Integrand size = 12, antiderivative size = 109 \[ \int \frac {e^{\text {arccosh}(a+b x)}}{x^2} \, dx=-\frac {a}{x}-\frac {\sqrt {-1+a+b x} \sqrt {1+a+b x}}{x}+2 b \text {arcsinh}\left (\frac {\sqrt {-1+a+b x}}{\sqrt {2}}\right )-\frac {2 a b \arctan \left (\frac {\sqrt {1-a} \sqrt {1+a+b x}}{\sqrt {1+a} \sqrt {-1+a+b x}}\right )}{\sqrt {1-a^2}}+b \log (x) \] Output:
-a/x-(b*x+a-1)^(1/2)*(b*x+a+1)^(1/2)/x+2*b*arcsinh(1/2*(b*x+a-1)^(1/2)*2^( 1/2))-2*a*b*arctan((1-a)^(1/2)*(b*x+a+1)^(1/2)/(1+a)^(1/2)/(b*x+a-1)^(1/2) )/(-a^2+1)^(1/2)+b*ln(x)
Result contains complex when optimal does not.
Time = 0.18 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.28 \[ \int \frac {e^{\text {arccosh}(a+b x)}}{x^2} \, dx=-\frac {a}{x}-\frac {\sqrt {-1+a+b x} \sqrt {1+a+b x}}{x}+b \log (x)+b \log \left (a+b x+\sqrt {-1+a+b x} \sqrt {1+a+b x}\right )-\frac {i a b \log \left (\frac {2 \left (\sqrt {-1+a+b x} \sqrt {1+a+b x}+\frac {i \left (-1+a^2+a b x\right )}{\sqrt {1-a^2}}\right )}{a b x}\right )}{\sqrt {1-a^2}} \] Input:
Integrate[E^ArcCosh[a + b*x]/x^2,x]
Output:
-(a/x) - (Sqrt[-1 + a + b*x]*Sqrt[1 + a + b*x])/x + b*Log[x] + b*Log[a + b *x + Sqrt[-1 + a + b*x]*Sqrt[1 + a + b*x]] - (I*a*b*Log[(2*(Sqrt[-1 + a + b*x]*Sqrt[1 + a + b*x] + (I*(-1 + a^2 + a*b*x))/Sqrt[1 - a^2]))/(a*b*x)])/ Sqrt[1 - a^2]
Time = 0.55 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {6435, 2010, 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.
\(\displaystyle \int \frac {e^{\text {arccosh}(a+b x)}}{x^2} \, dx\) |
\(\Big \downarrow \) 6435 |
\(\displaystyle \int \frac {\sqrt {a+b x-1} \sqrt {a+b x+1}+a+b x}{x^2}dx\) |
\(\Big \downarrow \) 2010 |
\(\displaystyle \int \left (\frac {\sqrt {a+b x-1} \sqrt {a+b x+1}}{x^2}+\frac {a}{x^2}+\frac {b}{x}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {2 a b \arctan \left (\frac {\sqrt {1-a} \sqrt {a+b x+1}}{\sqrt {a+1} \sqrt {a+b x-1}}\right )}{\sqrt {1-a^2}}+2 b \text {arcsinh}\left (\frac {\sqrt {a+b x-1}}{\sqrt {2}}\right )-\frac {\sqrt {a+b x-1} \sqrt {a+b x+1}}{x}-\frac {a}{x}+b \log (x)\) |
Input:
Int[E^ArcCosh[a + b*x]/x^2,x]
Output:
-(a/x) - (Sqrt[-1 + a + b*x]*Sqrt[1 + a + b*x])/x + 2*b*ArcSinh[Sqrt[-1 + a + b*x]/Sqrt[2]] - (2*a*b*ArcTan[(Sqrt[1 - a]*Sqrt[1 + a + b*x])/(Sqrt[1 + a]*Sqrt[-1 + a + b*x])])/Sqrt[1 - a^2] + b*Log[x]
Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x] , x] /; FreeQ[{c, m}, x] && SumQ[u] && !LinearQ[u, x] && !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]
Int[E^(ArcCosh[u_]*(n_.))*(x_)^(m_.), x_Symbol] :> Int[x^m*(u + Sqrt[-1 + u ]*Sqrt[1 + u])^n, x] /; RationalQ[m] && IntegerQ[n] && PolyQ[u, x]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.19 (sec) , antiderivative size = 237, normalized size of antiderivative = 2.17
method | result | size |
default | \(\frac {\left (-\sqrt {a^{2}-1}\, \ln \left (\frac {2 a^{2}-2+2 a b x +2 \sqrt {a^{2}-1}\, \sqrt {b^{2} x^{2}+2 a b x +a^{2}-1}}{x}\right ) \operatorname {csgn}\left (b \right ) a b x +\ln \left (\left (\sqrt {b^{2} x^{2}+2 a b x +a^{2}-1}\, \operatorname {csgn}\left (b \right )+b x +a \right ) \operatorname {csgn}\left (b \right )\right ) a^{2} b x -\sqrt {b^{2} x^{2}+2 a b x +a^{2}-1}\, \operatorname {csgn}\left (b \right ) a^{2}-\ln \left (\left (\sqrt {b^{2} x^{2}+2 a b x +a^{2}-1}\, \operatorname {csgn}\left (b \right )+b x +a \right ) \operatorname {csgn}\left (b \right )\right ) b x +\sqrt {b^{2} x^{2}+2 a b x +a^{2}-1}\, \operatorname {csgn}\left (b \right )\right ) \sqrt {b x +a -1}\, \sqrt {b x +a +1}\, \operatorname {csgn}\left (b \right )}{\sqrt {b^{2} x^{2}+2 a b x +a^{2}-1}\, \left (a^{2}-1\right ) x}-\frac {a}{x}+b \ln \left (x \right )\) | \(237\) |
Input:
int((b*x+a+(b*x+a-1)^(1/2)*(b*x+a+1)^(1/2))/x^2,x,method=_RETURNVERBOSE)
Output:
(-(a^2-1)^(1/2)*ln(2*(a*b*x+(a^2-1)^(1/2)*(b^2*x^2+2*a*b*x+a^2-1)^(1/2)+a^ 2-1)/x)*csgn(b)*a*b*x+ln(((b^2*x^2+2*a*b*x+a^2-1)^(1/2)*csgn(b)+b*x+a)*csg n(b))*a^2*b*x-(b^2*x^2+2*a*b*x+a^2-1)^(1/2)*csgn(b)*a^2-ln(((b^2*x^2+2*a*b *x+a^2-1)^(1/2)*csgn(b)+b*x+a)*csgn(b))*b*x+(b^2*x^2+2*a*b*x+a^2-1)^(1/2)* csgn(b))*(b*x+a-1)^(1/2)*(b*x+a+1)^(1/2)*csgn(b)/(b^2*x^2+2*a*b*x+a^2-1)^( 1/2)/(a^2-1)/x-a/x+b*ln(x)
Time = 0.09 (sec) , antiderivative size = 334, normalized size of antiderivative = 3.06 \[ \int \frac {e^{\text {arccosh}(a+b x)}}{x^2} \, dx=\left [\frac {\sqrt {a^{2} - 1} a b x \log \left (\frac {a^{2} b x + a^{3} + {\left (a^{2} - \sqrt {a^{2} - 1} a - 1\right )} \sqrt {b x + a + 1} \sqrt {b x + a - 1} - {\left (a b x + a^{2} - 1\right )} \sqrt {a^{2} - 1} - a}{x}\right ) - {\left (a^{2} - 1\right )} b x \log \left (-b x + \sqrt {b x + a + 1} \sqrt {b x + a - 1} - a\right ) + {\left (a^{2} - 1\right )} b x \log \left (x\right ) - a^{3} - {\left (a^{2} - 1\right )} b x - {\left (a^{2} - 1\right )} \sqrt {b x + a + 1} \sqrt {b x + a - 1} + a}{{\left (a^{2} - 1\right )} x}, \frac {2 \, \sqrt {-a^{2} + 1} a b x \arctan \left (-\frac {\sqrt {-a^{2} + 1} b x - \sqrt {-a^{2} + 1} \sqrt {b x + a + 1} \sqrt {b x + a - 1}}{a^{2} - 1}\right ) - {\left (a^{2} - 1\right )} b x \log \left (-b x + \sqrt {b x + a + 1} \sqrt {b x + a - 1} - a\right ) + {\left (a^{2} - 1\right )} b x \log \left (x\right ) - a^{3} - {\left (a^{2} - 1\right )} b x - {\left (a^{2} - 1\right )} \sqrt {b x + a + 1} \sqrt {b x + a - 1} + a}{{\left (a^{2} - 1\right )} x}\right ] \] Input:
integrate((b*x+a+(b*x+a-1)^(1/2)*(b*x+a+1)^(1/2))/x^2,x, algorithm="fricas ")
Output:
[(sqrt(a^2 - 1)*a*b*x*log((a^2*b*x + a^3 + (a^2 - sqrt(a^2 - 1)*a - 1)*sqr t(b*x + a + 1)*sqrt(b*x + a - 1) - (a*b*x + a^2 - 1)*sqrt(a^2 - 1) - a)/x) - (a^2 - 1)*b*x*log(-b*x + sqrt(b*x + a + 1)*sqrt(b*x + a - 1) - a) + (a^ 2 - 1)*b*x*log(x) - a^3 - (a^2 - 1)*b*x - (a^2 - 1)*sqrt(b*x + a + 1)*sqrt (b*x + a - 1) + a)/((a^2 - 1)*x), (2*sqrt(-a^2 + 1)*a*b*x*arctan(-(sqrt(-a ^2 + 1)*b*x - sqrt(-a^2 + 1)*sqrt(b*x + a + 1)*sqrt(b*x + a - 1))/(a^2 - 1 )) - (a^2 - 1)*b*x*log(-b*x + sqrt(b*x + a + 1)*sqrt(b*x + a - 1) - a) + ( a^2 - 1)*b*x*log(x) - a^3 - (a^2 - 1)*b*x - (a^2 - 1)*sqrt(b*x + a + 1)*sq rt(b*x + a - 1) + a)/((a^2 - 1)*x)]
\[ \int \frac {e^{\text {arccosh}(a+b x)}}{x^2} \, dx=\int \frac {a + b x + \sqrt {a + b x - 1} \sqrt {a + b x + 1}}{x^{2}}\, dx \] Input:
integrate((b*x+a+(b*x+a-1)**(1/2)*(b*x+a+1)**(1/2))/x**2,x)
Output:
Integral((a + b*x + sqrt(a + b*x - 1)*sqrt(a + b*x + 1))/x**2, x)
Exception generated. \[ \int \frac {e^{\text {arccosh}(a+b x)}}{x^2} \, dx=\text {Exception raised: ValueError} \] Input:
integrate((b*x+a+(b*x+a-1)^(1/2)*(b*x+a+1)^(1/2))/x^2,x, algorithm="maxima ")
Output:
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(a-1>0)', see `assume?` for more details)Is
Time = 0.16 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.56 \[ \int \frac {e^{\text {arccosh}(a+b x)}}{x^2} \, dx=-{\left (\frac {2 \, a \arctan \left (\frac {{\left (\sqrt {b x + a + 1} - \sqrt {b x + a - 1}\right )}^{2} - 2 \, a}{2 \, \sqrt {-a^{2} + 1}}\right )}{\sqrt {-a^{2} + 1}} - \frac {4 \, {\left (a {\left (\sqrt {b x + a + 1} - \sqrt {b x + a - 1}\right )}^{2} - 2\right )}}{{\left (\sqrt {b x + a + 1} - \sqrt {b x + a - 1}\right )}^{4} - 4 \, a {\left (\sqrt {b x + a + 1} - \sqrt {b x + a - 1}\right )}^{2} + 4} + \frac {b x + a}{b x} + \log \left ({\left (\sqrt {b x + a + 1} - \sqrt {b x + a - 1}\right )}^{2}\right ) - \log \left ({\left | b x \right |}\right )\right )} b \] Input:
