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

Optimal. Leaf size=321 \[ -\frac {\sqrt {\pi } e^2 e^{a/b} \text {erf}\left (\frac {\sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{6 b^{5/2} d}+\frac {\sqrt {3 \pi } e^2 e^{\frac {3 a}{b}} \text {erf}\left (\frac {\sqrt {3} \sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{2 b^{5/2} d}-\frac {\sqrt {\pi } e^2 e^{-\frac {a}{b}} \text {erfi}\left (\frac {\sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{6 b^{5/2} d}+\frac {\sqrt {3 \pi } e^2 e^{-\frac {3 a}{b}} \text {erfi}\left (\frac {\sqrt {3} \sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{2 b^{5/2} d}-\frac {4 e^2 (c+d x)^3}{b^2 d \sqrt {a+b \sinh ^{-1}(c+d x)}}-\frac {8 e^2 (c+d x)}{3 b^2 d \sqrt {a+b \sinh ^{-1}(c+d x)}}-\frac {2 e^2 \sqrt {(c+d x)^2+1} (c+d x)^2}{3 b d \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}} \]

[Out]

-1/6*e^2*exp(a/b)*erf((a+b*arcsinh(d*x+c))^(1/2)/b^(1/2))*Pi^(1/2)/b^(5/2)/d-1/6*e^2*erfi((a+b*arcsinh(d*x+c))
^(1/2)/b^(1/2))*Pi^(1/2)/b^(5/2)/d/exp(a/b)+1/2*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)/b^(5/2)/d+1/2*e^2*erfi(3^(1/2)*(a+b*arcsinh(d*x+c))^(1/2)/b^(1/2))*3^(1/2)*Pi^(1/2)/b^(5/2)/
d/exp(3*a/b)-2/3*e^2*(d*x+c)^2*(1+(d*x+c)^2)^(1/2)/b/d/(a+b*arcsinh(d*x+c))^(3/2)-8/3*e^2*(d*x+c)/b^2/d/(a+b*a
rcsinh(d*x+c))^(1/2)-4*e^2*(d*x+c)^3/b^2/d/(a+b*arcsinh(d*x+c))^(1/2)

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Rubi [A]  time = 0.96, antiderivative size = 321, normalized size of antiderivative = 1.00, number of steps used = 24, number of rules used = 11, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.440, Rules used = {5865, 12, 5667, 5774, 5669, 5448, 3307, 2180, 2204, 2205, 5657} \[ -\frac {\sqrt {\pi } e^2 e^{a/b} \text {Erf}\left (\frac {\sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{6 b^{5/2} d}+\frac {\sqrt {3 \pi } e^2 e^{\frac {3 a}{b}} \text {Erf}\left (\frac {\sqrt {3} \sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{2 b^{5/2} d}-\frac {\sqrt {\pi } e^2 e^{-\frac {a}{b}} \text {Erfi}\left (\frac {\sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{6 b^{5/2} d}+\frac {\sqrt {3 \pi } e^2 e^{-\frac {3 a}{b}} \text {Erfi}\left (\frac {\sqrt {3} \sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{2 b^{5/2} d}-\frac {4 e^2 (c+d x)^3}{b^2 d \sqrt {a+b \sinh ^{-1}(c+d x)}}-\frac {8 e^2 (c+d x)}{3 b^2 d \sqrt {a+b \sinh ^{-1}(c+d x)}}-\frac {2 e^2 \sqrt {(c+d x)^2+1} (c+d x)^2}{3 b d \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*e^2*(c + d*x)^2*Sqrt[1 + (c + d*x)^2])/(3*b*d*(a + b*ArcSinh[c + d*x])^(3/2)) - (8*e^2*(c + d*x))/(3*b^2*d
*Sqrt[a + b*ArcSinh[c + d*x]]) - (4*e^2*(c + d*x)^3)/(b^2*d*Sqrt[a + b*ArcSinh[c + d*x]]) - (e^2*E^(a/b)*Sqrt[
Pi]*Erf[Sqrt[a + b*ArcSinh[c + d*x]]/Sqrt[b]])/(6*b^(5/2)*d) + (e^2*E^((3*a)/b)*Sqrt[3*Pi]*Erf[(Sqrt[3]*Sqrt[a
+ b*ArcSinh[c + d*x]])/Sqrt[b]])/(2*b^(5/2)*d) - (e^2*Sqrt[Pi]*Erfi[Sqrt[a + b*ArcSinh[c + d*x]]/Sqrt[b]])/(6
*b^(5/2)*d*E^(a/b)) + (e^2*Sqrt[3*Pi]*Erfi[(Sqrt[3]*Sqrt[a + b*ArcSinh[c + d*x]])/Sqrt[b]])/(2*b^(5/2)*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 3307

