3.188 \(\int (c e+d e x)^2 (a+b \sinh ^{-1}(c+d x))^{3/2} \, dx\)

Optimal. Leaf size=328 \[ -\frac {3 \sqrt {\pi } b^{3/2} e^2 e^{a/b} \text {erf}\left (\frac {\sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{32 d}+\frac {\sqrt {\frac {\pi }{3}} b^{3/2} e^2 e^{\frac {3 a}{b}} \text {erf}\left (\frac {\sqrt {3} \sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{96 d}-\frac {3 \sqrt {\pi } b^{3/2} e^2 e^{-\frac {a}{b}} \text {erfi}\left (\frac {\sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{32 d}+\frac {\sqrt {\frac {\pi }{3}} b^{3/2} e^2 e^{-\frac {3 a}{b}} \text {erfi}\left (\frac {\sqrt {3} \sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{96 d}+\frac {e^2 (c+d x)^3 \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}{3 d}-\frac {b e^2 \sqrt {(c+d x)^2+1} (c+d x)^2 \sqrt {a+b \sinh ^{-1}(c+d x)}}{6 d}+\frac {b e^2 \sqrt {(c+d x)^2+1} \sqrt {a+b \sinh ^{-1}(c+d x)}}{3 d} \]

[Out]

1/3*e^2*(d*x+c)^3*(a+b*arcsinh(d*x+c))^(3/2)/d+1/288*b^(3/2)*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+1/288*b^(3/2)*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)-3/32*b^(3/2)*e^2*exp(a/b)*erf((a+b*arcsinh(d*x+c))^(1/2)/b^(1/2))*Pi^(1/2)/d-3/32*b^(3/2)*
e^2*erfi((a+b*arcsinh(d*x+c))^(1/2)/b^(1/2))*Pi^(1/2)/d/exp(a/b)+1/3*b*e^2*(1+(d*x+c)^2)^(1/2)*(a+b*arcsinh(d*
x+c))^(1/2)/d-1/6*b*e^2*(d*x+c)^2*(1+(d*x+c)^2)^(1/2)*(a+b*arcsinh(d*x+c))^(1/2)/d

