\(\int \frac {1}{x \sqrt {d x^2} (a+b x^2)} \, dx\) [674]

Optimal result

Integrand size = 22, antiderivative size = 50 \[ \int \frac {1}{x \sqrt {d x^2} \left (a+b x^2\right )} \, dx=-\frac {1}{a \sqrt {d x^2}}-\frac {\sqrt {b} x \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{a^{3/2} \sqrt {d x^2}} \]

[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)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {15, 331, 211} \[ \int \frac {1}{x \sqrt {d x^2} \left (a+b x^2\right )} \, dx=-\frac {\sqrt {b} x \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{a^{3/2} \sqrt {d x^2}}-\frac {1}{a \sqrt {d x^2}} \]

[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 211

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

Rule 331

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*} \text {integral}& = \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*}

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.26 \[ \int \frac {1}{x \sqrt {d x^2} \left (a+b x^2\right )} \, dx=d \left (-\frac {x^2}{a \left (d x^2\right )^{3/2}}-\frac {\sqrt {b} \arctan \left (\frac {\sqrt {b} \sqrt {d x^2}}{\sqrt {a} \sqrt {d}}\right )}{a^{3/2} d^{3/2}}\right ) \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 2.88 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.72

method	result	size
default	\(-\frac {b \arctan \left (\frac {b x}{\sqrt {a b}}\right ) x +\sqrt {a b}}{\sqrt {d \,x^{2}}\, a \sqrt {a b}}\)	\(36\)
pseudoelliptic	\(-\frac {b \arctan \left (\frac {b \sqrt {d \,x^{2}}}{\sqrt {a b d}}\right ) \sqrt {d \,x^{2}}+\sqrt {a b d}}{a \sqrt {d \,x^{2}}\, \sqrt {a b d}}\)	\(51\)
risch	\(-\frac {1}{a \sqrt {d \,x^{2}}}+\frac {x \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (a^{3} \textit {\_Z}^{2}+b \right )}{\sum }\textit {\_R} \ln \left (\left (3 \textit {\_R}^{2} a^{3}+2 b \right ) x +a^{2} \textit {\_R} \right )\right )}{2 \sqrt {d \,x^{2}}}\)	\(60\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 132, normalized size of antiderivative = 2.64 \[ \int \frac {1}{x \sqrt {d x^2} \left (a+b x^2\right )} \, dx=\left [\frac {d x^{2} \sqrt {-\frac {b}{a d}} \log \left (\frac {b x^{2} - 2 \, \sqrt {d x^{2}} a \sqrt {-\frac {b}{a d}} - a}{b x^{2} + a}\right ) - 2 \, \sqrt {d x^{2}}}{2 \, a d x^{2}}, -\frac {d x^{2} \sqrt {\frac {b}{a d}} \arctan \left (\sqrt {d x^{2}} \sqrt {\frac {b}{a d}}\right ) + \sqrt {d x^{2}}}{a d x^{2}}\right ] \]

[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)]

Sympy [A] (verification not implemented)

Time = 1.29 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.98 \[ \int \frac {1}{x \sqrt {d x^2} \left (a+b x^2\right )} \, dx=\begin {cases} \frac {2 \left (- \frac {d \operatorname {atan}{\left (\frac {\sqrt {d x^{2}}}{\sqrt {\frac {a d}{b}}} \right )}}{2 a \sqrt {\frac {a d}{b}}} - \frac {d}{2 a \sqrt {d x^{2}}}\right )}{d} & \text {for}\: d \neq 0 \\\tilde {\infty } x^{2} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((2*(-d*atan(sqrt(d*x**2)/sqrt(a*d/b))/(2*a*sqrt(a*d/b)) - d/(2*a*sqrt(d*x**2)))/d, Ne(d, 0)), (zoo*x
**2, True))

Maxima [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.70 \[ \int \frac {1}{x \sqrt {d x^2} \left (a+b x^2\right )} \, dx=-\frac {b \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{\sqrt {a b} a \sqrt {d}} - \frac {1}{a \sqrt {d} x} \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.86 \[ \int \frac {1}{x \sqrt {d x^2} \left (a+b x^2\right )} \, dx=-\frac {b \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{\sqrt {a b} a \sqrt {d} \mathrm {sgn}\left (x\right )} - \frac {1}{a \sqrt {d} x \mathrm {sgn}\left (x\right )} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 5.34 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.76 \[ \int \frac {1}{x \sqrt {d x^2} \left (a+b x^2\right )} \, dx=-\frac {1}{a\,\sqrt {d}\,\sqrt {x^2}}-\frac {\sqrt {b}\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {x^2}}{\sqrt {a}}\right )}{a^{3/2}\,\sqrt {d}} \]

[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))