3.276 \(\int e^{\cosh ^{-1}(a+b x)} x^2 \, dx\)

Optimal. Leaf size=115 \[ \frac{\left (4 a^2+1\right ) e^{2 \cosh ^{-1}(a+b x)}}{16 b^3}-\frac{\left (4 a^2+1\right ) \cosh ^{-1}(a+b x)}{8 b^3}-\frac{a e^{-\cosh ^{-1}(a+b x)}}{2 b^3}-\frac{a e^{3 \cosh ^{-1}(a+b x)}}{6 b^3}+\frac{e^{-2 \cosh ^{-1}(a+b x)}}{16 b^3}+\frac{e^{4 \cosh ^{-1}(a+b x)}}{32 b^3} \]

[Out]

1/(16*b^3*E^(2*ArcCosh[a + b*x])) - a/(2*b^3*E^ArcCosh[a + b*x]) + ((1 + 4*a^2)*E^(2*ArcCosh[a + b*x]))/(16*b^
3) - (a*E^(3*ArcCosh[a + b*x]))/(6*b^3) + E^(4*ArcCosh[a + b*x])/(32*b^3) - ((1 + 4*a^2)*ArcCosh[a + b*x])/(8*
b^3)

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Rubi [A]  time = 0.12071, antiderivative size = 115, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {5899, 2282, 12, 1628} \[ \frac{\left (4 a^2+1\right ) e^{2 \cosh ^{-1}(a+b x)}}{16 b^3}-\frac{\left (4 a^2+1\right ) \cosh ^{-1}(a+b x)}{8 b^3}-\frac{a e^{-\cosh ^{-1}(a+b x)}}{2 b^3}-\frac{a e^{3 \cosh ^{-1}(a+b x)}}{6 b^3}+\frac{e^{-2 \cosh ^{-1}(a+b x)}}{16 b^3}+\frac{e^{4 \cosh ^{-1}(a+b x)}}{32 b^3} \]

Antiderivative was successfully verified.

[In]

Int[E^ArcCosh[a + b*x]*x^2,x]

[Out]

1/(16*b^3*E^(2*ArcCosh[a + b*x])) - a/(2*b^3*E^ArcCosh[a + b*x]) + ((1 + 4*a^2)*E^(2*ArcCosh[a + b*x]))/(16*b^
3) - (a*E^(3*ArcCosh[a + b*x]))/(6*b^3) + E^(4*ArcCosh[a + b*x])/(32*b^3) - ((1 + 4*a^2)*ArcCosh[a + b*x])/(8*
b^3)

Rule 5899

Int[(f_)^(ArcCosh[(a_.) + (b_.)*(x_)]^(n_.)*(c_.))*(x_)^(m_.), x_Symbol] :> Dist[1/b, Subst[Int[(-(a/b) + Cosh
[x]/b)^m*f^(c*x^n)*Sinh[x], x], x, ArcCosh[a + b*x]], x] /; FreeQ[{a, b, c, f}, x] && IGtQ[m, 0] && IGtQ[n, 0]

Rule 2282

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 1628

Int[(Pq_)*((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegra
nd[(d + e*x)^m*Pq*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

