Integrand size = 14, antiderivative size = 158 \[ \int \frac {1}{(a+b \text {arcsinh}(c+d x))^{5/2}} \, dx=-\frac {2 \sqrt {1+(c+d x)^2}}{3 b d (a+b \text {arcsinh}(c+d x))^{3/2}}-\frac {4 (c+d x)}{3 b^2 d \sqrt {a+b \text {arcsinh}(c+d x)}}+\frac {2 e^{a/b} \sqrt {\pi } \text {erf}\left (\frac {\sqrt {a+b \text {arcsinh}(c+d x)}}{\sqrt {b}}\right )}{3 b^{5/2} d}+\frac {2 e^{-\frac {a}{b}} \sqrt {\pi } \text {erfi}\left (\frac {\sqrt {a+b \text {arcsinh}(c+d x)}}{\sqrt {b}}\right )}{3 b^{5/2} d} \]
2/3*exp(a/b)*erf((a+b*arcsinh(d*x+c))^(1/2)/b^(1/2))*Pi^(1/2)/b^(5/2)/d+2/ 3*erfi((a+b*arcsinh(d*x+c))^(1/2)/b^(1/2))*Pi^(1/2)/b^(5/2)/d/exp(a/b)-2/3 *(1+(d*x+c)^2)^(1/2)/b/d/(a+b*arcsinh(d*x+c))^(3/2)-4/3*(d*x+c)/b^2/d/(a+b *arcsinh(d*x+c))^(1/2)
Time = 0.43 (sec) , antiderivative size = 207, normalized size of antiderivative = 1.31 \[ \int \frac {1}{(a+b \text {arcsinh}(c+d x))^{5/2}} \, dx=\frac {e^{-\frac {a+b \text {arcsinh}(c+d x)}{b}} \left (-e^{a/b} \left (b+2 a \left (-1+e^{2 \text {arcsinh}(c+d x)}\right )-2 b \text {arcsinh}(c+d x)+b e^{2 \text {arcsinh}(c+d x)} (1+2 \text {arcsinh}(c+d x))\right )-2 e^{\frac {2 a}{b}+\text {arcsinh}(c+d x)} \sqrt {\frac {a}{b}+\text {arcsinh}(c+d x)} (a+b \text {arcsinh}(c+d x)) \Gamma \left (\frac {1}{2},\frac {a}{b}+\text {arcsinh}(c+d x)\right )-2 b e^{\text {arcsinh}(c+d x)} \left (-\frac {a+b \text {arcsinh}(c+d x)}{b}\right )^{3/2} \Gamma \left (\frac {1}{2},-\frac {a+b \text {arcsinh}(c+d x)}{b}\right )\right )}{3 b^2 d (a+b \text {arcsinh}(c+d x))^{3/2}} \]
(-(E^(a/b)*(b + 2*a*(-1 + E^(2*ArcSinh[c + d*x])) - 2*b*ArcSinh[c + d*x] + b*E^(2*ArcSinh[c + d*x])*(1 + 2*ArcSinh[c + d*x]))) - 2*E^((2*a)/b + ArcS inh[c + d*x])*Sqrt[a/b + ArcSinh[c + d*x]]*(a + b*ArcSinh[c + d*x])*Gamma[ 1/2, a/b + ArcSinh[c + d*x]] - 2*b*E^ArcSinh[c + d*x]*(-((a + b*ArcSinh[c + d*x])/b))^(3/2)*Gamma[1/2, -((a + b*ArcSinh[c + d*x])/b)])/(3*b^2*d*E^(( a + b*ArcSinh[c + d*x])/b)*(a + b*ArcSinh[c + d*x])^(3/2))
Time = 0.76 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.03, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.714, Rules used = {6273, 6188, 6233, 6189, 3042, 3788, 26, 2611, 2633, 2634}
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.
