\(\int \frac {\sqrt {x}}{(a+b x)^{3/2}} \, dx\) [579]

Optimal result

Integrand size = 15, antiderivative size = 48 \[ \int \frac {\sqrt {x}}{(a+b x)^{3/2}} \, dx=-\frac {2 \sqrt {x}}{b \sqrt {a+b x}}+\frac {2 \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )}{b^{3/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {49, 65, 223, 212} \[ \int \frac {\sqrt {x}}{(a+b x)^{3/2}} \, dx=\frac {2 \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )}{b^{3/2}}-\frac {2 \sqrt {x}}{b \sqrt {a+b x}} \]

[In]

Int[Sqrt[x]/(a + b*x)^(3/2),x]

[Out]

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

Rule 49

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 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 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 223

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

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

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.21 \[ \int \frac {\sqrt {x}}{(a+b x)^{3/2}} \, dx=-\frac {2 \sqrt {x}}{b \sqrt {a+b x}}+\frac {4 \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{-\sqrt {a}+\sqrt {a+b x}}\right )}{b^{3/2}} \]

[In]

Integrate[Sqrt[x]/(a + b*x)^(3/2),x]

[Out]

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

Maple [F]

\[\int \frac {\sqrt {x}}{\left (b x +a \right )^{\frac {3}{2}}}d x\]

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 119, normalized size of antiderivative = 2.48 \[ \int \frac {\sqrt {x}}{(a+b x)^{3/2}} \, dx=\left [\frac {{\left (b x + a\right )} \sqrt {b} \log \left (2 \, b x + 2 \, \sqrt {b x + a} \sqrt {b} \sqrt {x} + a\right ) - 2 \, \sqrt {b x + a} b \sqrt {x}}{b^{3} x + a b^{2}}, -\frac {2 \, {\left ({\left (b x + a\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-b}}{b \sqrt {x}}\right ) + \sqrt {b x + a} b \sqrt {x}\right )}}{b^{3} x + a b^{2}}\right ] \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 1.19 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.96 \[ \int \frac {\sqrt {x}}{(a+b x)^{3/2}} \, dx=\frac {2 \operatorname {asinh}{\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}} \right )}}{b^{\frac {3}{2}}} - \frac {2 \sqrt {x}}{\sqrt {a} b \sqrt {1 + \frac {b x}{a}}} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.19 \[ \int \frac {\sqrt {x}}{(a+b x)^{3/2}} \, dx=-\frac {\log \left (-\frac {\sqrt {b} - \frac {\sqrt {b x + a}}{\sqrt {x}}}{\sqrt {b} + \frac {\sqrt {b x + a}}{\sqrt {x}}}\right )}{b^{\frac {3}{2}}} - \frac {2 \, \sqrt {x}}{\sqrt {b x + a} b} \]

[In]

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

[Out]

-log(-(sqrt(b) - sqrt(b*x + a)/sqrt(x))/(sqrt(b) + sqrt(b*x + a)/sqrt(x)))/b^(3/2) - 2*sqrt(x)/(sqrt(b*x + a)*
b)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 85 vs. \(2 (36) = 72\).

Time = 16.25 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.77 \[ \int \frac {\sqrt {x}}{(a+b x)^{3/2}} \, dx=-\frac {{\left (\frac {4 \, a \sqrt {b}}{{\left (\sqrt {b x + a} \sqrt {b} - \sqrt {{\left (b x + a\right )} b - a b}\right )}^{2} + a b} + \frac {\log \left ({\left (\sqrt {b x + a} \sqrt {b} - \sqrt {{\left (b x + a\right )} b - a b}\right )}^{2}\right )}{\sqrt {b}}\right )} {\left | b \right |}}{b^{2}} \]

[In]

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

[Out]

-(4*a*sqrt(b)/((sqrt(b*x + a)*sqrt(b) - sqrt((b*x + a)*b - a*b))^2 + a*b) + log((sqrt(b*x + a)*sqrt(b) - sqrt(
(b*x + a)*b - a*b))^2)/sqrt(b))*abs(b)/b^2

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt {x}}{(a+b x)^{3/2}} \, dx=\int \frac {\sqrt {x}}{{\left (a+b\,x\right )}^{3/2}} \,d x \]

[In]

int(x^(1/2)/(a + b*x)^(3/2),x)

[Out]

int(x^(1/2)/(a + b*x)^(3/2), x)