3.2 \(\int \frac {\sinh ^{-1}(c x)^2}{d+e x} \, dx\)

Optimal. Leaf size=260 \[ \frac {2 \sinh ^{-1}(c x) \text {Li}_2\left (-\frac {e e^{\sinh ^{-1}(c x)}}{c d-\sqrt {c^2 d^2+e^2}}\right )}{e}+\frac {2 \sinh ^{-1}(c x) \text {Li}_2\left (-\frac {e e^{\sinh ^{-1}(c x)}}{c d+\sqrt {c^2 d^2+e^2}}\right )}{e}-\frac {2 \text {Li}_3\left (-\frac {e e^{\sinh ^{-1}(c x)}}{c d-\sqrt {c^2 d^2+e^2}}\right )}{e}-\frac {2 \text {Li}_3\left (-\frac {e e^{\sinh ^{-1}(c x)}}{c d+\sqrt {c^2 d^2+e^2}}\right )}{e}+\frac {\sinh ^{-1}(c x)^2 \log \left (\frac {e e^{\sinh ^{-1}(c x)}}{c d-\sqrt {c^2 d^2+e^2}}+1\right )}{e}+\frac {\sinh ^{-1}(c x)^2 \log \left (\frac {e e^{\sinh ^{-1}(c x)}}{\sqrt {c^2 d^2+e^2}+c d}+1\right )}{e}-\frac {\sinh ^{-1}(c x)^3}{3 e} \]

[Out]

-1/3*arcsinh(c*x)^3/e+arcsinh(c*x)^2*ln(1+e*(c*x+(c^2*x^2+1)^(1/2))/(c*d-(c^2*d^2+e^2)^(1/2)))/e+arcsinh(c*x)^
2*ln(1+e*(c*x+(c^2*x^2+1)^(1/2))/(c*d+(c^2*d^2+e^2)^(1/2)))/e+2*arcsinh(c*x)*polylog(2,-e*(c*x+(c^2*x^2+1)^(1/
2))/(c*d-(c^2*d^2+e^2)^(1/2)))/e+2*arcsinh(c*x)*polylog(2,-e*(c*x+(c^2*x^2+1)^(1/2))/(c*d+(c^2*d^2+e^2)^(1/2))
)/e-2*polylog(3,-e*(c*x+(c^2*x^2+1)^(1/2))/(c*d-(c^2*d^2+e^2)^(1/2)))/e-2*polylog(3,-e*(c*x+(c^2*x^2+1)^(1/2))
/(c*d+(c^2*d^2+e^2)^(1/2)))/e

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Rubi [A]  time = 0.40, antiderivative size = 260, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 6, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {5799, 5561, 2190, 2531, 2282, 6589} \[ \frac {2 \sinh ^{-1}(c x) \text {PolyLog}\left (2,-\frac {e e^{\sinh ^{-1}(c x)}}{c d-\sqrt {c^2 d^2+e^2}}\right )}{e}+\frac {2 \sinh ^{-1}(c x) \text {PolyLog}\left (2,-\frac {e e^{\sinh ^{-1}(c x)}}{\sqrt {c^2 d^2+e^2}+c d}\right )}{e}-\frac {2 \text {PolyLog}\left (3,-\frac {e e^{\sinh ^{-1}(c x)}}{c d-\sqrt {c^2 d^2+e^2}}\right )}{e}-\frac {2 \text {PolyLog}\left (3,-\frac {e e^{\sinh ^{-1}(c x)}}{\sqrt {c^2 d^2+e^2}+c d}\right )}{e}+\frac {\sinh ^{-1}(c x)^2 \log \left (\frac {e e^{\sinh ^{-1}(c x)}}{c d-\sqrt {c^2 d^2+e^2}}+1\right )}{e}+\frac {\sinh ^{-1}(c x)^2 \log \left (\frac {e e^{\sinh ^{-1}(c x)}}{\sqrt {c^2 d^2+e^2}+c d}+1\right )}{e}-\frac {\sinh ^{-1}(c x)^3}{3 e} \]

Antiderivative was successfully verified.

