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

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

[Out]

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

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Rubi [A]  time = 0.12525, 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 = {5898, 2282, 12, 1628} \[ -\frac{\left (1-4 a^2\right ) e^{2 \sinh ^{-1}(a+b x)}}{16 b^3}-\frac{\left (1-4 a^2\right ) \sinh ^{-1}(a+b x)}{8 b^3}-\frac{a e^{-\sinh ^{-1}(a+b x)}}{2 b^3}-\frac{a e^{3 \sinh ^{-1}(a+b x)}}{6 b^3}-\frac{e^{-2 \sinh ^{-1}(a+b x)}}{16 b^3}+\frac{e^{4 \sinh ^{-1}(a+b x)}}{32 b^3} \]

Antiderivative was successfully verified.

[In]

Int[E^ArcSinh[a + b*x]*x^2,x]

[Out]

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

Rule 5898

Int[(f_)^(ArcSinh[(a_.) + (b_.)*(x_)]^(n_.)*(c_.))*(x_)^(m_.), x_Symbol] :> Dist[1/b, Subst[Int[(-(a/b) + Sinh
[x]/b)^m*f^(c*x^n)*Cosh[x], x], x, ArcSinh[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^{\sinh ^{-1}(a+b x)} x^2 \, dx &=\frac{\operatorname{Subst}\left (\int e^x \cosh (x) \left (-\frac{a}{b}+\frac{\sinh (x)}{b}\right )^2 \, dx,x,\sinh ^{-1}(a+b x)\right )}{b}\\ &=\frac{\operatorname{Subst}\left (\int \frac{\left (1+2 a x-x^2\right )^2 \left (1+x^2\right )}{8 b^2 x^3} \, dx,x,e^{\sinh ^{-1}(a+b x)}\right )}{b}\\ &=\frac{\operatorname{Subst}\left (\int \frac{\left (1+2 a x-x^2\right )^2 \left (1+x^2\right )}{x^3} \, dx,x,e^{\sinh ^{-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^{\sinh ^{-1}(a+b x)}\right )}{8 b^3}\\ &=-\frac{e^{-2 \sinh ^{-1}(a+b x)}}{16 b^3}-\frac{a e^{-\sinh ^{-1}(a+b x)}}{2 b^3}-\frac{\left (1-4 a^2\right ) e^{2 \sinh ^{-1}(a+b x)}}{16 b^3}-\frac{a e^{3 \sinh ^{-1}(a+b x)}}{6 b^3}+\frac{e^{4 \sinh ^{-1}(a+b x)}}{32 b^3}-\frac{\left (1-4 a^2\right ) \sinh ^{-1}(a+b x)}{8 b^3}\\ \end{align*}

Mathematica [A]  time = 0.105329, size = 102, normalized size = 0.89 \[ \frac{\sqrt{a^2+2 a b x+b^2 x^2+1} \left (-2 a^2 b x+2 a^3+a \left (2 b^2 x^2-13\right )+6 b^3 x^3+3 b x\right )+8 a b^3 x^3+3 (2 a-1) (2 a+1) \sinh ^{-1}(a+b x)+6 b^4 x^4}{24 b^3} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[E^ArcSinh[a + b*x]*x^2,x]

[Out]

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

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Maple [A]  time = 0.005, size = 264, normalized size = 2.3 \begin{align*}{\frac{x}{4\,{b}^{2}} \left ({b}^{2}{x}^{2}+2\,xab+{a}^{2}+1 \right ) ^{{\frac{3}{2}}}}-{\frac{5\,a}{12\,{b}^{3}} \left ({b}^{2}{x}^{2}+2\,xab+{a}^{2}+1 \right ) ^{{\frac{3}{2}}}}+{\frac{{a}^{2}x}{2\,{b}^{2}}\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}+1}}+{\frac{{a}^{3}}{2\,{b}^{3}}\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}+1}}+{\frac{{a}^{2}}{2\,{b}^{2}}\ln \left ({({b}^{2}x+ab){\frac{1}{\sqrt{{b}^{2}}}}}+\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}+1} \right ){\frac{1}{\sqrt{{b}^{2}}}}}-{\frac{x}{8\,{b}^{2}}\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}+1}}-{\frac{a}{8\,{b}^{3}}\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}+1}}-{\frac{1}{8\,{b}^{2}}\ln \left ({({b}^{2}x+ab){\frac{1}{\sqrt{{b}^{2}}}}}+\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}+1} \right ){\frac{1}{\sqrt{{b}^{2}}}}}+{\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+(1+(b*x+a)^2)^(1/2))*x^2,x)

[Out]

1/4*x*(b^2*x^2+2*a*b*x+a^2+1)^(3/2)/b^2-5/12*a/b^3*(b^2*x^2+2*a*b*x+a^2+1)^(3/2)+1/2*a^2/b^2*(b^2*x^2+2*a*b*x+
a^2+1)^(1/2)*x+1/2*a^3/b^3*(b^2*x^2+2*a*b*x+a^2+1)^(1/2)+1/2*a^2/b^2*ln((b^2*x+a*b)/(b^2)^(1/2)+(b^2*x^2+2*a*b
*x+a^2+1)^(1/2))/(b^2)^(1/2)-1/8/b^2*(b^2*x^2+2*a*b*x+a^2+1)^(1/2)*x-1/8/b^3*(b^2*x^2+2*a*b*x+a^2+1)^(1/2)*a-1
/8/b^2*ln((b^2*x+a*b)/(b^2)^(1/2)+(b^2*x^2+2*a*b*x+a^2+1)^(1/2))/(b^2)^(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+(1+(b*x+a)^2)^(1/2))*x^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.89097, size = 266, normalized size = 2.31 \begin{align*} \frac{6 \, b^{4} x^{4} + 8 \, a b^{3} x^{3} - 3 \,{\left (4 \, a^{2} - 1\right )} \log \left (-b x - a + \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right ) +{\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^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{24 \, b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(6*b^4*x^4 + 8*a*b^3*x^3 - 3*(4*a^2 - 1)*log(-b*x - a + sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)) + (6*b^3*x^3 +
2*a*b^2*x^2 + 2*a^3 - (2*a^2 - 3)*b*x - 13*a)*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1))/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^{2} + 2 a b x + b^{2} x^{2} + 1}\right )\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.36013, size = 189, normalized size = 1.64 \begin{align*} \frac{1}{4} \, b x^{4} + \frac{1}{3} \, a x^{3} + \frac{1}{24} \, \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1}{\left ({\left (2 \,{\left (3 \, x + \frac{a}{b}\right )} x - \frac{2 \, a^{2} b^{3} - 3 \, b^{3}}{b^{5}}\right )} x + \frac{2 \, a^{3} b^{2} - 13 \, a b^{2}}{b^{5}}\right )} - \frac{{\left (4 \, a^{2} - 1\right )} \log \left (-a b -{\left (x{\left | b \right |} - \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right )}{\left | b \right |}\right )}{8 \, b^{2}{\left | b \right |}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*b*x^4 + 1/3*a*x^3 + 1/24*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)*((2*(3*x + a/b)*x - (2*a^2*b^3 - 3*b^3)/b^5)*x
+ (2*a^3*b^2 - 13*a*b^2)/b^5) - 1/8*(4*a^2 - 1)*log(-a*b - (x*abs(b) - sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1))*abs(
b))/(b^2*abs(b))