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

Optimal. Leaf size=346 \[ -\frac{5 \sqrt{\pi } d^2 e^{a/b} \text{Erf}\left (\frac{\sqrt{a+b \sinh ^{-1}(c x)}}{\sqrt{b}}\right )}{8 b^{3/2} c}-\frac{5 \sqrt{3 \pi } d^2 e^{\frac{3 a}{b}} \text{Erf}\left (\frac{\sqrt{3} \sqrt{a+b \sinh ^{-1}(c x)}}{\sqrt{b}}\right )}{16 b^{3/2} c}-\frac{\sqrt{5 \pi } d^2 e^{\frac{5 a}{b}} \text{Erf}\left (\frac{\sqrt{5} \sqrt{a+b \sinh ^{-1}(c x)}}{\sqrt{b}}\right )}{16 b^{3/2} c}+\frac{5 \sqrt{\pi } d^2 e^{-\frac{a}{b}} \text{Erfi}\left (\frac{\sqrt{a+b \sinh ^{-1}(c x)}}{\sqrt{b}}\right )}{8 b^{3/2} c}+\frac{5 \sqrt{3 \pi } d^2 e^{-\frac{3 a}{b}} \text{Erfi}\left (\frac{\sqrt{3} \sqrt{a+b \sinh ^{-1}(c x)}}{\sqrt{b}}\right )}{16 b^{3/2} c}+\frac{\sqrt{5 \pi } d^2 e^{-\frac{5 a}{b}} \text{Erfi}\left (\frac{\sqrt{5} \sqrt{a+b \sinh ^{-1}(c x)}}{\sqrt{b}}\right )}{16 b^{3/2} c}-\frac{2 d^2 \left (c^2 x^2+1\right )^{5/2}}{b c \sqrt{a+b \sinh ^{-1}(c x)}} \]

[Out]

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

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Rubi [A]  time = 0.728844, antiderivative size = 346, normalized size of antiderivative = 1., number of steps used = 19, number of rules used = 7, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.28, Rules used = {5696, 5779, 5448, 3308, 2180, 2204, 2205} \[ -\frac{5 \sqrt{\pi } d^2 e^{a/b} \text{Erf}\left (\frac{\sqrt{a+b \sinh ^{-1}(c x)}}{\sqrt{b}}\right )}{8 b^{3/2} c}-\frac{5 \sqrt{3 \pi } d^2 e^{\frac{3 a}{b}} \text{Erf}\left (\frac{\sqrt{3} \sqrt{a+b \sinh ^{-1}(c x)}}{\sqrt{b}}\right )}{16 b^{3/2} c}-\frac{\sqrt{5 \pi } d^2 e^{\frac{5 a}{b}} \text{Erf}\left (\frac{\sqrt{5} \sqrt{a+b \sinh ^{-1}(c x)}}{\sqrt{b}}\right )}{16 b^{3/2} c}+\frac{5 \sqrt{\pi } d^2 e^{-\frac{a}{b}} \text{Erfi}\left (\frac{\sqrt{a+b \sinh ^{-1}(c x)}}{\sqrt{b}}\right )}{8 b^{3/2} c}+\frac{5 \sqrt{3 \pi } d^2 e^{-\frac{3 a}{b}} \text{Erfi}\left (\frac{\sqrt{3} \sqrt{a+b \sinh ^{-1}(c x)}}{\sqrt{b}}\right )}{16 b^{3/2} c}+\frac{\sqrt{5 \pi } d^2 e^{-\frac{5 a}{b}} \text{Erfi}\left (\frac{\sqrt{5} \sqrt{a+b \sinh ^{-1}(c x)}}{\sqrt{b}}\right )}{16 b^{3/2} c}-\frac{2 d^2 \left (c^2 x^2+1\right )^{5/2}}{b c \sqrt{a+b \sinh ^{-1}(c x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 5696

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(Sqrt[1 + c^2*x^2]
*(d + e*x^2)^p*(a + b*ArcSinh[c*x])^(n + 1))/(b*c*(n + 1)), x] - Dist[(c*(2*p + 1)*d^IntPart[p]*(d + e*x^2)^Fr
acPart[p])/(b*(n + 1)*(1 + c^2*x^2)^FracPart[p]), Int[x*(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] && LtQ[n, -1]

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])

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 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 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]

