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

Optimal. Leaf size=305 \[ \frac{b^2 c \sqrt{c^2 x^2+1} \text{PolyLog}\left (2,-e^{2 \sinh ^{-1}(c x)}\right )}{d \sqrt{c^2 d x^2+d}}+\frac{b^2 c \sqrt{c^2 x^2+1} \text{PolyLog}\left (2,e^{2 \sinh ^{-1}(c x)}\right )}{d \sqrt{c^2 d x^2+d}}-\frac{2 c^2 x \left (a+b \sinh ^{-1}(c x)\right )^2}{d \sqrt{c^2 d x^2+d}}-\frac{2 c \sqrt{c^2 x^2+1} \left (a+b \sinh ^{-1}(c x)\right )^2}{d \sqrt{c^2 d x^2+d}}-\frac{\left (a+b \sinh ^{-1}(c x)\right )^2}{d x \sqrt{c^2 d x^2+d}}+\frac{4 b c \sqrt{c^2 x^2+1} \log \left (e^{2 \sinh ^{-1}(c x)}+1\right ) \left (a+b \sinh ^{-1}(c x)\right )}{d \sqrt{c^2 d x^2+d}}-\frac{4 b c \sqrt{c^2 x^2+1} \tanh ^{-1}\left (e^{2 \sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{d \sqrt{c^2 d x^2+d}} \]

[Out]

-((a + b*ArcSinh[c*x])^2/(d*x*Sqrt[d + c^2*d*x^2])) - (2*c^2*x*(a + b*ArcSinh[c*x])^2)/(d*Sqrt[d + c^2*d*x^2])
- (2*c*Sqrt[1 + c^2*x^2]*(a + b*ArcSinh[c*x])^2)/(d*Sqrt[d + c^2*d*x^2]) - (4*b*c*Sqrt[1 + c^2*x^2]*(a + b*Ar
cSinh[c*x])*ArcTanh[E^(2*ArcSinh[c*x])])/(d*Sqrt[d + c^2*d*x^2]) + (4*b*c*Sqrt[1 + c^2*x^2]*(a + b*ArcSinh[c*x
])*Log[1 + E^(2*ArcSinh[c*x])])/(d*Sqrt[d + c^2*d*x^2]) + (b^2*c*Sqrt[1 + c^2*x^2]*PolyLog[2, -E^(2*ArcSinh[c*
x])])/(d*Sqrt[d + c^2*d*x^2]) + (b^2*c*Sqrt[1 + c^2*x^2]*PolyLog[2, E^(2*ArcSinh[c*x])])/(d*Sqrt[d + c^2*d*x^2
])

________________________________________________________________________________________

Rubi [A]  time = 0.4642, antiderivative size = 305, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 10, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.357, Rules used = {5747, 5687, 5714, 3718, 2190, 2279, 2391, 5720, 5461, 4182} \[ \frac{b^2 c \sqrt{c^2 x^2+1} \text{PolyLog}\left (2,-e^{2 \sinh ^{-1}(c x)}\right )}{d \sqrt{c^2 d x^2+d}}+\frac{b^2 c \sqrt{c^2 x^2+1} \text{PolyLog}\left (2,e^{2 \sinh ^{-1}(c x)}\right )}{d \sqrt{c^2 d x^2+d}}-\frac{2 c^2 x \left (a+b \sinh ^{-1}(c x)\right )^2}{d \sqrt{c^2 d x^2+d}}-\frac{2 c \sqrt{c^2 x^2+1} \left (a+b \sinh ^{-1}(c x)\right )^2}{d \sqrt{c^2 d x^2+d}}-\frac{\left (a+b \sinh ^{-1}(c x)\right )^2}{d x \sqrt{c^2 d x^2+d}}+\frac{4 b c \sqrt{c^2 x^2+1} \log \left (e^{2 \sinh ^{-1}(c x)}+1\right ) \left (a+b \sinh ^{-1}(c x)\right )}{d \sqrt{c^2 d x^2+d}}-\frac{4 b c \sqrt{c^2 x^2+1} \tanh ^{-1}\left (e^{2 \sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{d \sqrt{c^2 d x^2+d}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a + b*ArcSinh[c*x])^2/(d*x*Sqrt[d + c^2*d*x^2])) - (2*c^2*x*(a + b*ArcSinh[c*x])^2)/(d*Sqrt[d + c^2*d*x^2])
- (2*c*Sqrt[1 + c^2*x^2]*(a + b*ArcSinh[c*x])^2)/(d*Sqrt[d + c^2*d*x^2]) - (4*b*c*Sqrt[1 + c^2*x^2]*(a + b*Ar
cSinh[c*x])*ArcTanh[E^(2*ArcSinh[c*x])])/(d*Sqrt[d + c^2*d*x^2]) + (4*b*c*Sqrt[1 + c^2*x^2]*(a + b*ArcSinh[c*x
])*Log[1 + E^(2*ArcSinh[c*x])])/(d*Sqrt[d + c^2*d*x^2]) + (b^2*c*Sqrt[1 + c^2*x^2]*PolyLog[2, -E^(2*ArcSinh[c*
x])])/(d*Sqrt[d + c^2*d*x^2]) + (b^2*c*Sqrt[1 + c^2*x^2]*PolyLog[2, E^(2*ArcSinh[c*x])])/(d*Sqrt[d + c^2*d*x^2
])

