3.4.41 \(\int \frac {1}{x^3 \sqrt {a+b x}} \, dx\)

Optimal. Leaf size=68 \[ -\frac {3 b^2 \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{4 a^{5/2}}+\frac {3 b \sqrt {a+b x}}{4 a^2 x}-\frac {\sqrt {a+b x}}{2 a x^2} \]

________________________________________________________________________________________

Rubi [A]  time = 0.02, antiderivative size = 68, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {51, 63, 208} \begin {gather*} -\frac {3 b^2 \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{4 a^{5/2}}+\frac {3 b \sqrt {a+b x}}{4 a^2 x}-\frac {\sqrt {a+b x}}{2 a x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x^3*Sqrt[a + b*x]),x]

[Out]

-Sqrt[a + b*x]/(2*a*x^2) + (3*b*Sqrt[a + b*x])/(4*a^2*x) - (3*b^2*ArcTanh[Sqrt[a + b*x]/Sqrt[a]])/(4*a^(5/2))

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 208

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

Rubi steps

\begin {align*} \int \frac {1}{x^3 \sqrt {a+b x}} \, dx &=-\frac {\sqrt {a+b x}}{2 a x^2}-\frac {(3 b) \int \frac {1}{x^2 \sqrt {a+b x}} \, dx}{4 a}\\ &=-\frac {\sqrt {a+b x}}{2 a x^2}+\frac {3 b \sqrt {a+b x}}{4 a^2 x}+\frac {\left (3 b^2\right ) \int \frac {1}{x \sqrt {a+b x}} \, dx}{8 a^2}\\ &=-\frac {\sqrt {a+b x}}{2 a x^2}+\frac {3 b \sqrt {a+b x}}{4 a^2 x}+\frac {(3 b) \operatorname {Subst}\left (\int \frac {1}{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+b x}\right )}{4 a^2}\\ &=-\frac {\sqrt {a+b x}}{2 a x^2}+\frac {3 b \sqrt {a+b x}}{4 a^2 x}-\frac {3 b^2 \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{4 a^{5/2}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C]  time = 0.01, size = 33, normalized size = 0.49 \begin {gather*} -\frac {2 b^2 \sqrt {a+b x} \, _2F_1\left (\frac {1}{2},3;\frac {3}{2};\frac {b x}{a}+1\right )}{a^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(x^3*Sqrt[a + b*x]),x]

[Out]

(-2*b^2*Sqrt[a + b*x]*Hypergeometric2F1[1/2, 3, 3/2, 1 + (b*x)/a])/a^3

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 0.08, size = 63, normalized size = 0.93 \begin {gather*} -\frac {3 b^2 \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{4 a^{5/2}}-\frac {5 a \sqrt {a+b x}-3 (a+b x)^{3/2}}{4 a^2 x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/4*(5*a*Sqrt[a + b*x] - 3*(a + b*x)^(3/2))/(a^2*x^2) - (3*b^2*ArcTanh[Sqrt[a + b*x]/Sqrt[a]])/(4*a^(5/2))

________________________________________________________________________________________

fricas [A]  time = 1.12, size = 123, normalized size = 1.81 \begin {gather*} \left [\frac {3 \, \sqrt {a} b^{2} x^{2} \log \left (\frac {b x - 2 \, \sqrt {b x + a} \sqrt {a} + 2 \, a}{x}\right ) + 2 \, {\left (3 \, a b x - 2 \, a^{2}\right )} \sqrt {b x + a}}{8 \, a^{3} x^{2}}, \frac {3 \, \sqrt {-a} b^{2} x^{2} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-a}}{a}\right ) + {\left (3 \, a b x - 2 \, a^{2}\right )} \sqrt {b x + a}}{4 \, a^{3} x^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/8*(3*sqrt(a)*b^2*x^2*log((b*x - 2*sqrt(b*x + a)*sqrt(a) + 2*a)/x) + 2*(3*a*b*x - 2*a^2)*sqrt(b*x + a))/(a^3
*x^2), 1/4*(3*sqrt(-a)*b^2*x^2*arctan(sqrt(b*x + a)*sqrt(-a)/a) + (3*a*b*x - 2*a^2)*sqrt(b*x + a))/(a^3*x^2)]