[In]

Int[ArcSinh[c*x]^2/(d + e*x),x]

[Out]

-ArcSinh[c*x]^3/(3*e) + (ArcSinh[c*x]^2*Log[1 + (e*E^ArcSinh[c*x])/(c*d - Sqrt[c^2*d^2 + e^2])])/e + (ArcSinh[
c*x]^2*Log[1 + (e*E^ArcSinh[c*x])/(c*d + Sqrt[c^2*d^2 + e^2])])/e + (2*ArcSinh[c*x]*PolyLog[2, -((e*E^ArcSinh[
c*x])/(c*d - Sqrt[c^2*d^2 + e^2]))])/e + (2*ArcSinh[c*x]*PolyLog[2, -((e*E^ArcSinh[c*x])/(c*d + Sqrt[c^2*d^2 +
e^2]))])/e - (2*PolyLog[3, -((e*E^ArcSinh[c*x])/(c*d - Sqrt[c^2*d^2 + e^2]))])/e - (2*PolyLog[3, -((e*E^ArcSi
nh[c*x])/(c*d + Sqrt[c^2*d^2 + e^2]))])/e

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 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 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 5561

Int[(Cosh[(c_.) + (d_.)*(x_)]*((e_.) + (f_.)*(x_))^(m_.))/((a_) + (b_.)*Sinh[(c_.) + (d_.)*(x_)]), x_Symbol] :
> -Simp[(e + f*x)^(m + 1)/(b*f*(m + 1)), x] + (Int[((e + f*x)^m*E^(c + d*x))/(a - Rt[a^2 + b^2, 2] + b*E^(c +
d*x)), x] + Int[((e + f*x)^m*E^(c + d*x))/(a + Rt[a^2 + b^2, 2] + b*E^(c + d*x)), x]) /; FreeQ[{a, b, c, d, e,
f}, x] && IGtQ[m, 0] && NeQ[a^2 + b^2, 0]

Rule 5799

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

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

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Mathematica [A]  time = 0.14, size = 240, normalized size = 0.92 \[ -\frac {-6 \sinh ^{-1}(c x) \text {Li}_2\left (\frac {e e^{\sinh ^{-1}(c x)}}{\sqrt {c^2 d^2+e^2}-c d}\right )-6 \sinh ^{-1}(c x) \text {Li}_2\left (-\frac {e e^{\sinh ^{-1}(c x)}}{c d+\sqrt {c^2 d^2+e^2}}\right )+6 \text {Li}_3\left (\frac {e e^{\sinh ^{-1}(c x)}}{\sqrt {c^2 d^2+e^2}-c d}\right )+6 \text {Li}_3\left (-\frac {e e^{\sinh ^{-1}(c x)}}{c d+\sqrt {c^2 d^2+e^2}}\right )-3 \sinh ^{-1}(c x)^2 \log \left (\frac {e e^{\sinh ^{-1}(c x)}}{c d-\sqrt {c^2 d^2+e^2}}+1\right )-3 \sinh ^{-1}(c x)^2 \log \left (\frac {e e^{\sinh ^{-1}(c x)}}{\sqrt {c^2 d^2+e^2}+c d}+1\right )+\sinh ^{-1}(c x)^3}{3 e} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcSinh[c*x]^2/(d + e*x),x]

[Out]

-1/3*(ArcSinh[c*x]^3 - 3*ArcSinh[c*x]^2*Log[1 + (e*E^ArcSinh[c*x])/(c*d - Sqrt[c^2*d^2 + e^2])] - 3*ArcSinh[c*
x]^2*Log[1 + (e*E^ArcSinh[c*x])/(c*d + Sqrt[c^2*d^2 + e^2])] - 6*ArcSinh[c*x]*PolyLog[2, (e*E^ArcSinh[c*x])/(-
(c*d) + Sqrt[c^2*d^2 + e^2])] - 6*ArcSinh[c*x]*PolyLog[2, -((e*E^ArcSinh[c*x])/(c*d + Sqrt[c^2*d^2 + e^2]))] +
6*PolyLog[3, (e*E^ArcSinh[c*x])/(-(c*d) + Sqrt[c^2*d^2 + e^2])] + 6*PolyLog[3, -((e*E^ArcSinh[c*x])/(c*d + Sq
rt[c^2*d^2 + e^2]))])/e

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fricas [F]  time = 0.53, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\operatorname {arsinh}\left (c x\right )^{2}}{e x + d}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(arcsinh(c*x)^2/(e*x + d), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {arsinh}\left (c x\right )^{2}}{e x + d}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.20, size = 0, normalized size = 0.00 \[ \int \frac {\arcsinh \left (c x \right )^{2}}{e x +d}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {arsinh}\left (c x\right )^{2}}{e x + d}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\mathrm {asinh}\left (c\,x\right )}^2}{d+e\,x} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(asinh(c*x)^2/(d + e*x),x)

[Out]

int(asinh(c*x)^2/(d + e*x), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {asinh}^{2}{\left (c x \right )}}{d + e x}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(asinh(c*x)**2/(e*x+d),x)

[Out]

Integral(asinh(c*x)**2/(d + e*x), x)

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