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

Optimal. Leaf size=90 \[ \frac{\left (-11 a^2+5 a b x+4\right ) \sqrt{(a+b x)^2+1}}{18 b^3}-\frac{a \left (3-2 a^2\right ) \sinh ^{-1}(a+b x)}{6 b^3}-\frac{x^2 \sqrt{(a+b x)^2+1}}{9 b}+\frac{1}{3} x^3 \sinh ^{-1}(a+b x) \]

[Out]

-(x^2*Sqrt[1 + (a + b*x)^2])/(9*b) + ((4 - 11*a^2 + 5*a*b*x)*Sqrt[1 + (a + b*x)^2])/(18*b^3) - (a*(3 - 2*a^2)*
ArcSinh[a + b*x])/(6*b^3) + (x^3*ArcSinh[a + b*x])/3

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Rubi [A]  time = 0.112533, antiderivative size = 90, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {5865, 5801, 743, 780, 215} \[ \frac{\left (-11 a^2+5 a b x+4\right ) \sqrt{(a+b x)^2+1}}{18 b^3}-\frac{a \left (3-2 a^2\right ) \sinh ^{-1}(a+b x)}{6 b^3}-\frac{x^2 \sqrt{(a+b x)^2+1}}{9 b}+\frac{1}{3} x^3 \sinh ^{-1}(a+b x) \]

Antiderivative was successfully verified.

[In]

Int[x^2*ArcSinh[a + b*x],x]

[Out]

-(x^2*Sqrt[1 + (a + b*x)^2])/(9*b) + ((4 - 11*a^2 + 5*a*b*x)*Sqrt[1 + (a + b*x)^2])/(18*b^3) - (a*(3 - 2*a^2)*
ArcSinh[a + b*x])/(6*b^3) + (x^3*ArcSinh[a + b*x])/3

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 743

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)^(m - 1)*(a + c*x^2)^(p
+ 1))/(c*(m + 2*p + 1)), x] + Dist[1/(c*(m + 2*p + 1)), Int[(d + e*x)^(m - 2)*Simp[c*d^2*(m + 2*p + 1) - a*e^
2*(m - 1) + 2*c*d*e*(m + p)*x, x]*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, m, p}, x] && NeQ[c*d^2 + a*e^2,
0] && If[RationalQ[m], GtQ[m, 1], SumSimplerQ[m, -2]] && NeQ[m + 2*p + 1, 0] && IntQuadraticQ[a, 0, c, d, e, m
, p, x]

Rule 780

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(((e*f + d*g)*(2*p
+ 3) + 2*e*g*(p + 1)*x)*(a + c*x^2)^(p + 1))/(2*c*(p + 1)*(2*p + 3)), x] - Dist[(a*e*g - c*d*f*(2*p + 3))/(c*
(2*p + 3)), Int[(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, p}, x] &&  !LeQ[p, -1]

Rule 215

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSinh[(Rt[b, 2]*x)/Sqrt[a]]/Rt[b, 2], x] /; FreeQ[{a, b},
x] && GtQ[a, 0] && PosQ[b]

Rubi steps

\begin{align*} \int x^2 \sinh ^{-1}(a+b x) \, dx &=\frac{\operatorname{Subst}\left (\int \left (-\frac{a}{b}+\frac{x}{b}\right )^2 \sinh ^{-1}(x) \, dx,x,a+b x\right )}{b}\\ &=\frac{1}{3} x^3 \sinh ^{-1}(a+b x)-\frac{1}{3} \operatorname{Subst}\left (\int \frac{\left (-\frac{a}{b}+\frac{x}{b}\right )^3}{\sqrt{1+x^2}} \, dx,x,a+b x\right )\\ &=-\frac{x^2 \sqrt{1+(a+b x)^2}}{9 b}+\frac{1}{3} x^3 \sinh ^{-1}(a+b x)-\frac{1}{9} \operatorname{Subst}\left (\int \frac{\left (-\frac{2-3 a^2}{b^2}-\frac{5 a x}{b^2}\right ) \left (-\frac{a}{b}+\frac{x}{b}\right )}{\sqrt{1+x^2}} \, dx,x,a+b x\right )\\ &=-\frac{x^2 \sqrt{1+(a+b x)^2}}{9 b}+\frac{\left (4-11 a^2+5 a b x\right ) \sqrt{1+(a+b x)^2}}{18 b^3}+\frac{1}{3} x^3 \sinh ^{-1}(a+b x)-\frac{\left (a \left (3-2 a^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{1+x^2}} \, dx,x,a+b x\right )}{6 b^3}\\ &=-\frac{x^2 \sqrt{1+(a+b x)^2}}{9 b}+\frac{\left (4-11 a^2+5 a b x\right ) \sqrt{1+(a+b x)^2}}{18 b^3}-\frac{a \left (3-2 a^2\right ) \sinh ^{-1}(a+b x)}{6 b^3}+\frac{1}{3} x^3 \sinh ^{-1}(a+b x)\\ \end{align*}

Mathematica [A]  time = 0.0562991, size = 74, normalized size = 0.82 \[ \frac{\sqrt{a^2+2 a b x+b^2 x^2+1} \left (-11 a^2+5 a b x-2 b^2 x^2+4\right )+\left (6 a^3-9 a+6 b^3 x^3\right ) \sinh ^{-1}(a+b x)}{18 b^3} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2*ArcSinh[a + b*x],x]

