3.2.54 \(\int \frac {x^2}{(a+b \sinh ^{-1}(c x))^{7/2}} \, dx\) [154]

Optimal. Leaf size=346 \[ -\frac {2 x^2 \sqrt {1+c^2 x^2}}{5 b c \left (a+b \sinh ^{-1}(c x)\right )^{5/2}}-\frac {8 x}{15 b^2 c^2 \left (a+b \sinh ^{-1}(c x)\right )^{3/2}}-\frac {4 x^3}{5 b^2 \left (a+b \sinh ^{-1}(c x)\right )^{3/2}}-\frac {16 \sqrt {1+c^2 x^2}}{15 b^3 c^3 \sqrt {a+b \sinh ^{-1}(c x)}}-\frac {24 x^2 \sqrt {1+c^2 x^2}}{5 b^3 c \sqrt {a+b \sinh ^{-1}(c x)}}+\frac {e^{a/b} \sqrt {\pi } \text {Erf}\left (\frac {\sqrt {a+b \sinh ^{-1}(c x)}}{\sqrt {b}}\right )}{15 b^{7/2} c^3}-\frac {3 e^{\frac {3 a}{b}} \sqrt {3 \pi } \text {Erf}\left (\frac {\sqrt {3} \sqrt {a+b \sinh ^{-1}(c x)}}{\sqrt {b}}\right )}{5 b^{7/2} c^3}-\frac {e^{-\frac {a}{b}} \sqrt {\pi } \text {Erfi}\left (\frac {\sqrt {a+b \sinh ^{-1}(c x)}}{\sqrt {b}}\right )}{15 b^{7/2} c^3}+\frac {3 e^{-\frac {3 a}{b}} \sqrt {3 \pi } \text {Erfi}\left (\frac {\sqrt {3} \sqrt {a+b \sinh ^{-1}(c x)}}{\sqrt {b}}\right )}{5 b^{7/2} c^3} \]

[Out]

-8/15*x/b^2/c^2/(a+b*arcsinh(c*x))^(3/2)-4/5*x^3/b^2/(a+b*arcsinh(c*x))^(3/2)+1/15*exp(a/b)*erf((a+b*arcsinh(c
*x))^(1/2)/b^(1/2))*Pi^(1/2)/b^(7/2)/c^3-1/15*erfi((a+b*arcsinh(c*x))^(1/2)/b^(1/2))*Pi^(1/2)/b^(7/2)/c^3/exp(
a/b)-3/5*exp(3*a/b)*erf(3^(1/2)*(a+b*arcsinh(c*x))^(1/2)/b^(1/2))*3^(1/2)*Pi^(1/2)/b^(7/2)/c^3+3/5*erfi(3^(1/2
)*(a+b*arcsinh(c*x))^(1/2)/b^(1/2))*3^(1/2)*Pi^(1/2)/b^(7/2)/c^3/exp(3*a/b)-2/5*x^2*(c^2*x^2+1)^(1/2)/b/c/(a+b
*arcsinh(c*x))^(5/2)-16/15*(c^2*x^2+1)^(1/2)/b^3/c^3/(a+b*arcsinh(c*x))^(1/2)-24/5*x^2*(c^2*x^2+1)^(1/2)/b^3/c
/(a+b*arcsinh(c*x))^(1/2)

