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

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Rubi [A]  time = 0.12, 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 = {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 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 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]

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*}

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Mathematica [A]  time = 0.14, size = 119, normalized size = 1.03 \[ \frac {-3 \left (4 a^2+1\right ) \log \left (\sqrt {a+b x-1} \sqrt {a+b x+1}+a+b x\right )+\sqrt {a+b x-1} \sqrt {a+b x+1} \left (2 a^3-2 a^2 b x+a \left (2 b^2 x^2+13\right )+6 b^3 x^3-3 b x\right )+8 a b^3 x^3+6 b^4 x^4}{24 b^3} \]

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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fricas [A]  time = 0.55, size = 112, normalized size = 0.97 \[ \frac {6 \, b^{4} x^{4} + 8 \, a b^{3} x^{3} + {\left (6 \, b^{3} x^{3} + 2 \, a b^{2} x^{2} + 2 \, a^{3} - {\left (2 \, a^{2} + 3\right )} b x + 13 \, a\right )} \sqrt {b x + a + 1} \sqrt {b x + a - 1} + 3 \, {\left (4 \, a^{2} + 1\right )} \log \left (-b x + \sqrt {b x + a + 1} \sqrt {b x + a - 1} - a\right )}{24 \, b^{3}} \]

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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giac [B]  time = 0.25, size = 417, normalized size = 3.63 \[ \frac {6 \, b^{2} x^{4} + 8 \, a b x^{3} + 4 \, \sqrt {b x + a + 1} \sqrt {b x + a - 1} {\left ({\left (b x + a + 1\right )} {\left (\frac {2 \, {\left (b x + a + 1\right )}}{b^{2}} - \frac {6 \, a b^{6} + 7 \, b^{6}}{b^{8}}\right )} + \frac {3 \, {\left (2 \, a^{2} b^{6} + 6 \, a b^{6} + 3 \, b^{6}\right )}}{b^{8}}\right )} + 4 \, {\left (\sqrt {b x + a + 1} \sqrt {b x + a - 1} {\left ({\left (b x + a + 1\right )} {\left (\frac {2 \, {\left (b x + a + 1\right )}}{b^{2}} - \frac {6 \, a b^{6} + 7 \, b^{6}}{b^{8}}\right )} + \frac {3 \, {\left (2 \, a^{2} b^{6} + 6 \, a b^{6} + 3 \, b^{6}\right )}}{b^{8}}\right )} + \frac {6 \, {\left (2 \, a^{2} + 2 \, a + 1\right )} \log \left (\sqrt {b x + a + 1} - \sqrt {b x + a - 1}\right )}{b^{2}}\right )} a + {\left ({\left ({\left (b x + a + 1\right )} {\left (2 \, {\left (b x + a + 1\right )} {\left (\frac {3 \, {\left (b x + a + 1\right )}}{b^{3}} - \frac {12 \, a b^{12} + 13 \, b^{12}}{b^{15}}\right )} + \frac {36 \, a^{2} b^{12} + 84 \, a b^{12} + 43 \, b^{12}}{b^{15}}\right )} - \frac {3 \, {\left (8 \, a^{3} b^{12} + 36 \, a^{2} b^{12} + 36 \, a b^{12} + 13 \, b^{12}\right )}}{b^{15}}\right )} \sqrt {b x + a + 1} \sqrt {b x + a - 1} - \frac {6 \, {\left (8 \, a^{3} + 12 \, a^{2} + 12 \, a + 3\right )} \log \left (\sqrt {b x + a + 1} - \sqrt {b x + a - 1}\right )}{b^{3}}\right )} b + \frac {24 \, {\left (2 \, a^{2} + 2 \, a + 1\right )} \log \left (\sqrt {b x + a + 1} - \sqrt {b x + a - 1}\right )}{b^{2}}}{24 \, b} \]

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*(6*b^2*x^4 + 8*a*b*x^3 + 4*sqrt(b*x + a + 1)*sqrt(b*x + a - 1)*((b*x + a + 1)*(2*(b*x + a + 1)/b^2 - (6*a
*b^6 + 7*b^6)/b^8) + 3*(2*a^2*b^6 + 6*a*b^6 + 3*b^6)/b^8) + 4*(sqrt(b*x + a + 1)*sqrt(b*x + a - 1)*((b*x + a +
1)*(2*(b*x + a + 1)/b^2 - (6*a*b^6 + 7*b^6)/b^8) + 3*(2*a^2*b^6 + 6*a*b^6 + 3*b^6)/b^8) + 6*(2*a^2 + 2*a + 1)
*log(sqrt(b*x + a + 1) - sqrt(b*x + a - 1))/b^2)*a + (((b*x + a + 1)*(2*(b*x + a + 1)*(3*(b*x + a + 1)/b^3 - (
12*a*b^12 + 13*b^12)/b^15) + (36*a^2*b^12 + 84*a*b^12 + 43*b^12)/b^15) - 3*(8*a^3*b^12 + 36*a^2*b^12 + 36*a*b^
12 + 13*b^12)/b^15)*sqrt(b*x + a + 1)*sqrt(b*x + a - 1) - 6*(8*a^3 + 12*a^2 + 12*a + 3)*log(sqrt(b*x + a + 1)
- sqrt(b*x + a - 1))/b^3)*b + 24*(2*a^2 + 2*a + 1)*log(sqrt(b*x + a + 1) - sqrt(b*x + a - 1))/b^2)/b