\(\displaystyle \int \frac {1}{(a+b \text {arcsinh}(c+d x))^{5/2}} \, dx\) |
\(\Big \downarrow \) 6273 |
\(\displaystyle \frac {\int \frac {1}{(a+b \text {arcsinh}(c+d x))^{5/2}}d(c+d x)}{d}\) |
\(\Big \downarrow \) 6188 |
\(\displaystyle \frac {\frac {2 \int \frac {c+d x}{\sqrt {(c+d x)^2+1} (a+b \text {arcsinh}(c+d x))^{3/2}}d(c+d x)}{3 b}-\frac {2 \sqrt {(c+d x)^2+1}}{3 b (a+b \text {arcsinh}(c+d x))^{3/2}}}{d}\) |
\(\Big \downarrow \) 6233 |
\(\displaystyle \frac {\frac {2 \left (\frac {2 \int \frac {1}{\sqrt {a+b \text {arcsinh}(c+d x)}}d(c+d x)}{b}-\frac {2 (c+d x)}{b \sqrt {a+b \text {arcsinh}(c+d x)}}\right )}{3 b}-\frac {2 \sqrt {(c+d x)^2+1}}{3 b (a+b \text {arcsinh}(c+d x))^{3/2}}}{d}\) |
\(\Big \downarrow \) 6189 |
\(\displaystyle \frac {\frac {2 \left (\frac {2 \int \frac {\cosh \left (\frac {a}{b}-\frac {a+b \text {arcsinh}(c+d x)}{b}\right )}{\sqrt {a+b \text {arcsinh}(c+d x)}}d(a+b \text {arcsinh}(c+d x))}{b^2}-\frac {2 (c+d x)}{b \sqrt {a+b \text {arcsinh}(c+d x)}}\right )}{3 b}-\frac {2 \sqrt {(c+d x)^2+1}}{3 b (a+b \text {arcsinh}(c+d x))^{3/2}}}{d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {-\frac {2 \sqrt {(c+d x)^2+1}}{3 b (a+b \text {arcsinh}(c+d x))^{3/2}}+\frac {2 \left (-\frac {2 (c+d x)}{b \sqrt {a+b \text {arcsinh}(c+d x)}}+\frac {2 \int \frac {\sin \left (\frac {i a}{b}-\frac {i (a+b \text {arcsinh}(c+d x))}{b}+\frac {\pi }{2}\right )}{\sqrt {a+b \text {arcsinh}(c+d x)}}d(a+b \text {arcsinh}(c+d x))}{b^2}\right )}{3 b}}{d}\) |
\(\Big \downarrow \) 3788 |
\(\displaystyle \frac {-\frac {2 \sqrt {(c+d x)^2+1}}{3 b (a+b \text {arcsinh}(c+d x))^{3/2}}+\frac {2 \left (-\frac {2 (c+d x)}{b \sqrt {a+b \text {arcsinh}(c+d x)}}+\frac {2 \left (\frac {1}{2} i \int -\frac {i e^{\frac {a-c-d x}{b}}}{\sqrt {a+b \text {arcsinh}(c+d x)}}d(a+b \text {arcsinh}(c+d x))-\frac {1}{2} i \int \frac {i e^{-\frac {a-c-d x}{b}}}{\sqrt {a+b \text {arcsinh}(c+d x)}}d(a+b \text {arcsinh}(c+d x))\right )}{b^2}\right )}{3 b}}{d}\) |
\(\Big \downarrow \) 26 |
\(\displaystyle \frac {\frac {2 \left (\frac {2 \left (\frac {1}{2} \int \frac {e^{-\frac {a-c-d x}{b}}}{\sqrt {a+b \text {arcsinh}(c+d x)}}d(a+b \text {arcsinh}(c+d x))+\frac {1}{2} \int \frac {e^{\frac {a-c-d x}{b}}}{\sqrt {a+b \text {arcsinh}(c+d x)}}d(a+b \text {arcsinh}(c+d x))\right )}{b^2}-\frac {2 (c+d x)}{b \sqrt {a+b \text {arcsinh}(c+d x)}}\right )}{3 b}-\frac {2 \sqrt {(c+d x)^2+1}}{3 b (a+b \text {arcsinh}(c+d x))^{3/2}}}{d}\) |
