Integrand size = 24, antiderivative size = 149 \[ \int \frac {\left (1+c^2 x^2\right )^{3/2}}{(a+b \text {arcsinh}(c x))^2} \, dx=-\frac {\left (1+c^2 x^2\right )^2}{b c (a+b \text {arcsinh}(c x))}-\frac {\text {Chi}\left (\frac {2 (a+b \text {arcsinh}(c x))}{b}\right ) \sinh \left (\frac {2 a}{b}\right )}{b^2 c}-\frac {\text {Chi}\left (\frac {4 (a+b \text {arcsinh}(c x))}{b}\right ) \sinh \left (\frac {4 a}{b}\right )}{2 b^2 c}+\frac {\cosh \left (\frac {2 a}{b}\right ) \text {Shi}\left (\frac {2 (a+b \text {arcsinh}(c x))}{b}\right )}{b^2 c}+\frac {\cosh \left (\frac {4 a}{b}\right ) \text {Shi}\left (\frac {4 (a+b \text {arcsinh}(c x))}{b}\right )}{2 b^2 c} \]
-(c^2*x^2+1)^2/b/c/(a+b*arcsinh(c*x))+cosh(2*a/b)*Shi(2*(a+b*arcsinh(c*x)) /b)/b^2/c+1/2*cosh(4*a/b)*Shi(4*(a+b*arcsinh(c*x))/b)/b^2/c-Chi(2*(a+b*arc sinh(c*x))/b)*sinh(2*a/b)/b^2/c-1/2*Chi(4*(a+b*arcsinh(c*x))/b)*sinh(4*a/b )/b^2/c
Time = 0.41 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.82 \[ \int \frac {\left (1+c^2 x^2\right )^{3/2}}{(a+b \text {arcsinh}(c x))^2} \, dx=\frac {-\frac {2 b \left (1+c^2 x^2\right )^2}{a+b \text {arcsinh}(c x)}-2 \text {Chi}\left (2 \left (\frac {a}{b}+\text {arcsinh}(c x)\right )\right ) \sinh \left (\frac {2 a}{b}\right )-\text {Chi}\left (4 \left (\frac {a}{b}+\text {arcsinh}(c x)\right )\right ) \sinh \left (\frac {4 a}{b}\right )+2 \cosh \left (\frac {2 a}{b}\right ) \text {Shi}\left (2 \left (\frac {a}{b}+\text {arcsinh}(c x)\right )\right )+\cosh \left (\frac {4 a}{b}\right ) \text {Shi}\left (4 \left (\frac {a}{b}+\text {arcsinh}(c x)\right )\right )}{2 b^2 c} \]
((-2*b*(1 + c^2*x^2)^2)/(a + b*ArcSinh[c*x]) - 2*CoshIntegral[2*(a/b + Arc Sinh[c*x])]*Sinh[(2*a)/b] - CoshIntegral[4*(a/b + ArcSinh[c*x])]*Sinh[(4*a )/b] + 2*Cosh[(2*a)/b]*SinhIntegral[2*(a/b + ArcSinh[c*x])] + Cosh[(4*a)/b ]*SinhIntegral[4*(a/b + ArcSinh[c*x])])/(2*b^2*c)
Time = 0.59 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.93, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {6205, 6234, 25, 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.
