3.58 \(\int x^3 \sinh ^{-1}(a+b x) \, dx\)

Optimal. Leaf size=131 \[ -\frac{\left (4 a \left (16-19 a^2\right )-\left (9-26 a^2\right ) (a+b x)\right ) \sqrt{(a+b x)^2+1}}{96 b^4}-\frac{\left (8 a^4-24 a^2+3\right ) \sinh ^{-1}(a+b x)}{32 b^4}+\frac{7 a x^2 \sqrt{(a+b x)^2+1}}{48 b^2}-\frac{x^3 \sqrt{(a+b x)^2+1}}{16 b}+\frac{1}{4} x^4 \sinh ^{-1}(a+b x) \]

[Out]

(7*a*x^2*Sqrt[1 + (a + b*x)^2])/(48*b^2) - (x^3*Sqrt[1 + (a + b*x)^2])/(16*b) - ((4*a*(16 - 19*a^2) - (9 - 26*
a^2)*(a + b*x))*Sqrt[1 + (a + b*x)^2])/(96*b^4) - ((3 - 24*a^2 + 8*a^4)*ArcSinh[a + b*x])/(32*b^4) + (x^4*ArcS
inh[a + b*x])/4

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Rubi [A]  time = 0.1722, antiderivative size = 131, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.6, Rules used = {5865, 5801, 743, 833, 780, 215} \[ -\frac{\left (4 a \left (16-19 a^2\right )-\left (9-26 a^2\right ) (a+b x)\right ) \sqrt{(a+b x)^2+1}}{96 b^4}-\frac{\left (8 a^4-24 a^2+3\right ) \sinh ^{-1}(a+b x)}{32 b^4}+\frac{7 a x^2 \sqrt{(a+b x)^2+1}}{48 b^2}-\frac{x^3 \sqrt{(a+b x)^2+1}}{16 b}+\frac{1}{4} x^4 \sinh ^{-1}(a+b x) \]

Antiderivative was successfully verified.

[In]

Int[x^3*ArcSinh[a + b*x],x]

[Out]

(7*a*x^2*Sqrt[1 + (a + b*x)^2])/(48*b^2) - (x^3*Sqrt[1 + (a + b*x)^2])/(16*b) - ((4*a*(16 - 19*a^2) - (9 - 26*
a^2)*(a + b*x))*Sqrt[1 + (a + b*x)^2])/(96*b^4) - ((3 - 24*a^2 + 8*a^4)*ArcSinh[a + b*x])/(32*b^4) + (x^4*ArcS
inh[a + b*x])/4

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 833

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(g*(d + e*x)
^m*(a + c*x^2)^(p + 1))/(c*(m + 2*p + 2)), x] + Dist[1/(c*(m + 2*p + 2)), Int[(d + e*x)^(m - 1)*(a + c*x^2)^p*
Simp[c*d*f*(m + 2*p + 2) - a*e*g*m + c*(e*f*(m + 2*p + 2) + d*g*m)*x, x], x], x] /; FreeQ[{a, c, d, e, f, g, p
}, x] && NeQ[c*d^2 + a*e^2, 0] && GtQ[m, 0] && NeQ[m + 2*p + 2, 0] && (IntegerQ[m] || IntegerQ[p] || IntegersQ
[2*m, 2*p]) &&  !(IGtQ[m, 0] && EqQ[f, 0])

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^3 \sinh ^{-1}(a+b x) \, dx &=\frac{\operatorname{Subst}\left (\int \left (-\frac{a}{b}+\frac{x}{b}\right )^3 \sinh ^{-1}(x) \, dx,x,a+b x\right )}{b}\\ &=\frac{1}{4} x^4 \sinh ^{-1}(a+b x)-\frac{1}{4} \operatorname{Subst}\left (\int \frac{\left (-\frac{a}{b}+\frac{x}{b}\right )^4}{\sqrt{1+x^2}} \, dx,x,a+b x\right )\\ &=-\frac{x^3 \sqrt{1+(a+b x)^2}}{16 b}+\frac{1}{4} x^4 \sinh ^{-1}(a+b x)-\frac{1}{16} \operatorname{Subst}\left (\int \frac{\left (-\frac{3-4 a^2}{b^2}-\frac{7 a x}{b^2}\right ) \left (-\frac{a}{b}+\frac{x}{b}\right )^2}{\sqrt{1+x^2}} \, dx,x,a+b x\right )\\ &=\frac{7 a x^2 \sqrt{1+(a+b x)^2}}{48 b^2}-\frac{x^3 \sqrt{1+(a+b x)^2}}{16 b}+\frac{1}{4} x^4 \sinh ^{-1}(a+b x)-\frac{1}{48} \operatorname{Subst}\left (\int \frac{\left (\frac{a \left (23-12 a^2\right )}{b^3}-\frac{\left (9-26 a^2\right ) x}{b^3}\right ) \left (-\frac{a}{b}+\frac{x}{b}\right )}{\sqrt{1+x^2}} \, dx,x,a+b x\right )\\ &=\frac{7 a x^2 \sqrt{1+(a+b x)^2}}{48 b^2}-\frac{x^3 \sqrt{1+(a+b x)^2}}{16 b}-\frac{\left (4 a \left (16-19 a^2\right )-\left (9-26 a^2\right ) (a+b x)\right ) \sqrt{1+(a+b x)^2}}{96 b^4}+\frac{1}{4} x^4 \sinh ^{-1}(a+b x)-\frac{\left (3-24 a^2+8 a^4\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{1+x^2}} \, dx,x,a+b x\right )}{32 b^4}\\ &=\frac{7 a x^2 \sqrt{1+(a+b x)^2}}{48 b^2}-\frac{x^3 \sqrt{1+(a+b x)^2}}{16 b}-\frac{\left (4 a \left (16-19 a^2\right )-\left (9-26 a^2\right ) (a+b x)\right ) \sqrt{1+(a+b x)^2}}{96 b^4}-\frac{\left (3-24 a^2+8 a^4\right ) \sinh ^{-1}(a+b x)}{32 b^4}+\frac{1}{4} x^4 \sinh ^{-1}(a+b x)\\ \end{align*}

