3.260 \(\int \sqrt{1+a^2+2 a b x+b^2 x^2} \sinh ^{-1}(a+b x)^3 \, dx\)

Optimal. Leaf size=131 \[ -\frac{3 (a+b x)^2}{8 b}+\frac{\sinh ^{-1}(a+b x)^4}{8 b}+\frac{(a+b x) \sqrt{(a+b x)^2+1} \sinh ^{-1}(a+b x)^3}{2 b}-\frac{3 (a+b x)^2 \sinh ^{-1}(a+b x)^2}{4 b}-\frac{3 \sinh ^{-1}(a+b x)^2}{8 b}+\frac{3 (a+b x) \sqrt{(a+b x)^2+1} \sinh ^{-1}(a+b x)}{4 b} \]

[Out]

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

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Rubi [A]  time = 0.188993, antiderivative size = 131, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {5867, 5682, 5675, 5661, 5758, 30} \[ -\frac{3 (a+b x)^2}{8 b}+\frac{\sinh ^{-1}(a+b x)^4}{8 b}+\frac{(a+b x) \sqrt{(a+b x)^2+1} \sinh ^{-1}(a+b x)^3}{2 b}-\frac{3 (a+b x)^2 \sinh ^{-1}(a+b x)^2}{4 b}-\frac{3 \sinh ^{-1}(a+b x)^2}{8 b}+\frac{3 (a+b x) \sqrt{(a+b x)^2+1} \sinh ^{-1}(a+b x)}{4 b} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[1 + a^2 + 2*a*b*x + b^2*x^2]*ArcSinh[a + b*x]^3,x]

[Out]

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

Rule 5867

Int[((a_.) + ArcSinh[(c_) + (d_.)*(x_)]*(b_.))^(n_.)*((A_.) + (B_.)*(x_) + (C_.)*(x_)^2)^(p_.), x_Symbol] :> D
ist[1/d, Subst[Int[(C/d^2 + (C*x^2)/d^2)^p*(a + b*ArcSinh[x])^n, x], x, c + d*x], x] /; FreeQ[{a, b, c, d, A,
B, C, n, p}, x] && EqQ[B*(1 + c^2) - 2*A*c*d, 0] && EqQ[2*c*C - B*d, 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 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 5758

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

Rule 30

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

Rubi steps

\begin{align*} \int \sqrt{1+a^2+2 a b x+b^2 x^2} \sinh ^{-1}(a+b x)^3 \, dx &=\frac{\operatorname{Subst}\left (\int \sqrt{1+x^2} \sinh ^{-1}(x)^3 \, dx,x,a+b x\right )}{b}\\ &=\frac{(a+b x) \sqrt{1+(a+b x)^2} \sinh ^{-1}(a+b x)^3}{2 b}+\frac{\operatorname{Subst}\left (\int \frac{\sinh ^{-1}(x)^3}{\sqrt{1+x^2}} \, dx,x,a+b x\right )}{2 b}-\frac{3 \operatorname{Subst}\left (\int x \sinh ^{-1}(x)^2 \, dx,x,a+b x\right )}{2 b}\\ &=-\frac{3 (a+b x)^2 \sinh ^{-1}(a+b x)^2}{4 b}+\frac{(a+b x) \sqrt{1+(a+b x)^2} \sinh ^{-1}(a+b x)^3}{2 b}+\frac{\sinh ^{-1}(a+b x)^4}{8 b}+\frac{3 \operatorname{Subst}\left (\int \frac{x^2 \sinh ^{-1}(x)}{\sqrt{1+x^2}} \, dx,x,a+b x\right )}{2 b}\\ &=\frac{3 (a+b x) \sqrt{1+(a+b x)^2} \sinh ^{-1}(a+b x)}{4 b}-\frac{3 (a+b x)^2 \sinh ^{-1}(a+b x)^2}{4 b}+\frac{(a+b x) \sqrt{1+(a+b x)^2} \sinh ^{-1}(a+b x)^3}{2 b}+\frac{\sinh ^{-1}(a+b x)^4}{8 b}-\frac{3 \operatorname{Subst}(\int x \, dx,x,a+b x)}{4 b}-\frac{3 \operatorname{Subst}\left (\int \frac{\sinh ^{-1}(x)}{\sqrt{1+x^2}} \, dx,x,a+b x\right )}{4 b}\\ &=-\frac{3 (a+b x)^2}{8 b}+\frac{3 (a+b x) \sqrt{1+(a+b x)^2} \sinh ^{-1}(a+b x)}{4 b}-\frac{3 \sinh ^{-1}(a+b x)^2}{8 b}-\frac{3 (a+b x)^2 \sinh ^{-1}(a+b x)^2}{4 b}+\frac{(a+b x) \sqrt{1+(a+b x)^2} \sinh ^{-1}(a+b x)^3}{2 b}+\frac{\sinh ^{-1}(a+b x)^4}{8 b}\\ \end{align*}