Rubi steps

\begin{align*} \int \frac{\left (d+c^2 d x^2\right )^2}{\left (a+b \sinh ^{-1}(c x)\right )^{3/2}} \, dx &=-\frac{2 d^2 \left (1+c^2 x^2\right )^{5/2}}{b c \sqrt{a+b \sinh ^{-1}(c x)}}+\frac{\left (10 c d^2\right ) \int \frac{x \left (1+c^2 x^2\right )^{3/2}}{\sqrt{a+b \sinh ^{-1}(c x)}} \, dx}{b}\\ &=-\frac{2 d^2 \left (1+c^2 x^2\right )^{5/2}}{b c \sqrt{a+b \sinh ^{-1}(c x)}}+\frac{\left (10 d^2\right ) \operatorname{Subst}\left (\int \frac{\cosh ^4(x) \sinh (x)}{\sqrt{a+b x}} \, dx,x,\sinh ^{-1}(c x)\right )}{b c}\\ &=-\frac{2 d^2 \left (1+c^2 x^2\right )^{5/2}}{b c \sqrt{a+b \sinh ^{-1}(c x)}}+\frac{\left (10 d^2\right ) \operatorname{Subst}\left (\int \left (\frac{\sinh (x)}{8 \sqrt{a+b x}}+\frac{3 \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}\\ &=-\frac{2 d^2 \left (1+c^2 x^2\right )^{5/2}}{b c \sqrt{a+b \sinh ^{-1}(c x)}}+\frac{\left (5 d^2\right ) \operatorname{Subst}\left (\int \frac{\sinh (5 x)}{\sqrt{a+b x}} \, dx,x,\sinh ^{-1}(c x)\right )}{8 b c}+\frac{\left (5 d^2\right ) \operatorname{Subst}\left (\int \frac{\sinh (x)}{\sqrt{a+b x}} \, dx,x,\sinh ^{-1}(c x)\right )}{4 b c}+\frac{\left (15 d^2\right ) \operatorname{Subst}\left (\int \frac{\sinh (3 x)}{\sqrt{a+b x}} \, dx,x,\sinh ^{-1}(c x)\right )}{8 b c}\\ &=-\frac{2 d^2 \left (1+c^2 x^2\right )^{5/2}}{b c \sqrt{a+b \sinh ^{-1}(c x)}}-\frac{\left (5 d^2\right ) \operatorname{Subst}\left (\int \frac{e^{-5 x}}{\sqrt{a+b x}} \, dx,x,\sinh ^{-1}(c x)\right )}{16 b c}+\frac{\left (5 d^2\right ) \operatorname{Subst}\left (\int \frac{e^{5 x}}{\sqrt{a+b x}} \, dx,x,\sinh ^{-1}(c x)\right )}{16 b c}-\frac{\left (5 d^2\right ) \operatorname{Subst}\left (\int \frac{e^{-x}}{\sqrt{a+b x}} \, dx,x,\sinh ^{-1}(c x)\right )}{8 b c}+\frac{\left (5 d^2\right ) \operatorname{Subst}\left (\int \frac{e^x}{\sqrt{a+b x}} \, dx,x,\sinh ^{-1}(c x)\right )}{8 b c}-\frac{\left (15 d^2\right ) \operatorname{Subst}\left (\int \frac{e^{-3 x}}{\sqrt{a+b x}} \, dx,x,\sinh ^{-1}(c x)\right )}{16 b c}+\frac{\left (15 d^2\right ) \operatorname{Subst}\left (\int \frac{e^{3 x}}{\sqrt{a+b x}} \, dx,x,\sinh ^{-1}(c x)\right )}{16 b c}\\ &=-\frac{2 d^2 \left (1+c^2 x^2\right )^{5/2}}{b c \sqrt{a+b \sinh ^{-1}(c x)}}-\frac{\left (5 d^2\right ) \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}+\frac{\left (5 d^2\right ) \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}-\frac{\left (5 d^2\right ) \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}+\frac{\left (5 d^2\right ) \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}-\frac{\left (15 d^2\right ) \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}+\frac{\left (15 d^2\right ) \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}\\ &=-\frac{2 d^2 \left (1+c^2 x^2\right )^{5/2}}{b c \sqrt{a+b \sinh ^{-1}(c x)}}-\frac{5 d^2 e^{a/b} \sqrt{\pi } \text{erf}\left (\frac{\sqrt{a+b \sinh ^{-1}(c x)}}{\sqrt{b}}\right )}{8 b^{3/2} c}-\frac{5 d^2 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}-\frac{d^2 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}+\frac{5 d^2 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}+\frac{5 d^2 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}+\frac{d^2 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}\\ \end{align*}

