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]

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

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Rubi [A]  time = 0.10, antiderivative size = 106, normalized size of antiderivative = 1.00, 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 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]]

Rule 30

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

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 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 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 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]

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*}

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Mathematica [A]  time = 0.10, size = 124, normalized size = 1.17 \[ \frac {-b x \left (4 a^3+6 a^2 b x+4 a b^2 x^2+10 a+b^3 x^3+5 b x\right )+2 \sqrt {a^2+2 a b x+b^2 x^2+1} \left (2 a^3+6 a^2 b x+6 a b^2 x^2+5 a+2 b^3 x^3+5 b x\right ) \sinh ^{-1}(a+b x)+3 \sinh ^{-1}(a+b x)^2}{16 b} \]

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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fricas [A]  time = 0.43, size = 160, normalized size = 1.51 \[ -\frac {b^{4} x^{4} + 4 \, a b^{3} x^{3} + {\left (6 \, a^{2} + 5\right )} b^{2} x^{2} + 2 \, {\left (2 \, a^{3} + 5 \, a\right )} b x - 2 \, {\left (2 \, b^{3} x^{3} + 6 \, a b^{2} x^{2} + 2 \, a^{3} + {\left (6 \, a^{2} + 5\right )} b x + 5 \, a\right )} \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} \log \left (b x + a + \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right ) - 3 \, \log \left (b x + a + \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right )^{2}}{16 \, b} \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )}^{\frac {3}{2}} \operatorname {arsinh}\left (b x + a\right )\,{d x} \]

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)

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maple [B]  time = 0.11, size = 262, normalized size = 2.47 \[ \frac {4 \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\, \arcsinh \left (b x +a \right ) x^{3} b^{3}-x^{4} b^{4}+12 \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\, \arcsinh \left (b x +a \right ) x^{2} a \,b^{2}-4 x^{3} a \,b^{3}+12 \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\, \arcsinh \left (b x +a \right ) x \,a^{2} b -6 x^{2} a^{2} b^{2}+4 \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\, \arcsinh \left (b x +a \right ) a^{3}-4 x \,a^{3} b +10 \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\, \arcsinh \left (b x +a \right ) x b -5 b^{2} x^{2}-a^{4}+10 \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\, \arcsinh \left (b x +a \right ) a -10 a b x +3 \arcsinh \left (b x +a \right )^{2}-5 a^{2}-4}{16 b} \]

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*(b^2*x^2+2*a*b*x+a^2+1)^(1/2)*arcsinh(b*x+a)*x^3*b^3-x^4*b^4+12*(b^2*x^2+2*a*b*x+a^2+1)^(1/2)*arcsinh(
b*x+a)*x^2*a*b^2-4*x^3*a*b^3+12*(b^2*x^2+2*a*b*x+a^2+1)^(1/2)*arcsinh(b*x+a)*x*a^2*b-6*x^2*a^2*b^2+4*(b^2*x^2+
2*a*b*x+a^2+1)^(1/2)*arcsinh(b*x+a)*a^3-4*x*a^3*b+10*(b^2*x^2+2*a*b*x+a^2+1)^(1/2)*arcsinh(b*x+a)*x*b-5*b^2*x^
2-a^4+10*(b^2*x^2+2*a*b*x+a^2+1)^(1/2)*arcsinh(b*x+a)*a-10*a*b*x+3*arcsinh(b*x+a)^2-5*a^2-4)/b

