3.131 \(\int (a+b \sinh ^{-1}(c+d x))^2 \, dx\)

Optimal. Leaf size=57 \[ -\frac{2 b \sqrt{(c+d x)^2+1} \left (a+b \sinh ^{-1}(c+d x)\right )}{d}+\frac{(c+d x) \left (a+b \sinh ^{-1}(c+d x)\right )^2}{d}+2 b^2 x \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0701418, antiderivative size = 57, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {5863, 5653, 5717, 8} \[ -\frac{2 b \sqrt{(c+d x)^2+1} \left (a+b \sinh ^{-1}(c+d x)\right )}{d}+\frac{(c+d x) \left (a+b \sinh ^{-1}(c+d x)\right )^2}{d}+2 b^2 x \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcSinh[c + d*x])^2,x]

[Out]

2*b^2*x - (2*b*Sqrt[1 + (c + d*x)^2]*(a + b*ArcSinh[c + d*x]))/d + ((c + d*x)*(a + b*ArcSinh[c + d*x])^2)/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 )^2 \, dx &=\frac{\operatorname{Subst}\left (\int \left (a+b \sinh ^{-1}(x)\right )^2 \, dx,x,c+d x\right )}{d}\\ &=\frac{(c+d x) \left (a+b \sinh ^{-1}(c+d x)\right )^2}{d}-\frac{(2 b) \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{2 b \sqrt{1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )}{d}+\frac{(c+d x) \left (a+b \sinh ^{-1}(c+d x)\right )^2}{d}+\frac{\left (2 b^2\right ) \operatorname{Subst}(\int 1 \, dx,x,c+d x)}{d}\\ &=2 b^2 x-\frac{2 b \sqrt{1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )}{d}+\frac{(c+d x) \left (a+b \sinh ^{-1}(c+d x)\right )^2}{d}\\ \end{align*}

Mathematica [A]  time = 0.0956592, size = 87, normalized size = 1.53 \[ \frac{\left (a^2+2 b^2\right ) (c+d x)-2 a b \sqrt{(c+d x)^2+1}+2 b \sinh ^{-1}(c+d x) \left (a c+a d x+b \left (-\sqrt{(c+d x)^2+1}\right )\right )+b^2 (c+d x) \sinh ^{-1}(c+d x)^2}{d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcSinh[c + d*x])^2,x]

[Out]

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

________________________________________________________________________________________

Maple [A]  time = 0.028, size = 90, normalized size = 1.6 \begin{align*}{\frac{1}{d} \left ( \left ( dx+c \right ){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 ) +2\,ab \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))^2,x)

[Out]

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

________________________________________________________________________________________

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))^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [B]  time = 2.66711, size = 333, normalized size = 5.84 \begin{align*} \frac{{\left (a^{2} + 2 \, b^{2}\right )} d x +{\left (b^{2} d x + b^{2} c\right )} \log \left (d x + c + \sqrt{d^{2} x^{2} + 2 \, c d x + c^{2} + 1}\right )^{2} - 2 \, \sqrt{d^{2} x^{2} + 2 \, c d x + c^{2} + 1} a b + 2 \,{\left (a b d x + a b c - \sqrt{d^{2} x^{2} + 2 \, c d x + c^{2} + 1} b^{2}\right )} \log \left (d x + c + \sqrt{d^{2} x^{2} + 2 \, c d x + c^{2} + 1}\right )}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsinh(d*x+c))^2,x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

Sympy [A]  time = 0.401433, size = 143, normalized size = 2.51 \begin{align*} \begin{cases} a^{2} x + \frac{2 a b c \operatorname{asinh}{\left (c + d x \right )}}{d} + 2 a b x \operatorname{asinh}{\left (c + d x \right )} - \frac{2 a b \sqrt{c^{2} + 2 c d x + d^{2} x^{2} + 1}}{d} + \frac{b^{2} c \operatorname{asinh}^{2}{\left (c + d x \right )}}{d} + b^{2} x \operatorname{asinh}^{2}{\left (c + d x \right )} + 2 b^{2} x - \frac{2 b^{2} \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 )^{2} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*asinh(d*x+c))**2,x)

[Out]

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

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \operatorname{arsinh}\left (d x + c\right ) + a\right )}^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsinh(d*x+c))^2,x, algorithm="giac")

[Out]

integrate((b*arcsinh(d*x + c) + a)^2, x)