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

Optimal. Leaf size=135 \[ -\frac{b d x \left (c^2 x^2+1\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )}{8 c}-\frac{3 b d x \sqrt{c^2 x^2+1} \left (a+b \sinh ^{-1}(c x)\right )}{16 c}+\frac{d \left (c^2 x^2+1\right )^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{4 c^2}-\frac{3 d \left (a+b \sinh ^{-1}(c x)\right )^2}{32 c^2}+\frac{1}{32} b^2 c^2 d x^4+\frac{5}{32} b^2 d x^2 \]

[Out]

(5*b^2*d*x^2)/32 + (b^2*c^2*d*x^4)/32 - (3*b*d*x*Sqrt[1 + c^2*x^2]*(a + b*ArcSinh[c*x]))/(16*c) - (b*d*x*(1 +
c^2*x^2)^(3/2)*(a + b*ArcSinh[c*x]))/(8*c) - (3*d*(a + b*ArcSinh[c*x])^2)/(32*c^2) + (d*(1 + c^2*x^2)^2*(a + b
*ArcSinh[c*x])^2)/(4*c^2)

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Rubi [A]  time = 0.134246, antiderivative size = 135, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.273, Rules used = {5717, 5684, 5682, 5675, 30, 14} \[ -\frac{b d x \left (c^2 x^2+1\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )}{8 c}-\frac{3 b d x \sqrt{c^2 x^2+1} \left (a+b \sinh ^{-1}(c x)\right )}{16 c}+\frac{d \left (c^2 x^2+1\right )^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{4 c^2}-\frac{3 d \left (a+b \sinh ^{-1}(c x)\right )^2}{32 c^2}+\frac{1}{32} b^2 c^2 d x^4+\frac{5}{32} b^2 d x^2 \]

Antiderivative was successfully verified.

[In]

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

[Out]

(5*b^2*d*x^2)/32 + (b^2*c^2*d*x^4)/32 - (3*b*d*x*Sqrt[1 + c^2*x^2]*(a + b*ArcSinh[c*x]))/(16*c) - (b*d*x*(1 +
c^2*x^2)^(3/2)*(a + b*ArcSinh[c*x]))/(8*c) - (3*d*(a + b*ArcSinh[c*x])^2)/(32*c^2) + (d*(1 + c^2*x^2)^2*(a + b
*ArcSinh[c*x])^2)/(4*c^2)

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 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 5682

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

Rule 5675

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

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rubi steps

\begin{align*} \int x \left (d+c^2 d x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )^2 \, dx &=\frac{d \left (1+c^2 x^2\right )^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{4 c^2}-\frac{(b d) \int \left (1+c^2 x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right ) \, dx}{2 c}\\ &=-\frac{b d x \left (1+c^2 x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )}{8 c}+\frac{d \left (1+c^2 x^2\right )^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{4 c^2}+\frac{1}{8} \left (b^2 d\right ) \int x \left (1+c^2 x^2\right ) \, dx-\frac{(3 b d) \int \sqrt{1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right ) \, dx}{8 c}\\ &=-\frac{3 b d x \sqrt{1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{16 c}-\frac{b d x \left (1+c^2 x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )}{8 c}+\frac{d \left (1+c^2 x^2\right )^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{4 c^2}+\frac{1}{8} \left (b^2 d\right ) \int \left (x+c^2 x^3\right ) \, dx+\frac{1}{16} \left (3 b^2 d\right ) \int x \, dx-\frac{(3 b d) \int \frac{a+b \sinh ^{-1}(c x)}{\sqrt{1+c^2 x^2}} \, dx}{16 c}\\ &=\frac{5}{32} b^2 d x^2+\frac{1}{32} b^2 c^2 d x^4-\frac{3 b d x \sqrt{1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{16 c}-\frac{b d x \left (1+c^2 x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )}{8 c}-\frac{3 d \left (a+b \sinh ^{-1}(c x)\right )^2}{32 c^2}+\frac{d \left (1+c^2 x^2\right )^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{4 c^2}\\ \end{align*}

