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+1/4*(b*x+a)^2/b^2+1/4*arcsinh(b*x+a)^2/b^2-1/2*a^2*arcsinh(b*x+a)^2/b^2+1/2*x^2*arcsinh(b*x+a)^2+2*a*
arcsinh(b*x+a)*(1+(b*x+a)^2)^(1/2)/b^2-1/2*(b*x+a)*arcsinh(b*x+a)*(1+(b*x+a)^2)^(1/2)/b^2

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Rubi [A]  time = 0.24, antiderivative size = 126, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 8, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.800, 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 8

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

Rule 30

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

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 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 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 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 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*}

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Mathematica [A]  time = 0.08, 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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fricas [A]  time = 0.57, size = 114, normalized size = 0.90 \[ \frac {b^{2} x^{2} - 6 \, a b x + {\left (2 \, b^{2} x^{2} - 2 \, a^{2} + 1\right )} \log \left (b x + a + \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right )^{2} - 2 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} {\left (b x - 3 \, a\right )} \log \left (b x + a + \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right )}{4 \, b^{2}} \]

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

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)

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maple [A]  time = 0.08, size = 113, normalized size = 0.90 \[ \frac {\frac {\arcsinh \left (b x +a \right )^{2} \left (1+\left (b x +a \right )^{2}\right )}{2}-\frac {\arcsinh \left (b x +a \right ) \sqrt {1+\left (b x +a \right )^{2}}\, \left (b x +a \right )}{2}-\frac {\arcsinh \left (b x +a \right )^{2}}{4}+\frac {\left (b x +a \right )^{2}}{4}+\frac {1}{4}-a \left (\arcsinh \left (b x +a \right )^{2} \left (b x +a \right )-2 \arcsinh \left (b x +a \right ) \sqrt {1+\left (b x +a \right )^{2}}+2 b x +2 a \right )}{b^{2}} \]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*x^2*log(b*x + a + sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1))^2 - integrate((b^3*x^4 + 2*a*b^2*x^3 + (a^2*b + b)*x^
2 + (b^2*x^3 + a*b*x^2)*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1))*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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x\,{\mathrm {asinh}\left (a+b\,x\right )}^2 \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*asinh(a + b*x)^2,x)

[Out]

int(x*asinh(a + b*x)^2, x)

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sympy [A]  time = 0.68, size = 138, normalized size = 1.10 \[ \begin {cases} - \frac {a^{2} \operatorname {asinh}^{2}{\left (a + b x \right )}}{2 b^{2}} - \frac {3 a x}{2 b} + \frac {3 a \sqrt {a^{2} + 2 a b x + b^{2} x^{2} + 1} \operatorname {asinh}{\left (a + b x \right )}}{2 b^{2}} + \frac {x^{2} \operatorname {asinh}^{2}{\left (a + b x \right )}}{2} + \frac {x^{2}}{4} - \frac {x \sqrt {a^{2} + 2 a b x + b^{2} x^{2} + 1} \operatorname {asinh}{\left (a + b x \right )}}{2 b} + \frac {\operatorname {asinh}^{2}{\left (a + b x \right )}}{4 b^{2}} & \text {for}\: b \neq 0 \\\frac {x^{2} \operatorname {asinh}^{2}{\relax (a )}}{2} & \text {otherwise} \end {cases} \]

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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