\(\int \sqrt {x} \sqrt {2+b x} \, dx\) [507]

Optimal result

Integrand size = 15, antiderivative size = 64 \[ \int \sqrt {x} \sqrt {2+b x} \, dx=\frac {\sqrt {x} \sqrt {2+b x}}{2 b}+\frac {1}{2} x^{3/2} \sqrt {2+b x}-\frac {\text {arcsinh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right )}{b^{3/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {52, 56, 221} \[ \int \sqrt {x} \sqrt {2+b x} \, dx=-\frac {\text {arcsinh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right )}{b^{3/2}}+\frac {1}{2} x^{3/2} \sqrt {b x+2}+\frac {\sqrt {x} \sqrt {b x+2}}{2 b} \]

[In]

Int[Sqrt[x]*Sqrt[2 + b*x],x]

[Out]

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

Rule 52

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 56

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

Rule 221

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

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

Mathematica [A] (verified)

Time = 0.16 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.02 \[ \int \sqrt {x} \sqrt {2+b x} \, dx=\frac {\sqrt {x} (1+b x) \sqrt {2+b x}}{2 b}+\frac {2 \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}-\sqrt {2+b x}}\right )}{b^{3/2}} \]

[In]

Integrate[Sqrt[x]*Sqrt[2 + b*x],x]

[Out]

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

Maple [A] (verified)

Time = 0.07 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.86

method	result	size
meijerg	\(-\frac {2 \left (-\frac {\sqrt {\pi }\, \sqrt {x}\, \sqrt {2}\, \sqrt {b}\, \left (3 b x +3\right ) \sqrt {\frac {b x}{2}+1}}{12}+\frac {\sqrt {\pi }\, \operatorname {arcsinh}\left (\frac {\sqrt {b}\, \sqrt {x}\, \sqrt {2}}{2}\right )}{2}\right )}{b^{\frac {3}{2}} \sqrt {\pi }}\)	\(55\)
risch	\(\frac {\left (b x +1\right ) \sqrt {x}\, \sqrt {b x +2}}{2 b}-\frac {\ln \left (\frac {b x +1}{\sqrt {b}}+\sqrt {b \,x^{2}+2 x}\right ) \sqrt {x \left (b x +2\right )}}{2 b^{\frac {3}{2}} \sqrt {x}\, \sqrt {b x +2}}\)	\(68\)
default	\(\frac {\sqrt {x}\, \left (b x +2\right )^{\frac {3}{2}}}{2 b}-\frac {\sqrt {x}\, \sqrt {b x +2}+\frac {\sqrt {x \left (b x +2\right )}\, \ln \left (\frac {b x +1}{\sqrt {b}}+\sqrt {b \,x^{2}+2 x}\right )}{\sqrt {b x +2}\, \sqrt {x}\, \sqrt {b}}}{2 b}\)	\(79\)

[In]

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

[Out]

-2/b^(3/2)/Pi^(1/2)*(-1/12*Pi^(1/2)*x^(1/2)*2^(1/2)*b^(1/2)*(3*b*x+3)*(1/2*b*x+1)^(1/2)+1/2*Pi^(1/2)*arcsinh(1
/2*b^(1/2)*x^(1/2)*2^(1/2)))

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.58 \[ \int \sqrt {x} \sqrt {2+b x} \, dx=\left [\frac {{\left (b^{2} x + b\right )} \sqrt {b x + 2} \sqrt {x} + \sqrt {b} \log \left (b x - \sqrt {b x + 2} \sqrt {b} \sqrt {x} + 1\right )}{2 \, b^{2}}, \frac {{\left (b^{2} x + b\right )} \sqrt {b x + 2} \sqrt {x} + 2 \, \sqrt {-b} \arctan \left (\frac {\sqrt {b x + 2} \sqrt {-b}}{b \sqrt {x}}\right )}{2 \, b^{2}}\right ] \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 2.15 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.11 \[ \int \sqrt {x} \sqrt {2+b x} \, dx=\frac {b x^{\frac {5}{2}}}{2 \sqrt {b x + 2}} + \frac {3 x^{\frac {3}{2}}}{2 \sqrt {b x + 2}} + \frac {\sqrt {x}}{b \sqrt {b x + 2}} - \frac {\operatorname {asinh}{\left (\frac {\sqrt {2} \sqrt {b} \sqrt {x}}{2} \right )}}{b^{\frac {3}{2}}} \]

[In]

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

[Out]

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 98 vs. \(2 (45) = 90\).

