\(\int x^{3/2} \sqrt {2+b x} \, dx\) [506]

Optimal result

Integrand size = 15, antiderivative size = 84 \[ \int x^{3/2} \sqrt {2+b x} \, dx=-\frac {\sqrt {x} \sqrt {2+b x}}{2 b^2}+\frac {x^{3/2} \sqrt {2+b x}}{6 b}+\frac {1}{3} x^{5/2} \sqrt {2+b x}+\frac {\text {arcsinh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right )}{b^{5/2}} \]

[Out]

arcsinh(1/2*b^(1/2)*x^(1/2)*2^(1/2))/b^(5/2)+1/6*x^(3/2)*(b*x+2)^(1/2)/b+1/3*x^(5/2)*(b*x+2)^(1/2)-1/2*x^(1/2)
*(b*x+2)^(1/2)/b^2

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {52, 56, 221} \[ \int x^{3/2} \sqrt {2+b x} \, dx=\frac {\text {arcsinh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right )}{b^{5/2}}-\frac {\sqrt {x} \sqrt {b x+2}}{2 b^2}+\frac {1}{3} x^{5/2} \sqrt {b x+2}+\frac {x^{3/2} \sqrt {b x+2}}{6 b} \]

[In]

Int[x^(3/2)*Sqrt[2 + b*x],x]

[Out]

-1/2*(Sqrt[x]*Sqrt[2 + b*x])/b^2 + (x^(3/2)*Sqrt[2 + b*x])/(6*b) + (x^(5/2)*Sqrt[2 + b*x])/3 + ArcSinh[(Sqrt[b
]*Sqrt[x])/Sqrt[2]]/b^(5/2)

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}{3} x^{5/2} \sqrt {2+b x}+\frac {1}{3} \int \frac {x^{3/2}}{\sqrt {2+b x}} \, dx \\ & = \frac {x^{3/2} \sqrt {2+b x}}{6 b}+\frac {1}{3} x^{5/2} \sqrt {2+b x}-\frac {\int \frac {\sqrt {x}}{\sqrt {2+b x}} \, dx}{2 b} \\ & = -\frac {\sqrt {x} \sqrt {2+b x}}{2 b^2}+\frac {x^{3/2} \sqrt {2+b x}}{6 b}+\frac {1}{3} x^{5/2} \sqrt {2+b x}+\frac {\int \frac {1}{\sqrt {x} \sqrt {2+b x}} \, dx}{2 b^2} \\ & = -\frac {\sqrt {x} \sqrt {2+b x}}{2 b^2}+\frac {x^{3/2} \sqrt {2+b x}}{6 b}+\frac {1}{3} x^{5/2} \sqrt {2+b x}+\frac {\text {Subst}\left (\int \frac {1}{\sqrt {2+b x^2}} \, dx,x,\sqrt {x}\right )}{b^2} \\ & = -\frac {\sqrt {x} \sqrt {2+b x}}{2 b^2}+\frac {x^{3/2} \sqrt {2+b x}}{6 b}+\frac {1}{3} x^{5/2} \sqrt {2+b x}+\frac {\sinh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right )}{b^{5/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.23 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.87 \[ \int x^{3/2} \sqrt {2+b x} \, dx=\frac {\sqrt {x} \sqrt {2+b x} \left (-3+b x+2 b^2 x^2\right )}{6 b^2}-\frac {2 \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}-\sqrt {2+b x}}\right )}{b^{5/2}} \]

[In]

Integrate[x^(3/2)*Sqrt[2 + b*x],x]

[Out]

(Sqrt[x]*Sqrt[2 + b*x]*(-3 + b*x + 2*b^2*x^2))/(6*b^2) - (2*ArcTanh[(Sqrt[b]*Sqrt[x])/(Sqrt[2] - Sqrt[2 + b*x]
)])/b^(5/2)

Maple [A] (verified)

