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

Optimal. Leaf size=126 \[ -\frac{a^2 \sinh ^{-1}(a+b x)^2}{2 b^2}+\frac{(a+b x)^2}{4 b^2}-\frac{\sqrt{(a+b x)^2+1} (a+b x) \sinh ^{-1}(a+b x)}{2 b^2}+\frac{\sinh ^{-1}(a+b x)^2}{4 b^2}+\frac{2 a \sqrt{(a+b x)^2+1} \sinh ^{-1}(a+b x)}{b^2}+\frac{1}{2} x^2 \sinh ^{-1}(a+b x)^2-\frac{2 a x}{b} \]

[Out]

(-2*a*x)/b + (a + b*x)^2/(4*b^2) + (2*a*Sqrt[1 + (a + b*x)^2]*ArcSinh[a + b*x])/b^2 - ((a + b*x)*Sqrt[1 + (a +
b*x)^2]*ArcSinh[a + b*x])/(2*b^2) + ArcSinh[a + b*x]^2/(4*b^2) - (a^2*ArcSinh[a + b*x]^2)/(2*b^2) + (x^2*ArcS
inh[a + b*x]^2)/2

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Rubi [A]  time = 0.236425, antiderivative size = 126, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 8, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.8, Rules used = {5865, 5801, 5821, 5675, 5717, 8, 5758, 30} \[ -\frac{a^2 \sinh ^{-1}(a+b x)^2}{2 b^2}+\frac{(a+b x)^2}{4 b^2}-\frac{\sqrt{(a+b x)^2+1} (a+b x) \sinh ^{-1}(a+b x)}{2 b^2}+\frac{\sinh ^{-1}(a+b x)^2}{4 b^2}+\frac{2 a \sqrt{(a+b x)^2+1} \sinh ^{-1}(a+b x)}{b^2}+\frac{1}{2} x^2 \sinh ^{-1}(a+b x)^2-\frac{2 a x}{b} \]

Antiderivative was successfully verified.

[In]

Int[x*ArcSinh[a + b*x]^2,x]

[Out]

(-2*a*x)/b + (a + b*x)^2/(4*b^2) + (2*a*Sqrt[1 + (a + b*x)^2]*ArcSinh[a + b*x])/b^2 - ((a + b*x)*Sqrt[1 + (a +
b*x)^2]*ArcSinh[a + b*x])/(2*b^2) + ArcSinh[a + b*x]^2/(4*b^2) - (a^2*ArcSinh[a + b*x]^2)/(2*b^2) + (x^2*ArcS
inh[a + b*x]^2)/2

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
]

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 5821

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

Rule 5675

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

Rule 5717

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*d^IntPart[p]*(d + e*x^2)^FracPart[p])/(2*c*(p +
1)*(1 + c^2*x^2)^FracPart[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 8

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

Rule 5758

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

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rubi steps

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

Mathematica [A]  time = 0.0795007, size = 79, normalized size = 0.63 \[ \frac{\left (-2 a^2+2 b^2 x^2+1\right ) \sinh ^{-1}(a+b x)^2+2 (3 a-b x) \sqrt{a^2+2 a b x+b^2 x^2+1} \sinh ^{-1}(a+b x)+b x (b x-6 a)}{4 b^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x*ArcSinh[a + b*x]^2,x]

[Out]

(b*x*(-6*a + b*x) + 2*(3*a - b*x)*Sqrt[1 + a^2 + 2*a*b*x + b^2*x^2]*ArcSinh[a + b*x] + (1 - 2*a^2 + 2*b^2*x^2)
*ArcSinh[a + b*x]^2)/(4*b^2)

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Maple [A]  time = 0.04, size = 113, normalized size = 0.9 \begin{align*}{\frac{1}{{b}^{2}} \left ({\frac{ \left ({\it Arcsinh} \left ( bx+a \right ) \right ) ^{2} \left ( 1+ \left ( bx+a \right ) ^{2} \right ) }{2}}-{\frac{{\it Arcsinh} \left ( bx+a \right ) \left ( bx+a \right ) }{2}\sqrt{1+ \left ( bx+a \right ) ^{2}}}-{\frac{ \left ({\it Arcsinh} \left ( bx+a \right ) \right ) ^{2}}{4}}+{\frac{ \left ( bx+a \right ) ^{2}}{4}}+{\frac{1}{4}}-a \left ( \left ({\it Arcsinh} \left ( bx+a \right ) \right ) ^{2} \left ( bx+a \right ) -2\,{\it Arcsinh} \left ( bx+a \right ) \sqrt{1+ \left ( bx+a \right ) ^{2}}+2\,bx+2\,a \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/b^2*(1/2*arcsinh(b*x+a)^2*(1+(b*x+a)^2)-1/2*arcsinh(b*x+a)*(1+(b*x+a)^2)^(1/2)*(b*x+a)-1/4*arcsinh(b*x+a)^2+
1/4*(b*x+a)^2+1/4-a*(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(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.54278, size = 277, normalized size = 2.2 \begin{align*} \frac{b^{2} x^{2} - 6 \, a b x +{\left (2 \, b^{2} x^{2} - 2 \, a^{2} + 1\right )} \log \left (b x + a + \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right )^{2} - 2 \, \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1}{\left (b x - 3 \, a\right )} \log \left (b x + a + \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right )}{4 \, b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(b^2*x^2 - 6*a*b*x + (2*b^2*x^2 - 2*a^2 + 1)*log(b*x + a + sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1))^2 - 2*sqrt(b
^2*x^2 + 2*a*b*x + a^2 + 1)*(b*x - 3*a)*log(b*x + a + sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)))/b^2

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Sympy [A]  time = 0.785134, size = 138, normalized size = 1.1 \begin{align*} \begin{cases} - \frac{a^{2} \operatorname{asinh}^{2}{\left (a + b x \right )}}{2 b^{2}} - \frac{3 a x}{2 b} + \frac{3 a \sqrt{a^{2} + 2 a b x + b^{2} x^{2} + 1} \operatorname{asinh}{\left (a + b x \right )}}{2 b^{2}} + \frac{x^{2} \operatorname{asinh}^{2}{\left (a + b x \right )}}{2} + \frac{x^{2}}{4} - \frac{x \sqrt{a^{2} + 2 a b x + b^{2} x^{2} + 1} \operatorname{asinh}{\left (a + b x \right )}}{2 b} + \frac{\operatorname{asinh}^{2}{\left (a + b x \right )}}{4 b^{2}} & \text{for}\: b \neq 0 \\\frac{x^{2} \operatorname{asinh}^{2}{\left (a \right )}}{2} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-a**2*asinh(a + b*x)**2/(2*b**2) - 3*a*x/(2*b) + 3*a*sqrt(a**2 + 2*a*b*x + b**2*x**2 + 1)*asinh(a +
b*x)/(2*b**2) + x**2*asinh(a + b*x)**2/2 + x**2/4 - x*sqrt(a**2 + 2*a*b*x + b**2*x**2 + 1)*asinh(a + b*x)/(2*
b) + asinh(a + b*x)**2/(4*b**2), Ne(b, 0)), (x**2*asinh(a)**2/2, True))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x*arcsinh(b*x + a)^2, x)