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

Optimal. Leaf size=15 \[ -\frac{1}{2 b \sinh ^{-1}(a+b x)^2} \]

[Out]

-1/(2*b*ArcSinh[a + b*x]^2)

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Rubi [A]  time = 0.0702373, antiderivative size = 15, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.067, Rules used = {5867, 5675} \[ -\frac{1}{2 b \sinh ^{-1}(a+b x)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/(2*b*ArcSinh[a + b*x]^2)

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 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
]

Rubi steps

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

Mathematica [A]  time = 0.0133136, size = 15, normalized size = 1. \[ -\frac{1}{2 b \sinh ^{-1}(a+b x)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/(2*b*ArcSinh[a + b*x]^2)

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Maple [A]  time = 0.043, size = 14, normalized size = 0.9 \begin{align*} -{\frac{1}{2\,b \left ({\it Arcsinh} \left ( bx+a \right ) \right ) ^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2/b/arcsinh(b*x+a)^2

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(b^7*x^7 + 7*a*b^6*x^6 + a^7 + 3*(7*a^2*b^5 + b^5)*x^5 + 3*a^5 + 5*(7*a^3*b^4 + 3*a*b^4)*x^4 + (35*a^4*b^
3 + 30*a^2*b^3 + 3*b^3)*x^3 + 3*a^3 + 3*(7*a^5*b^2 + 10*a^3*b^2 + 3*a*b^2)*x^2 + (b^4*x^4 + 4*a*b^3*x^3 + a^4
+ (6*a^2*b^2 + b^2)*x^2 + a^2 + 2*(2*a^3*b + a*b)*x)*(b^2*x^2 + 2*a*b*x + a^2 + 1)^(3/2) + (3*b^5*x^5 + 15*a*b
^4*x^4 + 3*a^5 + 5*(6*a^2*b^3 + b^3)*x^3 + 5*a^3 + 15*(2*a^3*b^2 + a*b^2)*x^2 + (15*a^4*b + 15*a^2*b + 2*b)*x
+ 2*a)*(b^2*x^2 + 2*a*b*x + a^2 + 1) + (7*a^6*b + 15*a^4*b + 9*a^2*b + b)*x + (b^5*x^5 + 5*a*b^4*x^4 + a^5 + 2
*(5*a^2*b^3 + b^3)*x^3 + 2*a^3 + 2*(5*a^3*b^2 + 3*a*b^2)*x^2 - (b^2*x^2 + 2*a*b*x + a^2 + 1)^(5/2) - (b^3*x^3
+ 3*a*b^2*x^2 + a^3 + (3*a^2*b + b)*x + a)*(b^2*x^2 + 2*a*b*x + a^2 + 1) + (5*a^4*b + 6*a^2*b + b)*x + (b^4*x^
4 + 4*a*b^3*x^3 + a^4 + 2*(3*a^2*b^2 + b^2)*x^2 + 2*a^2 + 4*(a^3*b + a*b)*x + 1)*sqrt(b^2*x^2 + 2*a*b*x + a^2
+ 1) + a)*log(b*x + a + sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)) + (3*b^6*x^6 + 18*a*b^5*x^5 + 3*a^6 + (45*a^2*b^4 +
7*b^4)*x^4 + 7*a^4 + 4*(15*a^3*b^3 + 7*a*b^3)*x^3 + (45*a^4*b^2 + 42*a^2*b^2 + 5*b^2)*x^2 + 5*a^2 + 2*(9*a^5*
b + 14*a^3*b + 5*a*b)*x + 1)*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1) + a)/(((b^4*x^3 + 3*a*b^3*x^2 + 3*a^2*b^2*x + a
^3*b)*(b^2*x^2 + 2*a*b*x + a^2 + 1)^2 + 3*(b^5*x^4 + 4*a*b^4*x^3 + a^4*b + a^2*b + (6*a^2*b^3 + b^3)*x^2 + 2*(
2*a^3*b^2 + a*b^2)*x)*(b^2*x^2 + 2*a*b*x + a^2 + 1)^(3/2) + 3*(b^6*x^5 + 5*a*b^5*x^4 + a^5*b + 2*a^3*b + 2*(5*
a^2*b^4 + b^4)*x^3 + 2*(5*a^3*b^3 + 3*a*b^3)*x^2 + a*b + (5*a^4*b^2 + 6*a^2*b^2 + b^2)*x)*(b^2*x^2 + 2*a*b*x +
a^2 + 1) + (b^7*x^6 + 6*a*b^6*x^5 + a^6*b + 3*a^4*b + 3*(5*a^2*b^5 + b^5)*x^4 + 4*(5*a^3*b^4 + 3*a*b^4)*x^3 +
