3.2.93 \(\int (c e+d e x)^3 (a+b \sinh ^{-1}(c+d x))^{5/2} \, dx\) [193]

Optimal. Leaf size=455 \[ -\frac {225 b^2 e^3 \sqrt {a+b \sinh ^{-1}(c+d x)}}{2048 d}-\frac {45 b^2 e^3 (c+d x)^2 \sqrt {a+b \sinh ^{-1}(c+d x)}}{256 d}+\frac {15 b^2 e^3 (c+d x)^4 \sqrt {a+b \sinh ^{-1}(c+d x)}}{256 d}+\frac {15 b e^3 (c+d x) \sqrt {1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}{64 d}-\frac {5 b e^3 (c+d x)^3 \sqrt {1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}{32 d}-\frac {3 e^3 \left (a+b \sinh ^{-1}(c+d x)\right )^{5/2}}{32 d}+\frac {e^3 (c+d x)^4 \left (a+b \sinh ^{-1}(c+d x)\right )^{5/2}}{4 d}-\frac {15 b^{5/2} e^3 e^{\frac {4 a}{b}} \sqrt {\pi } \text {Erf}\left (\frac {2 \sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{16384 d}+\frac {15 b^{5/2} e^3 e^{\frac {2 a}{b}} \sqrt {\frac {\pi }{2}} \text {Erf}\left (\frac {\sqrt {2} \sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{512 d}-\frac {15 b^{5/2} e^3 e^{-\frac {4 a}{b}} \sqrt {\pi } \text {Erfi}\left (\frac {2 \sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{16384 d}+\frac {15 b^{5/2} e^3 e^{-\frac {2 a}{b}} \sqrt {\frac {\pi }{2}} \text {Erfi}\left (\frac {\sqrt {2} \sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{512 d} \]

[Out]

-3/32*e^3*(a+b*arcsinh(d*x+c))^(5/2)/d+1/4*e^3*(d*x+c)^4*(a+b*arcsinh(d*x+c))^(5/2)/d+15/1024*b^(5/2)*e^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)/d+15/1024*b^(5/2)*e^3*erfi(2^(1/2)*(a+
b*arcsinh(d*x+c))^(1/2)/b^(1/2))*2^(1/2)*Pi^(1/2)/d/exp(2*a/b)-15/16384*b^(5/2)*e^3*exp(4*a/b)*erf(2*(a+b*arcs
inh(d*x+c))^(1/2)/b^(1/2))*Pi^(1/2)/d-15/16384*b^(5/2)*e^3*erfi(2*(a+b*arcsinh(d*x+c))^(1/2)/b^(1/2))*Pi^(1/2)
/d/exp(4*a/b)+15/64*b*e^3*(d*x+c)*(a+b*arcsinh(d*x+c))^(3/2)*(1+(d*x+c)^2)^(1/2)/d-5/32*b*e^3*(d*x+c)^3*(a+b*a
rcsinh(d*x+c))^(3/2)*(1+(d*x+c)^2)^(1/2)/d-225/2048*b^2*e^3*(a+b*arcsinh(d*x+c))^(1/2)/d-45/256*b^2*e^3*(d*x+c
)^2*(a+b*arcsinh(d*x+c))^(1/2)/d+15/256*b^2*e^3*(d*x+c)^4*(a+b*arcsinh(d*x+c))^(1/2)/d

________________________________________________________________________________________

Rubi [A]
time = 1.02, antiderivative size = 455, normalized size of antiderivative = 1.00, number of steps used = 29, number of rules used = 11, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.440, Rules used = {5859, 12, 5777, 5812, 5783, 5819, 3393, 3388, 2211, 2236, 2235} \begin {gather*} -\frac {15 \sqrt {\pi } b^{5/2} e^3 e^{\frac {4 a}{b}} \text {Erf}\left (\frac {2 \sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{16384 d}+\frac {15 \sqrt {\frac {\pi }{2}} b^{5/2} e^3 e^{\frac {2 a}{b}} \text {Erf}\left (\frac {\sqrt {2} \sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{512 d}-\frac {15 \sqrt {\pi } b^{5/2} e^3 e^{-\frac {4 a}{b}} \text {Erfi}\left (\frac {2 \sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{16384 d}+\frac {15 \sqrt {\frac {\pi }{2}} b^{5/2} e^3 e^{-\frac {2 a}{b}} \text {Erfi}\left (\frac {\sqrt {2} \sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{512 d}+\frac {15 b^2 e^3 (c+d x)^4 \sqrt {a+b \sinh ^{-1}(c+d x)}}{256 d}-\frac {45 b^2 e^3 (c+d x)^2 \sqrt {a+b \sinh ^{-1}(c+d x)}}{256 d}-\frac {225 b^2 e^3 \sqrt {a+b \sinh ^{-1}(c+d x)}}{2048 d}+\frac {e^3 (c+d x)^4 \left (a+b \sinh ^{-1}(c+d x)\right )^{5/2}}{4 d}-\frac {5 b e^3 \sqrt {(c+d x)^2+1} (c+d x)^3 \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}{32 d}+\frac {15 b e^3 \sqrt {(c+d x)^2+1} (c+d x) \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}{64 d}-\frac {3 e^3 \left (a+b \sinh ^{-1}(c+d x)\right )^{5/2}}{32 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(c*e + d*e*x)^3*(a + b*ArcSinh[c + d*x])^(5/2),x]

