\(\int \frac {\sqrt {e x}}{(a+b x) (a c-b c x)} \, dx\) [52]

Optimal result

Integrand size = 26, antiderivative size = 88 \[ \int \frac {\sqrt {e x}}{(a+b x) (a c-b c x)} \, dx=-\frac {\sqrt {e} \arctan \left (\frac {\sqrt {b} \sqrt {e x}}{\sqrt {a} \sqrt {e}}\right )}{\sqrt {a} b^{3/2} c}+\frac {\sqrt {e} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {e x}}{\sqrt {a} \sqrt {e}}\right )}{\sqrt {a} b^{3/2} c} \]

[Out]

-arctan(b^(1/2)*(e*x)^(1/2)/a^(1/2)/e^(1/2))*e^(1/2)/b^(3/2)/c/a^(1/2)+arctanh(b^(1/2)*(e*x)^(1/2)/a^(1/2)/e^(
1/2))*e^(1/2)/b^(3/2)/c/a^(1/2)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {74, 335, 304, 211, 214} \[ \int \frac {\sqrt {e x}}{(a+b x) (a c-b c x)} \, dx=\frac {\sqrt {e} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {e x}}{\sqrt {a} \sqrt {e}}\right )}{\sqrt {a} b^{3/2} c}-\frac {\sqrt {e} \arctan \left (\frac {\sqrt {b} \sqrt {e x}}{\sqrt {a} \sqrt {e}}\right )}{\sqrt {a} b^{3/2} c} \]

[In]

Int[Sqrt[e*x]/((a + b*x)*(a*c - b*c*x)),x]

[Out]

-((Sqrt[e]*ArcTan[(Sqrt[b]*Sqrt[e*x])/(Sqrt[a]*Sqrt[e])])/(Sqrt[a]*b^(3/2)*c)) + (Sqrt[e]*ArcTanh[(Sqrt[b]*Sqr
t[e*x])/(Sqrt[a]*Sqrt[e])])/(Sqrt[a]*b^(3/2)*c)

Rule 74

Int[((a_) + (b_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[(a*c + b*
d*x^2)^m*(e + f*x)^p, x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && EqQ[b*c + a*d, 0] && EqQ[n, m] && Integer
Q[m] && (NeQ[m, -1] || (EqQ[e, 0] && (EqQ[p, 1] ||  !IntegerQ[p])))

Rule 211

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/Rt[a/b, 2]], x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

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 304

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2]], s = Denominator[Rt[-a/b, 2]]}
, Dist[s/(2*b), Int[1/(r + s*x^2), x], x] - Dist[s/(2*b), Int[1/(r - s*x^2), x], x]] /; FreeQ[{a, b}, x] &&  !
GtQ[a/b, 0]

Rule 335

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + b*(x^(k*n)/c^n))^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rubi steps \begin{align*} \text {integral}& = \int \frac {\sqrt {e x}}{a^2 c-b^2 c x^2} \, dx \\ & = \frac {2 \text {Subst}\left (\int \frac {x^2}{a^2 c-\frac {b^2 c x^4}{e^2}} \, dx,x,\sqrt {e x}\right )}{e} \\ & = \frac {e \text {Subst}\left (\int \frac {1}{a e-b x^2} \, dx,x,\sqrt {e x}\right )}{b c}-\frac {e \text {Subst}\left (\int \frac {1}{a e+b x^2} \, dx,x,\sqrt {e x}\right )}{b c} \\ & = -\frac {\sqrt {e} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {e x}}{\sqrt {a} \sqrt {e}}\right )}{\sqrt {a} b^{3/2} c}+\frac {\sqrt {e} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {e x}}{\sqrt {a} \sqrt {e}}\right )}{\sqrt {a} b^{3/2} c} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.72 \[ \int \frac {\sqrt {e x}}{(a+b x) (a c-b c x)} \, dx=\frac {\sqrt {e x} \left (-\arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )+\text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )\right )}{\sqrt {a} b^{3/2} c \sqrt {x}} \]

[In]

Integrate[Sqrt[e*x]/((a + b*x)*(a*c - b*c*x)),x]

[Out]

(Sqrt[e*x]*(-ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[a]] + ArcTanh[(Sqrt[b]*Sqrt[x])/Sqrt[a]]))/(Sqrt[a]*b^(3/2)*c*Sqrt[
x])

Maple [A] (verified)

Time = 1.08 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.52

method	result	size
pseudoelliptic	\(\frac {e \left (\operatorname {arctanh}\left (\frac {b \sqrt {e x}}{\sqrt {a e b}}\right )-\arctan \left (\frac {b \sqrt {e x}}{\sqrt {a e b}}\right )\right )}{c b \sqrt {a e b}}\)	\(46\)
derivativedivides	\(-\frac {2 e \left (-\frac {\operatorname {arctanh}\left (\frac {b \sqrt {e x}}{\sqrt {a e b}}\right )}{2 b \sqrt {a e b}}+\frac {\arctan \left (\frac {b \sqrt {e x}}{\sqrt {a e b}}\right )}{2 b \sqrt {a e b}}\right )}{c}\)	\(58\)
default	\(\frac {2 e \left (\frac {\operatorname {arctanh}\left (\frac {b \sqrt {e x}}{\sqrt {a e b}}\right )}{2 b \sqrt {a e b}}-\frac {\arctan \left (\frac {b \sqrt {e x}}{\sqrt {a e b}}\right )}{2 b \sqrt {a e b}}\right )}{c}\)	\(58\)

