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

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Rubi [A]  time = 0.11, antiderivative size = 100, normalized size of antiderivative = 1.00, 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 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]

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 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]

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*}

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Mathematica [A]  time = 0.17, size = 147, normalized size = 1.47 \[ \frac {a \left (a^2+6 b^2\right ) (c+d x)-3 b \left (a^2+2 b^2\right ) \sqrt {(c+d x)^2+1}-3 b \sinh ^{-1}(c+d x) \left (-\left (a^2 (c+d x)\right )+2 a b \sqrt {(c+d x)^2+1}-2 b^2 (c+d x)\right )-3 b^2 \sinh ^{-1}(c+d x)^2 \left (b \sqrt {(c+d x)^2+1}-a (c+d x)\right )+b^3 (c+d x) \sinh ^{-1}(c+d x)^3}{d} \]

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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fricas [B]  time = 0.66, size = 239, normalized size = 2.39 \[ \frac {{\left (b^{3} d x + b^{3} c\right )} \log \left (d x + c + \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1}\right )^{3} + {\left (a^{3} + 6 \, a b^{2}\right )} d x + 3 \, {\left (a b^{2} d x + a b^{2} c - \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1} b^{3}\right )} \log \left (d x + c + \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1}\right )^{2} - 3 \, {\left (2 \, \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1} a b^{2} - {\left (a^{2} b + 2 \, b^{3}\right )} d x - {\left (a^{2} b + 2 \, b^{3}\right )} c\right )} \log \left (d x + c + \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1}\right ) - 3 \, \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1} {\left (a^{2} b + 2 \, b^{3}\right )}}{d} \]

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

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)

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maple [A]  time = 0.07, size = 160, normalized size = 1.60 \[ \frac {a^{3} \left (d x +c \right )+b^{3} \left (\left (d x +c \right ) \arcsinh \left (d x +c \right )^{3}-3 \arcsinh \left (d x +c \right )^{2} \sqrt {1+\left (d x +c \right )^{2}}+6 \left (d x +c \right ) \arcsinh \left (d x +c \right )-6 \sqrt {1+\left (d x +c \right )^{2}}\right )+3 a \,b^{2} \left (\left (d x +c \right ) \arcsinh \left (d x +c \right )^{2}-2 \arcsinh \left (d x +c \right ) \sqrt {1+\left (d x +c \right )^{2}}+2 d x +2 c \right )+3 a^{2} b \left (\left (d x +c \right ) \arcsinh \left (d x +c \right )-\sqrt {1+\left (d x +c \right )^{2}}\right )}{d} \]

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*((d*x+c)*arcsinh(d*x+c)^3-3*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*((d*x+c)*arcsinh(d*x+c)^2-2*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]  time = 0.00, size = 0, normalized size = 0.00 \[ b^{3} x \log \left (d x + c + \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1}\right )^{3} + a^{3} x + \frac {3 \, {\left ({\left (d x + c\right )} \operatorname {arsinh}\left (d x + c\right ) - \sqrt {{\left (d x + c\right )}^{2} + 1}\right )} a^{2} b}{d} + \int \frac {3 \, {\left ({\left (c^{3} + c\right )} a b^{2} + {\left (a b^{2} d^{3} - b^{3} d^{3}\right )} x^{3} + {\left (3 \, a b^{2} c d^{2} - 2 \, b^{3} c d^{2}\right )} x^{2} + {\left ({\left (3 \, c^{2} d + d\right )} a b^{2} - {\left (c^{2} d + d\right )} b^{3}\right )} x + {\left ({\left (c^{2} + 1\right )} a b^{2} + {\left (a b^{2} d^{2} - b^{3} d^{2}\right )} x^{2} + {\left (2 \, a b^{2} c d - b^{3} c d\right )} x\right )} \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1}\right )} \log \left (d x + c + \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1}\right )^{2}}{d^{3} x^{3} + 3 \, c d^{2} x^{2} + c^{3} + {\left (3 \, c^{2} d + d\right )} x + {\left (d^{2} x^{2} + 2 \, c d x + c^{2} + 1\right )}^{\frac {3}{2}} + c}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (a+b\,\mathrm {asinh}\left (c+d\,x\right )\right )}^3 \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*asinh(c + d*x))^3,x)

[Out]

int((a + b*asinh(c + d*x))^3, x)

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sympy [A]  time = 0.80, size = 282, normalized size = 2.82 \[ \begin {cases} a^{3} x + \frac {3 a^{2} b c \operatorname {asinh}{\left (c + d x \right )}}{d} + 3 a^{2} b x \operatorname {asinh}{\left (c + d x \right )} - \frac {3 a^{2} b \sqrt {c^{2} + 2 c d x + d^{2} x^{2} + 1}}{d} + \frac {3 a b^{2} c \operatorname {asinh}^{2}{\left (c + d x \right )}}{d} + 3 a b^{2} x \operatorname {asinh}^{2}{\left (c + d x \right )} + 6 a b^{2} x - \frac {6 a b^{2} \sqrt {c^{2} + 2 c d x + d^{2} x^{2} + 1} \operatorname {asinh}{\left (c + d x \right )}}{d} + \frac {b^{3} c \operatorname {asinh}^{3}{\left (c + d x \right )}}{d} + \frac {6 b^{3} c \operatorname {asinh}{\left (c + d x \right )}}{d} + b^{3} x \operatorname {asinh}^{3}{\left (c + d x \right )} + 6 b^{3} x \operatorname {asinh}{\left (c + d x \right )} - \frac {3 b^{3} \sqrt {c^{2} + 2 c d x + d^{2} x^{2} + 1} \operatorname {asinh}^{2}{\left (c + d x \right )}}{d} - \frac {6 b^{3} \sqrt {c^{2} + 2 c d x + d^{2} x^{2} + 1}}{d} & \text {for}\: d \neq 0 \\x \left (a + b \operatorname {asinh}{\relax (c )}\right )^{3} & \text {otherwise} \end {cases} \]

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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