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

Optimal. Leaf size=88 \[ -\frac {\sqrt {e} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {e x}}{\sqrt {a} \sqrt {e}}\right )}{\sqrt {a} b^{3/2} c}+\frac {\sqrt {e} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {e x}}{\sqrt {a} \sqrt {e}}\right )}{\sqrt {a} b^{3/2} c} \]

[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)

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Rubi [A]
time = 0.03, antiderivative size = 88, 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, 304, 211, 214} \begin {gather*} \frac {\sqrt {e} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {e x}}{\sqrt {a} \sqrt {e}}\right )}{\sqrt {a} b^{3/2} c}-\frac {\sqrt {e} \text {ArcTan}\left (\frac {\sqrt {b} \sqrt {e x}}{\sqrt {a} \sqrt {e}}\right )}{\sqrt {a} b^{3/2} c} \end {gather*}

Antiderivative was successfully verified.

[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*} \int \frac {\sqrt {e x}}{(a+b x) (a c-b c x)} \, dx &=\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*}

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Mathematica [A]
time = 0.04, size = 63, normalized size = 0.72 \begin {gather*} \frac {\sqrt {e 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 )}{\sqrt {a} b^{3/2} c \sqrt {x}} \end {gather*}

Antiderivative was successfully verified.

[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])

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Maple [A]
time = 0.06, size = 58, normalized size = 0.66

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.48, size = 69, normalized size = 0.78 \begin {gather*} -\frac {1}{2} \, {\left (\frac {2 \, \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} b c} + \frac {\log \left (\frac {b \sqrt {x} - \sqrt {a b}}{b \sqrt {x} + \sqrt {a b}}\right )}{\sqrt {a b} b c}\right )} e^{\frac {1}{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((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)*b*c) + log((b*sqrt(x) - sqrt(a*b))/(b*sqrt(x) + sqrt(a*b)))/(sq
rt(a*b)*b*c))*e^(1/2)

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Fricas [A]
time = 1.93, size = 141, normalized size = 1.60 \begin {gather*} \left [\frac {2 \, \sqrt {a b} \arctan \left (\frac {\sqrt {a b}}{b \sqrt {x}}\right ) e^{\frac {1}{2}} + \sqrt {a b} e^{\frac {1}{2}} \log \left (\frac {b x + a + 2 \, \sqrt {a b} \sqrt {x}}{b x - a}\right )}{2 \, a b^{2} c}, -\frac {2 \, \sqrt {-a b} \arctan \left (\frac {\sqrt {-a b}}{b \sqrt {x}}\right ) e^{\frac {1}{2}} + \sqrt {-a b} e^{\frac {1}{2}} \log \left (\frac {b x - a + 2 \, \sqrt {-a b} \sqrt {x}}{b x + a}\right )}{2 \, a b^{2} c}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 170 vs. \(2 (80) = 160\).
time = 0.91, size = 170, normalized size = 1.93 \begin {gather*} \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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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Giac [A]
time = 0.74, size = 53, normalized size = 0.60 \begin {gather*} -{\left (\frac {\arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} b c} + \frac {\arctan \left (\frac {b \sqrt {x}}{\sqrt {-a b}}\right )}{\sqrt {-a b} b c}\right )} e^{\frac {1}{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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Mupad [B]
time = 0.14, size = 53, normalized size = 0.60 \begin {gather*} -\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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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