3.355 \(\int \frac {e^{\sinh ^{-1}(a+b x)}}{x^3} \, dx\)

Optimal. Leaf size=116 \[ -\frac {\left (a^2+a b x+1\right ) \sqrt {a^2+2 a b x+b^2 x^2+1}}{2 \left (a^2+1\right ) x^2}-\frac {b^2 \tanh ^{-1}\left (\frac {a^2+a b x+1}{\sqrt {a^2+1} \sqrt {a^2+2 a b x+b^2 x^2+1}}\right )}{2 \left (a^2+1\right )^{3/2}}-\frac {a}{2 x^2}-\frac {b}{x} \]

[Out]

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

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Rubi [A]  time = 0.09, antiderivative size = 116, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {5907, 14, 720, 724, 206} \[ -\frac {\left (a^2+a b x+1\right ) \sqrt {a^2+2 a b x+b^2 x^2+1}}{2 \left (a^2+1\right ) x^2}-\frac {b^2 \tanh ^{-1}\left (\frac {a^2+a b x+1}{\sqrt {a^2+1} \sqrt {a^2+2 a b x+b^2 x^2+1}}\right )}{2 \left (a^2+1\right )^{3/2}}-\frac {a}{2 x^2}-\frac {b}{x} \]

Antiderivative was successfully verified.

[In]

Int[E^ArcSinh[a + b*x]/x^3,x]

[Out]

-a/(2*x^2) - b/x - ((1 + a^2 + a*b*x)*Sqrt[1 + a^2 + 2*a*b*x + b^2*x^2])/(2*(1 + a^2)*x^2) - (b^2*ArcTanh[(1 +
a^2 + a*b*x)/(Sqrt[1 + a^2]*Sqrt[1 + a^2 + 2*a*b*x + b^2*x^2])])/(2*(1 + a^2)^(3/2))

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 720

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((d + e*x)^(m + 1)*
(d*b - 2*a*e + (2*c*d - b*e)*x)*(a + b*x + c*x^2)^p)/(2*(m + 1)*(c*d^2 - b*d*e + a*e^2)), x] + Dist[(p*(b^2 -
4*a*c))/(2*(m + 1)*(c*d^2 - b*d*e + a*e^2)), Int[(d + e*x)^(m + 2)*(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[
{a, b, c, d, e}, 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] && EqQ[m +
2*p + 2, 0] && GtQ[p, 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 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^3} \, dx &=\int \frac {a+b x+\sqrt {1+(a+b x)^2}}{x^3} \, dx\\ &=\int \left (\frac {a}{x^3}+\frac {b}{x^2}+\frac {\sqrt {1+a^2+2 a b x+b^2 x^2}}{x^3}\right ) \, dx\\ &=-\frac {a}{2 x^2}-\frac {b}{x}+\int \frac {\sqrt {1+a^2+2 a b x+b^2 x^2}}{x^3} \, dx\\ &=-\frac {a}{2 x^2}-\frac {b}{x}-\frac {\left (1+a^2+a b x\right ) \sqrt {1+a^2+2 a b x+b^2 x^2}}{2 \left (1+a^2\right ) x^2}+\frac {b^2 \int \frac {1}{x \sqrt {1+a^2+2 a b x+b^2 x^2}} \, dx}{2 \left (1+a^2\right )}\\ &=-\frac {a}{2 x^2}-\frac {b}{x}-\frac {\left (1+a^2+a b x\right ) \sqrt {1+a^2+2 a b x+b^2 x^2}}{2 \left (1+a^2\right ) x^2}-\frac {b^2 \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 )}{1+a^2}\\ &=-\frac {a}{2 x^2}-\frac {b}{x}-\frac {\left (1+a^2+a b x\right ) \sqrt {1+a^2+2 a b x+b^2 x^2}}{2 \left (1+a^2\right ) x^2}-\frac {b^2 \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 )}{2 \left (1+a^2\right )^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 0.22, size = 129, normalized size = 1.11 \[ \frac {1}{2} \left (-\frac {\left (a^2+a b x+1\right ) \sqrt {a^2+2 a b x+b^2 x^2+1}}{\left (a^2+1\right ) x^2}-\frac {b^2 \log \left (\sqrt {a^2+1} \sqrt {a^2+2 a b x+b^2 x^2+1}+a^2+a b x+1\right )}{\left (a^2+1\right )^{3/2}}+\frac {b^2 \log (x)}{\left (a^2+1\right )^{3/2}}-\frac {a}{x^2}-\frac {2 b}{x}\right ) \]

Warning: Unable to verify antiderivative.

