3.10.55 \(\int \frac {1}{\sqrt {(a+b x) (c+d x)}} \, dx\) [955]

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

[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 = 0.02, 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 = {1976, 635, 212} \begin {gather*} \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}} \end {gather*}

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 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 635

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 1976

Int[(u_.)*((e_.)*((a_.) + (b_.)*(x_)^(n_.))*((c_) + (d_.)*(x_)^(n_.)))^(p_), x_Symbol] :> Int[u*(a*c*e + (b*c
+ a*d)*e*x^n + b*d*e*x^(2*n))^p, x] /; FreeQ[{a, b, c, d, e, n, p}, 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 \text {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 = 0.02, size = 75, normalized size = 1.23 \begin {gather*} \frac {2 \sqrt {a+b x} \sqrt {c+d x} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {a+b x}}\right )}{\sqrt {b} \sqrt {d} \sqrt {(a+b x) (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.31, size = 49, normalized size = 0.80

method	result	size
default	\(\frac {\ln \left (\frac {\frac {1}{2} a d +\frac {1}{2} b c +b d x}{\sqrt {b d}}+\sqrt {a c +\left (a d +b c \right ) x +b d \,x^{2}}\right )}{\sqrt {b d}}\)	\(49\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

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 detail

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Fricas [A]
time = 0.37, size = 192, normalized size = 3.15 \begin {gather*} \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 ] \end {gather*}

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

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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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 123 vs. \(2 (49) = 98\).
time = 3.91, size = 123, normalized size = 2.02 \begin {gather*} \frac {1}{4} \, \sqrt {b d x^{2} + b c x + a d x + a c} {\left (2 \, x + \frac {b c + a d}{b d}\right )} + \frac {{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \log \left ({\left | -b c - a d - 2 \, \sqrt {b d} {\left (\sqrt {b d} x - \sqrt {b d x^{2} + b c x + a d x + a c}\right )} \right |}\right )}{8 \, \sqrt {b d} b d} \end {gather*}

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]

1/4*sqrt(b*d*x^2 + b*c*x + a*d*x + a*c)*(2*x + (b*c + a*d)/(b*d)) + 1/8*(b^2*c^2 - 2*a*b*c*d + a^2*d^2)*log(ab
s(-b*c - a*d - 2*sqrt(b*d)*(sqrt(b*d)*x - sqrt(b*d*x^2 + b*c*x + a*d*x + a*c))))/(sqrt(b*d)*b*d)

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

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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