3.5.52 \(\int \frac {1}{\sqrt {x} (a+b x)} \, dx\)

Optimal. Leaf size=29 \[ \frac {2 \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{\sqrt {a} \sqrt {b}} \]

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Rubi [A]  time = 0.01, antiderivative size = 29, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {63, 205} \begin {gather*} \frac {2 \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{\sqrt {a} \sqrt {b}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(Sqrt[x]*(a + b*x)),x]

[Out]

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

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

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Mathematica [A]  time = 0.01, size = 29, normalized size = 1.00 \begin {gather*} \frac {2 \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{\sqrt {a} \sqrt {b}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(Sqrt[x]*(a + b*x)),x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.02, size = 29, normalized size = 1.00 \begin {gather*} \frac {2 \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{\sqrt {a} \sqrt {b}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[1/(Sqrt[x]*(a + b*x)),x]

[Out]

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

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fricas [A]  time = 0.91, size = 68, normalized size = 2.34 \begin {gather*} \left [-\frac {\sqrt {-a b} \log \left (\frac {b x - a - 2 \, \sqrt {-a b} \sqrt {x}}{b x + a}\right )}{a b}, -\frac {2 \, \sqrt {a b} \arctan \left (\frac {\sqrt {a b}}{b \sqrt {x}}\right )}{a b}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.95, size = 18, normalized size = 0.62 \begin {gather*} \frac {2 \, \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.01, size = 19, normalized size = 0.66 \begin {gather*} \frac {2 \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 2.93, size = 18, normalized size = 0.62 \begin {gather*} \frac {2 \, \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.04, size = 19, normalized size = 0.66 \begin {gather*} \frac {2\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {x}}{\sqrt {a}}\right )}{\sqrt {a}\,\sqrt {b}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^(1/2)*(a + b*x)),x)

[Out]

(2*atan((b^(1/2)*x^(1/2))/a^(1/2)))/(a^(1/2)*b^(1/2))

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sympy [A]  time = 1.29, size = 94, normalized size = 3.24 \begin {gather*} \begin {cases} \frac {\tilde {\infty }}{\sqrt {x}} & \text {for}\: a = 0 \wedge b = 0 \\- \frac {2}{b \sqrt {x}} & \text {for}\: a = 0 \\\frac {2 \sqrt {x}}{a} & \text {for}\: b = 0 \\- \frac {i \log {\left (- i \sqrt {a} \sqrt {\frac {1}{b}} + \sqrt {x} \right )}}{\sqrt {a} b \sqrt {\frac {1}{b}}} + \frac {i \log {\left (i \sqrt {a} \sqrt {\frac {1}{b}} + \sqrt {x} \right )}}{\sqrt {a} b \sqrt {\frac {1}{b}}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x+a)/x**(1/2),x)

[Out]

Piecewise((zoo/sqrt(x), Eq(a, 0) & Eq(b, 0)), (-2/(b*sqrt(x)), Eq(a, 0)), (2*sqrt(x)/a, Eq(b, 0)), (-I*log(-I*
sqrt(a)*sqrt(1/b) + sqrt(x))/(sqrt(a)*b*sqrt(1/b)) + I*log(I*sqrt(a)*sqrt(1/b) + sqrt(x))/(sqrt(a)*b*sqrt(1/b)
), True))

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