Time = 0.29 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.53 \[ \int \sqrt {x} \sqrt {2+b x} \, dx=\frac {\frac {\sqrt {b x + 2} b}{\sqrt {x}} + \frac {{\left (b x + 2\right )}^{\frac {3}{2}}}{x^{\frac {3}{2}}}}{b^{3} - \frac {2 \, {\left (b x + 2\right )} b^{2}}{x} + \frac {{\left (b x + 2\right )}^{2} b}{x^{2}}} + \frac {\log \left (-\frac {\sqrt {b} - \frac {\sqrt {b x + 2}}{\sqrt {x}}}{\sqrt {b} + \frac {\sqrt {b x + 2}}{\sqrt {x}}}\right )}{2 \, b^{\frac {3}{2}}} \]

[In]

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

[Out]

(sqrt(b*x + 2)*b/sqrt(x) + (b*x + 2)^(3/2)/x^(3/2))/(b^3 - 2*(b*x + 2)*b^2/x + (b*x + 2)^2*b/x^2) + 1/2*log(-(
sqrt(b) - sqrt(b*x + 2)/sqrt(x))/(sqrt(b) + sqrt(b*x + 2)/sqrt(x)))/b^(3/2)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 134 vs. \(2 (45) = 90\).

Time = 11.16 (sec) , antiderivative size = 134, normalized size of antiderivative = 2.09 \[ \int \sqrt {x} \sqrt {2+b x} \, dx=\frac {\frac {{\left (\sqrt {{\left (b x + 2\right )} b - 2 \, b} \sqrt {b x + 2} {\left (b x - 3\right )} - 6 \, \sqrt {b} \log \left ({\left | -\sqrt {b x + 2} \sqrt {b} + \sqrt {{\left (b x + 2\right )} b - 2 \, b} \right |}\right )\right )} {\left | b \right |}}{b^{2}} + \frac {4 \, {\left (2 \, \sqrt {b} \log \left ({\left | -\sqrt {b x + 2} \sqrt {b} + \sqrt {{\left (b x + 2\right )} b - 2 \, b} \right |}\right ) + \sqrt {{\left (b x + 2\right )} b - 2 \, b} \sqrt {b x + 2}\right )} {\left | b \right |}}{b^{2}}}{2 \, b} \]

[In]

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

[Out]

1/2*((sqrt((b*x + 2)*b - 2*b)*sqrt(b*x + 2)*(b*x - 3) - 6*sqrt(b)*log(abs(-sqrt(b*x + 2)*sqrt(b) + sqrt((b*x +
2)*b - 2*b))))*abs(b)/b^2 + 4*(2*sqrt(b)*log(abs(-sqrt(b*x + 2)*sqrt(b) + sqrt((b*x + 2)*b - 2*b))) + sqrt((b
*x + 2)*b - 2*b)*sqrt(b*x + 2))*abs(b)/b^2)/b

Mupad [B] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.72 \[ \int \sqrt {x} \sqrt {2+b x} \, dx=\sqrt {x}\,\left (\frac {x}{2}+\frac {1}{2\,b}\right )\,\sqrt {b\,x+2}-\frac {\ln \left (b\,x+\sqrt {b}\,\sqrt {x}\,\sqrt {b\,x+2}+1\right )}{2\,b^{3/2}} \]

[In]

int(x^(1/2)*(b*x + 2)^(1/2),x)

[Out]

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