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

Optimal. Leaf size=48 \[ \frac {2 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )}{b^{3/2}}-\frac {2 \sqrt {x}}{b \sqrt {a+b x}} \]

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Rubi [A]  time = 0.02, antiderivative size = 48, 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 = {47, 63, 217, 206} \begin {gather*} \frac {2 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )}{b^{3/2}}-\frac {2 \sqrt {x}}{b \sqrt {a+b x}} \end {gather*}

Antiderivative was successfully verified.

[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 47

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 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 206

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

Rule 217

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*} \int \frac {\sqrt {x}}{(a+b x)^{3/2}} \, dx &=-\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 \operatorname {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 \operatorname {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*}

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Mathematica [A]  time = 0.07, size = 64, normalized size = 1.33 \begin {gather*} \frac {2 \left (\sqrt {a} \sqrt {\frac {b x}{a}+1} \sinh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )-\sqrt {b} \sqrt {x}\right )}{b^{3/2} \sqrt {a+b x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[x]/(a + b*x)^(3/2),x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.08, size = 50, normalized size = 1.04 \begin {gather*} -\frac {2 \log \left (\sqrt {a+b x}-\sqrt {b} \sqrt {x}\right )}{b^{3/2}}-\frac {2 \sqrt {x}}{b \sqrt {a+b x}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[Sqrt[x]/(a + b*x)^(3/2),x]

[Out]

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

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fricas [A]  time = 1.33, size = 119, normalized size = 2.48 \begin {gather*} \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 ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)]

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giac [B]  time = 94.86, size = 85, normalized size = 1.77 \begin {gather*} -\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}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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maple [F]  time = 0.04, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {x}}{\left (b x +a \right )^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 2.98, size = 57, normalized size = 1.19 \begin {gather*} -\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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {\sqrt {x}}{{\left (a+b\,x\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(1/2)/(a + b*x)^(3/2),x)

[Out]

int(x^(1/2)/(a + b*x)^(3/2), x)

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sympy [A]  time = 1.78, size = 46, normalized size = 0.96 \begin {gather*} \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}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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