________________________________________________________________________________________

Rubi [A]  time = 0.88, antiderivative size = 328, normalized size of antiderivative = 1.00, number of steps used = 24, 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, 5657, 3307, 2180, 2205, 2204, 5669, 5448} \[ -\frac {3 \sqrt {\pi } b^{3/2} e^2 e^{a/b} \text {Erf}\left (\frac {\sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{32 d}+\frac {\sqrt {\frac {\pi }{3}} b^{3/2} e^2 e^{\frac {3 a}{b}} \text {Erf}\left (\frac {\sqrt {3} \sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{96 d}-\frac {3 \sqrt {\pi } b^{3/2} e^2 e^{-\frac {a}{b}} \text {Erfi}\left (\frac {\sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{32 d}+\frac {\sqrt {\frac {\pi }{3}} b^{3/2} e^2 e^{-\frac {3 a}{b}} \text {Erfi}\left (\frac {\sqrt {3} \sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{96 d}+\frac {e^2 (c+d x)^3 \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}{3 d}-\frac {b e^2 \sqrt {(c+d x)^2+1} (c+d x)^2 \sqrt {a+b \sinh ^{-1}(c+d x)}}{6 d}+\frac {b e^2 \sqrt {(c+d x)^2+1} \sqrt {a+b \sinh ^{-1}(c+d x)}}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*e^2*Sqrt[1 + (c + d*x)^2]*Sqrt[a + b*ArcSinh[c + d*x]])/(3*d) - (b*e^2*(c + d*x)^2*Sqrt[1 + (c + d*x)^2]*Sq
rt[a + b*ArcSinh[c + d*x]])/(6*d) + (e^2*(c + d*x)^3*(a + b*ArcSinh[c + d*x])^(3/2))/(3*d) - (3*b^(3/2)*e^2*E^
(a/b)*Sqrt[Pi]*Erf[Sqrt[a + b*ArcSinh[c + d*x]]/Sqrt[b]])/(32*d) + (b^(3/2)*e^2*E^((3*a)/b)*Sqrt[Pi/3]*Erf[(Sq
rt[3]*Sqrt[a + b*ArcSinh[c + d*x]])/Sqrt[b]])/(96*d) - (3*b^(3/2)*e^2*Sqrt[Pi]*Erfi[Sqrt[a + b*ArcSinh[c + d*x
]]/Sqrt[b]])/(32*d*E^(a/b)) + (b^(3/2)*e^2*Sqrt[Pi/3]*Erfi[(Sqrt[3]*Sqrt[a + b*ArcSinh[c + d*x]])/Sqrt[b]])/(9
6*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 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 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 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 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 )^{3/2} \, dx &=\frac {\operatorname {Subst}\left (\int e^2 x^2 \left (a+b \sinh ^{-1}(x)\right )^{3/2} \, dx,x,c+d x\right )}{d}\\ &=\frac {e^2 \operatorname {Subst}\left (\int x^2 \left (a+b \sinh ^{-1}(x)\right )^{3/2} \, dx,x,c+d x\right )}{d}\\ &=\frac {e^2 (c+d x)^3 \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}{3 d}-\frac {\left (b e^2\right ) \operatorname {Subst}\left (\int \frac {x^3 \sqrt {a+b \sinh ^{-1}(x)}}{\sqrt {1+x^2}} \, dx,x,c+d x\right )}{2 d}\\ &=-\frac {b e^2 (c+d x)^2 \sqrt {1+(c+d x)^2} \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 )^{3/2}}{3 d}+\frac {\left (b e^2\right ) \operatorname {Subst}\left (\int \frac {x \sqrt {a+b \sinh ^{-1}(x)}}{\sqrt {1+x^2}} \, dx,x,c+d x\right )}{3 d}+\frac {\left (b^2 e^2\right ) \operatorname {Subst}\left (\int \frac {x^2}{\sqrt {a+b \sinh ^{-1}(x)}} \, dx,x,c+d x\right )}{12 d}\\ &=\frac {b e^2 \sqrt {1+(c+d x)^2} \sqrt {a+b \sinh ^{-1}(c+d x)}}{3 d}-\frac {b e^2 (c+d x)^2 \sqrt {1+(c+d x)^2} \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 )^{3/2}}{3 d}+\frac {\left (b^2 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 )}{12 d}-\frac {\left (b^2 e^2\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+b \sinh ^{-1}(x)}} \, dx,x,c+d x\right )}{6 d}\\ &=\frac {b e^2 \sqrt {1+(c+d x)^2} \sqrt {a+b \sinh ^{-1}(c+d x)}}{3 d}-\frac {b e^2 (c+d x)^2 \sqrt {1+(c+d x)^2} \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 )^{3/2}}{3 d}-\frac {\left (b 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 )}{6 d}+\frac {\left (b^2 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 )}{12 d}\\ &=\frac {b e^2 \sqrt {1+(c+d x)^2} \sqrt {a+b \sinh ^{-1}(c+d x)}}{3 d}-\frac {b e^2 (c+d x)^2 \sqrt {1+(c+d x)^2} \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 )^{3/2}}{3 d}-\frac {\left (b 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 )}{12 d}-\frac {\left (b 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 )}{12 d}-\frac {\left (b^2 e^2\right ) \operatorname {Subst}\left (\int \frac {\cosh (x)}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{48 d}+\frac {\left (b^2 e^2\right ) \operatorname {Subst}\left (\int \frac {\cosh (3 x)}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{48 d}\\ &=\frac {b e^2 \sqrt {1+(c+d x)^2} \sqrt {a+b \sinh ^{-1}(c+d x)}}{3 d}-\frac {b e^2 (c+d x)^2 \sqrt {1+(c+d x)^2} \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 )^{3/2}}{3 d}-\frac {\left (b 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 )}{6 d}-\frac {\left (b 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 )}{6 d}+\frac {\left (b^2 e^2\right ) \operatorname {Subst}\left (\int \frac {e^{-3 x}}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{96 d}-\frac {\left (b^2 e^2\right ) \operatorname {Subst}\left (\int \frac {e^{-x}}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{96 d}-\frac {\left (b^2 e^2\right ) \operatorname {Subst}\left (\int \frac {e^x}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{96 d}+\frac {\left (b^2 e^2\right ) \operatorname {Subst}\left (\int \frac {e^{3 x}}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{96 d}\\ &=\frac {b e^2 \sqrt {1+(c+d x)^2} \sqrt {a+b \sinh ^{-1}(c+d x)}}{3 d}-\frac {b e^2 (c+d x)^2 \sqrt {1+(c+d x)^2} \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 )^{3/2}}{3 d}-\frac {b^{3/2} e^2 e^{a/b} \sqrt {\pi } \text {erf}\left (\frac {\sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{12 d}-\frac {b^{3/2} e^2 e^{-\frac {a}{b}} \sqrt {\pi } \text {erfi}\left (\frac {\sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{12 d}+\frac {\left (b 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 )}{48 d}-\frac {\left (b 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 )}{48 d}-\frac {\left (b 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 )}{48 d}+\frac {\left (b 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 )}{48 d}\\ &=\frac {b e^2 \sqrt {1+(c+d x)^2} \sqrt {a+b \sinh ^{-1}(c+d x)}}{3 d}-\frac {b e^2 (c+d x)^2 \sqrt {1+(c+d x)^2} \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 )^{3/2}}{3 d}-\frac {3 b^{3/2} e^2 e^{a/b} \sqrt {\pi } \text {erf}\left (\frac {\sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{32 d}+\frac {b^{3/2} 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 )}{96 d}-\frac {3 b^{3/2} e^2 e^{-\frac {a}{b}} \sqrt {\pi } \text {erfi}\left (\frac {\sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{32 d}+\frac {b^{3/2} 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 )}{96 d}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.30, size = 238, normalized size = 0.73 \[ -\frac {b e^2 e^{-\frac {3 a}{b}} \sqrt {a+b \sinh ^{-1}(c+d x)} \left (-27 e^{\frac {4 a}{b}} \sqrt {-\frac {a+b \sinh ^{-1}(c+d x)}{b}} \Gamma \left (\frac {5}{2},\frac {a}{b}+\sinh ^{-1}(c+d x)\right )+\sqrt {3} \sqrt {\frac {a}{b}+\sinh ^{-1}(c+d x)} \Gamma \left (\frac {5}{2},-\frac {3 \left (a+b \sinh ^{-1}(c+d x)\right )}{b}\right )-27 e^{\frac {2 a}{b}} \sqrt {\frac {a}{b}+\sinh ^{-1}(c+d x)} \Gamma \left (\frac {5}{2},-\frac {a+b \sinh ^{-1}(c+d x)}{b}\right )+\sqrt {3} e^{\frac {6 a}{b}} \sqrt {-\frac {a+b \sinh ^{-1}(c+d x)}{b}} \Gamma \left (\frac {5}{2},\frac {3 \left (a+b \sinh ^{-1}(c+d x)\right )}{b}\right )\right )}{216 d \sqrt {-\frac {\left (a+b \sinh ^{-1}(c+d x)\right )^2}{b^2}}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-1/216*(b*e^2*Sqrt[a + b*ArcSinh[c + d*x]]*(-27*E^((4*a)/b)*Sqrt[-((a + b*ArcSinh[c + d*x])/b)]*Gamma[5/2, a/b
 + ArcSinh[c + d*x]] + Sqrt[3]*Sqrt[a/b + ArcSinh[c + d*x]]*Gamma[5/2, (-3*(a + b*ArcSinh[c + d*x]))/b] - 27*E
^((2*a)/b)*Sqrt[a/b + ArcSinh[c + d*x]]*Gamma[5/2, -((a + b*ArcSinh[c + d*x])/b)] + Sqrt[3]*E^((6*a)/b)*Sqrt[-
((a + b*ArcSinh[c + d*x])/b)]*Gamma[5/2, (3*(a + b*ArcSinh[c + d*x]))/b]))/(d*E^((3*a)/b)*Sqrt[-((a + b*ArcSin
h[c + d*x])^2/b^2)])