\(\Big \downarrow \) 2611 |
\(\displaystyle \frac {\frac {2 \left (\frac {2 \left (\int e^{\frac {a}{b}-\frac {a+b \text {arcsinh}(c+d x)}{b}}d\sqrt {a+b \text {arcsinh}(c+d x)}+\int e^{\frac {a+b \text {arcsinh}(c+d x)}{b}-\frac {a}{b}}d\sqrt {a+b \text {arcsinh}(c+d x)}\right )}{b^2}-\frac {2 (c+d x)}{b \sqrt {a+b \text {arcsinh}(c+d x)}}\right )}{3 b}-\frac {2 \sqrt {(c+d x)^2+1}}{3 b (a+b \text {arcsinh}(c+d x))^{3/2}}}{d}\) |
\(\Big \downarrow \) 2633 |
\(\displaystyle \frac {\frac {2 \left (\frac {2 \left (\int e^{\frac {a}{b}-\frac {a+b \text {arcsinh}(c+d x)}{b}}d\sqrt {a+b \text {arcsinh}(c+d x)}+\frac {1}{2} \sqrt {\pi } \sqrt {b} e^{-\frac {a}{b}} \text {erfi}\left (\frac {\sqrt {a+b \text {arcsinh}(c+d x)}}{\sqrt {b}}\right )\right )}{b^2}-\frac {2 (c+d x)}{b \sqrt {a+b \text {arcsinh}(c+d x)}}\right )}{3 b}-\frac {2 \sqrt {(c+d x)^2+1}}{3 b (a+b \text {arcsinh}(c+d x))^{3/2}}}{d}\) |
\(\Big \downarrow \) 2634 |
\(\displaystyle \frac {\frac {2 \left (\frac {2 \left (\frac {1}{2} \sqrt {\pi } \sqrt {b} e^{a/b} \text {erf}\left (\frac {\sqrt {a+b \text {arcsinh}(c+d x)}}{\sqrt {b}}\right )+\frac {1}{2} \sqrt {\pi } \sqrt {b} e^{-\frac {a}{b}} \text {erfi}\left (\frac {\sqrt {a+b \text {arcsinh}(c+d x)}}{\sqrt {b}}\right )\right )}{b^2}-\frac {2 (c+d x)}{b \sqrt {a+b \text {arcsinh}(c+d x)}}\right )}{3 b}-\frac {2 \sqrt {(c+d x)^2+1}}{3 b (a+b \text {arcsinh}(c+d x))^{3/2}}}{d}\) |
((-2*Sqrt[1 + (c + d*x)^2])/(3*b*(a + b*ArcSinh[c + d*x])^(3/2)) + (2*((-2 *(c + d*x))/(b*Sqrt[a + b*ArcSinh[c + d*x]]) + (2*((Sqrt[b]*E^(a/b)*Sqrt[P i]*Erf[Sqrt[a + b*ArcSinh[c + d*x]]/Sqrt[b]])/2 + (Sqrt[b]*Sqrt[Pi]*Erfi[S qrt[a + b*ArcSinh[c + d*x]]/Sqrt[b]])/(2*E^(a/b))))/b^2))/(3*b))/d
3.3.20.3.1 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[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/Sqrt[(c_.) + (d_.)*(x_)], x_Symbol] : > Simp[2/d Subst[Int[F^(g*(e - c*(f/d)) + f*g*(x^2/d)), x], x, Sqrt[c + d *x]], x] /; FreeQ[{F, c, d, e, f, g}, x] && !TrueQ[$UseGamma]
Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^2), x_Symbol] :> Simp[F^a*Sqrt [Pi]*(Erfi[(c + d*x)*Rt[b*Log[F], 2]]/(2*d*Rt[b*Log[F], 2])), x] /; FreeQ[{ F, a, b, c, d}, x] && PosQ[b]
Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^2), x_Symbol] :> Simp[F^a*Sqrt [Pi]*(Erf[(c + d*x)*Rt[(-b)*Log[F], 2]]/(2*d*Rt[(-b)*Log[F], 2])), x] /; Fr eeQ[{F, a, b, c, d}, x] && NegQ[b]
Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + Pi*(k_.) + (f_.)*(x_)], x_Symbol ] :> Simp[I/2 Int[(c + d*x)^m/(E^(I*k*Pi)*E^(I*(e + f*x))), x], x] - Simp [I/2 Int[(c + d*x)^m*E^(I*k*Pi)*E^(I*(e + f*x)), x], x] /; FreeQ[{c, d, e , f, m}, x] && IntegerQ[2*k]
Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[Sqrt[1 + c^ 2*x^2]*((a + b*ArcSinh[c*x])^(n + 1)/(b*c*(n + 1))), x] - Simp[c/(b*(n + 1) ) Int[x*((a + b*ArcSinh[c*x])^(n + 1)/Sqrt[1 + c^2*x^2]), x], x] /; FreeQ [{a, b, c}, x] && LtQ[n, -1]
Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[1/(b*c) S ubst[Int[x^n*Cosh[-a/b + x/b], x], x, a + b*ArcSinh[c*x]], x] /; FreeQ[{a, b, c, n}, x]
Int[(((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_)*((f_.)*(x_))^(m_.))/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp[((f*x)^m/(b*c*(n + 1)))*Simp[Sqrt[1 + c ^2*x^2]/Sqrt[d + e*x^2]]*(a + b*ArcSinh[c*x])^(n + 1), x] - Simp[f*(m/(b*c* (n + 1)))*Simp[Sqrt[1 + c^2*x^2]/Sqrt[d + e*x^2]] Int[(f*x)^(m - 1)*(a + b*ArcSinh[c*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[e , c^2*d] && LtQ[n, -1]
Int[((a_.) + ArcSinh[(c_) + (d_.)*(x_)]*(b_.))^(n_.), x_Symbol] :> Simp[1/d Subst[Int[(a + b*ArcSinh[x])^n, x], x, c + d*x], x] /; FreeQ[{a, b, c, d , n}, x]
\[\int \frac {1}{\left (a +b \,\operatorname {arcsinh}\left (d x +c \right )\right )^{\frac {5}{2}}}d x\]
Exception generated. \[ \int \frac {1}{(a+b \text {arcsinh}(c+d x))^{5/2}} \, dx=\text {Exception raised: TypeError} \]
Exception raised: TypeError >> Error detected within library code: inte grate: implementation incomplete (constant residues)
\[ \int \frac {1}{(a+b \text {arcsinh}(c+d x))^{5/2}} \, dx=\int \frac {1}{\left (a + b \operatorname {asinh}{\left (c + d x \right )}\right )^{\frac {5}{2}}}\, dx \]
\[ \int \frac {1}{(a+b \text {arcsinh}(c+d x))^{5/2}} \, dx=\int { \frac {1}{{\left (b \operatorname {arsinh}\left (d x + c\right ) + a\right )}^{\frac {5}{2}}} \,d x } \]
\[ \int \frac {1}{(a+b \text {arcsinh}(c+d x))^{5/2}} \, dx=\int { \frac {1}{{\left (b \operatorname {arsinh}\left (d x + c\right ) + a\right )}^{\frac {5}{2}}} \,d x } \]
Timed out. \[ \int \frac {1}{(a+b \text {arcsinh}(c+d x))^{5/2}} \, dx=\int \frac {1}{{\left (a+b\,\mathrm {asinh}\left (c+d\,x\right )\right )}^{5/2}} \,d x \]