3.279 \(\int \frac {\sinh ^{-1}(a+b x)^2}{(1+a^2+2 a b x+b^2 x^2)^{3/2}} \, dx\)

Optimal. Leaf size=86 \[ -\frac {\text {Li}_2\left (-e^{2 \sinh ^{-1}(a+b x)}\right )}{b}+\frac {(a+b x) \sinh ^{-1}(a+b x)^2}{b \sqrt {(a+b x)^2+1}}+\frac {\sinh ^{-1}(a+b x)^2}{b}-\frac {2 \sinh ^{-1}(a+b x) \log \left (e^{2 \sinh ^{-1}(a+b x)}+1\right )}{b} \]

[Out]

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

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Rubi [A]  time = 0.16, antiderivative size = 86, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.233, Rules used = {5867, 5687, 5714, 3718, 2190, 2279, 2391} \[ -\frac {\text {PolyLog}\left (2,-e^{2 \sinh ^{-1}(a+b x)}\right )}{b}+\frac {(a+b x) \sinh ^{-1}(a+b x)^2}{b \sqrt {(a+b x)^2+1}}+\frac {\sinh ^{-1}(a+b x)^2}{b}-\frac {2 \sinh ^{-1}(a+b x) \log \left (e^{2 \sinh ^{-1}(a+b x)}+1\right )}{b} \]

Antiderivative was successfully verified.

[In]

Int[ArcSinh[a + b*x]^2/(1 + a^2 + 2*a*b*x + b^2*x^2)^(3/2),x]

[Out]

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

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 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 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 5687

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)/((d_) + (e_.)*(x_)^2)^(3/2), x_Symbol] :> Simp[(x*(a + b*ArcSinh
[c*x])^n)/(d*Sqrt[d + e*x^2]), x] - Dist[(b*c*n*Sqrt[1 + c^2*x^2])/(d*Sqrt[d + e*x^2]), Int[(x*(a + b*ArcSinh[
c*x])^(n - 1))/(1 + c^2*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*d] && GtQ[n, 0]

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 5867

Int[((a_.) + ArcSinh[(c_) + (d_.)*(x_)]*(b_.))^(n_.)*((A_.) + (B_.)*(x_) + (C_.)*(x_)^2)^(p_.), x_Symbol] :> D
ist[1/d, Subst[Int[(C/d^2 + (C*x^2)/d^2)^p*(a + b*ArcSinh[x])^n, x], x, c + d*x], x] /; FreeQ[{a, b, c, d, A,
B, C, n, p}, x] && EqQ[B*(1 + c^2) - 2*A*c*d, 0] && EqQ[2*c*C - B*d, 0]

Rubi steps

\begin {align*} \int \frac {\sinh ^{-1}(a+b x)^2}{\left (1+a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\sinh ^{-1}(x)^2}{\left (1+x^2\right )^{3/2}} \, dx,x,a+b x\right )}{b}\\ &=\frac {(a+b x) \sinh ^{-1}(a+b x)^2}{b \sqrt {1+(a+b x)^2}}-\frac {2 \operatorname {Subst}\left (\int \frac {x \sinh ^{-1}(x)}{1+x^2} \, dx,x,a+b x\right )}{b}\\ &=\frac {(a+b x) \sinh ^{-1}(a+b x)^2}{b \sqrt {1+(a+b x)^2}}-\frac {2 \operatorname {Subst}\left (\int x \tanh (x) \, dx,x,\sinh ^{-1}(a+b x)\right )}{b}\\ &=\frac {\sinh ^{-1}(a+b x)^2}{b}+\frac {(a+b x) \sinh ^{-1}(a+b x)^2}{b \sqrt {1+(a+b x)^2}}-\frac {4 \operatorname {Subst}\left (\int \frac {e^{2 x} x}{1+e^{2 x}} \, dx,x,\sinh ^{-1}(a+b x)\right )}{b}\\ &=\frac {\sinh ^{-1}(a+b x)^2}{b}+\frac {(a+b x) \sinh ^{-1}(a+b x)^2}{b \sqrt {1+(a+b x)^2}}-\frac {2 \sinh ^{-1}(a+b x) \log \left (1+e^{2 \sinh ^{-1}(a+b x)}\right )}{b}+\frac {2 \operatorname {Subst}\left (\int \log \left (1+e^{2 x}\right ) \, dx,x,\sinh ^{-1}(a+b x)\right )}{b}\\ &=\frac {\sinh ^{-1}(a+b x)^2}{b}+\frac {(a+b x) \sinh ^{-1}(a+b x)^2}{b \sqrt {1+(a+b x)^2}}-\frac {2 \sinh ^{-1}(a+b x) \log \left (1+e^{2 \sinh ^{-1}(a+b x)}\right )}{b}+\frac {\operatorname {Subst}\left (\int \frac {\log (1+x)}{x} \, dx,x,e^{2 \sinh ^{-1}(a+b x)}\right )}{b}\\ &=\frac {\sinh ^{-1}(a+b x)^2}{b}+\frac {(a+b x) \sinh ^{-1}(a+b x)^2}{b \sqrt {1+(a+b x)^2}}-\frac {2 \sinh ^{-1}(a+b x) \log \left (1+e^{2 \sinh ^{-1}(a+b x)}\right )}{b}-\frac {\text {Li}_2\left (-e^{2 \sinh ^{-1}(a+b x)}\right )}{b}\\ \end {align*}