[Out]

(-225*b^2*e^3*Sqrt[a + b*ArcSinh[c + d*x]])/(2048*d) - (45*b^2*e^3*(c + d*x)^2*Sqrt[a + b*ArcSinh[c + d*x]])/(
256*d) + (15*b^2*e^3*(c + d*x)^4*Sqrt[a + b*ArcSinh[c + d*x]])/(256*d) + (15*b*e^3*(c + d*x)*Sqrt[1 + (c + d*x
)^2]*(a + b*ArcSinh[c + d*x])^(3/2))/(64*d) - (5*b*e^3*(c + d*x)^3*Sqrt[1 + (c + d*x)^2]*(a + b*ArcSinh[c + d*
x])^(3/2))/(32*d) - (3*e^3*(a + b*ArcSinh[c + d*x])^(5/2))/(32*d) + (e^3*(c + d*x)^4*(a + b*ArcSinh[c + d*x])^
(5/2))/(4*d) - (15*b^(5/2)*e^3*E^((4*a)/b)*Sqrt[Pi]*Erf[(2*Sqrt[a + b*ArcSinh[c + d*x]])/Sqrt[b]])/(16384*d) +
(15*b^(5/2)*e^3*E^((2*a)/b)*Sqrt[Pi/2]*Erf[(Sqrt[2]*Sqrt[a + b*ArcSinh[c + d*x]])/Sqrt[b]])/(512*d) - (15*b^(
5/2)*e^3*Sqrt[Pi]*Erfi[(2*Sqrt[a + b*ArcSinh[c + d*x]])/Sqrt[b]])/(16384*d*E^((4*a)/b)) + (15*b^(5/2)*e^3*Sqrt
[Pi/2]*Erfi[(Sqrt[2]*Sqrt[a + b*ArcSinh[c + d*x]])/Sqrt[b]])/(512*d*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 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 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 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 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)^3 \left (a+b \sinh ^{-1}(c+d x)\right )^{5/2} \, dx &=\frac {\text {Subst}\left (\int e^3 x^3 \left (a+b \sinh ^{-1}(x)\right )^{5/2} \, dx,x,c+d x\right )}{d}\\ &=\frac {e^3 \text {Subst}\left (\int x^3 \left (a+b \sinh ^{-1}(x)\right )^{5/2} \, dx,x,c+d x\right )}{d}\\ &=\frac {e^3 (c+d x)^4 \left (a+b \sinh ^{-1}(c+d x)\right )^{5/2}}{4 d}-\frac {\left (5 b e^3\right ) \text {Subst}\left (\int \frac {x^4 \left (a+b \sinh ^{-1}(x)\right )^{3/2}}{\sqrt {1+x^2}} \, dx,x,c+d x\right )}{8 d}\\ &=-\frac {5 b e^3 (c+d x)^3 \sqrt {1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}{32 d}+\frac {e^3 (c+d x)^4 \left (a+b \sinh ^{-1}(c+d x)\right )^{5/2}}{4 d}+\frac {\left (15 b e^3\right ) \text {Subst}\left (\int \frac {x^2 \left (a+b \sinh ^{-1}(x)\right )^{3/2}}{\sqrt {1+x^2}} \, dx,x,c+d x\right )}{32 d}+\frac {\left (15 b^2 e^3\right ) \text {Subst}\left (\int x^3 \sqrt {a+b \sinh ^{-1}(x)} \, dx,x,c+d x\right )}{64 d}\\ &=\frac {15 b^2 e^3 (c+d x)^4 \sqrt {a+b \sinh ^{-1}(c+d x)}}{256 d}+\frac {15 b e^3 (c+d x) \sqrt {1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}{64 d}-\frac {5 b e^3 (c+d x)^3 \sqrt {1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}{32 d}+\frac {e^3 (c+d x)^4 \left (a+b \sinh ^{-1}(c+d x)\right )^{5/2}}{4 d}-\frac {\left (15 b e^3\right ) \text {Subst}\left (\int \frac {\left (a+b \sinh ^{-1}(x)\right )^{3/2}}{\sqrt {1+x^2}} \, dx,x,c+d x\right )}{64 d}-\frac {\left (45 b^2 e^3\right ) \text {Subst}\left (\int x \sqrt {a+b \sinh ^{-1}(x)} \, dx,x,c+d x\right )}{128 d}-\frac {\left (15 b^3 e^3\right ) \text {Subst}\left (\int \frac {x^4}{\sqrt {1+x^2} \sqrt {a+b \sinh ^{-1}(x)}} \, dx,x,c+d x\right )}{512 d}\\ &=-\frac {45 b^2 e^3 (c+d x)^2 \sqrt {a+b \sinh ^{-1}(c+d x)}}{256 d}+\frac {15 b^2 e^3 (c+d x)^4 \sqrt {a+b \sinh ^{-1}(c+d x)}}{256 d}+\frac {15 b e^3 (c+d x) \sqrt {1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}{64 d}-\frac {5 b e^3 (c+d x)^3 \sqrt {1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}{32 d}-\frac {3 e^3 \left (a+b \sinh ^{-1}(c+d x)\right )^{5/2}}{32 d}+\frac {e^3 (c+d x)^4 \left (a+b \sinh ^{-1}(c+d x)\right )^{5/2}}{4 d}-\frac {\left (15 b^3 e^3\right ) \text {Subst}\left (\int \frac {\sinh ^4(x)}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{512 d}+\frac {\left (45 b^3 e^3\right ) \text {Subst}\left (\int \frac {x^2}{\sqrt {1+x^2} \sqrt {a+b \sinh ^{-1}(x)}} \, dx,x,c+d x\right )}{512 d}\\ &=-\frac {45 b^2 e^3 (c+d x)^2 \sqrt {a+b \sinh ^{-1}(c+d x)}}{256 d}+\frac {15 b^2 e^3 (c+d x)^4 \sqrt {a+b \sinh ^{-1}(c+d x)}}{256 d}+\frac {15 b e^3 (c+d x) \sqrt {1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}{64 d}-\frac {5 b e^3 (c+d x)^3 \sqrt {1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}{32 d}-\frac {3 e^3 \left (a+b \sinh ^{-1}(c+d x)\right )^{5/2}}{32 d}+\frac {e^3 (c+d x)^4 \left (a+b \sinh ^{-1}(c+d x)\right )^{5/2}}{4 d}-\frac {\left (15 b^3 e^3\right ) \text {Subst}\left (\int \left (\frac {3}{8 \sqrt {a+b x}}-\frac {\cosh (2 x)}{2 \sqrt {a+b x}}+\frac {\cosh (4 x)}{8 \sqrt {a+b x}}\right ) \, dx,x,\sinh ^{-1}(c+d x)\right )}{512 d}+\frac {\left (45 b^3 e^3\right ) \text {Subst}\left (\int \frac {\sinh ^2(x)}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{512 d}\\ &=-\frac {45 b^2 e^3 \sqrt {a+b \sinh ^{-1}(c+d x)}}{2048 d}-\frac {45 b^2 e^3 (c+d x)^2 \sqrt {a+b \sinh ^{-1}(c+d x)}}{256 d}+\frac {15 b^2 e^3 (c+d x)^4 \sqrt {a+b \sinh ^{-1}(c+d x)}}{256 d}+\frac {15 b e^3 (c+d x) \sqrt {1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}{64 d}-\frac {5 b e^3 (c+d x)^3 \sqrt {1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}{32 d}-\frac {3 e^3 \left (a+b \sinh ^{-1}(c+d x)\right )^{5/2}}{32 d}+\frac {e^3 (c+d x)^4 \left (a+b \sinh ^{-1}(c+d x)\right )^{5/2}}{4 d}-\frac {\left (15 b^3 e^3\right ) \text {Subst}\left (\int \frac {\cosh (4 x)}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{4096 d}+\frac {\left (15 b^3 e^3\right ) \text {Subst}\left (\int \frac {\cosh (2 x)}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{1024 d}-\frac {\left (45 b^3 e^3\right ) \text {Subst}\left (\int \left (\frac {1}{2 \sqrt {a+b x}}-\frac {\cosh (2 x)}{2 \sqrt {a+b x}}\right ) \, dx,x,\sinh ^{-1}(c+d x)\right )}{512 d}\\ &=-\frac {225 b^2 e^3 \sqrt {a+b \sinh ^{-1}(c+d x)}}{2048 d}-\frac {45 b^2 e^3 (c+d x)^2 \sqrt {a+b \sinh ^{-1}(c+d x)}}{256 d}+\frac {15 b^2 e^3 (c+d x)^4 \sqrt {a+b \sinh ^{-1}(c+d x)}}{256 d}+\frac {15 b e^3 (c+d x) \sqrt {1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}{64 d}-\frac {5 b e^3 (c+d x)^3 \sqrt {1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}{32 d}-\frac {3 e^3 \left (a+b \sinh ^{-1}(c+d x)\right )^{5/2}}{32 d}+\frac {e^3 (c+d x)^4 \left (a+b \sinh ^{-1}(c+d x)\right )^{5/2}}{4 d}-\frac {\left (15 b^3 e^3\right ) \text {Subst}\left (\int \frac {e^{-4 x}}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{8192 d}-\frac {\left (15 b^3 e^3\right ) \text {Subst}\left (\int \frac {e^{4 x}}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{8192 d}+\frac {\left (15 b^3 e^3\right ) \text {Subst}\left (\int \frac {e^{-2 x}}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{2048 d}+\frac {\left (15 b^3 e^3\right ) \text {Subst}\left (\int \frac {e^{2 x}}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{2048 d}+\frac {\left (45 b^3 e^3\right ) \text {Subst}\left (\int \frac {\cosh (2 x)}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{1024 d}\\ &=-\frac {225 b^2 e^3 \sqrt {a+b \sinh ^{-1}(c+d x)}}{2048 d}-\frac {45 b^2 e^3 (c+d x)^2 \sqrt {a+b \sinh ^{-1}(c+d x)}}{256 d}+\frac {15 b^2 e^3 (c+d x)^4 \sqrt {a+b \sinh ^{-1}(c+d x)}}{256 d}+\frac {15 b e^3 (c+d x) \sqrt {1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}{64 d}-\frac {5 b e^3 (c+d x)^3 \sqrt {1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}{32 d}-\frac {3 e^3 \left (a+b \sinh ^{-1}(c+d x)\right )^{5/2}}{32 d}+\frac {e^3 (c+d x)^4 \left (a+b \sinh ^{-1}(c+d x)\right )^{5/2}}{4 d}-\frac {\left (15 b^2 e^3\right ) \text {Subst}\left (\int e^{\frac {4 a}{b}-\frac {4 x^2}{b}} \, dx,x,\sqrt {a+b \sinh ^{-1}(c+d x)}\right )}{4096 d}-\frac {\left (15 b^2 e^3\right ) \text {Subst}\left (\int e^{-\frac {4 a}{b}+\frac {4 x^2}{b}} \, dx,x,\sqrt {a+b \sinh ^{-1}(c+d x)}\right )}{4096 d}+\frac {\left (15 b^2 e^3\right ) \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 )}{1024 d}+\frac {\left (15 b^2 e^3\right ) \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 )}{1024 d}+\frac {\left (45 b^3 e^3\right ) \text {Subst}\left (\int \frac {e^{-2 x}}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{2048 d}+\frac {\left (45 b^3 e^3\right ) \text {Subst}\left (\int \frac {e^{2 x}}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{2048 d}\\ &=-\frac {225 b^2 e^3 \sqrt {a+b \sinh ^{-1}(c+d x)}}{2048 d}-\frac {45 b^2 e^3 (c+d x)^2 \sqrt {a+b \sinh ^{-1}(c+d x)}}{256 d}+\frac {15 b^2 e^3 (c+d x)^4 \sqrt {a+b \sinh ^{-1}(c+d x)}}{256 d}+\frac {15 b e^3 (c+d x) \sqrt {1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}{64 d}-\frac {5 b e^3 (c+d x)^3 \sqrt {1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}{32 d}-\frac {3 e^3 \left (a+b \sinh ^{-1}(c+d x)\right )^{5/2}}{32 d}+\frac {e^3 (c+d x)^4 \left (a+b \sinh ^{-1}(c+d x)\right )^{5/2}}{4 d}-\frac {15 b^{5/2} e^3 e^{\frac {4 a}{b}} \sqrt {\pi } \text {erf}\left (\frac {2 \sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{16384 d}+\frac {15 b^{5/2} e^3 e^{\frac {2 a}{b}} \sqrt {\frac {\pi }{2}} \text {erf}\left (\frac {\sqrt {2} \sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{2048 d}-\frac {15 b^{5/2} e^3 e^{-\frac {4 a}{b}} \sqrt {\pi } \text {erfi}\left (\frac {2 \sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{16384 d}+\frac {15 b^{5/2} e^3 e^{-\frac {2 a}{b}} \sqrt {\frac {\pi }{2}} \text {erfi}\left (\frac {\sqrt {2} \sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{2048 d}+\frac {\left (45 b^2 e^3\right ) \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 )}{1024 d}+\frac {\left (45 b^2 e^3\right ) \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 )}{1024 d}\\ &=-\frac {225 b^2 e^3 \sqrt {a+b \sinh ^{-1}(c+d x)}}{2048 d}-\frac {45 b^2 e^3 (c+d x)^2 \sqrt {a+b \sinh ^{-1}(c+d x)}}{256 d}+\frac {15 b^2 e^3 (c+d x)^4 \sqrt {a+b \sinh ^{-1}(c+d x)}}{256 d}+\frac {15 b e^3 (c+d x) \sqrt {1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}{64 d}-\frac {5 b e^3 (c+d x)^3 \sqrt {1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}{32 d}-\frac {3 e^3 \left (a+b \sinh ^{-1}(c+d x)\right )^{5/2}}{32 d}+\frac {e^3 (c+d x)^4 \left (a+b \sinh ^{-1}(c+d x)\right )^{5/2}}{4 d}-\frac {15 b^{5/2} e^3 e^{\frac {4 a}{b}} \sqrt {\pi } \text {erf}\left (\frac {2 \sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{16384 d}+\frac {15 b^{5/2} e^3 e^{\frac {2 a}{b}} \sqrt {\frac {\pi }{2}} \text {erf}\left (\frac {\sqrt {2} \sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{512 d}-\frac {15 b^{5/2} e^3 e^{-\frac {4 a}{b}} \sqrt {\pi } \text {erfi}\left (\frac {2 \sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{16384 d}+\frac {15 b^{5/2} e^3 e^{-\frac {2 a}{b}} \sqrt {\frac {\pi }{2}} \text {erfi}\left (\frac {\sqrt {2} \sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{512 d}\\ \end {align*}