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Rubi [A]
time = 0.69, antiderivative size = 346, normalized size of antiderivative = 1.00, number of steps used = 22, number of rules used = 9, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.562, Rules used = {5779, 5818, 5778, 3389, 2211, 2236, 2235, 5773, 5819} \begin {gather*} \frac {\sqrt {\pi } e^{a/b} \text {Erf}\left (\frac {\sqrt {a+b \sinh ^{-1}(c x)}}{\sqrt {b}}\right )}{15 b^{7/2} c^3}-\frac {3 \sqrt {3 \pi } e^{\frac {3 a}{b}} \text {Erf}\left (\frac {\sqrt {3} \sqrt {a+b \sinh ^{-1}(c x)}}{\sqrt {b}}\right )}{5 b^{7/2} c^3}-\frac {\sqrt {\pi } e^{-\frac {a}{b}} \text {Erfi}\left (\frac {\sqrt {a+b \sinh ^{-1}(c x)}}{\sqrt {b}}\right )}{15 b^{7/2} c^3}+\frac {3 \sqrt {3 \pi } e^{-\frac {3 a}{b}} \text {Erfi}\left (\frac {\sqrt {3} \sqrt {a+b \sinh ^{-1}(c x)}}{\sqrt {b}}\right )}{5 b^{7/2} c^3}-\frac {24 x^2 \sqrt {c^2 x^2+1}}{5 b^3 c \sqrt {a+b \sinh ^{-1}(c x)}}-\frac {16 \sqrt {c^2 x^2+1}}{15 b^3 c^3 \sqrt {a+b \sinh ^{-1}(c x)}}-\frac {8 x}{15 b^2 c^2 \left (a+b \sinh ^{-1}(c x)\right )^{3/2}}-\frac {4 x^3}{5 b^2 \left (a+b \sinh ^{-1}(c x)\right )^{3/2}}-\frac {2 x^2 \sqrt {c^2 x^2+1}}{5 b c \left (a+b \sinh ^{-1}(c x)\right )^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^2/(a + b*ArcSinh[c*x])^(7/2),x]

[Out]

(-2*x^2*Sqrt[1 + c^2*x^2])/(5*b*c*(a + b*ArcSinh[c*x])^(5/2)) - (8*x)/(15*b^2*c^2*(a + b*ArcSinh[c*x])^(3/2))
- (4*x^3)/(5*b^2*(a + b*ArcSinh[c*x])^(3/2)) - (16*Sqrt[1 + c^2*x^2])/(15*b^3*c^3*Sqrt[a + b*ArcSinh[c*x]]) -
(24*x^2*Sqrt[1 + c^2*x^2])/(5*b^3*c*Sqrt[a + b*ArcSinh[c*x]]) + (E^(a/b)*Sqrt[Pi]*Erf[Sqrt[a + b*ArcSinh[c*x]]
/Sqrt[b]])/(15*b^(7/2)*c^3) - (3*E^((3*a)/b)*Sqrt[3*Pi]*Erf[(Sqrt[3]*Sqrt[a + b*ArcSinh[c*x]])/Sqrt[b]])/(5*b^
(7/2)*c^3) - (Sqrt[Pi]*Erfi[Sqrt[a + b*ArcSinh[c*x]]/Sqrt[b]])/(15*b^(7/2)*c^3*E^(a/b)) + (3*Sqrt[3*Pi]*Erfi[(
Sqrt[3]*Sqrt[a + b*ArcSinh[c*x]])/Sqrt[b]])/(5*b^(7/2)*c^3*E^((3*a)/b))

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 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 5778

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[1/(b^2*c^(m + 1)*(n + 1)), Subst[Int[ExpandTrigReduce[x^(n + 1), Si
nh[-a/b + x/b]^(m - 1)*(m + (m + 1)*Sinh[-a/b + x/b]^2), x], x], x, a + b*ArcSinh[c*x]], x] /; FreeQ[{a, b, c}
, x] && IGtQ[m, 0] && GeQ[n, -2] && LtQ[n, -1]

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 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 5819

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[(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]