________________________________________________________________________________________

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))^(3/2),x, algorithm="fricas")

[Out]

Exception raised: TypeError >>  Error detected within library code:   integrate: implementation incomplete (co
nstant residues)

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (d e x + c e\right )}^{2} {\left (b \operatorname {arsinh}\left (d x + c\right ) + a\right )}^{\frac {3}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*e*x+c*e)^2*(a+b*arcsinh(d*x+c))^(3/2),x, algorithm="giac")

[Out]

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

________________________________________________________________________________________

maple [F(-2)]  time = 180.00, size = 0, normalized size = 0.00 \[ \int \left (d e x +c e \right )^{2} \left (a +b \arcsinh \left (d x +c \right )\right )^{\frac {3}{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (d e x + c e\right )}^{2} {\left (b \operatorname {arsinh}\left (d x + c\right ) + a\right )}^{\frac {3}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int {\left (c\,e+d\,e\,x\right )}^2\,{\left (a+b\,\mathrm {asinh}\left (c+d\,x\right )\right )}^{3/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*e*x+c*e)**2*(a+b*asinh(d*x+c))**(3/2),x)

[Out]

e**2*(Integral(a*c**2*sqrt(a + b*asinh(c + d*x)), x) + Integral(a*d**2*x**2*sqrt(a + b*asinh(c + d*x)), x) + I
ntegral(b*c**2*sqrt(a + b*asinh(c + d*x))*asinh(c + d*x), x) + Integral(2*a*c*d*x*sqrt(a + b*asinh(c + d*x)),
x) + Integral(b*d**2*x**2*sqrt(a + b*asinh(c + d*x))*asinh(c + d*x), x) + Integral(2*b*c*d*x*sqrt(a + b*asinh(
c + d*x))*asinh(c + d*x), x))

________________________________________________________________________________________