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

Optimal. Leaf size=198 \[ \frac{2 i b \text{PolyLog}\left (2,-i e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{c^3 d}-\frac{2 i b \text{PolyLog}\left (2,i e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{c^3 d}-\frac{2 i b^2 \text{PolyLog}\left (3,-i e^{\sinh ^{-1}(c x)}\right )}{c^3 d}+\frac{2 i b^2 \text{PolyLog}\left (3,i e^{\sinh ^{-1}(c x)}\right )}{c^3 d}-\frac{2 b \sqrt{c^2 x^2+1} \left (a+b \sinh ^{-1}(c x)\right )}{c^3 d}+\frac{x \left (a+b \sinh ^{-1}(c x)\right )^2}{c^2 d}-\frac{2 \tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{c^3 d}+\frac{2 b^2 x}{c^2 d} \]

[Out]

(2*b^2*x)/(c^2*d) - (2*b*Sqrt[1 + c^2*x^2]*(a + b*ArcSinh[c*x]))/(c^3*d) + (x*(a + b*ArcSinh[c*x])^2)/(c^2*d)
- (2*(a + b*ArcSinh[c*x])^2*ArcTan[E^ArcSinh[c*x]])/(c^3*d) + ((2*I)*b*(a + b*ArcSinh[c*x])*PolyLog[2, (-I)*E^
ArcSinh[c*x]])/(c^3*d) - ((2*I)*b*(a + b*ArcSinh[c*x])*PolyLog[2, I*E^ArcSinh[c*x]])/(c^3*d) - ((2*I)*b^2*Poly
Log[3, (-I)*E^ArcSinh[c*x]])/(c^3*d) + ((2*I)*b^2*PolyLog[3, I*E^ArcSinh[c*x]])/(c^3*d)

________________________________________________________________________________________

Rubi [A]  time = 0.293411, antiderivative size = 198, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 8, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.308, Rules used = {5767, 5693, 4180, 2531, 2282, 6589, 5717, 8} \[ \frac{2 i b \text{PolyLog}\left (2,-i e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{c^3 d}-\frac{2 i b \text{PolyLog}\left (2,i e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{c^3 d}-\frac{2 i b^2 \text{PolyLog}\left (3,-i e^{\sinh ^{-1}(c x)}\right )}{c^3 d}+\frac{2 i b^2 \text{PolyLog}\left (3,i e^{\sinh ^{-1}(c x)}\right )}{c^3 d}-\frac{2 b \sqrt{c^2 x^2+1} \left (a+b \sinh ^{-1}(c x)\right )}{c^3 d}+\frac{x \left (a+b \sinh ^{-1}(c x)\right )^2}{c^2 d}-\frac{2 \tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{c^3 d}+\frac{2 b^2 x}{c^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*b^2*x)/(c^2*d) - (2*b*Sqrt[1 + c^2*x^2]*(a + b*ArcSinh[c*x]))/(c^3*d) + (x*(a + b*ArcSinh[c*x])^2)/(c^2*d)
- (2*(a + b*ArcSinh[c*x])^2*ArcTan[E^ArcSinh[c*x]])/(c^3*d) + ((2*I)*b*(a + b*ArcSinh[c*x])*PolyLog[2, (-I)*E^
ArcSinh[c*x]])/(c^3*d) - ((2*I)*b*(a + b*ArcSinh[c*x])*PolyLog[2, I*E^ArcSinh[c*x]])/(c^3*d) - ((2*I)*b^2*Poly
Log[3, (-I)*E^ArcSinh[c*x]])/(c^3*d) + ((2*I)*b^2*PolyLog[3, I*E^ArcSinh[c*x]])/(c^3*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 5693

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

Rule 4180

Int[csc[(e_.) + Pi*(k_.) + (Complex[0, fz_])*(f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(-2*(c
+ d*x)^m*ArcTanh[E^(-(I*e) + f*fz*x)/E^(I*k*Pi)])/(f*fz*I), x] + (-Dist[(d*m)/(f*fz*I), Int[(c + d*x)^(m - 1)*
Log[1 - E^(-(I*e) + f*fz*x)/E^(I*k*Pi)], x], x] + Dist[(d*m)/(f*fz*I), Int[(c + d*x)^(m - 1)*Log[1 + E^(-(I*e)
+ f*fz*x)/E^(I*k*Pi)], x], x]) /; FreeQ[{c, d, e, f, fz}, x] && IntegerQ[2*k] && 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 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 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rubi steps

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

Mathematica [A]  time = 0.570927, size = 293, normalized size = 1.48 \[ \frac{2 a b \left (i \left (\text{PolyLog}\left (2,-i e^{-\sinh ^{-1}(c x)}\right )-\text{PolyLog}\left (2,i e^{-\sinh ^{-1}(c x)}\right )\right )-\sqrt{c^2 x^2+1}+c x \sinh ^{-1}(c x)+i \sinh ^{-1}(c x) \left (\log \left (1-i e^{-\sinh ^{-1}(c x)}\right )-\log \left (1+i e^{-\sinh ^{-1}(c x)}\right )\right )\right )}{c^3 d}+\frac{b^2 \left (-i \left (-2 \sinh ^{-1}(c x) \left (\text{PolyLog}\left (2,-i e^{-\sinh ^{-1}(c x)}\right )-\text{PolyLog}\left (2,i e^{-\sinh ^{-1}(c x)}\right )\right )-2 \left (\text{PolyLog}\left (3,-i e^{-\sinh ^{-1}(c x)}\right )-\text{PolyLog}\left (3,i e^{-\sinh ^{-1}(c x)}\right )\right )+\sinh ^{-1}(c x)^2 \left (-\left (\log \left (1-i e^{-\sinh ^{-1}(c x)}\right )-\log \left (1+i e^{-\sinh ^{-1}(c x)}\right )\right )\right )\right )-2 \sqrt{c^2 x^2+1} \sinh ^{-1}(c x)+c x \left (\sinh ^{-1}(c x)^2+2\right )\right )}{c^3 d}+\frac{a^2 x}{c^2 d}-\frac{a^2 \tan ^{-1}(c x)}{c^3 d} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(a^2*x)/(c^2*d) - (a^2*ArcTan[c*x])/(c^3*d) + (2*a*b*(-Sqrt[1 + c^2*x^2] + c*x*ArcSinh[c*x] + I*ArcSinh[c*x]*(
Log[1 - I/E^ArcSinh[c*x]] - Log[1 + I/E^ArcSinh[c*x]]) + I*(PolyLog[2, (-I)/E^ArcSinh[c*x]] - PolyLog[2, I/E^A
rcSinh[c*x]])))/(c^3*d) + (b^2*(-2*Sqrt[1 + c^2*x^2]*ArcSinh[c*x] + c*x*(2 + ArcSinh[c*x]^2) - I*(-(ArcSinh[c*
x]^2*(Log[1 - I/E^ArcSinh[c*x]] - Log[1 + I/E^ArcSinh[c*x]])) - 2*ArcSinh[c*x]*(PolyLog[2, (-I)/E^ArcSinh[c*x]
] - PolyLog[2, I/E^ArcSinh[c*x]]) - 2*(PolyLog[3, (-I)/E^ArcSinh[c*x]] - PolyLog[3, I/E^ArcSinh[c*x]]))))/(c^3
*d)

________________________________________________________________________________________

Maple [F]  time = 0.097, size = 0, normalized size = 0. \begin{align*} \int{\frac{{x}^{2} \left ( a+b{\it Arcsinh} \left ( cx \right ) \right ) ^{2}}{{c}^{2}d{x}^{2}+d}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

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^{2}}{c^{2} d x^{2} + d}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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