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

Optimal. Leaf size=203 \[ -\frac {a^2 \sinh ^{-1}(a+b x)^3}{2 b^2}-\frac {3 (a+b x) \sqrt {(a+b x)^2+1}}{8 b^2}+\frac {6 a \sqrt {(a+b x)^2+1}}{b^2}+\frac {\sinh ^{-1}(a+b x)^3}{4 b^2}-\frac {3 (a+b x) \sqrt {(a+b x)^2+1} \sinh ^{-1}(a+b x)^2}{4 b^2}+\frac {3 a \sqrt {(a+b x)^2+1} \sinh ^{-1}(a+b x)^2}{b^2}+\frac {3 (a+b x)^2 \sinh ^{-1}(a+b x)}{4 b^2}-\frac {6 a (a+b x) \sinh ^{-1}(a+b x)}{b^2}+\frac {3 \sinh ^{-1}(a+b x)}{8 b^2}+\frac {1}{2} x^2 \sinh ^{-1}(a+b x)^3 \]

[Out]

3/8*arcsinh(b*x+a)/b^2-6*a*(b*x+a)*arcsinh(b*x+a)/b^2+3/4*(b*x+a)^2*arcsinh(b*x+a)/b^2+1/4*arcsinh(b*x+a)^3/b^
2-1/2*a^2*arcsinh(b*x+a)^3/b^2+1/2*x^2*arcsinh(b*x+a)^3+6*a*(1+(b*x+a)^2)^(1/2)/b^2-3/8*(b*x+a)*(1+(b*x+a)^2)^
(1/2)/b^2+3*a*arcsinh(b*x+a)^2*(1+(b*x+a)^2)^(1/2)/b^2-3/4*(b*x+a)*arcsinh(b*x+a)^2*(1+(b*x+a)^2)^(1/2)/b^2

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Rubi [A]  time = 0.30, antiderivative size = 203, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 10, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.000, Rules used = {5865, 5801, 5831, 3317, 3296, 2638, 3311, 30, 2635, 8} \[ -\frac {a^2 \sinh ^{-1}(a+b x)^3}{2 b^2}-\frac {3 (a+b x) \sqrt {(a+b x)^2+1}}{8 b^2}+\frac {6 a \sqrt {(a+b x)^2+1}}{b^2}+\frac {\sinh ^{-1}(a+b x)^3}{4 b^2}-\frac {3 (a+b x) \sqrt {(a+b x)^2+1} \sinh ^{-1}(a+b x)^2}{4 b^2}+\frac {3 a \sqrt {(a+b x)^2+1} \sinh ^{-1}(a+b x)^2}{b^2}+\frac {3 (a+b x)^2 \sinh ^{-1}(a+b x)}{4 b^2}-\frac {6 a (a+b x) \sinh ^{-1}(a+b x)}{b^2}+\frac {3 \sinh ^{-1}(a+b x)}{8 b^2}+\frac {1}{2} x^2 \sinh ^{-1}(a+b x)^3 \]

Antiderivative was successfully verified.

[In]

Int[x*ArcSinh[a + b*x]^3,x]

[Out]

(6*a*Sqrt[1 + (a + b*x)^2])/b^2 - (3*(a + b*x)*Sqrt[1 + (a + b*x)^2])/(8*b^2) + (3*ArcSinh[a + b*x])/(8*b^2) -
(6*a*(a + b*x)*ArcSinh[a + b*x])/b^2 + (3*(a + b*x)^2*ArcSinh[a + b*x])/(4*b^2) + (3*a*Sqrt[1 + (a + b*x)^2]*
ArcSinh[a + b*x]^2)/b^2 - (3*(a + b*x)*Sqrt[1 + (a + b*x)^2]*ArcSinh[a + b*x]^2)/(4*b^2) + ArcSinh[a + b*x]^3/
(4*b^2) - (a^2*ArcSinh[a + b*x]^3)/(2*b^2) + (x^2*ArcSinh[a + b*x]^3)/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 2635

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]

Rule 2638

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Cos[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 3296

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> -Simp[((c + d*x)^m*Cos[e + f*x])/f, x] +
Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Cos[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]

