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

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Rubi [A]  time = 0.07, antiderivative size = 57, normalized size of antiderivative = 1.00, 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 8

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

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Mathematica [A]  time = 0.10, 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

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fricas [B]  time = 0.56, size = 141, normalized size = 2.47 \[ \frac {{\left (a^{2} + 2 \, b^{2}\right )} d x + {\left (b^{2} d x + b^{2} c\right )} \log \left (d x + c + \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1}\right )^{2} - 2 \, \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1} a b + 2 \, {\left (a b d x + a b c - \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1} b^{2}\right )} \log \left (d x + c + \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1}\right )}{d} \]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \operatorname {arsinh}\left (d x + c\right ) + a\right )}^{2}\,{d x} \]

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)

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maple [A]  time = 0.05, size = 90, normalized size = 1.58 \[ \frac {a^{2} \left (d x +c \right )+b^{2} \left (\left (d x +c \right ) \arcsinh \left (d x +c \right )^{2}-2 \arcsinh \left (d x +c \right ) \sqrt {1+\left (d x +c \right )^{2}}+2 d x +2 c \right )+2 a b \left (\left (d x +c \right ) \arcsinh \left (d x +c \right )-\sqrt {1+\left (d x +c \right )^{2}}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(a^2*(d*x+c)+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)+2*a*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 \[ {\left (x \log \left (d x + c + \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1}\right )^{2} - \int \frac {2 \, {\left (d^{3} x^{3} + 2 \, c d^{2} x^{2} + {\left (c^{2} d + d\right )} x + \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1} {\left (d^{2} x^{2} + c d x\right )}\right )} \log \left (d x + c + \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1}\right )}{d^{3} x^{3} + 3 \, c d^{2} x^{2} + c^{3} + {\left (3 \, c^{2} d + d\right )} x + {\left (d^{2} x^{2} + 2 \, c d x + c^{2} + 1\right )}^{\frac {3}{2}} + c}\,{d x}\right )} b^{2} + a^{2} x + \frac {2 \, {\left ({\left (d x + c\right )} \operatorname {arsinh}\left (d x + c\right ) - \sqrt {{\left (d x + c\right )}^{2} + 1}\right )} a b}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int {\left (a+b\,\mathrm {asinh}\left (c+d\,x\right )\right )}^2 \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*asinh(c + d*x))^2,x)

[Out]

int((a + b*asinh(c + d*x))^2, x)

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sympy [A]  time = 0.32, size = 143, normalized size = 2.51 \[ \begin {cases} a^{2} x + \frac {2 a b c \operatorname {asinh}{\left (c + d x \right )}}{d} + 2 a b x \operatorname {asinh}{\left (c + d x \right )} - \frac {2 a b \sqrt {c^{2} + 2 c d x + d^{2} x^{2} + 1}}{d} + \frac {b^{2} c \operatorname {asinh}^{2}{\left (c + d x \right )}}{d} + b^{2} x \operatorname {asinh}^{2}{\left (c + d x \right )} + 2 b^{2} x - \frac {2 b^{2} \sqrt {c^{2} + 2 c d x + d^{2} x^{2} + 1} \operatorname {asinh}{\left (c + d x \right )}}{d} & \text {for}\: d \neq 0 \\x \left (a + b \operatorname {asinh}{\relax (c )}\right )^{2} & \text {otherwise} \end {cases} \]

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))

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