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

Optimal. Leaf size=199 \[ -\frac{b \text{PolyLog}\left (2,-e^{2 \sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{c^4 d}+\frac{b^2 \text{PolyLog}\left (3,-e^{2 \sinh ^{-1}(c x)}\right )}{2 c^4 d}+\frac{x^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{2 c^2 d}-\frac{b x \sqrt{c^2 x^2+1} \left (a+b \sinh ^{-1}(c x)\right )}{2 c^3 d}+\frac{\left (a+b \sinh ^{-1}(c x)\right )^3}{3 b c^4 d}+\frac{\left (a+b \sinh ^{-1}(c x)\right )^2}{4 c^4 d}-\frac{\log \left (e^{2 \sinh ^{-1}(c x)}+1\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{c^4 d}+\frac{b^2 x^2}{4 c^2 d} \]

[Out]

(b^2*x^2)/(4*c^2*d) - (b*x*Sqrt[1 + c^2*x^2]*(a + b*ArcSinh[c*x]))/(2*c^3*d) + (a + b*ArcSinh[c*x])^2/(4*c^4*d
) + (x^2*(a + b*ArcSinh[c*x])^2)/(2*c^2*d) + (a + b*ArcSinh[c*x])^3/(3*b*c^4*d) - ((a + b*ArcSinh[c*x])^2*Log[
1 + E^(2*ArcSinh[c*x])])/(c^4*d) - (b*(a + b*ArcSinh[c*x])*PolyLog[2, -E^(2*ArcSinh[c*x])])/(c^4*d) + (b^2*Pol
yLog[3, -E^(2*ArcSinh[c*x])])/(2*c^4*d)

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Rubi [A]  time = 0.407942, antiderivative size = 199, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 10, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.385, Rules used = {5767, 5714, 3718, 2190, 2531, 2282, 6589, 5758, 5675, 30} \[ -\frac{b \text{PolyLog}\left (2,-e^{2 \sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{c^4 d}+\frac{b^2 \text{PolyLog}\left (3,-e^{2 \sinh ^{-1}(c x)}\right )}{2 c^4 d}+\frac{x^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{2 c^2 d}-\frac{b x \sqrt{c^2 x^2+1} \left (a+b \sinh ^{-1}(c x)\right )}{2 c^3 d}+\frac{\left (a+b \sinh ^{-1}(c x)\right )^3}{3 b c^4 d}+\frac{\left (a+b \sinh ^{-1}(c x)\right )^2}{4 c^4 d}-\frac{\log \left (e^{2 \sinh ^{-1}(c x)}+1\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{c^4 d}+\frac{b^2 x^2}{4 c^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b^2*x^2)/(4*c^2*d) - (b*x*Sqrt[1 + c^2*x^2]*(a + b*ArcSinh[c*x]))/(2*c^3*d) + (a + b*ArcSinh[c*x])^2/(4*c^4*d
) + (x^2*(a + b*ArcSinh[c*x])^2)/(2*c^2*d) + (a + b*ArcSinh[c*x])^3/(3*b*c^4*d) - ((a + b*ArcSinh[c*x])^2*Log[
1 + E^(2*ArcSinh[c*x])])/(c^4*d) - (b*(a + b*ArcSinh[c*x])*PolyLog[2, -E^(2*ArcSinh[c*x])])/(c^4*d) + (b^2*Pol
yLog[3, -E^(2*ArcSinh[c*x])])/(2*c^4*d)

Rule 5767

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

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 2531

Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.)*(x_))^(m_.), x_Symbol] :> -Simp[((
f + g*x)^m*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)])/(b*c*n*Log[F]), x] + Dist[(g*m)/(b*c*n*Log[F]), Int[(f + g*x)
^(m - 1)*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)], x], x] /; FreeQ[{F, a, b, c, e, f, g, n}, x] && GtQ[m, 0]

Rule 2282

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 6589

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[PolyLog[n + 1, c*(a
+ b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d, e, n, p}, x] && EqQ[b*d, a*e]

