\(\int \frac {e^{\text {arcsinh}(a+b x)}}{x} \, dx\) [353]

Optimal result

Integrand size = 12, antiderivative size = 89 \[ \int \frac {e^{\text {arcsinh}(a+b x)}}{x} \, dx=b x+\sqrt {1+a^2+2 a b x+b^2 x^2}+a \text {arcsinh}(a+b x)-\sqrt {1+a^2} \text {arctanh}\left (\frac {1+a^2+a b x}{\sqrt {1+a^2} \sqrt {1+a^2+2 a b x+b^2 x^2}}\right )+a \log (x) \]

[Out]

b*x+a*arcsinh(b*x+a)+a*ln(x)-arctanh((a*b*x+a^2+1)/(a^2+1)^(1/2)/(b^2*x^2+2*a*b*x+a^2+1)^(1/2))*(a^2+1)^(1/2)+
(b^2*x^2+2*a*b*x+a^2+1)^(1/2)

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.667, Rules used = {5878, 14, 748, 857, 633, 221, 738, 212} \[ \int \frac {e^{\text {arcsinh}(a+b x)}}{x} \, dx=-\sqrt {a^2+1} \text {arctanh}\left (\frac {a^2+a b x+1}{\sqrt {a^2+1} \sqrt {a^2+2 a b x+b^2 x^2+1}}\right )+\sqrt {a^2+2 a b x+b^2 x^2+1}+a \text {arcsinh}(a+b x)+a \log (x)+b x \]

[In]

Int[E^ArcSinh[a + b*x]/x,x]

[Out]

b*x + Sqrt[1 + a^2 + 2*a*b*x + b^2*x^2] + a*ArcSinh[a + b*x] - Sqrt[1 + a^2]*ArcTanh[(1 + a^2 + a*b*x)/(Sqrt[1
+ a^2]*Sqrt[1 + a^2 + 2*a*b*x + b^2*x^2])] + a*Log[x]

Rule 14

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]]

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 221

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSinh[Rt[b, 2]*(x/Sqrt[a])]/Rt[b, 2], x] /; FreeQ[{a, b},
x] && GtQ[a, 0] && PosQ[b]

Rule 633

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[1/(2*c*(-4*(c/(b^2 - 4*a*c)))^p), Subst[Int[Si
mp[1 - x^2/(b^2 - 4*a*c), x]^p, x], x, b + 2*c*x], x] /; FreeQ[{a, b, c, p}, x] && GtQ[4*a - b^2/c, 0]

Rule 738

Int[1/(((d_.) + (e_.)*(x_))*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Symbol] :> Dist[-2, Subst[Int[1/(4*c*d
^2 - 4*b*d*e + 4*a*e^2 - x^2), x], x, (2*a*e - b*d - (2*c*d - b*e)*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a,
b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[2*c*d - b*e, 0]

Rule 748

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(d + e*x)^(m + 1)*((
a + b*x + c*x^2)^p/(e*(m + 2*p + 1))), x] - Dist[p/(e*(m + 2*p + 1)), Int[(d + e*x)^m*Simp[b*d - 2*a*e + (2*c*
d - b*e)*x, x]*(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ
[c*d^2 - b*d*e + a*e^2, 0] && NeQ[2*c*d - b*e, 0] && GtQ[p, 0] && NeQ[m + 2*p + 1, 0] && ( !RationalQ[m] || Lt
Q[m, 1]) &&  !ILtQ[m + 2*p, 0] && IntQuadraticQ[a, b, c, d, e, m, p, x]

Rule 857

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dis
t[g/e, Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x], x] + Dist[(e*f - d*g)/e, Int[(d + e*x)^m*(a + b*x + c*x^
2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0]
&&  !IGtQ[m, 0]

Rule 5878

Int[E^(ArcSinh[u_]*(n_.))*(x_)^(m_.), x_Symbol] :> Int[x^m*(u + Sqrt[1 + u^2])^n, x] /; RationalQ[m] && Intege
rQ[n] && PolyQ[u, x]