Rubi steps

\begin {align*} \int \frac {x^2}{\left (a+b \sinh ^{-1}(c x)\right )^{7/2}} \, dx &=-\frac {2 x^2 \sqrt {1+c^2 x^2}}{5 b c \left (a+b \sinh ^{-1}(c x)\right )^{5/2}}+\frac {4 \int \frac {x}{\sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )^{5/2}} \, dx}{5 b c}+\frac {(6 c) \int \frac {x^3}{\sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )^{5/2}} \, dx}{5 b}\\ &=-\frac {2 x^2 \sqrt {1+c^2 x^2}}{5 b c \left (a+b \sinh ^{-1}(c x)\right )^{5/2}}-\frac {8 x}{15 b^2 c^2 \left (a+b \sinh ^{-1}(c x)\right )^{3/2}}-\frac {4 x^3}{5 b^2 \left (a+b \sinh ^{-1}(c x)\right )^{3/2}}+\frac {12 \int \frac {x^2}{\left (a+b \sinh ^{-1}(c x)\right )^{3/2}} \, dx}{5 b^2}+\frac {8 \int \frac {1}{\left (a+b \sinh ^{-1}(c x)\right )^{3/2}} \, dx}{15 b^2 c^2}\\ &=-\frac {2 x^2 \sqrt {1+c^2 x^2}}{5 b c \left (a+b \sinh ^{-1}(c x)\right )^{5/2}}-\frac {8 x}{15 b^2 c^2 \left (a+b \sinh ^{-1}(c x)\right )^{3/2}}-\frac {4 x^3}{5 b^2 \left (a+b \sinh ^{-1}(c x)\right )^{3/2}}-\frac {16 \sqrt {1+c^2 x^2}}{15 b^3 c^3 \sqrt {a+b \sinh ^{-1}(c x)}}-\frac {24 x^2 \sqrt {1+c^2 x^2}}{5 b^3 c \sqrt {a+b \sinh ^{-1}(c x)}}+\frac {24 \text {Subst}\left (\int \left (-\frac {\sinh (x)}{4 \sqrt {a+b x}}+\frac {3 \sinh (3 x)}{4 \sqrt {a+b x}}\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{5 b^3 c^3}+\frac {16 \int \frac {x}{\sqrt {1+c^2 x^2} \sqrt {a+b \sinh ^{-1}(c x)}} \, dx}{15 b^3 c}\\ &=-\frac {2 x^2 \sqrt {1+c^2 x^2}}{5 b c \left (a+b \sinh ^{-1}(c x)\right )^{5/2}}-\frac {8 x}{15 b^2 c^2 \left (a+b \sinh ^{-1}(c x)\right )^{3/2}}-\frac {4 x^3}{5 b^2 \left (a+b \sinh ^{-1}(c x)\right )^{3/2}}-\frac {16 \sqrt {1+c^2 x^2}}{15 b^3 c^3 \sqrt {a+b \sinh ^{-1}(c x)}}-\frac {24 x^2 \sqrt {1+c^2 x^2}}{5 b^3 c \sqrt {a+b \sinh ^{-1}(c x)}}+\frac {16 \text {Subst}\left (\int \frac {\sinh (x)}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c x)\right )}{15 b^3 c^3}-\frac {6 \text {Subst}\left (\int \frac {\sinh (x)}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c x)\right )}{5 b^3 c^3}+\frac {18 \text {Subst}\left (\int \frac {\sinh (3 x)}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c x)\right )}{5 b^3 c^3}\\ &=-\frac {2 x^2 \sqrt {1+c^2 x^2}}{5 b c \left (a+b \sinh ^{-1}(c x)\right )^{5/2}}-\frac {8 x}{15 b^2 c^2 \left (a+b \sinh ^{-1}(c x)\right )^{3/2}}-\frac {4 x^3}{5 b^2 \left (a+b \sinh ^{-1}(c x)\right )^{3/2}}-\frac {16 \sqrt {1+c^2 x^2}}{15 b^3 c^3 \sqrt {a+b \sinh ^{-1}(c x)}}-\frac {24 x^2 \sqrt {1+c^2 x^2}}{5 b^3 c \sqrt {a+b \sinh ^{-1}(c x)}}-\frac {8 \text {Subst}\left (\int \frac {e^{-x}}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c x)\right )}{15 b^3 c^3}+\frac {8 \text {Subst}\left (\int \frac {e^x}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c x)\right )}{15 b^3 c^3}+\frac {3 \text {Subst}\left (\int \frac {e^{-x}}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c x)\right )}{5 b^3 c^3}-\frac {3 \text {Subst}\left (\int \frac {e^x}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c x)\right )}{5 b^3 c^3}-\frac {9 \text {Subst}\left (\int \frac {e^{-3 x}}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c x)\right )}{5 b^3 c^3}+\frac {9 \text {Subst}\left (\int \frac {e^{3 x}}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c x)\right )}{5 b^3 c^3}\\ &=-\frac {2 x^2 \sqrt {1+c^2 x^2}}{5 b c \left (a+b \sinh ^{-1}(c x)\right )^{5/2}}-\frac {8 x}{15 b^2 c^2 \left (a+b \sinh ^{-1}(c x)\right )^{3/2}}-\frac {4 x^3}{5 b^2 \left (a+b \sinh ^{-1}(c x)\right )^{3/2}}-\frac {16 \sqrt {1+c^2 x^2}}{15 b^3 c^3 \sqrt {a+b \sinh ^{-1}(c x)}}-\frac {24 x^2 \sqrt {1+c^2 x^2}}{5 b^3 c \sqrt {a+b \sinh ^{-1}(c x)}}-\frac {16 \text {Subst}\left (\int e^{\frac {a}{b}-\frac {x^2}{b}} \, dx,x,\sqrt {a+b \sinh ^{-1}(c x)}\right )}{15 b^4 c^3}+\frac {16 \text {Subst}\left (\int e^{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+b \sinh ^{-1}(c x)}\right )}{15 b^4 c^3}+\frac {6 \text {Subst}\left (\int e^{\frac {a}{b}-\frac {x^2}{b}} \, dx,x,\sqrt {a+b \sinh ^{-1}(c x)}\right )}{5 b^4 c^3}-\frac {6 \text {Subst}\left (\int e^{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+b \sinh ^{-1}(c x)}\right )}{5 b^4 c^3}-\frac {18 \text {Subst}\left (\int e^{\frac {3 a}{b}-\frac {3 x^2}{b}} \, dx,x,\sqrt {a+b \sinh ^{-1}(c x)}\right )}{5 b^4 c^3}+\frac {18 \text {Subst}\left (\int e^{-\frac {3 a}{b}+\frac {3 x^2}{b}} \, dx,x,\sqrt {a+b \sinh ^{-1}(c x)}\right )}{5 b^4 c^3}\\ &=-\frac {2 x^2 \sqrt {1+c^2 x^2}}{5 b c \left (a+b \sinh ^{-1}(c x)\right )^{5/2}}-\frac {8 x}{15 b^2 c^2 \left (a+b \sinh ^{-1}(c x)\right )^{3/2}}-\frac {4 x^3}{5 b^2 \left (a+b \sinh ^{-1}(c x)\right )^{3/2}}-\frac {16 \sqrt {1+c^2 x^2}}{15 b^3 c^3 \sqrt {a+b \sinh ^{-1}(c x)}}-\frac {24 x^2 \sqrt {1+c^2 x^2}}{5 b^3 c \sqrt {a+b \sinh ^{-1}(c x)}}+\frac {e^{a/b} \sqrt {\pi } \text {erf}\left (\frac {\sqrt {a+b \sinh ^{-1}(c x)}}{\sqrt {b}}\right )}{15 b^{7/2} c^3}-\frac {3 e^{\frac {3 a}{b}} \sqrt {3 \pi } \text {erf}\left (\frac {\sqrt {3} \sqrt {a+b \sinh ^{-1}(c x)}}{\sqrt {b}}\right )}{5 b^{7/2} c^3}-\frac {e^{-\frac {a}{b}} \sqrt {\pi } \text {erfi}\left (\frac {\sqrt {a+b \sinh ^{-1}(c x)}}{\sqrt {b}}\right )}{15 b^{7/2} c^3}+\frac {3 e^{-\frac {3 a}{b}} \sqrt {3 \pi } \text {erfi}\left (\frac {\sqrt {3} \sqrt {a+b \sinh ^{-1}(c x)}}{\sqrt {b}}\right )}{5 b^{7/2} c^3}\\ \end {align*}