[In]

Integrate[E^ArcSinh[a + b*x]/x^3,x]

[Out]

(-(a/x^2) - (2*b)/x - ((1 + a^2 + a*b*x)*Sqrt[1 + a^2 + 2*a*b*x + b^2*x^2])/((1 + a^2)*x^2) + (b^2*Log[x])/(1
+ a^2)^(3/2) - (b^2*Log[1 + a^2 + a*b*x + Sqrt[1 + a^2]*Sqrt[1 + a^2 + 2*a*b*x + b^2*x^2]])/(1 + a^2)^(3/2))/2

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fricas [A]  time = 0.71, size = 181, normalized size = 1.56 \[ \frac {\sqrt {a^{2} + 1} b^{2} x^{2} \log \left (-\frac {a^{2} b x + a^{3} + \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} {\left (a^{2} - \sqrt {a^{2} + 1} a + 1\right )} - {\left (a b x + a^{2} + 1\right )} \sqrt {a^{2} + 1} + a}{x}\right ) - a^{5} - {\left (a^{3} + a\right )} b^{2} x^{2} - 2 \, a^{3} - 2 \, {\left (a^{4} + 2 \, a^{2} + 1\right )} b x - {\left (a^{4} + {\left (a^{3} + a\right )} b x + 2 \, a^{2} + 1\right )} \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} - a}{2 \, {\left (a^{4} + 2 \, a^{2} + 1\right )} x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a+(1+(b*x+a)^2)^(1/2))/x^3,x, algorithm="fricas")

[Out]

1/2*(sqrt(a^2 + 1)*b^2*x^2*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^5 - (a^3 + a)*b^2*x^2 - 2*a^3 - 2*(a^4 + 2*a^2 + 1)*b*x - (a^4
+ (a^3 + a)*b*x + 2*a^2 + 1)*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1) - a)/((a^4 + 2*a^2 + 1)*x^2)

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giac [B]  time = 0.38, size = 384, normalized size = 3.31 \[ \frac {b^{2} \log \left (\frac {{\left | -2 \, x {\left | b \right |} + 2 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} - 2 \, \sqrt {a^{2} + 1} \right |}}{{\left | -2 \, x {\left | b \right |} + 2 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} + 2 \, \sqrt {a^{2} + 1} \right |}}\right )}{2 \, {\left (a^{2} + 1\right )}^{\frac {3}{2}}} - \frac {2 \, b x + a}{2 \, x^{2}} + \frac {2 \, {\left (x {\left | b \right |} - \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right )}^{3} a^{2} b^{2} + 2 \, {\left (x {\left | b \right |} - \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right )} a^{4} b^{2} + 4 \, {\left (x {\left | b \right |} - \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right )}^{2} a^{3} b {\left | b \right |} + {\left (x {\left | b \right |} - \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right )}^{3} b^{2} + 3 \, {\left (x {\left | b \right |} - \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right )} a^{2} b^{2} + 4 \, {\left (x {\left | b \right |} - \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right )}^{2} a b {\left | b \right |} + {\left (x {\left | b \right |} - \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right )} b^{2}}{{\left ({\left (x {\left | b \right |} - \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right )}^{2} - a^{2} - 1\right )}^{2} {\left (a^{2} + 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a+(1+(b*x+a)^2)^(1/2))/x^3,x, algorithm="giac")

[Out]

1/2*b^2*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)))/(a^2 + 1)^(3/2) - 1/2*(2*b*x + a)/x^2 + (2*(x*abs(b) - sqrt(b
^2*x^2 + 2*a*b*x + a^2 + 1))^3*a^2*b^2 + 2*(x*abs(b) - sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1))*a^4*b^2 + 4*(x*abs(b
) - sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1))^2*a^3*b*abs(b) + (x*abs(b) - sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1))^3*b^2 +
3*(x*abs(b) - sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1))*a^2*b^2 + 4*(x*abs(b) - sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1))^2
*a*b*abs(b) + (x*abs(b) - sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1))*b^2)/(((x*abs(b) - sqrt(b^2*x^2 + 2*a*b*x + a^2 +
1))^2 - a^2 - 1)^2*(a^2 + 1))

