3.15.5 \(\int \frac {1}{x \sqrt {a x+\sqrt {b^2+a^2 x^2}}} \, dx\)

Optimal. Leaf size=100 \[ \frac {2}{\sqrt {\sqrt {a^2 x^2+b^2}+a x}}+\frac {2 \tan ^{-1}\left (\frac {\sqrt {\sqrt {a^2 x^2+b^2}+a x}}{\sqrt {b}}\right )}{\sqrt {b}}-\frac {2 \tanh ^{-1}\left (\frac {\sqrt {\sqrt {a^2 x^2+b^2}+a x}}{\sqrt {b}}\right )}{\sqrt {b}} \]

________________________________________________________________________________________

Rubi [A]  time = 0.12, antiderivative size = 100, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {2119, 453, 329, 298, 203, 206} \begin {gather*} \frac {2}{\sqrt {\sqrt {a^2 x^2+b^2}+a x}}+\frac {2 \tan ^{-1}\left (\frac {\sqrt {\sqrt {a^2 x^2+b^2}+a x}}{\sqrt {b}}\right )}{\sqrt {b}}-\frac {2 \tanh ^{-1}\left (\frac {\sqrt {\sqrt {a^2 x^2+b^2}+a x}}{\sqrt {b}}\right )}{\sqrt {b}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x*Sqrt[a*x + Sqrt[b^2 + a^2*x^2]]),x]

[Out]

2/Sqrt[a*x + Sqrt[b^2 + a^2*x^2]] + (2*ArcTan[Sqrt[a*x + Sqrt[b^2 + a^2*x^2]]/Sqrt[b]])/Sqrt[b] - (2*ArcTanh[S
qrt[a*x + Sqrt[b^2 + a^2*x^2]]/Sqrt[b]])/Sqrt[b]

Rule 203

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

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 298

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-(a/b), 2]], s = Denominator[Rt[-(a/b),
2]]}, Dist[s/(2*b), Int[1/(r + s*x^2), x], x] - Dist[s/(2*b), Int[1/(r - s*x^2), x], x]] /; FreeQ[{a, b}, x] &
&  !GtQ[a/b, 0]

Rule 329

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + (b*x^(k*n))/c^n)^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 453

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(c*(e*x)^(m
+ 1)*(a + b*x^n)^(p + 1))/(a*e*(m + 1)), x] + Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*e^n*(m + 1)), In
t[(e*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && (IntegerQ[n] ||
GtQ[e, 0]) && ((GtQ[n, 0] && LtQ[m, -1]) || (LtQ[n, 0] && GtQ[m + n, -1])) &&  !ILtQ[p, -1]

Rule 2119

Int[((g_.) + (h_.)*(x_))^(m_.)*((e_.)*(x_) + (f_.)*Sqrt[(a_.) + (c_.)*(x_)^2])^(n_.), x_Symbol] :> Dist[1/(2^(
m + 1)*e^(m + 1)), Subst[Int[x^(n - m - 2)*(a*f^2 + x^2)*(-(a*f^2*h) + 2*e*g*x + h*x^2)^m, x], x, e*x + f*Sqrt
[a + c*x^2]], x] /; FreeQ[{a, c, e, f, g, h, n}, x] && EqQ[e^2 - c*f^2, 0] && IntegerQ[m]

Rubi steps

\begin {align*} \int \frac {1}{x \sqrt {a x+\sqrt {b^2+a^2 x^2}}} \, dx &=\operatorname {Subst}\left (\int \frac {b^2+x^2}{x^{3/2} \left (-b^2+x^2\right )} \, dx,x,a x+\sqrt {b^2+a^2 x^2}\right )\\ &=\frac {2}{\sqrt {a x+\sqrt {b^2+a^2 x^2}}}+2 \operatorname {Subst}\left (\int \frac {\sqrt {x}}{-b^2+x^2} \, dx,x,a x+\sqrt {b^2+a^2 x^2}\right )\\ &=\frac {2}{\sqrt {a x+\sqrt {b^2+a^2 x^2}}}+4 \operatorname {Subst}\left (\int \frac {x^2}{-b^2+x^4} \, dx,x,\sqrt {a x+\sqrt {b^2+a^2 x^2}}\right )\\ &=\frac {2}{\sqrt {a x+\sqrt {b^2+a^2 x^2}}}-2 \operatorname {Subst}\left (\int \frac {1}{b-x^2} \, dx,x,\sqrt {a x+\sqrt {b^2+a^2 x^2}}\right )+2 \operatorname {Subst}\left (\int \frac {1}{b+x^2} \, dx,x,\sqrt {a x+\sqrt {b^2+a^2 x^2}}\right )\\ &=\frac {2}{\sqrt {a x+\sqrt {b^2+a^2 x^2}}}+\frac {2 \tan ^{-1}\left (\frac {\sqrt {a x+\sqrt {b^2+a^2 x^2}}}{\sqrt {b}}\right )}{\sqrt {b}}-\frac {2 \tanh ^{-1}\left (\frac {\sqrt {a x+\sqrt {b^2+a^2 x^2}}}{\sqrt {b}}\right )}{\sqrt {b}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C]  time = 2.19, size = 204, normalized size = 2.04 \begin {gather*} \frac {2}{3} \sqrt {\sqrt {a^2 x^2+b^2}+a x} \left (\frac {\sqrt {a^2 x^2+b^2}-2 a x}{b^2}-\frac {\left (a^2 x^2+b^2\right ) \left (\sqrt {a^2 x^2+b^2}+a x\right ) \left (2 \left (2 a x \left (\sqrt {a^2 x^2+b^2}+a x\right )+b^2\right ) \, _2F_1\left (\frac {3}{4},1;\frac {7}{4};\frac {\left (a x+\sqrt {b^2+a^2 x^2}\right )^2}{b^2}\right )-a x \left (\sqrt {a^2 x^2+b^2}+a x\right )-2 b^2\right )}{\left (a b x \left (\sqrt {a^2 x^2+b^2}+a x\right )+b^3\right )^2}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(x*Sqrt[a*x + Sqrt[b^2 + a^2*x^2]]),x]

