3.1.70 \(\int \sinh ^{-1}(a+b x)^2 \, dx\) [70]

Optimal. Leaf size=45 \[ 2 x-\frac {2 \sqrt {1+(a+b x)^2} \sinh ^{-1}(a+b x)}{b}+\frac {(a+b x) \sinh ^{-1}(a+b x)^2}{b} \]

[Out]

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

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Rubi [A]
time = 0.04, antiderivative size = 45, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {5858, 5772, 5798, 8} \begin {gather*} \frac {(a+b x) \sinh ^{-1}(a+b x)^2}{b}-\frac {2 \sqrt {(a+b x)^2+1} \sinh ^{-1}(a+b x)}{b}+2 x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[ArcSinh[a + b*x]^2,x]

[Out]

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

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 5772

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

Rule 5798

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

Rule 5858

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

Rubi steps

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

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Mathematica [A]
time = 0.02, size = 47, normalized size = 1.04 \begin {gather*} \frac {2 (a+b x)-2 \sqrt {1+(a+b x)^2} \sinh ^{-1}(a+b x)+(a+b x) \sinh ^{-1}(a+b x)^2}{b} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[ArcSinh[a + b*x]^2,x]

[Out]

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

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Maple [A]
time = 1.86, size = 46, normalized size = 1.02

method	result	size
derivativedivides	\(\frac {\arcsinh \left (b x +a \right )^{2} \left (b x +a \right )-2 \arcsinh \left (b x +a \right ) \sqrt {1+\left (b x +a \right )^{2}}+2 b x +2 a}{b}\)	\(46\)
default	\(\frac {\arcsinh \left (b x +a \right )^{2} \left (b x +a \right )-2 \arcsinh \left (b x +a \right ) \sqrt {1+\left (b x +a \right )^{2}}+2 b x +2 a}{b}\)	\(46\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arcsinh(b*x+a)^2,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 88 vs. \(2 (43) = 86\).
time = 0.39, size = 88, normalized size = 1.96 \begin {gather*} \frac {{\left (b x + a\right )} \log \left (b x + a + \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right )^{2} + 2 \, b x - 2 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} \log \left (b x + a + \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right )}{b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((b*x + a)*log(b*x + a + sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1))^2 + 2*b*x - 2*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)*lo
g(b*x + a + sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)))/b

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Sympy [A]
time = 0.09, size = 63, normalized size = 1.40 \begin {gather*} \begin {cases} \frac {a \operatorname {asinh}^{2}{\left (a + b x \right )}}{b} + x \operatorname {asinh}^{2}{\left (a + b x \right )} + 2 x - \frac {2 \sqrt {a^{2} + 2 a b x + b^{2} x^{2} + 1} \operatorname {asinh}{\left (a + b x \right )}}{b} & \text {for}\: b \neq 0 \\x \operatorname {asinh}^{2}{\left (a \right )} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((a*asinh(a + b*x)**2/b + x*asinh(a + b*x)**2 + 2*x - 2*sqrt(a**2 + 2*a*b*x + b**2*x**2 + 1)*asinh(a
+ b*x)/b, Ne(b, 0)), (x*asinh(a)**2, True))

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(arcsinh(b*x + a)^2, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int {\mathrm {asinh}\left (a+b\,x\right )}^2 \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(asinh(a + b*x)^2,x)

[Out]

int(asinh(a + b*x)^2, x)

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