Mathematica [A]  time = 0.258247, size = 155, normalized size = 1.15 \[ \frac{d \left (c x \left (8 a^2 c x \left (c^2 x^2+2\right )-2 a b \sqrt{c^2 x^2+1} \left (2 c^2 x^2+5\right )+b^2 c x \left (c^2 x^2+5\right )\right )+2 b \sinh ^{-1}(c x) \left (a \left (8 c^4 x^4+16 c^2 x^2+5\right )-b c x \sqrt{c^2 x^2+1} \left (2 c^2 x^2+5\right )\right )+b^2 \left (8 c^4 x^4+16 c^2 x^2+5\right ) \sinh ^{-1}(c x)^2\right )}{32 c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(d*(c*x*(8*a^2*c*x*(2 + c^2*x^2) + b^2*c*x*(5 + c^2*x^2) - 2*a*b*Sqrt[1 + c^2*x^2]*(5 + 2*c^2*x^2)) + 2*b*(-(b
*c*x*Sqrt[1 + c^2*x^2]*(5 + 2*c^2*x^2)) + a*(5 + 16*c^2*x^2 + 8*c^4*x^4))*ArcSinh[c*x] + b^2*(5 + 16*c^2*x^2 +
8*c^4*x^4)*ArcSinh[c*x]^2))/(32*c^2)

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Maple [A]  time = 0.036, size = 216, normalized size = 1.6 \begin{align*}{\frac{1}{{c}^{2}} \left ( d{a}^{2} \left ({\frac{{c}^{4}{x}^{4}}{4}}+{\frac{{c}^{2}{x}^{2}}{2}} \right ) +d{b}^{2} \left ({\frac{ \left ({\it Arcsinh} \left ( cx \right ) \right ) ^{2}{c}^{2}{x}^{2} \left ({c}^{2}{x}^{2}+1 \right ) }{4}}+{\frac{ \left ({\it Arcsinh} \left ( cx \right ) \right ) ^{2} \left ({c}^{2}{x}^{2}+1 \right ) }{4}}-{\frac{{\it Arcsinh} \left ( cx \right ) cx}{8} \left ({c}^{2}{x}^{2}+1 \right ) ^{{\frac{3}{2}}}}-{\frac{3\,{\it Arcsinh} \left ( cx \right ) cx}{16}\sqrt{{c}^{2}{x}^{2}+1}}-{\frac{3\, \left ({\it Arcsinh} \left ( cx \right ) \right ) ^{2}}{32}}+{\frac{{c}^{2}{x}^{2} \left ({c}^{2}{x}^{2}+1 \right ) }{32}}+{\frac{{c}^{2}{x}^{2}}{8}}+{\frac{1}{8}} \right ) +2\,dab \left ( 1/4\,{\it Arcsinh} \left ( cx \right ){c}^{4}{x}^{4}+1/2\,{\it Arcsinh} \left ( cx \right ){c}^{2}{x}^{2}-1/16\,{c}^{3}{x}^{3}\sqrt{{c}^{2}{x}^{2}+1}-{\frac{5\,cx\sqrt{{c}^{2}{x}^{2}+1}}{32}}+{\frac{5\,{\it Arcsinh} \left ( cx \right ) }{32}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/c^2*(d*a^2*(1/4*c^4*x^4+1/2*c^2*x^2)+d*b^2*(1/4*arcsinh(c*x)^2*c^2*x^2*(c^2*x^2+1)+1/4*arcsinh(c*x)^2*(c^2*x
^2+1)-1/8*arcsinh(c*x)*c*x*(c^2*x^2+1)^(3/2)-3/16*arcsinh(c*x)*(c^2*x^2+1)^(1/2)*c*x-3/32*arcsinh(c*x)^2+1/32*
c^2*x^2*(c^2*x^2+1)+1/8*c^2*x^2+1/8)+2*d*a*b*(1/4*arcsinh(c*x)*c^4*x^4+1/2*arcsinh(c*x)*c^2*x^2-1/16*c^3*x^3*(
c^2*x^2+1)^(1/2)-5/32*c*x*(c^2*x^2+1)^(1/2)+5/32*arcsinh(c*x)))

