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

Optimal. Leaf size=125 \[ \frac{1}{3} d x \left (c^2 x^2+1\right ) \left (a+b \sinh ^{-1}(c x)\right )^2-\frac{2 b d \left (c^2 x^2+1\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )}{9 c}-\frac{4 b d \sqrt{c^2 x^2+1} \left (a+b \sinh ^{-1}(c x)\right )}{3 c}+\frac{2}{3} d x \left (a+b \sinh ^{-1}(c x)\right )^2+\frac{2}{27} b^2 c^2 d x^3+\frac{14}{9} b^2 d x \]

[Out]

(14*b^2*d*x)/9 + (2*b^2*c^2*d*x^3)/27 - (4*b*d*Sqrt[1 + c^2*x^2]*(a + b*ArcSinh[c*x]))/(3*c) - (2*b*d*(1 + c^2
*x^2)^(3/2)*(a + b*ArcSinh[c*x]))/(9*c) + (2*d*x*(a + b*ArcSinh[c*x])^2)/3 + (d*x*(1 + c^2*x^2)*(a + b*ArcSinh
[c*x])^2)/3

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Rubi [A]  time = 0.144216, antiderivative size = 125, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {5684, 5653, 5717, 8} \[ \frac{1}{3} d x \left (c^2 x^2+1\right ) \left (a+b \sinh ^{-1}(c x)\right )^2-\frac{2 b d \left (c^2 x^2+1\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )}{9 c}-\frac{4 b d \sqrt{c^2 x^2+1} \left (a+b \sinh ^{-1}(c x)\right )}{3 c}+\frac{2}{3} d x \left (a+b \sinh ^{-1}(c x)\right )^2+\frac{2}{27} b^2 c^2 d x^3+\frac{14}{9} b^2 d x \]

Antiderivative was successfully verified.

[In]

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

[Out]

(14*b^2*d*x)/9 + (2*b^2*c^2*d*x^3)/27 - (4*b*d*Sqrt[1 + c^2*x^2]*(a + b*ArcSinh[c*x]))/(3*c) - (2*b*d*(1 + c^2
*x^2)^(3/2)*(a + b*ArcSinh[c*x]))/(9*c) + (2*d*x*(a + b*ArcSinh[c*x])^2)/3 + (d*x*(1 + c^2*x^2)*(a + b*ArcSinh
[c*x])^2)/3

Rule 5684

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

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 (d+c^2 d x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )^2 \, dx &=\frac{1}{3} d x \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )^2+\frac{1}{3} (2 d) \int \left (a+b \sinh ^{-1}(c x)\right )^2 \, dx-\frac{1}{3} (2 b c d) \int x \sqrt{1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right ) \, dx\\ &=-\frac{2 b d \left (1+c^2 x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )}{9 c}+\frac{2}{3} d x \left (a+b \sinh ^{-1}(c x)\right )^2+\frac{1}{3} d x \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )^2+\frac{1}{9} \left (2 b^2 d\right ) \int \left (1+c^2 x^2\right ) \, dx-\frac{1}{3} (4 b c d) \int \frac{x \left (a+b \sinh ^{-1}(c x)\right )}{\sqrt{1+c^2 x^2}} \, dx\\ &=\frac{2}{9} b^2 d x+\frac{2}{27} b^2 c^2 d x^3-\frac{4 b d \sqrt{1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{3 c}-\frac{2 b d \left (1+c^2 x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )}{9 c}+\frac{2}{3} d x \left (a+b \sinh ^{-1}(c x)\right )^2+\frac{1}{3} d x \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )^2+\frac{1}{3} \left (4 b^2 d\right ) \int 1 \, dx\\ &=\frac{14}{9} b^2 d x+\frac{2}{27} b^2 c^2 d x^3-\frac{4 b d \sqrt{1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{3 c}-\frac{2 b d \left (1+c^2 x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )}{9 c}+\frac{2}{3} d x \left (a+b \sinh ^{-1}(c x)\right )^2+\frac{1}{3} d x \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )^2\\ \end{align*}

