3.150 \(\int (a+b \sinh ^{-1}(c+d x))^4 \, dx\)

Optimal. Leaf size=115 \[ -\frac{24 b^3 \sqrt{(c+d x)^2+1} \left (a+b \sinh ^{-1}(c+d x)\right )}{d}+\frac{12 b^2 (c+d x) \left (a+b \sinh ^{-1}(c+d x)\right )^2}{d}-\frac{4 b \sqrt{(c+d x)^2+1} \left (a+b \sinh ^{-1}(c+d x)\right )^3}{d}+\frac{(c+d x) \left (a+b \sinh ^{-1}(c+d x)\right )^4}{d}+24 b^4 x \]

[Out]

24*b^4*x - (24*b^3*Sqrt[1 + (c + d*x)^2]*(a + b*ArcSinh[c + d*x]))/d + (12*b^2*(c + d*x)*(a + b*ArcSinh[c + d*
x])^2)/d - (4*b*Sqrt[1 + (c + d*x)^2]*(a + b*ArcSinh[c + d*x])^3)/d + ((c + d*x)*(a + b*ArcSinh[c + d*x])^4)/d

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Rubi [A]  time = 0.159742, antiderivative size = 115, 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, 8} \[ -\frac{24 b^3 \sqrt{(c+d x)^2+1} \left (a+b \sinh ^{-1}(c+d x)\right )}{d}+\frac{12 b^2 (c+d x) \left (a+b \sinh ^{-1}(c+d x)\right )^2}{d}-\frac{4 b \sqrt{(c+d x)^2+1} \left (a+b \sinh ^{-1}(c+d x)\right )^3}{d}+\frac{(c+d x) \left (a+b \sinh ^{-1}(c+d x)\right )^4}{d}+24 b^4 x \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcSinh[c + d*x])^4,x]

[Out]

24*b^4*x - (24*b^3*Sqrt[1 + (c + d*x)^2]*(a + b*ArcSinh[c + d*x]))/d + (12*b^2*(c + d*x)*(a + b*ArcSinh[c + d*
x])^2)/d - (4*b*Sqrt[1 + (c + d*x)^2]*(a + b*ArcSinh[c + d*x])^3)/d + ((c + d*x)*(a + b*ArcSinh[c + d*x])^4)/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 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rubi steps

\begin{align*} \int \left (a+b \sinh ^{-1}(c+d x)\right )^4 \, dx &=\frac{\operatorname{Subst}\left (\int \left (a+b \sinh ^{-1}(x)\right )^4 \, dx,x,c+d x\right )}{d}\\ &=\frac{(c+d x) \left (a+b \sinh ^{-1}(c+d x)\right )^4}{d}-\frac{(4 b) \operatorname{Subst}\left (\int \frac{x \left (a+b \sinh ^{-1}(x)\right )^3}{\sqrt{1+x^2}} \, dx,x,c+d x\right )}{d}\\ &=-\frac{4 b \sqrt{1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^3}{d}+\frac{(c+d x) \left (a+b \sinh ^{-1}(c+d x)\right )^4}{d}+\frac{\left (12 b^2\right ) \operatorname{Subst}\left (\int \left (a+b \sinh ^{-1}(x)\right )^2 \, dx,x,c+d x\right )}{d}\\ &=\frac{12 b^2 (c+d x) \left (a+b \sinh ^{-1}(c+d x)\right )^2}{d}-\frac{4 b \sqrt{1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^3}{d}+\frac{(c+d x) \left (a+b \sinh ^{-1}(c+d x)\right )^4}{d}-\frac{\left (24 b^3\right ) \operatorname{Subst}\left (\int \frac{x \left (a+b \sinh ^{-1}(x)\right )}{\sqrt{1+x^2}} \, dx,x,c+d x\right )}{d}\\ &=-\frac{24 b^3 \sqrt{1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )}{d}+\frac{12 b^2 (c+d x) \left (a+b \sinh ^{-1}(c+d x)\right )^2}{d}-\frac{4 b \sqrt{1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^3}{d}+\frac{(c+d x) \left (a+b \sinh ^{-1}(c+d x)\right )^4}{d}+\frac{\left (24 b^4\right ) \operatorname{Subst}(\int 1 \, dx,x,c+d x)}{d}\\ &=24 b^4 x-\frac{24 b^3 \sqrt{1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )}{d}+\frac{12 b^2 (c+d x) \left (a+b \sinh ^{-1}(c+d x)\right )^2}{d}-\frac{4 b \sqrt{1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^3}{d}+\frac{(c+d x) \left (a+b \sinh ^{-1}(c+d x)\right )^4}{d}\\ \end{align*}