Rule 5747

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

Rule 5687

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)/((d_) + (e_.)*(x_)^2)^(3/2), x_Symbol] :> Simp[(x*(a + b*ArcSinh
[c*x])^n)/(d*Sqrt[d + e*x^2]), x] - Dist[(b*c*n*Sqrt[1 + c^2*x^2])/(d*Sqrt[d + e*x^2]), Int[(x*(a + b*ArcSinh[
c*x])^(n - 1))/(1 + c^2*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*d] && GtQ[n, 0]

Rule 5714

Int[(((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*(x_))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Dist[1/e, Subst[Int[(
a + b*x)^n*Tanh[x], x], x, ArcSinh[c*x]], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*d] && IGtQ[n, 0]

Rule 3718

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + (Complex[0, fz_])*(f_.)*(x_)], x_Symbol] :> -Simp[(I*(c + d*x)^(m +
1))/(d*(m + 1)), x] + Dist[2*I, Int[((c + d*x)^m*E^(2*(-(I*e) + f*fz*x)))/(1 + E^(2*(-(I*e) + f*fz*x))), x],
x] /; FreeQ[{c, d, e, f, fz}, x] && IGtQ[m, 0]

Rule 2190

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/((a_) + (b_.)*((F_)^((g_.)*((e_.) +
(f_.)*(x_))))^(n_.)), x_Symbol] :> Simp[((c + d*x)^m*Log[1 + (b*(F^(g*(e + f*x)))^n)/a])/(b*f*g*n*Log[F]), x]
- Dist[(d*m)/(b*f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*Log[1 + (b*(F^(g*(e + f*x)))^n)/a], x], x] /; FreeQ[{F,
a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]

Rule 2279

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int
[Log[a + b*x]/x, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rule 5720

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)/((x_)*((d_) + (e_.)*(x_)^2)), x_Symbol] :> Dist[1/d, Subst[Int[(
a + b*x)^n/(Cosh[x]*Sinh[x]), x], x, ArcSinh[c*x]], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*d] && IGtQ[n
, 0]

Rule 5461

Int[Csch[(a_.) + (b_.)*(x_)]^(n_.)*((c_.) + (d_.)*(x_))^(m_.)*Sech[(a_.) + (b_.)*(x_)]^(n_.), x_Symbol] :> Dis
t[2^n, Int[(c + d*x)^m*Csch[2*a + 2*b*x]^n, x], x] /; FreeQ[{a, b, c, d}, x] && RationalQ[m] && IntegerQ[n]

Rule 4182

Int[csc[(e_.) + (Complex[0, fz_])*(f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(-2*(c + d*x)^m*Ar
cTanh[E^(-(I*e) + f*fz*x)])/(f*fz*I), x] + (-Dist[(d*m)/(f*fz*I), Int[(c + d*x)^(m - 1)*Log[1 - E^(-(I*e) + f*
fz*x)], x], x] + Dist[(d*m)/(f*fz*I), Int[(c + d*x)^(m - 1)*Log[1 + E^(-(I*e) + f*fz*x)], x], x]) /; FreeQ[{c,
d, e, f, fz}, x] && IGtQ[m, 0]

