Integrand size = 30, antiderivative size = 54 \[ \int \frac {\left (1+a^2+2 a b x+b^2 x^2\right )^{3/2}}{\text {arcsinh}(a+b x)^2} \, dx=-\frac {\left (1+(a+b x)^2\right )^2}{b \text {arcsinh}(a+b x)}+\frac {\text {Shi}(2 \text {arcsinh}(a+b x))}{b}+\frac {\text {Shi}(4 \text {arcsinh}(a+b x))}{2 b} \]
Time = 0.22 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.30 \[ \int \frac {\left (1+a^2+2 a b x+b^2 x^2\right )^{3/2}}{\text {arcsinh}(a+b x)^2} \, dx=\frac {-2 \left (1+a^2+2 a b x+b^2 x^2\right )^2+2 \text {arcsinh}(a+b x) \text {Shi}(2 \text {arcsinh}(a+b x))+\text {arcsinh}(a+b x) \text {Shi}(4 \text {arcsinh}(a+b x))}{2 b \text {arcsinh}(a+b x)} \]
(-2*(1 + a^2 + 2*a*b*x + b^2*x^2)^2 + 2*ArcSinh[a + b*x]*SinhIntegral[2*Ar cSinh[a + b*x]] + ArcSinh[a + b*x]*SinhIntegral[4*ArcSinh[a + b*x]])/(2*b* ArcSinh[a + b*x])
Time = 0.49 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.02, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {6275, 6205, 6234, 5971, 2009}
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.
| \(\displaystyle \int \frac {\left (a^2+2 a b x+b^2 x^2+1\right )^{3/2}}{\text {arcsinh}(a+b x)^2} \, dx\) |
| \(\Big \downarrow \) 6275 |
| \(\displaystyle \frac {\int \frac {\left ((a+b x)^2+1\right )^{3/2}}{\text {arcsinh}(a+b x)^2}d(a+b x)}{b}\) |
| \(\Big \downarrow \) 6205 |
| \(\displaystyle \frac {4 \int \frac {(a+b x) \left ((a+b x)^2+1\right )}{\text {arcsinh}(a+b x)}d(a+b x)-\frac {\left ((a+b x)^2+1\right )^2}{\text {arcsinh}(a+b x)}}{b}\) |
| \(\Big \downarrow \) 6234 |
| \(\displaystyle \frac {4 \int \frac {(a+b x) \left ((a+b x)^2+1\right )^{3/2}}{\text {arcsinh}(a+b x)}d\text {arcsinh}(a+b x)-\frac {\left ((a+b x)^2+1\right )^2}{\text {arcsinh}(a+b x)}}{b}\) |
| \(\Big \downarrow \) 5971 |
| \(\displaystyle \frac {4 \int \left (\frac {\sinh (2 \text {arcsinh}(a+b x))}{4 \text {arcsinh}(a+b x)}+\frac {\sinh (4 \text {arcsinh}(a+b x))}{8 \text {arcsinh}(a+b x)}\right )d\text {arcsinh}(a+b x)-\frac {\left ((a+b x)^2+1\right )^2}{\text {arcsinh}(a+b x)}}{b}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {4 \left (\frac {1}{4} \text {Shi}(2 \text {arcsinh}(a+b x))+\frac {1}{8} \text {Shi}(4 \text {arcsinh}(a+b x))\right )-\frac {\left ((a+b x)^2+1\right )^2}{\text {arcsinh}(a+b x)}}{b}\) |
(-((1 + (a + b*x)^2)^2/ArcSinh[a + b*x]) + 4*(SinhIntegral[2*ArcSinh[a + b *x]]/4 + SinhIntegral[4*ArcSinh[a + b*x]]/8))/b
3.3.70.3.1 Defintions of rubi rules used
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_)*((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_.)*(x_)]*(b_.))^(n_.)*(x_)^(m_.)*((d_) + (e_.)*(x_) ^2)^(p_.), x_Symbol] :> Simp[(1/(b*c^(m + 1)))*Simp[(d + e*x^2)^p/(1 + c^2* x^2)^p] Subst[Int[x^n*Sinh[-a/b + x/b]^m*Cosh[-a/b + x/b]^(2*p + 1), x], x, a + b*ArcSinh[c*x]], x] /; FreeQ[{a, b, c, d, e, n}, x] && EqQ[e, c^2*d] && IGtQ[2*p + 2, 0] && IGtQ[m, 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 = 0.70 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.33
