Integrand size = 16, antiderivative size = 259 \[ \int x \sqrt {a+b \text {arcsinh}(c+d x)} \, dx=-\frac {c (c+d x) \sqrt {a+b \text {arcsinh}(c+d x)}}{d^2}+\frac {\sqrt {a+b \text {arcsinh}(c+d x)} \cosh (2 \text {arcsinh}(c+d x))}{4 d^2}-\frac {\sqrt {b} c e^{a/b} \sqrt {\pi } \text {erf}\left (\frac {\sqrt {a+b \text {arcsinh}(c+d x)}}{\sqrt {b}}\right )}{4 d^2}-\frac {\sqrt {b} e^{\frac {2 a}{b}} \sqrt {\frac {\pi }{2}} \text {erf}\left (\frac {\sqrt {2} \sqrt {a+b \text {arcsinh}(c+d x)}}{\sqrt {b}}\right )}{16 d^2}+\frac {\sqrt {b} c e^{-\frac {a}{b}} \sqrt {\pi } \text {erfi}\left (\frac {\sqrt {a+b \text {arcsinh}(c+d x)}}{\sqrt {b}}\right )}{4 d^2}-\frac {\sqrt {b} e^{-\frac {2 a}{b}} \sqrt {\frac {\pi }{2}} \text {erfi}\left (\frac {\sqrt {2} \sqrt {a+b \text {arcsinh}(c+d x)}}{\sqrt {b}}\right )}{16 d^2} \]
-1/32*exp(2*a/b)*erf(2^(1/2)*(a+b*arcsinh(d*x+c))^(1/2)/b^(1/2))*b^(1/2)*2 ^(1/2)*Pi^(1/2)/d^2-1/32*erfi(2^(1/2)*(a+b*arcsinh(d*x+c))^(1/2)/b^(1/2))* b^(1/2)*2^(1/2)*Pi^(1/2)/d^2/exp(2*a/b)-1/4*c*exp(a/b)*erf((a+b*arcsinh(d* x+c))^(1/2)/b^(1/2))*b^(1/2)*Pi^(1/2)/d^2+1/4*c*erfi((a+b*arcsinh(d*x+c))^ (1/2)/b^(1/2))*b^(1/2)*Pi^(1/2)/d^2/exp(a/b)-c*(d*x+c)*(a+b*arcsinh(d*x+c) )^(1/2)/d^2+1/4*cosh(2*arcsinh(d*x+c))*(a+b*arcsinh(d*x+c))^(1/2)/d^2
Time = 2.27 (sec) , antiderivative size = 250, normalized size of antiderivative = 0.97 \[ \int x \sqrt {a+b \text {arcsinh}(c+d x)} \, dx=-\frac {-8 \sqrt {a+b \text {arcsinh}(c+d x)} \cosh (2 \text {arcsinh}(c+d x))+16 c e^{-\frac {a}{b}} \sqrt {a+b \text {arcsinh}(c+d x)} \left (-\frac {e^{\frac {2 a}{b}} \Gamma \left (\frac {3}{2},\frac {a}{b}+\text {arcsinh}(c+d x)\right )}{\sqrt {\frac {a}{b}+\text {arcsinh}(c+d x)}}+\frac {\Gamma \left (\frac {3}{2},-\frac {a+b \text {arcsinh}(c+d x)}{b}\right )}{\sqrt {-\frac {a+b \text {arcsinh}(c+d x)}{b}}}\right )+\sqrt {b} \sqrt {2 \pi } \text {erfi}\left (\frac {\sqrt {2} \sqrt {a+b \text {arcsinh}(c+d x)}}{\sqrt {b}}\right ) \left (\cosh \left (\frac {2 a}{b}\right )-\sinh \left (\frac {2 a}{b}\right )\right )+\sqrt {b} \sqrt {2 \pi } \text {erf}\left (\frac {\sqrt {2} \sqrt {a+b \text {arcsinh}(c+d x)}}{\sqrt {b}}\right ) \left (\cosh \left (\frac {2 a}{b}\right )+\sinh \left (\frac {2 a}{b}\right )\right )}{32 d^2} \]
