3.194 \(\int (c e+d e x)^2 (a+b \sinh ^{-1}(c+d x))^{5/2} \, dx\)

Optimal. Leaf size=394 \[ -\frac {15 \sqrt {\pi } b^{5/2} e^2 e^{a/b} \text {erf}\left (\frac {\sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{64 d}+\frac {5 \sqrt {\frac {\pi }{3}} b^{5/2} e^2 e^{\frac {3 a}{b}} \text {erf}\left (\frac {\sqrt {3} \sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{576 d}+\frac {15 \sqrt {\pi } b^{5/2} e^2 e^{-\frac {a}{b}} \text {erfi}\left (\frac {\sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{64 d}-\frac {5 \sqrt {\frac {\pi }{3}} b^{5/2} e^2 e^{-\frac {3 a}{b}} \text {erfi}\left (\frac {\sqrt {3} \sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{576 d}+\frac {5 b^2 e^2 (c+d x)^3 \sqrt {a+b \sinh ^{-1}(c+d x)}}{36 d}-\frac {5 b^2 e^2 (c+d x) \sqrt {a+b \sinh ^{-1}(c+d x)}}{6 d}+\frac {e^2 (c+d x)^3 \left (a+b \sinh ^{-1}(c+d x)\right )^{5/2}}{3 d}-\frac {5 b e^2 \sqrt {(c+d x)^2+1} (c+d x)^2 \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}{18 d}+\frac {5 b e^2 \sqrt {(c+d x)^2+1} \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}{9 d} \]

[Out]

1/3*e^2*(d*x+c)^3*(a+b*arcsinh(d*x+c))^(5/2)/d+5/1728*b^(5/2)*e^2*exp(3*a/b)*erf(3^(1/2)*(a+b*arcsinh(d*x+c))^
(1/2)/b^(1/2))*3^(1/2)*Pi^(1/2)/d-5/1728*b^(5/2)*e^2*erfi(3^(1/2)*(a+b*arcsinh(d*x+c))^(1/2)/b^(1/2))*3^(1/2)*
Pi^(1/2)/d/exp(3*a/b)-15/64*b^(5/2)*e^2*exp(a/b)*erf((a+b*arcsinh(d*x+c))^(1/2)/b^(1/2))*Pi^(1/2)/d+15/64*b^(5
/2)*e^2*erfi((a+b*arcsinh(d*x+c))^(1/2)/b^(1/2))*Pi^(1/2)/d/exp(a/b)+5/9*b*e^2*(a+b*arcsinh(d*x+c))^(3/2)*(1+(
d*x+c)^2)^(1/2)/d-5/18*b*e^2*(d*x+c)^2*(a+b*arcsinh(d*x+c))^(3/2)*(1+(d*x+c)^2)^(1/2)/d-5/6*b^2*e^2*(d*x+c)*(a
+b*arcsinh(d*x+c))^(1/2)/d+5/36*b^2*e^2*(d*x+c)^3*(a+b*arcsinh(d*x+c))^(1/2)/d

