\(\int \frac {1}{\sqrt {e x} (a+b x) (a c-b c x)} \, dx\) [53]

Optimal result

Integrand size = 26, antiderivative size = 87 \[ \int \frac {1}{\sqrt {e x} (a+b x) (a c-b c x)} \, dx=\frac {\arctan \left (\frac {\sqrt {b} \sqrt {e x}}{\sqrt {a} \sqrt {e}}\right )}{a^{3/2} \sqrt {b} c \sqrt {e}}+\frac {\text {arctanh}\left (\frac {\sqrt {b} \sqrt {e x}}{\sqrt {a} \sqrt {e}}\right )}{a^{3/2} \sqrt {b} c \sqrt {e}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 87, 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, 218, 214, 211} \[ \int \frac {1}{\sqrt {e x} (a+b x) (a c-b c x)} \, dx=\frac {\arctan \left (\frac {\sqrt {b} \sqrt {e x}}{\sqrt {a} \sqrt {e}}\right )}{a^{3/2} \sqrt {b} c \sqrt {e}}+\frac {\text {arctanh}\left (\frac {\sqrt {b} \sqrt {e x}}{\sqrt {a} \sqrt {e}}\right )}{a^{3/2} \sqrt {b} c \sqrt {e}} \]

[In]

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

[Out]

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

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 218

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2]], s = Denominator[Rt[-a/b, 2]]},
Dist[r/(2*a), Int[1/(r - s*x^2), x], x] + Dist[r/(2*a), Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] &&  !Gt
Q[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 {1}{\sqrt {e x} \left (a^2 c-b^2 c x^2\right )} \, dx \\ & = \frac {2 \text {Subst}\left (\int \frac {1}{a^2 c-\frac {b^2 c x^4}{e^2}} \, dx,x,\sqrt {e x}\right )}{e} \\ & = \frac {\text {Subst}\left (\int \frac {1}{a e-b x^2} \, dx,x,\sqrt {e x}\right )}{a c}+\frac {\text {Subst}\left (\int \frac {1}{a e+b x^2} \, dx,x,\sqrt {e x}\right )}{a c} \\ & = \frac {\tan ^{-1}\left (\frac {\sqrt {b} \sqrt {e x}}{\sqrt {a} \sqrt {e}}\right )}{a^{3/2} \sqrt {b} c \sqrt {e}}+\frac {\tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {e x}}{\sqrt {a} \sqrt {e}}\right )}{a^{3/2} \sqrt {b} c \sqrt {e}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.70 \[ \int \frac {1}{\sqrt {e x} (a+b x) (a c-b c x)} \, dx=\frac {\sqrt {x} \left (\arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )+\text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )\right )}{a^{3/2} \sqrt {b} c \sqrt {e x}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.42 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.49

method	result	size
pseudoelliptic	\(\frac {\operatorname {arctanh}\left (\frac {b \sqrt {e x}}{\sqrt {a e b}}\right )+\arctan \left (\frac {b \sqrt {e x}}{\sqrt {a e b}}\right )}{c a \sqrt {a e b}}\)	\(43\)
derivativedivides	\(-\frac {2 e \left (-\frac {\operatorname {arctanh}\left (\frac {b \sqrt {e x}}{\sqrt {a e b}}\right )}{2 a e \sqrt {a e b}}-\frac {\arctan \left (\frac {b \sqrt {e x}}{\sqrt {a e b}}\right )}{2 a e \sqrt {a e b}}\right )}{c}\)	\(64\)
default	\(\frac {2 e \left (\frac {\operatorname {arctanh}\left (\frac {b \sqrt {e x}}{\sqrt {a e b}}\right )}{2 a e \sqrt {a e b}}+\frac {\arctan \left (\frac {b \sqrt {e x}}{\sqrt {a e b}}\right )}{2 a e \sqrt {a e b}}\right )}{c}\)	\(64\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.99 \[ \int \frac {1}{\sqrt {e x} (a+b x) (a c-b c x)} \, dx=\left [-\frac {2 \, \sqrt {a b e} \arctan \left (\frac {\sqrt {a b e} \sqrt {e x}}{b e x}\right ) - \sqrt {a b e} \log \left (\frac {b e x + a e + 2 \, \sqrt {a b e} \sqrt {e x}}{b x - a}\right )}{2 \, a^{2} b c e}, -\frac {2 \, \sqrt {-a b e} \arctan \left (\frac {\sqrt {-a b e} \sqrt {e x}}{b e x}\right ) + \sqrt {-a b e} \log \left (\frac {b e x - a e - 2 \, \sqrt {-a b e} \sqrt {e x}}{b x + a}\right )}{2 \, a^{2} b c e}\right ] \]

[In]

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

[Out]

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

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 1.02 (sec) , antiderivative size = 291, normalized size of antiderivative = 3.34 \[ \int \frac {1}{\sqrt {e x} (a+b x) (a c-b c x)} \, dx=\begin {cases} \frac {1}{a b c \sqrt {e} \sqrt {x}} + \frac {\operatorname {acoth}{\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}} \right )}}{a^{\frac {3}{2}} \sqrt {b} c \sqrt {e}} + \frac {\operatorname {atan}{\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}} \right )}}{a^{\frac {3}{2}} \sqrt {b} c \sqrt {e}} & \text {for}\: \left |{\frac {b x}{a}}\right | > 1 \\- \frac {i \left (1 + i\right )}{2 a b c \sqrt {e} \sqrt {x}} + \frac {1 + i}{2 a b c \sqrt {e} \sqrt {x}} - \frac {i \left (1 + i\right ) \operatorname {atan}{\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}} \right )}}{2 a^{\frac {3}{2}} \sqrt {b} c \sqrt {e}} + \frac {\left (1 + i\right ) \operatorname {atan}{\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}} \right )}}{2 a^{\frac {3}{2}} \sqrt {b} c \sqrt {e}} - \frac {i \left (1 + i\right ) \operatorname {atanh}{\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}} \right )}}{2 a^{\frac {3}{2}} \sqrt {b} c \sqrt {e}} + \frac {\left (1 + i\right ) \operatorname {atanh}{\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}} \right )}}{2 a^{\frac {3}{2}} \sqrt {b} c \sqrt {e}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

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

Maxima [F(-2)]

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

[In]

integrate(1/(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 = 0.27 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.67 \[ \int \frac {1}{\sqrt {e x} (a+b x) (a c-b c x)} \, dx=\frac {\arctan \left (\frac {\sqrt {e x} b}{\sqrt {a b e}}\right )}{\sqrt {a b e} a c} - \frac {\arctan \left (\frac {\sqrt {e x} b}{\sqrt {-a b e}}\right )}{\sqrt {-a b e} a c} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.49 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.53 \[ \int \frac {1}{\sqrt {e x} (a+b x) (a c-b c x)} \, dx=\frac {\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {e\,x}}{\sqrt {a}\,\sqrt {e}}\right )+\mathrm {atanh}\left (\frac {\sqrt {b}\,\sqrt {e\,x}}{\sqrt {a}\,\sqrt {e}}\right )}{a^{3/2}\,\sqrt {b}\,c\,\sqrt {e}} \]

[In]

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

[Out]

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