Rule 3311

Int[((c_.) + (d_.)*(x_))^(m_)*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(d*m*(c + d*x)^(m - 1)*(
b*Sin[e + f*x])^n)/(f^2*n^2), x] + (Dist[(b^2*(n - 1))/n, Int[(c + d*x)^m*(b*Sin[e + f*x])^(n - 2), x], x] - D
ist[(d^2*m*(m - 1))/(f^2*n^2), Int[(c + d*x)^(m - 2)*(b*Sin[e + f*x])^n, x], x] - Simp[(b*(c + d*x)^m*Cos[e +
f*x]*(b*Sin[e + f*x])^(n - 1))/(f*n), x]) /; FreeQ[{b, c, d, e, f}, x] && GtQ[n, 1] && GtQ[m, 1]

Rule 3317

Int[((c_.) + (d_.)*(x_))^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Int[ExpandIntegrand[
(c + d*x)^m, (a + b*Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && IGtQ[n, 0] && (EqQ[n, 1] ||
IGtQ[m, 0] || NeQ[a^2 - b^2, 0])

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 5831

Int[(((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((f_) + (g_.)*(x_))^(m_.))/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol]
:> Dist[1/(c^(m + 1)*Sqrt[d]), Subst[Int[(a + b*x)^n*(c*f + g*Sinh[x])^m, x], x, ArcSinh[c*x]], x] /; FreeQ[{
a, b, c, d, e, f, g, n}, x] && EqQ[e, c^2*d] && IntegerQ[m] && GtQ[d, 0] && (GtQ[m, 0] || IGtQ[n, 0])

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)^3 \, dx &=\frac {\operatorname {Subst}\left (\int \left (-\frac {a}{b}+\frac {x}{b}\right ) \sinh ^{-1}(x)^3 \, dx,x,a+b x\right )}{b}\\ &=\frac {1}{2} x^2 \sinh ^{-1}(a+b x)^3-\frac {3}{2} \operatorname {Subst}\left (\int \frac {\left (-\frac {a}{b}+\frac {x}{b}\right )^2 \sinh ^{-1}(x)^2}{\sqrt {1+x^2}} \, dx,x,a+b x\right )\\ &=\frac {1}{2} x^2 \sinh ^{-1}(a+b x)^3-\frac {3}{2} \operatorname {Subst}\left (\int x^2 \left (-\frac {a}{b}+\frac {\sinh (x)}{b}\right )^2 \, dx,x,\sinh ^{-1}(a+b x)\right )\\ &=\frac {1}{2} x^2 \sinh ^{-1}(a+b x)^3-\frac {3}{2} \operatorname {Subst}\left (\int \left (\frac {a^2 x^2}{b^2}-\frac {2 a x^2 \sinh (x)}{b^2}+\frac {x^2 \sinh ^2(x)}{b^2}\right ) \, dx,x,\sinh ^{-1}(a+b x)\right )\\ &=-\frac {a^2 \sinh ^{-1}(a+b x)^3}{2 b^2}+\frac {1}{2} x^2 \sinh ^{-1}(a+b x)^3-\frac {3 \operatorname {Subst}\left (\int x^2 \sinh ^2(x) \, dx,x,\sinh ^{-1}(a+b x)\right )}{2 b^2}+\frac {(3 a) \operatorname {Subst}\left (\int x^2 \sinh (x) \, dx,x,\sinh ^{-1}(a+b x)\right )}{b^2}\\ &=\frac {3 (a+b x)^2 \sinh ^{-1}(a+b x)}{4 b^2}+\frac {3 a \sqrt {1+(a+b x)^2} \sinh ^{-1}(a+b x)^2}{b^2}-\frac {3 (a+b x) \sqrt {1+(a+b x)^2} \sinh ^{-1}(a+b x)^2}{4 b^2}-\frac {a^2 \sinh ^{-1}(a+b x)^3}{2 b^2}+\frac {1}{2} x^2 \sinh ^{-1}(a+b x)^3+\frac {3 \operatorname {Subst}\left (\int x^2 \, dx,x,\sinh ^{-1}(a+b x)\right )}{4 b^2}-\frac {3 \operatorname {Subst}\left (\int \sinh ^2(x) \, dx,x,\sinh ^{-1}(a+b x)\right )}{4 b^2}-\frac {(6 a) \operatorname {Subst}\left (\int x \cosh (x) \, dx,x,\sinh ^{-1}(a+b x)\right )}{b^2}\\ &=-\frac {3 (a+b x) \sqrt {1+(a+b x)^2}}{8 b^2}-\frac {6 a (a+b x) \sinh ^{-1}(a+b x)}{b^2}+\frac {3 (a+b x)^2 \sinh ^{-1}(a+b x)}{4 b^2}+\frac {3 a \sqrt {1+(a+b x)^2} \sinh ^{-1}(a+b x)^2}{b^2}-\frac {3 (a+b x) \sqrt {1+(a+b x)^2} \sinh ^{-1}(a+b x)^2}{4 b^2}+\frac {\sinh ^{-1}(a+b x)^3}{4 b^2}-\frac {a^2 \sinh ^{-1}(a+b x)^3}{2 b^2}+\frac {1}{2} x^2 \sinh ^{-1}(a+b x)^3+\frac {3 \operatorname {Subst}\left (\int 1 \, dx,x,\sinh ^{-1}(a+b x)\right )}{8 b^2}+\frac {(6 a) \operatorname {Subst}\left (\int \sinh (x) \, dx,x,\sinh ^{-1}(a+b x)\right )}{b^2}\\ &=\frac {6 a \sqrt {1+(a+b x)^2}}{b^2}-\frac {3 (a+b x) \sqrt {1+(a+b x)^2}}{8 b^2}+\frac {3 \sinh ^{-1}(a+b x)}{8 b^2}-\frac {6 a (a+b x) \sinh ^{-1}(a+b x)}{b^2}+\frac {3 (a+b x)^2 \sinh ^{-1}(a+b x)}{4 b^2}+\frac {3 a \sqrt {1+(a+b x)^2} \sinh ^{-1}(a+b x)^2}{b^2}-\frac {3 (a+b x) \sqrt {1+(a+b x)^2} \sinh ^{-1}(a+b x)^2}{4 b^2}+\frac {\sinh ^{-1}(a+b x)^3}{4 b^2}-\frac {a^2 \sinh ^{-1}(a+b x)^3}{2 b^2}+\frac {1}{2} x^2 \sinh ^{-1}(a+b x)^3\\ \end {align*}

