Integrand size = 12, antiderivative size = 115 \[ \int e^{\text {arccosh}(a+b x)} x^2 \, dx=\frac {e^{-2 \text {arccosh}(a+b x)}}{16 b^3}-\frac {a e^{-\text {arccosh}(a+b x)}}{2 b^3}+\frac {\left (1+4 a^2\right ) e^{2 \text {arccosh}(a+b x)}}{16 b^3}-\frac {a e^{3 \text {arccosh}(a+b x)}}{6 b^3}+\frac {e^{4 \text {arccosh}(a+b x)}}{32 b^3}-\frac {\left (1+4 a^2\right ) \text {arccosh}(a+b x)}{8 b^3} \] Output:
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
Time = 0.16 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.03 \[ \int e^{\text {arccosh}(a+b x)} x^2 \, dx=\frac {8 a b^3 x^3+6 b^4 x^4+\sqrt {-1+a+b x} \sqrt {1+a+b x} \left (2 a^3-3 b x-2 a^2 b x+6 b^3 x^3+a \left (13+2 b^2 x^2\right )\right )-3 \left (1+4 a^2\right ) \log \left (a+b x+\sqrt {-1+a+b x} \sqrt {1+a+b x}\right )}{24 b^3} \] Input:
Integrate[E^ArcCosh[a + b*x]*x^2,x]
Output:
(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)
Time = 0.62 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.89, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {6430, 2720, 27, 2159, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int x^2 e^{\text {arccosh}(a+b x)} \, dx\) |
\(\Big \downarrow \) 6430 |
\(\displaystyle \frac {\int e^{\text {arccosh}(a+b x)} \sqrt {\frac {a+b x-1}{a+b x+1}} (a+b x+1) \left (\frac {a}{b}-\frac {a+b x}{b}\right )^2d\text {arccosh}(a+b x)}{b}\) |
\(\Big \downarrow \) 2720 |
\(\displaystyle \frac {\int -\frac {e^{-3 \text {arccosh}(a+b x)} \left (1-e^{2 \text {arccosh}(a+b x)}\right ) \left (-2 e^{\text {arccosh}(a+b x)} a+e^{2 \text {arccosh}(a+b x)}+1\right )^2}{8 b^2}de^{\text {arccosh}(a+b x)}}{b}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {\int e^{-3 \text {arccosh}(a+b x)} \left (1-e^{2 \text {arccosh}(a+b x)}\right ) \left (-2 e^{\text {arccosh}(a+b x)} a+e^{2 \text {arccosh}(a+b x)}+1\right )^2de^{\text {arccosh}(a+b x)}}{8 b^3}\) |
\(\Big \downarrow \) 2159 |
\(\displaystyle -\frac {\int \left (-4 e^{-2 \text {arccosh}(a+b x)} a+4 e^{2 \text {arccosh}(a+b x)} a+e^{-3 \text {arccosh}(a+b x)}+\left (4 a^2+1\right ) e^{-\text {arccosh}(a+b x)}-\left (4 a^2+1\right ) e^{\text {arccosh}(a+b x)}-e^{3 \text {arccosh}(a+b x)}\right )de^{\text {arccosh}(a+b x)}}{8 b^3}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {-\frac {1}{2} \left (4 a^2+1\right ) e^{2 \text {arccosh}(a+b x)}+\left (4 a^2+1\right ) \log \left (e^{\text {arccosh}(a+b x)}\right )+4 a e^{-\text {arccosh}(a+b x)}+\frac {4}{3} a e^{3 \text {arccosh}(a+b x)}-\frac {1}{2} e^{-2 \text {arccosh}(a+b x)}-\frac {1}{4} e^{4 \text {arccosh}(a+b x)}}{8 b^3}\) |
Input:
Int[E^ArcCosh[a + b*x]*x^2,x]
Output:
-1/8*(-1/2*1/E^(2*ArcCosh[a + b*x]) + (4*a)/E^ArcCosh[a + b*x] - ((1 + 4*a ^2)*E^(2*ArcCosh[a + b*x]))/2 + (4*a*E^(3*ArcCosh[a + b*x]))/3 - E^(4*ArcC osh[a + b*x])/4 + (1 + 4*a^2)*Log[E^ArcCosh[a + b*x]])/b^3