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maple [B]  time = 0.01, size = 457, normalized size = 3.94 \[ -\frac {\left (b^{2} x^{2}+2 a b x +a^{2}+1\right )^{\frac {3}{2}}}{2 \left (a^{2}+1\right ) x^{2}}+\frac {a b \left (b^{2} x^{2}+2 a b x +a^{2}+1\right )^{\frac {3}{2}}}{2 \left (a^{2}+1\right )^{2} x}-\frac {a^{2} b^{2} \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{\left (a^{2}+1\right )^{2}}-\frac {a^{3} b^{3} \ln \left (\frac {b^{2} x +a b}{\sqrt {b^{2}}}+\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\right )}{2 \left (a^{2}+1\right )^{2} \sqrt {b^{2}}}+\frac {a^{2} b^{2} \ln \left (\frac {2 a^{2}+2+2 a b x +2 \sqrt {a^{2}+1}\, \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{x}\right )}{2 \left (a^{2}+1\right )^{\frac {3}{2}}}-\frac {a \,b^{3} \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\, x}{2 \left (a^{2}+1\right )^{2}}-\frac {a \,b^{3} \ln \left (\frac {b^{2} x +a b}{\sqrt {b^{2}}}+\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\right )}{2 \left (a^{2}+1\right )^{2} \sqrt {b^{2}}}+\frac {b^{2} \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{2 a^{2}+2}+\frac {b^{3} a \ln \left (\frac {b^{2} x +a b}{\sqrt {b^{2}}}+\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\right )}{2 \left (a^{2}+1\right ) \sqrt {b^{2}}}-\frac {b^{2} \ln \left (\frac {2 a^{2}+2+2 a b x +2 \sqrt {a^{2}+1}\, \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{x}\right )}{2 \sqrt {a^{2}+1}}-\frac {b}{x}-\frac {a}{2 x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a+(1+(b*x+a)^2)^(1/2))/x^3,x)

[Out]

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

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maxima [B]  time = 0.96, size = 313, normalized size = 2.70 \[ \frac {a^{2} b^{2} \operatorname {arsinh}\left (\frac {2 \, a b x}{\sqrt {-4 \, a^{2} b^{2} + 4 \, {\left (a^{2} + 1\right )} b^{2}} {\left | x \right |}} + \frac {2 \, a^{2}}{\sqrt {-4 \, a^{2} b^{2} + 4 \, {\left (a^{2} + 1\right )} b^{2}} {\left | x \right |}} + \frac {2}{\sqrt {-4 \, a^{2} b^{2} + 4 \, {\left (a^{2} + 1\right )} b^{2}} {\left | x \right |}}\right )}{2 \, {\left (a^{2} + 1\right )}^{\frac {3}{2}}} - \frac {b^{2} \operatorname {arsinh}\left (\frac {2 \, a b x}{\sqrt {-4 \, a^{2} b^{2} + 4 \, {\left (a^{2} + 1\right )} b^{2}} {\left | x \right |}} + \frac {2 \, a^{2}}{\sqrt {-4 \, a^{2} b^{2} + 4 \, {\left (a^{2} + 1\right )} b^{2}} {\left | x \right |}} + \frac {2}{\sqrt {-4 \, a^{2} b^{2} + 4 \, {\left (a^{2} + 1\right )} b^{2}} {\left | x \right |}}\right )}{2 \, \sqrt {a^{2} + 1}} + \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} b^{2}}{2 \, {\left (a^{2} + 1\right )}} + \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} a b}{2 \, {\left (a^{2} + 1\right )} x} - \frac {b}{x} - \frac {a}{2 \, x^{2}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )}^{\frac {3}{2}}}{2 \, {\left (a^{2} + 1\right )} x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a+(1+(b*x+a)^2)^(1/2))/x^3,x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {a+\sqrt {{\left (a+b\,x\right )}^2+1}+b\,x}{x^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + ((a + b*x)^2 + 1)^(1/2) + b*x)/x^3,x)

[Out]

int((a + ((a + b*x)^2 + 1)^(1/2) + b*x)/x^3, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {a + b x + \sqrt {a^{2} + 2 a b x + b^{2} x^{2} + 1}}{x^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a+(1+(b*x+a)**2)**(1/2))/x**3,x)

[Out]

Integral((a + b*x + sqrt(a**2 + 2*a*b*x + b**2*x**2 + 1))/x**3, x)

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