3.354 \(\int \frac {e^{\sinh ^{-1}(a+b x)}}{x^2} \, dx\)

Optimal. Leaf size=99 \[ -\frac {\sqrt {a^2+2 a b x+b^2 x^2+1}}{x}-\frac {a b \tanh ^{-1}\left (\frac {a^2+a b x+1}{\sqrt {a^2+1} \sqrt {a^2+2 a b x+b^2 x^2+1}}\right )}{\sqrt {a^2+1}}+b \sinh ^{-1}(a+b x)-\frac {a}{x}+b \log (x) \]

[Out]

-a/x+b*arcsinh(b*x+a)+b*ln(x)-a*b*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)/x

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Rubi [A]  time = 0.11, antiderivative size = 99, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.667, Rules used = {5907, 14, 732, 843, 619, 215, 724, 206} \[ -\frac {\sqrt {a^2+2 a b x+b^2 x^2+1}}{x}-\frac {a b \tanh ^{-1}\left (\frac {a^2+a b x+1}{\sqrt {a^2+1} \sqrt {a^2+2 a b x+b^2 x^2+1}}\right )}{\sqrt {a^2+1}}+b \sinh ^{-1}(a+b x)-\frac {a}{x}+b \log (x) \]

Antiderivative was successfully verified.

[In]

Int[E^ArcSinh[a + b*x]/x^2,x]

[Out]

-(a/x) - Sqrt[1 + a^2 + 2*a*b*x + b^2*x^2]/x + b*ArcSinh[a + b*x] - (a*b*ArcTanh[(1 + a^2 + a*b*x)/(Sqrt[1 + a
^2]*Sqrt[1 + a^2 + 2*a*b*x + b^2*x^2])])/Sqrt[1 + a^2] + b*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 206

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

Rule 215

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 619

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 724

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 732

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 + 1)), x] - Dist[p/(e*(m + 1)), Int[(d + e*x)^(m + 1)*(b + 2*c*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] && (IntegerQ[p] || LtQ[m, -1]) && NeQ[m, -1] &&  !ILtQ[m + 2*p + 1, 0] && IntQua
draticQ[a, b, c, d, e, m, p, x]

Rule 843

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 5907

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

Rubi steps

\begin {align*} \int \frac {e^{\sinh ^{-1}(a+b x)}}{x^2} \, dx &=\int \frac {a+b x+\sqrt {1+(a+b x)^2}}{x^2} \, dx\\ &=\int \left (\frac {a}{x^2}+\frac {b}{x}+\frac {\sqrt {1+a^2+2 a b x+b^2 x^2}}{x^2}\right ) \, dx\\ &=-\frac {a}{x}+b \log (x)+\int \frac {\sqrt {1+a^2+2 a b x+b^2 x^2}}{x^2} \, dx\\ &=-\frac {a}{x}-\frac {\sqrt {1+a^2+2 a b x+b^2 x^2}}{x}+b \log (x)+\frac {1}{2} \int \frac {2 a b+2 b^2 x}{x \sqrt {1+a^2+2 a b x+b^2 x^2}} \, dx\\ &=-\frac {a}{x}-\frac {\sqrt {1+a^2+2 a b x+b^2 x^2}}{x}+b \log (x)+(a b) \int \frac {1}{x \sqrt {1+a^2+2 a b x+b^2 x^2}} \, dx+b^2 \int \frac {1}{\sqrt {1+a^2+2 a b x+b^2 x^2}} \, dx\\ &=-\frac {a}{x}-\frac {\sqrt {1+a^2+2 a b x+b^2 x^2}}{x}+b \log (x)+\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{\sqrt {1+\frac {x^2}{4 b^2}}} \, dx,x,2 a b+2 b^2 x\right )-(2 a b) \operatorname {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}{x}-\frac {\sqrt {1+a^2+2 a b x+b^2 x^2}}{x}+b \sinh ^{-1}(a+b x)-\frac {a b \tanh ^{-1}\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 )}{\sqrt {1+a^2}}+b \log (x)\\ \end {align*}

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Mathematica [A]  time = 0.17, size = 110, normalized size = 1.11 \[ b \sinh ^{-1}(a+b x)-\frac {\sqrt {a^2+2 a b x+b^2 x^2+1}+\frac {a b x \log \left (\sqrt {a^2+1} \sqrt {a^2+2 a b x+b^2 x^2+1}+a^2+a b x+1\right )}{\sqrt {a^2+1}}+\left (-\frac {a}{\sqrt {a^2+1}}-1\right ) b x \log (x)+a}{x} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[E^ArcSinh[a + b*x]/x^2,x]

[Out]

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

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fricas [B]  time = 0.69, size = 183, normalized size = 1.85 \[ \frac {\sqrt {a^{2} + 1} a b x \log \left (-\frac {a^{2} b x + a^{3} + \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} {\left (a^{2} - \sqrt {a^{2} + 1} a + 1\right )} - {\left (a b x + a^{2} + 1\right )} \sqrt {a^{2} + 1} + a}{x}\right ) - {\left (a^{2} + 1\right )} b x \log \left (-b x - a + \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right ) + {\left (a^{2} + 1\right )} b x \log \relax (x) - a^{3} - {\left (a^{2} + 1\right )} b x - \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} {\left (a^{2} + 1\right )} - a}{{\left (a^{2} + 1\right )} x} \]

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]