integrate((b*x+a+(b*x+a-1)^(1/2)*(b*x+a+1)^(1/2))/x^2,x, algorithm="giac")
Output:
-(2*a*arctan(1/2*((sqrt(b*x + a + 1) - sqrt(b*x + a - 1))^2 - 2*a)/sqrt(-a ^2 + 1))/sqrt(-a^2 + 1) - 4*(a*(sqrt(b*x + a + 1) - sqrt(b*x + a - 1))^2 - 2)/((sqrt(b*x + a + 1) - sqrt(b*x + a - 1))^4 - 4*a*(sqrt(b*x + a + 1) - sqrt(b*x + a - 1))^2 + 4) + (b*x + a)/(b*x) + log((sqrt(b*x + a + 1) - sqr t(b*x + a - 1))^2) - log(abs(b*x)))*b
Time = 51.42 (sec) , antiderivative size = 10345, normalized size of antiderivative = 94.91 \[ \int \frac {e^{\text {arccosh}(a+b x)}}{x^2} \, dx=\text {Too large to display} \] Input:
int((a + (a + b*x - 1)^(1/2)*(a + b*x + 1)^(1/2) + b*x)/x^2,x)
Output:
((((a - 1)^(1/2) - (a + b*x - 1)^(1/2))^2*((5*b)/4 - (a^2*b)/4))/((a^2 - 1 )*((a + 1)^(1/2) - (a + b*x + 1)^(1/2))^2) - b/4 + (a*b*((a - 1)^(1/2) - ( a + b*x - 1)^(1/2))*(a - 1)^(1/2)*(a + 1)^(1/2))/(2*(a^2 - 1)*((a + 1)^(1/ 2) - (a + b*x + 1)^(1/2))))/(((a - 1)^(1/2) - (a + b*x - 1)^(1/2))/((a + 1 )^(1/2) - (a + b*x + 1)^(1/2)) + ((a - 1)^(1/2) - (a + b*x - 1)^(1/2))^3/( (a + 1)^(1/2) - (a + b*x + 1)^(1/2))^3 - (2*a*((a - 1)^(1/2) - (a + b*x - 1)^(1/2))^2*(a - 1)^(1/2)*(a + 1)^(1/2))/((a^2 - 1)*((a + 1)^(1/2) - (a + b*x + 1)^(1/2))^2)) - a/x + b*atan((b*(2*b*(2*b*((32*(2*b - 14*a^2*b + 42* a^4*b - 70*a^6*b + 70*a^8*b - 42*a^10*b + 14*a^12*b - 2*a^14*b + 2*a^2*b*( a - 1)*(a + 1) - 10*a^4*b*(a - 1)*(a + 1) + 20*a^6*b*(a - 1)*(a + 1) - 20* a^8*b*(a - 1)*(a + 1) + 10*a^10*b*(a - 1)*(a + 1) - 2*a^12*b*(a - 1)*(a + 1) - 4*a^4*b*(a - 1)^2*(a + 1)^2 + 12*a^6*b*(a - 1)^2*(a + 1)^2 - 12*a^8*b *(a - 1)^2*(a + 1)^2 + 4*a^10*b*(a - 1)^2*(a + 1)^2))/(7*a^2 - 21*a^4 + 35 *a^6 - 35*a^8 + 21*a^10 - 7*a^12 + a^14 - 1) - 2*b*((32*(a*(a - 1)^(1/2)*( a + 1)^(1/2) - 6*a^3*(a - 1)^(1/2)*(a + 1)^(1/2) + 15*a^5*(a - 1)^(1/2)*(a + 1)^(1/2) - 5*a^3*(a - 1)^(3/2)*(a + 1)^(3/2) - 20*a^7*(a - 1)^(1/2)*(a + 1)^(1/2) + 20*a^5*(a - 1)^(3/2)*(a + 1)^(3/2) + 15*a^9*(a - 1)^(1/2)*(a + 1)^(1/2) - 30*a^7*(a - 1)^(3/2)*(a + 1)^(3/2) - 6*a^11*(a - 1)^(1/2)*(a + 1)^(1/2) + 4*a^5*(a - 1)^(5/2)*(a + 1)^(5/2) + 20*a^9*(a - 1)^(3/2)*(a + 1)^(3/2) + a^13*(a - 1)^(1/2)*(a + 1)^(1/2) - 8*a^7*(a - 1)^(5/2)*(a +...