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Maxima [B]  time = 1.22708, size = 552, normalized size = 4.09 \begin{align*} \frac{1}{4} \, b^{2} c^{2} d x^{4} \operatorname{arsinh}\left (c x\right )^{2} + \frac{1}{4} \, a^{2} c^{2} d x^{4} + \frac{1}{2} \, b^{2} d x^{2} \operatorname{arsinh}\left (c x\right )^{2} + \frac{1}{16} \,{\left (8 \, x^{4} \operatorname{arsinh}\left (c x\right ) -{\left (\frac{2 \, \sqrt{c^{2} x^{2} + 1} x^{3}}{c^{2}} - \frac{3 \, \sqrt{c^{2} x^{2} + 1} x}{c^{4}} + \frac{3 \, \operatorname{arsinh}\left (\frac{c^{2} x}{\sqrt{c^{2}}}\right )}{\sqrt{c^{2}} c^{4}}\right )} c\right )} a b c^{2} d + \frac{1}{32} \,{\left ({\left (\frac{x^{4}}{c^{2}} - \frac{3 \, x^{2}}{c^{4}} + \frac{3 \, \log \left (\frac{c^{2} x}{\sqrt{c^{2}}} + \sqrt{c^{2} x^{2} + 1}\right )^{2}}{c^{6}}\right )} c^{2} - 2 \,{\left (\frac{2 \, \sqrt{c^{2} x^{2} + 1} x^{3}}{c^{2}} - \frac{3 \, \sqrt{c^{2} x^{2} + 1} x}{c^{4}} + \frac{3 \, \operatorname{arsinh}\left (\frac{c^{2} x}{\sqrt{c^{2}}}\right )}{\sqrt{c^{2}} c^{4}}\right )} c \operatorname{arsinh}\left (c x\right )\right )} b^{2} c^{2} d + \frac{1}{2} \, a^{2} d x^{2} + \frac{1}{2} \,{\left (2 \, x^{2} \operatorname{arsinh}\left (c x\right ) - c{\left (\frac{\sqrt{c^{2} x^{2} + 1} x}{c^{2}} - \frac{\operatorname{arsinh}\left (\frac{c^{2} x}{\sqrt{c^{2}}}\right )}{\sqrt{c^{2}} c^{2}}\right )}\right )} a b d + \frac{1}{4} \,{\left (c^{2}{\left (\frac{x^{2}}{c^{2}} - \frac{\log \left (\frac{c^{2} x}{\sqrt{c^{2}}} + \sqrt{c^{2} x^{2} + 1}\right )^{2}}{c^{4}}\right )} - 2 \, c{\left (\frac{\sqrt{c^{2} x^{2} + 1} x}{c^{2}} - \frac{\operatorname{arsinh}\left (\frac{c^{2} x}{\sqrt{c^{2}}}\right )}{\sqrt{c^{2}} c^{2}}\right )} \operatorname{arsinh}\left (c x\right )\right )} b^{2} d \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*b^2*c^2*d*x^4*arcsinh(c*x)^2 + 1/4*a^2*c^2*d*x^4 + 1/2*b^2*d*x^2*arcsinh(c*x)^2 + 1/16*(8*x^4*arcsinh(c*x)
- (2*sqrt(c^2*x^2 + 1)*x^3/c^2 - 3*sqrt(c^2*x^2 + 1)*x/c^4 + 3*arcsinh(c^2*x/sqrt(c^2))/(sqrt(c^2)*c^4))*c)*a
*b*c^2*d + 1/32*((x^4/c^2 - 3*x^2/c^4 + 3*log(c^2*x/sqrt(c^2) + sqrt(c^2*x^2 + 1))^2/c^6)*c^2 - 2*(2*sqrt(c^2*
x^2 + 1)*x^3/c^2 - 3*sqrt(c^2*x^2 + 1)*x/c^4 + 3*arcsinh(c^2*x/sqrt(c^2))/(sqrt(c^2)*c^4))*c*arcsinh(c*x))*b^2
*c^2*d + 1/2*a^2*d*x^2 + 1/2*(2*x^2*arcsinh(c*x) - c*(sqrt(c^2*x^2 + 1)*x/c^2 - arcsinh(c^2*x/sqrt(c^2))/(sqrt
(c^2)*c^2)))*a*b*d + 1/4*(c^2*(x^2/c^2 - log(c^2*x/sqrt(c^2) + sqrt(c^2*x^2 + 1))^2/c^4) - 2*c*(sqrt(c^2*x^2 +
1)*x/c^2 - arcsinh(c^2*x/sqrt(c^2))/(sqrt(c^2)*c^2))*arcsinh(c*x))*b^2*d