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Rubi [A]  time = 1.24, antiderivative size = 394, normalized size of antiderivative = 1.00, number of steps used = 26, number of rules used = 12, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.480, Rules used = {5865, 12, 5663, 5758, 5717, 5653, 5779, 3308, 2180, 2204, 2205, 3312} \[ -\frac {15 \sqrt {\pi } b^{5/2} e^2 e^{a/b} \text {Erf}\left (\frac {\sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{64 d}+\frac {5 \sqrt {\frac {\pi }{3}} b^{5/2} e^2 e^{\frac {3 a}{b}} \text {Erf}\left (\frac {\sqrt {3} \sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{576 d}+\frac {15 \sqrt {\pi } b^{5/2} e^2 e^{-\frac {a}{b}} \text {Erfi}\left (\frac {\sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{64 d}-\frac {5 \sqrt {\frac {\pi }{3}} b^{5/2} e^2 e^{-\frac {3 a}{b}} \text {Erfi}\left (\frac {\sqrt {3} \sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{576 d}+\frac {5 b^2 e^2 (c+d x)^3 \sqrt {a+b \sinh ^{-1}(c+d x)}}{36 d}-\frac {5 b^2 e^2 (c+d x) \sqrt {a+b \sinh ^{-1}(c+d x)}}{6 d}+\frac {e^2 (c+d x)^3 \left (a+b \sinh ^{-1}(c+d x)\right )^{5/2}}{3 d}-\frac {5 b e^2 \sqrt {(c+d x)^2+1} (c+d x)^2 \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}{18 d}+\frac {5 b e^2 \sqrt {(c+d x)^2+1} \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}{9 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-5*b^2*e^2*(c + d*x)*Sqrt[a + b*ArcSinh[c + d*x]])/(6*d) + (5*b^2*e^2*(c + d*x)^3*Sqrt[a + b*ArcSinh[c + d*x]
])/(36*d) + (5*b*e^2*Sqrt[1 + (c + d*x)^2]*(a + b*ArcSinh[c + d*x])^(3/2))/(9*d) - (5*b*e^2*(c + d*x)^2*Sqrt[1
+ (c + d*x)^2]*(a + b*ArcSinh[c + d*x])^(3/2))/(18*d) + (e^2*(c + d*x)^3*(a + b*ArcSinh[c + d*x])^(5/2))/(3*d
) - (15*b^(5/2)*e^2*E^(a/b)*Sqrt[Pi]*Erf[Sqrt[a + b*ArcSinh[c + d*x]]/Sqrt[b]])/(64*d) + (5*b^(5/2)*e^2*E^((3*
a)/b)*Sqrt[Pi/3]*Erf[(Sqrt[3]*Sqrt[a + b*ArcSinh[c + d*x]])/Sqrt[b]])/(576*d) + (15*b^(5/2)*e^2*Sqrt[Pi]*Erfi[
Sqrt[a + b*ArcSinh[c + d*x]]/Sqrt[b]])/(64*d*E^(a/b)) - (5*b^(5/2)*e^2*Sqrt[Pi/3]*Erfi[(Sqrt[3]*Sqrt[a + b*Arc
Sinh[c + d*x]])/Sqrt[b]])/(576*d*E^((3*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 2180

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] &&  !$UseGamma === True

Rule 2204

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 2205

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 3308

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 3312

Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)]^(n_), x_Symbol] :> Int[ExpandTrigReduce[(c + d*x)^m, Sin
[e + f*x]^n, x], x] /; FreeQ[{c, d, e, f, m}, x] && IGtQ[n, 1] && ( !RationalQ[m] || (GeQ[m, -1] && LtQ[m, 1])
)

Rule 5653

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

Rule 5663

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_)*(x_)^(m_.), x_Symbol] :> Simp[(x^(m + 1)*(a + b*ArcSinh[c*x])^n)/
(m + 1), x] - Dist[(b*c*n)/(m + 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] && GtQ[n, 0]

Rule 5717

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*(x_)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[((d + e*x^2)
^(p + 1)*(a + b*ArcSinh[c*x])^n)/(2*e*(p + 1)), x] - Dist[(b*n*d^IntPart[p]*(d + e*x^2)^FracPart[p])/(2*c*(p +
1)*(1 + c^2*x^2)^FracPart[p]), Int[(1 + c^2*x^2)^(p + 1/2)*(a + b*ArcSinh[c*x])^(n - 1), x], x] /; FreeQ[{a,
b, c, d, e, p}, x] && EqQ[e, c^2*d] && GtQ[n, 0] && NeQ[p, -1]

Rule 5758

Int[(((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_))/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp
[(f*(f*x)^(m - 1)*Sqrt[d + e*x^2]*(a + b*ArcSinh[c*x])^n)/(e*m), x] + (-Dist[(f^2*(m - 1))/(c^2*m), Int[((f*x)
^(m - 2)*(a + b*ArcSinh[c*x])^n)/Sqrt[d + e*x^2], x], x] - Dist[(b*f*n*Sqrt[1 + c^2*x^2])/(c*m*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}, x] && EqQ[e, c^2*d] &&
GtQ[n, 0] && GtQ[m, 1] && IntegerQ[m]

