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

Optimal. Leaf size=335 \[ \frac {\sqrt {\pi } d e^{a/b} \text {erf}\left (\frac {\sqrt {a+b \sinh ^{-1}(c x)}}{\sqrt {b}}\right )}{8 b^{3/2} c^3}-\frac {\sqrt {3 \pi } d e^{\frac {3 a}{b}} \text {erf}\left (\frac {\sqrt {3} \sqrt {a+b \sinh ^{-1}(c x)}}{\sqrt {b}}\right )}{16 b^{3/2} c^3}-\frac {\sqrt {5 \pi } d e^{\frac {5 a}{b}} \text {erf}\left (\frac {\sqrt {5} \sqrt {a+b \sinh ^{-1}(c x)}}{\sqrt {b}}\right )}{16 b^{3/2} c^3}-\frac {\sqrt {\pi } d e^{-\frac {a}{b}} \text {erfi}\left (\frac {\sqrt {a+b \sinh ^{-1}(c x)}}{\sqrt {b}}\right )}{8 b^{3/2} c^3}+\frac {\sqrt {3 \pi } d e^{-\frac {3 a}{b}} \text {erfi}\left (\frac {\sqrt {3} \sqrt {a+b \sinh ^{-1}(c x)}}{\sqrt {b}}\right )}{16 b^{3/2} c^3}+\frac {\sqrt {5 \pi } d e^{-\frac {5 a}{b}} \text {erfi}\left (\frac {\sqrt {5} \sqrt {a+b \sinh ^{-1}(c x)}}{\sqrt {b}}\right )}{16 b^{3/2} c^3}-\frac {2 d x^2 \left (c^2 x^2+1\right )^{3/2}}{b c \sqrt {a+b \sinh ^{-1}(c x)}} \]

[Out]

1/8*d*exp(a/b)*erf((a+b*arcsinh(c*x))^(1/2)/b^(1/2))*Pi^(1/2)/b^(3/2)/c^3-1/8*d*erfi((a+b*arcsinh(c*x))^(1/2)/
b^(1/2))*Pi^(1/2)/b^(3/2)/c^3/exp(a/b)-1/16*d*exp(3*a/b)*erf(3^(1/2)*(a+b*arcsinh(c*x))^(1/2)/b^(1/2))*3^(1/2)
*Pi^(1/2)/b^(3/2)/c^3+1/16*d*erfi(3^(1/2)*(a+b*arcsinh(c*x))^(1/2)/b^(1/2))*3^(1/2)*Pi^(1/2)/b^(3/2)/c^3/exp(3
*a/b)-1/16*d*exp(5*a/b)*erf(5^(1/2)*(a+b*arcsinh(c*x))^(1/2)/b^(1/2))*5^(1/2)*Pi^(1/2)/b^(3/2)/c^3+1/16*d*erfi
(5^(1/2)*(a+b*arcsinh(c*x))^(1/2)/b^(1/2))*5^(1/2)*Pi^(1/2)/b^(3/2)/c^3/exp(5*a/b)-2*d*x^2*(c^2*x^2+1)^(3/2)/b
/c/(a+b*arcsinh(c*x))^(1/2)

