3.5.58 \(\int \frac {\sqrt {x}}{(a+b x)^2} \, dx\)

Optimal. Leaf size=46 \[ \frac {\tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{\sqrt {a} b^{3/2}}-\frac {\sqrt {x}}{b (a+b x)} \]

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Rubi [A]  time = 0.01, antiderivative size = 46, 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 = {47, 63, 205} \begin {gather*} \frac {\tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{\sqrt {a} b^{3/2}}-\frac {\sqrt {x}}{b (a+b x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[x]/(a + b*x)^2,x]

[Out]

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

Rule 47

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + 1)), x] - Dist[(d*n)/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && 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 {\sqrt {x}}{(a+b x)^2} \, dx &=-\frac {\sqrt {x}}{b (a+b x)}+\frac {\int \frac {1}{\sqrt {x} (a+b x)} \, dx}{2 b}\\ &=-\frac {\sqrt {x}}{b (a+b x)}+\frac {\operatorname {Subst}\left (\int \frac {1}{a+b x^2} \, dx,x,\sqrt {x}\right )}{b}\\ &=-\frac {\sqrt {x}}{b (a+b x)}+\frac {\tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{\sqrt {a} b^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 46, normalized size = 1.00 \begin {gather*} \frac {\tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{\sqrt {a} b^{3/2}}-\frac {\sqrt {x}}{b (a+b x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[x]/(a + b*x)^2,x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.06, size = 46, normalized size = 1.00 \begin {gather*} \frac {\tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{\sqrt {a} b^{3/2}}-\frac {\sqrt {x}}{b (a+b x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.89, size = 115, normalized size = 2.50 \begin {gather*} \left [-\frac {2 \, a b \sqrt {x} + \sqrt {-a b} {\left (b x + a\right )} \log \left (\frac {b x - a - 2 \, \sqrt {-a b} \sqrt {x}}{b x + a}\right )}{2 \, {\left (a b^{3} x + a^{2} b^{2}\right )}}, -\frac {a b \sqrt {x} + \sqrt {a b} {\left (b x + a\right )} \arctan \left (\frac {\sqrt {a b}}{b \sqrt {x}}\right )}{a b^{3} x + a^{2} b^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/2*(2*a*b*sqrt(x) + sqrt(-a*b)*(b*x + a)*log((b*x - a - 2*sqrt(-a*b)*sqrt(x))/(b*x + a)))/(a*b^3*x + a^2*b^
2), -(a*b*sqrt(x) + sqrt(a*b)*(b*x + a)*arctan(sqrt(a*b)/(b*sqrt(x))))/(a*b^3*x + a^2*b^2)]

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giac [A]  time = 0.90, size = 36, normalized size = 0.78 \begin {gather*} \frac {\arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} b} - \frac {\sqrt {x}}{{\left (b x + a\right )} b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.01, size = 37, normalized size = 0.80 \begin {gather*} \frac {\arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b}\, b}-\frac {\sqrt {x}}{\left (b x +a \right ) b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 2.94, size = 37, normalized size = 0.80 \begin {gather*} -\frac {\sqrt {x}}{b^{2} x + a b} + \frac {\arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.04, size = 34, normalized size = 0.74 \begin {gather*} \frac {\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {x}}{\sqrt {a}}\right )}{\sqrt {a}\,b^{3/2}}-\frac {\sqrt {x}}{b\,\left (a+b\,x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(1/2)/(a + b*x)^2,x)

[Out]

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

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sympy [A]  time = 4.45, size = 337, normalized size = 7.33 \begin {gather*} \begin {cases} \frac {\tilde {\infty }}{\sqrt {x}} & \text {for}\: a = 0 \wedge b = 0 \\\frac {2 x^{\frac {3}{2}}}{3 a^{2}} & \text {for}\: b = 0 \\- \frac {2}{b^{2} \sqrt {x}} & \text {for}\: a = 0 \\- \frac {2 i \sqrt {a} b \sqrt {x} \sqrt {\frac {1}{b}}}{2 i a^{\frac {3}{2}} b^{2} \sqrt {\frac {1}{b}} + 2 i \sqrt {a} b^{3} x \sqrt {\frac {1}{b}}} + \frac {a \log {\left (- i \sqrt {a} \sqrt {\frac {1}{b}} + \sqrt {x} \right )}}{2 i a^{\frac {3}{2}} b^{2} \sqrt {\frac {1}{b}} + 2 i \sqrt {a} b^{3} x \sqrt {\frac {1}{b}}} - \frac {a \log {\left (i \sqrt {a} \sqrt {\frac {1}{b}} + \sqrt {x} \right )}}{2 i a^{\frac {3}{2}} b^{2} \sqrt {\frac {1}{b}} + 2 i \sqrt {a} b^{3} x \sqrt {\frac {1}{b}}} + \frac {b x \log {\left (- i \sqrt {a} \sqrt {\frac {1}{b}} + \sqrt {x} \right )}}{2 i a^{\frac {3}{2}} b^{2} \sqrt {\frac {1}{b}} + 2 i \sqrt {a} b^{3} x \sqrt {\frac {1}{b}}} - \frac {b x \log {\left (i \sqrt {a} \sqrt {\frac {1}{b}} + \sqrt {x} \right )}}{2 i a^{\frac {3}{2}} b^{2} \sqrt {\frac {1}{b}} + 2 i \sqrt {a} b^{3} x \sqrt {\frac {1}{b}}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((zoo/sqrt(x), Eq(a, 0) & Eq(b, 0)), (2*x**(3/2)/(3*a**2), Eq(b, 0)), (-2/(b**2*sqrt(x)), Eq(a, 0)),
(-2*I*sqrt(a)*b*sqrt(x)*sqrt(1/b)/(2*I*a**(3/2)*b**2*sqrt(1/b) + 2*I*sqrt(a)*b**3*x*sqrt(1/b)) + a*log(-I*sqrt
(a)*sqrt(1/b) + sqrt(x))/(2*I*a**(3/2)*b**2*sqrt(1/b) + 2*I*sqrt(a)*b**3*x*sqrt(1/b)) - a*log(I*sqrt(a)*sqrt(1
/b) + sqrt(x))/(2*I*a**(3/2)*b**2*sqrt(1/b) + 2*I*sqrt(a)*b**3*x*sqrt(1/b)) + b*x*log(-I*sqrt(a)*sqrt(1/b) + s
qrt(x))/(2*I*a**(3/2)*b**2*sqrt(1/b) + 2*I*sqrt(a)*b**3*x*sqrt(1/b)) - b*x*log(I*sqrt(a)*sqrt(1/b) + sqrt(x))/
(2*I*a**(3/2)*b**2*sqrt(1/b) + 2*I*sqrt(a)*b**3*x*sqrt(1/b)), True))

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