Rubi steps \begin{align*} \text {integral}& = \int \frac {a+b x+\sqrt {1+(a+b x)^2}}{x} \, dx \\ & = \int \left (b+\frac {a}{x}+\frac {\sqrt {1+a^2+2 a b x+b^2 x^2}}{x}\right ) \, dx \\ & = b x+a \log (x)+\int \frac {\sqrt {1+a^2+2 a b x+b^2 x^2}}{x} \, dx \\ & = b x+\sqrt {1+a^2+2 a b x+b^2 x^2}+a \log (x)-\frac {1}{2} \int \frac {-2 \left (1+a^2\right )-2 a b x}{x \sqrt {1+a^2+2 a b x+b^2 x^2}} \, dx \\ & = b x+\sqrt {1+a^2+2 a b x+b^2 x^2}+a \log (x)-\left (-1-a^2\right ) \int \frac {1}{x \sqrt {1+a^2+2 a b x+b^2 x^2}} \, dx+(a b) \int \frac {1}{\sqrt {1+a^2+2 a b x+b^2 x^2}} \, dx \\ & = b x+\sqrt {1+a^2+2 a b x+b^2 x^2}+a \log (x)-\left (2 \left (1+a^2\right )\right ) \text {Subst}\left (\int \frac {1}{4 \left (1+a^2\right )-x^2} \, dx,x,\frac {2 \left (1+a^2\right )+2 a b x}{\sqrt {1+a^2+2 a b x+b^2 x^2}}\right )+\frac {a \text {Subst}\left (\int \frac {1}{\sqrt {1+\frac {x^2}{4 b^2}}} \, dx,x,2 a b+2 b^2 x\right )}{2 b} \\ & = b x+\sqrt {1+a^2+2 a b x+b^2 x^2}+a \text {arcsinh}(a+b x)-\sqrt {1+a^2} \text {arctanh}\left (\frac {1+a^2+a b x}{\sqrt {1+a^2} \sqrt {1+a^2+2 a b x+b^2 x^2}}\right )+a \log (x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.11 \[ \int \frac {e^{\text {arcsinh}(a+b x)}}{x} \, dx=b x+\sqrt {1+a^2+2 a b x+b^2 x^2}+a \text {arcsinh}(a+b x)+\left (a+\sqrt {1+a^2}\right ) \log (x)-\sqrt {1+a^2} \log \left (1+a^2+a b x+\sqrt {1+a^2} \sqrt {1+a^2+2 a b x+b^2 x^2}\right ) \]

[In]

Integrate[E^ArcSinh[a + b*x]/x,x]

[Out]

b*x + Sqrt[1 + a^2 + 2*a*b*x + b^2*x^2] + a*ArcSinh[a + b*x] + (a + Sqrt[1 + a^2])*Log[x] - Sqrt[1 + a^2]*Log[
1 + a^2 + a*b*x + Sqrt[1 + a^2]*Sqrt[1 + a^2 + 2*a*b*x + b^2*x^2]]

Maple [A] (verified)

Time = 0.89 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.42

method	result	size
default	\(\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}+\frac {a b \ln \left (\frac {b^{2} x +a b}{\sqrt {b^{2}}}+\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\right )}{\sqrt {b^{2}}}-\sqrt {a^{2}+1}\, \ln \left (\frac {2 a^{2}+2+2 a b x +2 \sqrt {a^{2}+1}\, \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{x}\right )+b x +a \ln \left (x \right )\)	\(126\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.53 \[ \int \frac {e^{\text {arcsinh}(a+b x)}}{x} \, dx=b x - a \log \left (-b x - a + \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right ) + a \log \left (x\right ) + \sqrt {a^{2} + 1} \log \left (-\frac {a^{2} b x + a^{3} + \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} {\left (a^{2} - \sqrt {a^{2} + 1} a + 1\right )} - {\left (a b x + a^{2} + 1\right )} \sqrt {a^{2} + 1} + a}{x}\right ) + \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} \]

[In]

integrate((b*x+a+(1+(b*x+a)^2)^(1/2))/x,x, algorithm="fricas")

[Out]

b*x - a*log(-b*x - a + sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)) + a*log(x) + sqrt(a^2 + 1)*log(-(a^2*b*x + a^3 + sqr
t(b^2*x^2 + 2*a*b*x + a^2 + 1)*(a^2 - sqrt(a^2 + 1)*a + 1) - (a*b*x + a^2 + 1)*sqrt(a^2 + 1) + a)/x) + sqrt(b^
2*x^2 + 2*a*b*x + a^2 + 1)

