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

Optimal. Leaf size=100 \[ 6 a b^2 x-\frac{3 b \sqrt{(c+d x)^2+1} \left (a+b \sinh ^{-1}(c+d x)\right )^2}{d}+\frac{(c+d x) \left (a+b \sinh ^{-1}(c+d x)\right )^3}{d}-\frac{6 b^3 \sqrt{(c+d x)^2+1}}{d}+\frac{6 b^3 (c+d x) \sinh ^{-1}(c+d x)}{d} \]

[Out]

6*a*b^2*x - (6*b^3*Sqrt[1 + (c + d*x)^2])/d + (6*b^3*(c + d*x)*ArcSinh[c + d*x])/d - (3*b*Sqrt[1 + (c + d*x)^2
]*(a + b*ArcSinh[c + d*x])^2)/d + ((c + d*x)*(a + b*ArcSinh[c + d*x])^3)/d

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Rubi [A]  time = 0.108284, antiderivative size = 100, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {5863, 5653, 5717, 261} \[ 6 a b^2 x-\frac{3 b \sqrt{(c+d x)^2+1} \left (a+b \sinh ^{-1}(c+d x)\right )^2}{d}+\frac{(c+d x) \left (a+b \sinh ^{-1}(c+d x)\right )^3}{d}-\frac{6 b^3 \sqrt{(c+d x)^2+1}}{d}+\frac{6 b^3 (c+d x) \sinh ^{-1}(c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcSinh[c + d*x])^3,x]

[Out]

6*a*b^2*x - (6*b^3*Sqrt[1 + (c + d*x)^2])/d + (6*b^3*(c + d*x)*ArcSinh[c + d*x])/d - (3*b*Sqrt[1 + (c + d*x)^2
]*(a + b*ArcSinh[c + d*x])^2)/d + ((c + d*x)*(a + b*ArcSinh[c + d*x])^3)/d

Rule 5863

Int[((a_.) + ArcSinh[(c_) + (d_.)*(x_)]*(b_.))^(n_.), x_Symbol] :> Dist[1/d, Subst[Int[(a + b*ArcSinh[x])^n, x
], x, c + d*x], x] /; FreeQ[{a, b, c, d, n}, x]

Rule 5653

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.), x_Symbol] :> Simp[x*(a + b*ArcSinh[c*x])^n, x] - Dist[b*c*n, In
t[(x*(a + b*ArcSinh[c*x])^(n - 1))/Sqrt[1 + c^2*x^2], x], x] /; FreeQ[{a, b, c}, x] && GtQ[n, 0]

Rule 5717

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*(x_)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[((d + e*x^2)
^(p + 1)*(a + b*ArcSinh[c*x])^n)/(2*e*(p + 1)), x] - Dist[(b*n*d^IntPart[p]*(d + e*x^2)^FracPart[p])/(2*c*(p +
1)*(1 + c^2*x^2)^FracPart[p]), Int[(1 + c^2*x^2)^(p + 1/2)*(a + b*ArcSinh[c*x])^(n - 1), x], x] /; FreeQ[{a,
b, c, d, e, p}, x] && EqQ[e, c^2*d] && GtQ[n, 0] && NeQ[p, -1]

Rule 261

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a + b*x^n)^(p + 1)/(b*n*(p + 1)), x] /; FreeQ
[{a, b, m, n, p}, x] && EqQ[m, n - 1] && NeQ[p, -1]

Rubi steps

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

Mathematica [A]  time = 0.15662, size = 147, normalized size = 1.47 \[ \frac{a \left (a^2+6 b^2\right ) (c+d x)-3 b \left (a^2+2 b^2\right ) \sqrt{(c+d x)^2+1}-3 b \sinh ^{-1}(c+d x) \left (a^2 (-(c+d x))+2 a b \sqrt{(c+d x)^2+1}-2 b^2 (c+d x)\right )-3 b^2 \sinh ^{-1}(c+d x)^2 \left (b \sqrt{(c+d x)^2+1}-a (c+d x)\right )+b^3 (c+d x) \sinh ^{-1}(c+d x)^3}{d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcSinh[c + d*x])^3,x]

[Out]

(a*(a^2 + 6*b^2)*(c + d*x) - 3*b*(a^2 + 2*b^2)*Sqrt[1 + (c + d*x)^2] - 3*b*(-(a^2*(c + d*x)) - 2*b^2*(c + d*x)
+ 2*a*b*Sqrt[1 + (c + d*x)^2])*ArcSinh[c + d*x] - 3*b^2*(-(a*(c + d*x)) + b*Sqrt[1 + (c + d*x)^2])*ArcSinh[c
+ d*x]^2 + b^3*(c + d*x)*ArcSinh[c + d*x]^3)/d

