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

Optimal. Leaf size=184 \[ \frac{\sqrt{c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )^3}{6 b c \sqrt{c^2 x^2+1}}+\frac{1}{2} x \sqrt{c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )^2-\frac{b c x^2 \sqrt{c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )}{2 \sqrt{c^2 x^2+1}}+\frac{1}{4} b^2 x \sqrt{c^2 d x^2+d}-\frac{b^2 \sqrt{c^2 d x^2+d} \sinh ^{-1}(c x)}{4 c \sqrt{c^2 x^2+1}} \]

[Out]

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

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Rubi [A]  time = 0.120337, antiderivative size = 184, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {5682, 5675, 5661, 321, 215} \[ \frac{\sqrt{c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )^3}{6 b c \sqrt{c^2 x^2+1}}+\frac{1}{2} x \sqrt{c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )^2-\frac{b c x^2 \sqrt{c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )}{2 \sqrt{c^2 x^2+1}}+\frac{1}{4} b^2 x \sqrt{c^2 d x^2+d}-\frac{b^2 \sqrt{c^2 d x^2+d} \sinh ^{-1}(c x)}{4 c \sqrt{c^2 x^2+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 5661

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*ArcS
inh[c*x])^n)/(d*(m + 1)), x] - Dist[(b*c*n)/(d*(m + 1)), Int[((d*x)^(m + 1)*(a + b*ArcSinh[c*x])^(n - 1))/Sqrt
[1 + c^2*x^2], x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0] && NeQ[m, -1]

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 215

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSinh[(Rt[b, 2]*x)/Sqrt[a]]/Rt[b, 2], x] /; FreeQ[{a, b},
x] && GtQ[a, 0] && PosQ[b]

Rubi steps

\begin{align*} \int \sqrt{d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )^2 \, dx &=\frac{1}{2} x \sqrt{d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )^2+\frac{\sqrt{d+c^2 d x^2} \int \frac{\left (a+b \sinh ^{-1}(c x)\right )^2}{\sqrt{1+c^2 x^2}} \, dx}{2 \sqrt{1+c^2 x^2}}-\frac{\left (b c \sqrt{d+c^2 d x^2}\right ) \int x \left (a+b \sinh ^{-1}(c x)\right ) \, dx}{\sqrt{1+c^2 x^2}}\\ &=-\frac{b c x^2 \sqrt{d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )}{2 \sqrt{1+c^2 x^2}}+\frac{1}{2} x \sqrt{d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )^2+\frac{\sqrt{d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )^3}{6 b c \sqrt{1+c^2 x^2}}+\frac{\left (b^2 c^2 \sqrt{d+c^2 d x^2}\right ) \int \frac{x^2}{\sqrt{1+c^2 x^2}} \, dx}{2 \sqrt{1+c^2 x^2}}\\ &=\frac{1}{4} b^2 x \sqrt{d+c^2 d x^2}-\frac{b c x^2 \sqrt{d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )}{2 \sqrt{1+c^2 x^2}}+\frac{1}{2} x \sqrt{d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )^2+\frac{\sqrt{d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )^3}{6 b c \sqrt{1+c^2 x^2}}-\frac{\left (b^2 \sqrt{d+c^2 d x^2}\right ) \int \frac{1}{\sqrt{1+c^2 x^2}} \, dx}{4 \sqrt{1+c^2 x^2}}\\ &=\frac{1}{4} b^2 x \sqrt{d+c^2 d x^2}-\frac{b^2 \sqrt{d+c^2 d x^2} \sinh ^{-1}(c x)}{4 c \sqrt{1+c^2 x^2}}-\frac{b c x^2 \sqrt{d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )}{2 \sqrt{1+c^2 x^2}}+\frac{1}{2} x \sqrt{d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )^2+\frac{\sqrt{d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )^3}{6 b c \sqrt{1+c^2 x^2}}\\ \end{align*}