________________________________________________________________________________________

giac [A]  time = 1.19, size = 69, normalized size = 1.01 \begin {gather*} \frac {\frac {3 \, b^{3} \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a^{2}} + \frac {3 \, {\left (b x + a\right )}^{\frac {3}{2}} b^{3} - 5 \, \sqrt {b x + a} a b^{3}}{a^{2} b^{2} x^{2}}}{4 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(3*b^3*arctan(sqrt(b*x + a)/sqrt(-a))/(sqrt(-a)*a^2) + (3*(b*x + a)^(3/2)*b^3 - 5*sqrt(b*x + a)*a*b^3)/(a^
2*b^2*x^2))/b

________________________________________________________________________________________

maple [A]  time = 0.01, size = 66, normalized size = 0.97 \begin {gather*} 2 \left (-\frac {3 \left (\frac {\arctanh \left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{2 a^{\frac {3}{2}}}-\frac {\sqrt {b x +a}}{2 a b x}\right )}{4 a}-\frac {\sqrt {b x +a}}{4 a \,b^{2} x^{2}}\right ) b^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 2.92, size = 92, normalized size = 1.35 \begin {gather*} \frac {3 \, b^{2} \log \left (\frac {\sqrt {b x + a} - \sqrt {a}}{\sqrt {b x + a} + \sqrt {a}}\right )}{8 \, a^{\frac {5}{2}}} + \frac {3 \, {\left (b x + a\right )}^{\frac {3}{2}} b^{2} - 5 \, \sqrt {b x + a} a b^{2}}{4 \, {\left ({\left (b x + a\right )}^{2} a^{2} - 2 \, {\left (b x + a\right )} a^{3} + a^{4}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/8*b^2*log((sqrt(b*x + a) - sqrt(a))/(sqrt(b*x + a) + sqrt(a)))/a^(5/2) + 1/4*(3*(b*x + a)^(3/2)*b^2 - 5*sqrt
(b*x + a)*a*b^2)/((b*x + a)^2*a^2 - 2*(b*x + a)*a^3 + a^4)

________________________________________________________________________________________

mupad [B]  time = 0.06, size = 51, normalized size = 0.75 \begin {gather*} \frac {3\,{\left (a+b\,x\right )}^{3/2}}{4\,a^2\,x^2}-\frac {5\,\sqrt {a+b\,x}}{4\,a\,x^2}-\frac {3\,b^2\,\mathrm {atanh}\left (\frac {\sqrt {a+b\,x}}{\sqrt {a}}\right )}{4\,a^{5/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^3*(a + b*x)^(1/2)),x)

[Out]

(3*(a + b*x)^(3/2))/(4*a^2*x^2) - (5*(a + b*x)^(1/2))/(4*a*x^2) - (3*b^2*atanh((a + b*x)^(1/2)/a^(1/2)))/(4*a^
(5/2))

________________________________________________________________________________________

sympy [A]  time = 4.37, size = 102, normalized size = 1.50 \begin {gather*} - \frac {1}{2 \sqrt {b} x^{\frac {5}{2}} \sqrt {\frac {a}{b x} + 1}} + \frac {\sqrt {b}}{4 a x^{\frac {3}{2}} \sqrt {\frac {a}{b x} + 1}} + \frac {3 b^{\frac {3}{2}}}{4 a^{2} \sqrt {x} \sqrt {\frac {a}{b x} + 1}} - \frac {3 b^{2} \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} \sqrt {x}} \right )}}{4 a^{\frac {5}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/(2*sqrt(b)*x**(5/2)*sqrt(a/(b*x) + 1)) + sqrt(b)/(4*a*x**(3/2)*sqrt(a/(b*x) + 1)) + 3*b**(3/2)/(4*a**2*sqrt
(x)*sqrt(a/(b*x) + 1)) - 3*b**2*asinh(sqrt(a)/(sqrt(b)*sqrt(x)))/(4*a**(5/2))

________________________________________________________________________________________