\(\int \frac {x}{(a+b \text {arcsinh}(c x))^{5/2}} \, dx\) [152]

Optimal result

Integrand size = 14, antiderivative size = 183 \[ \int \frac {x}{(a+b \text {arcsinh}(c x))^{5/2}} \, dx=-\frac {2 x \sqrt {1+c^2 x^2}}{3 b c (a+b \text {arcsinh}(c x))^{3/2}}-\frac {4}{3 b^2 c^2 \sqrt {a+b \text {arcsinh}(c x)}}-\frac {8 x^2}{3 b^2 \sqrt {a+b \text {arcsinh}(c x)}}-\frac {2 e^{\frac {2 a}{b}} \sqrt {2 \pi } \text {erf}\left (\frac {\sqrt {2} \sqrt {a+b \text {arcsinh}(c x)}}{\sqrt {b}}\right )}{3 b^{5/2} c^2}+\frac {2 e^{-\frac {2 a}{b}} \sqrt {2 \pi } \text {erfi}\left (\frac {\sqrt {2} \sqrt {a+b \text {arcsinh}(c x)}}{\sqrt {b}}\right )}{3 b^{5/2} c^2} \]

[Out]

-2/3*exp(2*a/b)*erf(2^(1/2)*(a+b*arcsinh(c*x))^(1/2)/b^(1/2))*2^(1/2)*Pi^(1/2)/b^(5/2)/c^2+2/3*erfi(2^(1/2)*(a
+b*arcsinh(c*x))^(1/2)/b^(1/2))*2^(1/2)*Pi^(1/2)/b^(5/2)/c^2/exp(2*a/b)-2/3*x*(c^2*x^2+1)^(1/2)/b/c/(a+b*arcsi
nh(c*x))^(3/2)-4/3/b^2/c^2/(a+b*arcsinh(c*x))^(1/2)-8/3*x^2/b^2/(a+b*arcsinh(c*x))^(1/2)

Rubi [A] (verified)

Time = 0.32 (sec) , antiderivative size = 183, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.714, Rules used = {5779, 5818, 5780, 5556, 12, 3389, 2211, 2236, 2235, 5783} \[ \int \frac {x}{(a+b \text {arcsinh}(c x))^{5/2}} \, dx=-\frac {2 \sqrt {2 \pi } e^{\frac {2 a}{b}} \text {erf}\left (\frac {\sqrt {2} \sqrt {a+b \text {arcsinh}(c x)}}{\sqrt {b}}\right )}{3 b^{5/2} c^2}+\frac {2 \sqrt {2 \pi } e^{-\frac {2 a}{b}} \text {erfi}\left (\frac {\sqrt {2} \sqrt {a+b \text {arcsinh}(c x)}}{\sqrt {b}}\right )}{3 b^{5/2} c^2}-\frac {4}{3 b^2 c^2 \sqrt {a+b \text {arcsinh}(c x)}}-\frac {8 x^2}{3 b^2 \sqrt {a+b \text {arcsinh}(c x)}}-\frac {2 x \sqrt {c^2 x^2+1}}{3 b c (a+b \text {arcsinh}(c x))^{3/2}} \]

[In]

Int[x/(a + b*ArcSinh[c*x])^(5/2),x]

[Out]

(-2*x*Sqrt[1 + c^2*x^2])/(3*b*c*(a + b*ArcSinh[c*x])^(3/2)) - 4/(3*b^2*c^2*Sqrt[a + b*ArcSinh[c*x]]) - (8*x^2)
/(3*b^2*Sqrt[a + b*ArcSinh[c*x]]) - (2*E^((2*a)/b)*Sqrt[2*Pi]*Erf[(Sqrt[2]*Sqrt[a + b*ArcSinh[c*x]])/Sqrt[b]])
/(3*b^(5/2)*c^2) + (2*Sqrt[2*Pi]*Erfi[(Sqrt[2]*Sqrt[a + b*ArcSinh[c*x]])/Sqrt[b]])/(3*b^(5/2)*c^2*E^((2*a)/b))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2211

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/Sqrt[(c_.) + (d_.)*(x_)], x_Symbol] :> Dist[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]