Mathematica [A]  time = 0.201368, size = 135, normalized size = 1.08 \[ \frac{d \left (9 a^2 c x \left (c^2 x^2+3\right )-6 a b \sqrt{c^2 x^2+1} \left (c^2 x^2+7\right )-6 b \sinh ^{-1}(c x) \left (b \sqrt{c^2 x^2+1} \left (c^2 x^2+7\right )-3 a c x \left (c^2 x^2+3\right )\right )+2 b^2 c x \left (c^2 x^2+21\right )+9 b^2 c x \left (c^2 x^2+3\right ) \sinh ^{-1}(c x)^2\right )}{27 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(d*(9*a^2*c*x*(3 + c^2*x^2) - 6*a*b*Sqrt[1 + c^2*x^2]*(7 + c^2*x^2) + 2*b^2*c*x*(21 + c^2*x^2) - 6*b*(-3*a*c*x
*(3 + c^2*x^2) + b*Sqrt[1 + c^2*x^2]*(7 + c^2*x^2))*ArcSinh[c*x] + 9*b^2*c*x*(3 + c^2*x^2)*ArcSinh[c*x]^2))/(2
7*c)

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Maple [A]  time = 0.031, size = 172, normalized size = 1.4 \begin{align*}{\frac{1}{c} \left ( d{a}^{2} \left ({\frac{{c}^{3}{x}^{3}}{3}}+cx \right ) +d{b}^{2} \left ({\frac{2\, \left ({\it Arcsinh} \left ( cx \right ) \right ) ^{2}cx}{3}}+{\frac{ \left ({\it Arcsinh} \left ( cx \right ) \right ) ^{2}cx \left ({c}^{2}{x}^{2}+1 \right ) }{3}}-{\frac{14\,{\it Arcsinh} \left ( cx \right ) }{9}\sqrt{{c}^{2}{x}^{2}+1}}+{\frac{40\,cx}{27}}-{\frac{2\,{\it Arcsinh} \left ( cx \right ){c}^{2}{x}^{2}}{9}\sqrt{{c}^{2}{x}^{2}+1}}+{\frac{2\,cx \left ({c}^{2}{x}^{2}+1 \right ) }{27}} \right ) +2\,dab \left ( 1/3\,{\it Arcsinh} \left ( cx \right ){c}^{3}{x}^{3}+{\it Arcsinh} \left ( cx \right ) cx-1/9\,{c}^{2}{x}^{2}\sqrt{{c}^{2}{x}^{2}+1}-{\frac{7\,\sqrt{{c}^{2}{x}^{2}+1}}{9}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/c*(d*a^2*(1/3*c^3*x^3+c*x)+d*b^2*(2/3*arcsinh(c*x)^2*c*x+1/3*arcsinh(c*x)^2*c*x*(c^2*x^2+1)-14/9*arcsinh(c*x
)*(c^2*x^2+1)^(1/2)+40/27*c*x-2/9*arcsinh(c*x)*c^2*x^2*(c^2*x^2+1)^(1/2)+2/27*c*x*(c^2*x^2+1))+2*d*a*b*(1/3*ar
csinh(c*x)*c^3*x^3+arcsinh(c*x)*c*x-1/9*c^2*x^2*(c^2*x^2+1)^(1/2)-7/9*(c^2*x^2+1)^(1/2)))