Mathematica [A]  time = 0.0817239, size = 95, normalized size = 0.73 \[ \frac{\sqrt{a^2+2 a b x+b^2 x^2+1} \left (-26 a^2 b x+50 a^3+a \left (14 b^2 x^2-55\right )-6 b^3 x^3+9 b x\right )-3 \left (8 a^4-24 a^2-8 b^4 x^4+3\right ) \sinh ^{-1}(a+b x)}{96 b^4} \]

Antiderivative was successfully verified.

[In]

Integrate[x^3*ArcSinh[a + b*x],x]

[Out]

(Sqrt[1 + a^2 + 2*a*b*x + b^2*x^2]*(50*a^3 + 9*b*x - 26*a^2*b*x - 6*b^3*x^3 + a*(-55 + 14*b^2*x^2)) - 3*(3 - 2
4*a^2 + 8*a^4 - 8*b^4*x^4)*ArcSinh[a + b*x])/(96*b^4)

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Maple [A]  time = 0.014, size = 200, normalized size = 1.5 \begin{align*}{\frac{1}{{b}^{4}} \left ({\frac{{\it Arcsinh} \left ( bx+a \right ) \left ( bx+a \right ) ^{4}}{4}}-{\it Arcsinh} \left ( bx+a \right ) \left ( bx+a \right ) ^{3}a+{\frac{3\,{\it Arcsinh} \left ( bx+a \right ) \left ( bx+a \right ) ^{2}{a}^{2}}{2}}-{\it Arcsinh} \left ( bx+a \right ) \left ( bx+a \right ){a}^{3}-{\frac{ \left ( bx+a \right ) ^{3}}{16}\sqrt{1+ \left ( bx+a \right ) ^{2}}}+{\frac{3\,bx+3\,a}{32}\sqrt{1+ \left ( bx+a \right ) ^{2}}}-{\frac{3\,{\it Arcsinh} \left ( bx+a \right ) }{32}}+a \left ({\frac{ \left ( bx+a \right ) ^{2}}{3}\sqrt{1+ \left ( bx+a \right ) ^{2}}}-{\frac{2}{3}\sqrt{1+ \left ( bx+a \right ) ^{2}}} \right ) -{\frac{3\,{a}^{2}}{2} \left ({\frac{bx+a}{2}\sqrt{1+ \left ( bx+a \right ) ^{2}}}-{\frac{{\it Arcsinh} \left ( bx+a \right ) }{2}} \right ) }+{a}^{3}\sqrt{1+ \left ( bx+a \right ) ^{2}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*arcsinh(b*x+a),x)

[Out]

1/b^4*(1/4*arcsinh(b*x+a)*(b*x+a)^4-arcsinh(b*x+a)*(b*x+a)^3*a+3/2*arcsinh(b*x+a)*(b*x+a)^2*a^2-arcsinh(b*x+a)
*(b*x+a)*a^3-1/16*(b*x+a)^3*(1+(b*x+a)^2)^(1/2)+3/32*(b*x+a)*(1+(b*x+a)^2)^(1/2)-3/32*arcsinh(b*x+a)+a*(1/3*(b
*x+a)^2*(1+(b*x+a)^2)^(1/2)-2/3*(1+(b*x+a)^2)^(1/2))-3/2*a^2*(1/2*(b*x+a)*(1+(b*x+a)^2)^(1/2)-1/2*arcsinh(b*x+
a))+a^3*(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^3*arcsinh(b*x+a),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.8298, size = 262, normalized size = 2. \begin{align*} \frac{3 \,{\left (8 \, b^{4} x^{4} - 8 \, a^{4} + 24 \, a^{2} - 3\right )} \log \left (b x + a + \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right ) -{\left (6 \, b^{3} x^{3} - 14 \, a b^{2} x^{2} - 50 \, a^{3} +{\left (26 \, a^{2} - 9\right )} b x + 55 \, a\right )} \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{96 \, b^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*arcsinh(b*x+a),x, algorithm="fricas")

