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

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

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Rubi [A]  time = 0.05, 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 = {73, 329, 212, 208, 205} \begin {gather*} \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 {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 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 212

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] &&
!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 {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 \operatorname {Subst}\left (\int \frac {1}{a^2 c-\frac {b^2 c x^4}{e^2}} \, dx,x,\sqrt {e x}\right )}{e}\\ &=\frac {\operatorname {Subst}\left (\int \frac {1}{a e-b x^2} \, dx,x,\sqrt {e x}\right )}{a c}+\frac {\operatorname {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*}

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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
])

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IntegrateAlgebraic [A]  time = 0.06, size = 87, normalized size = 1.00 \begin {gather*} \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 {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[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])

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fricas [A]  time = 1.40, size = 173, normalized size = 1.99 \begin {gather*} \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 ] \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*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)]

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giac [A]  time = 1.01, size = 58, normalized size = 0.67 \begin {gather*} \frac {\arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right ) e^{\left (-\frac {1}{2}\right )}}{\sqrt {a b} a c} - \frac {\arctan \left (\frac {b \sqrt {x} e^{\frac {1}{2}}}{\sqrt {-a b e}}\right )}{\sqrt {-a b e} a c} \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))*e^(-1/2)/(sqrt(a*b)*a*c) - arctan(b*sqrt(x)*e^(1/2)/sqrt(-a*b*e))/(sqrt(-a*b*e)*a*
c)

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maple [A]  time = 0.01, size = 56, normalized size = 0.64 \begin {gather*} \frac {\arctanh \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} \end {gather*}

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)

[Out]

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

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maxima [A]  time = 2.50, size = 84, normalized size = 0.97 \begin {gather*} \frac {\frac {2 \, e \arctan \left (\frac {\sqrt {e x} b}{\sqrt {a b e}}\right )}{\sqrt {a b e} a c} - \frac {e \log \left (\frac {\sqrt {e x} b - \sqrt {a b e}}{\sqrt {e x} b + \sqrt {a b e}}\right )}{\sqrt {a b e} a c}}{2 \, e} \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*e*arctan(sqrt(e*x)*b/sqrt(a*b*e))/(sqrt(a*b*e)*a*c) - e*log((sqrt(e*x)*b - sqrt(a*b*e))/(sqrt(e*x)*b +
sqrt(a*b*e)))/(sqrt(a*b*e)*a*c))/e

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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))

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sympy [A]  time = 2.66, 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 {1 - i}{2 a b c \sqrt {e} \sqrt {x}} + \frac {i \left (1 - i\right )}{2 a b c \sqrt {e} \sqrt {x}} + \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 {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 {atanh}{\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}} & \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), ((1 - I)/(2*a*b*c*sqrt(e)*sqrt(x)) + I*(1
- I)/(2*a*b*c*sqrt(e)*sqrt(x)) + (1 - I)*atan(sqrt(b)*sqrt(x)/sqrt(a))/(2*a**(3/2)*sqrt(b)*c*sqrt(e)) + I*(1 -
I)*atan(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)) + I*(1 - I)*atanh(sqrt(b)*sqrt(x)/sqrt(a))/(2*a**(3/2)*sqrt(b)*c*sqrt(e)), True))

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