Integrand size = 25, antiderivative size = 54 \[ \int \frac {x}{\sqrt {1+c^2 x^2} (a+b \text {arcsinh}(c x))} \, dx=-\frac {\text {Chi}\left (\frac {a+b \text {arcsinh}(c x)}{b}\right ) \sinh \left (\frac {a}{b}\right )}{b c^2}+\frac {\cosh \left (\frac {a}{b}\right ) \text {Shi}\left (\frac {a+b \text {arcsinh}(c x)}{b}\right )}{b c^2} \]
Time = 0.17 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.85 \[ \int \frac {x}{\sqrt {1+c^2 x^2} (a+b \text {arcsinh}(c x))} \, dx=-\frac {\text {Chi}\left (\frac {a}{b}+\text {arcsinh}(c x)\right ) \sinh \left (\frac {a}{b}\right )-\cosh \left (\frac {a}{b}\right ) \text {Shi}\left (\frac {a}{b}+\text {arcsinh}(c x)\right )}{b c^2} \]
-((CoshIntegral[a/b + ArcSinh[c*x]]*Sinh[a/b] - Cosh[a/b]*SinhIntegral[a/b + ArcSinh[c*x]])/(b*c^2))
Result contains complex when optimal does not.
Time = 0.48 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.06, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {6234, 25, 3042, 26, 3784, 26, 3042, 26, 3779, 3782}
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.
\(\displaystyle \int \frac {x}{\sqrt {c^2 x^2+1} (a+b \text {arcsinh}(c x))} \, dx\) |
\(\Big \downarrow \) 6234 |
\(\displaystyle \frac {\int -\frac {\sinh \left (\frac {a}{b}-\frac {a+b \text {arcsinh}(c x)}{b}\right )}{a+b \text {arcsinh}(c x)}d(a+b \text {arcsinh}(c x))}{b c^2}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {\int \frac {\sinh \left (\frac {a}{b}-\frac {a+b \text {arcsinh}(c x)}{b}\right )}{a+b \text {arcsinh}(c x)}d(a+b \text {arcsinh}(c x))}{b c^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {\int -\frac {i \sin \left (\frac {i a}{b}-\frac {i (a+b \text {arcsinh}(c x))}{b}\right )}{a+b \text {arcsinh}(c x)}d(a+b \text {arcsinh}(c x))}{b c^2}\) |
\(\Big \downarrow \) 26 |
\(\displaystyle \frac {i \int \frac {\sin \left (\frac {i a}{b}-\frac {i (a+b \text {arcsinh}(c x))}{b}\right )}{a+b \text {arcsinh}(c x)}d(a+b \text {arcsinh}(c x))}{b c^2}\) |
\(\Big \downarrow \) 3784 |
\(\displaystyle \frac {i \left (i \sinh \left (\frac {a}{b}\right ) \int \frac {\cosh \left (\frac {a+b \text {arcsinh}(c x)}{b}\right )}{a+b \text {arcsinh}(c x)}d(a+b \text {arcsinh}(c x))+\cosh \left (\frac {a}{b}\right ) \int -\frac {i \sinh \left (\frac {a+b \text {arcsinh}(c x)}{b}\right )}{a+b \text {arcsinh}(c x)}d(a+b \text {arcsinh}(c x))\right )}{b c^2}\) |
\(\Big \downarrow \) 26 |
\(\displaystyle \frac {i \left (i \sinh \left (\frac {a}{b}\right ) \int \frac {\cosh \left (\frac {a+b \text {arcsinh}(c x)}{b}\right )}{a+b \text {arcsinh}(c x)}d(a+b \text {arcsinh}(c x))-i \cosh \left (\frac {a}{b}\right ) \int \frac {\sinh \left (\frac {a+b \text {arcsinh}(c x)}{b}\right )}{a+b \text {arcsinh}(c x)}d(a+b \text {arcsinh}(c x))\right )}{b c^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {i \left (i \sinh \left (\frac {a}{b}\right ) \int \frac {\sin \left (\frac {i (a+b \text {arcsinh}(c x))}{b}+\frac {\pi }{2}\right )}{a+b \text {arcsinh}(c x)}d(a+b \text {arcsinh}(c x))-i \cosh \left (\frac {a}{b}\right ) \int -\frac {i \sin \left (\frac {i (a+b \text {arcsinh}(c x))}{b}\right )}{a+b \text {arcsinh}(c x)}d(a+b \text {arcsinh}(c x))\right )}{b c^2}\) |
