3.491 \(\int \sqrt{x} \sqrt{a+b x} \, dx\)

Optimal. Leaf size=74 \[ -\frac{a^2 \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a+b x}}\right )}{4 b^{3/2}}+\frac{1}{2} x^{3/2} \sqrt{a+b x}+\frac{a \sqrt{x} \sqrt{a+b x}}{4 b} \]

[Out]

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

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Rubi [A]  time = 0.0226927, antiderivative size = 74, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.267, Rules used = {50, 63, 217, 206} \[ -\frac{a^2 \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a+b x}}\right )}{4 b^{3/2}}+\frac{1}{2} x^{3/2} \sqrt{a+b x}+\frac{a \sqrt{x} \sqrt{a+b x}}{4 b} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[x]*Sqrt[a + b*x],x]

[Out]

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

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 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 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]

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])

Rubi steps

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

Mathematica [A]  time = 0.113867, size = 72, normalized size = 0.97 \[ \frac{\sqrt{a+b x} \left (\sqrt{b} \sqrt{x} (a+2 b x)-\frac{a^{3/2} \sinh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}}\right )}{\sqrt{\frac{b x}{a}+1}}\right )}{4 b^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[x]*Sqrt[a + b*x],x]

[Out]

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

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Maple [A]  time = 0.004, size = 81, normalized size = 1.1 \begin{align*}{\frac{1}{2}{x}^{{\frac{3}{2}}}\sqrt{bx+a}}+{\frac{a}{4\,b}\sqrt{x}\sqrt{bx+a}}-{\frac{{a}^{2}}{8}\sqrt{x \left ( bx+a \right ) }\ln \left ({ \left ({\frac{a}{2}}+bx \right ){\frac{1}{\sqrt{b}}}}+\sqrt{b{x}^{2}+ax} \right ){b}^{-{\frac{3}{2}}}{\frac{1}{\sqrt{x}}}{\frac{1}{\sqrt{bx+a}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*x^(3/2)*(b*x+a)^(1/2)+1/4*a*x^(1/2)*(b*x+a)^(1/2)/b-1/8*a^2/b^(3/2)*(x*(b*x+a))^(1/2)/x^(1/2)/(b*x+a)^(1/2
)*ln((1/2*a+b*x)/b^(1/2)+(b*x^2+a*x)^(1/2))

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.71643, size = 304, normalized size = 4.11 \begin{align*} \left [\frac{a^{2} \sqrt{b} \log \left (2 \, b x - 2 \, \sqrt{b x + a} \sqrt{b} \sqrt{x} + a\right ) + 2 \,{\left (2 \, b^{2} x + a b\right )} \sqrt{b x + a} \sqrt{x}}{8 \, b^{2}}, \frac{a^{2} \sqrt{-b} \arctan \left (\frac{\sqrt{b x + a} \sqrt{-b}}{b \sqrt{x}}\right ) +{\left (2 \, b^{2} x + a b\right )} \sqrt{b x + a} \sqrt{x}}{4 \, b^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 3.74969, size = 97, normalized size = 1.31 \begin{align*} \frac{a^{\frac{3}{2}} \sqrt{x}}{4 b \sqrt{1 + \frac{b x}{a}}} + \frac{3 \sqrt{a} x^{\frac{3}{2}}}{4 \sqrt{1 + \frac{b x}{a}}} - \frac{a^{2} \operatorname{asinh}{\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}} \right )}}{4 b^{\frac{3}{2}}} + \frac{b x^{\frac{5}{2}}}{2 \sqrt{a} \sqrt{1 + \frac{b x}{a}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out