Time = 0.07 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.75

method	result	size
meijerg	\(-\frac {4 \left (\frac {\sqrt {\pi }\, \sqrt {x}\, \sqrt {2}\, \sqrt {b}\, \left (-10 b^{2} x^{2}-5 b x +15\right ) \sqrt {\frac {b x}{2}+1}}{120}-\frac {\sqrt {\pi }\, \operatorname {arcsinh}\left (\frac {\sqrt {b}\, \sqrt {x}\, \sqrt {2}}{2}\right )}{4}\right )}{b^{\frac {5}{2}} \sqrt {\pi }}\)	\(63\)
risch	\(\frac {\left (2 b^{2} x^{2}+b x -3\right ) \sqrt {x}\, \sqrt {b x +2}}{6 b^{2}}+\frac {\ln \left (\frac {b x +1}{\sqrt {b}}+\sqrt {b \,x^{2}+2 x}\right ) \sqrt {x \left (b x +2\right )}}{2 b^{\frac {5}{2}} \sqrt {x}\, \sqrt {b x +2}}\)	\(76\)
default	\(\frac {x^{\frac {3}{2}} \left (b x +2\right )^{\frac {3}{2}}}{3 b}-\frac {\frac {\sqrt {x}\, \left (b x +2\right )^{\frac {3}{2}}}{2 b}-\frac {\sqrt {x}\, \sqrt {b x +2}+\frac {\sqrt {x \left (b x +2\right )}\, \ln \left (\frac {b x +1}{\sqrt {b}}+\sqrt {b \,x^{2}+2 x}\right )}{\sqrt {b x +2}\, \sqrt {x}\, \sqrt {b}}}{2 b}}{b}\)	\(100\)

[In]

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

[Out]

-4/b^(5/2)/Pi^(1/2)*(1/120*Pi^(1/2)*x^(1/2)*2^(1/2)*b^(1/2)*(-10*b^2*x^2-5*b*x+15)*(1/2*b*x+1)^(1/2)-1/4*Pi^(1
/2)*arcsinh(1/2*b^(1/2)*x^(1/2)*2^(1/2)))

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.44 \[ \int x^{3/2} \sqrt {2+b x} \, dx=\left [\frac {{\left (2 \, b^{3} x^{2} + b^{2} x - 3 \, b\right )} \sqrt {b x + 2} \sqrt {x} + 3 \, \sqrt {b} \log \left (b x + \sqrt {b x + 2} \sqrt {b} \sqrt {x} + 1\right )}{6 \, b^{3}}, \frac {{\left (2 \, b^{3} x^{2} + b^{2} x - 3 \, b\right )} \sqrt {b x + 2} \sqrt {x} - 6 \, \sqrt {-b} \arctan \left (\frac {\sqrt {b x + 2} \sqrt {-b}}{b \sqrt {x}}\right )}{6 \, b^{3}}\right ] \]

[In]

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

[Out]

[1/6*((2*b^3*x^2 + b^2*x - 3*b)*sqrt(b*x + 2)*sqrt(x) + 3*sqrt(b)*log(b*x + sqrt(b*x + 2)*sqrt(b)*sqrt(x) + 1)
)/b^3, 1/6*((2*b^3*x^2 + b^2*x - 3*b)*sqrt(b*x + 2)*sqrt(x) - 6*sqrt(-b)*arctan(sqrt(b*x + 2)*sqrt(-b)/(b*sqrt
(x))))/b^3]

Sympy [A] (verification not implemented)

Time = 5.44 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.07 \[ \int x^{3/2} \sqrt {2+b x} \, dx=\frac {b x^{\frac {7}{2}}}{3 \sqrt {b x + 2}} + \frac {5 x^{\frac {5}{2}}}{6 \sqrt {b x + 2}} - \frac {x^{\frac {3}{2}}}{6 b \sqrt {b x + 2}} - \frac {\sqrt {x}}{b^{2} \sqrt {b x + 2}} + \frac {\operatorname {asinh}{\left (\frac {\sqrt {2} \sqrt {b} \sqrt {x}}{2} \right )}}{b^{\frac {5}{2}}} \]

[In]

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

[Out]

b*x**(7/2)/(3*sqrt(b*x + 2)) + 5*x**(5/2)/(6*sqrt(b*x + 2)) - x**(3/2)/(6*b*sqrt(b*x + 2)) - sqrt(x)/(b**2*sqr
t(b*x + 2)) + asinh(sqrt(2)*sqrt(b)*sqrt(x)/2)/b**(5/2)

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 134 vs. \(2 (59) = 118\).