[Out]

1/96*(3*(8*b^4*x^4 - 8*a^4 + 24*a^2 - 3)*log(b*x + a + sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)) - (6*b^3*x^3 - 14*a*
b^2*x^2 - 50*a^3 + (26*a^2 - 9)*b*x + 55*a)*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1))/b^4

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Sympy [A]  time = 1.81439, size = 255, normalized size = 1.95 \begin{align*} \begin{cases} - \frac{a^{4} \operatorname{asinh}{\left (a + b x \right )}}{4 b^{4}} + \frac{25 a^{3} \sqrt{a^{2} + 2 a b x + b^{2} x^{2} + 1}}{48 b^{4}} - \frac{13 a^{2} x \sqrt{a^{2} + 2 a b x + b^{2} x^{2} + 1}}{48 b^{3}} + \frac{3 a^{2} \operatorname{asinh}{\left (a + b x \right )}}{4 b^{4}} + \frac{7 a x^{2} \sqrt{a^{2} + 2 a b x + b^{2} x^{2} + 1}}{48 b^{2}} - \frac{55 a \sqrt{a^{2} + 2 a b x + b^{2} x^{2} + 1}}{96 b^{4}} + \frac{x^{4} \operatorname{asinh}{\left (a + b x \right )}}{4} - \frac{x^{3} \sqrt{a^{2} + 2 a b x + b^{2} x^{2} + 1}}{16 b} + \frac{3 x \sqrt{a^{2} + 2 a b x + b^{2} x^{2} + 1}}{32 b^{3}} - \frac{3 \operatorname{asinh}{\left (a + b x \right )}}{32 b^{4}} & \text{for}\: b \neq 0 \\\frac{x^{4} \operatorname{asinh}{\left (a \right )}}{4} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*asinh(b*x+a),x)

[Out]

Piecewise((-a**4*asinh(a + b*x)/(4*b**4) + 25*a**3*sqrt(a**2 + 2*a*b*x + b**2*x**2 + 1)/(48*b**4) - 13*a**2*x*
sqrt(a**2 + 2*a*b*x + b**2*x**2 + 1)/(48*b**3) + 3*a**2*asinh(a + b*x)/(4*b**4) + 7*a*x**2*sqrt(a**2 + 2*a*b*x
+ b**2*x**2 + 1)/(48*b**2) - 55*a*sqrt(a**2 + 2*a*b*x + b**2*x**2 + 1)/(96*b**4) + x**4*asinh(a + b*x)/4 - x*
*3*sqrt(a**2 + 2*a*b*x + b**2*x**2 + 1)/(16*b) + 3*x*sqrt(a**2 + 2*a*b*x + b**2*x**2 + 1)/(32*b**3) - 3*asinh(
a + b*x)/(32*b**4), Ne(b, 0)), (x**4*asinh(a)/4, True))

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Giac [A]  time = 1.29298, size = 219, normalized size = 1.67 \begin{align*} \frac{1}{4} \, x^{4} \log \left (b x + a + \sqrt{{\left (b x + a\right )}^{2} + 1}\right ) - \frac{1}{96} \,{\left (\sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1}{\left ({\left (2 \, x{\left (\frac{3 \, x}{b^{2}} - \frac{7 \, a}{b^{3}}\right )} + \frac{26 \, a^{2} b^{3} - 9 \, b^{3}}{b^{7}}\right )} x - \frac{5 \,{\left (10 \, a^{3} b^{2} - 11 \, a b^{2}\right )}}{b^{7}}\right )} - \frac{3 \,{\left (8 \, a^{4} - 24 \, a^{2} + 3\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^{4}{\left | b \right |}}\right )} b \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*arcsinh(b*x+a),x, algorithm="giac")

[Out]

1/4*x^4*log(b*x + a + sqrt((b*x + a)^2 + 1)) - 1/96*(sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)*((2*x*(3*x/b^2 - 7*a/b^
3) + (26*a^2*b^3 - 9*b^3)/b^7)*x - 5*(10*a^3*b^2 - 11*a*b^2)/b^7) - 3*(8*a^4 - 24*a^2 + 3)*log(-a*b - (x*abs(b
) - sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1))*abs(b))/(b^4*abs(b)))*b