Rubi steps

\begin{align*} \int \frac{\left (a+b \sinh ^{-1}(c x)\right )^2}{x^2 \left (d+c^2 d x^2\right )^{3/2}} \, dx &=-\frac{\left (a+b \sinh ^{-1}(c x)\right )^2}{d x \sqrt{d+c^2 d x^2}}-\left (2 c^2\right ) \int \frac{\left (a+b \sinh ^{-1}(c x)\right )^2}{\left (d+c^2 d x^2\right )^{3/2}} \, dx+\frac{\left (2 b c \sqrt{1+c^2 x^2}\right ) \int \frac{a+b \sinh ^{-1}(c x)}{x \left (1+c^2 x^2\right )} \, dx}{d \sqrt{d+c^2 d x^2}}\\ &=-\frac{\left (a+b \sinh ^{-1}(c x)\right )^2}{d x \sqrt{d+c^2 d x^2}}-\frac{2 c^2 x \left (a+b \sinh ^{-1}(c x)\right )^2}{d \sqrt{d+c^2 d x^2}}+\frac{\left (2 b c \sqrt{1+c^2 x^2}\right ) \operatorname{Subst}\left (\int (a+b x) \text{csch}(x) \text{sech}(x) \, dx,x,\sinh ^{-1}(c x)\right )}{d \sqrt{d+c^2 d x^2}}+\frac{\left (4 b c^3 \sqrt{1+c^2 x^2}\right ) \int \frac{x \left (a+b \sinh ^{-1}(c x)\right )}{1+c^2 x^2} \, dx}{d \sqrt{d+c^2 d x^2}}\\ &=-\frac{\left (a+b \sinh ^{-1}(c x)\right )^2}{d x \sqrt{d+c^2 d x^2}}-\frac{2 c^2 x \left (a+b \sinh ^{-1}(c x)\right )^2}{d \sqrt{d+c^2 d x^2}}+\frac{\left (4 b c \sqrt{1+c^2 x^2}\right ) \operatorname{Subst}\left (\int (a+b x) \text{csch}(2 x) \, dx,x,\sinh ^{-1}(c x)\right )}{d \sqrt{d+c^2 d x^2}}+\frac{\left (4 b c \sqrt{1+c^2 x^2}\right ) \operatorname{Subst}\left (\int (a+b x) \tanh (x) \, dx,x,\sinh ^{-1}(c x)\right )}{d \sqrt{d+c^2 d x^2}}\\ &=-\frac{\left (a+b \sinh ^{-1}(c x)\right )^2}{d x \sqrt{d+c^2 d x^2}}-\frac{2 c^2 x \left (a+b \sinh ^{-1}(c x)\right )^2}{d \sqrt{d+c^2 d x^2}}-\frac{2 c \sqrt{1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )^2}{d \sqrt{d+c^2 d x^2}}-\frac{4 b c \sqrt{1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{2 \sinh ^{-1}(c x)}\right )}{d \sqrt{d+c^2 d x^2}}+\frac{\left (8 b c \sqrt{1+c^2 x^2}\right ) \operatorname{Subst}\left (\int \frac{e^{2 x} (a+b x)}{1+e^{2 x}} \, dx,x,\sinh ^{-1}(c x)\right )}{d \sqrt{d+c^2 d x^2}}-\frac{\left (2 b^2 c \sqrt{1+c^2 x^2}\right ) \operatorname{Subst}\left (\int \log \left (1-e^{2 x}\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{d \sqrt{d+c^2 d x^2}}+\frac{\left (2 b^2 c \sqrt{1+c^2 x^2}\right ) \operatorname{Subst}\left (\int \log \left (1+e^{2 x}\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{d \sqrt{d+c^2 d x^2}}\\ &=-\frac{\left (a+b \sinh ^{-1}(c x)\right )^2}{d x \sqrt{d+c^2 d x^2}}-\frac{2 c^2 x \left (a+b \sinh ^{-1}(c x)\right )^2}{d \sqrt{d+c^2 d x^2}}-\frac{2 c \sqrt{1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )^2}{d \sqrt{d+c^2 d x^2}}-\frac{4 b c \sqrt{1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{2 \sinh ^{-1}(c x)}\right )}{d \sqrt{d+c^2 d x^2}}+\frac{4 b c \sqrt{1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right ) \log \left (1+e^{2 \sinh ^{-1}(c x)}\right )}{d \sqrt{d+c^2 d x^2}}-\frac{\left (b^2 c \sqrt{1+c^2 x^2}\right ) \operatorname{Subst}\left (\int \frac{\log (1-x)}{x} \, dx,x,e^{2 \sinh ^{-1}(c x)}\right )}{d \sqrt{d+c^2 d x^2}}+\frac{\left (b^2 c \sqrt{1+c^2 x^2}\right ) \operatorname{Subst}\left (\int \frac{\log (1+x)}{x} \, dx,x,e^{2 \sinh ^{-1}(c x)}\right )}{d \sqrt{d+c^2 d x^2}}-\frac{\left (4 b^2 c \sqrt{1+c^2 x^2}\right ) \operatorname{Subst}\left (\int \log \left (1+e^{2 x}\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{d \sqrt{d+c^2 d x^2}}\\ &=-\frac{\left (a+b \sinh ^{-1}(c x)\right )^2}{d x \sqrt{d+c^2 d x^2}}-\frac{2 c^2 x \left (a+b \sinh ^{-1}(c x)\right )^2}{d \sqrt{d+c^2 d x^2}}-\frac{2 c \sqrt{1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )^2}{d \sqrt{d+c^2 d x^2}}-\frac{4 b c \sqrt{1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{2 \sinh ^{-1}(c x)}\right )}{d \sqrt{d+c^2 d x^2}}+\frac{4 b c \sqrt{1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right ) \log \left (1+e^{2 \sinh ^{-1}(c x)}\right )}{d \sqrt{d+c^2 d x^2}}-\frac{b^2 c \sqrt{1+c^2 x^2} \text{Li}_2\left (-e^{2 \sinh ^{-1}(c x)}\right )}{d \sqrt{d+c^2 d x^2}}+\frac{b^2 c \sqrt{1+c^2 x^2} \text{Li}_2\left (e^{2 \sinh ^{-1}(c x)}\right )}{d \sqrt{d+c^2 d x^2}}-\frac{\left (2 b^2 c \sqrt{1+c^2 x^2}\right ) \operatorname{Subst}\left (\int \frac{\log (1+x)}{x} \, dx,x,e^{2 \sinh ^{-1}(c x)}\right )}{d \sqrt{d+c^2 d x^2}}\\ &=-\frac{\left (a+b \sinh ^{-1}(c x)\right )^2}{d x \sqrt{d+c^2 d x^2}}-\frac{2 c^2 x \left (a+b \sinh ^{-1}(c x)\right )^2}{d \sqrt{d+c^2 d x^2}}-\frac{2 c \sqrt{1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )^2}{d \sqrt{d+c^2 d x^2}}-\frac{4 b c \sqrt{1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{2 \sinh ^{-1}(c x)}\right )}{d \sqrt{d+c^2 d x^2}}+\frac{4 b c \sqrt{1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right ) \log \left (1+e^{2 \sinh ^{-1}(c x)}\right )}{d \sqrt{d+c^2 d x^2}}+\frac{b^2 c \sqrt{1+c^2 x^2} \text{Li}_2\left (-e^{2 \sinh ^{-1}(c x)}\right )}{d \sqrt{d+c^2 d x^2}}+\frac{b^2 c \sqrt{1+c^2 x^2} \text{Li}_2\left (e^{2 \sinh ^{-1}(c x)}\right )}{d \sqrt{d+c^2 d x^2}}\\ \end{align*}

