\(\int \frac {1}{x \sqrt {a+b x^2} \sqrt {c+d x^2}} \, dx\) [972]

Optimal result

Integrand size = 26, antiderivative size = 46 \[ \int \frac {1}{x \sqrt {a+b x^2} \sqrt {c+d x^2}} \, dx=-\frac {\text {arctanh}\left (\frac {\sqrt {c} \sqrt {a+b x^2}}{\sqrt {a} \sqrt {c+d x^2}}\right )}{\sqrt {a} \sqrt {c}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {457, 95, 214} \[ \int \frac {1}{x \sqrt {a+b x^2} \sqrt {c+d x^2}} \, dx=-\frac {\text {arctanh}\left (\frac {\sqrt {c} \sqrt {a+b x^2}}{\sqrt {a} \sqrt {c+d x^2}}\right )}{\sqrt {a} \sqrt {c}} \]

[In]

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

[Out]

-(ArcTanh[(Sqrt[c]*Sqrt[a + b*x^2])/(Sqrt[a]*Sqrt[c + d*x^2])]/(Sqrt[a]*Sqrt[c]))

Rule 95

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> With[{q = Denomin
ator[m]}, Dist[q, Subst[Int[x^(q*(m + 1) - 1)/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^
(1/q)], x]] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && LtQ[-1, m, 0] && SimplerQ[
a + b*x, c + d*x]

Rule 214

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

Rule 457

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \text {Subst}\left (\int \frac {1}{x \sqrt {a+b x} \sqrt {c+d x}} \, dx,x,x^2\right ) \\ & = \text {Subst}\left (\int \frac {1}{-a+c x^2} \, dx,x,\frac {\sqrt {a+b x^2}}{\sqrt {c+d x^2}}\right ) \\ & = -\frac {\tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x^2}}{\sqrt {a} \sqrt {c+d x^2}}\right )}{\sqrt {a} \sqrt {c}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.48 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.00 \[ \int \frac {1}{x \sqrt {a+b x^2} \sqrt {c+d x^2}} \, dx=-\frac {\text {arctanh}\left (\frac {\sqrt {c} \sqrt {a+b x^2}}{\sqrt {a} \sqrt {c+d x^2}}\right )}{\sqrt {a} \sqrt {c}} \]

[In]

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

[Out]

-(ArcTanh[(Sqrt[c]*Sqrt[a + b*x^2])/(Sqrt[a]*Sqrt[c + d*x^2])]/(Sqrt[a]*Sqrt[c]))

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(88\) vs. \(2(34)=68\).

Time = 3.21 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.93

method	result	size
default	\(-\frac {\ln \left (\frac {a d \,x^{2}+c b \,x^{2}+2 \sqrt {a c}\, \sqrt {\left (b \,x^{2}+a \right ) \left (d \,x^{2}+c \right )}+2 a c}{x^{2}}\right ) \sqrt {d \,x^{2}+c}\, \sqrt {b \,x^{2}+a}}{2 \sqrt {a c}\, \sqrt {\left (b \,x^{2}+a \right ) \left (d \,x^{2}+c \right )}}\)	\(89\)
elliptic	\(-\frac {\sqrt {\left (b \,x^{2}+a \right ) \left (d \,x^{2}+c \right )}\, \ln \left (\frac {2 a c +\left (a d +b c \right ) x^{2}+2 \sqrt {a c}\, \sqrt {b d \,x^{4}+\left (a d +b c \right ) x^{2}+a c}}{x^{2}}\right )}{2 \sqrt {b \,x^{2}+a}\, \sqrt {d \,x^{2}+c}\, \sqrt {a c}}\)	\(94\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 92 vs. \(2 (34) = 68\).