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[(Pq_)*((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p _.), x_Symbol] :> Int[ExpandIntegrand[(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]
Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Simp[v/D[v, x] Subst[Int[FunctionOfExponentialFunction[u, x]/x, x], x, v], x]] /; Funct ionOfExponentialQ[u, x] && !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; FreeQ [{a, m, n}, x] && IntegerQ[m*n]] && !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x)) *(F_)[v_] /; FreeQ[{a, b, c}, x] && InverseFunctionQ[F[x]]]
Int[(f_)^(ArcCosh[(a_.) + (b_.)*(x_)]^(n_.)*(c_.))*(x_)^(m_.), x_Symbol] :> Simp[1/b Subst[Int[(-a/b + Cosh[x]/b)^m*f^(c*x^n)*Sinh[x], x], x, ArcCos h[a + b*x]], x] /; FreeQ[{a, b, c, f}, x] && IGtQ[m, 0] && IGtQ[n, 0]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.18 (sec) , antiderivative size = 288, normalized size of antiderivative = 2.50
method | result | size |
default | \(\frac {\sqrt {b x +a -1}\, \sqrt {b x +a +1}\, \left (6 \,\operatorname {csgn}\left (b \right ) b^{3} x^{3} \sqrt {b^{2} x^{2}+2 a b x +a^{2}-1}+2 \,\operatorname {csgn}\left (b \right ) a \,b^{2} x^{2} \sqrt {b^{2} x^{2}+2 a b x +a^{2}-1}-2 \sqrt {b^{2} x^{2}+2 a b x +a^{2}-1}\, \operatorname {csgn}\left (b \right ) a^{2} b x +2 \sqrt {b^{2} x^{2}+2 a b x +a^{2}-1}\, \operatorname {csgn}\left (b \right ) a^{3}-3 \sqrt {b^{2} x^{2}+2 a b x +a^{2}-1}\, \operatorname {csgn}\left (b \right ) b x +13 \sqrt {b^{2} x^{2}+2 a b x +a^{2}-1}\, \operatorname {csgn}\left (b \right ) a -12 \ln \left (\left (\sqrt {b^{2} x^{2}+2 a b x +a^{2}-1}\, \operatorname {csgn}\left (b \right )+b x +a \right ) \operatorname {csgn}\left (b \right )\right ) a^{2}-3 \ln \left (\left (\sqrt {b^{2} x^{2}+2 a b x +a^{2}-1}\, \operatorname {csgn}\left (b \right )+b x +a \right ) \operatorname {csgn}\left (b \right )\right )\right ) \operatorname {csgn}\left (b \right )}{24 b^{3} \sqrt {b^{2} x^{2}+2 a b x +a^{2}-1}}+\frac {b \,x^{4}}{4}+\frac {a \,x^{3}}{3}\) | \(288\) |
Input:
int((b*x+a+(b*x+a-1)^(1/2)*(b*x+a+1)^(1/2))*x^2,x,method=_RETURNVERBOSE)
Output:
1/24*(b*x+a-1)^(1/2)*(b*x+a+1)^(1/2)*(6*csgn(b)*b^3*x^3*(b^2*x^2+2*a*b*x+a ^2-1)^(1/2)+2*csgn(b)*a*b^2*x^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)*a^2*b*x+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)*b*x+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)*csg n(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)))*csg n(b)/b^3/(b^2*x^2+2*a*b*x+a^2-1)^(1/2)+1/4*b*x^4+1/3*a*x^3
Time = 0.08 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.97 \[ \int e^{\text {arccosh}(a+b x)} x^2 \, dx=\frac {6 \, b^{4} x^{4} + 8 \, a b^{3} x^{3} + {\left (6 \, b^{3} x^{3} + 2 \, a b^{2} x^{2} + 2 \, a^{3} - {\left (2 \, a^{2} + 3\right )} b x + 13 \, a\right )} \sqrt {b x + a + 1} \sqrt {b x + a - 1} + 3 \, {\left (4 \, a^{2} + 1\right )} \log \left (-b x + \sqrt {b x + a + 1} \sqrt {b x + a - 1} - a\right )}{24 \, b^{3}} \] Input:
