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

Optimal. Leaf size=138 \[ -\frac{2 i b \text{PolyLog}\left (2,-i e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{c d}+\frac{2 i b \text{PolyLog}\left (2,i e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{c d}+\frac{2 i b^2 \text{PolyLog}\left (3,-i e^{\sinh ^{-1}(c x)}\right )}{c d}-\frac{2 i b^2 \text{PolyLog}\left (3,i e^{\sinh ^{-1}(c x)}\right )}{c d}+\frac{2 \tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{c d} \]

[Out]

(2*(a + b*ArcSinh[c*x])^2*ArcTan[E^ArcSinh[c*x]])/(c*d) - ((2*I)*b*(a + b*ArcSinh[c*x])*PolyLog[2, (-I)*E^ArcS
inh[c*x]])/(c*d) + ((2*I)*b*(a + b*ArcSinh[c*x])*PolyLog[2, I*E^ArcSinh[c*x]])/(c*d) + ((2*I)*b^2*PolyLog[3, (
-I)*E^ArcSinh[c*x]])/(c*d) - ((2*I)*b^2*PolyLog[3, I*E^ArcSinh[c*x]])/(c*d)

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Rubi [A]  time = 0.123686, antiderivative size = 138, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 5, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.217, Rules used = {5693, 4180, 2531, 2282, 6589} \[ -\frac{2 i b \text{PolyLog}\left (2,-i e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{c d}+\frac{2 i b \text{PolyLog}\left (2,i e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{c d}+\frac{2 i b^2 \text{PolyLog}\left (3,-i e^{\sinh ^{-1}(c x)}\right )}{c d}-\frac{2 i b^2 \text{PolyLog}\left (3,i e^{\sinh ^{-1}(c x)}\right )}{c d}+\frac{2 \tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{c d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(a + b*ArcSinh[c*x])^2*ArcTan[E^ArcSinh[c*x]])/(c*d) - ((2*I)*b*(a + b*ArcSinh[c*x])*PolyLog[2, (-I)*E^ArcS
inh[c*x]])/(c*d) + ((2*I)*b*(a + b*ArcSinh[c*x])*PolyLog[2, I*E^ArcSinh[c*x]])/(c*d) + ((2*I)*b^2*PolyLog[3, (
-I)*E^ArcSinh[c*x]])/(c*d) - ((2*I)*b^2*PolyLog[3, I*E^ArcSinh[c*x]])/(c*d)

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]

Rubi steps

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

Mathematica [A]  time = 0.240249, size = 274, normalized size = 1.99 \[ -\frac{c \left (2 b c \text{PolyLog}\left (2,\frac{c e^{\sinh ^{-1}(c x)}}{\sqrt{-c^2}}\right ) \left (a+b \sinh ^{-1}(c x)\right )-2 b c \text{PolyLog}\left (2,\frac{\sqrt{-c^2} e^{\sinh ^{-1}(c x)}}{c}\right ) \left (a+b \sinh ^{-1}(c x)\right )-2 b^2 c \text{PolyLog}\left (3,\frac{c e^{\sinh ^{-1}(c x)}}{\sqrt{-c^2}}\right )+2 b^2 c \text{PolyLog}\left (3,\frac{\sqrt{-c^2} e^{\sinh ^{-1}(c x)}}{c}\right )+a^2 \sqrt{-c^2} \tan ^{-1}(c x)-2 a b c \sinh ^{-1}(c x) \log \left (\frac{c e^{\sinh ^{-1}(c x)}}{\sqrt{-c^2}}+1\right )+2 a b c \sinh ^{-1}(c x) \log \left (\frac{\sqrt{-c^2} e^{\sinh ^{-1}(c x)}}{c}+1\right )-b^2 c \sinh ^{-1}(c x)^2 \log \left (\frac{c e^{\sinh ^{-1}(c x)}}{\sqrt{-c^2}}+1\right )+b^2 c \sinh ^{-1}(c x)^2 \log \left (\frac{\sqrt{-c^2} e^{\sinh ^{-1}(c x)}}{c}+1\right )\right )}{\left (-c^2\right )^{3/2} d} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [F]  time = 0.128, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \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((a+b*arcsinh(c*x))^2/(c^2*d*x^2+d),x)

[Out]

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

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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