Mathematica [A]  time = 0.897617, size = 296, normalized size = 0.97 \[ -\frac{b^2 c x \sqrt{c^2 x^2+1} \text{PolyLog}\left (2,-e^{-2 \sinh ^{-1}(c x)}\right )+b^2 c x \sqrt{c^2 x^2+1} \text{PolyLog}\left (2,e^{-2 \sinh ^{-1}(c x)}\right )+2 a^2 c^2 x^2+a^2-2 a b c x \sqrt{c^2 x^2+1} \log (c x)-a b c x \sqrt{c^2 x^2+1} \log \left (c^2 x^2+1\right )+4 a b c^2 x^2 \sinh ^{-1}(c x)+2 a b \sinh ^{-1}(c x)+2 b^2 c^2 x^2 \sinh ^{-1}(c x)^2-2 b^2 c x \sqrt{c^2 x^2+1} \sinh ^{-1}(c x)^2-2 b^2 c x \sqrt{c^2 x^2+1} \sinh ^{-1}(c x) \log \left (1-e^{-2 \sinh ^{-1}(c x)}\right )-2 b^2 c x \sqrt{c^2 x^2+1} \sinh ^{-1}(c x) \log \left (e^{-2 \sinh ^{-1}(c x)}+1\right )+b^2 \sinh ^{-1}(c x)^2}{d x \sqrt{c^2 d x^2+d}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-((a^2 + 2*a^2*c^2*x^2 + 2*a*b*ArcSinh[c*x] + 4*a*b*c^2*x^2*ArcSinh[c*x] + b^2*ArcSinh[c*x]^2 + 2*b^2*c^2*x^2*
ArcSinh[c*x]^2 - 2*b^2*c*x*Sqrt[1 + c^2*x^2]*ArcSinh[c*x]^2 - 2*b^2*c*x*Sqrt[1 + c^2*x^2]*ArcSinh[c*x]*Log[1 -
E^(-2*ArcSinh[c*x])] - 2*b^2*c*x*Sqrt[1 + c^2*x^2]*ArcSinh[c*x]*Log[1 + E^(-2*ArcSinh[c*x])] - 2*a*b*c*x*Sqrt
[1 + c^2*x^2]*Log[c*x] - a*b*c*x*Sqrt[1 + c^2*x^2]*Log[1 + c^2*x^2] + b^2*c*x*Sqrt[1 + c^2*x^2]*PolyLog[2, -E^
(-2*ArcSinh[c*x])] + b^2*c*x*Sqrt[1 + c^2*x^2]*PolyLog[2, E^(-2*ArcSinh[c*x])])/(d*x*Sqrt[d + c^2*d*x^2]))