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Rubi [A]  time = 1.38, antiderivative size = 335, normalized size of antiderivative = 1.00, number of steps used = 32, number of rules used = 7, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.269, Rules used = {5777, 5779, 5448, 3308, 2180, 2204, 2205} \[ \frac {\sqrt {\pi } d e^{a/b} \text {Erf}\left (\frac {\sqrt {a+b \sinh ^{-1}(c x)}}{\sqrt {b}}\right )}{8 b^{3/2} c^3}-\frac {\sqrt {3 \pi } d e^{\frac {3 a}{b}} \text {Erf}\left (\frac {\sqrt {3} \sqrt {a+b \sinh ^{-1}(c x)}}{\sqrt {b}}\right )}{16 b^{3/2} c^3}-\frac {\sqrt {5 \pi } d e^{\frac {5 a}{b}} \text {Erf}\left (\frac {\sqrt {5} \sqrt {a+b \sinh ^{-1}(c x)}}{\sqrt {b}}\right )}{16 b^{3/2} c^3}-\frac {\sqrt {\pi } d e^{-\frac {a}{b}} \text {Erfi}\left (\frac {\sqrt {a+b \sinh ^{-1}(c x)}}{\sqrt {b}}\right )}{8 b^{3/2} c^3}+\frac {\sqrt {3 \pi } d e^{-\frac {3 a}{b}} \text {Erfi}\left (\frac {\sqrt {3} \sqrt {a+b \sinh ^{-1}(c x)}}{\sqrt {b}}\right )}{16 b^{3/2} c^3}+\frac {\sqrt {5 \pi } d e^{-\frac {5 a}{b}} \text {Erfi}\left (\frac {\sqrt {5} \sqrt {a+b \sinh ^{-1}(c x)}}{\sqrt {b}}\right )}{16 b^{3/2} c^3}-\frac {2 d x^2 \left (c^2 x^2+1\right )^{3/2}}{b c \sqrt {a+b \sinh ^{-1}(c x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*d*x^2*(1 + c^2*x^2)^(3/2))/(b*c*Sqrt[a + b*ArcSinh[c*x]]) + (d*E^(a/b)*Sqrt[Pi]*Erf[Sqrt[a + b*ArcSinh[c*x
]]/Sqrt[b]])/(8*b^(3/2)*c^3) - (d*E^((3*a)/b)*Sqrt[3*Pi]*Erf[(Sqrt[3]*Sqrt[a + b*ArcSinh[c*x]])/Sqrt[b]])/(16*
b^(3/2)*c^3) - (d*E^((5*a)/b)*Sqrt[5*Pi]*Erf[(Sqrt[5]*Sqrt[a + b*ArcSinh[c*x]])/Sqrt[b]])/(16*b^(3/2)*c^3) - (
d*Sqrt[Pi]*Erfi[Sqrt[a + b*ArcSinh[c*x]]/Sqrt[b]])/(8*b^(3/2)*c^3*E^(a/b)) + (d*Sqrt[3*Pi]*Erfi[(Sqrt[3]*Sqrt[
a + b*ArcSinh[c*x]])/Sqrt[b]])/(16*b^(3/2)*c^3*E^((3*a)/b)) + (d*Sqrt[5*Pi]*Erfi[(Sqrt[5]*Sqrt[a + b*ArcSinh[c
*x]])/Sqrt[b]])/(16*b^(3/2)*c^3*E^((5*a)/b))

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 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 5777

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_)*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Simp
[((f*x)^m*Sqrt[1 + c^2*x^2]*(d + e*x^2)^p*(a + b*ArcSinh[c*x])^(n + 1))/(b*c*(n + 1)), x] + (-Dist[(f*m*d^IntP
art[p]*(d + e*x^2)^FracPart[p])/(b*c*(n + 1)*(1 + c^2*x^2)^FracPart[p]), Int[(f*x)^(m - 1)*(1 + c^2*x^2)^(p -
1/2)*(a + b*ArcSinh[c*x])^(n + 1), x], x] - Dist[(c*(m + 2*p + 1)*d^IntPart[p]*(d + e*x^2)^FracPart[p])/(b*f*(
n + 1)*(1 + c^2*x^2)^FracPart[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}, x] && EqQ[e, c^2*d] && LtQ[n, -1] && IGtQ[m, -3] && IGtQ[2*p, 0]

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])

