3.349 \(\int e^{\sinh ^{-1}(a+b x)} x^3 \, dx\)

Optimal. Leaf size=165 \[ \frac{\left (3-4 a^2\right ) a e^{2 \sinh ^{-1}(a+b x)}}{16 b^4}+\frac{\left (3-4 a^2\right ) a \sinh ^{-1}(a+b x)}{8 b^4}-\frac{\left (1-6 a^2\right ) e^{-\sinh ^{-1}(a+b x)}}{8 b^4}-\frac{\left (1-6 a^2\right ) e^{3 \sinh ^{-1}(a+b x)}}{24 b^4}+\frac{3 a e^{-2 \sinh ^{-1}(a+b x)}}{16 b^4}-\frac{3 a e^{4 \sinh ^{-1}(a+b x)}}{32 b^4}+\frac{e^{-3 \sinh ^{-1}(a+b x)}}{48 b^4}+\frac{e^{5 \sinh ^{-1}(a+b x)}}{80 b^4} \]

[Out]

1/(48*b^4*E^(3*ArcSinh[a + b*x])) + (3*a)/(16*b^4*E^(2*ArcSinh[a + b*x])) - (1 - 6*a^2)/(8*b^4*E^ArcSinh[a + b
*x]) + (a*(3 - 4*a^2)*E^(2*ArcSinh[a + b*x]))/(16*b^4) - ((1 - 6*a^2)*E^(3*ArcSinh[a + b*x]))/(24*b^4) - (3*a*
E^(4*ArcSinh[a + b*x]))/(32*b^4) + E^(5*ArcSinh[a + b*x])/(80*b^4) + (a*(3 - 4*a^2)*ArcSinh[a + b*x])/(8*b^4)

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Rubi [A]  time = 0.172355, antiderivative size = 165, 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} \[ \frac{\left (3-4 a^2\right ) a e^{2 \sinh ^{-1}(a+b x)}}{16 b^4}+\frac{\left (3-4 a^2\right ) a \sinh ^{-1}(a+b x)}{8 b^4}-\frac{\left (1-6 a^2\right ) e^{-\sinh ^{-1}(a+b x)}}{8 b^4}-\frac{\left (1-6 a^2\right ) e^{3 \sinh ^{-1}(a+b x)}}{24 b^4}+\frac{3 a e^{-2 \sinh ^{-1}(a+b x)}}{16 b^4}-\frac{3 a e^{4 \sinh ^{-1}(a+b x)}}{32 b^4}+\frac{e^{-3 \sinh ^{-1}(a+b x)}}{48 b^4}+\frac{e^{5 \sinh ^{-1}(a+b x)}}{80 b^4} \]

Antiderivative was successfully verified.

[In]

Int[E^ArcSinh[a + b*x]*x^3,x]

[Out]

1/(48*b^4*E^(3*ArcSinh[a + b*x])) + (3*a)/(16*b^4*E^(2*ArcSinh[a + b*x])) - (1 - 6*a^2)/(8*b^4*E^ArcSinh[a + b
*x]) + (a*(3 - 4*a^2)*E^(2*ArcSinh[a + b*x]))/(16*b^4) - ((1 - 6*a^2)*E^(3*ArcSinh[a + b*x]))/(24*b^4) - (3*a*
E^(4*ArcSinh[a + b*x]))/(32*b^4) + E^(5*ArcSinh[a + b*x])/(80*b^4) + (a*(3 - 4*a^2)*ArcSinh[a + b*x])/(8*b^4)

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^3 \, dx &=\frac{\operatorname{Subst}\left (\int e^x \cosh (x) \left (-\frac{a}{b}+\frac{\sinh (x)}{b}\right )^3 \, dx,x,\sinh ^{-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 )^3}{16 b^3 x^4} \, dx,x,e^{\sinh ^{-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 )^3}{x^4} \, dx,x,e^{\sinh ^{-1}(a+b x)}\right )}{16 b^4}\\ &=\frac{\operatorname{Subst}\left (\int \left (-\frac{1}{x^4}-\frac{6 a}{x^3}-\frac{2 \left (-1+6 a^2\right )}{x^2}+\frac{2 a \left (3-4 a^2\right )}{x}+2 a \left (3-4 a^2\right ) x+2 \left (-1+6 a^2\right ) x^2-6 a x^3+x^4\right ) \, dx,x,e^{\sinh ^{-1}(a+b x)}\right )}{16 b^4}\\ &=\frac{e^{-3 \sinh ^{-1}(a+b x)}}{48 b^4}+\frac{3 a e^{-2 \sinh ^{-1}(a+b x)}}{16 b^4}-\frac{\left (1-6 a^2\right ) e^{-\sinh ^{-1}(a+b x)}}{8 b^4}+\frac{a \left (3-4 a^2\right ) e^{2 \sinh ^{-1}(a+b x)}}{16 b^4}-\frac{\left (1-6 a^2\right ) e^{3 \sinh ^{-1}(a+b x)}}{24 b^4}-\frac{3 a e^{4 \sinh ^{-1}(a+b x)}}{32 b^4}+\frac{e^{5 \sinh ^{-1}(a+b x)}}{80 b^4}+\frac{a \left (3-4 a^2\right ) \sinh ^{-1}(a+b x)}{8 b^4}\\ \end{align*}

Mathematica [A]  time = 0.0810424, size = 119, normalized size = 0.72 \[ \frac{-\sqrt{a^2+2 a b x+b^2 x^2+1} \left (2 \left (3 a^2-4\right ) b^2 x^2+\left (29-6 a^2\right ) a b x+6 a^4-83 a^2-6 a b^3 x^3-24 b^4 x^4+16\right )+15 a \left (3-4 a^2\right ) \sinh ^{-1}(a+b x)+30 a b^4 x^4+24 b^5 x^5}{120 b^4} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[E^ArcSinh[a + b*x]*x^3,x]