[In]

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

[Out]

e*(arctanh(b*(e*x)^(1/2)/(a*e*b)^(1/2))-arctan(b*(e*x)^(1/2)/(a*e*b)^(1/2)))/c/b/(a*e*b)^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 193, normalized size of antiderivative = 2.19 \[ \int \frac {\sqrt {e x}}{(a+b x) (a c-b c x)} \, dx=\left [\frac {2 \, \sqrt {\frac {e}{a b}} \arctan \left (\frac {\sqrt {e x} a \sqrt {\frac {e}{a b}}}{e x}\right ) + \sqrt {\frac {e}{a b}} \log \left (\frac {b e x + 2 \, \sqrt {e x} a b \sqrt {\frac {e}{a b}} + a e}{b x - a}\right )}{2 \, b c}, -\frac {2 \, \sqrt {-\frac {e}{a b}} \arctan \left (\frac {\sqrt {e x} a \sqrt {-\frac {e}{a b}}}{e x}\right ) - \sqrt {-\frac {e}{a b}} \log \left (\frac {b e x - 2 \, \sqrt {e x} a b \sqrt {-\frac {e}{a b}} - a e}{b x + a}\right )}{2 \, b c}\right ] \]

[In]

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

[Out]

[1/2*(2*sqrt(e/(a*b))*arctan(sqrt(e*x)*a*sqrt(e/(a*b))/(e*x)) + sqrt(e/(a*b))*log((b*e*x + 2*sqrt(e*x)*a*b*sqr
t(e/(a*b)) + a*e)/(b*x - a)))/(b*c), -1/2*(2*sqrt(-e/(a*b))*arctan(sqrt(e*x)*a*sqrt(-e/(a*b))/(e*x)) - sqrt(-e
/(a*b))*log((b*e*x - 2*sqrt(e*x)*a*b*sqrt(-e/(a*b)) - a*e)/(b*x + a)))/(b*c)]

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 170 vs. \(2 (80) = 160\).

Time = 0.99 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.93 \[ \int \frac {\sqrt {e x}}{(a+b x) (a c-b c x)} \, dx=\begin {cases} - \frac {\sqrt {e} \sqrt {x}}{a b c} + \frac {\sqrt {e} \operatorname {acoth}{\left (\frac {\sqrt {a}}{\sqrt {b} \sqrt {x}} \right )}}{\sqrt {a} b^{\frac {3}{2}} c} + \frac {\sqrt {e} \operatorname {atan}{\left (\frac {\sqrt {a}}{\sqrt {b} \sqrt {x}} \right )}}{\sqrt {a} b^{\frac {3}{2}} c} & \text {for}\: \left |{\frac {a}{b x}}\right | > 1 \\- \frac {\sqrt {e} \sqrt {x}}{a b c} + \frac {\sqrt {e} \operatorname {atan}{\left (\frac {\sqrt {a}}{\sqrt {b} \sqrt {x}} \right )}}{\sqrt {a} b^{\frac {3}{2}} c} + \frac {\sqrt {e} \operatorname {atanh}{\left (\frac {\sqrt {a}}{\sqrt {b} \sqrt {x}} \right )}}{\sqrt {a} b^{\frac {3}{2}} c} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((-sqrt(e)*sqrt(x)/(a*b*c) + sqrt(e)*acoth(sqrt(a)/(sqrt(b)*sqrt(x)))/(sqrt(a)*b**(3/2)*c) + sqrt(e)*
atan(sqrt(a)/(sqrt(b)*sqrt(x)))/(sqrt(a)*b**(3/2)*c), Abs(a/(b*x)) > 1), (-sqrt(e)*sqrt(x)/(a*b*c) + sqrt(e)*a
tan(sqrt(a)/(sqrt(b)*sqrt(x)))/(sqrt(a)*b**(3/2)*c) + sqrt(e)*atanh(sqrt(a)/(sqrt(b)*sqrt(x)))/(sqrt(a)*b**(3/
2)*c), True))

Maxima [F(-2)]

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

[In]

integrate((e*x)^(1/2)/(b*x+a)/(-b*c*x+a*c),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(e>0)', see `assume?` for more
details)Is e

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.77 \[ \int \frac {\sqrt {e x}}{(a+b x) (a c-b c x)} \, dx=-\frac {\frac {e^{2} \arctan \left (\frac {\sqrt {e x} b}{\sqrt {a b e}}\right )}{\sqrt {a b e} b c} + \frac {e^{2} \arctan \left (\frac {\sqrt {e x} b}{\sqrt {-a b e}}\right )}{\sqrt {-a b e} b c}}{e} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.60 \[ \int \frac {\sqrt {e x}}{(a+b x) (a c-b c x)} \, dx=-\frac {\sqrt {e}\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {e\,x}}{\sqrt {a}\,\sqrt {e}}\right )-\sqrt {e}\,\mathrm {atanh}\left (\frac {\sqrt {b}\,\sqrt {e\,x}}{\sqrt {a}\,\sqrt {e}}\right )}{\sqrt {a}\,b^{3/2}\,c} \]

[In]

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

[Out]

-(e^(1/2)*atan((b^(1/2)*(e*x)^(1/2))/(a^(1/2)*e^(1/2))) - e^(1/2)*atanh((b^(1/2)*(e*x)^(1/2))/(a^(1/2)*e^(1/2)
)))/(a^(1/2)*b^(3/2)*c)