Rubi steps

\begin {align*} \int \frac {x^2 \left (d+c^2 d x^2\right )}{\left (a+b \sinh ^{-1}(c x)\right )^{3/2}} \, dx &=-\frac {2 d x^2 \left (1+c^2 x^2\right )^{3/2}}{b c \sqrt {a+b \sinh ^{-1}(c x)}}+\frac {(4 d) \int \frac {x \sqrt {1+c^2 x^2}}{\sqrt {a+b \sinh ^{-1}(c x)}} \, dx}{b c}+\frac {(10 c d) \int \frac {x^3 \sqrt {1+c^2 x^2}}{\sqrt {a+b \sinh ^{-1}(c x)}} \, dx}{b}\\ &=-\frac {2 d x^2 \left (1+c^2 x^2\right )^{3/2}}{b c \sqrt {a+b \sinh ^{-1}(c x)}}+\frac {(4 d) \operatorname {Subst}\left (\int \frac {\cosh ^2(x) \sinh (x)}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c x)\right )}{b c^3}+\frac {(10 d) \operatorname {Subst}\left (\int \frac {\cosh ^2(x) \sinh ^3(x)}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c x)\right )}{b c^3}\\ &=-\frac {2 d x^2 \left (1+c^2 x^2\right )^{3/2}}{b c \sqrt {a+b \sinh ^{-1}(c x)}}+\frac {(4 d) \operatorname {Subst}\left (\int \left (\frac {\sinh (x)}{4 \sqrt {a+b x}}+\frac {\sinh (3 x)}{4 \sqrt {a+b x}}\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{b c^3}+\frac {(10 d) \operatorname {Subst}\left (\int \left (-\frac {\sinh (x)}{8 \sqrt {a+b x}}-\frac {\sinh (3 x)}{16 \sqrt {a+b x}}+\frac {\sinh (5 x)}{16 \sqrt {a+b x}}\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{b c^3}\\ &=-\frac {2 d x^2 \left (1+c^2 x^2\right )^{3/2}}{b c \sqrt {a+b \sinh ^{-1}(c x)}}-\frac {(5 d) \operatorname {Subst}\left (\int \frac {\sinh (3 x)}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c x)\right )}{8 b c^3}+\frac {(5 d) \operatorname {Subst}\left (\int \frac {\sinh (5 x)}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c x)\right )}{8 b c^3}+\frac {d \operatorname {Subst}\left (\int \frac {\sinh (x)}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c x)\right )}{b c^3}+\frac {d \operatorname {Subst}\left (\int \frac {\sinh (3 x)}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c x)\right )}{b c^3}-\frac {(5 d) \operatorname {Subst}\left (\int \frac {\sinh (x)}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c x)\right )}{4 b c^3}\\ &=-\frac {2 d x^2 \left (1+c^2 x^2\right )^{3/2}}{b c \sqrt {a+b \sinh ^{-1}(c x)}}-\frac {(5 d) \operatorname {Subst}\left (\int \frac {e^{-5 x}}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c x)\right )}{16 b c^3}+\frac {(5 d) \operatorname {Subst}\left (\int \frac {e^{-3 x}}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c x)\right )}{16 b c^3}-\frac {(5 d) \operatorname {Subst}\left (\int \frac {e^{3 x}}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c x)\right )}{16 b c^3}+\frac {(5 d) \operatorname {Subst}\left (\int \frac {e^{5 x}}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c x)\right )}{16 b c^3}-\frac {d \operatorname {Subst}\left (\int \frac {e^{-3 x}}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c x)\right )}{2 b c^3}-\frac {d \operatorname {Subst}\left (\int \frac {e^{-x}}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c x)\right )}{2 b c^3}+\frac {d \operatorname {Subst}\left (\int \frac {e^x}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c x)\right )}{2 b c^3}+\frac {d \operatorname {Subst}\left (\int \frac {e^{3 x}}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c x)\right )}{2 b c^3}+\frac {(5 d) \operatorname {Subst}\left (\int \frac {e^{-x}}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c x)\right )}{8 b c^3}-\frac {(5 d) \operatorname {Subst}\left (\int \frac {e^x}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c x)\right )}{8 b c^3}\\ &=-\frac {2 d x^2 \left (1+c^2 x^2\right )^{3/2}}{b c \sqrt {a+b \sinh ^{-1}(c x)}}-\frac {(5 d) \operatorname {Subst}\left (\int e^{\frac {5 a}{b}-\frac {5 x^2}{b}} \, dx,x,\sqrt {a+b \sinh ^{-1}(c x)}\right )}{8 b^2 c^3}+\frac {(5 d) \operatorname {Subst}\left (\int e^{\frac {3 a}{b}-\frac {3 x^2}{b}} \, dx,x,\sqrt {a+b \sinh ^{-1}(c x)}\right )}{8 b^2 c^3}-\frac {(5 d) \operatorname {Subst}\left (\int e^{-\frac {3 a}{b}+\frac {3 x^2}{b}} \, dx,x,\sqrt {a+b \sinh ^{-1}(c x)}\right )}{8 b^2 c^3}+\frac {(5 d) \operatorname {Subst}\left (\int e^{-\frac {5 a}{b}+\frac {5 x^2}{b}} \, dx,x,\sqrt {a+b \sinh ^{-1}(c x)}\right )}{8 b^2 c^3}-\frac {d \operatorname {Subst}\left (\int e^{\frac {3 a}{b}-\frac {3 x^2}{b}} \, dx,x,\sqrt {a+b \sinh ^{-1}(c x)}\right )}{b^2 c^3}-\frac {d \operatorname {Subst}\left (\int e^{\frac {a}{b}-\frac {x^2}{b}} \, dx,x,\sqrt {a+b \sinh ^{-1}(c x)}\right )}{b^2 c^3}+\frac {d \operatorname {Subst}\left (\int e^{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+b \sinh ^{-1}(c x)}\right )}{b^2 c^3}+\frac {d \operatorname {Subst}\left (\int e^{-\frac {3 a}{b}+\frac {3 x^2}{b}} \, dx,x,\sqrt {a+b \sinh ^{-1}(c x)}\right )}{b^2 c^3}+\frac {(5 d) \operatorname {Subst}\left (\int e^{\frac {a}{b}-\frac {x^2}{b}} \, dx,x,\sqrt {a+b \sinh ^{-1}(c x)}\right )}{4 b^2 c^3}-\frac {(5 d) \operatorname {Subst}\left (\int e^{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+b \sinh ^{-1}(c x)}\right )}{4 b^2 c^3}\\ &=-\frac {2 d x^2 \left (1+c^2 x^2\right )^{3/2}}{b c \sqrt {a+b \sinh ^{-1}(c x)}}+\frac {d e^{a/b} \sqrt {\pi } \text {erf}\left (\frac {\sqrt {a+b \sinh ^{-1}(c x)}}{\sqrt {b}}\right )}{8 b^{3/2} c^3}-\frac {d e^{\frac {3 a}{b}} \sqrt {3 \pi } \text {erf}\left (\frac {\sqrt {3} \sqrt {a+b \sinh ^{-1}(c x)}}{\sqrt {b}}\right )}{16 b^{3/2} c^3}-\frac {d e^{\frac {5 a}{b}} \sqrt {5 \pi } \text {erf}\left (\frac {\sqrt {5} \sqrt {a+b \sinh ^{-1}(c x)}}{\sqrt {b}}\right )}{16 b^{3/2} c^3}-\frac {d e^{-\frac {a}{b}} \sqrt {\pi } \text {erfi}\left (\frac {\sqrt {a+b \sinh ^{-1}(c x)}}{\sqrt {b}}\right )}{8 b^{3/2} c^3}+\frac {d e^{-\frac {3 a}{b}} \sqrt {3 \pi } \text {erfi}\left (\frac {\sqrt {3} \sqrt {a+b \sinh ^{-1}(c x)}}{\sqrt {b}}\right )}{16 b^{3/2} c^3}+\frac {d e^{-\frac {5 a}{b}} \sqrt {5 \pi } \text {erfi}\left (\frac {\sqrt {5} \sqrt {a+b \sinh ^{-1}(c x)}}{\sqrt {b}}\right )}{16 b^{3/2} c^3}\\ \end {align*}

