3.131 \(\int (d+c^2 d x^2)^{3/2} (a+b \sinh ^{-1}(c x)) \, dx\)

Optimal. Leaf size=180 \[ \frac{1}{4} x \left (c^2 d x^2+d\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )+\frac{3}{8} d x \sqrt{c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )+\frac{3 d \sqrt{c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )^2}{16 b c \sqrt{c^2 x^2+1}}-\frac{b c^3 d x^4 \sqrt{c^2 d x^2+d}}{16 \sqrt{c^2 x^2+1}}-\frac{5 b c d x^2 \sqrt{c^2 d x^2+d}}{16 \sqrt{c^2 x^2+1}} \]

[Out]

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

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Rubi [A]  time = 0.108896, antiderivative size = 180, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.217, Rules used = {5684, 5682, 5675, 30, 14} \[ \frac{1}{4} x \left (c^2 d x^2+d\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )+\frac{3}{8} d x \sqrt{c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )+\frac{3 d \sqrt{c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )^2}{16 b c \sqrt{c^2 x^2+1}}-\frac{b c^3 d x^4 \sqrt{c^2 d x^2+d}}{16 \sqrt{c^2 x^2+1}}-\frac{5 b c d x^2 \sqrt{c^2 d x^2+d}}{16 \sqrt{c^2 x^2+1}} \]

Antiderivative was successfully verified.

[In]

Int[(d + c^2*d*x^2)^(3/2)*(a + b*ArcSinh[c*x]),x]

[Out]

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

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

Mathematica [A]  time = 0.792922, size = 200, normalized size = 1.11 \[ \frac{3 a d^{3/2} \log \left (\sqrt{d} \sqrt{c^2 d x^2+d}+c d x\right )}{8 c}+\frac{1}{8} a d x \left (2 c^2 x^2+5\right ) \sqrt{c^2 d x^2+d}+\frac{b d \sqrt{c^2 d x^2+d} \left (2 \sinh ^{-1}(c x) \left (\sinh ^{-1}(c x)+\sinh \left (2 \sinh ^{-1}(c x)\right )\right )-\cosh \left (2 \sinh ^{-1}(c x)\right )\right )}{8 c \sqrt{c^2 x^2+1}}-\frac{b d \sqrt{c^2 d x^2+d} \left (8 \sinh ^{-1}(c x)^2-4 \sinh \left (4 \sinh ^{-1}(c x)\right ) \sinh ^{-1}(c x)+\cosh \left (4 \sinh ^{-1}(c x)\right )\right )}{128 c \sqrt{c^2 x^2+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[(d + c^2*d*x^2)^(3/2)*(a + b*ArcSinh[c*x]),x]

[Out]

(a*d*x*(5 + 2*c^2*x^2)*Sqrt[d + c^2*d*x^2])/8 + (3*a*d^(3/2)*Log[c*d*x + Sqrt[d]*Sqrt[d + c^2*d*x^2]])/(8*c) +
(b*d*Sqrt[d + c^2*d*x^2]*(-Cosh[2*ArcSinh[c*x]] + 2*ArcSinh[c*x]*(ArcSinh[c*x] + Sinh[2*ArcSinh[c*x]])))/(8*c
*Sqrt[1 + c^2*x^2]) - (b*d*Sqrt[d + c^2*d*x^2]*(8*ArcSinh[c*x]^2 + Cosh[4*ArcSinh[c*x]] - 4*ArcSinh[c*x]*Sinh[
4*ArcSinh[c*x]]))/(128*c*Sqrt[1 + c^2*x^2])