-1/32*(-8*Sqrt[a + b*ArcSinh[c + d*x]]*Cosh[2*ArcSinh[c + d*x]] + (16*c*Sq rt[a + b*ArcSinh[c + d*x]]*(-((E^((2*a)/b)*Gamma[3/2, a/b + ArcSinh[c + d* x]])/Sqrt[a/b + ArcSinh[c + d*x]]) + Gamma[3/2, -((a + b*ArcSinh[c + d*x]) /b)]/Sqrt[-((a + b*ArcSinh[c + d*x])/b)]))/E^(a/b) + Sqrt[b]*Sqrt[2*Pi]*Er fi[(Sqrt[2]*Sqrt[a + b*ArcSinh[c + d*x]])/Sqrt[b]]*(Cosh[(2*a)/b] - Sinh[( 2*a)/b]) + Sqrt[b]*Sqrt[2*Pi]*Erf[(Sqrt[2]*Sqrt[a + b*ArcSinh[c + d*x]])/S qrt[b]]*(Cosh[(2*a)/b] + Sinh[(2*a)/b]))/d^2
Time = 1.05 (sec) , antiderivative size = 284, normalized size of antiderivative = 1.10, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {6274, 25, 27, 6245, 7267, 7292, 7293, 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.
\(\displaystyle \int x \sqrt {a+b \text {arcsinh}(c+d x)} \, dx\) |
\(\Big \downarrow \) 6274 |
\(\displaystyle \frac {\int x \sqrt {a+b \text {arcsinh}(c+d x)}d(c+d x)}{d}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {\int -x \sqrt {a+b \text {arcsinh}(c+d x)}d(c+d x)}{d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {\int -d x \sqrt {a+b \text {arcsinh}(c+d x)}d(c+d x)}{d^2}\) |
\(\Big \downarrow \) 6245 |
\(\displaystyle -\frac {\int -d x \sqrt {(c+d x)^2+1} \sqrt {a+b \text {arcsinh}(c+d x)}d\text {arcsinh}(c+d x)}{d^2}\) |
\(\Big \downarrow \) 7267 |
\(\displaystyle -\frac {2 \int -d x \sqrt {(c+d x)^2+1} (a+b \text {arcsinh}(c+d x))d\sqrt {a+b \text {arcsinh}(c+d x)}}{b d^2}\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle -\frac {2 \int -d x (a+b \text {arcsinh}(c+d x)) \cosh \left (\frac {a}{b}-\frac {a+b \text {arcsinh}(c+d x)}{b}\right )d\sqrt {a+b \text {arcsinh}(c+d x)}}{b d^2}\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -\frac {2 \int \left (c (a+b \text {arcsinh}(c+d x)) \cosh \left (\frac {a}{b}-\frac {a+b \text {arcsinh}(c+d x)}{b}\right )+\frac {1}{2} (a+b \text {arcsinh}(c+d x)) \sinh \left (\frac {2 a}{b}-\frac {2 (a+b \text {arcsinh}(c+d x))}{b}\right )\right )d\sqrt {a+b \text {arcsinh}(c+d x)}}{b d^2}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {2 \left (\frac {1}{8} \sqrt {\pi } b^{3/2} c e^{a/b} \text {erf}\left (\frac {\sqrt {a+b \text {arcsinh}(c+d x)}}{\sqrt {b}}\right )+\frac {1}{32} \sqrt {\frac {\pi }{2}} b^{3/2} e^{\frac {2 a}{b}} \text {erf}\left (\frac {\sqrt {2} \sqrt {a+b \text {arcsinh}(c+d x)}}{\sqrt {b}}\right )-\frac {1}{8} \sqrt {\pi } b^{3/2} c e^{-\frac {a}{b}} \text {erfi}\left (\frac {\sqrt {a+b \text {arcsinh}(c+d x)}}{\sqrt {b}}\right )+\frac {1}{32} \sqrt {\frac {\pi }{2}} b^{3/2} e^{-\frac {2 