Mathematica [A]  time = 0.250068, size = 226, normalized size = 1.97 \[ \frac{\left (12 a^2 b^2+a^4+24 b^4\right ) (c+d x)-4 a b \left (a^2+6 b^2\right ) \sqrt{(c+d x)^2+1}+6 b^2 \sinh ^{-1}(c+d x)^2 \left (a^2 (c+d x)-2 a b \sqrt{(c+d x)^2+1}+2 b^2 (c+d x)\right )-4 b \sinh ^{-1}(c+d x) \left (3 a^2 b \sqrt{(c+d x)^2+1}+a^3 (-(c+d x))-6 a b^2 (c+d x)+6 b^3 \sqrt{(c+d x)^2+1}\right )-4 b^3 \sinh ^{-1}(c+d x)^3 \left (b \sqrt{(c+d x)^2+1}-a (c+d x)\right )+b^4 (c+d x) \sinh ^{-1}(c+d x)^4}{d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcSinh[c + d*x])^4,x]

[Out]

((a^4 + 12*a^2*b^2 + 24*b^4)*(c + d*x) - 4*a*b*(a^2 + 6*b^2)*Sqrt[1 + (c + d*x)^2] - 4*b*(-(a^3*(c + d*x)) - 6
*a*b^2*(c + d*x) + 3*a^2*b*Sqrt[1 + (c + d*x)^2] + 6*b^3*Sqrt[1 + (c + d*x)^2])*ArcSinh[c + d*x] + 6*b^2*(a^2*
(c + d*x) + 2*b^2*(c + d*x) - 2*a*b*Sqrt[1 + (c + d*x)^2])*ArcSinh[c + d*x]^2 - 4*b^3*(-(a*(c + d*x)) + b*Sqrt
[1 + (c + d*x)^2])*ArcSinh[c + d*x]^3 + b^4*(c + d*x)*ArcSinh[c + d*x]^4)/d

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Maple [B]  time = 0.03, size = 245, normalized size = 2.1 \begin{align*}{\frac{1}{d} \left ( \left ( dx+c \right ){a}^{4}+{b}^{4} \left ( \left ({\it Arcsinh} \left ( dx+c \right ) \right ) ^{4} \left ( dx+c \right ) -4\, \left ({\it Arcsinh} \left ( dx+c \right ) \right ) ^{3}\sqrt{1+ \left ( dx+c \right ) ^{2}}+12\, \left ({\it Arcsinh} \left ( dx+c \right ) \right ) ^{2} \left ( dx+c \right ) -24\,{\it Arcsinh} \left ( dx+c \right ) \sqrt{1+ \left ( dx+c \right ) ^{2}}+24\,dx+24\,c \right ) +4\,a{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 ) +6\,{a}^{2}{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 ) +4\,{a}^{3}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))^4,x)

[Out]

1/d*((d*x+c)*a^4+b^4*(arcsinh(d*x+c)^4*(d*x+c)-4*arcsinh(d*x+c)^3*(1+(d*x+c)^2)^(1/2)+12*arcsinh(d*x+c)^2*(d*x
+c)-24*arcsinh(d*x+c)*(1+(d*x+c)^2)^(1/2)+24*d*x+24*c)+4*a*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))+6*a^2*b^2*(arcsinh(d*x+c)^2*(d*x+c)-2*arcsin
h(d*x+c)*(1+(d*x+c)^2)^(1/2)+2*d*x+2*c)+4*a^3*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))^4,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.75134, size = 784, normalized size = 6.82 \begin{align*} \frac{{\left (b^{4} d x + b^{4} c\right )} \log \left (d x + c + \sqrt{d^{2} x^{2} + 2 \, c d x + c^{2} + 1}\right )^{4} + 4 \,{\left (a b^{3} d x + a b^{3} c - \sqrt{d^{2} x^{2} + 2 \, c d x + c^{2} + 1} b^{4}\right )} \log \left (d x + c + \sqrt{d^{2} x^{2} + 2 \, c d x + c^{2} + 1}\right )^{3} +{\left (a^{4} + 12 \, a^{2} b^{2} + 24 \, b^{4}\right )} d x - 6 \,{\left (2 \, \sqrt{d^{2} x^{2} + 2 \, c d x + c^{2} + 1} a b^{3} -{\left (a^{2} b^{2} + 2 \, b^{4}\right )} d x -{\left (a^{2} b^{2} + 2 \, b^{4}\right )} c\right )} \log \left (d x + c + \sqrt{d^{2} x^{2} + 2 \, c d x + c^{2} + 1}\right )^{2} + 4 \,{\left ({\left (a^{3} b + 6 \, a b^{3}\right )} d x +{\left (a^{3} b + 6 \, a b^{3}\right )} c - 3 \,{\left (a^{2} b^{2} + 2 \, b^{4}\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 ) - 4 \,{\left (a^{3} b + 6 \, a b^{3}\right )} \sqrt{d^{2} x^{2} + 2 \, c d x + c^{2} + 1}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsinh(d*x+c))^4,x, algorithm="fricas")

[Out]

