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

Optimal. Leaf size=103 \[ -\frac {1}{x \sqrt {\sqrt {a^2 x^2+b^2}+a x}}+\frac {a \tan ^{-1}\left (\frac {\sqrt {\sqrt {a^2 x^2+b^2}+a x}}{\sqrt {b}}\right )}{b^{3/2}}+\frac {a \tanh ^{-1}\left (\frac {\sqrt {\sqrt {a^2 x^2+b^2}+a x}}{\sqrt {b}}\right )}{b^{3/2}} \]

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Rubi [A]  time = 0.21, antiderivative size = 130, normalized size of antiderivative = 1.26, 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, 457, 329, 212, 206, 203} \begin {gather*} \frac {2 a \sqrt {\sqrt {a^2 x^2+b^2}+a x}}{b^2-\left (\sqrt {a^2 x^2+b^2}+a x\right )^2}+\frac {a \tan ^{-1}\left (\frac {\sqrt {\sqrt {a^2 x^2+b^2}+a x}}{\sqrt {b}}\right )}{b^{3/2}}+\frac {a \tanh ^{-1}\left (\frac {\sqrt {\sqrt {a^2 x^2+b^2}+a x}}{\sqrt {b}}\right )}{b^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 212

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-(a/b), 2]], s = Denominator[Rt[-(a/b), 2]
]}, Dist[r/(2*a), Int[1/(r - s*x^2), x], x] + Dist[r/(2*a), 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 457

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> -Simp[((b*c - a*d
)*(e*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*b*e*n*(p + 1)), x] - Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*b
*n*(p + 1)), Int[(e*x)^m*(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && NeQ[b*c - a*d, 0] &
& LtQ[p, -1] && (( !IntegerQ[p + 1/2] && NeQ[p, -5/4]) ||  !RationalQ[m] || (IGtQ[n, 0] && ILtQ[p + 1/2, 0] &&
LeQ[-1, m, -(n*(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^2 \sqrt {a x+\sqrt {b^2+a^2 x^2}}} \, dx &=(2 a) \operatorname {Subst}\left (\int \frac {b^2+x^2}{\sqrt {x} \left (-b^2+x^2\right )^2} \, dx,x,a x+\sqrt {b^2+a^2 x^2}\right )\\ &=\frac {2 a \sqrt {a x+\sqrt {b^2+a^2 x^2}}}{b^2-\left (a x+\sqrt {b^2+a^2 x^2}\right )^2}-a \operatorname {Subst}\left (\int \frac {1}{\sqrt {x} \left (-b^2+x^2\right )} \, dx,x,a x+\sqrt {b^2+a^2 x^2}\right )\\ &=\frac {2 a \sqrt {a x+\sqrt {b^2+a^2 x^2}}}{b^2-\left (a x+\sqrt {b^2+a^2 x^2}\right )^2}-(2 a) \operatorname {Subst}\left (\int \frac {1}{-b^2+x^4} \, dx,x,\sqrt {a x+\sqrt {b^2+a^2 x^2}}\right )\\ &=\frac {2 a \sqrt {a x+\sqrt {b^2+a^2 x^2}}}{b^2-\left (a x+\sqrt {b^2+a^2 x^2}\right )^2}+\frac {a \operatorname {Subst}\left (\int \frac {1}{b-x^2} \, dx,x,\sqrt {a x+\sqrt {b^2+a^2 x^2}}\right )}{b}+\frac {a \operatorname {Subst}\left (\int \frac {1}{b+x^2} \, dx,x,\sqrt {a x+\sqrt {b^2+a^2 x^2}}\right )}{b}\\ &=\frac {2 a \sqrt {a x+\sqrt {b^2+a^2 x^2}}}{b^2-\left (a x+\sqrt {b^2+a^2 x^2}\right )^2}+\frac {a \tan ^{-1}\left (\frac {\sqrt {a x+\sqrt {b^2+a^2 x^2}}}{\sqrt {b}}\right )}{b^{3/2}}+\frac {a \tanh ^{-1}\left (\frac {\sqrt {a x+\sqrt {b^2+a^2 x^2}}}{\sqrt {b}}\right )}{b^{3/2}}\\ \end {align*}

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Mathematica [C]  time = 23.02, size = 9150, normalized size = 88.83 \begin {gather*} \text {Result too large to show} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

Result too large to show

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IntegrateAlgebraic [A]  time = 0.18, size = 103, normalized size = 1.00 \begin {gather*} -\frac {1}{x \sqrt {a x+\sqrt {b^2+a^2 x^2}}}+\frac {a \tan ^{-1}\left (\frac {\sqrt {a x+\sqrt {b^2+a^2 x^2}}}{\sqrt {b}}\right )}{b^{3/2}}+\frac {a \tanh ^{-1}\left (\frac {\sqrt {a x+\sqrt {b^2+a^2 x^2}}}{\sqrt {b}}\right )}{b^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 1.03, size = 335, normalized size = 3.25 \begin {gather*} \left [\frac {2 \, a \sqrt {b} x \arctan \left (\frac {\sqrt {a x + \sqrt {a^{2} x^{2} + b^{2}}}}{\sqrt {b}}\right ) + a \sqrt {b} x \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 )}}{2 \, b^{2} x}, -\frac {2 \, a \sqrt {-b} x \arctan \left (\frac {\sqrt {a x + \sqrt {a^{2} x^{2} + b^{2}}} \sqrt {-b}}{b}\right ) + a \sqrt {-b} x \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 )}}{2 \, b^{2} x}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*(2*a*sqrt(b)*x*arctan(sqrt(a*x + sqrt(a^2*x^2 + b^2))/sqrt(b)) + a*sqrt(b)*x*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 + sq
rt(a^2*x^2 + b^2))*(a*x - sqrt(a^2*x^2 + b^2)))/(b^2*x), -1/2*(2*a*sqrt(-b)*x*arctan(sqrt(a*x + sqrt(a^2*x^2 +
b^2))*sqrt(-b)/b) + a*sqrt(-b)*x*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*x)]

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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^{2}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^2/(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^2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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^{2}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^2/(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^2), x)

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

[Out]

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

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

[Out]

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

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