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 5675

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

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rubi steps

\begin{align*} \int \frac{x^3 \left (a+b \sinh ^{-1}(c x)\right )^2}{d+c^2 d x^2} \, dx &=\frac{x^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{2 c^2 d}-\frac{\int \frac{x \left (a+b \sinh ^{-1}(c x)\right )^2}{d+c^2 d x^2} \, dx}{c^2}-\frac{b \int \frac{x^2 \left (a+b \sinh ^{-1}(c x)\right )}{\sqrt{1+c^2 x^2}} \, dx}{c d}\\ &=-\frac{b x \sqrt{1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{2 c^3 d}+\frac{x^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{2 c^2 d}-\frac{\operatorname{Subst}\left (\int (a+b x)^2 \tanh (x) \, dx,x,\sinh ^{-1}(c x)\right )}{c^4 d}+\frac{b \int \frac{a+b \sinh ^{-1}(c x)}{\sqrt{1+c^2 x^2}} \, dx}{2 c^3 d}+\frac{b^2 \int x \, dx}{2 c^2 d}\\ &=\frac{b^2 x^2}{4 c^2 d}-\frac{b x \sqrt{1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{2 c^3 d}+\frac{\left (a+b \sinh ^{-1}(c x)\right )^2}{4 c^4 d}+\frac{x^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{2 c^2 d}+\frac{\left (a+b \sinh ^{-1}(c x)\right )^3}{3 b c^4 d}-\frac{2 \operatorname{Subst}\left (\int \frac{e^{2 x} (a+b x)^2}{1+e^{2 x}} \, dx,x,\sinh ^{-1}(c x)\right )}{c^4 d}\\ &=\frac{b^2 x^2}{4 c^2 d}-\frac{b x \sqrt{1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{2 c^3 d}+\frac{\left (a+b \sinh ^{-1}(c x)\right )^2}{4 c^4 d}+\frac{x^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{2 c^2 d}+\frac{\left (a+b \sinh ^{-1}(c x)\right )^3}{3 b c^4 d}-\frac{\left (a+b \sinh ^{-1}(c x)\right )^2 \log \left (1+e^{2 \sinh ^{-1}(c x)}\right )}{c^4 d}+\frac{(2 b) \operatorname{Subst}\left (\int (a+b x) \log \left (1+e^{2 x}\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{c^4 d}\\ &=\frac{b^2 x^2}{4 c^2 d}-\frac{b x \sqrt{1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{2 c^3 d}+\frac{\left (a+b \sinh ^{-1}(c x)\right )^2}{4 c^4 d}+\frac{x^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{2 c^2 d}+\frac{\left (a+b \sinh ^{-1}(c x)\right )^3}{3 b c^4 d}-\frac{\left (a+b \sinh ^{-1}(c x)\right )^2 \log \left (1+e^{2 \sinh ^{-1}(c x)}\right )}{c^4 d}-\frac{b \left (a+b \sinh ^{-1}(c x)\right ) \text{Li}_2\left (-e^{2 \sinh ^{-1}(c x)}\right )}{c^4 d}+\frac{b^2 \operatorname{Subst}\left (\int \text{Li}_2\left (-e^{2 x}\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{c^4 d}\\ &=\frac{b^2 x^2}{4 c^2 d}-\frac{b x \sqrt{1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{2 c^3 d}+\frac{\left (a+b \sinh ^{-1}(c x)\right )^2}{4 c^4 d}+\frac{x^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{2 c^2 d}+\frac{\left (a+b \sinh ^{-1}(c x)\right )^3}{3 b c^4 d}-\frac{\left (a+b \sinh ^{-1}(c x)\right )^2 \log \left (1+e^{2 \sinh ^{-1}(c x)}\right )}{c^4 d}-\frac{b \left (a+b \sinh ^{-1}(c x)\right ) \text{Li}_2\left (-e^{2 \sinh ^{-1}(c x)}\right )}{c^4 d}+\frac{b^2 \operatorname{Subst}\left (\int \frac{\text{Li}_2(-x)}{x} \, dx,x,e^{2 \sinh ^{-1}(c x)}\right )}{2 c^4 d}\\ &=\frac{b^2 x^2}{4 c^2 d}-\frac{b x \sqrt{1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{2 c^3 d}+\frac{\left (a+b \sinh ^{-1}(c x)\right )^2}{4 c^4 d}+\frac{x^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{2 c^2 d}+\frac{\left (a+b \sinh ^{-1}(c x)\right )^3}{3 b c^4 d}-\frac{\left (a+b \sinh ^{-1}(c x)\right )^2 \log \left (1+e^{2 \sinh ^{-1}(c x)}\right )}{c^4 d}-\frac{b \left (a+b \sinh ^{-1}(c x)\right ) \text{Li}_2\left (-e^{2 \sinh ^{-1}(c x)}\right )}{c^4 d}+\frac{b^2 \text{Li}_3\left (-e^{2 \sinh ^{-1}(c x)}\right )}{2 c^4 d}\\ \end{align*}

