3.955 \(\int \frac {1}{\sqrt {(a+b x) (c+d x)}} \, dx\)

Optimal. Leaf size=61 \[ \frac {\tanh ^{-1}\left (\frac {a d+b c+2 b d x}{2 \sqrt {b} \sqrt {d} \sqrt {x (a d+b c)+a c+b d x^2}}\right )}{\sqrt {b} \sqrt {d}} \]

[Out]

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

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Rubi [A]  time = 0.03, antiderivative size = 61, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {1981, 621, 206} \[ \frac {\tanh ^{-1}\left (\frac {a d+b c+2 b d x}{2 \sqrt {b} \sqrt {d} \sqrt {x (a d+b c)+a c+b d x^2}}\right )}{\sqrt {b} \sqrt {d}} \]

Antiderivative was successfully verified.

[In]

Int[1/Sqrt[(a + b*x)*(c + d*x)],x]

[Out]

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

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 621

Int[1/Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[2, Subst[Int[1/(4*c - x^2), x], x, (b + 2*c*x)
/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 1981

Int[(u_)^(p_), x_Symbol] :> Int[ExpandToSum[u, x]^p, x] /; FreeQ[p, x] && QuadraticQ[u, x] &&  !QuadraticMatch
Q[u, x]

Rubi steps

\begin {align*} \int \frac {1}{\sqrt {(a+b x) (c+d x)}} \, dx &=\int \frac {1}{\sqrt {a c+(b c+a d) x+b d x^2}} \, dx\\ &=2 \operatorname {Subst}\left (\int \frac {1}{4 b d-x^2} \, dx,x,\frac {b c+a d+2 b d x}{\sqrt {a c+(b c+a d) x+b d x^2}}\right )\\ &=\frac {\tanh ^{-1}\left (\frac {b c+a d+2 b d x}{2 \sqrt {b} \sqrt {d} \sqrt {a c+(b c+a d) x+b d x^2}}\right )}{\sqrt {b} \sqrt {d}}\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 95, normalized size = 1.56 \[ \frac {2 \sqrt {a+b x} \sqrt {b c-a d} \sqrt {\frac {b (c+d x)}{b c-a d}} \sinh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b c-a d}}\right )}{b \sqrt {d} \sqrt {(a+b x) (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.71, size = 192, normalized size = 3.15 \[ \left [\frac {\sqrt {b d} \log \left (8 \, b^{2} d^{2} x^{2} + b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2} + 4 \, \sqrt {b d x^{2} + a c + {\left (b c + a d\right )} x} {\left (2 \, b d x + b c + a d\right )} \sqrt {b d} + 8 \, {\left (b^{2} c d + a b d^{2}\right )} x\right )}{2 \, b d}, -\frac {\sqrt {-b d} \arctan \left (\frac {\sqrt {b d x^{2} + a c + {\left (b c + a d\right )} x} {\left (2 \, b d x + b c + a d\right )} \sqrt {-b d}}{2 \, {\left (b^{2} d^{2} x^{2} + a b c d + {\left (b^{2} c d + a b d^{2}\right )} x\right )}}\right )}{b d}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.72, size = 68, normalized size = 1.11 \[ -\frac {\sqrt {b d} \log \left ({\left | -2 \, {\left (\sqrt {b d} x - \sqrt {b d x^{2} + b c x + a d x + a c}\right )} b d - \sqrt {b d} b c - \sqrt {b d} a d \right |}\right )}{b d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.01, size = 49, normalized size = 0.80 \[ \frac {\ln \left (\frac {b d x +\frac {1}{2} a d +\frac {1}{2} b c}{\sqrt {b d}}+\sqrt {b d \,x^{2}+a c +\left (a d +b c \right ) x}\right )}{\sqrt {b d}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/((b*x+a)*(d*x+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 details)Is a*d-b*c zero or nonzero?

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {1}{\sqrt {\left (a+b\,x\right )\,\left (c+d\,x\right )}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((a + b*x)*(c + d*x))^(1/2),x)

[Out]

int(1/((a + b*x)*(c + d*x))^(1/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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