Mathematica [A]  time = 1.08672, size = 200, normalized size = 1.09 \[ \frac{1}{24} \left (12 a^2 x \sqrt{c^2 d x^2+d}+\frac{12 a^2 \sqrt{d} \log \left (\sqrt{d} \sqrt{c^2 d x^2+d}+c d x\right )}{c}+\frac{6 a b \sqrt{c^2 d x^2+d} \left (2 \sinh ^{-1}(c x) \left (\sinh ^{-1}(c x)+\sinh \left (2 \sinh ^{-1}(c x)\right )\right )-\cosh \left (2 \sinh ^{-1}(c x)\right )\right )}{c \sqrt{c^2 x^2+1}}+\frac{b^2 \sqrt{c^2 d x^2+d} \left (4 \sinh ^{-1}(c x)^3+\left (6 \sinh ^{-1}(c x)^2+3\right ) \sinh \left (2 \sinh ^{-1}(c x)\right )-6 \sinh ^{-1}(c x) \cosh \left (2 \sinh ^{-1}(c x)\right )\right )}{c \sqrt{c^2 x^2+1}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.146, size = 482, normalized size = 2.6 \begin{align*}{\frac{x{a}^{2}}{2}\sqrt{{c}^{2}d{x}^{2}+d}}+{\frac{{a}^{2}d}{2}\ln \left ({{c}^{2}dx{\frac{1}{\sqrt{{c}^{2}d}}}}+\sqrt{{c}^{2}d{x}^{2}+d} \right ){\frac{1}{\sqrt{{c}^{2}d}}}}-{\frac{{b}^{2}c{\it Arcsinh} \left ( cx \right ){x}^{2}}{2}\sqrt{d \left ({c}^{2}{x}^{2}+1 \right ) }{\frac{1}{\sqrt{{c}^{2}{x}^{2}+1}}}}+{\frac{{b}^{2} \left ({\it Arcsinh} \left ( cx \right ) \right ) ^{3}}{6\,c}\sqrt{d \left ({c}^{2}{x}^{2}+1 \right ) }{\frac{1}{\sqrt{{c}^{2}{x}^{2}+1}}}}+{\frac{{b}^{2}{c}^{2} \left ({\it Arcsinh} \left ( cx \right ) \right ) ^{2}{x}^{3}}{2\,{c}^{2}{x}^{2}+2}\sqrt{d \left ({c}^{2}{x}^{2}+1 \right ) }}+{\frac{{b}^{2} \left ({\it Arcsinh} \left ( cx \right ) \right ) ^{2}x}{2\,{c}^{2}{x}^{2}+2}\sqrt{d \left ({c}^{2}{x}^{2}+1 \right ) }}-{\frac{{b}^{2}{\it Arcsinh} \left ( cx \right ) }{4\,c}\sqrt{d \left ({c}^{2}{x}^{2}+1 \right ) }{\frac{1}{\sqrt{{c}^{2}{x}^{2}+1}}}}+{\frac{{b}^{2}{c}^{2}{x}^{3}}{4\,{c}^{2}{x}^{2}+4}\sqrt{d \left ({c}^{2}{x}^{2}+1 \right ) }}+{\frac{{b}^{2}x}{4\,{c}^{2}{x}^{2}+4}\sqrt{d \left ({c}^{2}{x}^{2}+1 \right ) }}+{\frac{ab \left ({\it Arcsinh} \left ( cx \right ) \right ) ^{2}}{2\,c}\sqrt{d \left ({c}^{2}{x}^{2}+1 \right ) }{\frac{1}{\sqrt{{c}^{2}{x}^{2}+1}}}}+{\frac{ab{c}^{2}{\it Arcsinh} \left ( cx \right ){x}^{3}}{{c}^{2}{x}^{2}+1}\sqrt{d \left ({c}^{2}{x}^{2}+1 \right ) }}-{\frac{abc{x}^{2}}{2}\sqrt{d \left ({c}^{2}{x}^{2}+1 \right ) }{\frac{1}{\sqrt{{c}^{2}{x}^{2}+1}}}}+{\frac{ab{\it Arcsinh} \left ( cx \right ) x}{{c}^{2}{x}^{2}+1}\sqrt{d \left ({c}^{2}{x}^{2}+1 \right ) }}-{\frac{ab}{4\,c}\sqrt{d \left ({c}^{2}{x}^{2}+1 \right ) }{\frac{1}{\sqrt{{c}^{2}{x}^{2}+1}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*x*a^2*(c^2*d*x^2+d)^(1/2)+1/2*a^2*d*ln(x*c^2*d/(c^2*d)^(1/2)+(c^2*d*x^2+d)^(1/2))/(c^2*d)^(1/2)-1/2*b^2*(d
*(c^2*x^2+1))^(1/2)*c/(c^2*x^2+1)^(1/2)*arcsinh(c*x)*x^2+1/6*b^2*(d*(c^2*x^2+1))^(1/2)/(c^2*x^2+1)^(1/2)/c*arc
sinh(c*x)^3+1/2*b^2*(d*(c^2*x^2+1))^(1/2)*c^2/(c^2*x^2+1)*arcsinh(c*x)^2*x^3+1/2*b^2*(d*(c^2*x^2+1))^(1/2)/(c^
2*x^2+1)*arcsinh(c*x)^2*x-1/4*b^2*(d*(c^2*x^2+1))^(1/2)/c/(c^2*x^2+1)^(1/2)*arcsinh(c*x)+1/4*b^2*(d*(c^2*x^2+1
))^(1/2)*c^2/(c^2*x^2+1)*x^3+1/4*b^2*(d*(c^2*x^2+1))^(1/2)/(c^2*x^2+1)*x+1/2*a*b*(d*(c^2*x^2+1))^(1/2)/(c^2*x^
2+1)^(1/2)/c*arcsinh(c*x)^2+a*b*(d*(c^2*x^2+1))^(1/2)*c^2/(c^2*x^2+1)*arcsinh(c*x)*x^3-1/2*a*b*(d*(c^2*x^2+1))
^(1/2)*c/(c^2*x^2+1)^(1/2)*x^2+a*b*(d*(c^2*x^2+1))^(1/2)/(c^2*x^2+1)*arcsinh(c*x)*x-1/4*a*b*(d*(c^2*x^2+1))^(1
/2)/c/(c^2*x^2+1)^(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(c*x))^2*(c^2*d*x^2+d)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsinh(c*x))^2*(c^2*d*x^2+d)^(1/2),x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{d \left (c^{2} x^{2} + 1\right )} \left (a + b \operatorname{asinh}{\left (c x \right )}\right )^{2}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*asinh(c*x))**2*(c**2*d*x**2+d)**(1/2),x)

[Out]

Integral(sqrt(d*(c**2*x**2 + 1))*(a + b*asinh(c*x))**2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsinh(c*x))^2*(c^2*d*x^2+d)^(1/2),x, algorithm="giac")

[Out]

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