3.4.27 \(\int \frac {\sqrt {-a+b x}}{x^3} \, dx\) [327]

Optimal. Leaf size=71 \[ -\frac {\sqrt {-a+b x}}{2 x^2}+\frac {b \sqrt {-a+b x}}{4 a x}+\frac {b^2 \tan ^{-1}\left (\frac {\sqrt {-a+b x}}{\sqrt {a}}\right )}{4 a^{3/2}} \]

[Out]

1/4*b^2*arctan((b*x-a)^(1/2)/a^(1/2))/a^(3/2)-1/2*(b*x-a)^(1/2)/x^2+1/4*b*(b*x-a)^(1/2)/a/x

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Rubi [A]
time = 0.01, antiderivative size = 71, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {43, 44, 65, 211} \begin {gather*} \frac {b^2 \tan ^{-1}\left (\frac {\sqrt {b x-a}}{\sqrt {a}}\right )}{4 a^{3/2}}-\frac {\sqrt {b x-a}}{2 x^2}+\frac {b \sqrt {b x-a}}{4 a x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[-a + b*x]/x^3,x]

[Out]

-1/2*Sqrt[-a + b*x]/x^2 + (b*Sqrt[-a + b*x])/(4*a*x) + (b^2*ArcTan[Sqrt[-a + b*x]/Sqrt[a]])/(4*a^(3/2))

Rule 43

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, n
}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, -1] &&  !IntegerQ[n] && GtQ[n, 0]

Rule 44

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] && ILtQ[m, -1] &&  !IntegerQ[n] && LtQ[n, 0]

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 211

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/Rt[a/b, 2]], x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rubi steps

\begin {align*} \int \frac {\sqrt {-a+b x}}{x^3} \, dx &=-\frac {\sqrt {-a+b x}}{2 x^2}+\frac {1}{4} b \int \frac {1}{x^2 \sqrt {-a+b x}} \, dx\\ &=-\frac {\sqrt {-a+b x}}{2 x^2}+\frac {b \sqrt {-a+b x}}{4 a x}+\frac {b^2 \int \frac {1}{x \sqrt {-a+b x}} \, dx}{8 a}\\ &=-\frac {\sqrt {-a+b x}}{2 x^2}+\frac {b \sqrt {-a+b x}}{4 a x}+\frac {b \text {Subst}\left (\int \frac {1}{\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {-a+b x}\right )}{4 a}\\ &=-\frac {\sqrt {-a+b x}}{2 x^2}+\frac {b \sqrt {-a+b x}}{4 a x}+\frac {b^2 \tan ^{-1}\left (\frac {\sqrt {-a+b x}}{\sqrt {a}}\right )}{4 a^{3/2}}\\ \end {align*}

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Mathematica [A]
time = 0.08, size = 60, normalized size = 0.85 \begin {gather*} -\frac {(2 a-b x) \sqrt {-a+b x}}{4 a x^2}+\frac {b^2 \tan ^{-1}\left (\frac {\sqrt {-a+b x}}{\sqrt {a}}\right )}{4 a^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[-a + b*x]/x^3,x]

[Out]

-1/4*((2*a - b*x)*Sqrt[-a + b*x])/(a*x^2) + (b^2*ArcTan[Sqrt[-a + b*x]/Sqrt[a]])/(4*a^(3/2))

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Mathics [C] Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.
time = 24.37, size = 206, normalized size = 2.90 \begin {gather*} \text {Piecewise}\left [\left \{\left \{\frac {I \left (-2 a^{\frac {7}{2}} x \left (a-b x\right )+3 a^{\frac {5}{2}} b x^2 \left (a-b x\right )-a^{\frac {3}{2}} b^2 x^3 \left (a-b x\right )+a b^{\frac {7}{2}} x^{\frac {9}{2}} \text {ArcCosh}\left [\frac {\sqrt {a}}{\sqrt {b} \sqrt {x}}\right ] \left (\frac {a-b x}{b x}\right )^{\frac {3}{2}}\right )}{4 a^{\frac {5}{2}} b^{\frac {3}{2}} x^{\frac {9}{2}} \left (\frac {a-b x}{b x}\right )^{\frac {3}{2}}},\text {Abs}\left [\frac {a}{b x}\right ]>1\right \}\right \},\frac {a}{2 \sqrt {b} x^{\frac {5}{2}} \sqrt {1-\frac {a}{b x}}}+\frac {b^{\frac {3}{2}}}{4 a \sqrt {x} \sqrt {1-\frac {a}{b x}}}-\frac {b^2 \text {ArcSin}\left [\frac {\sqrt {a}}{\sqrt {b} \sqrt {x}}\right ]}{4 a^{\frac {3}{2}}}-\frac {3 \sqrt {b}}{4 x^{\frac {3}{2}} \sqrt {1-\frac {a}{b x}}}\right ] \end {gather*}

Warning: Unable to verify antiderivative.