Rule 2235

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]

Rule 2236

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] /; FreeQ[{F, a, b, c, d}, x] && NegQ[b]

Rule 3389

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> Dist[I/2, Int[(c + d*x)^m/E^(I*(e + f*x))
, x], x] - Dist[I/2, Int[(c + d*x)^m*E^(I*(e + f*x)), x], x] /; FreeQ[{c, d, e, f, m}, x]

Rule 5556

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]

Rule 5779

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_)*(x_)^(m_.), x_Symbol] :> Simp[x^m*Sqrt[1 + c^2*x^2]*((a + b*ArcSi
nh[c*x])^(n + 1)/(b*c*(n + 1))), x] + (-Dist[c*((m + 1)/(b*(n + 1))), Int[x^(m + 1)*((a + b*ArcSinh[c*x])^(n +
1)/Sqrt[1 + c^2*x^2]), x], x] - Dist[m/(b*c*(n + 1)), Int[x^(m - 1)*((a + b*ArcSinh[c*x])^(n + 1)/Sqrt[1 + c^
2*x^2]), x], x]) /; FreeQ[{a, b, c}, x] && IGtQ[m, 0] && LtQ[n, -2]

Rule 5780

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_)*(x_)^(m_.), x_Symbol] :> Dist[1/(b*c^(m + 1)), Subst[Int[x^n*Sinh
[-a/b + x/b]^m*Cosh[-a/b + x/b], x], x, a + b*ArcSinh[c*x]], x] /; FreeQ[{a, b, c, n}, x] && IGtQ[m, 0]

Rule 5783

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp[(1/(b*c*(n + 1)))*S
imp[Sqrt[1 + c^2*x^2]/Sqrt[d + e*x^2]]*(a + b*ArcSinh[c*x])^(n + 1), x] /; FreeQ[{a, b, c, d, e, n}, x] && EqQ
[e, c^2*d] && NeQ[n, -1]

Rule 5818

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] - Dist[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]

