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/(b*x+a+(1+(b*x+a)^2)^(1/2))^2-1/2*a/b^3/(b*x+a+(1+(b*x+a)^2)^(1/2))-1/16*(-4*a^2+1)*(b*x+a+(1+(b*x+a
)^2)^(1/2))^2/b^3-1/6*a*(b*x+a+(1+(b*x+a)^2)^(1/2))^3/b^3+1/32*(b*x+a+(1+(b*x+a)^2)^(1/2))^4/b^3-1/8*(-4*a^2+1
)*arcsinh(b*x+a)/b^3

________________________________________________________________________________________

Rubi [A]  time = 0.13, antiderivative size = 115, normalized size of antiderivative = 1.00, 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 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]

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 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]

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 = 0.12, size = 102, normalized size = 0.89 \[ \frac {\sqrt {a^2+2 a b x+b^2 x^2+1} \left (2 a^3-2 a^2 b x+a \left (2 b^2 x^2-13\right )+6 b^3 x^3+3 b x\right )+8 a b^3 x^3+3 (2 a-1) (2 a+1) \sinh ^{-1}(a+b x)+6 b^4 x^4}{24 b^3} \]

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)

________________________________________________________________________________________

fricas [A]  time = 0.63, size = 117, normalized size = 1.02 \[ \frac {6 \, b^{4} x^{4} + 8 \, a b^{3} x^{3} - 3 \, {\left (4 \, a^{2} - 1\right )} \log \left (-b x - a + \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right ) + {\left (6 \, b^{3} x^{3} + 2 \, a b^{2} x^{2} + 2 \, a^{3} - {\left (2 \, a^{2} - 3\right )} b x - 13 \, a\right )} \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{24 \, b^{3}} \]

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

________________________________________________________________________________________

giac [A]  time = 0.61, size = 140, normalized size = 1.22 \[ \frac {1}{4} \, b x^{4} + \frac {1}{3} \, a x^{3} + \frac {1}{24} \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} {\left ({\left (2 \, {\left (3 \, x + \frac {a}{b}\right )} x - \frac {2 \, a^{2} b^{3} - 3 \, b^{3}}{b^{5}}\right )} x + \frac {2 \, a^{3} b^{2} - 13 \, a b^{2}}{b^{5}}\right )} - \frac {{\left (4 \, a^{2} - 1\right )} \log \left (-a b - {\left (x {\left | b \right |} - \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right )} {\left | b \right |}\right )}{8 \, b^{2} {\left | b \right |}} \]

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))

________________________________________________________________________________________

maple [A]  time = 0.00, size = 264, normalized size = 2.30 \[ \frac {x \left (b^{2} x^{2}+2 a b x +a^{2}+1\right )^{\frac {3}{2}}}{4 b^{2}}-\frac {5 a \left (b^{2} x^{2}+2 a b x +a^{2}+1\right )^{\frac {3}{2}}}{12 b^{3}}+\frac {a^{2} \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\, x}{2 b^{2}}+\frac {a^{3} \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{2 b^{3}}+\frac {a^{2} \ln \left (\frac {b^{2} x +a b}{\sqrt {b^{2}}}+\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\right )}{2 b^{2} \sqrt {b^{2}}}-\frac {\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\, x}{8 b^{2}}-\frac {\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\, a}{8 b^{3}}-\frac {\ln \left (\frac {b^{2} x +a b}{\sqrt {b^{2}}}+\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\right )}{8 b^{2} \sqrt {b^{2}}}+\frac {b \,x^{4}}{4}+\frac {x^{3} a}{3} \]

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

________________________________________________________________________________________

maxima [A]  time = 0.69, size = 273, normalized size = 2.37 \[ \frac {1}{4} \, b x^{4} + \frac {1}{3} \, a x^{3} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )}^{\frac {3}{2}} x}{4 \, b^{2}} - \frac {5 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )}^{\frac {3}{2}} a}{12 \, b^{3}} - \frac {{\left (5 \, a^{2} b^{2} - {\left (a^{2} + 1\right )} b^{2}\right )} a^{2} \operatorname {arsinh}\left (\frac {2 \, {\left (b^{2} x + a b\right )}}{\sqrt {-4 \, a^{2} b^{2} + 4 \, {\left (a^{2} + 1\right )} b^{2}}}\right )}{8 \, b^{5}} + \frac {{\left (5 \, a^{2} b^{2} - {\left (a^{2} + 1\right )} b^{2}\right )} \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} x}{8 \, b^{4}} + \frac {{\left (5 \, a^{2} b^{2} - {\left (a^{2} + 1\right )} b^{2}\right )} {\left (a^{2} + 1\right )} \operatorname {arsinh}\left (\frac {2 \, {\left (b^{2} x + a b\right )}}{\sqrt {-4 \, a^{2} b^{2} + 4 \, {\left (a^{2} + 1\right )} b^{2}}}\right )}{8 \, b^{5}} + \frac {{\left (5 \, a^{2} b^{2} - {\left (a^{2} + 1\right )} b^{2}\right )} \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} a}{8 \, b^{5}} \]

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]

1/4*b*x^4 + 1/3*a*x^3 + 1/4*(b^2*x^2 + 2*a*b*x + a^2 + 1)^(3/2)*x/b^2 - 5/12*(b^2*x^2 + 2*a*b*x + a^2 + 1)^(3/
2)*a/b^3 - 1/8*(5*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^5
+ 1/8*(5*a^2*b^2 - (a^2 + 1)*b^2)*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)*x/b^4 + 1/8*(5*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^5 + 1/8*(5*a^2*b^2 - (a^2 + 1)*b^2)*sqr
t(b^2*x^2 + 2*a*b*x + a^2 + 1)*a/b^5

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x^2\,\left (a+\sqrt {{\left (a+b\,x\right )}^2+1}+b\,x\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{2} \left (a + b x + \sqrt {a^{2} + 2 a b x + b^{2} x^{2} + 1}\right )\, dx \]

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)

________________________________________________________________________________________