Rubi steps

\begin{align*} \int e^{\cosh ^{-1}(a+b x)} x^2 \, dx &=\frac{\operatorname{Subst}\left (\int e^x \left (-\frac{a}{b}+\frac{\cosh (x)}{b}\right )^2 \sinh (x) \, dx,x,\cosh ^{-1}(a+b x)\right )}{b}\\ &=\frac{\operatorname{Subst}\left (\int \frac{\left (-1+x^2\right ) \left (1-2 a x+x^2\right )^2}{8 b^2 x^3} \, dx,x,e^{\cosh ^{-1}(a+b x)}\right )}{b}\\ &=\frac{\operatorname{Subst}\left (\int \frac{\left (-1+x^2\right ) \left (1-2 a x+x^2\right )^2}{x^3} \, dx,x,e^{\cosh ^{-1}(a+b x)}\right )}{8 b^3}\\ &=\frac{\operatorname{Subst}\left (\int \left (-\frac{1}{x^3}+\frac{4 a}{x^2}+\frac{-1-4 a^2}{x}+\left (1+4 a^2\right ) x-4 a x^2+x^3\right ) \, dx,x,e^{\cosh ^{-1}(a+b x)}\right )}{8 b^3}\\ &=\frac{e^{-2 \cosh ^{-1}(a+b x)}}{16 b^3}-\frac{a e^{-\cosh ^{-1}(a+b x)}}{2 b^3}+\frac{\left (1+4 a^2\right ) e^{2 \cosh ^{-1}(a+b x)}}{16 b^3}-\frac{a e^{3 \cosh ^{-1}(a+b x)}}{6 b^3}+\frac{e^{4 \cosh ^{-1}(a+b x)}}{32 b^3}-\frac{\left (1+4 a^2\right ) \cosh ^{-1}(a+b x)}{8 b^3}\\ \end{align*}

Mathematica [A]  time = 0.123629, size = 119, normalized size = 1.03 \[ \frac{\sqrt{a+b x-1} \sqrt{a+b x+1} \left (-2 a^2 b x+2 a^3+a \left (2 b^2 x^2+13\right )+6 b^3 x^3-3 b x\right )-3 \left (4 a^2+1\right ) \log \left (\sqrt{a+b x-1} \sqrt{a+b x+1}+a+b x\right )+8 a b^3 x^3+6 b^4 x^4}{24 b^3} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[E^ArcCosh[a + b*x]*x^2,x]

[Out]

(8*a*b^3*x^3 + 6*b^4*x^4 + Sqrt[-1 + a + b*x]*Sqrt[1 + a + b*x]*(2*a^3 - 3*b*x - 2*a^2*b*x + 6*b^3*x^3 + a*(13
+ 2*b^2*x^2)) - 3*(1 + 4*a^2)*Log[a + b*x + Sqrt[-1 + a + b*x]*Sqrt[1 + a + b*x]])/(24*b^3)

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Maple [C]  time = 0.01, size = 288, normalized size = 2.5 \begin{align*}{\frac{{\it csgn} \left ( b \right ) }{24\,{b}^{3}}\sqrt{bx+a-1}\sqrt{bx+a+1} \left ( 6\,{\it csgn} \left ( b \right ){x}^{3}{b}^{3}\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}-1}+2\,{\it csgn} \left ( b \right ){x}^{2}a{b}^{2}\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}-1}-2\,\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}-1}{\it csgn} \left ( b \right ) x{a}^{2}b+2\,\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}-1}{\it csgn} \left ( b \right ){a}^{3}-3\,\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}-1}{\it csgn} \left ( b \right ) xb+13\,\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}-1}{\it csgn} \left ( b \right ) a-12\,\ln \left ( \left ( \sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}-1}{\it csgn} \left ( b \right ) +bx+a \right ){\it csgn} \left ( b \right ) \right ){a}^{2}-3\,\ln \left ( \left ( \sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}-1}{\it csgn} \left ( b \right ) +bx+a \right ){\it csgn} \left ( b \right ) \right ) \right ){\frac{1}{\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}-1}}}}+{\frac{b{x}^{4}}{4}}+{\frac{{x}^{3}a}{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(b*x+a-1)^(1/2)*(b*x+a+1)^(1/2)*(6*csgn(b)*x^3*b^3*(b^2*x^2+2*a*b*x+a^2-1)^(1/2)+2*csgn(b)*x^2*a*b^2*(b^2
*x^2+2*a*b*x+a^2-1)^(1/2)-2*(b^2*x^2+2*a*b*x+a^2-1)^(1/2)*csgn(b)*x*a^2*b+2*(b^2*x^2+2*a*b*x+a^2-1)^(1/2)*csgn
(b)*a^3-3*(b^2*x^2+2*a*b*x+a^2-1)^(1/2)*csgn(b)*x*b+13*(b^2*x^2+2*a*b*x+a^2-1)^(1/2)*csgn(b)*a-12*ln(((b^2*x^2
+2*a*b*x+a^2-1)^(1/2)*csgn(b)+b*x+a)*csgn(b))*a^2-3*ln(((b^2*x^2+2*a*b*x+a^2-1)^(1/2)*csgn(b)+b*x+a)*csgn(b)))
*csgn(b)/b^3/(b^2*x^2+2*a*b*x+a^2-1)^(1/2)+1/4*b*x^4+1/3*x^3*a