Sympy [F]

\[ \int \frac {e^{\text {arcsinh}(a+b x)}}{x} \, dx=\int \frac {a + b x + \sqrt {a^{2} + 2 a b x + b^{2} x^{2} + 1}}{x}\, dx \]

[In]

integrate((b*x+a+(1+(b*x+a)**2)**(1/2))/x,x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.80 \[ \int \frac {e^{\text {arcsinh}(a+b x)}}{x} \, dx=b x + a \operatorname {arsinh}\left (\frac {2 \, {\left (b^{2} x + a b\right )}}{\sqrt {-4 \, a^{2} b^{2} + 4 \, {\left (a^{2} + 1\right )} b^{2}}}\right ) + a \log \left (x\right ) - \sqrt {a^{2} + 1} \operatorname {arsinh}\left (\frac {2 \, a b x}{\sqrt {-4 \, a^{2} b^{2} + 4 \, {\left (a^{2} + 1\right )} b^{2}} {\left | x \right |}} + \frac {2 \, a^{2}}{\sqrt {-4 \, a^{2} b^{2} + 4 \, {\left (a^{2} + 1\right )} b^{2}} {\left | x \right |}} + \frac {2}{\sqrt {-4 \, a^{2} b^{2} + 4 \, {\left (a^{2} + 1\right )} b^{2}} {\left | x \right |}}\right ) + \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} \]

[In]

integrate((b*x+a+(1+(b*x+a)^2)^(1/2))/x,x, algorithm="maxima")

[Out]

b*x + a*arcsinh(2*(b^2*x + a*b)/sqrt(-4*a^2*b^2 + 4*(a^2 + 1)*b^2)) + a*log(x) - sqrt(a^2 + 1)*arcsinh(2*a*b*x
/(sqrt(-4*a^2*b^2 + 4*(a^2 + 1)*b^2)*abs(x)) + 2*a^2/(sqrt(-4*a^2*b^2 + 4*(a^2 + 1)*b^2)*abs(x)) + 2/(sqrt(-4*
a^2*b^2 + 4*(a^2 + 1)*b^2)*abs(x))) + sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.78 \[ \int \frac {e^{\text {arcsinh}(a+b x)}}{x} \, dx=b x - \frac {a b \log \left (-a b - {\left (x {\left | b \right |} - \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right )} {\left | b \right |}\right )}{{\left | b \right |}} + a \log \left ({\left | x \right |}\right ) + \sqrt {a^{2} + 1} \log \left (\frac {{\left | -2 \, x {\left | b \right |} + 2 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} - 2 \, \sqrt {a^{2} + 1} \right |}}{{\left | -2 \, x {\left | b \right |} + 2 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} + 2 \, \sqrt {a^{2} + 1} \right |}}\right ) + \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} \]

[In]

integrate((b*x+a+(1+(b*x+a)^2)^(1/2))/x,x, algorithm="giac")

[Out]

b*x - a*b*log(-a*b - (x*abs(b) - sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1))*abs(b))/abs(b) + a*log(abs(x)) + sqrt(a^2
+ 1)*log(abs(-2*x*abs(b) + 2*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1) - 2*sqrt(a^2 + 1))/abs(-2*x*abs(b) + 2*sqrt(b^2
*x^2 + 2*a*b*x + a^2 + 1) + 2*sqrt(a^2 + 1))) + sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)

Mupad [B] (verification not implemented)

Time = 3.42 (sec) , antiderivative size = 180, normalized size of antiderivative = 2.02 \[ \int \frac {e^{\text {arcsinh}(a+b x)}}{x} \, dx=\sqrt {a^2+2\,a\,b\,x+b^2\,x^2+1}+b\,x+a\,\ln \left (x\right )-\frac {\ln \left (a\,b+\frac {a^2+1}{x}+\frac {\sqrt {a^2+1}\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2+1}}{x}\right )}{\sqrt {a^2+1}}-\frac {a^2\,\ln \left (a\,b+\frac {a^2+1}{x}+\frac {\sqrt {a^2+1}\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2+1}}{x}\right )}{\sqrt {a^2+1}}+\frac {a\,b\,\ln \left (\sqrt {a^2+2\,a\,b\,x+b^2\,x^2+1}+\frac {x\,b^2+a\,b}{\sqrt {b^2}}\right )}{\sqrt {b^2}} \]

[In]

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

[Out]

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