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

Optimal. Leaf size=105 \[ \frac{b \text{PolyLog}\left (2,-e^{2 \sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{c^2 d}-\frac{b^2 \text{PolyLog}\left (3,-e^{2 \sinh ^{-1}(c x)}\right )}{2 c^2 d}-\frac{\left (a+b \sinh ^{-1}(c x)\right )^3}{3 b c^2 d}+\frac{\log \left (e^{2 \sinh ^{-1}(c x)}+1\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{c^2 d} \]

[Out]

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

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Rubi [A]  time = 0.183706, antiderivative size = 105, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {5714, 3718, 2190, 2531, 2282, 6589} \[ \frac{b \text{PolyLog}\left (2,-e^{2 \sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{c^2 d}-\frac{b^2 \text{PolyLog}\left (3,-e^{2 \sinh ^{-1}(c x)}\right )}{2 c^2 d}-\frac{\left (a+b \sinh ^{-1}(c x)\right )^3}{3 b c^2 d}+\frac{\log \left (e^{2 \sinh ^{-1}(c x)}+1\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{c^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

Rubi steps

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

Mathematica [B]  time = 0.238463, size = 281, normalized size = 2.68 \[ \frac{12 b \text{PolyLog}\left (2,\frac{c e^{\sinh ^{-1}(c x)}}{\sqrt{-c^2}}\right ) \left (a+b \sinh ^{-1}(c x)\right )+12 b \text{PolyLog}\left (2,\frac{\sqrt{-c^2} e^{\sinh ^{-1}(c x)}}{c}\right ) \left (a+b \sinh ^{-1}(c x)\right )-12 b^2 \text{PolyLog}\left (3,\frac{c e^{\sinh ^{-1}(c x)}}{\sqrt{-c^2}}\right )-12 b^2 \text{PolyLog}\left (3,\frac{\sqrt{-c^2} e^{\sinh ^{-1}(c x)}}{c}\right )+3 a^2 \log \left (c^2 x^2+1\right )+12 a b \sinh ^{-1}(c x) \log \left (\frac{c e^{\sinh ^{-1}(c x)}}{\sqrt{-c^2}}+1\right )+12 a b \sinh ^{-1}(c x) \log \left (\frac{\sqrt{-c^2} e^{\sinh ^{-1}(c x)}}{c}+1\right )-6 a b \sinh ^{-1}(c x)^2+6 b^2 \sinh ^{-1}(c x)^2 \log \left (\frac{c e^{\sinh ^{-1}(c x)}}{\sqrt{-c^2}}+1\right )+6 b^2 \sinh ^{-1}(c x)^2 \log \left (\frac{\sqrt{-c^2} e^{\sinh ^{-1}(c x)}}{c}+1\right )-2 b^2 \sinh ^{-1}(c x)^3}{6 c^2 d} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-6*a*b*ArcSinh[c*x]^2 - 2*b^2*ArcSinh[c*x]^3 + 12*a*b*ArcSinh[c*x]*Log[1 + (c*E^ArcSinh[c*x])/Sqrt[-c^2]] + 6
*b^2*ArcSinh[c*x]^2*Log[1 + (c*E^ArcSinh[c*x])/Sqrt[-c^2]] + 12*a*b*ArcSinh[c*x]*Log[1 + (Sqrt[-c^2]*E^ArcSinh
[c*x])/c] + 6*b^2*ArcSinh[c*x]^2*Log[1 + (Sqrt[-c^2]*E^ArcSinh[c*x])/c] + 3*a^2*Log[1 + c^2*x^2] + 12*b*(a + b
*ArcSinh[c*x])*PolyLog[2, (c*E^ArcSinh[c*x])/Sqrt[-c^2]] + 12*b*(a + b*ArcSinh[c*x])*PolyLog[2, (Sqrt[-c^2]*E^
ArcSinh[c*x])/c] - 12*b^2*PolyLog[3, (c*E^ArcSinh[c*x])/Sqrt[-c^2]] - 12*b^2*PolyLog[3, (Sqrt[-c^2]*E^ArcSinh[
c*x])/c])/(6*c^2*d)