integrate((b*x+a+(b*x+a-1)^(1/2)*(b*x+a+1)^(1/2))*x^2,x, algorithm="fricas ")
Output:
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)*sqrt(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
\[ \int e^{\text {arccosh}(a+b x)} x^2 \, dx=\int x^{2} \left (a + b x + \sqrt {a + b x - 1} \sqrt {a + b x + 1}\right )\, dx \] Input:
integrate((b*x+a+(b*x+a-1)**(1/2)*(b*x+a+1)**(1/2))*x**2,x)
Output:
Integral(x**2*(a + b*x + sqrt(a + b*x - 1)*sqrt(a + b*x + 1)), x)
Time = 0.11 (sec) , antiderivative size = 275, normalized size of antiderivative = 2.39 \[ \int e^{\text {arccosh}(a+b x)} x^2 \, dx=\frac {1}{4} \, b x^{4} + \frac {1}{3} \, a x^{3} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2} - 1\right )}^{\frac {3}{2}} x}{4 \, b^{2}} - \frac {5 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2} - 1\right )}^{\frac {3}{2}} a}{12 \, b^{3}} - \frac {{\left (5 \, a^{2} b^{2} - {\left (a^{2} - 1\right )} b^{2}\right )} a^{2} \log \left (2 \, b^{2} x + 2 \, a b + 2 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} - 1} b\right )}{8 \, b^{5}} + \frac {{\left (5 \, a^{2} b^{2} - {\left (a^{2} - 1\right )} b^{2}\right )} \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} - 1} x}{8 \, b^{4}} + \frac {{\left (5 \, a^{2} b^{2} - {\left (a^{2} - 1\right )} b^{2}\right )} {\left (a^{2} - 1\right )} \log \left (2 \, b^{2} x + 2 \, a b + 2 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} - 1} b\right )}{8 \, b^{5}} + \frac {{\left (5 \, a^{2} b^{2} - {\left (a^{2} - 1\right )} b^{2}\right )} \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} - 1} a}{8 \, b^{5}} \] Input:
integrate((b*x+a+(b*x+a-1)^(1/2)*(b*x+a+1)^(1/2))*x^2,x, algorithm="maxima ")
Output:
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)*sqrt( b^2*x^2 + 2*a*b*x + a^2 - 1)*a/b^5
Leaf count of result is larger than twice the leaf count of optimal. 347 vs. \(2 (173) = 346\).
Time = 0.14 (sec) , antiderivative size = 347, normalized size of antiderivative = 3.02 \[ \int e^{\text {arccosh}(a+b x)} x^2 \, dx=\frac {6 \, b^{2} x^{4} + 8 \, a b x^{3} + {\left ({\left ({\left (b x + a + 1\right )} {\left (2 \, {\left (b x + a + 1\right )} {\left (\frac {3 \, {\left (b x + a + 1\right )}}{b^{3}} - \frac {12 \, a b^{12} + 13 \, b^{12}}{b^{15}}\right )} + \frac {36 \, a^{2} b^{12} + 84 \, a b^{12} + 43 \, b^{12}}{b^{15}}\right )} - \frac {3 \, {\left (8 \, a^{3} b^{12} + 36 \, a^{2} b^{12} + 36 \, a b^{12} + 13 \, b^{12}\right )}}{b^{15}}\right )} \sqrt {b x + a + 1} \sqrt {b x + a - 1} - \frac {6 \, {\left (8 \, a^{3} + 12 \, a^{2} + 12 \, a + 3\right )} \log \left (\sqrt {b x + a + 1} - \sqrt {b x + a - 1}\right )}{b^{3}}\right )} b + \frac {4 \, {\left ({\left ({\left (2 \, b x - 4 \, a - 5\right )} {\left (b x + a + 1\right )} + 6 \, a^{2} + 18 \, a + 9\right )} \sqrt {b x + a + 1} \sqrt {b x + a - 1} + 6 \, {\left (2 \, a^{2} + 2 \, a + 1\right )} \log \left (\sqrt {b x + a + 1} - \sqrt {b x + a - 1}\right )\right )} a}{b^{2}} + \frac {4 \, {\left ({\left ({\left (2 \, b x - 4 \, a - 5\right )} {\left (b x + a + 1\right )} + 6 \, a^{2} + 18 \, a + 9\right )} \sqrt {b x + a + 1} \sqrt {b x + a - 1} + 6 \, {\left (2 \, a^{2} + 2 \, a + 1\right )} \log \left (\sqrt {b x + a + 1} - \sqrt {b x + a - 1}\right )\right )}}{b^{2}}}{24 \, b} \] Input:
integrate((b*x+a+(b*x+a-1)^(1/2)*(b*x+a+1)^(1/2))*x^2,x, algorithm="giac")
Output:
1/24*(6*b^2*x^4 + 8*a*b*x^3 + (((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 + 4*(((2*b*x - 4*a - 5)*(b*x + a + 1) + 6*a^2 + 18*a + 9)*sqrt(b*x + a + 1)*sqrt(b*x + a - 1) + 6*(2*a^2 + 2* a + 1)*log(sqrt(b*x + a + 1) - sqrt(b*x + a - 1)))*a/b^2 + 4*(((2*b*x - 4* a - 5)*(b*x + a + 1) + 6*a^2 + 18*a + 9)*sqrt(b*x + a + 1)*sqrt(b*x + a - 1) + 6*(2*a^2 + 2*a + 1)*log(sqrt(b*x + a + 1) - sqrt(b*x + a - 1)))/b^2)/ b
Time = 116.32 (sec) , antiderivative size = 1067, normalized size of antiderivative = 9.28 \[ \int e^{\text {arccosh}(a+b x)} x^2 \, dx=\text {Too large to display} \] Input:
int(x^2*(a + (a + b*x - 1)^(1/2)*(a + b*x + 1)^(1/2) + b*x),x)
Output:
(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 + 3 5/2))/(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*((98 56*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...
Time = 0.17 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.67 \[ \int e^{\text {arccosh}(a+b x)} x^2 \, dx=\frac {2 \sqrt {b x +a +1}\, \sqrt {b x +a -1}\, a^{3}-2 \sqrt {b x +a +1}\, \sqrt {b x +a -1}\, a^{2} b x +2 \sqrt {b x +a +1}\, \sqrt {b x +a -1}\, a \,b^{2} x^{2}+13 \sqrt {b x +a +1}\, \sqrt {b x +a -1}\, a +6 \sqrt {b x +a +1}\, \sqrt {b x +a -1}\, b^{3} x^{3}-3 \sqrt {b x +a +1}\, \sqrt {b x +a -1}\, b x -24 \,\mathrm {log}\left (\frac {\sqrt {b x +a -1}+\sqrt {b x +a +1}}{\sqrt {2}}\right ) a^{2}-6 \,\mathrm {log}\left (\frac {\sqrt {b x +a -1}+\sqrt {b x +a +1}}{\sqrt {2}}\right )+8 a \,b^{3} x^{3}+6 b^{4} x^{4}}{24 b^{3}} \] Input:
int((b*x+a+(b*x+a-1)^(1/2)*(b*x+a+1)^(1/2))*x^2,x)
Output:
(2*sqrt(a + b*x + 1)*sqrt(a + b*x - 1)*a**3 - 2*sqrt(a + b*x + 1)*sqrt(a + b*x - 1)*a**2*b*x + 2*sqrt(a + b*x + 1)*sqrt(a + b*x - 1)*a*b**2*x**2 + 1 3*sqrt(a + b*x + 1)*sqrt(a + b*x - 1)*a + 6*sqrt(a + b*x + 1)*sqrt(a + b*x - 1)*b**3*x**3 - 3*sqrt(a + b*x + 1)*sqrt(a + b*x - 1)*b*x - 24*log((sqrt (a + b*x - 1) + sqrt(a + b*x + 1))/sqrt(2))*a**2 - 6*log((sqrt(a + b*x - 1 ) + sqrt(a + b*x + 1))/sqrt(2)) + 8*a*b**3*x**3 + 6*b**4*x**4)/(24*b**3)