\(\int \frac {1}{(a+b x) \sqrt {c+d x}} \, dx\) [1419]

Optimal result

Integrand size = 17, antiderivative size = 47 \[ \int \frac {1}{(a+b x) \sqrt {c+d x}} \, dx=-\frac {2 \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{\sqrt {b} \sqrt {b c-a d}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {65, 214} \[ \int \frac {1}{(a+b x) \sqrt {c+d x}} \, dx=-\frac {2 \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{\sqrt {b} \sqrt {b c-a d}} \]

[In]

Int[1/((a + b*x)*Sqrt[c + d*x]),x]

[Out]

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

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

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.00 \[ \int \frac {1}{(a+b x) \sqrt {c+d x}} \, dx=\frac {2 \arctan \left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {-b c+a d}}\right )}{\sqrt {b} \sqrt {-b c+a d}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.48 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.79

method	result	size
derivativedivides	\(\frac {2 \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{\sqrt {\left (a d -b c \right ) b}}\)	\(37\)
default	\(\frac {2 \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{\sqrt {\left (a d -b c \right ) b}}\)	\(37\)
pseudoelliptic	\(\frac {2 \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{\sqrt {\left (a d -b c \right ) b}}\)	\(37\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 119, normalized size of antiderivative = 2.53 \[ \int \frac {1}{(a+b x) \sqrt {c+d x}} \, dx=\left [\frac {\log \left (\frac {b d x + 2 \, b c - a d - 2 \, \sqrt {b^{2} c - a b d} \sqrt {d x + c}}{b x + a}\right )}{\sqrt {b^{2} c - a b d}}, \frac {2 \, \sqrt {-b^{2} c + a b d} \arctan \left (\frac {\sqrt {-b^{2} c + a b d} \sqrt {d x + c}}{b d x + b c}\right )}{b^{2} c - a b d}\right ] \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 1.99 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.13 \[ \int \frac {1}{(a+b x) \sqrt {c+d x}} \, dx=\begin {cases} \frac {2 \operatorname {atan}{\left (\frac {\sqrt {c + d x}}{\sqrt {\frac {a d - b c}{b}}} \right )}}{b \sqrt {\frac {a d - b c}{b}}} & \text {for}\: d \neq 0 \\\frac {\begin {cases} \frac {x}{a} & \text {for}\: b = 0 \\\frac {\log {\left (a + b x \right )}}{b} & \text {otherwise} \end {cases}}{\sqrt {c}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((2*atan(sqrt(c + d*x)/sqrt((a*d - b*c)/b))/(b*sqrt((a*d - b*c)/b)), Ne(d, 0)), (Piecewise((x/a, Eq(b
, 0)), (log(a + b*x)/b, True))/sqrt(c), True))

Maxima [F(-2)]

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

[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

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.81 \[ \int \frac {1}{(a+b x) \sqrt {c+d x}} \, dx=\frac {2 \, \arctan \left (\frac {\sqrt {d x + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{\sqrt {-b^{2} c + a b d}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.81 \[ \int \frac {1}{(a+b x) \sqrt {c+d x}} \, dx=\frac {2\,\mathrm {atan}\left (\frac {b\,\sqrt {c+d\,x}}{\sqrt {a\,b\,d-b^2\,c}}\right )}{\sqrt {a\,b\,d-b^2\,c}} \]

[In]

int(1/((a + b*x)*(c + d*x)^(1/2)),x)

[Out]

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