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Mathematica [A]
time = 0.23, size = 223, normalized size = 0.49 \begin {gather*} -\frac {e^3 e^{-\frac {4 a}{b}} \left (a+b \sinh ^{-1}(c+d x)\right )^{5/2} \left (\sqrt {\frac {a}{b}+\sinh ^{-1}(c+d x)} \Gamma \left (\frac {7}{2},-\frac {4 \left (a+b \sinh ^{-1}(c+d x)\right )}{b}\right )-16 \sqrt {2} e^{\frac {2 a}{b}} \sqrt {\frac {a}{b}+\sinh ^{-1}(c+d x)} \Gamma \left (\frac {7}{2},-\frac {2 \left (a+b \sinh ^{-1}(c+d x)\right )}{b}\right )+e^{\frac {6 a}{b}} \sqrt {-\frac {a+b \sinh ^{-1}(c+d x)}{b}} \left (-16 \sqrt {2} \Gamma \left (\frac {7}{2},\frac {2 \left (a+b \sinh ^{-1}(c+d x)\right )}{b}\right )+e^{\frac {2 a}{b}} \Gamma \left (\frac {7}{2},\frac {4 \left (a+b \sinh ^{-1}(c+d x)\right )}{b}\right )\right )\right )}{2048 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)^3*(a + b*ArcSinh[c + d*x])^(5/2),x]

