\(\int x^{3/2} (2+b x)^{3/2} \, dx\) [534]

Optimal result

Integrand size = 15, antiderivative size = 105 \[ \int x^{3/2} (2+b x)^{3/2} \, dx=-\frac {3 \sqrt {x} \sqrt {2+b x}}{8 b^2}+\frac {x^{3/2} \sqrt {2+b x}}{8 b}+\frac {1}{4} x^{5/2} \sqrt {2+b x}+\frac {1}{4} x^{5/2} (2+b x)^{3/2}+\frac {3 \text {arcsinh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right )}{4 b^{5/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {52, 56, 221} \[ \int x^{3/2} (2+b x)^{3/2} \, dx=\frac {3 \text {arcsinh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right )}{4 b^{5/2}}-\frac {3 \sqrt {x} \sqrt {b x+2}}{8 b^2}+\frac {1}{4} x^{5/2} (b x+2)^{3/2}+\frac {1}{4} x^{5/2} \sqrt {b x+2}+\frac {x^{3/2} \sqrt {b x+2}}{8 b} \]

[In]

Int[x^(3/2)*(2 + b*x)^(3/2),x]

[Out]

(-3*Sqrt[x]*Sqrt[2 + b*x])/(8*b^2) + (x^(3/2)*Sqrt[2 + b*x])/(8*b) + (x^(5/2)*Sqrt[2 + b*x])/4 + (x^(5/2)*(2 +
b*x)^(3/2))/4 + (3*ArcSinh[(Sqrt[b]*Sqrt[x])/Sqrt[2]])/(4*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}{4} x^{5/2} (2+b x)^{3/2}+\frac {3}{4} \int x^{3/2} \sqrt {2+b x} \, dx \\ & = \frac {1}{4} x^{5/2} \sqrt {2+b x}+\frac {1}{4} x^{5/2} (2+b x)^{3/2}+\frac {1}{4} \int \frac {x^{3/2}}{\sqrt {2+b x}} \, dx \\ & = \frac {x^{3/2} \sqrt {2+b x}}{8 b}+\frac {1}{4} x^{5/2} \sqrt {2+b x}+\frac {1}{4} x^{5/2} (2+b x)^{3/2}-\frac {3 \int \frac {\sqrt {x}}{\sqrt {2+b x}} \, dx}{8 b} \\ & = -\frac {3 \sqrt {x} \sqrt {2+b x}}{8 b^2}+\frac {x^{3/2} \sqrt {2+b x}}{8 b}+\frac {1}{4} x^{5/2} \sqrt {2+b x}+\frac {1}{4} x^{5/2} (2+b x)^{3/2}+\frac {3 \int \frac {1}{\sqrt {x} \sqrt {2+b x}} \, dx}{8 b^2} \\ & = -\frac {3 \sqrt {x} \sqrt {2+b x}}{8 b^2}+\frac {x^{3/2} \sqrt {2+b x}}{8 b}+\frac {1}{4} x^{5/2} \sqrt {2+b x}+\frac {1}{4} x^{5/2} (2+b x)^{3/2}+\frac {3 \text {Subst}\left (\int \frac {1}{\sqrt {2+b x^2}} \, dx,x,\sqrt {x}\right )}{4 b^2} \\ & = -\frac {3 \sqrt {x} \sqrt {2+b x}}{8 b^2}+\frac {x^{3/2} \sqrt {2+b x}}{8 b}+\frac {1}{4} x^{5/2} \sqrt {2+b x}+\frac {1}{4} x^{5/2} (2+b x)^{3/2}+\frac {3 \sinh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right )}{4 b^{5/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.29 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.79 \[ \int x^{3/2} (2+b x)^{3/2} \, dx=\frac {\sqrt {x} \sqrt {2+b x} \left (-3+b x+6 b^2 x^2+2 b^3 x^3\right )}{8 b^2}-\frac {3 \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}-\sqrt {2+b x}}\right )}{2 b^{5/2}} \]

[In]

Integrate[x^(3/2)*(2 + b*x)^(3/2),x]

[Out]

(Sqrt[x]*Sqrt[2 + b*x]*(-3 + b*x + 6*b^2*x^2 + 2*b^3*x^3))/(8*b^2) - (3*ArcTanh[(Sqrt[b]*Sqrt[x])/(Sqrt[2] - S
qrt[2 + b*x])])/(2*b^(5/2))

Maple [A] (verified)

