])*(f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbo l] :> Simp[I*(SinhIntegral[c*f*(fz/d) + f*fz*x]/d), x] /; FreeQ[{c, d, e, f , fz}, x] && EqQ[d*e - c*f*fz*I, 0]
Int[Cosh[(a_.) + (b_.)*(x_)]^(p_.)*((c_.) + (d_.)*(x_))^(m_.)*Sinh[(a_.) + (b_.)*(x_)]^(n_.), x_Symbol] :> Int[ExpandTrigReduce[(c + d*x)^m, Sinh[a + b*x]^n*Cosh[a + b*x]^p, x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0] & & IGtQ[p, 0]
Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_)*(x_)^(m_.), x_Symbol] :> Simp[ 1/(b*c^(m + 1)) Subst[Int[x^n*Sinh[-a/b + x/b]^m*Cosh[-a/b + x/b], x], x, a + b*ArcSinh[c*x]], x] /; FreeQ[{a, b, c, n}, x] && IGtQ[m, 0]
Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_)*((d_) + (e_.)*(x_)^2)^(p_.), x _Symbol] :> Simp[Simp[Sqrt[1 + c^2*x^2]*(d + e*x^2)^p]*((a + b*ArcSinh[c*x] )^(n + 1)/(b*c*(n + 1))), x] - Simp[c*((2*p + 1)/(b*(n + 1)))*Simp[(d + e*x ^2)^p/(1 + c^2*x^2)^p] Int[x*(1 + c^2*x^2)^(p - 1/2)*(a + b*ArcSinh[c*x]) ^(n + 1), x], x] /; FreeQ[{a, b, c, d, e, p}, x] && EqQ[e, c^2*d] && LtQ[n, -1]
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 = 0.85 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.22
method | result | size |
default | \(\frac {2 \,\operatorname {Shi}\left (2 \,\operatorname {arcsinh}\left (b x +a \right )\right ) \operatorname {arcsinh}\left (b x +a \right )-\cosh \left (2 \,\operatorname {arcsinh}\left (b x +a \right )\right )-1}{2 b \,\operatorname {arcsinh}\left (b x +a \right )}\) | \(44\) |
\[ \int \frac {\sqrt {1+a^2+2 a b x+b^2 x^2}}{\text {arcsinh}(a+b x)^2} \, dx=\int { \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{\operatorname {arsinh}\left (b x + a\right )^{2}} \,d x } \]
\[ \int \frac {\sqrt {1+a^2+2 a b x+b^2 x^2}}{\text {arcsinh}(a+b x)^2} \, dx=\int \frac {\sqrt {a^{2} + 2 a b x + b^{2} x^{2} + 1}}{\operatorname {asinh}^{2}{\left (a + b x \right )}}\, dx \]
\[ \int \frac {\sqrt {1+a^2+2 a b x+b^2 x^2}}{\text {arcsinh}(a+b x)^2} \, dx=\int { \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{\operatorname {arsinh}\left (b x + a\right )^{2}} \,d x } \]
-((b^2*x^2 + 2*a*b*x + a^2 + 1)^2 + (b^3*x^3 + 3*a*b^2*x^2 + a^3 + (3*a^2* b + b)*x + a)*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1))/((b^3*x^2 + 2*a*b^2*x + a ^2*b + sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)*(b^2*x + a*b) + b)*log(b*x + a + sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1))) + integrate(((2*b^2*x^2 + 4*a*b*x + 2* a^2 - 1)*(b^2*x^2 + 2*a*b*x + a^2 + 1)^(3/2) + 2*(2*b^3*x^3 + 6*a*b^2*x^2 + 2*a^3 + (6*a^2*b + b)*x + a)*(b^2*x^2 + 2*a*b*x + a^2 + 1) + (2*b^4*x^4 + 8*a*b^3*x^3 + 2*a^4 + 3*(4*a^2*b^2 + b^2)*x^2 + 3*a^2 + 2*(4*a^3*b + 3*a *b)*x + 1)*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1))/((b^4*x^4 + 4*a*b^3*x^3 + a^ 4 + 2*(3*a^2*b^2 + b^2)*x^2 + (b^2*x^2 + 2*a*b*x + a^2 + 1)*(b^2*x^2 + 2*a *b*x + a^2) + 2*a^2 + 4*(a^3*b + a*b)*x + 2*(b^3*x^3 + 3*a*b^2*x^2 + a^3 + (3*a^2*b + b)*x + a)*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1) + 1)*log(b*x + a + sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1))), x)
\[ \int \frac {\sqrt {1+a^2+2 a b x+b^2 x^2}}{\text {arcsinh}(a+b x)^2} \, dx=\int { \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{\operatorname {arsinh}\left (b x + a\right )^{2}} \,d x } \]
Timed out. \[ \int \frac {\sqrt {1+a^2+2 a b x+b^2 x^2}}{\text {arcsinh}(a+b x)^2} \, dx=\int \frac {\sqrt {a^2+2\,a\,b\,x+b^2\,x^2+1}}{{\mathrm {asinh}\left (a+b\,x\right )}^2} \,d x \]