\(\displaystyle \int \frac {\left (c^2 x^2+1\right )^{3/2}}{(a+b \text {arcsinh}(c x))^2} \, dx\) |
\(\Big \downarrow \) 6205 |
\(\displaystyle \frac {4 c \int \frac {x \left (c^2 x^2+1\right )}{a+b \text {arcsinh}(c x)}dx}{b}-\frac {\left (c^2 x^2+1\right )^2}{b c (a+b \text {arcsinh}(c x))}\) |
\(\Big \downarrow \) 6234 |
\(\displaystyle \frac {4 \int -\frac {\cosh ^3\left (\frac {a}{b}-\frac {a+b \text {arcsinh}(c x)}{b}\right ) \sinh \left (\frac {a}{b}-\frac {a+b \text {arcsinh}(c x)}{b}\right )}{a+b \text {arcsinh}(c x)}d(a+b \text {arcsinh}(c x))}{b^2 c}-\frac {\left (c^2 x^2+1\right )^2}{b c (a+b \text {arcsinh}(c x))}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {4 \int \frac {\cosh ^3\left (\frac {a}{b}-\frac {a+b \text {arcsinh}(c x)}{b}\right ) \sinh \left (\frac {a}{b}-\frac {a+b \text {arcsinh}(c x)}{b}\right )}{a+b \text {arcsinh}(c x)}d(a+b \text {arcsinh}(c x))}{b^2 c}-\frac {\left (c^2 x^2+1\right )^2}{b c (a+b \text {arcsinh}(c x))}\) |
\(\Big \downarrow \) 5971 |
\(\displaystyle -\frac {4 \int \left (\frac {\sinh \left (\frac {4 a}{b}-\frac {4 (a+b \text {arcsinh}(c x))}{b}\right )}{8 (a+b \text {arcsinh}(c x))}+\frac {\sinh \left (\frac {2 a}{b}-\frac {2 (a+b \text {arcsinh}(c x))}{b}\right )}{4 (a+b \text {arcsinh}(c x))}\right )d(a+b \text {arcsinh}(c x))}{b^2 c}-\frac {\left (c^2 x^2+1\right )^2}{b c (a+b \text {arcsinh}(c x))}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {4 \left (-\frac {1}{4} \sinh \left (\frac {2 a}{b}\right ) \text {Chi}\left (\frac {2 (a+b \text {arcsinh}(c x))}{b}\right )-\frac {1}{8} \sinh \left (\frac {4 a}{b}\right ) \text {Chi}\left (\frac {4 (a+b \text {arcsinh}(c x))}{b}\right )+\frac {1}{4} \cosh \left (\frac {2 a}{b}\right ) \text {Shi}\left (\frac {2 (a+b \text {arcsinh}(c x))}{b}\right )+\frac {1}{8} \cosh \left (\frac {4 a}{b}\right ) \text {Shi}\left (\frac {4 (a+b \text {arcsinh}(c x))}{b}\right )\right )}{b^2 c}-\frac {\left (c^2 x^2+1\right )^2}{b c (a+b \text {arcsinh}(c x))}\) |
-((1 + c^2*x^2)^2/(b*c*(a + b*ArcSinh[c*x]))) + (4*(-1/4*(CoshIntegral[(2* (a + b*ArcSinh[c*x]))/b]*Sinh[(2*a)/b]) - (CoshIntegral[(4*(a + b*ArcSinh[ c*x]))/b]*Sinh[(4*a)/b])/8 + (Cosh[(2*a)/b]*SinhIntegral[(2*(a + b*ArcSinh [c*x]))/b])/4 + (Cosh[(4*a)/b]*SinhIntegral[(4*(a + b*ArcSinh[c*x]))/b])/8 ))/(b^2*c)
3.5.22.3.1 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]
Time = 0.49 (sec) , antiderivative size = 255, normalized size of antiderivative = 1.71
method | result | size |
default | \(-\frac {4 b \,c^{4} x^{4}+8 b \,c^{2} x^{2}+{\mathrm e}^{-\frac {4 a}{b}} \operatorname {Ei}_{1}\left (-4 \,\operatorname {arcsinh}\left (c x \right )-\frac {4 a}{b}\right ) b \,\operatorname {arcsinh}\left (c x \right )+2 \,{\mathrm e}^{-\frac {2 a}{b}} \operatorname {Ei}_{1}\left (-2 \,\operatorname {arcsinh}\left (c x \right )-\frac {2 a}{b}\right ) b \,\operatorname {arcsinh}\left (c x \right )-{\mathrm e}^{\frac {4 a}{b}} \operatorname {Ei}_{1}\left (4 \,\operatorname {arcsinh}\left (c x \right )+\frac {4 a}{b}\right ) b \,\operatorname {arcsinh}\left (c x \right )-2 \,{\mathrm e}^{\frac {2 a}{b}} \operatorname {Ei}_{1}\left (2 \,\operatorname {arcsinh}\left (c x \right )+\frac {2 a}{b}\right ) b \,\operatorname {arcsinh}\left (c x \right )+{\mathrm e}^{-\frac {4 a}{b}} \operatorname {Ei}_{1}\left (-4 \,\operatorname {arcsinh}\left (c x \right )-\frac {4 a}{b}\right ) a +2 \,{\mathrm e}^{-\frac {2 a}{b}} \operatorname {Ei}_{1}\left (-2 \,\operatorname {arcsinh}\left (c x \right )-\frac {2 a}{b}\right ) a -{\mathrm e}^{\frac {4 a}{b}} \operatorname {Ei}_{1}\left (4 \,\operatorname {arcsinh}\left (c x \right )+\frac {4 a}{b}\right ) a -2 \,{\mathrm e}^{\frac {2 a}{b}} \operatorname {Ei}_{1}\left (2 \,\operatorname {arcsinh}\left (c x \right )+\frac {2 a}{b}\right ) a +4 b}{4 c \,b^{2} \left (a +b \,\operatorname {arcsinh}\left (c x \right )\right )}\) | \(255\) |
-1/4*(4*b*c^4*x^4+8*b*c^2*x^2+exp(-4*a/b)*Ei(1,-4*arcsinh(c*x)-4*a/b)*b*ar csinh(c*x)+2*exp(-2*a/b)*Ei(1,-2*arcsinh(c*x)-2*a/b)*b*arcsinh(c*x)-exp(4* a/b)*Ei(1,4*arcsinh(c*x)+4*a/b)*b*arcsinh(c*x)-2*exp(2*a/b)*Ei(1,2*arcsinh (c*x)+2*a/b)*b*arcsinh(c*x)+exp(-4*a/b)*Ei(1,-4*arcsinh(c*x)-4*a/b)*a+2*ex p(-2*a/b)*Ei(1,-2*arcsinh(c*x)-2*a/b)*a-exp(4*a/b)*Ei(1,4*arcsinh(c*x)+4*a /b)*a-2*exp(2*a/b)*Ei(1,2*arcsinh(c*x)+2*a/b)*a+4*b)/c/b^2/(a+b*arcsinh(c* x))
\[ \int \frac {\left (1+c^2 x^2\right )^{3/2}}{(a+b \text {arcsinh}(c x))^2} \, dx=\int { \frac {{\left (c^{2} x^{2} + 1\right )}^{\frac {3}{2}}}{{\left (b \operatorname {arsinh}\left (c x\right ) + a\right )}^{2}} \,d x } \]
\[ \int \frac {\left (1+c^2 x^2\right )^{3/2}}{(a+b \text {arcsinh}(c x))^2} \, dx=\int \frac {\left (c^{2} x^{2} + 1\right )^{\frac {3}{2}}}{\left (a + b \operatorname {asinh}{\left (c x \right )}\right )^{2}}\, dx \]
\[ \int \frac {\left (1+c^2 x^2\right )^{3/2}}{(a+b \text {arcsinh}(c x))^2} \, dx=\int { \frac {{\left (c^{2} x^{2} + 1\right )}^{\frac {3}{2}}}{{\left (b \operatorname {arsinh}\left (c x\right ) + a\right )}^{2}} \,d x } \]
-((c^4*x^4 + 2*c^2*x^2 + 1)*(c^2*x^2 + 1) + (c^5*x^5 + 2*c^3*x^3 + c*x)*sq rt(c^2*x^2 + 1))/(a*b*c^3*x^2 + sqrt(c^2*x^2 + 1)*a*b*c^2*x + a*b*c + (b^2 *c^3*x^2 + sqrt(c^2*x^2 + 1)*b^2*c^2*x + b^2*c)*log(c*x + sqrt(c^2*x^2 + 1 ))) + integrate(((4*c^4*x^4 + 3*c^2*x^2 - 1)*(c^2*x^2 + 1)^(3/2) + 4*(2*c^ 5*x^5 + 3*c^3*x^3 + c*x)*(c^2*x^2 + 1) + (4*c^6*x^6 + 9*c^4*x^4 + 6*c^2*x^ 2 + 1)*sqrt(c^2*x^2 + 1))/(a*b*c^4*x^4 + (c^2*x^2 + 1)*a*b*c^2*x^2 + 2*a*b *c^2*x^2 + a*b + (b^2*c^4*x^4 + (c^2*x^2 + 1)*b^2*c^2*x^2 + 2*b^2*c^2*x^2 + b^2 + 2*(b^2*c^3*x^3 + b^2*c*x)*sqrt(c^2*x^2 + 1))*log(c*x + sqrt(c^2*x^ 2 + 1)) + 2*(a*b*c^3*x^3 + a*b*c*x)*sqrt(c^2*x^2 + 1)), x)
\[ \int \frac {\left (1+c^2 x^2\right )^{3/2}}{(a+b \text {arcsinh}(c x))^2} \, dx=\int { \frac {{\left (c^{2} x^{2} + 1\right )}^{\frac {3}{2}}}{{\left (b \operatorname {arsinh}\left (c x\right ) + a\right )}^{2}} \,d x } \]
Timed out. \[ \int \frac {\left (1+c^2 x^2\right )^{3/2}}{(a+b \text {arcsinh}(c x))^2} \, dx=\int \frac {{\left (c^2\,x^2+1\right )}^{3/2}}{{\left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )}^2} \,d x \]