[In]

mathics('Integrate[Sqrt[-a + b*x]/x^3,x]')

[Out]

Piecewise[{{I / 4 (-2 a ^ (7 / 2) x (a - b x) + 3 a ^ (5 / 2) b x ^ 2 (a - b x) - a ^ (3 / 2) b ^ 2 x ^ 3 (a -
b x) + a b ^ (7 / 2) x ^ (9 / 2) ArcCosh[Sqrt[a] / (Sqrt[b] Sqrt[x])] ((a - b x) / (b x)) ^ (3 / 2)) / (a ^ (
5 / 2) b ^ (3 / 2) x ^ (9 / 2) ((a - b x) / (b x)) ^ (3 / 2)), Abs[a / (b x)] > 1}}, a / (2 Sqrt[b] x ^ (5 / 2
) Sqrt[1 - a / (b x)]) + b ^ (3 / 2) / (4 a Sqrt[x] Sqrt[1 - a / (b x)]) - b ^ 2 ArcSin[Sqrt[a] / (Sqrt[b] Sqr
t[x])] / (4 a ^ (3 / 2)) - 3 Sqrt[b] / (4 x ^ (3 / 2) Sqrt[1 - a / (b x)])]

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Maple [A]
time = 0.11, size = 59, normalized size = 0.83

method	result	size
risch	\(\frac {\left (-b x +a \right ) \left (-b x +2 a \right )}{4 x^{2} \sqrt {b x -a}\, a}+\frac {b^{2} \arctan \left (\frac {\sqrt {b x -a}}{\sqrt {a}}\right )}{4 a^{\frac {3}{2}}}\)	\(55\)
derivativedivides	\(2 b^{2} \left (\frac {\frac {\left (b x -a \right )^{\frac {3}{2}}}{8 a}-\frac {\sqrt {b x -a}}{8}}{b^{2} x^{2}}+\frac {\arctan \left (\frac {\sqrt {b x -a}}{\sqrt {a}}\right )}{8 a^{\frac {3}{2}}}\right )\)	\(59\)
default	\(2 b^{2} \left (\frac {\frac {\left (b x -a \right )^{\frac {3}{2}}}{8 a}-\frac {\sqrt {b x -a}}{8}}{b^{2} x^{2}}+\frac {\arctan \left (\frac {\sqrt {b x -a}}{\sqrt {a}}\right )}{8 a^{\frac {3}{2}}}\right )\)	\(59\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*b^2*((1/8/a*(b*x-a)^(3/2)-1/8*(b*x-a)^(1/2))/b^2/x^2+1/8*arctan((b*x-a)^(1/2)/a^(1/2))/a^(3/2))

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Maxima [A]
time = 0.35, size = 83, normalized size = 1.17 \begin {gather*} \frac {b^{2} \arctan \left (\frac {\sqrt {b x - a}}{\sqrt {a}}\right )}{4 \, a^{\frac {3}{2}}} + \frac {{\left (b x - a\right )}^{\frac {3}{2}} b^{2} - \sqrt {b x - a} a b^{2}}{4 \, {\left ({\left (b x - a\right )}^{2} a + 2 \, {\left (b x - a\right )} a^{2} + a^{3}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*b^2*arctan(sqrt(b*x - a)/sqrt(a))/a^(3/2) + 1/4*((b*x - a)^(3/2)*b^2 - sqrt(b*x - a)*a*b^2)/((b*x - a)^2*a
+ 2*(b*x - a)*a^2 + a^3)