Mathematica [A]  time = 2.02187, size = 440, normalized size = 1.27 \[ \frac{d^2 e^{-5 \left (\frac{a}{b}+\sinh ^{-1}(c x)\right )} \left (10 e^{\frac{6 a}{b}+5 \sinh ^{-1}(c x)} \sqrt{\frac{a}{b}+\sinh ^{-1}(c x)} \text{Gamma}\left (\frac{1}{2},\frac{a}{b}+\sinh ^{-1}(c x)\right )+\sqrt{5} e^{5 \sinh ^{-1}(c x)} \sqrt{-\frac{a+b \sinh ^{-1}(c x)}{b}} \text{Gamma}\left (\frac{1}{2},-\frac{5 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )+5 \sqrt{3} e^{\frac{2 a}{b}+5 \sinh ^{-1}(c x)} \sqrt{-\frac{a+b \sinh ^{-1}(c x)}{b}} \text{Gamma}\left (\frac{1}{2},-\frac{3 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )+10 e^{\frac{4 a}{b}+5 \sinh ^{-1}(c x)} \sqrt{-\frac{a+b \sinh ^{-1}(c x)}{b}} \text{Gamma}\left (\frac{1}{2},-\frac{a+b \sinh ^{-1}(c x)}{b}\right )+5 \sqrt{3} e^{\frac{8 a}{b}+5 \sinh ^{-1}(c x)} \sqrt{\frac{a}{b}+\sinh ^{-1}(c x)} \text{Gamma}\left (\frac{1}{2},\frac{3 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )+\sqrt{5} e^{5 \left (\frac{2 a}{b}+\sinh ^{-1}(c x)\right )} \sqrt{\frac{a}{b}+\sinh ^{-1}(c x)} \text{Gamma}\left (\frac{1}{2},\frac{5 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )-5 e^{\frac{5 a}{b}+2 \sinh ^{-1}(c x)}-10 e^{\frac{5 a}{b}+4 \sinh ^{-1}(c x)}-10 e^{\frac{5 a}{b}+6 \sinh ^{-1}(c x)}-5 e^{\frac{5 a}{b}+8 \sinh ^{-1}(c x)}-e^{\frac{5 a}{b}+10 \sinh ^{-1}(c x)}-e^{\frac{5 a}{b}}\right )}{16 b c \sqrt{a+b \sinh ^{-1}(c x)}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(d^2*(-E^((5*a)/b) - 5*E^((5*a)/b + 2*ArcSinh[c*x]) - 10*E^((5*a)/b + 4*ArcSinh[c*x]) - 10*E^((5*a)/b + 6*ArcS
inh[c*x]) - 5*E^((5*a)/b + 8*ArcSinh[c*x]) - E^((5*a)/b + 10*ArcSinh[c*x]) + 10*E^((6*a)/b + 5*ArcSinh[c*x])*S
qrt[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)]*Gamma[1/2, (-5*(a + b*ArcSinh[c*x]))/b] + 5*Sqrt[3]*E^((2*a)/b + 5*ArcSinh[c*x])*Sqrt[-((a + b*ArcSinh[c
*x])/b)]*Gamma[1/2, (-3*(a + b*ArcSinh[c*x]))/b] + 10*E^((4*a)/b + 5*ArcSinh[c*x])*Sqrt[-((a + b*ArcSinh[c*x])
/b)]*Gamma[1/2, -((a + b*ArcSinh[c*x])/b)] + 5*Sqrt[3]*E^((8*a)/b + 5*ArcSinh[c*x])*Sqrt[a/b + ArcSinh[c*x]]*G
amma[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*E^(5*(a/b + ArcSinh[c*x]))*Sqrt[a + b*ArcSinh[c*x]])

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Maple [F]  time = 0.204, size = 0, normalized size = 0. \begin{align*} \int{ \left ({c}^{2}d{x}^{2}+d \right ) ^{2} \left ( a+b{\it Arcsinh} \left ( cx \right ) \right ) ^{-{\frac{3}{2}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (c^{2} d x^{2} + d\right )}^{2}}{{\left (b \operatorname{arsinh}\left (c x\right ) + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: UnboundLocalError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: UnboundLocalError

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} d^{2} \left (\int \frac{2 c^{2} 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^{4} 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 + \int \frac{1}{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 ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

d**2*(Integral(2*c**2*x**2/(a*sqrt(a + b*asinh(c*x)) + b*sqrt(a + b*asinh(c*x))*asinh(c*x)), x) + Integral(c**
4*x**4/(a*sqrt(a + b*asinh(c*x)) + b*sqrt(a + b*asinh(c*x))*asinh(c*x)), x) + Integral(1/(a*sqrt(a + b*asinh(c
*x)) + b*sqrt(a + b*asinh(c*x))*asinh(c*x)), x))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (c^{2} d x^{2} + d\right )}^{2}}{{\left (b \operatorname{arsinh}\left (c x\right ) + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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