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Mathematica [A]  time = 0.14, size = 129, normalized size = 0.64 \[ \frac {3 (15 a-b x) \sqrt {a^2+2 a b x+b^2 x^2+1}+\left (-4 a^2+4 b^2 x^2+2\right ) \sinh ^{-1}(a+b x)^3+6 (3 a-b x) \sqrt {a^2+2 a b x+b^2 x^2+1} \sinh ^{-1}(a+b x)^2+\left (-42 a^2-36 a b x+6 b^2 x^2+3\right ) \sinh ^{-1}(a+b x)}{8 b^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x*ArcSinh[a + b*x]^3,x]

[Out]

(3*(15*a - b*x)*Sqrt[1 + a^2 + 2*a*b*x + b^2*x^2] + (3 - 42*a^2 - 36*a*b*x + 6*b^2*x^2)*ArcSinh[a + b*x] + 6*(
3*a - b*x)*Sqrt[1 + a^2 + 2*a*b*x + b^2*x^2]*ArcSinh[a + b*x]^2 + (2 - 4*a^2 + 4*b^2*x^2)*ArcSinh[a + b*x]^3)/
(8*b^2)

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fricas [A]  time = 0.50, size = 180, normalized size = 0.89 \[ \frac {2 \, {\left (2 \, b^{2} x^{2} - 2 \, a^{2} + 1\right )} \log \left (b x + a + \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right )^{3} - 6 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} {\left (b x - 3 \, a\right )} \log \left (b x + a + \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right )^{2} + 3 \, {\left (2 \, b^{2} x^{2} - 12 \, a b x - 14 \, a^{2} + 1\right )} \log \left (b x + a + \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right ) - 3 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} {\left (b x - 15 \, a\right )}}{8 \, b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(2*(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))^3 - 6*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))^2 + 3*(2*b^2*x^2 - 12*a*b*x - 14*a^2 +
1)*log(b*x + a + sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)) - 3*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)*(b*x - 15*a))/b^2

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x \operatorname {arsinh}\left (b x + a\right )^{3}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x*arcsinh(b*x + a)^3, x)