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Mathematica [A]
time = 1.07, size = 417, normalized size = 1.21 \begin {gather*} \frac {3 b^2 e^{\sinh ^{-1}(c x)}+e^{-\sinh ^{-1}(c x)} \left (4 a^2-2 a b+3 b^2+2 (4 a-b) b \sinh ^{-1}(c x)+4 b^2 \sinh ^{-1}(c x)^2-4 e^{\frac {a}{b}+\sinh ^{-1}(c x)} \sqrt {\frac {a}{b}+\sinh ^{-1}(c x)} \left (a+b \sinh ^{-1}(c x)\right )^2 \Gamma \left (\frac {1}{2},\frac {a}{b}+\sinh ^{-1}(c x)\right )\right )-3 \left (b^2 e^{3 \sinh ^{-1}(c x)}+2 e^{-\frac {3 a}{b}} \left (a+b \sinh ^{-1}(c x)\right ) \left (e^{3 \left (\frac {a}{b}+\sinh ^{-1}(c x)\right )} \left (6 a+b+6 b \sinh ^{-1}(c x)\right )+6 \sqrt {3} b \left (-\frac {a+b \sinh ^{-1}(c x)}{b}\right )^{3/2} \Gamma \left (\frac {1}{2},-\frac {3 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )\right )\right )+2 e^{-\frac {a}{b}} \left (a+b \sinh ^{-1}(c x)\right ) \left (e^{\frac {a}{b}+\sinh ^{-1}(c x)} \left (2 a+b+2 b \sinh ^{-1}(c x)\right )+2 b \left (-\frac {a+b \sinh ^{-1}(c x)}{b}\right )^{3/2} \Gamma \left (\frac {1}{2},-\frac {a+b \sinh ^{-1}(c x)}{b}\right )\right )-3 e^{-3 \sinh ^{-1}(c x)} \left (b^2+2 \left (a+b \sinh ^{-1}(c x)\right ) \left (6 a-b+6 b \sinh ^{-1}(c x)-6 \sqrt {3} e^{3 \left (\frac {a}{b}+\sinh ^{-1}(c x)\right )} \sqrt {\frac {a}{b}+\sinh ^{-1}(c x)} \left (a+b \sinh ^{-1}(c x)\right ) \Gamma \left (\frac {1}{2},\frac {3 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )\right )\right )}{60 b^3 c^3 \left (a+b \sinh ^{-1}(c x)\right )^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^2/(a + b*ArcSinh[c*x])^(7/2),x]