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maple [C]  time = 0.01, size = 288, normalized size = 2.50 \[ \frac {\sqrt {b x +a -1}\, \sqrt {b x +a +1}\, \left (6 \,\mathrm {csgn}\relax (b ) x^{3} b^{3} \sqrt {b^{2} x^{2}+2 a b x +a^{2}-1}+2 \,\mathrm {csgn}\relax (b ) x^{2} a \,b^{2} \sqrt {b^{2} x^{2}+2 a b x +a^{2}-1}-2 \,\mathrm {csgn}\relax (b ) \sqrt {b^{2} x^{2}+2 a b x +a^{2}-1}\, x \,a^{2} b +2 \,\mathrm {csgn}\relax (b ) \sqrt {b^{2} x^{2}+2 a b x +a^{2}-1}\, a^{3}-3 \,\mathrm {csgn}\relax (b ) \sqrt {b^{2} x^{2}+2 a b x +a^{2}-1}\, x b +13 \,\mathrm {csgn}\relax (b ) \sqrt {b^{2} x^{2}+2 a b x +a^{2}-1}\, a -12 \ln \left (\left (\sqrt {b^{2} x^{2}+2 a b x +a^{2}-1}\, \mathrm {csgn}\relax (b )+b x +a \right ) \mathrm {csgn}\relax (b )\right ) a^{2}-3 \ln \left (\left (\sqrt {b^{2} x^{2}+2 a b x +a^{2}-1}\, \mathrm {csgn}\relax (b )+b x +a \right ) \mathrm {csgn}\relax (b )\right )\right ) \mathrm {csgn}\relax (b )}{24 b^{3} \sqrt {b^{2} x^{2}+2 a b x +a^{2}-1}}+\frac {b \,x^{4}}{4}+\frac {x^{3} a}{3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a+(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*csgn(b)*(b^2*x^2+2*a*b*x+a^2-1)^(1/2)*x*a^2*b+2*csgn(b)*(b^2*x^2+2*a*b*x+a^2-1)^(1
/2)*a^3-3*csgn(b)*(b^2*x^2+2*a*b*x+a^2-1)^(1/2)*x*b+13*csgn(b)*(b^2*x^2+2*a*b*x+a^2-1)^(1/2)*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 [A]  time = 0.32, size = 275, normalized size = 2.39 \[ \frac {1}{4} \, b x^{4} + \frac {1}{3} \, a x^{3} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2} - 1\right )}^{\frac {3}{2}} x}{4 \, b^{2}} - \frac {5 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2} - 1\right )}^{\frac {3}{2}} a}{12 \, b^{3}} - \frac {{\left (5 \, a^{2} b^{2} - {\left (a^{2} - 1\right )} b^{2}\right )} a^{2} \log \left (2 \, b^{2} x + 2 \, a b + 2 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} - 1} b\right )}{8 \, b^{5}} + \frac {{\left (5 \, a^{2} b^{2} - {\left (a^{2} - 1\right )} b^{2}\right )} \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} - 1} x}{8 \, b^{4}} + \frac {{\left (5 \, a^{2} b^{2} - {\left (a^{2} - 1\right )} b^{2}\right )} {\left (a^{2} - 1\right )} \log \left (2 \, b^{2} x + 2 \, a b + 2 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} - 1} b\right )}{8 \, b^{5}} + \frac {{\left (5 \, a^{2} b^{2} - {\left (a^{2} - 1\right )} b^{2}\right )} \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} - 1} a}{8 \, b^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a+(b*x+a-1)^(1/2)*(b*x+a+1)^(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*log(2*b^2*x + 2*a*b + 2*sqrt(b^2*x^2 + 2*a*b*x + a^2 - 1)*b)/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)*log(2*b^2*x + 2*a*b + 2*sqrt(b^2*x^2 + 2*a*b*x + a^2 - 1)*b)/b^5 + 1/8*(5*a^2*b^2 - (a^2 - 1)*b^2)*s
qrt(b^2*x^2 + 2*a*b*x + a^2 - 1)*a/b^5

