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

Optimal. Leaf size=178 \[ -\frac {2 b \text {Li}_2\left (\frac {e^{\sinh ^{-1}(a+b x)}}{a-\sqrt {a^2+1}}\right )}{\sqrt {a^2+1}}+\frac {2 b \text {Li}_2\left (\frac {e^{\sinh ^{-1}(a+b x)}}{a+\sqrt {a^2+1}}\right )}{\sqrt {a^2+1}}-\frac {2 b \sinh ^{-1}(a+b x) \log \left (1-\frac {e^{\sinh ^{-1}(a+b x)}}{a-\sqrt {a^2+1}}\right )}{\sqrt {a^2+1}}+\frac {2 b \sinh ^{-1}(a+b x) \log \left (1-\frac {e^{\sinh ^{-1}(a+b x)}}{\sqrt {a^2+1}+a}\right )}{\sqrt {a^2+1}}-\frac {\sinh ^{-1}(a+b x)^2}{x} \]

[Out]

-arcsinh(b*x+a)^2/x-2*b*arcsinh(b*x+a)*ln(1-(b*x+a+(1+(b*x+a)^2)^(1/2))/(a-(a^2+1)^(1/2)))/(a^2+1)^(1/2)+2*b*a
rcsinh(b*x+a)*ln(1-(b*x+a+(1+(b*x+a)^2)^(1/2))/(a+(a^2+1)^(1/2)))/(a^2+1)^(1/2)-2*b*polylog(2,(b*x+a+(1+(b*x+a
)^2)^(1/2))/(a-(a^2+1)^(1/2)))/(a^2+1)^(1/2)+2*b*polylog(2,(b*x+a+(1+(b*x+a)^2)^(1/2))/(a+(a^2+1)^(1/2)))/(a^2
+1)^(1/2)

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Rubi [A]  time = 0.38, antiderivative size = 178, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 8, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.667, Rules used = {5865, 5801, 5831, 3322, 2264, 2190, 2279, 2391} \[ -\frac {2 b \text {PolyLog}\left (2,\frac {e^{\sinh ^{-1}(a+b x)}}{a-\sqrt {a^2+1}}\right )}{\sqrt {a^2+1}}+\frac {2 b \text {PolyLog}\left (2,\frac {e^{\sinh ^{-1}(a+b x)}}{\sqrt {a^2+1}+a}\right )}{\sqrt {a^2+1}}-\frac {2 b \sinh ^{-1}(a+b x) \log \left (1-\frac {e^{\sinh ^{-1}(a+b x)}}{a-\sqrt {a^2+1}}\right )}{\sqrt {a^2+1}}+\frac {2 b \sinh ^{-1}(a+b x) \log \left (1-\frac {e^{\sinh ^{-1}(a+b x)}}{\sqrt {a^2+1}+a}\right )}{\sqrt {a^2+1}}-\frac {\sinh ^{-1}(a+b x)^2}{x} \]

Antiderivative was successfully verified.

[In]

Int[ArcSinh[a + b*x]^2/x^2,x]

[Out]

-(ArcSinh[a + b*x]^2/x) - (2*b*ArcSinh[a + b*x]*Log[1 - E^ArcSinh[a + b*x]/(a - Sqrt[1 + a^2])])/Sqrt[1 + a^2]
+ (2*b*ArcSinh[a + b*x]*Log[1 - E^ArcSinh[a + b*x]/(a + Sqrt[1 + a^2])])/Sqrt[1 + a^2] - (2*b*PolyLog[2, E^Ar
cSinh[a + b*x]/(a - Sqrt[1 + a^2])])/Sqrt[1 + a^2] + (2*b*PolyLog[2, E^ArcSinh[a + b*x]/(a + Sqrt[1 + a^2])])/
Sqrt[1 + a^2]

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 2264

Int[((F_)^(u_)*((f_.) + (g_.)*(x_))^(m_.))/((a_.) + (b_.)*(F_)^(u_) + (c_.)*(F_)^(v_)), x_Symbol] :> With[{q =
Rt[b^2 - 4*a*c, 2]}, Dist[(2*c)/q, Int[((f + g*x)^m*F^u)/(b - q + 2*c*F^u), x], x] - Dist[(2*c)/q, Int[((f +
g*x)^m*F^u)/(b + q + 2*c*F^u), x], x]] /; FreeQ[{F, a, b, c, f, g}, x] && EqQ[v, 2*u] && LinearQ[u, x] && NeQ[
b^2 - 4*a*c, 0] && IGtQ[m, 0]

Rule 2279

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

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rule 3322

Int[((c_.) + (d_.)*(x_))^(m_.)/((a_) + (b_.)*sin[(e_.) + (Complex[0, fz_])*(f_.)*(x_)]), x_Symbol] :> Dist[2,
Int[((c + d*x)^m*E^(-(I*e) + f*fz*x))/(-(I*b) + 2*a*E^(-(I*e) + f*fz*x) + I*b*E^(2*(-(I*e) + f*fz*x))), x], x]
/; FreeQ[{a, b, c, d, e, f, fz}, x] && NeQ[a^2 - b^2, 0] && IGtQ[m, 0]