| method | result | size |
| default | \(\frac {8 \,\operatorname {Shi}\left (2 \,\operatorname {arcsinh}\left (b x +a \right )\right ) \operatorname {arcsinh}\left (b x +a \right )+4 \,\operatorname {Shi}\left (4 \,\operatorname {arcsinh}\left (b x +a \right )\right ) \operatorname {arcsinh}\left (b x +a \right )-4 \cosh \left (2 \,\operatorname {arcsinh}\left (b x +a \right )\right )-\cosh \left (4 \,\operatorname {arcsinh}\left (b x +a \right )\right )-3}{8 b \,\operatorname {arcsinh}\left (b x +a \right )}\) | \(72\) |
1/8/b*(8*Shi(2*arcsinh(b*x+a))*arcsinh(b*x+a)+4*Shi(4*arcsinh(b*x+a))*arcs inh(b*x+a)-4*cosh(2*arcsinh(b*x+a))-cosh(4*arcsinh(b*x+a))-3)/arcsinh(b*x+ a)
\[ \int \frac {\left (1+a^2+2 a b x+b^2 x^2\right )^{3/2}}{\text {arcsinh}(a+b x)^2} \, dx=\int { \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )}^{\frac {3}{2}}}{\operatorname {arsinh}\left (b x + a\right )^{2}} \,d x } \]
\[ \int \frac {\left (1+a^2+2 a b x+b^2 x^2\right )^{3/2}}{\text {arcsinh}(a+b x)^2} \, dx=\int \frac {\left (a^{2} + 2 a b x + b^{2} x^{2} + 1\right )^{\frac {3}{2}}}{\operatorname {asinh}^{2}{\left (a + b x \right )}}\, dx \]
\[ \int \frac {\left (1+a^2+2 a b x+b^2 x^2\right )^{3/2}}{\text {arcsinh}(a+b x)^2} \, dx=\int { \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )}^{\frac {3}{2}}}{\operatorname {arsinh}\left (b x + a\right )^{2}} \,d x } \]
-((b^4*x^4 + 4*a*b^3*x^3 + a^4 + 2*(3*a^2*b^2 + b^2)*x^2 + 2*a^2 + 4*(a^3* b + a*b)*x + 1)*(b^2*x^2 + 2*a*b*x + a^2 + 1) + (b^5*x^5 + 5*a*b^4*x^4 + a ^5 + 2*(5*a^2*b^3 + b^3)*x^3 + 2*a^3 + 2*(5*a^3*b^2 + 3*a*b^2)*x^2 + (5*a^ 4*b + 6*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)*lo g(b*x + a + sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1))) + integrate(((4*b^4*x^4 + 16*a*b^3*x^3 + 4*a^4 + 3*(8*a^2*b^2 + b^2)*x^2 + 3*a^2 + 2*(8*a^3*b + 3*a* b)*x - 1)*(b^2*x^2 + 2*a*b*x + a^2 + 1)^(3/2) + 4*(2*b^5*x^5 + 10*a*b^4*x^ 4 + 2*a^5 + (20*a^2*b^3 + 3*b^3)*x^3 + 3*a^3 + (20*a^3*b^2 + 9*a*b^2)*x^2 + (10*a^4*b + 9*a^2*b + b)*x + a)*(b^2*x^2 + 2*a*b*x + a^2 + 1) + (4*b^6*x ^6 + 24*a*b^5*x^5 + 4*a^6 + 3*(20*a^2*b^4 + 3*b^4)*x^4 + 9*a^4 + 4*(20*a^3 *b^3 + 9*a*b^3)*x^3 + 6*(10*a^4*b^2 + 9*a^2*b^2 + b^2)*x^2 + 6*a^2 + 12*(2 *a^5*b + 3*a^3*b + 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)
\[ \int \frac {\left (1+a^2+2 a b x+b^2 x^2\right )^{3/2}}{\text {arcsinh}(a+b x)^2} \, dx=\int { \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )}^{\frac {3}{2}}}{\operatorname {arsinh}\left (b x + a\right )^{2}} \,d x } \]
Timed out. \[ \int \frac {\left (1+a^2+2 a b x+b^2 x^2\right )^{3/2}}{\text {arcsinh}(a+b x)^2} \, dx=\int \frac {{\left (a^2+2\,a\,b\,x+b^2\,x^2+1\right )}^{3/2}}{{\mathrm {asinh}\left (a+b\,x\right )}^2} \,d x \]