3.2.10 \(\int \frac {x}{(a+b \sinh ^{-1}(c+d x))^{5/2}} \, dx\) [110]

Optimal. Leaf size=365 \[ \frac {2 c \sqrt {1+(c+d x)^2}}{3 b d^2 \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}-\frac {2 (c+d x) \sqrt {1+(c+d x)^2}}{3 b d^2 \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}-\frac {4}{3 b^2 d^2 \sqrt {a+b \sinh ^{-1}(c+d x)}}+\frac {4 c (c+d x)}{3 b^2 d^2 \sqrt {a+b \sinh ^{-1}(c+d x)}}-\frac {8 (c+d x)^2}{3 b^2 d^2 \sqrt {a+b \sinh ^{-1}(c+d x)}}-\frac {2 c e^{a/b} \sqrt {\pi } \text {Erf}\left (\frac {\sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{3 b^{5/2} d^2}-\frac {2 e^{\frac {2 a}{b}} \sqrt {2 \pi } \text {Erf}\left (\frac {\sqrt {2} \sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{3 b^{5/2} d^2}-\frac {2 c e^{-\frac {a}{b}} \sqrt {\pi } \text {Erfi}\left (\frac {\sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{3 b^{5/2} d^2}+\frac {2 e^{-\frac {2 a}{b}} \sqrt {2 \pi } \text {Erfi}\left (\frac {\sqrt {2} \sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{3 b^{5/2} d^2} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.61, antiderivative size = 365, normalized size of antiderivative = 1.00, number of steps used = 22, number of rules used = 15, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.938, Rules used = {5859, 5829, 5773, 5818, 5774, 3388, 2211, 2236, 2235, 5779, 5780, 5556, 12, 3389, 5783} \begin {gather*} -\frac {2 \sqrt {\pi } c e^{a/b} \text {Erf}\left (\frac {\sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{3 b^{5/2} d^2}-\frac {2 \sqrt {2 \pi } e^{\frac {2 a}{b}} \text {Erf}\left (\frac {\sqrt {2} \sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{3 b^{5/2} d^2}-\frac {2 \sqrt {\pi } c e^{-\frac {a}{b}} \text {Erfi}\left (\frac {\sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{3 b^{5/2} d^2}+\frac {2 \sqrt {2 \pi } e^{-\frac {2 a}{b}} \text {Erfi}\left (\frac {\sqrt {2} \sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{3 b^{5/2} d^2}-\frac {8 (c+d x)^2}{3 b^2 d^2 \sqrt {a+b \sinh ^{-1}(c+d x)}}+\frac {4 c (c+d x)}{3 b^2 d^2 \sqrt {a+b \sinh ^{-1}(c+d x)}}-\frac {4}{3 b^2 d^2 \sqrt {a+b \sinh ^{-1}(c+d x)}}-\frac {2 \sqrt {(c+d x)^2+1} (c+d x)}{3 b d^2 \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}+\frac {2 c \sqrt {(c+d x)^2+1}}{3 b d^2 \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*c*Sqrt[1 + (c + d*x)^2])/(3*b*d^2*(a + b*ArcSinh[c + d*x])^(3/2)) - (2*(c + d*x)*Sqrt[1 + (c + d*x)^2])/(3*
b*d^2*(a + b*ArcSinh[c + d*x])^(3/2)) - 4/(3*b^2*d^2*Sqrt[a + b*ArcSinh[c + d*x]]) + (4*c*(c + d*x))/(3*b^2*d^
2*Sqrt[a + b*ArcSinh[c + d*x]]) - (8*(c + d*x)^2)/(3*b^2*d^2*Sqrt[a + b*ArcSinh[c + d*x]]) - (2*c*E^(a/b)*Sqrt
[Pi]*Erf[Sqrt[a + b*ArcSinh[c + d*x]]/Sqrt[b]])/(3*b^(5/2)*d^2) - (2*E^((2*a)/b)*Sqrt[2*Pi]*Erf[(Sqrt[2]*Sqrt[
a + b*ArcSinh[c + d*x]])/Sqrt[b]])/(3*b^(5/2)*d^2) - (2*c*Sqrt[Pi]*Erfi[Sqrt[a + b*ArcSinh[c + d*x]]/Sqrt[b]])
/(3*b^(5/2)*d^2*E^(a/b)) + (2*Sqrt[2*Pi]*Erfi[(Sqrt[2]*Sqrt[a + b*ArcSinh[c + d*x]])/Sqrt[b]])/(3*b^(5/2)*d^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 3388