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Fricas [A]
time = 0.32, size = 124, normalized size = 1.75 \begin {gather*} \left [-\frac {\sqrt {-a} b^{2} x^{2} \log \left (\frac {b x - 2 \, \sqrt {b x - a} \sqrt {-a} - 2 \, a}{x}\right ) - 2 \, {\left (a b x - 2 \, a^{2}\right )} \sqrt {b x - a}}{8 \, a^{2} x^{2}}, \frac {\sqrt {a} b^{2} x^{2} \arctan \left (\frac {\sqrt {b x - a}}{\sqrt {a}}\right ) + {\left (a b x - 2 \, a^{2}\right )} \sqrt {b x - a}}{4 \, a^{2} x^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/8*(sqrt(-a)*b^2*x^2*log((b*x - 2*sqrt(b*x - a)*sqrt(-a) - 2*a)/x) - 2*(a*b*x - 2*a^2)*sqrt(b*x - a))/(a^2*
x^2), 1/4*(sqrt(a)*b^2*x^2*arctan(sqrt(b*x - a)/sqrt(a)) + (a*b*x - 2*a^2)*sqrt(b*x - a))/(a^2*x^2)]

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Sympy [A]
time = 2.05, size = 207, normalized size = 2.92 \begin {gather*} \begin {cases} - \frac {i a}{2 \sqrt {b} x^{\frac {5}{2}} \sqrt {\frac {a}{b x} - 1}} + \frac {3 i \sqrt {b}}{4 x^{\frac {3}{2}} \sqrt {\frac {a}{b x} - 1}} - \frac {i b^{\frac {3}{2}}}{4 a \sqrt {x} \sqrt {\frac {a}{b x} - 1}} + \frac {i b^{2} \operatorname {acosh}{\left (\frac {\sqrt {a}}{\sqrt {b} \sqrt {x}} \right )}}{4 a^{\frac {3}{2}}} & \text {for}\: \left |{\frac {a}{b x}}\right | > 1 \\\frac {a}{2 \sqrt {b} x^{\frac {5}{2}} \sqrt {- \frac {a}{b x} + 1}} - \frac {3 \sqrt {b}}{4 x^{\frac {3}{2}} \sqrt {- \frac {a}{b x} + 1}} + \frac {b^{\frac {3}{2}}}{4 a \sqrt {x} \sqrt {- \frac {a}{b x} + 1}} - \frac {b^{2} \operatorname {asin}{\left (\frac {\sqrt {a}}{\sqrt {b} \sqrt {x}} \right )}}{4 a^{\frac {3}{2}}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-I*a/(2*sqrt(b)*x**(5/2)*sqrt(a/(b*x) - 1)) + 3*I*sqrt(b)/(4*x**(3/2)*sqrt(a/(b*x) - 1)) - I*b**(3/
2)/(4*a*sqrt(x)*sqrt(a/(b*x) - 1)) + I*b**2*acosh(sqrt(a)/(sqrt(b)*sqrt(x)))/(4*a**(3/2)), Abs(a/(b*x)) > 1),
(a/(2*sqrt(b)*x**(5/2)*sqrt(-a/(b*x) + 1)) - 3*sqrt(b)/(4*x**(3/2)*sqrt(-a/(b*x) + 1)) + b**(3/2)/(4*a*sqrt(x)
*sqrt(-a/(b*x) + 1)) - b**2*asin(sqrt(a)/(sqrt(b)*sqrt(x)))/(4*a**(3/2)), True))

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Giac [A]
time = 0.00, size = 90, normalized size = 1.27 \begin {gather*} \frac {\frac {\sqrt {-a+b x} \left (-a+b x\right ) b^{3}-\sqrt {-a+b x} a b^{3}}{4 a \left (-a+b x+a\right )^{2}}+\frac {b^{3} \arctan \left (\frac {\sqrt {-a+b x}}{\sqrt {a}}\right )}{2 a\cdot 2 \sqrt {a}}}{b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(b^3*arctan(sqrt(b*x - a)/sqrt(a))/a^(3/2) + ((b*x - a)^(3/2)*b^3 - sqrt(b*x - a)*a*b^3)/(a*b^2*x^2))/b

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Mupad [B]
time = 0.10, size = 54, normalized size = 0.76 \begin {gather*} \frac {b^2\,\mathrm {atan}\left (\frac {\sqrt {b\,x-a}}{\sqrt {a}}\right )}{4\,a^{3/2}}-\frac {\sqrt {b\,x-a}}{4\,x^2}+\frac {{\left (b\,x-a\right )}^{3/2}}{4\,a\,x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x - a)^(1/2)/x^3,x)

[Out]

(b^2*atan((b*x - a)^(1/2)/a^(1/2)))/(4*a^(3/2)) - (b*x - a)^(1/2)/(4*x^2) + (b*x - a)^(3/2)/(4*a*x^2)

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