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

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

[Out]

-1/6*a*(-2*a^2+3)*arcsinh(b*x+a)/b^3+1/3*x^3*arcsinh(b*x+a)-1/9*x^2*(1+(b*x+a)^2)^(1/2)/b+1/18*(5*a*b*x-11*a^2
+4)*(1+(b*x+a)^2)^(1/2)/b^3

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Rubi [A]  time = 0.11, antiderivative size = 90, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, 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 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]

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 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 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^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*}

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Mathematica [A]  time = 0.06, size = 74, normalized size = 0.82 \[ \frac {\left (6 a^3-9 a+6 b^3 x^3\right ) \sinh ^{-1}(a+b x)+\sqrt {a^2+2 a b x+b^2 x^2+1} \left (-11 a^2+5 a b x-2 b^2 x^2+4\right )}{18 b^3} \]

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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fricas [A]  time = 0.55, size = 91, normalized size = 1.01 \[ \frac {3 \, {\left (2 \, b^{3} x^{3} + 2 \, a^{3} - 3 \, a\right )} \log \left (b x + a + \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right ) - {\left (2 \, b^{2} x^{2} - 5 \, a b x + 11 \, a^{2} - 4\right )} \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{18 \, b^{3}} \]

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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giac [A]  time = 0.33, size = 131, normalized size = 1.46 \[ \frac {1}{3} \, x^{3} \log \left (b x + a + \sqrt {{\left (b x + a\right )}^{2} + 1}\right ) - \frac {1}{18} \, {\left (\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} {\left (x {\left (\frac {2 \, x}{b^{2}} - \frac {5 \, a}{b^{3}}\right )} + \frac {11 \, a^{2} b - 4 \, b}{b^{5}}\right )} + \frac {3 \, {\left (2 \, a^{3} - 3 \, a\right )} \log \left (-a b - {\left (x {\left | b \right |} - \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right )} {\left | b \right |}\right )}{b^{3} {\left | b \right |}}\right )} b \]

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

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maple [A]  time = 0.00, size = 130, normalized size = 1.44 \[ \frac {\frac {\arcsinh \left (b x +a \right ) \left (b x +a \right )^{3}}{3}-\arcsinh \left (b x +a \right ) \left (b x +a \right )^{2} a +\arcsinh \left (b x +a \right ) \left (b x +a \right ) a^{2}-\frac {\left (b x +a \right )^{2} \sqrt {1+\left (b x +a \right )^{2}}}{9}+\frac {2 \sqrt {1+\left (b x +a \right )^{2}}}{9}+a \left (\frac {\left (b x +a \right ) \sqrt {1+\left (b x +a \right )^{2}}}{2}-\frac {\arcsinh \left (b x +a \right )}{2}\right )-a^{2} \sqrt {1+\left (b x +a \right )^{2}}}{b^{3}} \]

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 [B]  time = 0.36, size = 210, normalized size = 2.33 \[ \frac {1}{3} \, x^{3} \operatorname {arsinh}\left (b x + a\right ) - \frac {1}{18} \, b {\left (\frac {2 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} x^{2}}{b^{2}} - \frac {15 \, a^{3} \operatorname {arsinh}\left (\frac {2 \, {\left (b^{2} x + a b\right )}}{\sqrt {-4 \, a^{2} b^{2} + 4 \, {\left (a^{2} + 1\right )} b^{2}}}\right )}{b^{4}} - \frac {5 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} a x}{b^{3}} + \frac {9 \, {\left (a^{2} + 1\right )} a \operatorname {arsinh}\left (\frac {2 \, {\left (b^{2} x + a b\right )}}{\sqrt {-4 \, a^{2} b^{2} + 4 \, {\left (a^{2} + 1\right )} b^{2}}}\right )}{b^{4}} + \frac {15 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} a^{2}}{b^{4}} - \frac {4 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} {\left (a^{2} + 1\right )}}{b^{4}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*x^3*arcsinh(b*x + a) - 1/18*b*(2*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)*x^2/b^2 - 15*a^3*arcsinh(2*(b^2*x + a*b
)/sqrt(-4*a^2*b^2 + 4*(a^2 + 1)*b^2))/b^4 - 5*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)*a*x/b^3 + 9*(a^2 + 1)*a*arcsin
h(2*(b^2*x + a*b)/sqrt(-4*a^2*b^2 + 4*(a^2 + 1)*b^2))/b^4 + 15*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)*a^2/b^4 - 4*s
qrt(b^2*x^2 + 2*a*b*x + a^2 + 1)*(a^2 + 1)/b^4)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*asinh(a + b*x),x)

[Out]

int(x^2*asinh(a + b*x), x)

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sympy [A]  time = 0.67, size = 170, normalized size = 1.89 \[ \begin {cases} \frac {a^{3} \operatorname {asinh}{\left (a + b x \right )}}{3 b^{3}} - \frac {11 a^{2} \sqrt {a^{2} + 2 a b x + b^{2} x^{2} + 1}}{18 b^{3}} + \frac {5 a x \sqrt {a^{2} + 2 a b x + b^{2} x^{2} + 1}}{18 b^{2}} - \frac {a \operatorname {asinh}{\left (a + b x \right )}}{2 b^{3}} + \frac {x^{3} \operatorname {asinh}{\left (a + b x \right )}}{3} - \frac {x^{2} \sqrt {a^{2} + 2 a b x + b^{2} x^{2} + 1}}{9 b} + \frac {2 \sqrt {a^{2} + 2 a b x + b^{2} x^{2} + 1}}{9 b^{3}} & \text {for}\: b \neq 0 \\\frac {x^{3} \operatorname {asinh}{\relax (a )}}{3} & \text {otherwise} \end {cases} \]

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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