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

Optimal. Leaf size=50 \[ -\frac {\sqrt {b} x \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{a^{3/2} \sqrt {d x^2}}-\frac {1}{a \sqrt {d x^2}} \]

[Out]

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

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Rubi [A]  time = 0.02, antiderivative size = 50, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {15, 325, 205} \[ -\frac {\sqrt {b} x \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{a^{3/2} \sqrt {d x^2}}-\frac {1}{a \sqrt {d x^2}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x*Sqrt[d*x^2]*(a + b*x^2)),x]

[Out]

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

Rule 15

Int[(u_.)*((a_.)*(x_)^(n_))^(m_), x_Symbol] :> Dist[(a^IntPart[m]*(a*x^n)^FracPart[m])/x^(n*FracPart[m]), Int[
u*x^(m*n), x], x] /; FreeQ[{a, m, n}, x] &&  !IntegerQ[m]

Rule 205

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

Rule 325

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*
c*(m + 1)), x] - Dist[(b*(m + n*(p + 1) + 1))/(a*c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rubi steps

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

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Mathematica [A]  time = 0.02, size = 46, normalized size = 0.92 \[ -\frac {d x^2 \left (\sqrt {b} x \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )+\sqrt {a}\right )}{a^{3/2} \left (d x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.76, size = 132, normalized size = 2.64 \[ \left [\frac {d x^{2} \sqrt {-\frac {b}{a d}} \log \left (\frac {b x^{2} - 2 \, \sqrt {d x^{2}} a \sqrt {-\frac {b}{a d}} - a}{b x^{2} + a}\right ) - 2 \, \sqrt {d x^{2}}}{2 \, a d x^{2}}, -\frac {d x^{2} \sqrt {\frac {b}{a d}} \arctan \left (\sqrt {d x^{2}} \sqrt {\frac {b}{a d}}\right ) + \sqrt {d x^{2}}}{a d x^{2}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.28, size = 41, normalized size = 0.82 \[ -\frac {b \arctan \left (\frac {\sqrt {d x^{2}} b}{\sqrt {a b d}}\right )}{\sqrt {a b d} a} - \frac {1}{\sqrt {d x^{2}} a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-b*arctan(sqrt(d*x^2)*b/sqrt(a*b*d))/(sqrt(a*b*d)*a) - 1/(sqrt(d*x^2)*a)

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maple [A]  time = 0.01, size = 36, normalized size = 0.72 \[ -\frac {b x \arctan \left (\frac {b x}{\sqrt {a b}}\right )+\sqrt {a b}}{\sqrt {d \,x^{2}}\, \sqrt {a b}\, a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.99, size = 35, normalized size = 0.70 \[ -\frac {b \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{\sqrt {a b} a \sqrt {d}} - \frac {1}{a \sqrt {d} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-b*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*a*sqrt(d)) - 1/(a*sqrt(d)*x)

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mupad [B]  time = 0.62, size = 38, normalized size = 0.76 \[ -\frac {1}{a\,\sqrt {d}\,\sqrt {x^2}}-\frac {\sqrt {b}\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {x^2}}{\sqrt {a}}\right )}{a^{3/2}\,\sqrt {d}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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