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

Optimal. Leaf size=76 \[ \frac {\left (1-2 a^2\right ) \sinh ^{-1}(a+b x)}{4 b^2}+\frac {3 a \sqrt {(a+b x)^2+1}}{4 b^2}+\frac {1}{2} x^2 \sinh ^{-1}(a+b x)-\frac {x \sqrt {(a+b x)^2+1}}{4 b} \]

[Out]

1/4*(-2*a^2+1)*arcsinh(b*x+a)/b^2+1/2*x^2*arcsinh(b*x+a)+3/4*a*(1+(b*x+a)^2)^(1/2)/b^2-1/4*x*(1+(b*x+a)^2)^(1/
2)/b

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Rubi [A]  time = 0.07, antiderivative size = 76, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.625, Rules used = {5865, 5801, 743, 641, 215} \[ \frac {\left (1-2 a^2\right ) \sinh ^{-1}(a+b x)}{4 b^2}+\frac {3 a \sqrt {(a+b x)^2+1}}{4 b^2}+\frac {1}{2} x^2 \sinh ^{-1}(a+b x)-\frac {x \sqrt {(a+b x)^2+1}}{4 b} \]

Antiderivative was successfully verified.

[In]

Int[x*ArcSinh[a + b*x],x]

[Out]

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

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 641

Int[((d_) + (e_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(e*(a + c*x^2)^(p + 1))/(2*c*(p + 1)),
x] + Dist[d, Int[(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, p}, x] && NeQ[p, -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 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 \sinh ^{-1}(a+b x) \, dx &=\frac {\operatorname {Subst}\left (\int \left (-\frac {a}{b}+\frac {x}{b}\right ) \sinh ^{-1}(x) \, dx,x,a+b x\right )}{b}\\ &=\frac {1}{2} x^2 \sinh ^{-1}(a+b x)-\frac {1}{2} \operatorname {Subst}\left (\int \frac {\left (-\frac {a}{b}+\frac {x}{b}\right )^2}{\sqrt {1+x^2}} \, dx,x,a+b x\right )\\ &=-\frac {x \sqrt {1+(a+b x)^2}}{4 b}+\frac {1}{2} x^2 \sinh ^{-1}(a+b x)-\frac {1}{4} \operatorname {Subst}\left (\int \frac {-\frac {1-2 a^2}{b^2}-\frac {3 a x}{b^2}}{\sqrt {1+x^2}} \, dx,x,a+b x\right )\\ &=\frac {3 a \sqrt {1+(a+b x)^2}}{4 b^2}-\frac {x \sqrt {1+(a+b x)^2}}{4 b}+\frac {1}{2} x^2 \sinh ^{-1}(a+b x)+\frac {\left (1-2 a^2\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {1+x^2}} \, dx,x,a+b x\right )}{4 b^2}\\ &=\frac {3 a \sqrt {1+(a+b x)^2}}{4 b^2}-\frac {x \sqrt {1+(a+b x)^2}}{4 b}+\frac {\left (1-2 a^2\right ) \sinh ^{-1}(a+b x)}{4 b^2}+\frac {1}{2} x^2 \sinh ^{-1}(a+b x)\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 60, normalized size = 0.79 \[ \frac {\sqrt {a^2+2 a b x+b^2 x^2+1} (3 a-b x)+\left (-2 a^2+2 b^2 x^2+1\right ) \sinh ^{-1}(a+b x)}{4 b^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x*ArcSinh[a + b*x],x]

[Out]

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

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fricas [A]  time = 0.53, size = 75, normalized size = 0.99 \[ \frac {{\left (2 \, b^{2} x^{2} - 2 \, a^{2} + 1\right )} \log \left (b x + a + \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right ) - \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} {\left (b x - 3 \, a\right )}}{4 \, b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.36, size = 111, normalized size = 1.46 \[ \frac {1}{2} \, x^{2} \log \left (b x + a + \sqrt {{\left (b x + a\right )}^{2} + 1}\right ) - \frac {1}{4} \, {\left (\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} {\left (\frac {x}{b^{2}} - \frac {3 \, a}{b^{3}}\right )} - \frac {{\left (2 \, a^{2} - 1\right )} \log \left (-a b - {\left (x {\left | b \right |} - \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right )} {\left | b \right |}\right )}{b^{2} {\left | b \right |}}\right )} b \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.00, size = 74, normalized size = 0.97 \[ \frac {\frac {\arcsinh \left (b x +a \right ) \left (b x +a \right )^{2}}{2}-\arcsinh \left (b x +a \right ) a \left (b x +a \right )-\frac {\left (b x +a \right ) \sqrt {1+\left (b x +a \right )^{2}}}{4}+\frac {\arcsinh \left (b x +a \right )}{4}+a \sqrt {1+\left (b x +a \right )^{2}}}{b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.37, size = 149, normalized size = 1.96 \[ \frac {1}{2} \, x^{2} \operatorname {arsinh}\left (b x + a\right ) - \frac {1}{4} \, b {\left (\frac {3 \, a^{2} \operatorname {arsinh}\left (\frac {2 \, {\left (b^{2} x + a b\right )}}{\sqrt {-4 \, a^{2} b^{2} + 4 \, {\left (a^{2} + 1\right )} b^{2}}}\right )}{b^{3}} + \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} x}{b^{2}} - \frac {{\left (a^{2} + 1\right )} \operatorname {arsinh}\left (\frac {2 \, {\left (b^{2} x + a b\right )}}{\sqrt {-4 \, a^{2} b^{2} + 4 \, {\left (a^{2} + 1\right )} b^{2}}}\right )}{b^{3}} - \frac {3 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} a}{b^{3}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*asinh(a + b*x),x)

[Out]

int(x*asinh(a + b*x), x)

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sympy [A]  time = 0.30, size = 104, normalized size = 1.37 \[ \begin {cases} - \frac {a^{2} \operatorname {asinh}{\left (a + b x \right )}}{2 b^{2}} + \frac {3 a \sqrt {a^{2} + 2 a b x + b^{2} x^{2} + 1}}{4 b^{2}} + \frac {x^{2} \operatorname {asinh}{\left (a + b x \right )}}{2} - \frac {x \sqrt {a^{2} + 2 a b x + b^{2} x^{2} + 1}}{4 b} + \frac {\operatorname {asinh}{\left (a + b x \right )}}{4 b^{2}} & \text {for}\: b \neq 0 \\\frac {x^{2} \operatorname {asinh}{\relax (a )}}{2} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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