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mupad [B]  time = 33.48, size = 1067, normalized size = 9.28 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a*x^3)/3 + (b*x^4)/4 + ((((a - 1)^(1/2) - (a + b*x - 1)^(1/2))*(2*a^2 + 1/2))/(b^3*((a + 1)^(1/2) - (a + b*x
+ 1)^(1/2))) + (((a - 1)^(1/2) - (a + b*x - 1)^(1/2))^15*(2*a^2 + 1/2))/(b^3*((a + 1)^(1/2) - (a + b*x + 1)^(1
/2))^15) + (((a - 1)^(1/2) - (a + b*x - 1)^(1/2))^3*((64*a^4)/3 - 58*a^2 + 35/2))/(b^3*((a + 1)^(1/2) - (a + b
*x + 1)^(1/2))^3) + (((a - 1)^(1/2) - (a + b*x - 1)^(1/2))^13*((64*a^4)/3 - 58*a^2 + 35/2))/(b^3*((a + 1)^(1/2
) - (a + b*x + 1)^(1/2))^13) + (((a - 1)^(1/2) - (a + b*x - 1)^(1/2))^5*((2368*a^4)/3 - 862*a^2 + 273/2))/(b^3
*((a + 1)^(1/2) - (a + b*x + 1)^(1/2))^5) + (((a - 1)^(1/2) - (a + b*x - 1)^(1/2))^11*((2368*a^4)/3 - 862*a^2
+ 273/2))/(b^3*((a + 1)^(1/2) - (a + b*x + 1)^(1/2))^11) + (((a - 1)^(1/2) - (a + b*x - 1)^(1/2))^7*((9856*a^4
)/3 - 3178*a^2 + 715/2))/(b^3*((a + 1)^(1/2) - (a + b*x + 1)^(1/2))^7) + (((a - 1)^(1/2) - (a + b*x - 1)^(1/2)
)^9*((9856*a^4)/3 - 3178*a^2 + 715/2))/(b^3*((a + 1)^(1/2) - (a + b*x + 1)^(1/2))^9) + (((a - 1)^(1/2) - (a +
b*x - 1)^(1/2))^4*(192*a - 192*a^3)*(a - 1)^(1/2)*(a + 1)^(1/2))/(b^3*((a + 1)^(1/2) - (a + b*x + 1)^(1/2))^4)
+ (((a - 1)^(1/2) - (a + b*x - 1)^(1/2))^12*(192*a - 192*a^3)*(a - 1)^(1/2)*(a + 1)^(1/2))/(b^3*((a + 1)^(1/2
) - (a + b*x + 1)^(1/2))^12) + (((a - 1)^(1/2) - (a + b*x - 1)^(1/2))^6*((2816*a)/3 - (5888*a^3)/3)*(a - 1)^(1
/2)*(a + 1)^(1/2))/(b^3*((a + 1)^(1/2) - (a + b*x + 1)^(1/2))^6) + (((a - 1)^(1/2) - (a + b*x - 1)^(1/2))^10*(
(2816*a)/3 - (5888*a^3)/3)*(a - 1)^(1/2)*(a + 1)^(1/2))/(b^3*((a + 1)^(1/2) - (a + b*x + 1)^(1/2))^10) + (((a
- 1)^(1/2) - (a + b*x - 1)^(1/2))^8*((5504*a)/3 - (11648*a^3)/3)*(a - 1)^(1/2)*(a + 1)^(1/2))/(b^3*((a + 1)^(1
/2) - (a + b*x + 1)^(1/2))^8))/((28*((a - 1)^(1/2) - (a + b*x - 1)^(1/2))^4)/((a + 1)^(1/2) - (a + b*x + 1)^(1
/2))^4 - (8*((a - 1)^(1/2) - (a + b*x - 1)^(1/2))^2)/((a + 1)^(1/2) - (a + b*x + 1)^(1/2))^2 - (56*((a - 1)^(1
/2) - (a + b*x - 1)^(1/2))^6)/((a + 1)^(1/2) - (a + b*x + 1)^(1/2))^6 + (70*((a - 1)^(1/2) - (a + b*x - 1)^(1/
2))^8)/((a + 1)^(1/2) - (a + b*x + 1)^(1/2))^8 - (56*((a - 1)^(1/2) - (a + b*x - 1)^(1/2))^10)/((a + 1)^(1/2)
- (a + b*x + 1)^(1/2))^10 + (28*((a - 1)^(1/2) - (a + b*x - 1)^(1/2))^12)/((a + 1)^(1/2) - (a + b*x + 1)^(1/2)
)^12 - (8*((a - 1)^(1/2) - (a + b*x - 1)^(1/2))^14)/((a + 1)^(1/2) - (a + b*x + 1)^(1/2))^14 + ((a - 1)^(1/2)
- (a + b*x - 1)^(1/2))^16/((a + 1)^(1/2) - (a + b*x + 1)^(1/2))^16 + 1) - (2*atanh(((a - 1)^(1/2) - (a + b*x -
1)^(1/2))/((a + 1)^(1/2) - (a + b*x + 1)^(1/2)))*(a^2 + 1/4))/b^3

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

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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