Rule 5779

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[d^p/c^
(m + 1), Subst[Int[(a + b*x)^n*Sinh[x]^m*Cosh[x]^(2*p + 1), x], x, ArcSinh[c*x]], x] /; FreeQ[{a, b, c, d, e,
n}, x] && EqQ[e, c^2*d] && IntegerQ[2*p] && GtQ[p, -1] && IGtQ[m, 0] && (IntegerQ[p] || GtQ[d, 0])

Rule 5865

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 (c e+d e x)^2 \left (a+b \sinh ^{-1}(c+d x)\right )^{5/2} \, dx &=\frac {\operatorname {Subst}\left (\int e^2 x^2 \left (a+b \sinh ^{-1}(x)\right )^{5/2} \, dx,x,c+d x\right )}{d}\\ &=\frac {e^2 \operatorname {Subst}\left (\int x^2 \left (a+b \sinh ^{-1}(x)\right )^{5/2} \, dx,x,c+d x\right )}{d}\\ &=\frac {e^2 (c+d x)^3 \left (a+b \sinh ^{-1}(c+d x)\right )^{5/2}}{3 d}-\frac {\left (5 b e^2\right ) \operatorname {Subst}\left (\int \frac {x^3 \left (a+b \sinh ^{-1}(x)\right )^{3/2}}{\sqrt {1+x^2}} \, dx,x,c+d x\right )}{6 d}\\ &=-\frac {5 b e^2 (c+d x)^2 \sqrt {1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}{18 d}+\frac {e^2 (c+d x)^3 \left (a+b \sinh ^{-1}(c+d x)\right )^{5/2}}{3 d}+\frac {\left (5 b e^2\right ) \operatorname {Subst}\left (\int \frac {x \left (a+b \sinh ^{-1}(x)\right )^{3/2}}{\sqrt {1+x^2}} \, dx,x,c+d x\right )}{9 d}+\frac {\left (5 b^2 e^2\right ) \operatorname {Subst}\left (\int x^2 \sqrt {a+b \sinh ^{-1}(x)} \, dx,x,c+d x\right )}{12 d}\\ &=\frac {5 b^2 e^2 (c+d x)^3 \sqrt {a+b \sinh ^{-1}(c+d x)}}{36 d}+\frac {5 b e^2 \sqrt {1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}{9 d}-\frac {5 b e^2 (c+d x)^2 \sqrt {1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}{18 d}+\frac {e^2 (c+d x)^3 \left (a+b \sinh ^{-1}(c+d x)\right )^{5/2}}{3 d}-\frac {\left (5 b^2 e^2\right ) \operatorname {Subst}\left (\int \sqrt {a+b \sinh ^{-1}(x)} \, dx,x,c+d x\right )}{6 d}-\frac {\left (5 b^3 e^2\right ) \operatorname {Subst}\left (\int \frac {x^3}{\sqrt {1+x^2} \sqrt {a+b \sinh ^{-1}(x)}} \, dx,x,c+d x\right )}{72 d}\\ &=-\frac {5 b^2 e^2 (c+d x) \sqrt {a+b \sinh ^{-1}(c+d x)}}{6 d}+\frac {5 b^2 e^2 (c+d x)^3 \sqrt {a+b \sinh ^{-1}(c+d x)}}{36 d}+\frac {5 b e^2 \sqrt {1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}{9 d}-\frac {5 b e^2 (c+d x)^2 \sqrt {1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}{18 d}+\frac {e^2 (c+d x)^3 \left (a+b \sinh ^{-1}(c+d x)\right )^{5/2}}{3 d}-\frac {\left (5 b^3 e^2\right ) \operatorname {Subst}\left (\int \frac {\sinh ^3(x)}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{72 d}+\frac {\left (5 b^3 e^2\right ) \operatorname {Subst}\left (\int \frac {x}{\sqrt {1+x^2} \sqrt {a+b \sinh ^{-1}(x)}} \, dx,x,c+d x\right )}{12 d}\\ &=-\frac {5 b^2 e^2 (c+d x) \sqrt {a+b \sinh ^{-1}(c+d x)}}{6 d}+\frac {5 b^2 e^2 (c+d x)^3 \sqrt {a+b \sinh ^{-1}(c+d x)}}{36 d}+\frac {5 b e^2 \sqrt {1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}{9 d}-\frac {5 b e^2 (c+d x)^2 \sqrt {1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}{18 d}+\frac {e^2 (c+d x)^3 \left (a+b \sinh ^{-1}(c+d x)\right )^{5/2}}{3 d}-\frac {\left (5 i b^3 e^2\right ) \operatorname {Subst}\left (\int \left (\frac {3 i \sinh (x)}{4 \sqrt {a+b x}}-\frac {i \sinh (3 x)}{4 \sqrt {a+b x}}\right ) \, dx,x,\sinh ^{-1}(c+d x)\right )}{72 d}+\frac {\left (5 b^3 e^2\right ) \operatorname {Subst}\left (\int \frac {\sinh (x)}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{12 d}\\ &=-\frac {5 b^2 e^2 (c+d x) \sqrt {a+b \sinh ^{-1}(c+d x)}}{6 d}+\frac {5 b^2 e^2 (c+d x)^3 \sqrt {a+b \sinh ^{-1}(c+d x)}}{36 d}+\frac {5 b e^2 \sqrt {1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}{9 d}-\frac {5 b e^2 (c+d x)^2 \sqrt {1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}{18 d}+\frac {e^2 (c+d x)^3 \left (a+b \sinh ^{-1}(c+d x)\right )^{5/2}}{3 d}-\frac {\left (5 b^3 e^2\right ) \operatorname {Subst}\left (\int \frac {\sinh (3 x)}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{288 d}+\frac {\left (5 b^3 e^2\right ) \operatorname {Subst}\left (\int \frac {\sinh (x)}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{96 d}-\frac {\left (5 b^3 e^2\right ) \operatorname {Subst}\left (\int \frac {e^{-x}}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{24 d}+\frac {\left (5 b^3 e^2\right ) \operatorname {Subst}\left (\int \frac {e^x}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{24 d}\\ &=-\frac {5 b^2 e^2 (c+d x) \sqrt {a+b \sinh ^{-1}(c+d x)}}{6 d}+\frac {5 b^2 e^2 (c+d x)^3 \sqrt {a+b \sinh ^{-1}(c+d x)}}{36 d}+\frac {5 b e^2 \sqrt {1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}{9 d}-\frac {5 b e^2 (c+d x)^2 \sqrt {1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}{18 d}+\frac {e^2 (c+d x)^3 \left (a+b \sinh ^{-1}(c+d x)\right )^{5/2}}{3 d}-\frac {\left (5 b^2 e^2\right ) \operatorname {Subst}\left (\int e^{\frac {a}{b}-\frac {x^2}{b}} \, dx,x,\sqrt {a+b \sinh ^{-1}(c+d x)}\right )}{12 d}+\frac {\left (5 b^2 e^2\right ) \operatorname {Subst}\left (\int e^{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+b \sinh ^{-1}(c+d x)}\right )}{12 d}+\frac {\left (5 b^3 e^2\right ) \operatorname {Subst}\left (\int \frac {e^{-3 x}}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{576 d}-\frac {\left (5 b^3 e^2\right ) \operatorname {Subst}\left (\int \frac {e^{3 x}}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{576 d}-\frac {\left (5 b^3 e^2\right ) \operatorname {Subst}\left (\int \frac {e^{-x}}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{192 d}+\frac {\left (5 b^3 e^2\right ) \operatorname {Subst}\left (\int \frac {e^x}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{192 d}\\ &=-\frac {5 b^2 e^2 (c+d x) \sqrt {a+b \sinh ^{-1}(c+d x)}}{6 d}+\frac {5 b^2 e^2 (c+d x)^3 \sqrt {a+b \sinh ^{-1}(c+d x)}}{36 d}+\frac {5 b e^2 \sqrt {1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}{9 d}-\frac {5 b e^2 (c+d x)^2 \sqrt {1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}{18 d}+\frac {e^2 (c+d x)^3 \left (a+b \sinh ^{-1}(c+d x)\right )^{5/2}}{3 d}-\frac {5 b^{5/2} e^2 e^{a/b} \sqrt {\pi } \text {erf}\left (\frac {\sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{24 d}+\frac {5 b^{5/2} e^2 e^{-\frac {a}{b}} \sqrt {\pi } \text {erfi}\left (\frac {\sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{24 d}+\frac {\left (5 b^2 e^2\right ) \operatorname {Subst}\left (\int e^{\frac {3 a}{b}-\frac {3 x^2}{b}} \, dx,x,\sqrt {a+b \sinh ^{-1}(c+d x)}\right )}{288 d}-\frac {\left (5 b^2 e^2\right ) \operatorname {Subst}\left (\int e^{-\frac {3 a}{b}+\frac {3 x^2}{b}} \, dx,x,\sqrt {a+b \sinh ^{-1}(c+d x)}\right )}{288 d}-\frac {\left (5 b^2 e^2\right ) \operatorname {Subst}\left (\int e^{\frac {a}{b}-\frac {x^2}{b}} \, dx,x,\sqrt {a+b \sinh ^{-1}(c+d x)}\right )}{96 d}+\frac {\left (5 b^2 e^2\right ) \operatorname {Subst}\left (\int e^{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+b \sinh ^{-1}(c+d x)}\right )}{96 d}\\ &=-\frac {5 b^2 e^2 (c+d x) \sqrt {a+b \sinh ^{-1}(c+d x)}}{6 d}+\frac {5 b^2 e^2 (c+d x)^3 \sqrt {a+b \sinh ^{-1}(c+d x)}}{36 d}+\frac {5 b e^2 \sqrt {1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}{9 d}-\frac {5 b e^2 (c+d x)^2 \sqrt {1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}{18 d}+\frac {e^2 (c+d x)^3 \left (a+b \sinh ^{-1}(c+d x)\right )^{5/2}}{3 d}-\frac {15 b^{5/2} e^2 e^{a/b} \sqrt {\pi } \text {erf}\left (\frac {\sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{64 d}+\frac {5 b^{5/2} e^2 e^{\frac {3 a}{b}} \sqrt {\frac {\pi }{3}} \text {erf}\left (\frac {\sqrt {3} \sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{576 d}+\frac {15 b^{5/2} e^2 e^{-\frac {a}{b}} \sqrt {\pi } \text {erfi}\left (\frac {\sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{64 d}-\frac {5 b^{5/2} e^2 e^{-\frac {3 a}{b}} \sqrt {\frac {\pi }{3}} \text {erfi}\left (\frac {\sqrt {3} \sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{576 d}\\ \end {align*}