[Out]

(30*a*b^4*x^4 + 24*b^5*x^5 - Sqrt[1 + a^2 + 2*a*b*x + b^2*x^2]*(16 - 83*a^2 + 6*a^4 + a*(29 - 6*a^2)*b*x + 2*(
-4 + 3*a^2)*b^2*x^2 - 6*a*b^3*x^3 - 24*b^4*x^4) + 15*a*(3 - 4*a^2)*ArcSinh[a + b*x])/(120*b^4)

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Maple [A]  time = 0.006, size = 322, normalized size = 2. \begin{align*}{\frac{{x}^{2}}{5\,{b}^{2}} \left ({b}^{2}{x}^{2}+2\,xab+{a}^{2}+1 \right ) ^{{\frac{3}{2}}}}-{\frac{7\,ax}{20\,{b}^{3}} \left ({b}^{2}{x}^{2}+2\,xab+{a}^{2}+1 \right ) ^{{\frac{3}{2}}}}+{\frac{9\,{a}^{2}}{20\,{b}^{4}} \left ({b}^{2}{x}^{2}+2\,xab+{a}^{2}+1 \right ) ^{{\frac{3}{2}}}}-{\frac{{a}^{3}x}{2\,{b}^{3}}\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}+1}}-{\frac{{a}^{4}}{2\,{b}^{4}}\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}+1}}-{\frac{{a}^{3}}{2\,{b}^{3}}\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{3\,ax}{8\,{b}^{3}}\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}+1}}+{\frac{3\,{a}^{2}}{8\,{b}^{4}}\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}+1}}+{\frac{3\,a}{8\,{b}^{3}}\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{2}{15\,{b}^{4}} \left ({b}^{2}{x}^{2}+2\,xab+{a}^{2}+1 \right ) ^{{\frac{3}{2}}}}+{\frac{b{x}^{5}}{5}}+{\frac{{x}^{4}a}{4}} \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^3,x)

[Out]

1/5*x^2*(b^2*x^2+2*a*b*x+a^2+1)^(3/2)/b^2-7/20*a/b^3*x*(b^2*x^2+2*a*b*x+a^2+1)^(3/2)+9/20*a^2/b^4*(b^2*x^2+2*a
*b*x+a^2+1)^(3/2)-1/2*a^3/b^3*(b^2*x^2+2*a*b*x+a^2+1)^(1/2)*x-1/2*a^4/b^4*(b^2*x^2+2*a*b*x+a^2+1)^(1/2)-1/2*a^
3/b^3*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)+3/8*a/b^3*(b^2*x^2+2*a*b*x+a^2+1)^
(1/2)*x+3/8*a^2/b^4*(b^2*x^2+2*a*b*x+a^2+1)^(1/2)+3/8*a/b^3*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)-2/15/b^4*(b^2*x^2+2*a*b*x+a^2+1)^(3/2)+1/5*b*x^5+1/4*x^4*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^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.84998, size = 321, normalized size = 1.95 \begin{align*} \frac{24 \, b^{5} x^{5} + 30 \, a b^{4} x^{4} + 15 \,{\left (4 \, a^{3} - 3 \, a\right )} \log \left (-b x - a + \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right ) +{\left (24 \, b^{4} x^{4} + 6 \, a b^{3} x^{3} - 2 \,{\left (3 \, a^{2} - 4\right )} b^{2} x^{2} - 6 \, a^{4} +{\left (6 \, a^{3} - 29 \, a\right )} b x + 83 \, a^{2} - 16\right )} \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{120 \, b^{4}} \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^3,x, algorithm="fricas")

[Out]

1/120*(24*b^5*x^5 + 30*a*b^4*x^4 + 15*(4*a^3 - 3*a)*log(-b*x - a + sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)) + (24*b^
4*x^4 + 6*a*b^3*x^3 - 2*(3*a^2 - 4)*b^2*x^2 - 6*a^4 + (6*a^3 - 29*a)*b*x + 83*a^2 - 16)*sqrt(b^2*x^2 + 2*a*b*x
+ a^2 + 1))/b^4

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{3} \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**3,x)

[Out]

Integral(x**3*(a + b*x + sqrt(a**2 + 2*a*b*x + b**2*x**2 + 1)), x)

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Giac [A]  time = 1.32659, size = 234, normalized size = 1.42 \begin{align*} \frac{1}{5} \, b x^{5} + \frac{1}{4} \, a x^{4} + \frac{1}{120} \, \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1}{\left ({\left (2 \,{\left (3 \,{\left (4 \, x + \frac{a}{b}\right )} x - \frac{3 \, a^{2} b^{5} - 4 \, b^{5}}{b^{7}}\right )} x + \frac{6 \, a^{3} b^{4} - 29 \, a b^{4}}{b^{7}}\right )} x - \frac{6 \, a^{4} b^{3} - 83 \, a^{2} b^{3} + 16 \, b^{3}}{b^{7}}\right )} + \frac{{\left (4 \, a^{3} - 3 \, a\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^{3}{\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^3,x, algorithm="giac")

[Out]

1/5*b*x^5 + 1/4*a*x^4 + 1/120*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)*((2*(3*(4*x + a/b)*x - (3*a^2*b^5 - 4*b^5)/b^7
)*x + (6*a^3*b^4 - 29*a*b^4)/b^7)*x - (6*a^4*b^3 - 83*a^2*b^3 + 16*b^3)/b^7) + 1/8*(4*a^3 - 3*a)*log(-a*b - (x
*abs(b) - sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1))*abs(b))/(b^3*abs(b))