[Out]

(3*b^2*E^ArcSinh[c*x] + (4*a^2 - 2*a*b + 3*b^2 + 2*(4*a - b)*b*ArcSinh[c*x] + 4*b^2*ArcSinh[c*x]^2 - 4*E^(a/b
+ ArcSinh[c*x])*Sqrt[a/b + ArcSinh[c*x]]*(a + b*ArcSinh[c*x])^2*Gamma[1/2, a/b + ArcSinh[c*x]])/E^ArcSinh[c*x]
- 3*(b^2*E^(3*ArcSinh[c*x]) + (2*(a + b*ArcSinh[c*x])*(E^(3*(a/b + ArcSinh[c*x]))*(6*a + b + 6*b*ArcSinh[c*x]
) + 6*Sqrt[3]*b*(-((a + b*ArcSinh[c*x])/b))^(3/2)*Gamma[1/2, (-3*(a + b*ArcSinh[c*x]))/b]))/E^((3*a)/b)) + (2*
(a + b*ArcSinh[c*x])*(E^(a/b + ArcSinh[c*x])*(2*a + b + 2*b*ArcSinh[c*x]) + 2*b*(-((a + b*ArcSinh[c*x])/b))^(3
/2)*Gamma[1/2, -((a + b*ArcSinh[c*x])/b)]))/E^(a/b) - (3*(b^2 + 2*(a + b*ArcSinh[c*x])*(6*a - b + 6*b*ArcSinh[
c*x] - 6*Sqrt[3]*E^(3*(a/b + ArcSinh[c*x]))*Sqrt[a/b + ArcSinh[c*x]]*(a + b*ArcSinh[c*x])*Gamma[1/2, (3*(a + b
*ArcSinh[c*x]))/b])))/E^(3*ArcSinh[c*x]))/(60*b^3*c^3*(a + b*ArcSinh[c*x])^(5/2))

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Maple [F]
time = 180.00, size = 0, normalized size = 0.00 \[\int \frac {x^{2}}{\left (a +b \arcsinh \left (c x \right )\right )^{\frac {7}{2}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/(a+b*arcsinh(c*x))^(7/2),x)

[Out]

int(x^2/(a+b*arcsinh(c*x))^(7/2),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^2/(a+b*arcsinh(c*x))^(7/2),x, algorithm="maxima")

[Out]

integrate(x^2/(b*arcsinh(c*x) + a)^(7/2), 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^2/(a+b*arcsinh(c*x))^(7/2),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^{2}}{\left (a + b \operatorname {asinh}{\left (c x \right )}\right )^{\frac {7}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2/(a+b*asinh(c*x))**(7/2),x)

[Out]

Integral(x**2/(a + b*asinh(c*x))**(7/2), 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^2/(a+b*arcsinh(c*x))^(7/2),x, algorithm="giac")

[Out]

integrate(x^2/(b*arcsinh(c*x) + a)^(7/2), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {x^2}{{\left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )}^{7/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/(a + b*asinh(c*x))^(7/2),x)

[Out]

int(x^2/(a + b*asinh(c*x))^(7/2), x)

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