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Maxima [B]  time = 1.10435, size = 311, normalized size = 2.49 \begin{align*} \frac{1}{3} \, b^{2} c^{2} d x^{3} \operatorname{arsinh}\left (c x\right )^{2} + \frac{1}{3} \, a^{2} c^{2} d x^{3} + \frac{2}{9} \,{\left (3 \, x^{3} \operatorname{arsinh}\left (c x\right ) - c{\left (\frac{\sqrt{c^{2} x^{2} + 1} x^{2}}{c^{2}} - \frac{2 \, \sqrt{c^{2} x^{2} + 1}}{c^{4}}\right )}\right )} a b c^{2} d - \frac{2}{27} \,{\left (3 \, c{\left (\frac{\sqrt{c^{2} x^{2} + 1} x^{2}}{c^{2}} - \frac{2 \, \sqrt{c^{2} x^{2} + 1}}{c^{4}}\right )} \operatorname{arsinh}\left (c x\right ) - \frac{c^{2} x^{3} - 6 \, x}{c^{2}}\right )} b^{2} c^{2} d + b^{2} d x \operatorname{arsinh}\left (c x\right )^{2} + 2 \, b^{2} d{\left (x - \frac{\sqrt{c^{2} x^{2} + 1} \operatorname{arsinh}\left (c x\right )}{c}\right )} + a^{2} d x + \frac{2 \,{\left (c x \operatorname{arsinh}\left (c x\right ) - \sqrt{c^{2} x^{2} + 1}\right )} a b d}{c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*b^2*c^2*d*x^3*arcsinh(c*x)^2 + 1/3*a^2*c^2*d*x^3 + 2/9*(3*x^3*arcsinh(c*x) - c*(sqrt(c^2*x^2 + 1)*x^2/c^2
- 2*sqrt(c^2*x^2 + 1)/c^4))*a*b*c^2*d - 2/27*(3*c*(sqrt(c^2*x^2 + 1)*x^2/c^2 - 2*sqrt(c^2*x^2 + 1)/c^4)*arcsin
h(c*x) - (c^2*x^3 - 6*x)/c^2)*b^2*c^2*d + b^2*d*x*arcsinh(c*x)^2 + 2*b^2*d*(x - sqrt(c^2*x^2 + 1)*arcsinh(c*x)
/c) + a^2*d*x + 2*(c*x*arcsinh(c*x) - sqrt(c^2*x^2 + 1))*a*b*d/c

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Fricas [A]  time = 2.71419, size = 387, normalized size = 3.1 \begin{align*} \frac{{\left (9 \, a^{2} + 2 \, b^{2}\right )} c^{3} d x^{3} + 3 \,{\left (9 \, a^{2} + 14 \, b^{2}\right )} c d x + 9 \,{\left (b^{2} c^{3} d x^{3} + 3 \, b^{2} c d x\right )} \log \left (c x + \sqrt{c^{2} x^{2} + 1}\right )^{2} + 6 \,{\left (3 \, a b c^{3} d x^{3} + 9 \, a b c d x -{\left (b^{2} c^{2} d x^{2} + 7 \, b^{2} d\right )} \sqrt{c^{2} x^{2} + 1}\right )} \log \left (c x + \sqrt{c^{2} x^{2} + 1}\right ) - 6 \,{\left (a b c^{2} d x^{2} + 7 \, a b d\right )} \sqrt{c^{2} x^{2} + 1}}{27 \, c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/27*((9*a^2 + 2*b^2)*c^3*d*x^3 + 3*(9*a^2 + 14*b^2)*c*d*x + 9*(b^2*c^3*d*x^3 + 3*b^2*c*d*x)*log(c*x + sqrt(c^
2*x^2 + 1))^2 + 6*(3*a*b*c^3*d*x^3 + 9*a*b*c*d*x - (b^2*c^2*d*x^2 + 7*b^2*d)*sqrt(c^2*x^2 + 1))*log(c*x + sqrt
(c^2*x^2 + 1)) - 6*(a*b*c^2*d*x^2 + 7*a*b*d)*sqrt(c^2*x^2 + 1))/c