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

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 5773

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] - Dist[c/(b*(n + 1)), Int[x*((a + b*ArcSinh[c*x])^(n + 1)/Sqrt[1 + c^2*x^2]), x], x] /; F
reeQ[{a, b, c}, x] && LtQ[n, -1]

Rule 5774

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

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]

Rule 5829

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_)*((d_) + (e_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(d +
e*x)^m*(a + b*ArcSinh[c*x])^n, x], x] /; FreeQ[{a, b, c, d, e}, x] && IGtQ[m, 0] && LtQ[n, -1]

Rule 5859

Int[((a_.) + ArcSinh[(c_) + (d_.)*(x_)]*(b_.))^(n_.)*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Dist[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
]

Rubi steps

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

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Mathematica [A]
time = 1.81, size = 375, normalized size = 1.03 \begin {gather*} \frac {\frac {\sqrt {b} c e^{-\frac {a+b \sinh ^{-1}(c+d x)}{b}} \left (e^{a/b} \left (-2 a+b+2 a e^{2 \sinh ^{-1}(c+d x)}+b e^{2 \sinh ^{-1}(c+d x)}+2 b \left (-1+e^{2 \sinh ^{-1}(c+d x)}\right ) \sinh ^{-1}(c+d x)\right )+2 e^{\frac {2 a}{b}+\sinh ^{-1}(c+d x)} \sqrt {\frac {a}{b}+\sinh ^{-1}(c+d x)} \left (a+b \sinh ^{-1}(c+d x)\right ) \Gamma \left (\frac {1}{2},\frac {a}{b}+\sinh ^{-1}(c+d x)\right )+2 b e^{\sinh ^{-1}(c+d x)} \left (-\frac {a+b \sinh ^{-1}(c+d x)}{b}\right )^{3/2} \Gamma \left (\frac {1}{2},-\frac {a+b \sinh ^{-1}(c+d x)}{b}\right )\right )}{\left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}+2 \sqrt {2 \pi } \text {Erfi}\left (\frac {\sqrt {2} \sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right ) \left (\cosh \left (\frac {2 a}{b}\right )-\sinh \left (\frac {2 a}{b}\right )\right )-2 \sqrt {2 \pi } \text {Erf}\left (\frac {\sqrt {2} \sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right ) \left (\cosh \left (\frac {2 a}{b}\right )+\sinh \left (\frac {2 a}{b}\right )\right )-\frac {\sqrt {b} \left (4 \left (a+b \sinh ^{-1}(c+d x)\right ) \cosh \left (2 \sinh ^{-1}(c+d x)\right )+b \sinh \left (2 \sinh ^{-1}(c+d x)\right )\right )}{\left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}}{3 b^{5/2} d^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((Sqrt[b]*c*(E^(a/b)*(-2*a + b + 2*a*E^(2*ArcSinh[c + d*x]) + b*E^(2*ArcSinh[c + d*x]) + 2*b*(-1 + E^(2*ArcSin
h[c + d*x]))*ArcSinh[c + d*x]) + 2*E^((2*a)/b + ArcSinh[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)*Ga
mma[1/2, -((a + b*ArcSinh[c + d*x])/b)]))/(E^((a + b*ArcSinh[c + d*x])/b)*(a + b*ArcSinh[c + d*x])^(3/2)) + 2*
Sqrt[2*Pi]*Erfi[(Sqrt[2]*Sqrt[a + b*ArcSinh[c + d*x]])/Sqrt[b]]*(Cosh[(2*a)/b] - Sinh[(2*a)/b]) - 2*Sqrt[2*Pi]
*Erf[(Sqrt[2]*Sqrt[a + b*ArcSinh[c + d*x]])/Sqrt[b]]*(Cosh[(2*a)/b] + Sinh[(2*a)/b]) - (Sqrt[b]*(4*(a + b*ArcS
inh[c + d*x])*Cosh[2*ArcSinh[c + d*x]] + b*Sinh[2*ArcSinh[c + d*x]]))/(a + b*ArcSinh[c + d*x])^(3/2))/(3*b^(5/
2)*d^2)

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Maple [F]
time = 180.00, size = 0, normalized size = 0.00 \[\int \frac {x}{\left (a +b \arcsinh \left (d x +c \right )\right )^{\frac {5}{2}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x}{\left (a + b \operatorname {asinh}{\left (c + d x \right )}\right )^{\frac {5}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {x}{{\left (a+b\,\mathrm {asinh}\left (c+d\,x\right )\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/(a + b*asinh(c + d*x))^(5/2),x)

[Out]

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

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