\(\int \frac {1}{\sqrt {a+b x^2} (c+d x^2)} \, dx\) [78]

Optimal result

Integrand size = 21, antiderivative size = 49 \[ \int \frac {1}{\sqrt {a+b x^2} \left (c+d x^2\right )} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {b c-a d} x}{\sqrt {c} \sqrt {a+b x^2}}\right )}{\sqrt {c} \sqrt {b c-a d}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {385, 214} \[ \int \frac {1}{\sqrt {a+b x^2} \left (c+d x^2\right )} \, dx=\frac {\text {arctanh}\left (\frac {x \sqrt {b c-a d}}{\sqrt {c} \sqrt {a+b x^2}}\right )}{\sqrt {c} \sqrt {b c-a d}} \]

[In]

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

[Out]

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

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 385

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

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

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.37 \[ \int \frac {1}{\sqrt {a+b x^2} \left (c+d x^2\right )} \, dx=-\frac {\arctan \left (\frac {-d x \sqrt {a+b x^2}+\sqrt {b} \left (c+d x^2\right )}{\sqrt {c} \sqrt {-b c+a d}}\right )}{\sqrt {c} \sqrt {-b c+a d}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 2.33 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.86

method	result	size
pseudoelliptic	\(-\frac {\arctan \left (\frac {c \sqrt {b \,x^{2}+a}}{x \sqrt {\left (a d -b c \right ) c}}\right )}{\sqrt {\left (a d -b c \right ) c}}\)	\(42\)
default	\(\frac {\ln \left (\frac {\frac {2 a d -2 b c}{d}-\frac {2 b \sqrt {-c d}\, \left (x +\frac {\sqrt {-c d}}{d}\right )}{d}+2 \sqrt {\frac {a d -b c}{d}}\, \sqrt {\left (x +\frac {\sqrt {-c d}}{d}\right )^{2} b -\frac {2 b \sqrt {-c d}\, \left (x +\frac {\sqrt {-c d}}{d}\right )}{d}+\frac {a d -b c}{d}}}{x +\frac {\sqrt {-c d}}{d}}\right )}{2 \sqrt {-c d}\, \sqrt {\frac {a d -b c}{d}}}-\frac {\ln \left (\frac {\frac {2 a d -2 b c}{d}+\frac {2 b \sqrt {-c d}\, \left (x -\frac {\sqrt {-c d}}{d}\right )}{d}+2 \sqrt {\frac {a d -b c}{d}}\, \sqrt {\left (x -\frac {\sqrt {-c d}}{d}\right )^{2} b +\frac {2 b \sqrt {-c d}\, \left (x -\frac {\sqrt {-c d}}{d}\right )}{d}+\frac {a d -b c}{d}}}{x -\frac {\sqrt {-c d}}{d}}\right )}{2 \sqrt {-c d}\, \sqrt {\frac {a d -b c}{d}}}\)	\(300\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 107 vs. \(2 (39) = 78\).

Time = 0.30 (sec) , antiderivative size = 241, normalized size of antiderivative = 4.92 \[ \int \frac {1}{\sqrt {a+b x^2} \left (c+d x^2\right )} \, dx=\left [\frac {\log \left (\frac {{\left (8 \, b^{2} c^{2} - 8 \, a b c d + a^{2} d^{2}\right )} x^{4} + a^{2} c^{2} + 2 \, {\left (4 \, a b c^{2} - 3 \, a^{2} c d\right )} x^{2} + 4 \, {\left ({\left (2 \, b c - a d\right )} x^{3} + a c x\right )} \sqrt {b c^{2} - a c d} \sqrt {b x^{2} + a}}{d^{2} x^{4} + 2 \, c d x^{2} + c^{2}}\right )}{4 \, \sqrt {b c^{2} - a c d}}, -\frac {\sqrt {-b c^{2} + a c d} \arctan \left (\frac {\sqrt {-b c^{2} + a c d} {\left ({\left (2 \, b c - a d\right )} x^{2} + a c\right )} \sqrt {b x^{2} + a}}{2 \, {\left ({\left (b^{2} c^{2} - a b c d\right )} x^{3} + {\left (a b c^{2} - a^{2} c d\right )} x\right )}}\right )}{2 \, {\left (b c^{2} - a c d\right )}}\right ] \]

[In]

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

[Out]

[1/4*log(((8*b^2*c^2 - 8*a*b*c*d + a^2*d^2)*x^4 + a^2*c^2 + 2*(4*a*b*c^2 - 3*a^2*c*d)*x^2 + 4*((2*b*c - a*d)*x
^3 + a*c*x)*sqrt(b*c^2 - a*c*d)*sqrt(b*x^2 + a))/(d^2*x^4 + 2*c*d*x^2 + c^2))/sqrt(b*c^2 - a*c*d), -1/2*sqrt(-
b*c^2 + a*c*d)*arctan(1/2*sqrt(-b*c^2 + a*c*d)*((2*b*c - a*d)*x^2 + a*c)*sqrt(b*x^2 + a)/((b^2*c^2 - a*b*c*d)*
x^3 + (a*b*c^2 - a^2*c*d)*x))/(b*c^2 - a*c*d)]

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.43 \[ \int \frac {1}{\sqrt {a+b x^2} \left (c+d x^2\right )} \, dx=-\frac {\sqrt {b} \arctan \left (\frac {{\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} d + 2 \, b c - a d}{2 \, \sqrt {-b^{2} c^{2} + a b c d}}\right )}{\sqrt {-b^{2} c^{2} + a b c d}} \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\sqrt {a+b x^2} \left (c+d x^2\right )} \, dx=\left \{\begin {array}{cl} \frac {\mathrm {atan}\left (\frac {x\,\sqrt {a\,d-b\,c}}{\sqrt {c}\,\sqrt {b\,x^2+a}}\right )}{\sqrt {c\,\left (a\,d-b\,c\right )}} & \text {\ if\ \ }0<a\,d-b\,c\\ \frac {\ln \left (\frac {\sqrt {c\,\left (b\,x^2+a\right )}+x\,\sqrt {b\,c-a\,d}}{\sqrt {c\,\left (b\,x^2+a\right )}-x\,\sqrt {b\,c-a\,d}}\right )}{2\,\sqrt {-c\,\left (a\,d-b\,c\right )}} & \text {\ if\ \ }a\,d-b\,c<0\\ \int \frac {1}{\sqrt {b\,x^2+a}\,\left (d\,x^2+c\right )} \,d x & \text {\ if\ \ }a\,d-b\,c\notin \mathbb {R}\vee a\,d=b\,c \end {array}\right . \]

[In]

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

[Out]

piecewise(0 < a*d - b*c, atan((x*(a*d - b*c)^(1/2))/(c^(1/2)*(a + b*x^2)^(1/2)))/(c*(a*d - b*c))^(1/2), a*d -
b*c < 0, log(((c*(a + b*x^2))^(1/2) + x*(- a*d + b*c)^(1/2))/((c*(a + b*x^2))^(1/2) - x*(- a*d + b*c)^(1/2)))/
(2*(-c*(a*d - b*c))^(1/2)), ~in(a*d - b*c, 'real') | a*d == b*c, int(1/((a + b*x^2)^(1/2)*(c + d*x^2)), x))