Mathematica [C]  time = 0.457684, size = 279, normalized size = 1.4 \[ \frac{-48 a b \text{PolyLog}\left (2,-i e^{\sinh ^{-1}(c x)}\right )-48 a b \text{PolyLog}\left (2,i e^{\sinh ^{-1}(c x)}\right )+24 b^2 \sinh ^{-1}(c x) \text{PolyLog}\left (2,-e^{-2 \sinh ^{-1}(c x)}\right )+12 b^2 \text{PolyLog}\left (3,-e^{-2 \sinh ^{-1}(c x)}\right )+12 a^2 c^2 x^2-12 a^2 \log \left (c^2 x^2+1\right )-12 a b c x \sqrt{c^2 x^2+1}+24 a b c^2 x^2 \sinh ^{-1}(c x)+24 a b \sinh ^{-1}(c x)^2+12 a b \sinh ^{-1}(c x)-48 a b \sinh ^{-1}(c x) \log \left (1-i e^{\sinh ^{-1}(c x)}\right )-48 a b \sinh ^{-1}(c x) \log \left (1+i e^{\sinh ^{-1}(c x)}\right )-8 b^2 \sinh ^{-1}(c x)^3-6 b^2 \sinh \left (2 \sinh ^{-1}(c x)\right ) \sinh ^{-1}(c x)-24 b^2 \sinh ^{-1}(c x)^2 \log \left (e^{-2 \sinh ^{-1}(c x)}+1\right )+6 b^2 \sinh ^{-1}(c x)^2 \cosh \left (2 \sinh ^{-1}(c x)\right )+3 b^2 \cosh \left (2 \sinh ^{-1}(c x)\right )}{24 c^4 d} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(12*a^2*c^2*x^2 - 12*a*b*c*x*Sqrt[1 + c^2*x^2] + 12*a*b*ArcSinh[c*x] + 24*a*b*c^2*x^2*ArcSinh[c*x] + 24*a*b*Ar
cSinh[c*x]^2 - 8*b^2*ArcSinh[c*x]^3 + 3*b^2*Cosh[2*ArcSinh[c*x]] + 6*b^2*ArcSinh[c*x]^2*Cosh[2*ArcSinh[c*x]] -
24*b^2*ArcSinh[c*x]^2*Log[1 + E^(-2*ArcSinh[c*x])] - 48*a*b*ArcSinh[c*x]*Log[1 - I*E^ArcSinh[c*x]] - 48*a*b*A
rcSinh[c*x]*Log[1 + I*E^ArcSinh[c*x]] - 12*a^2*Log[1 + c^2*x^2] + 24*b^2*ArcSinh[c*x]*PolyLog[2, -E^(-2*ArcSin
h[c*x])] - 48*a*b*PolyLog[2, (-I)*E^ArcSinh[c*x]] - 48*a*b*PolyLog[2, I*E^ArcSinh[c*x]] + 12*b^2*PolyLog[3, -E
^(-2*ArcSinh[c*x])] - 6*b^2*ArcSinh[c*x]*Sinh[2*ArcSinh[c*x]])/(24*c^4*d)