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

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giac [B]  time = 0.94, size = 234, normalized size = 2.36 \[ \frac {a b \log \left (\frac {{\left | -2 \, x {\left | b \right |} + 2 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} - 2 \, \sqrt {a^{2} + 1} \right |}}{{\left | -2 \, x {\left | b \right |} + 2 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} + 2 \, \sqrt {a^{2} + 1} \right |}}\right )}{\sqrt {a^{2} + 1}} - \frac {b^{2} \log \left (-a b - {\left (x {\left | b \right |} - \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right )} {\left | b \right |}\right )}{{\left | b \right |}} + b \log \left ({\left | x \right |}\right ) - \frac {a}{x} + \frac {2 \, {\left ({\left (x {\left | b \right |} - \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right )} a b^{5} + a^{2} b^{4} {\left | b \right |} + b^{4} {\left | b \right |}\right )}}{{\left ({\left (x {\left | b \right |} - \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right )}^{2} - a^{2} - 1\right )} b^{4}} \]

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]

a*b*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(a^2 + 1) - b^2*log(-a*b - (x*abs(b) - sqrt(b^2*x^2 + 2*a*b*x
+ a^2 + 1))*abs(b))/abs(b) + b*log(abs(x)) - a/x + 2*((x*abs(b) - sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1))*a*b^5 +
a^2*b^4*abs(b) + b^4*abs(b))/(((x*abs(b) - sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1))^2 - a^2 - 1)*b^4)

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maple [B]  time = 0.01, size = 267, normalized size = 2.70 \[ -\frac {\left (b^{2} x^{2}+2 a b x +a^{2}+1\right )^{\frac {3}{2}}}{\left (a^{2}+1\right ) x}+\frac {2 a b \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{a^{2}+1}+\frac {a^{2} b^{2} \ln \left (\frac {b^{2} x +a b}{\sqrt {b^{2}}}+\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\right )}{\left (a^{2}+1\right ) \sqrt {b^{2}}}-\frac {a b \ln \left (\frac {2 a^{2}+2+2 a b x +2 \sqrt {a^{2}+1}\, \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{x}\right )}{\sqrt {a^{2}+1}}+\frac {b^{2} \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\, x}{a^{2}+1}+\frac {b^{2} \ln \left (\frac {b^{2} x +a b}{\sqrt {b^{2}}}+\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\right )}{\left (a^{2}+1\right ) \sqrt {b^{2}}}+b \ln \relax (x )-\frac {a}{x} \]

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/(a^2+1)/x*(b^2*x^2+2*a*b*x+a^2+1)^(3/2)+2*a*b/(a^2+1)*(b^2*x^2+2*a*b*x+a^2+1)^(1/2)+a^2*b^2/(a^2+1)*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*b/(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^2/(a^2+1)*(b^2*x^2+2*a*b*x+a^2+1)^(1/2)*x+b^2/(a^2+1)*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)+b*ln(x)-a/x

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maxima [A]  time = 0.36, size = 170, normalized size = 1.72 \[ -\frac {a b \operatorname {arsinh}\left (\frac {2 \, a b x}{\sqrt {-4 \, a^{2} b^{2} + 4 \, {\left (a^{2} + 1\right )} b^{2}} {\left | x \right |}} + \frac {2 \, a^{2}}{\sqrt {-4 \, a^{2} b^{2} + 4 \, {\left (a^{2} + 1\right )} b^{2}} {\left | x \right |}} + \frac {2}{\sqrt {-4 \, a^{2} b^{2} + 4 \, {\left (a^{2} + 1\right )} b^{2}} {\left | x \right |}}\right )}{\sqrt {a^{2} + 1}} + b \operatorname {arsinh}\left (\frac {2 \, {\left (b^{2} x + a b\right )}}{\sqrt {-4 \, a^{2} b^{2} + 4 \, {\left (a^{2} + 1\right )} b^{2}}}\right ) + b \log \relax (x) - \frac {a}{x} - \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{x} \]

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]

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

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mupad [B]  time = 1.26, size = 269, normalized size = 2.72 \[ b\,\ln \relax (x)-\frac {a}{x}+\ln \left (\sqrt {a^2+2\,a\,b\,x+b^2\,x^2+1}+\frac {x\,b^2+a\,b}{\sqrt {b^2}}\right )\,\sqrt {b^2}-\frac {\sqrt {a^2+2\,a\,b\,x+b^2\,x^2+1}}{x\,\left (a^2+1\right )}+\frac {a^3\,b\,\mathrm {atanh}\left (\frac {a^2+b\,x\,a+1}{\sqrt {a^2+1}\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2+1}}\right )}{{\left (a^2+1\right )}^{3/2}}-\frac {a^2\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2+1}}{x\,\left (a^2+1\right )}+\frac {a\,b\,\mathrm {atanh}\left (\frac {a^2+b\,x\,a+1}{\sqrt {a^2+1}\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2+1}}\right )}{{\left (a^2+1\right )}^{3/2}}-\frac {2\,a\,b\,\ln \left (a\,b+\frac {a^2+1}{x}+\frac {\sqrt {a^2+1}\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2+1}}{x}\right )}{\sqrt {a^2+1}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b*log(x) - a/x + 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) - (a^2 + b^2
*x^2 + 2*a*b*x + 1)^(1/2)/(x*(a^2 + 1)) + (a^3*b*atanh((a^2 + a*b*x + 1)/((a^2 + 1)^(1/2)*(a^2 + b^2*x^2 + 2*a
*b*x + 1)^(1/2))))/(a^2 + 1)^(3/2) - (a^2*(a^2 + b^2*x^2 + 2*a*b*x + 1)^(1/2))/(x*(a^2 + 1)) + (a*b*atanh((a^2
+ a*b*x + 1)/((a^2 + 1)^(1/2)*(a^2 + b^2*x^2 + 2*a*b*x + 1)^(1/2))))/(a^2 + 1)^(3/2) - (2*a*b*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)

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

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

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