a}{b}} \text {erfi}\left (\frac {\sqrt {2} \sqrt {a+b \text {arcsinh}(c+d x)}}{\sqrt {b}}\right )-\frac {1}{2} b c \sqrt {a+b \text {arcsinh}(c+d x)} \sinh \left (\frac {a}{b}-\frac {a+b \text {arcsinh}(c+d x)}{b}\right )-\frac {1}{8} b \sqrt {a+b \text {arcsinh}(c+d x)} \cosh \left (\frac {2 a}{b}-\frac {2 (a+b \text {arcsinh}(c+d x))}{b}\right )\right )}{b d^2}\) |
(-2*(-1/8*(b*Sqrt[a + b*ArcSinh[c + d*x]]*Cosh[(2*a)/b - (2*(a + b*ArcSinh [c + d*x]))/b]) + (b^(3/2)*c*E^(a/b)*Sqrt[Pi]*Erf[Sqrt[a + b*ArcSinh[c + d *x]]/Sqrt[b]])/8 + (b^(3/2)*E^((2*a)/b)*Sqrt[Pi/2]*Erf[(Sqrt[2]*Sqrt[a + b *ArcSinh[c + d*x]])/Sqrt[b]])/32 - (b^(3/2)*c*Sqrt[Pi]*Erfi[Sqrt[a + b*Arc Sinh[c + d*x]]/Sqrt[b]])/(8*E^(a/b)) + (b^(3/2)*Sqrt[Pi/2]*Erfi[(Sqrt[2]*S qrt[a + b*ArcSinh[c + d*x]])/Sqrt[b]])/(32*E^((2*a)/b)) - (b*c*Sqrt[a + b* ArcSinh[c + d*x]]*Sinh[a/b - (a + b*ArcSinh[c + d*x])/b])/2))/(b*d^2)
3.1.99.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((d_.) + (e_.)*(x_))^(m_.), x _Symbol] :> Simp[1/c^(m + 1) Subst[Int[(a + b*x)^n*Cosh[x]*(c*d + e*Sinh[ x])^m, x], x, ArcSinh[c*x]], x] /; FreeQ[{a, b, c, d, e, n}, x] && IGtQ[m, 0]
Int[((a_.) + ArcSinh[(c_) + (d_.)*(x_)]*(b_.))^(n_.)*((e_.) + (f_.)*(x_))^( m_.), x_Symbol] :> Simp[1/d Subst[Int[((d*e - c*f)/d + f*(x/d))^m*(a + b* ArcSinh[x])^n, x], x, c + d*x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x]
Int[u_, x_Symbol] :> With[{lst = SubstForFractionalPowerOfLinear[u, x]}, Si mp[lst[[2]]*lst[[4]] Subst[Int[lst[[1]], x], x, lst[[3]]^(1/lst[[2]])], x ] /; !FalseQ[lst] && SubstForFractionalPowerQ[u, lst[[3]], x]]
\[\int x \sqrt {a +b \,\operatorname {arcsinh}\left (d x +c \right )}d x\]
Exception generated. \[ \int x \sqrt {a+b \text {arcsinh}(c+d x)} \, dx=\text {Exception raised: TypeError} \]
Exception raised: TypeError >> Error detected within library code: inte grate: implementation incomplete (constant residues)
\[ \int x \sqrt {a+b \text {arcsinh}(c+d x)} \, dx=\int x \sqrt {a + b \operatorname {asinh}{\left (c + d x \right )}}\, dx \]
\[ \int x \sqrt {a+b \text {arcsinh}(c+d x)} \, dx=\int { \sqrt {b \operatorname {arsinh}\left (d x + c\right ) + a} x \,d x } \]
\[ \int x \sqrt {a+b \text {arcsinh}(c+d x)} \, dx=\int { \sqrt {b \operatorname {arsinh}\left (d x + c\right ) + a} x \,d x } \]
Timed out. \[ \int x \sqrt {a+b \text {arcsinh}(c+d x)} \, dx=\int x\,\sqrt {a+b\,\mathrm {asinh}\left (c+d\,x\right )} \,d x \]