[Out]

-1/2048*(e^3*(a + b*ArcSinh[c + d*x])^(5/2)*(Sqrt[a/b + ArcSinh[c + d*x]]*Gamma[7/2, (-4*(a + b*ArcSinh[c + d*
x]))/b] - 16*Sqrt[2]*E^((2*a)/b)*Sqrt[a/b + ArcSinh[c + d*x]]*Gamma[7/2, (-2*(a + b*ArcSinh[c + d*x]))/b] + E^
((6*a)/b)*Sqrt[-((a + b*ArcSinh[c + d*x])/b)]*(-16*Sqrt[2]*Gamma[7/2, (2*(a + b*ArcSinh[c + d*x]))/b] + E^((2*
a)/b)*Gamma[7/2, (4*(a + b*ArcSinh[c + d*x]))/b])))/(d*E^((4*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 \[\int \left (d e x +c e \right )^{3} \left (a +b \arcsinh \left (d x +c \right )\right )^{\frac {5}{2}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*e*x+c*e)^3*(a+b*arcsinh(d*x+c))^(5/2),x)

[Out]

int((d*e*x+c*e)^3*(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)^3*(a+b*arcsinh(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

integrate((d*x*e + c*e)^3*(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)^3*(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^{3} \left (\int a^{2} c^{3} \sqrt {a + b \operatorname {asinh}{\left (c + d x \right )}}\, dx + \int a^{2} d^{3} x^{3} \sqrt {a + b \operatorname {asinh}{\left (c + d x \right )}}\, dx + \int b^{2} c^{3} \sqrt {a + b \operatorname {asinh}{\left (c + d x \right )}} \operatorname {asinh}^{2}{\left (c + d x \right )}\, dx + \int 2 a b c^{3} \sqrt {a + b \operatorname {asinh}{\left (c + d x \right )}} \operatorname {asinh}{\left (c + d x \right )}\, dx + \int 3 a^{2} c d^{2} x^{2} \sqrt {a + b \operatorname {asinh}{\left (c + d x \right )}}\, dx + \int 3 a^{2} c^{2} d x \sqrt {a + b \operatorname {asinh}{\left (c + d x \right )}}\, dx + \int b^{2} d^{3} x^{3} \sqrt {a + b \operatorname {asinh}{\left (c + d x \right )}} \operatorname {asinh}^{2}{\left (c + d x \right )}\, dx + \int 2 a b d^{3} x^{3} \sqrt {a + b \operatorname {asinh}{\left (c + d x \right )}} \operatorname {asinh}{\left (c + d x \right )}\, dx + \int 3 b^{2} c d^{2} x^{2} \sqrt {a + b \operatorname {asinh}{\left (c + d x \right )}} \operatorname {asinh}^{2}{\left (c + d x \right )}\, dx + \int 3 b^{2} c^{2} d x \sqrt {a + b \operatorname {asinh}{\left (c + d x \right )}} \operatorname {asinh}^{2}{\left (c + d x \right )}\, dx + \int 6 a b c d^{2} x^{2} \sqrt {a + b \operatorname {asinh}{\left (c + d x \right )}} \operatorname {asinh}{\left (c + d x \right )}\, dx + \int 6 a b c^{2} 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)**3*(a+b*asinh(d*x+c))**(5/2),x)

[Out]

e**3*(Integral(a**2*c**3*sqrt(a + b*asinh(c + d*x)), x) + Integral(a**2*d**3*x**3*sqrt(a + b*asinh(c + d*x)),
x) + Integral(b**2*c**3*sqrt(a + b*asinh(c + d*x))*asinh(c + d*x)**2, x) + Integral(2*a*b*c**3*sqrt(a + b*asin
h(c + d*x))*asinh(c + d*x), x) + Integral(3*a**2*c*d**2*x**2*sqrt(a + b*asinh(c + d*x)), x) + Integral(3*a**2*
c**2*d*x*sqrt(a + b*asinh(c + d*x)), x) + Integral(b**2*d**3*x**3*sqrt(a + b*asinh(c + d*x))*asinh(c + d*x)**2
, x) + Integral(2*a*b*d**3*x**3*sqrt(a + b*asinh(c + d*x))*asinh(c + d*x), x) + Integral(3*b**2*c*d**2*x**2*sq
rt(a + b*asinh(c + d*x))*asinh(c + d*x)**2, x) + Integral(3*b**2*c**2*d*x*sqrt(a + b*asinh(c + d*x))*asinh(c +
d*x)**2, x) + Integral(6*a*b*c*d**2*x**2*sqrt(a + b*asinh(c + d*x))*asinh(c + d*x), x) + Integral(6*a*b*c**2*
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)^3*(a+b*arcsinh(d*x+c))^(5/2),x, algorithm="giac")

[Out]

integrate((d*e*x + c*e)^3*(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 )}^3\,{\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)^3*(a + b*asinh(c + d*x))^(5/2),x)

[Out]

int((c*e + d*e*x)^3*(a + b*asinh(c + d*x))^(5/2), x)

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