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Mathematica [A]  time = 0.42, size = 238, normalized size = 0.60 \[ -\frac {e^2 e^{-\frac {3 a}{b}} \left (a+b \sinh ^{-1}(c+d x)\right )^{5/2} \left (81 e^{\frac {4 a}{b}} \sqrt {-\frac {a+b \sinh ^{-1}(c+d x)}{b}} \Gamma \left (\frac {7}{2},\frac {a}{b}+\sinh ^{-1}(c+d x)\right )+\sqrt {3} \sqrt {\frac {a}{b}+\sinh ^{-1}(c+d x)} \Gamma \left (\frac {7}{2},-\frac {3 \left (a+b \sinh ^{-1}(c+d x)\right )}{b}\right )-81 e^{\frac {2 a}{b}} \sqrt {\frac {a}{b}+\sinh ^{-1}(c+d x)} \Gamma \left (\frac {7}{2},-\frac {a+b \sinh ^{-1}(c+d x)}{b}\right )-\sqrt {3} e^{\frac {6 a}{b}} \sqrt {-\frac {a+b \sinh ^{-1}(c+d x)}{b}} \Gamma \left (\frac {7}{2},\frac {3 \left (a+b \sinh ^{-1}(c+d x)\right )}{b}\right )\right )}{648 d \left (-\frac {\left (a+b \sinh ^{-1}(c+d x)\right )^2}{b^2}\right )^{3/2}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-1/648*(e^2*(a + b*ArcSinh[c + d*x])^(5/2)*(81*E^((4*a)/b)*Sqrt[-((a + b*ArcSinh[c + d*x])/b)]*Gamma[7/2, a/b
+ ArcSinh[c + d*x]] + Sqrt[3]*Sqrt[a/b + ArcSinh[c + d*x]]*Gamma[7/2, (-3*(a + b*ArcSinh[c + d*x]))/b] - 81*E^
((2*a)/b)*Sqrt[a/b + ArcSinh[c + d*x]]*Gamma[7/2, -((a + b*ArcSinh[c + d*x])/b)] - Sqrt[3]*E^((6*a)/b)*Sqrt[-(
(a + b*ArcSinh[c + d*x])/b)]*Gamma[7/2, (3*(a + b*ArcSinh[c + d*x]))/b]))/(d*E^((3*a)/b)*(-((a + b*ArcSinh[c +
d*x])^2/b^2))^(3/2))