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.9775, size = 271, normalized size = 2.36 \begin{align*} \frac{6 \, b^{4} x^{4} + 8 \, a b^{3} x^{3} +{\left (6 \, b^{3} x^{3} + 2 \, a b^{2} x^{2} + 2 \, a^{3} -{\left (2 \, a^{2} + 3\right )} b x + 13 \, a\right )} \sqrt{b x + a + 1} \sqrt{b x + a - 1} + 3 \,{\left (4 \, a^{2} + 1\right )} \log \left (-b x + \sqrt{b x + a + 1} \sqrt{b x + a - 1} - a\right )}{24 \, b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(6*b^4*x^4 + 8*a*b^3*x^3 + (6*b^3*x^3 + 2*a*b^2*x^2 + 2*a^3 - (2*a^2 + 3)*b*x + 13*a)*sqrt(b*x + a + 1)*s
qrt(b*x + a - 1) + 3*(4*a^2 + 1)*log(-b*x + sqrt(b*x + a + 1)*sqrt(b*x + a - 1) - a))/b^3

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**2*(a + b*x + sqrt(a + b*x - 1)*sqrt(a + b*x + 1)), x)

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Giac [A]  time = 1.2641, size = 271, normalized size = 2.36 \begin{align*} \frac{{\left ({\left (b x + a + 1\right )}{\left (2 \,{\left (b x + a + 1\right )}{\left (\frac{3 \,{\left (b x + a + 1\right )}}{b^{2}} - \frac{8 \, a b^{6} + 9 \, b^{6}}{b^{8}}\right )} + \frac{12 \, a^{2} b^{6} + 32 \, a b^{6} + 15 \, b^{6}}{b^{8}}\right )} - \frac{3 \,{\left (4 \, a^{2} b^{6} + b^{6}\right )}}{b^{8}}\right )} \sqrt{b x + a + 1} \sqrt{b x + a - 1} + 8 \,{\left (b x^{3} + \frac{a^{3} + 3 \, a^{2} + 3 \, a + 1}{b^{2}}\right )} a + 6 \,{\left (b x^{4} - \frac{a^{4} + 4 \, a^{3} + 6 \, a^{2} + 4 \, a + 1}{b^{3}}\right )} b + \frac{6 \,{\left (4 \, a^{2} + 1\right )} \log \left ({\left | -\sqrt{b x + a + 1} + \sqrt{b x + a - 1} \right |}\right )}{b^{2}}}{24 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(((b*x + a + 1)*(2*(b*x + a + 1)*(3*(b*x + a + 1)/b^2 - (8*a*b^6 + 9*b^6)/b^8) + (12*a^2*b^6 + 32*a*b^6 +
15*b^6)/b^8) - 3*(4*a^2*b^6 + b^6)/b^8)*sqrt(b*x + a + 1)*sqrt(b*x + a - 1) + 8*(b*x^3 + (a^3 + 3*a^2 + 3*a +
1)/b^2)*a + 6*(b*x^4 - (a^4 + 4*a^3 + 6*a^2 + 4*a + 1)/b^3)*b + 6*(4*a^2 + 1)*log(abs(-sqrt(b*x + a + 1) + sq
rt(b*x + a - 1)))/b^2)/b