3.3.60 \(\int \sqrt {1+a^2+2 a b x+b^2 x^2} \sinh ^{-1}(a+b x)^3 \, dx\) [260]

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

[Out]

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

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Rubi [A]
time = 0.12, antiderivative size = 131, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {5860, 5785, 5783, 5776, 5812, 30} \begin {gather*} -\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} \end {gather*}

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 30

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

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 5812

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol] :> Simp[
f*(f*x)^(m - 1)*(d + e*x^2)^(p + 1)*((a + b*ArcSinh[c*x])^n/(e*(m + 2*p + 1))), x] + (-Dist[f^2*((m - 1)/(c^2*
(m + 2*p + 1))), Int[(f*x)^(m - 2)*(d + e*x^2)^p*(a + b*ArcSinh[c*x])^n, x], x] - Dist[b*f*(n/(c*(m + 2*p + 1)
))*Simp[(d + e*x^2)^p/(1 + c^2*x^2)^p], Int[(f*x)^(m - 1)*(1 + c^2*x^2)^(p + 1/2)*(a + b*ArcSinh[c*x])^(n - 1)
, x], x]) /; FreeQ[{a, b, c, d, e, f, p}, x] && EqQ[e, c^2*d] && GtQ[n, 0] && IGtQ[m, 1] && NeQ[m + 2*p + 1, 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)^3 \, dx &=\frac {\text {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 {\text {Subst}\left (\int \frac {\sinh ^{-1}(x)^3}{\sqrt {1+x^2}} \, dx,x,a+b x\right )}{2 b}-\frac {3 \text {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 \text {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 \text {Subst}(\int x \, dx,x,a+b x)}{4 b}-\frac {3 \text {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*}

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Mathematica [A]
time = 0.08, size = 127, normalized size = 0.97 \begin {gather*} \frac {-3 b x (2 a+b x)+6 (a+b x) \sqrt {1+a^2+2 a b x+b^2 x^2} \sinh ^{-1}(a+b x)-3 \left (1+2 a^2+4 a b x+2 b^2 x^2\right ) \sinh ^{-1}(a+b x)^2+4 (a+b x) \sqrt {1+a^2+2 a b x+b^2 x^2} \sinh ^{-1}(a+b x)^3+\sinh ^{-1}(a+b x)^4}{8 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]^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 = 3.50, size = 204, normalized size = 1.56

method	result	size
default	\(\frac {4 \arcsinh \left (b x +a \right )^{3} \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\, b x -6 \arcsinh \left (b x +a \right )^{2} b^{2} x^{2}+4 \arcsinh \left (b x +a \right )^{3} \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\, a -12 \arcsinh \left (b x +a \right )^{2} a b x +\arcsinh \left (b x +a \right )^{4}-6 \arcsinh \left (b x +a \right )^{2} a^{2}+6 \arcsinh \left (b x +a \right ) \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\, b x -3 b^{2} x^{2}+6 \arcsinh \left (b x +a \right ) \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\, a -6 a b x -3 \arcsinh \left (b x +a \right )^{2}-3 a^{2}-3}{8 b}\)	\(204\)

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,method=_RETURNVERBOSE)

[Out]

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

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Fricas [A]
time = 0.39, size = 199, normalized size = 1.52 \begin {gather*} \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 {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \sqrt {a^{2} + 2 a b x + b^{2} x^{2} + 1} \operatorname {asinh}^{3}{\left (a + b x \right )}\, dx \end {gather*}

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 = 0.00, 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)^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)

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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 )}^3\,\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)^3*(a^2 + b^2*x^2 + 2*a*b*x + 1)^(1/2),x)

[Out]

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

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