Mathematica [A]  time = 0.108373, size = 127, normalized size = 0.97 \[ \frac{4 (a+b x) \sqrt{a^2+2 a b x+b^2 x^2+1} \sinh ^{-1}(a+b x)^3-3 \left (2 a^2+4 a b x+2 b^2 x^2+1\right ) \sinh ^{-1}(a+b x)^2+6 (a+b x) \sqrt{a^2+2 a b x+b^2 x^2+1} \sinh ^{-1}(a+b x)-3 b x (2 a+b x)+\sinh ^{-1}(a+b x)^4}{8 b} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[1 + a^2 + 2*a*b*x + b^2*x^2]*ArcSinh[a + b*x]^3,x]

[Out]

(-3*b*x*(2*a + b*x) + 6*(a + b*x)*Sqrt[1 + a^2 + 2*a*b*x + b^2*x^2]*ArcSinh[a + b*x] - 3*(1 + 2*a^2 + 4*a*b*x
+ 2*b^2*x^2)*ArcSinh[a + b*x]^2 + 4*(a + b*x)*Sqrt[1 + a^2 + 2*a*b*x + b^2*x^2]*ArcSinh[a + b*x]^3 + ArcSinh[a
+ b*x]^4)/(8*b)

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Maple [A]  time = 0.084, size = 204, normalized size = 1.6 \begin{align*}{\frac{1}{8\,b} \left ( 4\, \left ({\it Arcsinh} \left ( bx+a \right ) \right ) ^{3}\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}+1}xb-6\, \left ({\it Arcsinh} \left ( bx+a \right ) \right ) ^{2}{x}^{2}{b}^{2}+4\, \left ({\it Arcsinh} \left ( bx+a \right ) \right ) ^{3}\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}+1}a-12\, \left ({\it Arcsinh} \left ( bx+a \right ) \right ) ^{2}xab+ \left ({\it Arcsinh} \left ( bx+a \right ) \right ) ^{4}-6\, \left ({\it Arcsinh} \left ( bx+a \right ) \right ) ^{2}{a}^{2}+6\,{\it Arcsinh} \left ( bx+a \right ) \sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}+1}xb-3\,{b}^{2}{x}^{2}+6\,{\it Arcsinh} \left ( bx+a \right ) \sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}+1}a-6\,xab-3\, \left ({\it Arcsinh} \left ( bx+a \right ) \right ) ^{2}-3\,{a}^{2}-3 \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arcsinh(b*x+a)^3*(b^2*x^2+2*a*b*x+a^2+1)^(1/2),x)

[Out]

1/8*(4*arcsinh(b*x+a)^3*(b^2*x^2+2*a*b*x+a^2+1)^(1/2)*x*b-6*arcsinh(b*x+a)^2*x^2*b^2+4*arcsinh(b*x+a)^3*(b^2*x
^2+2*a*b*x+a^2+1)^(1/2)*a-12*arcsinh(b*x+a)^2*x*a*b+arcsinh(b*x+a)^4-6*arcsinh(b*x+a)^2*a^2+6*arcsinh(b*x+a)*(
b^2*x^2+2*a*b*x+a^2+1)^(1/2)*x*b-3*b^2*x^2+6*arcsinh(b*x+a)*(b^2*x^2+2*a*b*x+a^2+1)^(1/2)*a-6*x*a*b-3*arcsinh(
b*x+a)^2-3*a^2-3)/b

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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(arcsinh(b*x+a)^3*(b^2*x^2+2*a*b*x+a^2+1)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.71438, size = 495, normalized size = 3.78 \begin{align*} \frac{4 \, \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1}{\left (b x + a\right )} \log \left (b x + a + \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right )^{3} - 3 \, b^{2} x^{2} + \log \left (b x + a + \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right )^{4} - 6 \, a b x - 3 \,{\left (2 \, b^{2} x^{2} + 4 \, a b x + 2 \, a^{2} + 1\right )} \log \left (b x + a + \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right )^{2} + 6 \, \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1}{\left (b x + a\right )} \log \left (b x + a + \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right )}{8 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsinh(b*x+a)^3*(b^2*x^2+2*a*b*x+a^2+1)^(1/2),x, algorithm="fricas")

[Out]

1/8*(4*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)*(b*x + a)*log(b*x + a + sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1))^3 - 3*b^2*
x^2 + log(b*x + a + sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1))^4 - 6*a*b*x - 3*(2*b^2*x^2 + 4*a*b*x + 2*a^2 + 1)*log(b
*x + a + sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1))^2 + 6*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)*(b*x + a)*log(b*x + a + sq
rt(b^2*x^2 + 2*a*b*x + a^2 + 1)))/b

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(asinh(b*x+a)**3*(b**2*x**2+2*a*b*x+a**2+1)**(1/2),x)

[Out]

Integral(sqrt(a**2 + 2*a*b*x + b**2*x**2 + 1)*asinh(a + b*x)**3, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsinh(b*x+a)^3*(b^2*x^2+2*a*b*x+a^2+1)^(1/2),x, algorithm="giac")

[Out]

integrate(sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)*arcsinh(b*x + a)^3, x)