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fricas [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*e*x+c*e)^2*(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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (d e x + c e\right )}^{2} {\left (b \operatorname {arsinh}\left (d x + c\right ) + a\right )}^{\frac {5}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (d e x + c e\right )}^{2} {\left (b \operatorname {arsinh}\left (d x + c\right ) + a\right )}^{\frac {5}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int {\left (c\,e+d\,e\,x\right )}^2\,{\left (a+b\,\mathrm {asinh}\left (c+d\,x\right )\right )}^{5/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ e^{2} \left (\int a^{2} c^{2} \sqrt {a + b \operatorname {asinh}{\left (c + d x \right )}}\, dx + \int a^{2} d^{2} x^{2} \sqrt {a + b \operatorname {asinh}{\left (c + d x \right )}}\, dx + \int b^{2} c^{2} \sqrt {a + b \operatorname {asinh}{\left (c + d x \right )}} \operatorname {asinh}^{2}{\left (c + d x \right )}\, dx + \int 2 a b c^{2} \sqrt {a + b \operatorname {asinh}{\left (c + d x \right )}} \operatorname {asinh}{\left (c + d x \right )}\, dx + \int 2 a^{2} c d x \sqrt {a + b \operatorname {asinh}{\left (c + d x \right )}}\, dx + \int b^{2} d^{2} x^{2} \sqrt {a + b \operatorname {asinh}{\left (c + d x \right )}} \operatorname {asinh}^{2}{\left (c + d x \right )}\, dx + \int 2 a b d^{2} x^{2} \sqrt {a + b \operatorname {asinh}{\left (c + d x \right )}} \operatorname {asinh}{\left (c + d x \right )}\, dx + \int 2 b^{2} c d x \sqrt {a + b \operatorname {asinh}{\left (c + d x \right )}} \operatorname {asinh}^{2}{\left (c + d x \right )}\, dx + \int 4 a b c d x \sqrt {a + b \operatorname {asinh}{\left (c + d x \right )}} \operatorname {asinh}{\left (c + d x \right )}\, dx\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

e**2*(Integral(a**2*c**2*sqrt(a + b*asinh(c + d*x)), x) + Integral(a**2*d**2*x**2*sqrt(a + b*asinh(c + d*x)),
x) + Integral(b**2*c**2*sqrt(a + b*asinh(c + d*x))*asinh(c + d*x)**2, x) + Integral(2*a*b*c**2*sqrt(a + b*asin
h(c + d*x))*asinh(c + d*x), x) + Integral(2*a**2*c*d*x*sqrt(a + b*asinh(c + d*x)), x) + Integral(b**2*d**2*x**
2*sqrt(a + b*asinh(c + d*x))*asinh(c + d*x)**2, x) + Integral(2*a*b*d**2*x**2*sqrt(a + b*asinh(c + d*x))*asinh
(c + d*x), x) + Integral(2*b**2*c*d*x*sqrt(a + b*asinh(c + d*x))*asinh(c + d*x)**2, x) + Integral(4*a*b*c*d*x*
sqrt(a + b*asinh(c + d*x))*asinh(c + d*x), x))

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