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Maple [A]  time = 0.114, size = 380, normalized size = 1.9 \begin{align*}{\frac{{a}^{2}{x}^{2}}{2\,{c}^{2}d}}-{\frac{{a}^{2}\ln \left ({c}^{2}{x}^{2}+1 \right ) }{2\,d{c}^{4}}}+{\frac{{b}^{2} \left ({\it Arcsinh} \left ( cx \right ) \right ) ^{3}}{3\,d{c}^{4}}}+{\frac{{b}^{2} \left ({\it Arcsinh} \left ( cx \right ) \right ) ^{2}{x}^{2}}{2\,{c}^{2}d}}-{\frac{{b}^{2}{\it Arcsinh} \left ( cx \right ) x}{2\,{c}^{3}d}\sqrt{{c}^{2}{x}^{2}+1}}+{\frac{{b}^{2} \left ({\it Arcsinh} \left ( cx \right ) \right ) ^{2}}{4\,d{c}^{4}}}+{\frac{{b}^{2}{x}^{2}}{4\,{c}^{2}d}}+{\frac{{b}^{2}}{8\,d{c}^{4}}}-{\frac{{b}^{2} \left ({\it Arcsinh} \left ( cx \right ) \right ) ^{2}}{d{c}^{4}}\ln \left ( 1+ \left ( cx+\sqrt{{c}^{2}{x}^{2}+1} \right ) ^{2} \right ) }-{\frac{{b}^{2}{\it Arcsinh} \left ( cx \right ) }{d{c}^{4}}{\it polylog} \left ( 2,- \left ( cx+\sqrt{{c}^{2}{x}^{2}+1} \right ) ^{2} \right ) }+{\frac{{b}^{2}}{2\,d{c}^{4}}{\it polylog} \left ( 3,- \left ( cx+\sqrt{{c}^{2}{x}^{2}+1} \right ) ^{2} \right ) }+{\frac{ab \left ({\it Arcsinh} \left ( cx \right ) \right ) ^{2}}{d{c}^{4}}}+{\frac{ab{\it Arcsinh} \left ( cx \right ){x}^{2}}{{c}^{2}d}}-{\frac{abx}{2\,{c}^{3}d}\sqrt{{c}^{2}{x}^{2}+1}}+{\frac{ab{\it Arcsinh} \left ( cx \right ) }{2\,d{c}^{4}}}-2\,{\frac{ab{\it Arcsinh} \left ( cx \right ) \ln \left ( 1+ \left ( cx+\sqrt{{c}^{2}{x}^{2}+1} \right ) ^{2} \right ) }{d{c}^{4}}}-{\frac{ab}{d{c}^{4}}{\it polylog} \left ( 2,- \left ( cx+\sqrt{{c}^{2}{x}^{2}+1} \right ) ^{2} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/c^2*a^2/d*x^2-1/2/c^4*a^2/d*ln(c^2*x^2+1)+1/3/c^4*b^2/d*arcsinh(c*x)^3+1/2/c^2*b^2/d*arcsinh(c*x)^2*x^2-1/
2/c^3*b^2/d*arcsinh(c*x)*(c^2*x^2+1)^(1/2)*x+1/4/c^4*b^2/d*arcsinh(c*x)^2+1/4*b^2*x^2/c^2/d+1/8/c^4*b^2/d-1/c^
4*b^2/d*arcsinh(c*x)^2*ln(1+(c*x+(c^2*x^2+1)^(1/2))^2)-1/c^4*b^2/d*arcsinh(c*x)*polylog(2,-(c*x+(c^2*x^2+1)^(1
/2))^2)+1/2*b^2*polylog(3,-(c*x+(c^2*x^2+1)^(1/2))^2)/c^4/d+1/c^4*a*b/d*arcsinh(c*x)^2+1/c^2*a*b/d*arcsinh(c*x
)*x^2-1/2/c^3*a*b/d*x*(c^2*x^2+1)^(1/2)+1/2/c^4*a*b/d*arcsinh(c*x)-2/c^4*a*b/d*arcsinh(c*x)*ln(1+(c*x+(c^2*x^2
+1)^(1/2))^2)-1/c^4*a*b/d*polylog(2,-(c*x+(c^2*x^2+1)^(1/2))^2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{1}{2} \, a^{2}{\left (\frac{x^{2}}{c^{2} d} - \frac{\log \left (c^{2} x^{2} + 1\right )}{c^{4} d}\right )} + \frac{{\left (b^{2} c^{2} x^{2} - b^{2} \log \left (c^{2} x^{2} + 1\right )\right )} \log \left (c x + \sqrt{c^{2} x^{2} + 1}\right )^{2}}{2 \, c^{4} d} + \int -\frac{{\left (b^{2} c^{2} x^{2} -{\left (2 \, a b c^{4} - b^{2} c^{4}\right )} x^{4} -{\left (b^{2} c^{2} x^{2} + b^{2}\right )} \log \left (c^{2} x^{2} + 1\right ) -{\left (b^{2} c x \log \left (c^{2} x^{2} + 1\right ) +{\left (2 \, a b c^{3} - b^{2} c^{3}\right )} x^{3}\right )} \sqrt{c^{2} x^{2} + 1}\right )} \log \left (c x + \sqrt{c^{2} x^{2} + 1}\right )}{c^{6} d x^{3} + c^{4} d x +{\left (c^{5} d x^{2} + c^{3} d\right )} \sqrt{c^{2} x^{2} + 1}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*a^2*(x^2/(c^2*d) - log(c^2*x^2 + 1)/(c^4*d)) + 1/2*(b^2*c^2*x^2 - b^2*log(c^2*x^2 + 1))*log(c*x + sqrt(c^2
*x^2 + 1))^2/(c^4*d) + integrate(-(b^2*c^2*x^2 - (2*a*b*c^4 - b^2*c^4)*x^4 - (b^2*c^2*x^2 + b^2)*log(c^2*x^2 +
1) - (b^2*c*x*log(c^2*x^2 + 1) + (2*a*b*c^3 - b^2*c^3)*x^3)*sqrt(c^2*x^2 + 1))*log(c*x + sqrt(c^2*x^2 + 1))/(
c^6*d*x^3 + c^4*d*x + (c^5*d*x^2 + c^3*d)*sqrt(c^2*x^2 + 1)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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