Rubi steps \begin{align*} \text {integral}& = -\frac {2 x \sqrt {1+c^2 x^2}}{3 b c (a+b \text {arcsinh}(c x))^{3/2}}+\frac {2 \int \frac {1}{\sqrt {1+c^2 x^2} (a+b \text {arcsinh}(c x))^{3/2}} \, dx}{3 b c}+\frac {(4 c) \int \frac {x^2}{\sqrt {1+c^2 x^2} (a+b \text {arcsinh}(c x))^{3/2}} \, dx}{3 b} \\ & = -\frac {2 x \sqrt {1+c^2 x^2}}{3 b c (a+b \text {arcsinh}(c x))^{3/2}}-\frac {4}{3 b^2 c^2 \sqrt {a+b \text {arcsinh}(c x)}}-\frac {8 x^2}{3 b^2 \sqrt {a+b \text {arcsinh}(c x)}}+\frac {16 \int \frac {x}{\sqrt {a+b \text {arcsinh}(c x)}} \, dx}{3 b^2} \\ & = -\frac {2 x \sqrt {1+c^2 x^2}}{3 b c (a+b \text {arcsinh}(c x))^{3/2}}-\frac {4}{3 b^2 c^2 \sqrt {a+b \text {arcsinh}(c x)}}-\frac {8 x^2}{3 b^2 \sqrt {a+b \text {arcsinh}(c x)}}-\frac {16 \text {Subst}\left (\int \frac {\cosh \left (\frac {a}{b}-\frac {x}{b}\right ) \sinh \left (\frac {a}{b}-\frac {x}{b}\right )}{\sqrt {x}} \, dx,x,a+b \text {arcsinh}(c x)\right )}{3 b^3 c^2} \\ & = -\frac {2 x \sqrt {1+c^2 x^2}}{3 b c (a+b \text {arcsinh}(c x))^{3/2}}-\frac {4}{3 b^2 c^2 \sqrt {a+b \text {arcsinh}(c x)}}-\frac {8 x^2}{3 b^2 \sqrt {a+b \text {arcsinh}(c x)}}-\frac {16 \text {Subst}\left (\int \frac {\sinh \left (\frac {2 a}{b}-\frac {2 x}{b}\right )}{2 \sqrt {x}} \, dx,x,a+b \text {arcsinh}(c x)\right )}{3 b^3 c^2} \\ & = -\frac {2 x \sqrt {1+c^2 x^2}}{3 b c (a+b \text {arcsinh}(c x))^{3/2}}-\frac {4}{3 b^2 c^2 \sqrt {a+b \text {arcsinh}(c x)}}-\frac {8 x^2}{3 b^2 \sqrt {a+b \text {arcsinh}(c x)}}-\frac {8 \text {Subst}\left (\int \frac {\sinh \left (\frac {2 a}{b}-\frac {2 x}{b}\right )}{\sqrt {x}} \, dx,x,a+b \text {arcsinh}(c x)\right )}{3 b^3 c^2} \\ & = -\frac {2 x \sqrt {1+c^2 x^2}}{3 b c (a+b \text {arcsinh}(c x))^{3/2}}-\frac {4}{3 b^2 c^2 \sqrt {a+b \text {arcsinh}(c x)}}-\frac {8 x^2}{3 b^2 \sqrt {a+b \text {arcsinh}(c x)}}-\frac {4 \text {Subst}\left (\int \frac {e^{-i \left (\frac {2 i a}{b}-\frac {2 i x}{b}\right )}}{\sqrt {x}} \, dx,x,a+b \text {arcsinh}(c x)\right )}{3 b^3 c^2}+\frac {4 \text {Subst}\left (\int \frac {e^{i \left (\frac {2 i a}{b}-\frac {2 i x}{b}\right )}}{\sqrt {x}} \, dx,x,a+b \text {arcsinh}(c x)\right )}{3 b^3 c^2} \\ & = -\frac {2 x \sqrt {1+c^2 x^2}}{3 b c (a+b \text {arcsinh}(c x))^{3/2}}-\frac {4}{3 b^2 c^2 \sqrt {a+b \text {arcsinh}(c x)}}-\frac {8 x^2}{3 b^2 \sqrt {a+b \text {arcsinh}(c x)}}-\frac {8 \text {Subst}\left (\int e^{\frac {2 a}{b}-\frac {2 x^2}{b}} \, dx,x,\sqrt {a+b \text {arcsinh}(c x)}\right )}{3 b^3 c^2}+\frac {8 \text {Subst}\left (\int e^{-\frac {2 a}{b}+\frac {2 x^2}{b}} \, dx,x,\sqrt {a+b \text {arcsinh}(c x)}\right )}{3 b^3 c^2} \\ & = -\frac {2 x \sqrt {1+c^2 x^2}}{3 b c (a+b \text {arcsinh}(c x))^{3/2}}-\frac {4}{3 b^2 c^2 \sqrt {a+b \text {arcsinh}(c x)}}-\frac {8 x^2}{3 b^2 \sqrt {a+b \text {arcsinh}(c x)}}-\frac {2 e^{\frac {2 a}{b}} \sqrt {2 \pi } \text {erf}\left (\frac {\sqrt {2} \sqrt {a+b \text {arcsinh}(c x)}}{\sqrt {b}}\right )}{3 b^{5/2} c^2}+\frac {2 e^{-\frac {2 a}{b}} \sqrt {2 \pi } \text {erfi}\left (\frac {\sqrt {2} \sqrt {a+b \text {arcsinh}(c x)}}{\sqrt {b}}\right )}{3 b^{5/2} c^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.55 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.09 \[ \int \frac {x}{(a+b \text {arcsinh}(c x))^{5/2}} \, dx=\frac {e^{-2 \left (\frac {a}{b}+\text {arcsinh}(c x)\right )} \left (-4 \sqrt {2} b e^{2 \text {arcsinh}(c x)} \left (-\frac {a+b \text {arcsinh}(c x)}{b}\right )^{3/2} \Gamma \left (\frac {1}{2},-\frac {2 (a+b \text {arcsinh}(c x))}{b}\right )+e^{\frac {2 a}{b}} \left (-4 a+b-4 a e^{4 \text {arcsinh}(c x)}-b e^{4 \text {arcsinh}(c x)}-4 b \left (1+e^{4 \text {arcsinh}(c x)}\right ) \text {arcsinh}(c x)+4 \sqrt {2} e^{2 \left (\frac {a}{b}+\text {arcsinh}(c x)\right )} \sqrt {\frac {a}{b}+\text {arcsinh}(c x)} (a+b \text {arcsinh}(c x)) \Gamma \left (\frac {1}{2},\frac {2 (a+b \text {arcsinh}(c x))}{b}\right )\right )\right )}{6 b^2 c^2 (a+b \text {arcsinh}(c x))^{3/2}} \]

