3.5.73 \(\int \frac {1}{x^{3/2} (-a+b x)} \, dx\) [473]

Optimal. Leaf size=40 \[ \frac {2}{a \sqrt {x}}-\frac {2 \sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{a^{3/2}} \]

[Out]

-2*arctanh(b^(1/2)*x^(1/2)/a^(1/2))*b^(1/2)/a^(3/2)+2/a/x^(1/2)

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Rubi [A]
time = 0.01, antiderivative size = 40, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {53, 65, 214} \begin {gather*} \frac {2}{a \sqrt {x}}-\frac {2 \sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{a^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x^(3/2)*(-a + b*x)),x]

[Out]

2/(a*Sqrt[x]) - (2*Sqrt[b]*ArcTanh[(Sqrt[b]*Sqrt[x])/Sqrt[a]])/a^(3/2)

Rule 53

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n + 1
)/((b*c - a*d)*(m + 1))), x] - Dist[d*((m + n + 2)/((b*c - a*d)*(m + 1))), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

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]

Rubi steps

\begin {align*} \int \frac {1}{x^{3/2} (-a+b x)} \, dx &=\frac {2}{a \sqrt {x}}+\frac {b \int \frac {1}{\sqrt {x} (-a+b x)} \, dx}{a}\\ &=\frac {2}{a \sqrt {x}}+\frac {(2 b) \text {Subst}\left (\int \frac {1}{-a+b x^2} \, dx,x,\sqrt {x}\right )}{a}\\ &=\frac {2}{a \sqrt {x}}-\frac {2 \sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{a^{3/2}}\\ \end {align*}

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Mathematica [A]
time = 0.03, size = 40, normalized size = 1.00 \begin {gather*} \frac {2}{a \sqrt {x}}-\frac {2 \sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{a^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(x^(3/2)*(-a + b*x)),x]

[Out]

2/(a*Sqrt[x]) - (2*Sqrt[b]*ArcTanh[(Sqrt[b]*Sqrt[x])/Sqrt[a]])/a^(3/2)

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Maple [A]
time = 0.13, size = 32, normalized size = 0.80

method	result	size
derivativedivides	\(\frac {2}{a \sqrt {x}}-\frac {2 b \arctanh \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{a \sqrt {a b}}\)	\(32\)
default	\(\frac {2}{a \sqrt {x}}-\frac {2 b \arctanh \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{a \sqrt {a b}}\)	\(32\)
risch	\(\frac {2}{a \sqrt {x}}-\frac {2 b \arctanh \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{a \sqrt {a b}}\)	\(32\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^(3/2)/(b*x-a),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.51, size = 47, normalized size = 1.18 \begin {gather*} \frac {b \log \left (\frac {b \sqrt {x} - \sqrt {a b}}{b \sqrt {x} + \sqrt {a b}}\right )}{\sqrt {a b} a} + \frac {2}{a \sqrt {x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^(3/2)/(b*x-a),x, algorithm="maxima")

[Out]

b*log((b*sqrt(x) - sqrt(a*b))/(b*sqrt(x) + sqrt(a*b)))/(sqrt(a*b)*a) + 2/(a*sqrt(x))

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Fricas [A]
time = 0.45, size = 91, normalized size = 2.28 \begin {gather*} \left [\frac {x \sqrt {\frac {b}{a}} \log \left (\frac {b x - 2 \, a \sqrt {x} \sqrt {\frac {b}{a}} + a}{b x - a}\right ) + 2 \, \sqrt {x}}{a x}, \frac {2 \, {\left (x \sqrt {-\frac {b}{a}} \arctan \left (\frac {a \sqrt {-\frac {b}{a}}}{b \sqrt {x}}\right ) + \sqrt {x}\right )}}{a x}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^(3/2)/(b*x-a),x, algorithm="fricas")

[Out]

[(x*sqrt(b/a)*log((b*x - 2*a*sqrt(x)*sqrt(b/a) + a)/(b*x - a)) + 2*sqrt(x))/(a*x), 2*(x*sqrt(-b/a)*arctan(a*sq
rt(-b/a)/(b*sqrt(x))) + sqrt(x))/(a*x)]

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 76 vs. \(2 (36) = 72\).
time = 1.02, size = 76, normalized size = 1.90 \begin {gather*} \begin {cases} \frac {\tilde {\infty }}{x^{\frac {3}{2}}} & \text {for}\: a = 0 \wedge b = 0 \\\frac {2}{a \sqrt {x}} & \text {for}\: b = 0 \\- \frac {2}{3 b x^{\frac {3}{2}}} & \text {for}\: a = 0 \\\frac {\log {\left (\sqrt {x} - \sqrt {\frac {a}{b}} \right )}}{a \sqrt {\frac {a}{b}}} - \frac {\log {\left (\sqrt {x} + \sqrt {\frac {a}{b}} \right )}}{a \sqrt {\frac {a}{b}}} + \frac {2}{a \sqrt {x}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**(3/2)/(b*x-a),x)

[Out]

Piecewise((zoo/x**(3/2), Eq(a, 0) & Eq(b, 0)), (2/(a*sqrt(x)), Eq(b, 0)), (-2/(3*b*x**(3/2)), Eq(a, 0)), (log(
sqrt(x) - sqrt(a/b))/(a*sqrt(a/b)) - log(sqrt(x) + sqrt(a/b))/(a*sqrt(a/b)) + 2/(a*sqrt(x)), True))

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Giac [A]
time = 0.59, size = 33, normalized size = 0.82 \begin {gather*} \frac {2 \, b \arctan \left (\frac {b \sqrt {x}}{\sqrt {-a b}}\right )}{\sqrt {-a b} a} + \frac {2}{a \sqrt {x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^(3/2)/(b*x-a),x, algorithm="giac")

[Out]

2*b*arctan(b*sqrt(x)/sqrt(-a*b))/(sqrt(-a*b)*a) + 2/(a*sqrt(x))

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Mupad [B]
time = 0.06, size = 28, normalized size = 0.70 \begin {gather*} \frac {2}{a\,\sqrt {x}}-\frac {2\,\sqrt {b}\,\mathrm {atanh}\left (\frac {\sqrt {b}\,\sqrt {x}}{\sqrt {a}}\right )}{a^{3/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-1/(x^(3/2)*(a - b*x)),x)

[Out]

2/(a*x^(1/2)) - (2*b^(1/2)*atanh((b^(1/2)*x^(1/2))/a^(1/2)))/a^(3/2)

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