\(\int \frac {1}{x^3 \sqrt {a+b x}} \, dx\) [341]

Optimal result

Integrand size = 13, antiderivative size = 68 \[ \int \frac {1}{x^3 \sqrt {a+b x}} \, dx=-\frac {\sqrt {a+b x}}{2 a x^2}+\frac {3 b \sqrt {a+b x}}{4 a^2 x}-\frac {3 b^2 \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{4 a^{5/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {44, 65, 214} \[ \int \frac {1}{x^3 \sqrt {a+b x}} \, dx=-\frac {3 b^2 \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{4 a^{5/2}}+\frac {3 b \sqrt {a+b x}}{4 a^2 x}-\frac {\sqrt {a+b x}}{2 a x^2} \]

[In]

Int[1/(x^3*Sqrt[a + b*x]),x]

[Out]

-1/2*Sqrt[a + b*x]/(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 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*} \text {integral}& = -\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) \text {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 [A] (verified)

Time = 0.07 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.82 \[ \int \frac {1}{x^3 \sqrt {a+b x}} \, dx=\frac {\sqrt {a+b x} (-2 a+3 b x)}{4 a^2 x^2}-\frac {3 b^2 \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{4 a^{5/2}} \]

[In]

Integrate[1/(x^3*Sqrt[a + b*x]),x]

[Out]

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

Maple [A] (verified)

Time = 0.07 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.66

method	result	size
risch	\(-\frac {\sqrt {b x +a}\, \left (-3 b x +2 a \right )}{4 a^{2} x^{2}}-\frac {3 b^{2} \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{4 a^{\frac {5}{2}}}\)	\(45\)
pseudoelliptic	\(\frac {-3 \,\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right ) b^{2} x^{2}-2 \sqrt {b x +a}\, a^{\frac {3}{2}}+3 b x \sqrt {b x +a}\, \sqrt {a}}{4 a^{\frac {5}{2}} x^{2}}\)	\(56\)
derivativedivides	\(2 b^{2} \left (-\frac {\sqrt {b x +a}}{4 a \,b^{2} x^{2}}-\frac {3 \left (-\frac {\sqrt {b x +a}}{2 a b x}+\frac {\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{2 a^{\frac {3}{2}}}\right )}{4 a}\right )\)	\(66\)
default	\(2 b^{2} \left (-\frac {\sqrt {b x +a}}{4 a \,b^{2} x^{2}}-\frac {3 \left (-\frac {\sqrt {b x +a}}{2 a b x}+\frac {\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{2 a^{\frac {3}{2}}}\right )}{4 a}\right )\)	\(66\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.81 \[ \int \frac {1}{x^3 \sqrt {a+b x}} \, dx=\left [\frac {3 \, \sqrt {a} b^{2} x^{2} \log \left (\frac {b x - 2 \, \sqrt {b x + a} \sqrt {a} + 2 \, a}{x}\right ) + 2 \, {\left (3 \, a b x - 2 \, a^{2}\right )} \sqrt {b x + a}}{8 \, a^{3} x^{2}}, \frac {3 \, \sqrt {-a} b^{2} x^{2} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-a}}{a}\right ) + {\left (3 \, a b x - 2 \, a^{2}\right )} \sqrt {b x + a}}{4 \, a^{3} x^{2}}\right ] \]

[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)]

Sympy [A] (verification not implemented)

Time = 3.38 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.50 \[ \int \frac {1}{x^3 \sqrt {a+b x}} \, dx=- \frac {1}{2 \sqrt {b} x^{\frac {5}{2}} \sqrt {\frac {a}{b x} + 1}} + \frac {\sqrt {b}}{4 a x^{\frac {3}{2}} \sqrt {\frac {a}{b x} + 1}} + \frac {3 b^{\frac {3}{2}}}{4 a^{2} \sqrt {x} \sqrt {\frac {a}{b x} + 1}} - \frac {3 b^{2} \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} \sqrt {x}} \right )}}{4 a^{\frac {5}{2}}} \]

[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))

Maxima [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.35 \[ \int \frac {1}{x^3 \sqrt {a+b x}} \, dx=\frac {3 \, b^{2} \log \left (\frac {\sqrt {b x + a} - \sqrt {a}}{\sqrt {b x + a} + \sqrt {a}}\right )}{8 \, a^{\frac {5}{2}}} + \frac {3 \, {\left (b x + a\right )}^{\frac {3}{2}} b^{2} - 5 \, \sqrt {b x + a} a b^{2}}{4 \, {\left ({\left (b x + a\right )}^{2} a^{2} - 2 \, {\left (b x + a\right )} a^{3} + a^{4}\right )}} \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.01 \[ \int \frac {1}{x^3 \sqrt {a+b x}} \, dx=\frac {\frac {3 \, b^{3} \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a^{2}} + \frac {3 \, {\left (b x + a\right )}^{\frac {3}{2}} b^{3} - 5 \, \sqrt {b x + a} a b^{3}}{a^{2} b^{2} x^{2}}}{4 \, b} \]

[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

Mupad [B] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.75 \[ \int \frac {1}{x^3 \sqrt {a+b x}} \, dx=\frac {3\,{\left (a+b\,x\right )}^{3/2}}{4\,a^2\,x^2}-\frac {5\,\sqrt {a+b\,x}}{4\,a\,x^2}-\frac {3\,b^2\,\mathrm {atanh}\left (\frac {\sqrt {a+b\,x}}{\sqrt {a}}\right )}{4\,a^{5/2}} \]

[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))