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Mathematica [A]  time = 0.75, size = 436, normalized size = 1.30 \[ \frac {d e^{-5 \left (\frac {a}{b}+\sinh ^{-1}(c x)\right )} \left (-e^{\frac {5 a}{b}+2 \sinh ^{-1}(c x)}+2 e^{\frac {5 a}{b}+4 \sinh ^{-1}(c x)}+2 e^{\frac {5 a}{b}+6 \sinh ^{-1}(c x)}-e^{\frac {5 a}{b}+8 \sinh ^{-1}(c x)}-e^{\frac {5 a}{b}+10 \sinh ^{-1}(c x)}-2 e^{\frac {6 a}{b}+5 \sinh ^{-1}(c x)} \sqrt {\frac {a}{b}+\sinh ^{-1}(c x)} \Gamma \left (\frac {1}{2},\frac {a}{b}+\sinh ^{-1}(c x)\right )+\sqrt {5} e^{5 \sinh ^{-1}(c x)} \sqrt {-\frac {a+b \sinh ^{-1}(c x)}{b}} \Gamma \left (\frac {1}{2},-\frac {5 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )+\sqrt {3} e^{\frac {2 a}{b}+5 \sinh ^{-1}(c x)} \sqrt {-\frac {a+b \sinh ^{-1}(c x)}{b}} \Gamma \left (\frac {1}{2},-\frac {3 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )-2 e^{\frac {4 a}{b}+5 \sinh ^{-1}(c x)} \sqrt {-\frac {a+b \sinh ^{-1}(c x)}{b}} \Gamma \left (\frac {1}{2},-\frac {a+b \sinh ^{-1}(c x)}{b}\right )+\sqrt {3} e^{\frac {8 a}{b}+5 \sinh ^{-1}(c x)} \sqrt {\frac {a}{b}+\sinh ^{-1}(c x)} \Gamma \left (\frac {1}{2},\frac {3 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )+\sqrt {5} e^{5 \left (\frac {2 a}{b}+\sinh ^{-1}(c x)\right )} \sqrt {\frac {a}{b}+\sinh ^{-1}(c x)} \Gamma \left (\frac {1}{2},\frac {5 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )-e^{\frac {5 a}{b}}\right )}{16 b c^3 \sqrt {a+b \sinh ^{-1}(c x)}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(d*(-E^((5*a)/b) - E^((5*a)/b + 2*ArcSinh[c*x]) + 2*E^((5*a)/b + 4*ArcSinh[c*x]) + 2*E^((5*a)/b + 6*ArcSinh[c*
x]) - E^((5*a)/b + 8*ArcSinh[c*x]) - E^((5*a)/b + 10*ArcSinh[c*x]) - 2*E^((6*a)/b + 5*ArcSinh[c*x])*Sqrt[a/b +
ArcSinh[c*x]]*Gamma[1/2, a/b + ArcSinh[c*x]] + Sqrt[5]*E^(5*ArcSinh[c*x])*Sqrt[-((a + b*ArcSinh[c*x])/b)]*Gam
ma[1/2, (-5*(a + b*ArcSinh[c*x]))/b] + Sqrt[3]*E^((2*a)/b + 5*ArcSinh[c*x])*Sqrt[-((a + b*ArcSinh[c*x])/b)]*Ga
mma[1/2, (-3*(a + b*ArcSinh[c*x]))/b] - 2*E^((4*a)/b + 5*ArcSinh[c*x])*Sqrt[-((a + b*ArcSinh[c*x])/b)]*Gamma[1
/2, -((a + b*ArcSinh[c*x])/b)] + Sqrt[3]*E^((8*a)/b + 5*ArcSinh[c*x])*Sqrt[a/b + ArcSinh[c*x]]*Gamma[1/2, (3*(
a + b*ArcSinh[c*x]))/b] + Sqrt[5]*E^(5*((2*a)/b + ArcSinh[c*x]))*Sqrt[a/b + ArcSinh[c*x]]*Gamma[1/2, (5*(a + b
*ArcSinh[c*x]))/b]))/(16*b*c^3*E^(5*(a/b + ArcSinh[c*x]))*Sqrt[a + b*ArcSinh[c*x]])

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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(x^2*(c^2*d*x^2+d)/(a+b*arcsinh(c*x))^(3/2),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 \[ \int \frac {{\left (c^{2} d x^{2} + d\right )} x^{2}}{{\left (b \operatorname {arsinh}\left (c x\right ) + a\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

d*(Integral(x**2/(a*sqrt(a + b*asinh(c*x)) + b*sqrt(a + b*asinh(c*x))*asinh(c*x)), x) + Integral(c**2*x**4/(a*
sqrt(a + b*asinh(c*x)) + b*sqrt(a + b*asinh(c*x))*asinh(c*x)), x))

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