Rule 5801

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

Rule 5831

Int[(((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((f_) + (g_.)*(x_))^(m_.))/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol]
:> Dist[1/(c^(m + 1)*Sqrt[d]), Subst[Int[(a + b*x)^n*(c*f + g*Sinh[x])^m, x], x, ArcSinh[c*x]], x] /; FreeQ[{
a, b, c, d, e, f, g, n}, x] && EqQ[e, c^2*d] && IntegerQ[m] && GtQ[d, 0] && (GtQ[m, 0] || IGtQ[n, 0])

Rule 5865

Int[((a_.) + ArcSinh[(c_) + (d_.)*(x_)]*(b_.))^(n_.)*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Dist[1/d, Subst[
Int[((d*e - c*f)/d + (f*x)/d)^m*(a + b*ArcSinh[x])^n, x], x, c + d*x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x
]

Rubi steps

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

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Mathematica [A]  time = 0.12, size = 178, normalized size = 1.00 \[ \frac {-2 b x \text {Li}_2\left (\frac {e^{\sinh ^{-1}(a+b x)}}{a-\sqrt {a^2+1}}\right )+2 b x \text {Li}_2\left (\frac {e^{\sinh ^{-1}(a+b x)}}{a+\sqrt {a^2+1}}\right )-\sinh ^{-1}(a+b x) \left (\sqrt {a^2+1} \sinh ^{-1}(a+b x)+2 b x \left (\log \left (\frac {\sqrt {a^2+1}+e^{\sinh ^{-1}(a+b x)}-a}{\sqrt {a^2+1}-a}\right )-\log \left (\frac {\sqrt {a^2+1}-e^{\sinh ^{-1}(a+b x)}+a}{\sqrt {a^2+1}+a}\right )\right )\right )}{\sqrt {a^2+1} x} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcSinh[a + b*x]^2/x^2,x]

[Out]

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

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fricas [F]  time = 0.52, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\operatorname {arsinh}\left (b x + a\right )^{2}}{x^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(arcsinh(b*x + a)^2/x^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(arcsinh(b*x + a)^2/x^2, x)

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maple [A]  time = 0.39, size = 217, normalized size = 1.22 \[ -\frac {\arcsinh \left (b x +a \right )^{2}}{x}+\frac {2 b \arcsinh \left (b x +a \right ) \ln \left (\frac {\sqrt {a^{2}+1}-b x -\sqrt {1+\left (b x +a \right )^{2}}}{a +\sqrt {a^{2}+1}}\right )}{\sqrt {a^{2}+1}}-\frac {2 b \arcsinh \left (b x +a \right ) \ln \left (\frac {\sqrt {a^{2}+1}+b x +\sqrt {1+\left (b x +a \right )^{2}}}{-a +\sqrt {a^{2}+1}}\right )}{\sqrt {a^{2}+1}}+\frac {2 b \dilog \left (\frac {\sqrt {a^{2}+1}-b x -\sqrt {1+\left (b x +a \right )^{2}}}{a +\sqrt {a^{2}+1}}\right )}{\sqrt {a^{2}+1}}-\frac {2 b \dilog \left (\frac {\sqrt {a^{2}+1}+b x +\sqrt {1+\left (b x +a \right )^{2}}}{-a +\sqrt {a^{2}+1}}\right )}{\sqrt {a^{2}+1}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-arcsinh(b*x+a)^2/x+2*b*arcsinh(b*x+a)/(a^2+1)^(1/2)*ln(((a^2+1)^(1/2)-b*x-(1+(b*x+a)^2)^(1/2))/(a+(a^2+1)^(1/
2)))-2*b*arcsinh(b*x+a)/(a^2+1)^(1/2)*ln(((a^2+1)^(1/2)+b*x+(1+(b*x+a)^2)^(1/2))/(-a+(a^2+1)^(1/2)))+2*b/(a^2+
1)^(1/2)*dilog(((a^2+1)^(1/2)-b*x-(1+(b*x+a)^2)^(1/2))/(a+(a^2+1)^(1/2)))-2*b/(a^2+1)^(1/2)*dilog(((a^2+1)^(1/
2)+b*x+(1+(b*x+a)^2)^(1/2))/(-a+(a^2+1)^(1/2)))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {\log \left (b x + a + \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right )^{2}}{x} + \int \frac {2 \, {\left (b^{3} x^{2} + 2 \, a b^{2} x + a^{2} b + \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} {\left (b^{2} x + a b\right )} + b\right )} \log \left (b x + a + \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right )}{b^{3} x^{4} + 3 \, a b^{2} x^{3} + {\left (3 \, a^{2} b + b\right )} x^{2} + {\left (a^{3} + a\right )} x + {\left (b^{2} x^{3} + 2 \, a b x^{2} + {\left (a^{2} + 1\right )} x\right )} \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\mathrm {asinh}\left (a+b\,x\right )}^2}{x^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(asinh(a + b*x)^2/x^2,x)

[Out]

int(asinh(a + b*x)^2/x^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(asinh(a + b*x)**2/x**2, x)

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