\(\Big \downarrow \) 26 |
\(\displaystyle \frac {i \left (i \sinh \left (\frac {a}{b}\right ) \int \frac {\sin \left (\frac {i (a+b \text {arcsinh}(c x))}{b}+\frac {\pi }{2}\right )}{a+b \text {arcsinh}(c x)}d(a+b \text {arcsinh}(c x))-\cosh \left (\frac {a}{b}\right ) \int \frac {\sin \left (\frac {i (a+b \text {arcsinh}(c x))}{b}\right )}{a+b \text {arcsinh}(c x)}d(a+b \text {arcsinh}(c x))\right )}{b c^2}\) |
\(\Big \downarrow \) 3779 |
\(\displaystyle \frac {i \left (i \sinh \left (\frac {a}{b}\right ) \int \frac {\sin \left (\frac {i (a+b \text {arcsinh}(c x))}{b}+\frac {\pi }{2}\right )}{a+b \text {arcsinh}(c x)}d(a+b \text {arcsinh}(c x))-i \cosh \left (\frac {a}{b}\right ) \text {Shi}\left (\frac {a+b \text {arcsinh}(c x)}{b}\right )\right )}{b c^2}\) |
\(\Big \downarrow \) 3782 |
\(\displaystyle \frac {i \left (i \sinh \left (\frac {a}{b}\right ) \text {Chi}\left (\frac {a+b \text {arcsinh}(c x)}{b}\right )-i \cosh \left (\frac {a}{b}\right ) \text {Shi}\left (\frac {a+b \text {arcsinh}(c x)}{b}\right )\right )}{b c^2}\) |
(I*(I*CoshIntegral[(a + b*ArcSinh[c*x])/b]*Sinh[a/b] - I*Cosh[a/b]*SinhInt egral[(a + b*ArcSinh[c*x])/b]))/(b*c^2)
3.4.92.3.1 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[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[sin[(e_.) + (Complex[0, fz_])*(f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbo l] :> Simp[CoshIntegral[c*f*(fz/d) + f*fz*x]/d, x] /; FreeQ[{c, d, e, f, fz }, x] && EqQ[d*(e - Pi/2) - c*f*fz*I, 0]
Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[Cos[(d* e - c*f)/d] Int[Sin[c*(f/d) + f*x]/(c + d*x), x], x] + Simp[Sin[(d*e - c* f)/d] Int[Cos[c*(f/d) + f*x]/(c + d*x), x], x] /; FreeQ[{c, d, e, f}, x] && NeQ[d*e - c*f, 0]
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 = 0.27 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.98
method | result | size |
default | \(\frac {{\mathrm e}^{\frac {a}{b}} \operatorname {Ei}_{1}\left (\operatorname {arcsinh}\left (c x \right )+\frac {a}{b}\right )-{\mathrm e}^{-\frac {a}{b}} \operatorname {Ei}_{1}\left (-\operatorname {arcsinh}\left (c x \right )-\frac {a}{b}\right )}{2 c^{2} b}\) | \(53\) |
\[ \int \frac {x}{\sqrt {1+c^2 x^2} (a+b \text {arcsinh}(c x))} \, dx=\int { \frac {x}{\sqrt {c^{2} x^{2} + 1} {\left (b \operatorname {arsinh}\left (c x\right ) + a\right )}} \,d x } \]
\[ \int \frac {x}{\sqrt {1+c^2 x^2} (a+b \text {arcsinh}(c x))} \, dx=\int \frac {x}{\left (a + b \operatorname {asinh}{\left (c x \right )}\right ) \sqrt {c^{2} x^{2} + 1}}\, dx \]
\[ \int \frac {x}{\sqrt {1+c^2 x^2} (a+b \text {arcsinh}(c x))} \, dx=\int { \frac {x}{\sqrt {c^{2} x^{2} + 1} {\left (b \operatorname {arsinh}\left (c x\right ) + a\right )}} \,d x } \]
\[ \int \frac {x}{\sqrt {1+c^2 x^2} (a+b \text {arcsinh}(c x))} \, dx=\int { \frac {x}{\sqrt {c^{2} x^{2} + 1} {\left (b \operatorname {arsinh}\left (c x\right ) + a\right )}} \,d x } \]
Timed out. \[ \int \frac {x}{\sqrt {1+c^2 x^2} (a+b \text {arcsinh}(c x))} \, dx=\int \frac {x}{\left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )\,\sqrt {c^2\,x^2+1}} \,d x \]