3*a^2*b + 3*(5*a^4*b^3 + 6*a^2*b^3 + b^3)*x^2 + 6*(a^5*b^2 + 2*a^3*b^2 + a*b^2)*x + b)*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))^2) + integrate(-1/2*(2*b^6*x^6 + 12*a*b^5*x^5 +
2*a^6 + 3*(10*a^2*b^4 + b^4)*x^4 + 3*a^4 + 4*(10*a^3*b^3 + 3*a*b^3)*x^3 - (2*b^2*x^2 + 4*a*b*x + 2*a^2 + 3)*(b
^2*x^2 + 2*a*b*x + a^2 + 1)^2 + 6*(5*a^4*b^2 + 3*a^2*b^2)*x^2 - 4*(b^3*x^3 + 3*a*b^2*x^2 + a^3 + (3*a^2*b + b)
*x + a)*(b^2*x^2 + 2*a*b*x + a^2 + 1)^(3/2) + 4*(b^2*x^2 + 2*a*b*x + a^2 + 1)^2 + 12*(a^5*b + a^3*b)*x + 4*(b^
5*x^5 + 5*a*b^4*x^4 + a^5 + 2*(5*a^2*b^3 + b^3)*x^3 + 2*a^3 + 2*(5*a^3*b^2 + 3*a*b^2)*x^2 + (5*a^4*b + 6*a^2*b
+ b)*x + a)*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1) - 1)/(((b^4*x^4 + 4*a*b^3*x^3 + 6*a^2*b^2*x^2 + 4*a^3*b*x + a^4
)*(b^2*x^2 + 2*a*b*x + a^2 + 1)^(5/2) + 4*(b^5*x^5 + 5*a*b^4*x^4 + a^5 + (10*a^2*b^3 + b^3)*x^3 + a^3 + (10*a^
3*b^2 + 3*a*b^2)*x^2 + (5*a^4*b + 3*a^2*b)*x)*(b^2*x^2 + 2*a*b*x + a^2 + 1)^2 + 6*(b^6*x^6 + 6*a*b^5*x^5 + a^6
+ (15*a^2*b^4 + 2*b^4)*x^4 + 2*a^4 + 4*(5*a^3*b^3 + 2*a*b^3)*x^3 + (15*a^4*b^2 + 12*a^2*b^2 + b^2)*x^2 + a^2
+ 2*(3*a^5*b + 4*a^3*b + a*b)*x)*(b^2*x^2 + 2*a*b*x + a^2 + 1)^(3/2) + 4*(b^7*x^7 + 7*a*b^6*x^6 + a^7 + 3*(7*a
^2*b^5 + b^5)*x^5 + 3*a^5 + 5*(7*a^3*b^4 + 3*a*b^4)*x^4 + (35*a^4*b^3 + 30*a^2*b^3 + 3*b^3)*x^3 + 3*a^3 + 3*(7
*a^5*b^2 + 10*a^3*b^2 + 3*a*b^2)*x^2 + (7*a^6*b + 15*a^4*b + 9*a^2*b + b)*x + a)*(b^2*x^2 + 2*a*b*x + a^2 + 1)
+ (b^8*x^8 + 8*a*b^7*x^7 + a^8 + 4*(7*a^2*b^6 + b^6)*x^6 + 4*a^6 + 8*(7*a^3*b^5 + 3*a*b^5)*x^5 + 2*(35*a^4*b^
4 + 30*a^2*b^4 + 3*b^4)*x^4 + 6*a^4 + 8*(7*a^5*b^3 + 10*a^3*b^3 + 3*a*b^3)*x^3 + 4*(7*a^6*b^2 + 15*a^4*b^2 + 9
*a^2*b^2 + b^2)*x^2 + 4*a^2 + 8*(a^7*b + 3*a^5*b + 3*a^3*b + a*b)*x + 1)*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1))*lo
g(b*x + a + sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1))), x)

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Fricas [B]  time = 2.61269, size = 82, normalized size = 5.47 \begin{align*} -\frac{1}{2 \, b \log \left (b x + a + \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right )^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2/(b*log(b*x + a + sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1))^2)

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Sympy [A]  time = 2.6884, size = 29, normalized size = 1.93 \begin{align*} \begin{cases} - \frac{1}{2 b \operatorname{asinh}^{2}{\left (a + b x \right )}} & \text{for}\: b \neq 0 \\\frac{x}{\sqrt{a^{2} + 1} \operatorname{asinh}^{3}{\left (a \right )}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.40197, size = 43, normalized size = 2.87 \begin{align*} -\frac{1}{2 \, b \log \left (b x + a + \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right )^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2/(b*log(b*x + a + sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1))^2)