3.3.61 \(\int \sqrt {1+a^2+2 a b x+b^2 x^2} \sinh ^{-1}(a+b x)^2 \, dx\) [261]

Optimal. Leaf size=107 \[ \frac {(a+b x) \sqrt {1+(a+b x)^2}}{4 b}-\frac {\sinh ^{-1}(a+b x)}{4 b}-\frac {(a+b x)^2 \sinh ^{-1}(a+b x)}{2 b}+\frac {(a+b x) \sqrt {1+(a+b x)^2} \sinh ^{-1}(a+b x)^2}{2 b}+\frac {\sinh ^{-1}(a+b x)^3}{6 b} \]

[Out]

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

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Rubi [A]
time = 0.09, antiderivative size = 107, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {5860, 5785, 5783, 5776, 327, 221} \begin {gather*} \frac {(a+b x) \sqrt {(a+b x)^2+1}}{4 b}+\frac {\sinh ^{-1}(a+b x)^3}{6 b}+\frac {(a+b x) \sqrt {(a+b x)^2+1} \sinh ^{-1}(a+b x)^2}{2 b}-\frac {(a+b x)^2 \sinh ^{-1}(a+b x)}{2 b}-\frac {\sinh ^{-1}(a+b x)}{4 b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 221

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSinh[Rt[b, 2]*(x/Sqrt[a])]/Rt[b, 2], x] /; FreeQ[{a, b},
x] && GtQ[a, 0] && PosQ[b]

Rule 327

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^(n - 1)*(c*x)^(m - n + 1)*((a + b*x^n
)^(p + 1)/(b*(m + n*p + 1))), x] - Dist[a*c^n*((m - n + 1)/(b*(m + n*p + 1))), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 5776

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 5783

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp[(1/(b*c*(n + 1)))*S
imp[Sqrt[1 + c^2*x^2]/Sqrt[d + e*x^2]]*(a + b*ArcSinh[c*x])^(n + 1), x] /; FreeQ[{a, b, c, d, e, n}, x] && EqQ
[e, c^2*d] && NeQ[n, -1]

Rule 5785

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[(1/2)*Simp[Sqrt[d + e*x^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/2)*Simp[Sqrt[d + e*x^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 5860

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/d^2)*x^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 \sqrt {1+a^2+2 a b x+b^2 x^2} \sinh ^{-1}(a+b x)^2 \, dx &=\frac {\text {Subst}\left (\int \sqrt {1+x^2} \sinh ^{-1}(x)^2 \, dx,x,a+b x\right )}{b}\\ &=\frac {(a+b x) \sqrt {1+(a+b x)^2} \sinh ^{-1}(a+b x)^2}{2 b}+\frac {\text {Subst}\left (\int \frac {\sinh ^{-1}(x)^2}{\sqrt {1+x^2}} \, dx,x,a+b x\right )}{2 b}-\frac {\text {Subst}\left (\int x \sinh ^{-1}(x) \, dx,x,a+b x\right )}{b}\\ &=-\frac {(a+b x)^2 \sinh ^{-1}(a+b x)}{2 b}+\frac {(a+b x) \sqrt {1+(a+b x)^2} \sinh ^{-1}(a+b x)^2}{2 b}+\frac {\sinh ^{-1}(a+b x)^3}{6 b}+\frac {\text {Subst}\left (\int \frac {x^2}{\sqrt {1+x^2}} \, dx,x,a+b x\right )}{2 b}\\ &=\frac {(a+b x) \sqrt {1+(a+b x)^2}}{4 b}-\frac {(a+b x)^2 \sinh ^{-1}(a+b x)}{2 b}+\frac {(a+b x) \sqrt {1+(a+b x)^2} \sinh ^{-1}(a+b x)^2}{2 b}+\frac {\sinh ^{-1}(a+b x)^3}{6 b}-\frac {\text {Subst}\left (\int \frac {1}{\sqrt {1+x^2}} \, dx,x,a+b x\right )}{4 b}\\ &=\frac {(a+b x) \sqrt {1+(a+b x)^2}}{4 b}-\frac {\sinh ^{-1}(a+b x)}{4 b}-\frac {(a+b x)^2 \sinh ^{-1}(a+b x)}{2 b}+\frac {(a+b x) \sqrt {1+(a+b x)^2} \sinh ^{-1}(a+b x)^2}{2 b}+\frac {\sinh ^{-1}(a+b x)^3}{6 b}\\ \end {align*}

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Mathematica [A]
time = 0.06, size = 110, normalized size = 1.03 \begin {gather*} \frac {3 (a+b x) \sqrt {1+a^2+2 a b x+b^2 x^2}-3 \left (1+2 a^2+4 a b x+2 b^2 x^2\right ) \sinh ^{-1}(a+b x)+6 (a+b x) \sqrt {1+a^2+2 a b x+b^2 x^2} \sinh ^{-1}(a+b x)^2+2 \sinh ^{-1}(a+b x)^3}{12 b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(3*(a + b*x)*Sqrt[1 + a^2 + 2*a*b*x + b^2*x^2] - 3*(1 + 2*a^2 + 4*a*b*x + 2*b^2*x^2)*ArcSinh[a + b*x] + 6*(a +
b*x)*Sqrt[1 + a^2 + 2*a*b*x + b^2*x^2]*ArcSinh[a + b*x]^2 + 2*ArcSinh[a + b*x]^3)/(12*b)

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Maple [A]
time = 3.50, size = 167, normalized size = 1.56

method	result	size
default	\(\frac {6 \arcsinh \left (b x +a \right )^{2} \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\, b x -6 \arcsinh \left (b x +a \right ) b^{2} x^{2}+6 \arcsinh \left (b x +a \right )^{2} \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\, a -12 \arcsinh \left (b x +a \right ) a b x +2 \arcsinh \left (b x +a \right )^{3}-6 a^{2} \arcsinh \left (b x +a \right )+3 \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\, b x +3 \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\, a -3 \arcsinh \left (b x +a \right )}{12 b}\)	\(167\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arcsinh(b*x+a)^2*(b^2*x^2+2*a*b*x+a^2+1)^(1/2),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.35, size = 161, normalized size = 1.50 \begin {gather*} \frac {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 )^{2} + 2 \, \log \left (b x + a + \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right )^{3} - 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 ) + 3 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} {\left (b x + a\right )}}{12 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(6*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))^2 + 2*log
(b*x + a + sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1))^3 - 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)) + 3*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)*(b*x + a))/b

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \sqrt {a^{2} + 2 a b x + b^{2} x^{2} + 1} \operatorname {asinh}^{2}{\left (a + b x \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(asinh(b*x+a)**2*(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)**2, x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsinh(b*x+a)^2*(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)^2, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\mathrm {asinh}\left (a+b\,x\right )}^2\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2+1} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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