[Out]

((4 - 11*a^2 + 5*a*b*x - 2*b^2*x^2)*Sqrt[1 + a^2 + 2*a*b*x + b^2*x^2] + (-9*a + 6*a^3 + 6*b^3*x^3)*ArcSinh[a +
b*x])/(18*b^3)

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Maple [A]  time = 0.004, size = 130, normalized size = 1.4 \begin{align*}{\frac{1}{{b}^{3}} \left ({\frac{{\it Arcsinh} \left ( bx+a \right ) \left ( bx+a \right ) ^{3}}{3}}-{\it Arcsinh} \left ( bx+a \right ) \left ( bx+a \right ) ^{2}a+{\it Arcsinh} \left ( bx+a \right ) \left ( bx+a \right ){a}^{2}-{\frac{ \left ( bx+a \right ) ^{2}}{9}\sqrt{1+ \left ( bx+a \right ) ^{2}}}+{\frac{2}{9}\sqrt{1+ \left ( bx+a \right ) ^{2}}}+a \left ({\frac{bx+a}{2}\sqrt{1+ \left ( bx+a \right ) ^{2}}}-{\frac{{\it Arcsinh} \left ( bx+a \right ) }{2}} \right ) -{a}^{2}\sqrt{1+ \left ( bx+a \right ) ^{2}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/b^3*(1/3*arcsinh(b*x+a)*(b*x+a)^3-arcsinh(b*x+a)*(b*x+a)^2*a+arcsinh(b*x+a)*(b*x+a)*a^2-1/9*(b*x+a)^2*(1+(b*
x+a)^2)^(1/2)+2/9*(1+(b*x+a)^2)^(1/2)+a*(1/2*(b*x+a)*(1+(b*x+a)^2)^(1/2)-1/2*arcsinh(b*x+a))-a^2*(1+(b*x+a)^2)
^(1/2))

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.56749, size = 216, normalized size = 2.4 \begin{align*} \frac{3 \,{\left (2 \, b^{3} x^{3} + 2 \, a^{3} - 3 \, a\right )} \log \left (b x + a + \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right ) -{\left (2 \, b^{2} x^{2} - 5 \, a b x + 11 \, a^{2} - 4\right )} \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{18 \, b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/18*(3*(2*b^3*x^3 + 2*a^3 - 3*a)*log(b*x + a + sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)) - (2*b^2*x^2 - 5*a*b*x + 11
*a^2 - 4)*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1))/b^3

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Sympy [A]  time = 0.844151, size = 170, normalized size = 1.89 \begin{align*} \begin{cases} \frac{a^{3} \operatorname{asinh}{\left (a + b x \right )}}{3 b^{3}} - \frac{11 a^{2} \sqrt{a^{2} + 2 a b x + b^{2} x^{2} + 1}}{18 b^{3}} + \frac{5 a x \sqrt{a^{2} + 2 a b x + b^{2} x^{2} + 1}}{18 b^{2}} - \frac{a \operatorname{asinh}{\left (a + b x \right )}}{2 b^{3}} + \frac{x^{3} \operatorname{asinh}{\left (a + b x \right )}}{3} - \frac{x^{2} \sqrt{a^{2} + 2 a b x + b^{2} x^{2} + 1}}{9 b} + \frac{2 \sqrt{a^{2} + 2 a b x + b^{2} x^{2} + 1}}{9 b^{3}} & \text{for}\: b \neq 0 \\\frac{x^{3} \operatorname{asinh}{\left (a \right )}}{3} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((a**3*asinh(a + b*x)/(3*b**3) - 11*a**2*sqrt(a**2 + 2*a*b*x + b**2*x**2 + 1)/(18*b**3) + 5*a*x*sqrt(
a**2 + 2*a*b*x + b**2*x**2 + 1)/(18*b**2) - a*asinh(a + b*x)/(2*b**3) + x**3*asinh(a + b*x)/3 - x**2*sqrt(a**2
+ 2*a*b*x + b**2*x**2 + 1)/(9*b) + 2*sqrt(a**2 + 2*a*b*x + b**2*x**2 + 1)/(9*b**3), Ne(b, 0)), (x**3*asinh(a)
/3, True))

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Giac [A]  time = 1.30377, size = 177, normalized size = 1.97 \begin{align*} \frac{1}{3} \, x^{3} \log \left (b x + a + \sqrt{{\left (b x + a\right )}^{2} + 1}\right ) - \frac{1}{18} \,{\left (\sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1}{\left (x{\left (\frac{2 \, x}{b^{2}} - \frac{5 \, a}{b^{3}}\right )} + \frac{11 \, a^{2} b - 4 \, b}{b^{5}}\right )} + \frac{3 \,{\left (2 \, a^{3} - 3 \, a\right )} \log \left (-a b -{\left (x{\left | b \right |} - \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right )}{\left | b \right |}\right )}{b^{3}{\left | b \right |}}\right )} b \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*x^3*log(b*x + a + sqrt((b*x + a)^2 + 1)) - 1/18*(sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)*(x*(2*x/b^2 - 5*a/b^3)
+ (11*a^2*b - 4*b)/b^5) + 3*(2*a^3 - 3*a)*log(-a*b - (x*abs(b) - sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1))*abs(b))/(b
^3*abs(b)))*b