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Maple [B]  time = 0.13, size = 318, normalized size = 1.8 \begin{align*}{\frac{ax}{4} \left ({c}^{2}d{x}^{2}+d \right ) ^{{\frac{3}{2}}}}+{\frac{3\,adx}{8}\sqrt{{c}^{2}d{x}^{2}+d}}+{\frac{3\,a{d}^{2}}{8}\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{bd{c}^{4}{\it Arcsinh} \left ( cx \right ){x}^{5}}{4\,{c}^{2}{x}^{2}+4}\sqrt{d \left ({c}^{2}{x}^{2}+1 \right ) }}-{\frac{bd{c}^{3}{x}^{4}}{16}\sqrt{d \left ({c}^{2}{x}^{2}+1 \right ) }{\frac{1}{\sqrt{{c}^{2}{x}^{2}+1}}}}+{\frac{7\,{c}^{2}db{\it Arcsinh} \left ( cx \right ){x}^{3}}{8\,{c}^{2}{x}^{2}+8}\sqrt{d \left ({c}^{2}{x}^{2}+1 \right ) }}-{\frac{5\,bdc{x}^{2}}{16}\sqrt{d \left ({c}^{2}{x}^{2}+1 \right ) }{\frac{1}{\sqrt{{c}^{2}{x}^{2}+1}}}}+{\frac{5\,bd{\it Arcsinh} \left ( cx \right ) x}{8\,{c}^{2}{x}^{2}+8}\sqrt{d \left ({c}^{2}{x}^{2}+1 \right ) }}+{\frac{3\,b \left ({\it Arcsinh} \left ( cx \right ) \right ) ^{2}d}{16\,c}\sqrt{d \left ({c}^{2}{x}^{2}+1 \right ) }{\frac{1}{\sqrt{{c}^{2}{x}^{2}+1}}}}-{\frac{17\,bd}{128\,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((c^2*d*x^2+d)^(3/2)*(a+b*arcsinh(c*x)),x)

[Out]

1/4*x*(c^2*d*x^2+d)^(3/2)*a+3/8*a*d*x*(c^2*d*x^2+d)^(1/2)+3/8*a*d^2*ln(x*c^2*d/(c^2*d)^(1/2)+(c^2*d*x^2+d)^(1/
2))/(c^2*d)^(1/2)+1/4*b*(d*(c^2*x^2+1))^(1/2)*d*c^4/(c^2*x^2+1)*arcsinh(c*x)*x^5-1/16*b*(d*(c^2*x^2+1))^(1/2)*
d*c^3/(c^2*x^2+1)^(1/2)*x^4+7/8*b*(d*(c^2*x^2+1))^(1/2)*d*c^2/(c^2*x^2+1)*arcsinh(c*x)*x^3-5/16*b*(d*(c^2*x^2+
1))^(1/2)*d*c/(c^2*x^2+1)^(1/2)*x^2+5/8*b*(d*(c^2*x^2+1))^(1/2)*d/(c^2*x^2+1)*arcsinh(c*x)*x+3/16*b*(d*(c^2*x^
2+1))^(1/2)/(c^2*x^2+1)^(1/2)/c*arcsinh(c*x)^2*d-17/128*b*(d*(c^2*x^2+1))^(1/2)*d/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((c^2*d*x^2+d)^(3/2)*(a+b*arcsinh(c*x)),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 ({\left (a c^{2} d x^{2} + a d +{\left (b c^{2} d x^{2} + b d\right )} \operatorname{arsinh}\left (c x\right )\right )} \sqrt{c^{2} d x^{2} + d}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c^2*d*x^2+d)^(3/2)*(a+b*arcsinh(c*x)),x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (d \left (c^{2} x^{2} + 1\right )\right )^{\frac{3}{2}} \left (a + b \operatorname{asinh}{\left (c x \right )}\right )\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c**2*d*x**2+d)**(3/2)*(a+b*asinh(c*x)),x)

[Out]

Integral((d*(c**2*x**2 + 1))**(3/2)*(a + b*asinh(c*x)), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (c^{2} d x^{2} + d\right )}^{\frac{3}{2}}{\left (b \operatorname{arsinh}\left (c x\right ) + a\right )}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c^2*d*x^2+d)^(3/2)*(a+b*arcsinh(c*x)),x, algorithm="giac")

[Out]

integrate((c^2*d*x^2 + d)^(3/2)*(b*arcsinh(c*x) + a), x)