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 5448

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 5657

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 5667

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 5669

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_)*(x_)^(m_.), x_Symbol] :> Dist[1/c^(m + 1), Subst[Int[(a + b*x)^n*
Sinh[x]^m*Cosh[x], x], x, ArcSinh[c*x]], x] /; FreeQ[{a, b, c, n}, x] && IGtQ[m, 0]

Rule 5774

Int[(((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_)*((f_.)*(x_))^(m_.))/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp
[((f*x)^m*(a + b*ArcSinh[c*x])^(n + 1))/(b*c*Sqrt[d]*(n + 1)), x] - Dist[(f*m)/(b*c*Sqrt[d]*(n + 1)), 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] && 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 \frac {(c e+d e x)^2}{\left (a+b \sinh ^{-1}(c+d x)\right )^{5/2}} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {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 \frac {x^2}{\left (a+b \sinh ^{-1}(x)\right )^{5/2}} \, dx,x,c+d x\right )}{d}\\ &=-\frac {2 e^2 (c+d x)^2 \sqrt {1+(c+d x)^2}}{3 b d \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}+\frac {\left (4 e^2\right ) \operatorname {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}+\frac {\left (2 e^2\right ) \operatorname {Subst}\left (\int \frac {x^3}{\sqrt {1+x^2} \left (a+b \sinh ^{-1}(x)\right )^{3/2}} \, dx,x,c+d x\right )}{b d}\\ &=-\frac {2 e^2 (c+d x)^2 \sqrt {1+(c+d x)^2}}{3 b d \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}-\frac {8 e^2 (c+d x)}{3 b^2 d \sqrt {a+b \sinh ^{-1}(c+d x)}}-\frac {4 e^2 (c+d x)^3}{b^2 d \sqrt {a+b \sinh ^{-1}(c+d x)}}+\frac {\left (8 e^2\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+b \sinh ^{-1}(x)}} \, dx,x,c+d x\right )}{3 b^2 d}+\frac {\left (12 e^2\right ) \operatorname {Subst}\left (\int \frac {x^2}{\sqrt {a+b \sinh ^{-1}(x)}} \, dx,x,c+d x\right )}{b^2 d}\\ &=-\frac {2 e^2 (c+d x)^2 \sqrt {1+(c+d x)^2}}{3 b d \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}-\frac {8 e^2 (c+d x)}{3 b^2 d \sqrt {a+b \sinh ^{-1}(c+d x)}}-\frac {4 e^2 (c+d x)^3}{b^2 d \sqrt {a+b \sinh ^{-1}(c+d x)}}+\frac {\left (8 e^2\right ) \operatorname {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}+\frac {\left (12 e^2\right ) \operatorname {Subst}\left (\int \frac {\cosh (x) \sinh ^2(x)}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{b^2 d}\\ &=-\frac {2 e^2 (c+d x)^2 \sqrt {1+(c+d x)^2}}{3 b d \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}-\frac {8 e^2 (c+d x)}{3 b^2 d \sqrt {a+b \sinh ^{-1}(c+d x)}}-\frac {4 e^2 (c+d x)^3}{b^2 d \sqrt {a+b \sinh ^{-1}(c+d x)}}+\frac {\left (4 e^2\right ) \operatorname {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}+\frac {\left (4 e^2\right ) \operatorname {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}+\frac {\left (12 e^2\right ) \operatorname {Subst}\left (\int \left (-\frac {\cosh (x)}{4 \sqrt {a+b x}}+\frac {\cosh (3 x)}{4 \sqrt {a+b x}}\right ) \, dx,x,\sinh ^{-1}(c+d x)\right )}{b^2 d}\\ &=-\frac {2 e^2 (c+d x)^2 \sqrt {1+(c+d x)^2}}{3 b d \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}-\frac {8 e^2 (c+d x)}{3 b^2 d \sqrt {a+b \sinh ^{-1}(c+d x)}}-\frac {4 e^2 (c+d x)^3}{b^2 d \sqrt {a+b \sinh ^{-1}(c+d x)}}+\frac {\left (8 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 )}{3 b^3 d}+\frac {\left (8 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 )}{3 b^3 d}-\frac {\left (3 e^2\right ) \operatorname {Subst}\left (\int \frac {\cosh (x)}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{b^2 d}+\frac {\left (3 e^2\right ) \operatorname {Subst}\left (\int \frac {\cosh (3 x)}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{b^2 d}\\ &=-\frac {2 e^2 (c+d x)^2 \sqrt {1+(c+d x)^2}}{3 b d \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}-\frac {8 e^2 (c+d x)}{3 b^2 d \sqrt {a+b \sinh ^{-1}(c+d x)}}-\frac {4 e^2 (c+d x)^3}{b^2 d \sqrt {a+b \sinh ^{-1}(c+d x)}}+\frac {4 e^2 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}+\frac {4 e^2 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}+\frac {\left (3 e^2\right ) \operatorname {Subst}\left (\int \frac {e^{-3 x}}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{2 b^2 d}-\frac {\left (3 e^2\right ) \operatorname {Subst}\left (\int \frac {e^{-x}}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{2 b^2 d}-\frac {\left (3 e^2\right ) \operatorname {Subst}\left (\int \frac {e^x}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{2 b^2 d}+\frac {\left (3 e^2\right ) \operatorname {Subst}\left (\int \frac {e^{3 x}}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{2 b^2 d}\\ &=-\frac {2 e^2 (c+d x)^2 \sqrt {1+(c+d x)^2}}{3 b d \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}-\frac {8 e^2 (c+d x)}{3 b^2 d \sqrt {a+b \sinh ^{-1}(c+d x)}}-\frac {4 e^2 (c+d x)^3}{b^2 d \sqrt {a+b \sinh ^{-1}(c+d x)}}+\frac {4 e^2 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}+\frac {4 e^2 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}+\frac {\left (3 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 )}{b^3 d}-\frac {\left (3 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 )}{b^3 d}-\frac {\left (3 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 )}{b^3 d}+\frac {\left (3 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 )}{b^3 d}\\ &=-\frac {2 e^2 (c+d x)^2 \sqrt {1+(c+d x)^2}}{3 b d \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}-\frac {8 e^2 (c+d x)}{3 b^2 d \sqrt {a+b \sinh ^{-1}(c+d x)}}-\frac {4 e^2 (c+d x)^3}{b^2 d \sqrt {a+b \sinh ^{-1}(c+d x)}}-\frac {e^2 e^{a/b} \sqrt {\pi } \text {erf}\left (\frac {\sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{6 b^{5/2} d}+\frac {e^2 e^{\frac {3 a}{b}} \sqrt {3 \pi } \text {erf}\left (\frac {\sqrt {3} \sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{2 b^{5/2} d}-\frac {e^2 e^{-\frac {a}{b}} \sqrt {\pi } \text {erfi}\left (\frac {\sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{6 b^{5/2} d}+\frac {e^2 e^{-\frac {3 a}{b}} \sqrt {3 \pi } \text {erfi}\left (\frac {\sqrt {3} \sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{2 b^{5/2} d}\\ \end {align*}