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maxima [B]  time = 0.79, size = 394, normalized size = 3.72 \[ -\frac {1}{16} \, {\left (b^{2} x^{4} + 4 \, a b x^{3} + 6 \, a^{2} x^{2} + \frac {4 \, a^{3} x}{b} + 5 \, x^{2} + \frac {10 \, a x}{b} + \frac {6 \, \operatorname {arsinh}\left (b x + a\right ) \operatorname {arsinh}\left (\frac {2 \, {\left (b^{2} x + a b\right )}}{\sqrt {-4 \, a^{2} b^{2} + 4 \, {\left (a^{2} + 1\right )} b^{2}}}\right )}{b^{2}} - \frac {3 \, \operatorname {arsinh}\left (\frac {2 \, {\left (b^{2} x + a b\right )}}{\sqrt {-4 \, a^{2} b^{2} + 4 \, {\left (a^{2} + 1\right )} b^{2}}}\right )^{2}}{b^{2}}\right )} b + \frac {1}{8} \, {\left (2 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )}^{\frac {3}{2}} x + \frac {2 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )}^{\frac {3}{2}} a}{b} + \frac {3 \, {\left (a^{2} b^{2} - {\left (a^{2} + 1\right )} b^{2}\right )} a^{2} \operatorname {arsinh}\left (\frac {2 \, {\left (b^{2} x + a b\right )}}{\sqrt {-4 \, a^{2} b^{2} + 4 \, {\left (a^{2} + 1\right )} b^{2}}}\right )}{b^{3}} - \frac {3 \, {\left (a^{2} b^{2} - {\left (a^{2} + 1\right )} b^{2}\right )} \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} x}{b^{2}} - \frac {3 \, {\left (a^{2} b^{2} - {\left (a^{2} + 1\right )} b^{2}\right )} {\left (a^{2} + 1\right )} \operatorname {arsinh}\left (\frac {2 \, {\left (b^{2} x + a b\right )}}{\sqrt {-4 \, a^{2} b^{2} + 4 \, {\left (a^{2} + 1\right )} b^{2}}}\right )}{b^{3}} - \frac {3 \, {\left (a^{2} b^{2} - {\left (a^{2} + 1\right )} b^{2}\right )} \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} a}{b^{3}}\right )} \operatorname {arsinh}\left (b x + a\right ) \]

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]

-1/16*(b^2*x^4 + 4*a*b*x^3 + 6*a^2*x^2 + 4*a^3*x/b + 5*x^2 + 10*a*x/b + 6*arcsinh(b*x + a)*arcsinh(2*(b^2*x +
a*b)/sqrt(-4*a^2*b^2 + 4*(a^2 + 1)*b^2))/b^2 - 3*arcsinh(2*(b^2*x + a*b)/sqrt(-4*a^2*b^2 + 4*(a^2 + 1)*b^2))^2
/b^2)*b + 1/8*(2*(b^2*x^2 + 2*a*b*x + a^2 + 1)^(3/2)*x + 2*(b^2*x^2 + 2*a*b*x + a^2 + 1)^(3/2)*a/b + 3*(a^2*b^
2 - (a^2 + 1)*b^2)*a^2*arcsinh(2*(b^2*x + a*b)/sqrt(-4*a^2*b^2 + 4*(a^2 + 1)*b^2))/b^3 - 3*(a^2*b^2 - (a^2 + 1
)*b^2)*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)*x/b^2 - 3*(a^2*b^2 - (a^2 + 1)*b^2)*(a^2 + 1)*arcsinh(2*(b^2*x + a*b)
/sqrt(-4*a^2*b^2 + 4*(a^2 + 1)*b^2))/b^3 - 3*(a^2*b^2 - (a^2 + 1)*b^2)*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)*a/b^3
)*arcsinh(b*x + a)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \mathrm {asinh}\left (a+b\,x\right )\,{\left (a^2+2\,a\,b\,x+b^2\,x^2+1\right )}^{3/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 5.28, size = 298, normalized size = 2.81 \[ \begin {cases} - \frac {a^{3} x}{4} + \frac {a^{3} \sqrt {a^{2} + 2 a b x + b^{2} x^{2} + 1} \operatorname {asinh}{\left (a + b x \right )}}{4 b} - \frac {3 a^{2} b x^{2}}{8} + \frac {3 a^{2} x \sqrt {a^{2} + 2 a b x + b^{2} x^{2} + 1} \operatorname {asinh}{\left (a + b x \right )}}{4} - \frac {a b^{2} x^{3}}{4} + \frac {3 a b x^{2} \sqrt {a^{2} + 2 a b x + b^{2} x^{2} + 1} \operatorname {asinh}{\left (a + b x \right )}}{4} - \frac {5 a x}{8} + \frac {5 a \sqrt {a^{2} + 2 a b x + b^{2} x^{2} + 1} \operatorname {asinh}{\left (a + b x \right )}}{8 b} - \frac {b^{3} x^{4}}{16} + \frac {b^{2} x^{3} \sqrt {a^{2} + 2 a b x + b^{2} x^{2} + 1} \operatorname {asinh}{\left (a + b x \right )}}{4} - \frac {5 b x^{2}}{16} + \frac {5 x \sqrt {a^{2} + 2 a b x + b^{2} x^{2} + 1} \operatorname {asinh}{\left (a + b x \right )}}{8} + \frac {3 \operatorname {asinh}^{2}{\left (a + b x \right )}}{16 b} & \text {for}\: b \neq 0 \\x \left (a^{2} + 1\right )^{\frac {3}{2}} \operatorname {asinh}{\relax (a )} & \text {otherwise} \end {cases} \]

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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