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

Optimal. Leaf size=87 \[ \frac {\tan ^{-1}\left (\frac {\sqrt {b} \sqrt {e x}}{\sqrt {a} \sqrt {e}}\right )}{a^{3/2} \sqrt {b} c \sqrt {e}}+\frac {\tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {e x}}{\sqrt {a} \sqrt {e}}\right )}{a^{3/2} \sqrt {b} c \sqrt {e}} \]

[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]
time = 0.03, antiderivative size = 87, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {74, 335, 218, 214, 211} \begin {gather*} \frac {\text {ArcTan}\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 {gather*}

Antiderivative was successfully verified.

[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*} \int \frac {1}{\sqrt {e x} (a+b x) (a c-b c x)} \, dx &=\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]
time = 0.04, size = 61, normalized size = 0.70 \begin {gather*} \frac {\sqrt {x} \left (\tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )+\tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )\right )}{a^{3/2} \sqrt {b} c \sqrt {e x}} \end {gather*}

Antiderivative was successfully verified.

[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]
time = 0.06, size = 64, normalized size = 0.74

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [A]
time = 0.49, size = 70, normalized size = 0.80 \begin {gather*} \frac {1}{2} \, {\left (\frac {2 \, \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} a c} - \frac {\log \left (\frac {b \sqrt {x} - \sqrt {a b}}{b \sqrt {x} + \sqrt {a b}}\right )}{\sqrt {a b} a c}\right )} e^{\left (-\frac {1}{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [A]
time = 1.33, size = 138, normalized size = 1.59 \begin {gather*} \left [-\frac {{\left (2 \, \sqrt {a b} \arctan \left (\frac {\sqrt {a b}}{b \sqrt {x}}\right ) - \sqrt {a b} \log \left (\frac {b x + a + 2 \, \sqrt {a b} \sqrt {x}}{b x - a}\right )\right )} e^{\left (-\frac {1}{2}\right )}}{2 \, a^{2} b c}, -\frac {{\left (2 \, \sqrt {-a b} \arctan \left (\frac {\sqrt {-a b}}{b \sqrt {x}}\right ) + \sqrt {-a b} \log \left (\frac {b x - a - 2 \, \sqrt {-a b} \sqrt {x}}{b x + a}\right )\right )} e^{\left (-\frac {1}{2}\right )}}{2 \, a^{2} b c}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

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

________________________________________________________________________________________

Sympy [C] Result contains complex when optimal does not.
time = 0.89, size = 291, normalized size = 3.34 \begin {gather*} \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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

________________________________________________________________________________________

Giac [A]
time = 0.94, size = 53, normalized size = 0.61 \begin {gather*} {\left (\frac {\arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} a c} - \frac {\arctan \left (\frac {b \sqrt {x}}{\sqrt {-a b}}\right )}{\sqrt {-a b} a c}\right )} e^{\left (-\frac {1}{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Mupad [B]
time = 0.40, size = 46, normalized size = 0.53 \begin {gather*} \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}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

________________________________________________________________________________________