3.5.53 \(\int \frac {1}{x^{3/2} (a+b x)} \, dx\)

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

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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 = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {51, 63, 205} \begin {gather*} -\frac {2 \sqrt {b} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{a^{3/2}}-\frac {2}{a \sqrt {x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x^(3/2)*(a + b*x)),x]

[Out]

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

Rule 51

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 63

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 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]

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

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Mathematica [C]  time = 0.00, size = 25, normalized size = 0.62 \begin {gather*} -\frac {2 \, _2F_1\left (-\frac {1}{2},1;\frac {1}{2};-\frac {b x}{a}\right )}{a \sqrt {x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(x^(3/2)*(a + b*x)),x]

[Out]

(-2*Hypergeometric2F1[-1/2, 1, 1/2, -((b*x)/a)])/(a*Sqrt[x])

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

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[1/(x^(3/2)*(a + b*x)),x]

[Out]

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

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fricas [A]  time = 0.95, size = 93, normalized size = 2.32 \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*s
qrt(b/a)/(b*sqrt(x))) - sqrt(x))/(a*x)]

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giac [A]  time = 1.00, size = 31, normalized size = 0.78 \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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maple [A]  time = 0.01, size = 32, normalized size = 0.80 \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]

int(1/x^(3/2)/(b*x+a),x)

[Out]

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

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maxima [A]  time = 2.88, size = 31, normalized size = 0.78 \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="maxima")

[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 = 0.04, size = 28, normalized size = 0.70 \begin {gather*} -\frac {2}{a\,\sqrt {x}}-\frac {2\,\sqrt {b}\,\mathrm {atan}\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)*atan((b^(1/2)*x^(1/2))/a^(1/2)))/a^(3/2)

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sympy [A]  time = 2.77, size = 102, normalized size = 2.55 \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 {2}{a \sqrt {x}} + \frac {i \log {\left (- i \sqrt {a} \sqrt {\frac {1}{b}} + \sqrt {x} \right )}}{a^{\frac {3}{2}} \sqrt {\frac {1}{b}}} - \frac {i \log {\left (i \sqrt {a} \sqrt {\frac {1}{b}} + \sqrt {x} \right )}}{a^{\frac {3}{2}} \sqrt {\frac {1}{b}}} & \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)), (-2/
(a*sqrt(x)) + I*log(-I*sqrt(a)*sqrt(1/b) + sqrt(x))/(a**(3/2)*sqrt(1/b)) - I*log(I*sqrt(a)*sqrt(1/b) + sqrt(x)
)/(a**(3/2)*sqrt(1/b)), True))

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