\(\int \frac {1}{\sqrt {a+b x} \sqrt {c+d x}} \, dx\) [736]

Optimal result

Integrand size = 19, antiderivative size = 42 \[ \int \frac {1}{\sqrt {a+b x} \sqrt {c+d x}} \, dx=\frac {2 \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{\sqrt {b} \sqrt {d}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {65, 223, 212} \[ \int \frac {1}{\sqrt {a+b x} \sqrt {c+d x}} \, dx=\frac {2 \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{\sqrt {b} \sqrt {d}} \]

[In]

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

[Out]

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

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

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 223

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

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.00 \[ \int \frac {1}{\sqrt {a+b x} \sqrt {c+d x}} \, dx=\frac {2 \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {a+b x}}\right )}{\sqrt {b} \sqrt {d}} \]

[In]

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

[Out]

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(75\) vs. \(2(30)=60\).

Time = 0.58 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.81

method	result	size
default	\(\frac {\sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \ln \left (\frac {\frac {1}{2} a d +\frac {1}{2} b c +b d x}{\sqrt {b d}}+\sqrt {b d \,x^{2}+\left (a d +b c \right ) x +a c}\right )}{\sqrt {b x +a}\, \sqrt {d x +c}\, \sqrt {b d}}\)	\(76\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 82 vs. \(2 (30) = 60\).

Time = 0.24 (sec) , antiderivative size = 178, normalized size of antiderivative = 4.24 \[ \int \frac {1}{\sqrt {a+b x} \sqrt {c+d x}} \, dx=\left [\frac {\sqrt {b d} \log \left (8 \, b^{2} d^{2} x^{2} + b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2} + 4 \, {\left (2 \, b d x + b c + a d\right )} \sqrt {b d} \sqrt {b x + a} \sqrt {d x + c} + 8 \, {\left (b^{2} c d + a b d^{2}\right )} x\right )}{2 \, b d}, -\frac {\sqrt {-b d} \arctan \left (\frac {{\left (2 \, b d x + b c + a d\right )} \sqrt {-b d} \sqrt {b x + a} \sqrt {d x + c}}{2 \, {\left (b^{2} d^{2} x^{2} + a b c d + {\left (b^{2} c d + a b d^{2}\right )} x\right )}}\right )}{b d}\right ] \]

[In]

integrate(1/(b*x+a)^(1/2)/(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*(2*b*d*x + b*c + a*d)*sqrt(b*d)*sqrt(b*x
+ a)*sqrt(d*x + c) + 8*(b^2*c*d + a*b*d^2)*x)/(b*d), -sqrt(-b*d)*arctan(1/2*(2*b*d*x + b*c + a*d)*sqrt(-b*d)*s
qrt(b*x + a)*sqrt(d*x + c)/(b^2*d^2*x^2 + a*b*c*d + (b^2*c*d + a*b*d^2)*x))/(b*d)]

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F(-2)]

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

[In]

integrate(1/(b*x+a)^(1/2)/(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

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.19 \[ \int \frac {1}{\sqrt {a+b x} \sqrt {c+d x}} \, dx=-\frac {2 \, b \log \left ({\left | -\sqrt {b d} \sqrt {b x + a} + \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d} \right |}\right )}{\sqrt {b d} {\left | b \right |}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 1.02 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.07 \[ \int \frac {1}{\sqrt {a+b x} \sqrt {c+d x}} \, dx=-\frac {4\,\mathrm {atan}\left (\frac {b\,\left (\sqrt {c+d\,x}-\sqrt {c}\right )}{\sqrt {-b\,d}\,\left (\sqrt {a+b\,x}-\sqrt {a}\right )}\right )}{\sqrt {-b\,d}} \]

[In]

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

[Out]

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