((b^4*d*x + b^4*c)*log(d*x + c + sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1))^4 + 4*(a*b^3*d*x + a*b^3*c - sqrt(d^2*x^2
+ 2*c*d*x + c^2 + 1)*b^4)*log(d*x + c + sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1))^3 + (a^4 + 12*a^2*b^2 + 24*b^4)*d*x
- 6*(2*sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1)*a*b^3 - (a^2*b^2 + 2*b^4)*d*x - (a^2*b^2 + 2*b^4)*c)*log(d*x + c + s
qrt(d^2*x^2 + 2*c*d*x + c^2 + 1))^2 + 4*((a^3*b + 6*a*b^3)*d*x + (a^3*b + 6*a*b^3)*c - 3*(a^2*b^2 + 2*b^4)*sqr
t(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)) - 4*(a^3*b + 6*a*b^3)*sqrt(d^
2*x^2 + 2*c*d*x + c^2 + 1))/d

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Sympy [A]  time = 2.20034, size = 444, normalized size = 3.86 \begin{align*} \begin{cases} a^{4} x + \frac{4 a^{3} b c \operatorname{asinh}{\left (c + d x \right )}}{d} + 4 a^{3} b x \operatorname{asinh}{\left (c + d x \right )} - \frac{4 a^{3} b \sqrt{c^{2} + 2 c d x + d^{2} x^{2} + 1}}{d} + \frac{6 a^{2} b^{2} c \operatorname{asinh}^{2}{\left (c + d x \right )}}{d} + 6 a^{2} b^{2} x \operatorname{asinh}^{2}{\left (c + d x \right )} + 12 a^{2} b^{2} x - \frac{12 a^{2} b^{2} \sqrt{c^{2} + 2 c d x + d^{2} x^{2} + 1} \operatorname{asinh}{\left (c + d x \right )}}{d} + \frac{4 a b^{3} c \operatorname{asinh}^{3}{\left (c + d x \right )}}{d} + \frac{24 a b^{3} c \operatorname{asinh}{\left (c + d x \right )}}{d} + 4 a b^{3} x \operatorname{asinh}^{3}{\left (c + d x \right )} + 24 a b^{3} x \operatorname{asinh}{\left (c + d x \right )} - \frac{12 a b^{3} \sqrt{c^{2} + 2 c d x + d^{2} x^{2} + 1} \operatorname{asinh}^{2}{\left (c + d x \right )}}{d} - \frac{24 a b^{3} \sqrt{c^{2} + 2 c d x + d^{2} x^{2} + 1}}{d} + \frac{b^{4} c \operatorname{asinh}^{4}{\left (c + d x \right )}}{d} + \frac{12 b^{4} c \operatorname{asinh}^{2}{\left (c + d x \right )}}{d} + b^{4} x \operatorname{asinh}^{4}{\left (c + d x \right )} + 12 b^{4} x \operatorname{asinh}^{2}{\left (c + d x \right )} + 24 b^{4} x - \frac{4 b^{4} \sqrt{c^{2} + 2 c d x + d^{2} x^{2} + 1} \operatorname{asinh}^{3}{\left (c + d x \right )}}{d} - \frac{24 b^{4} \sqrt{c^{2} + 2 c d x + d^{2} x^{2} + 1} \operatorname{asinh}{\left (c + d x \right )}}{d} & \text{for}\: d \neq 0 \\x \left (a + b \operatorname{asinh}{\left (c \right )}\right )^{4} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*asinh(d*x+c))**4,x)

[Out]

Piecewise((a**4*x + 4*a**3*b*c*asinh(c + d*x)/d + 4*a**3*b*x*asinh(c + d*x) - 4*a**3*b*sqrt(c**2 + 2*c*d*x + d
**2*x**2 + 1)/d + 6*a**2*b**2*c*asinh(c + d*x)**2/d + 6*a**2*b**2*x*asinh(c + d*x)**2 + 12*a**2*b**2*x - 12*a*
*2*b**2*sqrt(c**2 + 2*c*d*x + d**2*x**2 + 1)*asinh(c + d*x)/d + 4*a*b**3*c*asinh(c + d*x)**3/d + 24*a*b**3*c*a
sinh(c + d*x)/d + 4*a*b**3*x*asinh(c + d*x)**3 + 24*a*b**3*x*asinh(c + d*x) - 12*a*b**3*sqrt(c**2 + 2*c*d*x +
d**2*x**2 + 1)*asinh(c + d*x)**2/d - 24*a*b**3*sqrt(c**2 + 2*c*d*x + d**2*x**2 + 1)/d + b**4*c*asinh(c + d*x)*
*4/d + 12*b**4*c*asinh(c + d*x)**2/d + b**4*x*asinh(c + d*x)**4 + 12*b**4*x*asinh(c + d*x)**2 + 24*b**4*x - 4*
b**4*sqrt(c**2 + 2*c*d*x + d**2*x**2 + 1)*asinh(c + d*x)**3/d - 24*b**4*sqrt(c**2 + 2*c*d*x + d**2*x**2 + 1)*a
sinh(c + d*x)/d, Ne(d, 0)), (x*(a + b*asinh(c))**4, 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 )}^{4}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsinh(d*x+c))^4,x, algorithm="giac")

[Out]

integrate((b*arcsinh(d*x + c) + a)^4, x)