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

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

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Rubi [A]  time = 0.05, 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 = {73, 329, 298, 205, 208} \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} \tan ^{-1}\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 73

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]

Rule 205

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

Rule 208

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

Rule 298

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 329

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 \operatorname {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 \operatorname {Subst}\left (\int \frac {1}{a e-b x^2} \, dx,x,\sqrt {e x}\right )}{b c}-\frac {e \operatorname {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 = 0.02, size = 63, normalized size = 0.72 \begin {gather*} \frac {\sqrt {e x} \left (\tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )-\tan ^{-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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IntegrateAlgebraic [A]  time = 0.06, size = 88, normalized size = 1.00 \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} \tan ^{-1}\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]

IntegrateAlgebraic[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)

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fricas [A]  time = 1.71, size = 193, normalized size = 2.19 \begin {gather*} \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 ] \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(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)]

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giac [A]  time = 0.92, size = 63, normalized size = 0.72 \begin {gather*} -{\left (\frac {\arctan \left (\frac {b \sqrt {x} e^{\frac {1}{2}}}{\sqrt {-a b e}}\right ) e^{2}}{\sqrt {-a b e} b c} + \frac {\arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right ) e^{\frac {3}{2}}}{\sqrt {a b} b c}\right )} e^{\left (-1\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="giac")

[Out]

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

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maple [A]  time = 0.01, size = 59, normalized size = 0.67 \begin {gather*} \frac {e \arctanh \left (\frac {\sqrt {e x}\, b}{\sqrt {a b e}}\right )}{\sqrt {a b e}\, b c}-\frac {e \arctan \left (\frac {\sqrt {e x}\, b}{\sqrt {a b e}}\right )}{\sqrt {a b e}\, b c} \end {gather*}

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)

[Out]

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

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maxima [A]  time = 2.39, size = 87, normalized size = 0.99 \begin {gather*} -\frac {\frac {2 \, e^{2} \arctan \left (\frac {\sqrt {e x} b}{\sqrt {a b e}}\right )}{\sqrt {a b e} b c} + \frac {e^{2} \log \left (\frac {\sqrt {e x} b - \sqrt {a b e}}{\sqrt {e x} b + \sqrt {a b e}}\right )}{\sqrt {a b e} b c}}{2 \, e} \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*e^2*arctan(sqrt(e*x)*b/sqrt(a*b*e))/(sqrt(a*b*e)*b*c) + e^2*log((sqrt(e*x)*b - sqrt(a*b*e))/(sqrt(e*x)
*b + sqrt(a*b*e)))/(sqrt(a*b*e)*b*c))/e

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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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sympy [A]  time = 2.39, 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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