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

Optimal. Leaf size=106 \[ -\frac{(a+b x)^4}{16 b}-\frac{5 (a+b x)^2}{16 b}+\frac{\left ((a+b x)^2+1\right )^{3/2} (a+b x) \sinh ^{-1}(a+b x)}{4 b}+\frac{3 \sqrt{(a+b x)^2+1} (a+b x) \sinh ^{-1}(a+b x)}{8 b}+\frac{3 \sinh ^{-1}(a+b x)^2}{16 b} \]

[Out]

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

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Rubi [A]  time = 0.103627, antiderivative size = 106, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.214, Rules used = {5867, 5684, 5682, 5675, 30, 14} \[ -\frac{(a+b x)^4}{16 b}-\frac{5 (a+b x)^2}{16 b}+\frac{\left ((a+b x)^2+1\right )^{3/2} (a+b x) \sinh ^{-1}(a+b x)}{4 b}+\frac{3 \sqrt{(a+b x)^2+1} (a+b x) \sinh ^{-1}(a+b x)}{8 b}+\frac{3 \sinh ^{-1}(a+b x)^2}{16 b} \]

Antiderivative was successfully verified.

[In]

Int[(1 + a^2 + 2*a*b*x + b^2*x^2)^(3/2)*ArcSinh[a + b*x],x]

[Out]

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

Mathematica [A]  time = 0.090909, size = 124, normalized size = 1.17 \[ \frac{-b x \left (6 a^2 b x+4 a^3+4 a b^2 x^2+10 a+b^3 x^3+5 b x\right )+2 \sqrt{a^2+2 a b x+b^2 x^2+1} \left (6 a^2 b x+2 a^3+6 a b^2 x^2+5 a+2 b^3 x^3+5 b x\right ) \sinh ^{-1}(a+b x)+3 \sinh ^{-1}(a+b x)^2}{16 b} \]

Antiderivative was successfully verified.

[In]

Integrate[(1 + a^2 + 2*a*b*x + b^2*x^2)^(3/2)*ArcSinh[a + b*x],x]

[Out]

(-(b*x*(10*a + 4*a^3 + 5*b*x + 6*a^2*b*x + 4*a*b^2*x^2 + b^3*x^3)) + 2*Sqrt[1 + a^2 + 2*a*b*x + b^2*x^2]*(5*a
+ 2*a^3 + 5*b*x + 6*a^2*b*x + 6*a*b^2*x^2 + 2*b^3*x^3)*ArcSinh[a + b*x] + 3*ArcSinh[a + b*x]^2)/(16*b)

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Maple [B]  time = 0.055, size = 262, normalized size = 2.5 \begin{align*}{\frac{1}{16\,b} \left ( 4\,{\it Arcsinh} \left ( bx+a \right ) \sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}+1}{x}^{3}{b}^{3}-{x}^{4}{b}^{4}+12\,{\it Arcsinh} \left ( bx+a \right ) \sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}+1}{x}^{2}a{b}^{2}-4\,{x}^{3}a{b}^{3}+12\,{\it Arcsinh} \left ( bx+a \right ) \sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}+1}x{a}^{2}b-6\,{x}^{2}{a}^{2}{b}^{2}+4\,{\it Arcsinh} \left ( bx+a \right ) \sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}+1}{a}^{3}-4\,x{a}^{3}b+10\,{\it Arcsinh} \left ( bx+a \right ) \sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}+1}xb-5\,{b}^{2}{x}^{2}-{a}^{4}+10\,{\it Arcsinh} \left ( bx+a \right ) \sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}+1}a-10\,xab+3\, \left ({\it Arcsinh} \left ( bx+a \right ) \right ) ^{2}-5\,{a}^{2}-4 \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(4*arcsinh(b*x+a)*(b^2*x^2+2*a*b*x+a^2+1)^(1/2)*x^3*b^3-x^4*b^4+12*arcsinh(b*x+a)*(b^2*x^2+2*a*b*x+a^2+1)
^(1/2)*x^2*a*b^2-4*x^3*a*b^3+12*arcsinh(b*x+a)*(b^2*x^2+2*a*b*x+a^2+1)^(1/2)*x*a^2*b-6*x^2*a^2*b^2+4*arcsinh(b
*x+a)*(b^2*x^2+2*a*b*x+a^2+1)^(1/2)*a^3-4*x*a^3*b+10*arcsinh(b*x+a)*(b^2*x^2+2*a*b*x+a^2+1)^(1/2)*x*b-5*b^2*x^
2-a^4+10*arcsinh(b*x+a)*(b^2*x^2+2*a*b*x+a^2+1)^(1/2)*a-10*x*a*b+3*arcsinh(b*x+a)^2-5*a^2-4)/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((b^2*x^2+2*a*b*x+a^2+1)^(3/2)*arcsinh(b*x+a),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.69118, size = 374, normalized size = 3.53 \begin{align*} -\frac{b^{4} x^{4} + 4 \, a b^{3} x^{3} +{\left (6 \, a^{2} + 5\right )} b^{2} x^{2} + 2 \,{\left (2 \, a^{3} + 5 \, a\right )} b x - 2 \,{\left (2 \, b^{3} x^{3} + 6 \, a b^{2} x^{2} + 2 \, a^{3} +{\left (6 \, a^{2} + 5\right )} b x + 5 \, a\right )} \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1} \log \left (b x + a + \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right ) - 3 \, \log \left (b x + a + \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right )^{2}}{16 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*(b^4*x^4 + 4*a*b^3*x^3 + (6*a^2 + 5)*b^2*x^2 + 2*(2*a^3 + 5*a)*b*x - 2*(2*b^3*x^3 + 6*a*b^2*x^2 + 2*a^3
+ (6*a^2 + 5)*b*x + 5*a)*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)*log(b*x + a + sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)) -
3*log(b*x + a + sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1))^2)/b

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Sympy [A]  time = 7.3441, size = 298, normalized size = 2.81 \begin{align*} \begin{cases} - \frac{a^{3} x}{4} + \frac{a^{3} \sqrt{a^{2} + 2 a b x + b^{2} x^{2} + 1} \operatorname{asinh}{\left (a + b x \right )}}{4 b} - \frac{3 a^{2} b x^{2}}{8} + \frac{3 a^{2} x \sqrt{a^{2} + 2 a b x + b^{2} x^{2} + 1} \operatorname{asinh}{\left (a + b x \right )}}{4} - \frac{a b^{2} x^{3}}{4} + \frac{3 a b x^{2} \sqrt{a^{2} + 2 a b x + b^{2} x^{2} + 1} \operatorname{asinh}{\left (a + b x \right )}}{4} - \frac{5 a x}{8} + \frac{5 a \sqrt{a^{2} + 2 a b x + b^{2} x^{2} + 1} \operatorname{asinh}{\left (a + b x \right )}}{8 b} - \frac{b^{3} x^{4}}{16} + \frac{b^{2} x^{3} \sqrt{a^{2} + 2 a b x + b^{2} x^{2} + 1} \operatorname{asinh}{\left (a + b x \right )}}{4} - \frac{5 b x^{2}}{16} + \frac{5 x \sqrt{a^{2} + 2 a b x + b^{2} x^{2} + 1} \operatorname{asinh}{\left (a + b x \right )}}{8} + \frac{3 \operatorname{asinh}^{2}{\left (a + b x \right )}}{16 b} & \text{for}\: b \neq 0 \\x \left (a^{2} + 1\right )^{\frac{3}{2}} \operatorname{asinh}{\left (a \right )} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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