________________________________________________________________________________________

Maple [B]  time = 0.227, size = 660, normalized size = 2.2 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a^2/d/x/(c^2*d*x^2+d)^(1/2)-2*a^2*c^2/d*x/(c^2*d*x^2+d)^(1/2)-2*b^2*(d*(c^2*x^2+1))^(1/2)*arcsinh(c*x)^2/(c^2
*x^2+1)/d^2*x*c^2-2*b^2*(d*(c^2*x^2+1))^(1/2)*arcsinh(c*x)^2/(c^2*x^2+1)^(1/2)/d^2*c-b^2*(d*(c^2*x^2+1))^(1/2)
*arcsinh(c*x)^2/(c^2*x^2+1)/d^2/x+2*b^2*(d*(c^2*x^2+1))^(1/2)/(c^2*x^2+1)^(1/2)/d^2*arcsinh(c*x)*ln(1+c*x+(c^2
*x^2+1)^(1/2))*c+2*b^2*(d*(c^2*x^2+1))^(1/2)/(c^2*x^2+1)^(1/2)/d^2*polylog(2,-c*x-(c^2*x^2+1)^(1/2))*c+2*b^2*(
d*(c^2*x^2+1))^(1/2)/(c^2*x^2+1)^(1/2)/d^2*arcsinh(c*x)*ln(1+(c*x+(c^2*x^2+1)^(1/2))^2)*c+b^2*(d*(c^2*x^2+1))^
(1/2)/(c^2*x^2+1)^(1/2)/d^2*polylog(2,-(c*x+(c^2*x^2+1)^(1/2))^2)*c+2*b^2*(d*(c^2*x^2+1))^(1/2)/(c^2*x^2+1)^(1
/2)/d^2*arcsinh(c*x)*ln(1-c*x-(c^2*x^2+1)^(1/2))*c+2*b^2*(d*(c^2*x^2+1))^(1/2)/(c^2*x^2+1)^(1/2)/d^2*polylog(2
,c*x+(c^2*x^2+1)^(1/2))*c-4*a*b*(d*(c^2*x^2+1))^(1/2)/(c^2*x^2+1)^(1/2)/d^2*arcsinh(c*x)*c-4*a*b*(d*(c^2*x^2+1
))^(1/2)*arcsinh(c*x)/(c^2*x^2+1)/d^2*x*c^2-2*a*b*(d*(c^2*x^2+1))^(1/2)*arcsinh(c*x)/(c^2*x^2+1)/d^2/x+2*a*b*(
d*(c^2*x^2+1))^(1/2)/(c^2*x^2+1)^(1/2)/d^2*ln((c*x+(c^2*x^2+1)^(1/2))^4-1)*c

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{c^{2} d x^{2} + d}{\left (b^{2} \operatorname{arsinh}\left (c x\right )^{2} + 2 \, a b \operatorname{arsinh}\left (c x\right ) + a^{2}\right )}}{c^{4} d^{2} x^{6} + 2 \, c^{2} d^{2} x^{4} + d^{2} x^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(c^2*d*x^2 + d)*(b^2*arcsinh(c*x)^2 + 2*a*b*arcsinh(c*x) + a^2)/(c^4*d^2*x^6 + 2*c^2*d^2*x^4 + d^
2*x^2), x)

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (a + b \operatorname{asinh}{\left (c x \right )}\right )^{2}}{x^{2} \left (d \left (c^{2} x^{2} + 1\right )\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a + b*asinh(c*x))**2/(x**2*(d*(c**2*x**2 + 1))**(3/2)), x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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