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Mathematica [A]  time = 1.55, size = 389, normalized size = 1.21 \[ \frac {e^2 e^{-3 \left (\frac {a}{b}+\sinh ^{-1}(c+d x)\right )} \left (-6 \sqrt {3} b e^{3 \sinh ^{-1}(c+d x)} \left (-\frac {a+b \sinh ^{-1}(c+d x)}{b}\right )^{3/2} \Gamma \left (\frac {1}{2},-\frac {3 \left (a+b \sinh ^{-1}(c+d x)\right )}{b}\right )+2 b e^{\frac {2 a}{b}+3 \sinh ^{-1}(c+d x)} \left (-\frac {a+b \sinh ^{-1}(c+d x)}{b}\right )^{3/2} \Gamma \left (\frac {1}{2},-\frac {a+b \sinh ^{-1}(c+d x)}{b}\right )+2 e^{\frac {4 a}{b}+3 \sinh ^{-1}(c+d x)} \sqrt {\frac {a}{b}+\sinh ^{-1}(c+d x)} \left (a+b \sinh ^{-1}(c+d x)\right ) \Gamma \left (\frac {1}{2},\frac {a}{b}+\sinh ^{-1}(c+d x)\right )-e^{\frac {3 a}{b}} \left (\left (e^{2 \sinh ^{-1}(c+d x)}-1\right ) \left (a \left (4 e^{2 \sinh ^{-1}(c+d x)}+6 e^{4 \sinh ^{-1}(c+d x)}+6\right )+b \left (e^{4 \sinh ^{-1}(c+d x)}-1\right )+2 b \left (2 e^{2 \sinh ^{-1}(c+d x)}+3 e^{4 \sinh ^{-1}(c+d x)}+3\right ) \sinh ^{-1}(c+d x)\right )+6 \sqrt {3} e^{3 \left (\frac {a}{b}+\sinh ^{-1}(c+d x)\right )} \sqrt {\frac {a}{b}+\sinh ^{-1}(c+d x)} \left (a+b \sinh ^{-1}(c+d x)\right ) \Gamma \left (\frac {1}{2},\frac {3 \left (a+b \sinh ^{-1}(c+d x)\right )}{b}\right )\right )\right )}{12 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(e^2*(2*E^((4*a)/b + 3*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]] - 6*Sqrt[3]*b*E^(3*ArcSinh[c + d*x])*(-((a + b*ArcSinh[c + d*x])/b))^(3/2)*Gamma[1/2, (-3*
(a + b*ArcSinh[c + d*x]))/b] + 2*b*E^((2*a)/b + 3*ArcSinh[c + d*x])*(-((a + b*ArcSinh[c + d*x])/b))^(3/2)*Gamm
a[1/2, -((a + b*ArcSinh[c + d*x])/b)] - E^((3*a)/b)*((-1 + E^(2*ArcSinh[c + d*x]))*(b*(-1 + E^(4*ArcSinh[c + d
*x])) + a*(6 + 4*E^(2*ArcSinh[c + d*x]) + 6*E^(4*ArcSinh[c + d*x])) + 2*b*(3 + 2*E^(2*ArcSinh[c + d*x]) + 3*E^
(4*ArcSinh[c + d*x]))*ArcSinh[c + d*x]) + 6*Sqrt[3]*E^(3*(a/b + ArcSinh[c + d*x]))*Sqrt[a/b + ArcSinh[c + d*x]
]*(a + b*ArcSinh[c + d*x])*Gamma[1/2, (3*(a + b*ArcSinh[c + d*x]))/b])))/(12*b^2*d*E^(3*(a/b + ArcSinh[c + d*x
]))*(a + b*ArcSinh[c + d*x])^(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 \[ \int \frac {{\left (d e x + c e\right )}^{2}}{{\left (b \operatorname {arsinh}\left (d x + c\right ) + a\right )}^{\frac {5}{2}}}\,{d x} \]