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Sympy [A]  time = 1.56101, size = 224, normalized size = 1.79 \begin{align*} \begin{cases} \frac{a^{2} c^{2} d x^{3}}{3} + a^{2} d x + \frac{2 a b c^{2} d x^{3} \operatorname{asinh}{\left (c x \right )}}{3} - \frac{2 a b c d x^{2} \sqrt{c^{2} x^{2} + 1}}{9} + 2 a b d x \operatorname{asinh}{\left (c x \right )} - \frac{14 a b d \sqrt{c^{2} x^{2} + 1}}{9 c} + \frac{b^{2} c^{2} d x^{3} \operatorname{asinh}^{2}{\left (c x \right )}}{3} + \frac{2 b^{2} c^{2} d x^{3}}{27} - \frac{2 b^{2} c d x^{2} \sqrt{c^{2} x^{2} + 1} \operatorname{asinh}{\left (c x \right )}}{9} + b^{2} d x \operatorname{asinh}^{2}{\left (c x \right )} + \frac{14 b^{2} d x}{9} - \frac{14 b^{2} d \sqrt{c^{2} x^{2} + 1} \operatorname{asinh}{\left (c x \right )}}{9 c} & \text{for}\: c \neq 0 \\a^{2} d x & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((a**2*c**2*d*x**3/3 + a**2*d*x + 2*a*b*c**2*d*x**3*asinh(c*x)/3 - 2*a*b*c*d*x**2*sqrt(c**2*x**2 + 1)
/9 + 2*a*b*d*x*asinh(c*x) - 14*a*b*d*sqrt(c**2*x**2 + 1)/(9*c) + b**2*c**2*d*x**3*asinh(c*x)**2/3 + 2*b**2*c**
2*d*x**3/27 - 2*b**2*c*d*x**2*sqrt(c**2*x**2 + 1)*asinh(c*x)/9 + b**2*d*x*asinh(c*x)**2 + 14*b**2*d*x/9 - 14*b
**2*d*sqrt(c**2*x**2 + 1)*asinh(c*x)/(9*c), Ne(c, 0)), (a**2*d*x, True))

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Giac [B]  time = 2.06326, size = 383, normalized size = 3.06 \begin{align*} \frac{1}{3} \, a^{2} c^{2} d x^{3} + \frac{2}{9} \,{\left (3 \, x^{3} \log \left (c x + \sqrt{c^{2} x^{2} + 1}\right ) - \frac{{\left (c^{2} x^{2} + 1\right )}^{\frac{3}{2}} - 3 \, \sqrt{c^{2} x^{2} + 1}}{c^{3}}\right )} a b c^{2} d + \frac{1}{27} \,{\left (9 \, x^{3} \log \left (c x + \sqrt{c^{2} x^{2} + 1}\right )^{2} + 2 \, c{\left (\frac{c^{2} x^{3} - 6 \, x}{c^{3}} - \frac{3 \,{\left ({\left (c^{2} x^{2} + 1\right )}^{\frac{3}{2}} - 3 \, \sqrt{c^{2} x^{2} + 1}\right )} \log \left (c x + \sqrt{c^{2} x^{2} + 1}\right )}{c^{4}}\right )}\right )} b^{2} c^{2} d + 2 \,{\left (x \log \left (c x + \sqrt{c^{2} x^{2} + 1}\right ) - \frac{\sqrt{c^{2} x^{2} + 1}}{c}\right )} a b d +{\left (x \log \left (c x + \sqrt{c^{2} x^{2} + 1}\right )^{2} + 2 \, c{\left (\frac{x}{c} - \frac{\sqrt{c^{2} x^{2} + 1} \log \left (c x + \sqrt{c^{2} x^{2} + 1}\right )}{c^{2}}\right )}\right )} b^{2} d + a^{2} d x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*a^2*c^2*d*x^3 + 2/9*(3*x^3*log(c*x + sqrt(c^2*x^2 + 1)) - ((c^2*x^2 + 1)^(3/2) - 3*sqrt(c^2*x^2 + 1))/c^3)
*a*b*c^2*d + 1/27*(9*x^3*log(c*x + sqrt(c^2*x^2 + 1))^2 + 2*c*((c^2*x^3 - 6*x)/c^3 - 3*((c^2*x^2 + 1)^(3/2) -
3*sqrt(c^2*x^2 + 1))*log(c*x + sqrt(c^2*x^2 + 1))/c^4))*b^2*c^2*d + 2*(x*log(c*x + sqrt(c^2*x^2 + 1)) - sqrt(c
^2*x^2 + 1)/c)*a*b*d + (x*log(c*x + sqrt(c^2*x^2 + 1))^2 + 2*c*(x/c - sqrt(c^2*x^2 + 1)*log(c*x + sqrt(c^2*x^2
+ 1))/c^2))*b^2*d + a^2*d*x