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Mathematica [A]  time = 0.40, size = 98, normalized size = 1.14 \[ \frac {\sinh ^{-1}(a+b x) \left (\frac {\left (-\sqrt {a^2+2 a b x+b^2 x^2+1}+a+b x\right ) \sinh ^{-1}(a+b x)}{\sqrt {a^2+2 a b x+b^2 x^2+1}}-2 \log \left (e^{-2 \sinh ^{-1}(a+b x)}+1\right )\right )+\text {Li}_2\left (-e^{-2 \sinh ^{-1}(a+b x)}\right )}{b} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[ArcSinh[a + b*x]^2/(1 + a^2 + 2*a*b*x + b^2*x^2)^(3/2),x]

[Out]

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

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fricas [F]  time = 0.56, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} \operatorname {arsinh}\left (b x + a\right )^{2}}{b^{4} x^{4} + 4 \, a b^{3} x^{3} + 2 \, {\left (3 \, a^{2} + 1\right )} b^{2} x^{2} + a^{4} + 4 \, {\left (a^{3} + a\right )} b x + 2 \, a^{2} + 1}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsinh(b*x+a)^2/(b^2*x^2+2*a*b*x+a^2+1)^(3/2),x, algorithm="fricas")

[Out]

integral(sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)*arcsinh(b*x + a)^2/(b^4*x^4 + 4*a*b^3*x^3 + 2*(3*a^2 + 1)*b^2*x^2 +
a^4 + 4*(a^3 + a)*b*x + 2*a^2 + 1), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {arsinh}\left (b x + a\right )^{2}}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsinh(b*x+a)^2/(b^2*x^2+2*a*b*x+a^2+1)^(3/2),x, algorithm="giac")

[Out]

integrate(arcsinh(b*x + a)^2/(b^2*x^2 + 2*a*b*x + a^2 + 1)^(3/2), x)

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maple [A]  time = 0.25, size = 168, normalized size = 1.95 \[ -\frac {\left (b^{2} x^{2}-\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\, x b +2 a b x -\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\, a +a^{2}+1\right ) \arcsinh \left (b x +a \right )^{2}}{b \left (b^{2} x^{2}+2 a b x +a^{2}+1\right )}+\frac {2 \arcsinh \left (b x +a \right )^{2}}{b}-\frac {2 \arcsinh \left (b x +a \right ) \ln \left (1+\left (b x +a +\sqrt {1+\left (b x +a \right )^{2}}\right )^{2}\right )}{b}-\frac {\polylog \left (2, -\left (b x +a +\sqrt {1+\left (b x +a \right )^{2}}\right )^{2}\right )}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arcsinh(b*x+a)^2/(b^2*x^2+2*a*b*x+a^2+1)^(3/2),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsinh(b*x+a)^2/(b^2*x^2+2*a*b*x+a^2+1)^(3/2),x, algorithm="maxima")

[Out]

integrate(arcsinh(b*x + a)^2/(b^2*x^2 + 2*a*b*x + a^2 + 1)^(3/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\mathrm {asinh}\left (a+b\,x\right )}^2}{{\left (a^2+2\,a\,b\,x+b^2\,x^2+1\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(asinh(a + b*x)^2/(a^2 + b^2*x^2 + 2*a*b*x + 1)^(3/2),x)

[Out]

int(asinh(a + b*x)^2/(a^2 + b^2*x^2 + 2*a*b*x + 1)^(3/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {asinh}^{2}{\left (a + b x \right )}}{\left (a^{2} + 2 a b x + b^{2} x^{2} + 1\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(asinh(b*x+a)**2/(b**2*x**2+2*a*b*x+a**2+1)**(3/2),x)

[Out]

Integral(asinh(a + b*x)**2/(a**2 + 2*a*b*x + b**2*x**2 + 1)**(3/2), x)

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