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Maple [A]  time = 0.029, size = 160, normalized size = 1.6 \begin{align*}{\frac{1}{d} \left ({a}^{3} \left ( dx+c \right ) +{b}^{3} \left ( \left ({\it Arcsinh} \left ( dx+c \right ) \right ) ^{3} \left ( dx+c \right ) -3\, \left ({\it Arcsinh} \left ( dx+c \right ) \right ) ^{2}\sqrt{1+ \left ( dx+c \right ) ^{2}}+6\, \left ( dx+c \right ){\it Arcsinh} \left ( dx+c \right ) -6\,\sqrt{1+ \left ( dx+c \right ) ^{2}} \right ) +3\,a{b}^{2} \left ( \left ({\it Arcsinh} \left ( dx+c \right ) \right ) ^{2} \left ( dx+c \right ) -2\,{\it Arcsinh} \left ( dx+c \right ) \sqrt{1+ \left ( dx+c \right ) ^{2}}+2\,dx+2\,c \right ) +3\,{a}^{2}b \left ( \left ( dx+c \right ){\it Arcsinh} \left ( dx+c \right ) -\sqrt{1+ \left ( dx+c \right ) ^{2}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(a^3*(d*x+c)+b^3*(arcsinh(d*x+c)^3*(d*x+c)-3*arcsinh(d*x+c)^2*(1+(d*x+c)^2)^(1/2)+6*(d*x+c)*arcsinh(d*x+c)
-6*(1+(d*x+c)^2)^(1/2))+3*a*b^2*(arcsinh(d*x+c)^2*(d*x+c)-2*arcsinh(d*x+c)*(1+(d*x+c)^2)^(1/2)+2*d*x+2*c)+3*a^
2*b*((d*x+c)*arcsinh(d*x+c)-(1+(d*x+c)^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((a+b*arcsinh(d*x+c))^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.73792, size = 548, normalized size = 5.48 \begin{align*} \frac{{\left (b^{3} d x + b^{3} c\right )} \log \left (d x + c + \sqrt{d^{2} x^{2} + 2 \, c d x + c^{2} + 1}\right )^{3} +{\left (a^{3} + 6 \, a b^{2}\right )} d x + 3 \,{\left (a b^{2} d x + a b^{2} c - \sqrt{d^{2} x^{2} + 2 \, c d x + c^{2} + 1} b^{3}\right )} \log \left (d x + c + \sqrt{d^{2} x^{2} + 2 \, c d x + c^{2} + 1}\right )^{2} - 3 \,{\left (2 \, \sqrt{d^{2} x^{2} + 2 \, c d x + c^{2} + 1} a b^{2} -{\left (a^{2} b + 2 \, b^{3}\right )} d x -{\left (a^{2} b + 2 \, b^{3}\right )} c\right )} \log \left (d x + c + \sqrt{d^{2} x^{2} + 2 \, c d x + c^{2} + 1}\right ) - 3 \, \sqrt{d^{2} x^{2} + 2 \, c d x + c^{2} + 1}{\left (a^{2} b + 2 \, b^{3}\right )}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((b^3*d*x + b^3*c)*log(d*x + c + sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1))^3 + (a^3 + 6*a*b^2)*d*x + 3*(a*b^2*d*x + a
*b^2*c - sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1)*b^3)*log(d*x + c + sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1))^2 - 3*(2*sqrt
(d^2*x^2 + 2*c*d*x + c^2 + 1)*a*b^2 - (a^2*b + 2*b^3)*d*x - (a^2*b + 2*b^3)*c)*log(d*x + c + sqrt(d^2*x^2 + 2*
c*d*x + c^2 + 1)) - 3*sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1)*(a^2*b + 2*b^3))/d

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Sympy [A]  time = 1.06046, size = 282, normalized size = 2.82 \begin{align*} \begin{cases} a^{3} x + \frac{3 a^{2} b c \operatorname{asinh}{\left (c + d x \right )}}{d} + 3 a^{2} b x \operatorname{asinh}{\left (c + d x \right )} - \frac{3 a^{2} b \sqrt{c^{2} + 2 c d x + d^{2} x^{2} + 1}}{d} + \frac{3 a b^{2} c \operatorname{asinh}^{2}{\left (c + d x \right )}}{d} + 3 a b^{2} x \operatorname{asinh}^{2}{\left (c + d x \right )} + 6 a b^{2} x - \frac{6 a b^{2} \sqrt{c^{2} + 2 c d x + d^{2} x^{2} + 1} \operatorname{asinh}{\left (c + d x \right )}}{d} + \frac{b^{3} c \operatorname{asinh}^{3}{\left (c + d x \right )}}{d} + \frac{6 b^{3} c \operatorname{asinh}{\left (c + d x \right )}}{d} + b^{3} x \operatorname{asinh}^{3}{\left (c + d x \right )} + 6 b^{3} x \operatorname{asinh}{\left (c + d x \right )} - \frac{3 b^{3} \sqrt{c^{2} + 2 c d x + d^{2} x^{2} + 1} \operatorname{asinh}^{2}{\left (c + d x \right )}}{d} - \frac{6 b^{3} \sqrt{c^{2} + 2 c d x + d^{2} x^{2} + 1}}{d} & \text{for}\: d \neq 0 \\x \left (a + b \operatorname{asinh}{\left (c \right )}\right )^{3} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((a**3*x + 3*a**2*b*c*asinh(c + d*x)/d + 3*a**2*b*x*asinh(c + d*x) - 3*a**2*b*sqrt(c**2 + 2*c*d*x + d
**2*x**2 + 1)/d + 3*a*b**2*c*asinh(c + d*x)**2/d + 3*a*b**2*x*asinh(c + d*x)**2 + 6*a*b**2*x - 6*a*b**2*sqrt(c
**2 + 2*c*d*x + d**2*x**2 + 1)*asinh(c + d*x)/d + b**3*c*asinh(c + d*x)**3/d + 6*b**3*c*asinh(c + d*x)/d + b**
3*x*asinh(c + d*x)**3 + 6*b**3*x*asinh(c + d*x) - 3*b**3*sqrt(c**2 + 2*c*d*x + d**2*x**2 + 1)*asinh(c + d*x)**
2/d - 6*b**3*sqrt(c**2 + 2*c*d*x + d**2*x**2 + 1)/d, Ne(d, 0)), (x*(a + b*asinh(c))**3, True))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \operatorname{arsinh}\left (d x + c\right ) + a\right )}^{3}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*arcsinh(d*x + c) + a)^3, x)