Time = 0.30 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.60 \[ \int x^{3/2} \sqrt {2+b x} \, dx=-\frac {\frac {3 \, \sqrt {b x + 2} b^{2}}{\sqrt {x}} + \frac {8 \, {\left (b x + 2\right )}^{\frac {3}{2}} b}{x^{\frac {3}{2}}} - \frac {3 \, {\left (b x + 2\right )}^{\frac {5}{2}}}{x^{\frac {5}{2}}}}{3 \, {\left (b^{5} - \frac {3 \, {\left (b x + 2\right )} b^{4}}{x} + \frac {3 \, {\left (b x + 2\right )}^{2} b^{3}}{x^{2}} - \frac {{\left (b x + 2\right )}^{3} b^{2}}{x^{3}}\right )}} - \frac {\log \left (-\frac {\sqrt {b} - \frac {\sqrt {b x + 2}}{\sqrt {x}}}{\sqrt {b} + \frac {\sqrt {b x + 2}}{\sqrt {x}}}\right )}{2 \, b^{\frac {5}{2}}} \]

[In]

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

[Out]

-1/3*(3*sqrt(b*x + 2)*b^2/sqrt(x) + 8*(b*x + 2)^(3/2)*b/x^(3/2) - 3*(b*x + 2)^(5/2)/x^(5/2))/(b^5 - 3*(b*x + 2
)*b^4/x + 3*(b*x + 2)^2*b^3/x^2 - (b*x + 2)^3*b^2/x^3) - 1/2*log(-(sqrt(b) - sqrt(b*x + 2)/sqrt(x))/(sqrt(b) +
sqrt(b*x + 2)/sqrt(x)))/b^(5/2)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 162 vs. \(2 (59) = 118\).

Time = 11.25 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.93 \[ \int x^{3/2} \sqrt {2+b x} \, dx=\frac {\frac {{\left (\sqrt {{\left (b x + 2\right )} b - 2 \, b} \sqrt {b x + 2} {\left ({\left (b x + 2\right )} {\left (\frac {2 \, {\left (b x + 2\right )}}{b^{2}} - \frac {13}{b^{2}}\right )} + \frac {33}{b^{2}}\right )} + \frac {30 \, \log \left ({\left | -\sqrt {b x + 2} \sqrt {b} + \sqrt {{\left (b x + 2\right )} b - 2 \, b} \right |}\right )}{b^{\frac {3}{2}}}\right )} {\left | b \right |}}{b} + \frac {6 \, {\left (\sqrt {{\left (b x + 2\right )} b - 2 \, b} \sqrt {b x + 2} {\left (b x - 3\right )} - 6 \, \sqrt {b} \log \left ({\left | -\sqrt {b x + 2} \sqrt {b} + \sqrt {{\left (b x + 2\right )} b - 2 \, b} \right |}\right )\right )} {\left | b \right |}}{b^{3}}}{6 \, b} \]

[In]

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

[Out]

1/6*((sqrt((b*x + 2)*b - 2*b)*sqrt(b*x + 2)*((b*x + 2)*(2*(b*x + 2)/b^2 - 13/b^2) + 33/b^2) + 30*log(abs(-sqrt
(b*x + 2)*sqrt(b) + sqrt((b*x + 2)*b - 2*b)))/b^(3/2))*abs(b)/b + 6*(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^3)/b

Mupad [F(-1)]

Timed out. \[ \int x^{3/2} \sqrt {2+b x} \, dx=\int x^{3/2}\,\sqrt {b\,x+2} \,d x \]

[In]

int(x^(3/2)*(b*x + 2)^(1/2),x)

[Out]

int(x^(3/2)*(b*x + 2)^(1/2), x)