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maple [A]  time = 0.08, size = 169, normalized size = 0.83 \[ \frac {\frac {\arcsinh \left (b x +a \right )^{3} \left (1+\left (b x +a \right )^{2}\right )}{2}-\frac {3 \arcsinh \left (b x +a \right )^{2} \sqrt {1+\left (b x +a \right )^{2}}\, \left (b x +a \right )}{4}-\frac {\arcsinh \left (b x +a \right )^{3}}{4}+\frac {3 \arcsinh \left (b x +a \right ) \left (1+\left (b x +a \right )^{2}\right )}{4}-\frac {3 \left (b x +a \right ) \sqrt {1+\left (b x +a \right )^{2}}}{8}-\frac {3 \arcsinh \left (b x +a \right )}{8}-a \left (\arcsinh \left (b x +a \right )^{3} \left (b x +a \right )-3 \arcsinh \left (b x +a \right )^{2} \sqrt {1+\left (b x +a \right )^{2}}+6 \left (b x +a \right ) \arcsinh \left (b x +a \right )-6 \sqrt {1+\left (b x +a \right )^{2}}\right )}{b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/b^2*(1/2*arcsinh(b*x+a)^3*(1+(b*x+a)^2)-3/4*arcsinh(b*x+a)^2*(1+(b*x+a)^2)^(1/2)*(b*x+a)-1/4*arcsinh(b*x+a)^
3+3/4*arcsinh(b*x+a)*(1+(b*x+a)^2)-3/8*(b*x+a)*(1+(b*x+a)^2)^(1/2)-3/8*arcsinh(b*x+a)-a*(arcsinh(b*x+a)^3*(b*x
+a)-3*arcsinh(b*x+a)^2*(1+(b*x+a)^2)^(1/2)+6*(b*x+a)*arcsinh(b*x+a)-6*(1+(b*x+a)^2)^(1/2)))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*asinh(a + b*x)^3,x)

[Out]

int(x*asinh(a + b*x)^3, x)

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sympy [A]  time = 1.29, size = 248, normalized size = 1.22 \[ \begin {cases} - \frac {a^{2} \operatorname {asinh}^{3}{\left (a + b x \right )}}{2 b^{2}} - \frac {21 a^{2} \operatorname {asinh}{\left (a + b x \right )}}{4 b^{2}} - \frac {9 a x \operatorname {asinh}{\left (a + b x \right )}}{2 b} + \frac {9 a \sqrt {a^{2} + 2 a b x + b^{2} x^{2} + 1} \operatorname {asinh}^{2}{\left (a + b x \right )}}{4 b^{2}} + \frac {45 a \sqrt {a^{2} + 2 a b x + b^{2} x^{2} + 1}}{8 b^{2}} + \frac {x^{2} \operatorname {asinh}^{3}{\left (a + b x \right )}}{2} + \frac {3 x^{2} \operatorname {asinh}{\left (a + b x \right )}}{4} - \frac {3 x \sqrt {a^{2} + 2 a b x + b^{2} x^{2} + 1} \operatorname {asinh}^{2}{\left (a + b x \right )}}{4 b} - \frac {3 x \sqrt {a^{2} + 2 a b x + b^{2} x^{2} + 1}}{8 b} + \frac {\operatorname {asinh}^{3}{\left (a + b x \right )}}{4 b^{2}} + \frac {3 \operatorname {asinh}{\left (a + b x \right )}}{8 b^{2}} & \text {for}\: b \neq 0 \\\frac {x^{2} \operatorname {asinh}^{3}{\relax (a )}}{2} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-a**2*asinh(a + b*x)**3/(2*b**2) - 21*a**2*asinh(a + b*x)/(4*b**2) - 9*a*x*asinh(a + b*x)/(2*b) + 9
*a*sqrt(a**2 + 2*a*b*x + b**2*x**2 + 1)*asinh(a + b*x)**2/(4*b**2) + 45*a*sqrt(a**2 + 2*a*b*x + b**2*x**2 + 1)
/(8*b**2) + x**2*asinh(a + b*x)**3/2 + 3*x**2*asinh(a + b*x)/4 - 3*x*sqrt(a**2 + 2*a*b*x + b**2*x**2 + 1)*asin
h(a + b*x)**2/(4*b) - 3*x*sqrt(a**2 + 2*a*b*x + b**2*x**2 + 1)/(8*b) + asinh(a + b*x)**3/(4*b**2) + 3*asinh(a
+ b*x)/(8*b**2), Ne(b, 0)), (x**2*asinh(a)**3/2, True))

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