3.2.94 \(\int (c e+d e x)^2 (a+b \sinh ^{-1}(c+d x))^{5/2} \, dx\) [194]

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

[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

________________________________________________________________________________________

Rubi [A]
time = 0.84, 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 = {5859, 12, 5777, 5812, 5798, 5772, 5819, 3389, 2211, 2236, 2235, 3393} \begin {gather*} -\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} \end {gather*}

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 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 3393

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 5772

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 5777

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 5798

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/(2*c*(p + 1)))*Simp[(d + e*x^2)^p/(1 + c^2*x^2)
^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 5812

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

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]

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 (c e+d e x)^2 \left (a+b \sinh ^{-1}(c+d x)\right )^{5/2} \, dx &=\frac {\text {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 \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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 = 0.30, size = 238, normalized size = 0.60 \begin {gather*} -\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}} \end {gather*}

Antiderivative was successfully verified.

[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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Maple [F]
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 \begin {gather*} \text {Failed to integrate} \end {gather*}

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*x*e + c*e)^2*(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((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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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} 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 ) \end {gather*}

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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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((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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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int {\left (c\,e+d\,e\,x\right )}^2\,{\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((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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