3.3.90 \(\int \frac {\sqrt {a+b x}}{x^3} \, dx\) [290]

Optimal. Leaf size=65 \[ -\frac {\sqrt {a+b x}}{2 x^2}-\frac {b \sqrt {a+b x}}{4 a x}+\frac {b^2 \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{4 a^{3/2}} \]

[Out]

1/4*b^2*arctanh((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 = 65, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {43, 44, 65, 214} \begin {gather*} \frac {b^2 \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{4 a^{3/2}}-\frac {\sqrt {a+b x}}{2 x^2}-\frac {b \sqrt {a+b x}}{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*ArcTanh[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 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 {\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 \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{4 a^{3/2}}\\ \end {align*}

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Mathematica [A]
time = 0.07, size = 55, normalized size = 0.85 \begin {gather*} -\frac {\sqrt {a+b x} (2 a+b x)}{4 a x^2}+\frac {b^2 \tanh ^{-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*(Sqrt[a + b*x]*(2*a + b*x))/(a*x^2) + (b^2*ArcTanh[Sqrt[a + b*x]/Sqrt[a]])/(4*a^(3/2))

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Mathics [A]
time = 4.09, size = 100, normalized size = 1.54 \begin {gather*} \frac {\frac {-2 a^{\frac {7}{2}} x \left (a+b x\right )}{b}-3 a^{\frac {5}{2}} x^2 \left (a+b x\right )-a^{\frac {3}{2}} b x^3 \left (a+b x\right )+a b^{\frac {5}{2}} x^{\frac {9}{2}} \text {ArcSinh}\left [\frac {\sqrt {a}}{\sqrt {b} \sqrt {x}}\right ] \left (\frac {a+b x}{b x}\right )^{\frac {3}{2}}}{4 a^{\frac {5}{2}} \sqrt {b} x^{\frac {9}{2}} \left (\frac {a+b x}{b x}\right )^{\frac {3}{2}}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-2 a ^ (7 / 2) x (a + b x) / b - 3 a ^ (5 / 2) x ^ 2 (a + b x) - a ^ (3 / 2) b x ^ 3 (a + b x) + a b ^ (5 / 2
) x ^ (9 / 2) ArcSinh[Sqrt[a] / (Sqrt[b] Sqrt[x])] ((a + b x) / (b x)) ^ (3 / 2)) / (4 a ^ (5 / 2) Sqrt[b] x ^
(9 / 2) ((a + b x) / (b x)) ^ (3 / 2))

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Maple [A]
time = 0.10, size = 54, normalized size = 0.83

method	result	size
risch	\(-\frac {\sqrt {b x +a}\, \left (b x +2 a \right )}{4 x^{2} a}+\frac {b^{2} \arctanh \left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{4 a^{\frac {3}{2}}}\)	\(44\)
derivativedivides	\(2 b^{2} \left (-\frac {\frac {\left (b x +a \right )^{\frac {3}{2}}}{8 a}+\frac {\sqrt {b x +a}}{8}}{b^{2} x^{2}}+\frac {\arctanh \left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{8 a^{\frac {3}{2}}}\right )\)	\(54\)
default	\(2 b^{2} \left (-\frac {\frac {\left (b x +a \right )^{\frac {3}{2}}}{8 a}+\frac {\sqrt {b x +a}}{8}}{b^{2} x^{2}}+\frac {\arctanh \left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{8 a^{\frac {3}{2}}}\right )\)	\(54\)

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

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Maxima [A]
time = 0.35, size = 88, normalized size = 1.35 \begin {gather*} -\frac {b^{2} \log \left (\frac {\sqrt {b x + a} - \sqrt {a}}{\sqrt {b x + a} + \sqrt {a}}\right )}{8 \, 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]

-1/8*b^2*log((sqrt(b*x + a) - sqrt(a))/(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 = 0.33, size = 119, normalized size = 1.83 \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}}{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) + (a*b*x + 2*a^2)*sqrt(b*x + a))/(a^2*x^2)]

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Sympy [A]
time = 2.11, size = 97, normalized size = 1.49 \begin {gather*} - \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 {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} \sqrt {x}} \right )}}{4 a^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-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)*s
qrt(a/(b*x) + 1)) + b**2*asinh(sqrt(a)/(sqrt(b)*sqrt(x)))/(4*a**(3/2))

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Giac [A]
time = 0.00, size = 89, normalized size = 1.37 \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))/(sqrt(-a)*a) + ((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 = 0.07, size = 48, normalized size = 0.74 \begin {gather*} \frac {b^2\,\mathrm {atanh}\left (\frac {\sqrt {a+b\,x}}{\sqrt {a}}\right )}{4\,a^{3/2}}-\frac {{\left (a+b\,x\right )}^{3/2}}{4\,a\,x^2}-\frac {\sqrt {a+b\,x}}{4\,x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x)^(1/2)/x^3,x)

[Out]

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

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