[Out]

(2*Sqrt[a*x + Sqrt[b^2 + a^2*x^2]]*((-2*a*x + Sqrt[b^2 + a^2*x^2])/b^2 - ((b^2 + a^2*x^2)*(a*x + Sqrt[b^2 + a^
2*x^2])*(-2*b^2 - a*x*(a*x + Sqrt[b^2 + a^2*x^2]) + 2*(b^2 + 2*a*x*(a*x + Sqrt[b^2 + a^2*x^2]))*Hypergeometric
2F1[3/4, 1, 7/4, (a*x + Sqrt[b^2 + a^2*x^2])^2/b^2]))/(b^3 + a*b*x*(a*x + Sqrt[b^2 + a^2*x^2]))^2))/3

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 0.12, size = 100, normalized size = 1.00 \begin {gather*} \frac {2}{\sqrt {a x+\sqrt {b^2+a^2 x^2}}}+\frac {2 \tan ^{-1}\left (\frac {\sqrt {a x+\sqrt {b^2+a^2 x^2}}}{\sqrt {b}}\right )}{\sqrt {b}}-\frac {2 \tanh ^{-1}\left (\frac {\sqrt {a x+\sqrt {b^2+a^2 x^2}}}{\sqrt {b}}\right )}{\sqrt {b}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[1/(x*Sqrt[a*x + Sqrt[b^2 + a^2*x^2]]),x]

[Out]

2/Sqrt[a*x + Sqrt[b^2 + a^2*x^2]] + (2*ArcTan[Sqrt[a*x + Sqrt[b^2 + a^2*x^2]]/Sqrt[b]])/Sqrt[b] - (2*ArcTanh[S
qrt[a*x + Sqrt[b^2 + a^2*x^2]]/Sqrt[b]])/Sqrt[b]

________________________________________________________________________________________

fricas [A]  time = 0.67, size = 321, normalized size = 3.21 \begin {gather*} \left [\frac {2 \, b^{\frac {3}{2}} \arctan \left (\frac {\sqrt {a x + \sqrt {a^{2} x^{2} + b^{2}}}}{\sqrt {b}}\right ) + b^{\frac {3}{2}} \log \left (\frac {b^{2} + \sqrt {a x + \sqrt {a^{2} x^{2} + b^{2}}} {\left ({\left (a x - b\right )} \sqrt {b} - \sqrt {a^{2} x^{2} + b^{2}} \sqrt {b}\right )} + \sqrt {a^{2} x^{2} + b^{2}} b}{x}\right ) - 2 \, \sqrt {a x + \sqrt {a^{2} x^{2} + b^{2}}} {\left (a x - \sqrt {a^{2} x^{2} + b^{2}}\right )}}{b^{2}}, \frac {2 \, \sqrt {-b} b \arctan \left (\frac {\sqrt {a x + \sqrt {a^{2} x^{2} + b^{2}}} \sqrt {-b}}{b}\right ) - \sqrt {-b} b \log \left (-\frac {b^{2} + \sqrt {a x + \sqrt {a^{2} x^{2} + b^{2}}} {\left ({\left (a x + b\right )} \sqrt {-b} - \sqrt {a^{2} x^{2} + b^{2}} \sqrt {-b}\right )} - \sqrt {a^{2} x^{2} + b^{2}} b}{x}\right ) - 2 \, \sqrt {a x + \sqrt {a^{2} x^{2} + b^{2}}} {\left (a x - \sqrt {a^{2} x^{2} + b^{2}}\right )}}{b^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[(2*b^(3/2)*arctan(sqrt(a*x + sqrt(a^2*x^2 + b^2))/sqrt(b)) + b^(3/2)*log((b^2 + sqrt(a*x + sqrt(a^2*x^2 + b^2
))*((a*x - b)*sqrt(b) - sqrt(a^2*x^2 + b^2)*sqrt(b)) + sqrt(a^2*x^2 + b^2)*b)/x) - 2*sqrt(a*x + sqrt(a^2*x^2 +
b^2))*(a*x - sqrt(a^2*x^2 + b^2)))/b^2, (2*sqrt(-b)*b*arctan(sqrt(a*x + sqrt(a^2*x^2 + b^2))*sqrt(-b)/b) - sq
rt(-b)*b*log(-(b^2 + sqrt(a*x + sqrt(a^2*x^2 + b^2))*((a*x + b)*sqrt(-b) - sqrt(a^2*x^2 + b^2)*sqrt(-b)) - sqr
t(a^2*x^2 + b^2)*b)/x) - 2*sqrt(a*x + sqrt(a^2*x^2 + b^2))*(a*x - sqrt(a^2*x^2 + b^2)))/b^2]

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\sqrt {a x + \sqrt {a^{2} x^{2} + b^{2}}} x}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [F]  time = 180.00, size = 0, normalized size = 0.00 \[\int \frac {1}{x \sqrt {a x +\sqrt {a^{2} x^{2}+b^{2}}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\sqrt {a x + \sqrt {a^{2} x^{2} + b^{2}}} x}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{x\,\sqrt {a\,x+\sqrt {a^2\,x^2+b^2}}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________