Time = 0.26 (sec) , antiderivative size = 204, normalized size of antiderivative = 4.43 \[ \int \frac {1}{x \sqrt {a+b x^2} \sqrt {c+d x^2}} \, dx=\left [\frac {\sqrt {a c} \log \left (\frac {{\left (b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2}\right )} x^{4} + 8 \, a^{2} c^{2} + 8 \, {\left (a b c^{2} + a^{2} c d\right )} x^{2} - 4 \, {\left ({\left (b c + a d\right )} x^{2} + 2 \, a c\right )} \sqrt {b x^{2} + a} \sqrt {d x^{2} + c} \sqrt {a c}}{x^{4}}\right )}{4 \, a c}, \frac {\sqrt {-a c} \arctan \left (\frac {{\left ({\left (b c + a d\right )} x^{2} + 2 \, a c\right )} \sqrt {b x^{2} + a} \sqrt {d x^{2} + c} \sqrt {-a c}}{2 \, {\left (a b c d x^{4} + a^{2} c^{2} + {\left (a b c^{2} + a^{2} c d\right )} x^{2}\right )}}\right )}{2 \, a c}\right ] \]

[In]

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

[Out]

[1/4*sqrt(a*c)*log(((b^2*c^2 + 6*a*b*c*d + a^2*d^2)*x^4 + 8*a^2*c^2 + 8*(a*b*c^2 + a^2*c*d)*x^2 - 4*((b*c + a*
d)*x^2 + 2*a*c)*sqrt(b*x^2 + a)*sqrt(d*x^2 + c)*sqrt(a*c))/x^4)/(a*c), 1/2*sqrt(-a*c)*arctan(1/2*((b*c + a*d)*
x^2 + 2*a*c)*sqrt(b*x^2 + a)*sqrt(d*x^2 + c)*sqrt(-a*c)/(a*b*c*d*x^4 + a^2*c^2 + (a*b*c^2 + a^2*c*d)*x^2))/(a*
c)]

Sympy [F]

\[ \int \frac {1}{x \sqrt {a+b x^2} \sqrt {c+d x^2}} \, dx=\int \frac {1}{x \sqrt {a + b x^{2}} \sqrt {c + d x^{2}}}\, dx \]

[In]

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

[Out]

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

Maxima [F(-2)]

Exception generated. \[ \int \frac {1}{x \sqrt {a+b x^2} \sqrt {c+d x^2}} \, dx=\text {Exception raised: ValueError} \]

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(a*d-b*c>0)', see `assume?` for
more detail

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 89 vs. \(2 (34) = 68\).

Time = 0.32 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.93 \[ \int \frac {1}{x \sqrt {a+b x^2} \sqrt {c+d x^2}} \, dx=-\frac {\sqrt {b d} b \arctan \left (-\frac {b^{2} c + a b d - {\left (\sqrt {b x^{2} + a} \sqrt {b d} - \sqrt {b^{2} c + {\left (b x^{2} + a\right )} b d - a b d}\right )}^{2}}{2 \, \sqrt {-a b c d} b}\right )}{\sqrt {-a b c d} {\left | b \right |}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 8.29 (sec) , antiderivative size = 136, normalized size of antiderivative = 2.96 \[ \int \frac {1}{x \sqrt {a+b x^2} \sqrt {c+d x^2}} \, dx=-\frac {\ln \left (\frac {\sqrt {b\,x^2+a}-\sqrt {a}}{\sqrt {d\,x^2+c}-\sqrt {c}}\right )-\ln \left (\frac {\left (\sqrt {c}\,\sqrt {b\,x^2+a}-\sqrt {a}\,\sqrt {d\,x^2+c}\right )\,\left (b\,\sqrt {c}-\frac {\sqrt {a}\,d\,\left (\sqrt {b\,x^2+a}-\sqrt {a}\right )}{\sqrt {d\,x^2+c}-\sqrt {c}}\right )}{\sqrt {d\,x^2+c}-\sqrt {c}}\right )}{2\,\sqrt {a}\,\sqrt {c}} \]

[In]

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

[Out]

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