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 \[ \int \frac {\left (d e x +c e \right )^{2}}{\left (a +b \arcsinh \left (d x +c \right )\right )^{\frac {5}{2}}}\, dx \]

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 \[ \int \frac {{\left (d e x + c e\right )}^{2}}{{\left (b \operatorname {arsinh}\left (d x + c\right ) + a\right )}^{\frac {5}{2}}}\,{d x} \]

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 \[ \int \frac {{\left (c\,e+d\,e\,x\right )}^2}{{\left (a+b\,\mathrm {asinh}\left (c+d\,x\right )\right )}^{5/2}} \,d x \]

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 \[ e^{2} \left (\int \frac {c^{2}}{a^{2} \sqrt {a + b \operatorname {asinh}{\left (c + d x \right )}} + 2 a b \sqrt {a + b \operatorname {asinh}{\left (c + d x \right )}} \operatorname {asinh}{\left (c + d x \right )} + b^{2} \sqrt {a + b \operatorname {asinh}{\left (c + d x \right )}} \operatorname {asinh}^{2}{\left (c + d x \right )}}\, dx + \int \frac {d^{2} x^{2}}{a^{2} \sqrt {a + b \operatorname {asinh}{\left (c + d x \right )}} + 2 a b \sqrt {a + b \operatorname {asinh}{\left (c + d x \right )}} \operatorname {asinh}{\left (c + d x \right )} + b^{2} \sqrt {a + b \operatorname {asinh}{\left (c + d x \right )}} \operatorname {asinh}^{2}{\left (c + d x \right )}}\, dx + \int \frac {2 c d x}{a^{2} \sqrt {a + b \operatorname {asinh}{\left (c + d x \right )}} + 2 a b \sqrt {a + b \operatorname {asinh}{\left (c + d x \right )}} \operatorname {asinh}{\left (c + d x \right )} + b^{2} \sqrt {a + b \operatorname {asinh}{\left (c + d x \right )}} \operatorname {asinh}^{2}{\left (c + d x \right )}}\, dx\right ) \]

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(c**2/(a**2*sqrt(a + b*asinh(c + d*x)) + 2*a*b*sqrt(a + b*asinh(c + d*x))*asinh(c + d*x) + b**2*
sqrt(a + b*asinh(c + d*x))*asinh(c + d*x)**2), x) + Integral(d**2*x**2/(a**2*sqrt(a + b*asinh(c + d*x)) + 2*a*
b*sqrt(a + b*asinh(c + d*x))*asinh(c + d*x) + b**2*sqrt(a + b*asinh(c + d*x))*asinh(c + d*x)**2), x) + Integra
l(2*c*d*x/(a**2*sqrt(a + b*asinh(c + d*x)) + 2*a*b*sqrt(a + b*asinh(c + d*x))*asinh(c + d*x) + b**2*sqrt(a + b
*asinh(c + d*x))*asinh(c + d*x)**2), x))

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