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Maple [A]  time = 0.049, size = 223, normalized size = 2.1 \begin{align*}{\frac{{a}^{2}\ln \left ({c}^{2}{x}^{2}+1 \right ) }{2\,{c}^{2}d}}-{\frac{{b}^{2} \left ({\it Arcsinh} \left ( cx \right ) \right ) ^{3}}{3\,{c}^{2}d}}+{\frac{{b}^{2} \left ({\it Arcsinh} \left ( cx \right ) \right ) ^{2}}{{c}^{2}d}\ln \left ( 1+ \left ( cx+\sqrt{{c}^{2}{x}^{2}+1} \right ) ^{2} \right ) }+{\frac{{b}^{2}{\it Arcsinh} \left ( cx \right ) }{{c}^{2}d}{\it polylog} \left ( 2,- \left ( cx+\sqrt{{c}^{2}{x}^{2}+1} \right ) ^{2} \right ) }-{\frac{{b}^{2}}{2\,{c}^{2}d}{\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}}{{c}^{2}d}}+2\,{\frac{ab{\it Arcsinh} \left ( cx \right ) \ln \left ( 1+ \left ( cx+\sqrt{{c}^{2}{x}^{2}+1} \right ) ^{2} \right ) }{{c}^{2}d}}+{\frac{ab}{{c}^{2}d}{\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*(a+b*arcsinh(c*x))^2/(c^2*d*x^2+d),x)

[Out]

1/2/c^2*a^2/d*ln(c^2*x^2+1)-1/3/c^2*b^2/d*arcsinh(c*x)^3+1/c^2*b^2/d*arcsinh(c*x)^2*ln(1+(c*x+(c^2*x^2+1)^(1/2
))^2)+1/c^2*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^2/d-1/c^2*a*b/d*arcsinh(c*x)^2+2/c^2*a*b/d*arcsinh(c*x)*ln(1+(c*x+(c^2*x^2+1)^(1/2))^2)+1/c^2*a*b/d*pol
ylog(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{b^{2} \log \left (c^{2} x^{2} + 1\right ) \log \left (c x + \sqrt{c^{2} x^{2} + 1}\right )^{2}}{2 \, c^{2} d} + \frac{a^{2} \log \left (c^{2} d x^{2} + d\right )}{2 \, c^{2} d} - \int -\frac{{\left (2 \, a b c^{2} x^{2} -{\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 ) - 2 \, a b c x\right )} \sqrt{c^{2} x^{2} + 1}\right )} \log \left (c x + \sqrt{c^{2} x^{2} + 1}\right )}{c^{4} d x^{3} + c^{2} d x +{\left (c^{3} d x^{2} + c 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*(a+b*arcsinh(c*x))^2/(c^2*d*x^2+d),x, algorithm="maxima")

[Out]

1/2*b^2*log(c^2*x^2 + 1)*log(c*x + sqrt(c^2*x^2 + 1))^2/(c^2*d) + 1/2*a^2*log(c^2*d*x^2 + d)/(c^2*d) - integra
te(-(2*a*b*c^2*x^2 - (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*x)*sqrt(c^2*x^
2 + 1))*log(c*x + sqrt(c^2*x^2 + 1))/(c^4*d*x^3 + c^2*d*x + (c^3*d*x^2 + c*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 \operatorname{arsinh}\left (c x\right )^{2} + 2 \, a b x \operatorname{arsinh}\left (c x\right ) + a^{2} x}{c^{2} d x^{2} + d}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b^2*x*arcsinh(c*x)^2 + 2*a*b*x*arcsinh(c*x) + a^2*x)/(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}{c^{2} x^{2} + 1}\, dx + \int \frac{b^{2} x \operatorname{asinh}^{2}{\left (c x \right )}}{c^{2} x^{2} + 1}\, dx + \int \frac{2 a b x \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*(a+b*asinh(c*x))**2/(c**2*d*x**2+d),x)

[Out]

(Integral(a**2*x/(c**2*x**2 + 1), x) + Integral(b**2*x*asinh(c*x)**2/(c**2*x**2 + 1), x) + Integral(2*a*b*x*as
inh(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}{c^{2} d x^{2} + d}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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