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Fricas [A]  time = 2.69083, size = 447, normalized size = 3.31 \begin{align*} \frac{{\left (8 \, a^{2} + b^{2}\right )} c^{4} d x^{4} +{\left (16 \, a^{2} + 5 \, b^{2}\right )} c^{2} d x^{2} +{\left (8 \, b^{2} c^{4} d x^{4} + 16 \, b^{2} c^{2} d x^{2} + 5 \, b^{2} d\right )} \log \left (c x + \sqrt{c^{2} x^{2} + 1}\right )^{2} + 2 \,{\left (8 \, a b c^{4} d x^{4} + 16 \, a b c^{2} d x^{2} + 5 \, a b d -{\left (2 \, b^{2} c^{3} d x^{3} + 5 \, b^{2} c d x\right )} \sqrt{c^{2} x^{2} + 1}\right )} \log \left (c x + \sqrt{c^{2} x^{2} + 1}\right ) - 2 \,{\left (2 \, a b c^{3} d x^{3} + 5 \, a b c d x\right )} \sqrt{c^{2} x^{2} + 1}}{32 \, c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/32*((8*a^2 + b^2)*c^4*d*x^4 + (16*a^2 + 5*b^2)*c^2*d*x^2 + (8*b^2*c^4*d*x^4 + 16*b^2*c^2*d*x^2 + 5*b^2*d)*lo
g(c*x + sqrt(c^2*x^2 + 1))^2 + 2*(8*a*b*c^4*d*x^4 + 16*a*b*c^2*d*x^2 + 5*a*b*d - (2*b^2*c^3*d*x^3 + 5*b^2*c*d*
x)*sqrt(c^2*x^2 + 1))*log(c*x + sqrt(c^2*x^2 + 1)) - 2*(2*a*b*c^3*d*x^3 + 5*a*b*c*d*x)*sqrt(c^2*x^2 + 1))/c^2

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Sympy [A]  time = 3.39426, size = 269, normalized size = 1.99 \begin{align*} \begin{cases} \frac{a^{2} c^{2} d x^{4}}{4} + \frac{a^{2} d x^{2}}{2} + \frac{a b c^{2} d x^{4} \operatorname{asinh}{\left (c x \right )}}{2} - \frac{a b c d x^{3} \sqrt{c^{2} x^{2} + 1}}{8} + a b d x^{2} \operatorname{asinh}{\left (c x \right )} - \frac{5 a b d x \sqrt{c^{2} x^{2} + 1}}{16 c} + \frac{5 a b d \operatorname{asinh}{\left (c x \right )}}{16 c^{2}} + \frac{b^{2} c^{2} d x^{4} \operatorname{asinh}^{2}{\left (c x \right )}}{4} + \frac{b^{2} c^{2} d x^{4}}{32} - \frac{b^{2} c d x^{3} \sqrt{c^{2} x^{2} + 1} \operatorname{asinh}{\left (c x \right )}}{8} + \frac{b^{2} d x^{2} \operatorname{asinh}^{2}{\left (c x \right )}}{2} + \frac{5 b^{2} d x^{2}}{32} - \frac{5 b^{2} d x \sqrt{c^{2} x^{2} + 1} \operatorname{asinh}{\left (c x \right )}}{16 c} + \frac{5 b^{2} d \operatorname{asinh}^{2}{\left (c x \right )}}{32 c^{2}} & \text{for}\: c \neq 0 \\\frac{a^{2} d x^{2}}{2} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((a**2*c**2*d*x**4/4 + a**2*d*x**2/2 + a*b*c**2*d*x**4*asinh(c*x)/2 - a*b*c*d*x**3*sqrt(c**2*x**2 + 1
)/8 + a*b*d*x**2*asinh(c*x) - 5*a*b*d*x*sqrt(c**2*x**2 + 1)/(16*c) + 5*a*b*d*asinh(c*x)/(16*c**2) + b**2*c**2*
d*x**4*asinh(c*x)**2/4 + b**2*c**2*d*x**4/32 - b**2*c*d*x**3*sqrt(c**2*x**2 + 1)*asinh(c*x)/8 + b**2*d*x**2*as
inh(c*x)**2/2 + 5*b**2*d*x**2/32 - 5*b**2*d*x*sqrt(c**2*x**2 + 1)*asinh(c*x)/(16*c) + 5*b**2*d*asinh(c*x)**2/(
32*c**2), Ne(c, 0)), (a**2*d*x**2/2, True))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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