Time = 0.16 (sec) , antiderivative size = 276, normalized size of antiderivative = 2.53 \[ \int \frac {e^{\text {arccosh}(a+b x)}}{x^2} \, dx=\frac {-\sqrt {a +1}\, \sqrt {-a +1}\, \mathrm {log}\left (\sqrt {a +1}\, \sqrt {-a +1}\, i +\sqrt {b x +a +1}\, \sqrt {b x +a -1}+b x \right ) a b i x +\sqrt {a +1}\, \sqrt {-a +1}\, \mathrm {log}\left (\frac {-\sqrt {\sqrt {a +1}\, \sqrt {-a +1}\, i +a}\, \sqrt {2}+\sqrt {b x +a -1}+\sqrt {b x +a +1}}{\sqrt {2}}\right ) a b i x +\sqrt {a +1}\, \sqrt {-a +1}\, \mathrm {log}\left (\frac {\sqrt {\sqrt {a +1}\, \sqrt {-a +1}\, i +a}\, \sqrt {2}+\sqrt {b x +a -1}+\sqrt {b x +a +1}}{\sqrt {2}}\right ) a b i x -\sqrt {b x +a +1}\, \sqrt {b x +a -1}\, a^{2}+\sqrt {b x +a +1}\, \sqrt {b x +a -1}+2 \,\mathrm {log}\left (\frac {\sqrt {b x +a -1}+\sqrt {b x +a +1}}{\sqrt {2}}\right ) a^{2} b x -2 \,\mathrm {log}\left (\frac {\sqrt {b x +a -1}+\sqrt {b x +a +1}}{\sqrt {2}}\right ) b x +\mathrm {log}\left (x \right ) a^{2} b x -\mathrm {log}\left (x \right ) b x -a^{3}+a}{x \left (a^{2}-1\right )} \] Input:
int((b*x+a+(b*x+a-1)^(1/2)*(b*x+a+1)^(1/2))/x^2,x)
Output:
( - sqrt(a + 1)*sqrt( - a + 1)*log(sqrt(a + 1)*sqrt( - a + 1)*i + sqrt(a + b*x + 1)*sqrt(a + b*x - 1) + b*x)*a*b*i*x + sqrt(a + 1)*sqrt( - a + 1)*lo g(( - sqrt(sqrt(a + 1)*sqrt( - a + 1)*i + a)*sqrt(2) + sqrt(a + b*x - 1) + sqrt(a + b*x + 1))/sqrt(2))*a*b*i*x + sqrt(a + 1)*sqrt( - a + 1)*log((sqr t(sqrt(a + 1)*sqrt( - a + 1)*i + a)*sqrt(2) + sqrt(a + b*x - 1) + sqrt(a + b*x + 1))/sqrt(2))*a*b*i*x - sqrt(a + b*x + 1)*sqrt(a + b*x - 1)*a**2 + s qrt(a + b*x + 1)*sqrt(a + b*x - 1) + 2*log((sqrt(a + b*x - 1) + sqrt(a + b *x + 1))/sqrt(2))*a**2*b*x - 2*log((sqrt(a + b*x - 1) + sqrt(a + b*x + 1)) /sqrt(2))*b*x + log(x)*a**2*b*x - log(x)*b*x - a**3 + a)/(x*(a**2 - 1))