3.3.68 \(\int (1+a^2+2 a b x+b^2 x^2)^{3/2} \sinh ^{-1}(a+b x) \, dx\) [268]

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

[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 = 0.07, 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 = {5860, 5786, 5785, 5783, 30, 14} \begin {gather*} -\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} \end {gather*}

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 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 5786

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/(2*p + 1))*Simp[(d + e*x^2)^p/(1 + c^2*x^2)^p], Int[x*(1 + c^2*x^2)^(p - 1/2)*(a + b*A
rcSinh[c*x])^(n - 1), x], x]) /; FreeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*d] && GtQ[n, 0] && GtQ[p, 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 \left (1+a^2+2 a b x+b^2 x^2\right )^{3/2} \sinh ^{-1}(a+b x) \, dx &=\frac {\text {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 {\text {Subst}\left (\int x \left (1+x^2\right ) \, dx,x,a+b x\right )}{4 b}+\frac {3 \text {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 {\text {Subst}\left (\int \left (x+x^3\right ) \, dx,x,a+b x\right )}{4 b}-\frac {3 \text {Subst}(\int x \, dx,x,a+b x)}{8 b}+\frac {3 \text {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 = 0.07, size = 124, normalized size = 1.17 \begin {gather*} \frac {-b x \left (10 a+4 a^3+5 b x+6 a^2 b x+4 a b^2 x^2+b^3 x^3\right )+2 \sqrt {1+a^2+2 a b x+b^2 x^2} \left (5 a+2 a^3+5 b x+6 a^2 b x+6 a b^2 x^2+2 b^3 x^3\right ) \sinh ^{-1}(a+b x)+3 \sinh ^{-1}(a+b x)^2}{16 b} \end {gather*}

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 [A]
time = 2.88, size = 82, normalized size = 0.77 \[-\frac {-4 \arcsinh \left (b x +a \right ) \cosh \left (2 \arcsinh \left (b x +a \right )\right ) \sinh \left (2 \arcsinh \left (b x +a \right )\right )+\cosh ^{2}\left (2 \arcsinh \left (b x +a \right )\right )-16 \sinh \left (2 \arcsinh \left (b x +a \right )\right ) \arcsinh \left (b x +a \right )-12 \arcsinh \left (b x +a \right )^{2}+8 \cosh \left (2 \arcsinh \left (b x +a \right )\right )}{64 b}\]

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]

-1/64*(-4*arcsinh(b*x+a)*cosh(2*arcsinh(b*x+a))*sinh(2*arcsinh(b*x+a))+cosh(2*arcsinh(b*x+a))^2-16*sinh(2*arcs
inh(b*x+a))*arcsinh(b*x+a)-12*arcsinh(b*x+a)^2+8*cosh(2*arcsinh(b*x+a)))/b

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 394 vs. \(2 (92) = 184\).
time = 0.28, size = 394, normalized size = 3.72 \begin {gather*} -\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 ) \end {gather*}

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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Fricas [A]
time = 0.36, size = 160, normalized size = 1.51 \begin {gather*} -\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 {gather*}

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 [B] Leaf count of result is larger than twice the leaf count of optimal. 298 vs. \(2 (95) = 190\).
time = 0.61, size = 298, normalized size = 2.81 \begin {gather*} \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 {gather*}

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 = 0.00, 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((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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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \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 \end {gather*}

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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