[In]

Integrate[x/(a + b*ArcSinh[c*x])^(5/2),x]

[Out]

(-4*Sqrt[2]*b*E^(2*ArcSinh[c*x])*(-((a + b*ArcSinh[c*x])/b))^(3/2)*Gamma[1/2, (-2*(a + b*ArcSinh[c*x]))/b] + E
^((2*a)/b)*(-4*a + b - 4*a*E^(4*ArcSinh[c*x]) - b*E^(4*ArcSinh[c*x]) - 4*b*(1 + E^(4*ArcSinh[c*x]))*ArcSinh[c*
x] + 4*Sqrt[2]*E^(2*(a/b + ArcSinh[c*x]))*Sqrt[a/b + ArcSinh[c*x]]*(a + b*ArcSinh[c*x])*Gamma[1/2, (2*(a + b*A
rcSinh[c*x]))/b]))/(6*b^2*c^2*E^(2*(a/b + ArcSinh[c*x]))*(a + b*ArcSinh[c*x])^(3/2))

Maple [F]

\[\int \frac {x}{\left (a +b \,\operatorname {arcsinh}\left (c x \right )\right )^{\frac {5}{2}}}d x\]

[In]

int(x/(a+b*arcsinh(c*x))^(5/2),x)

[Out]

int(x/(a+b*arcsinh(c*x))^(5/2),x)

Fricas [F(-2)]

Exception generated. \[ \int \frac {x}{(a+b \text {arcsinh}(c x))^{5/2}} \, dx=\text {Exception raised: TypeError} \]

[In]

integrate(x/(a+b*arcsinh(c*x))^(5/2),x, algorithm="fricas")

[Out]

Exception raised: TypeError >>  Error detected within library code:   integrate: implementation incomplete (co
nstant residues)

Sympy [F]

\[ \int \frac {x}{(a+b \text {arcsinh}(c x))^{5/2}} \, dx=\int \frac {x}{\left (a + b \operatorname {asinh}{\left (c x \right )}\right )^{\frac {5}{2}}}\, dx \]

[In]

integrate(x/(a+b*asinh(c*x))**(5/2),x)

[Out]

Integral(x/(a + b*asinh(c*x))**(5/2), x)

Maxima [F]

\[ \int \frac {x}{(a+b \text {arcsinh}(c x))^{5/2}} \, dx=\int { \frac {x}{{\left (b \operatorname {arsinh}\left (c x\right ) + a\right )}^{\frac {5}{2}}} \,d x } \]

[In]

integrate(x/(a+b*arcsinh(c*x))^(5/2),x, algorithm="maxima")

[Out]

integrate(x/(b*arcsinh(c*x) + a)^(5/2), x)

Giac [F]

\[ \int \frac {x}{(a+b \text {arcsinh}(c x))^{5/2}} \, dx=\int { \frac {x}{{\left (b \operatorname {arsinh}\left (c x\right ) + a\right )}^{\frac {5}{2}}} \,d x } \]

[In]

integrate(x/(a+b*arcsinh(c*x))^(5/2),x, algorithm="giac")

[Out]

integrate(x/(b*arcsinh(c*x) + a)^(5/2), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {x}{(a+b \text {arcsinh}(c x))^{5/2}} \, dx=\int \frac {x}{{\left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )}^{5/2}} \,d x \]

[In]

int(x/(a + b*asinh(c*x))^(5/2),x)

[Out]

int(x/(a + b*asinh(c*x))^(5/2), x)