Time = 0.08 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.68

method	result	size
meijerg	\(\frac {-\frac {\sqrt {\pi }\, \sqrt {x}\, \sqrt {2}\, \sqrt {b}\, \left (-10 b^{3} x^{3}-30 b^{2} x^{2}-5 b x +15\right ) \sqrt {\frac {b x}{2}+1}}{40}+\frac {3 \sqrt {\pi }\, \operatorname {arcsinh}\left (\frac {\sqrt {b}\, \sqrt {x}\, \sqrt {2}}{2}\right )}{4}}{b^{\frac {5}{2}} \sqrt {\pi }}\)	\(71\)
risch	\(\frac {\left (2 b^{3} x^{3}+6 b^{2} x^{2}+b x -3\right ) \sqrt {x}\, \sqrt {b x +2}}{8 b^{2}}+\frac {3 \ln \left (\frac {b x +1}{\sqrt {b}}+\sqrt {b \,x^{2}+2 x}\right ) \sqrt {x \left (b x +2\right )}}{8 b^{\frac {5}{2}} \sqrt {x}\, \sqrt {b x +2}}\)	\(84\)
default	\(\frac {x^{\frac {3}{2}} \left (b x +2\right )^{\frac {5}{2}}}{4 b}-\frac {3 \left (\frac {\sqrt {x}\, \left (b x +2\right )^{\frac {5}{2}}}{3 b}-\frac {\frac {\left (b x +2\right )^{\frac {3}{2}} \sqrt {x}}{2}+\frac {3 \sqrt {x}\, \sqrt {b x +2}}{2}+\frac {3 \sqrt {x \left (b x +2\right )}\, \ln \left (\frac {b x +1}{\sqrt {b}}+\sqrt {b \,x^{2}+2 x}\right )}{2 \sqrt {b x +2}\, \sqrt {x}\, \sqrt {b}}}{3 b}\right )}{4 b}\)	\(114\)

[In]

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

[Out]

12/b^(5/2)/Pi^(1/2)*(-1/480*Pi^(1/2)*x^(1/2)*2^(1/2)*b^(1/2)*(-10*b^3*x^3-30*b^2*x^2-5*b*x+15)*(1/2*b*x+1)^(1/
2)+1/16*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 = 137, normalized size of antiderivative = 1.30 \[ \int x^{3/2} (2+b x)^{3/2} \, dx=\left [\frac {{\left (2 \, b^{4} x^{3} + 6 \, b^{3} x^{2} + b^{2} x - 3 \, b\right )} \sqrt {b x + 2} \sqrt {x} + 3 \, \sqrt {b} \log \left (b x + \sqrt {b x + 2} \sqrt {b} \sqrt {x} + 1\right )}{8 \, b^{3}}, \frac {{\left (2 \, b^{4} x^{3} + 6 \, b^{3} x^{2} + b^{2} x - 3 \, b\right )} \sqrt {b x + 2} \sqrt {x} - 6 \, \sqrt {-b} \arctan \left (\frac {\sqrt {b x + 2} \sqrt {-b}}{b \sqrt {x}}\right )}{8 \, b^{3}}\right ] \]

[In]

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

[Out]

[1/8*((2*b^4*x^3 + 6*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, 1/8*((2*b^4*x^3 + 6*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 = 16.73 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.11 \[ \int x^{3/2} (2+b x)^{3/2} \, dx=\frac {b^{2} x^{\frac {9}{2}}}{4 \sqrt {b x + 2}} + \frac {5 b x^{\frac {7}{2}}}{4 \sqrt {b x + 2}} + \frac {13 x^{\frac {5}{2}}}{8 \sqrt {b x + 2}} - \frac {x^{\frac {3}{2}}}{8 b \sqrt {b x + 2}} - \frac {3 \sqrt {x}}{4 b^{2} \sqrt {b x + 2}} + \frac {3 \operatorname {asinh}{\left (\frac {\sqrt {2} \sqrt {b} \sqrt {x}}{2} \right )}}{4 b^{\frac {5}{2}}} \]

[In]

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

[Out]

b**2*x**(9/2)/(4*sqrt(b*x + 2)) + 5*b*x**(7/2)/(4*sqrt(b*x + 2)) + 13*x**(5/2)/(8*sqrt(b*x + 2)) - x**(3/2)/(8
*b*sqrt(b*x + 2)) - 3*sqrt(x)/(4*b**2*sqrt(b*x + 2)) + 3*asinh(sqrt(2)*sqrt(b)*sqrt(x)/2)/(4*b**(5/2))

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 163 vs. \(2 (72) = 144\).

Time = 0.30 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.55 \[ \int x^{3/2} (2+b x)^{3/2} \, dx=-\frac {\frac {3 \, \sqrt {b x + 2} b^{3}}{\sqrt {x}} - \frac {11 \, {\left (b x + 2\right )}^{\frac {3}{2}} b^{2}}{x^{\frac {3}{2}}} - \frac {11 \, {\left (b x + 2\right )}^{\frac {5}{2}} b}{x^{\frac {5}{2}}} + \frac {3 \, {\left (b x + 2\right )}^{\frac {7}{2}}}{x^{\frac {7}{2}}}}{4 \, {\left (b^{6} - \frac {4 \, {\left (b x + 2\right )} b^{5}}{x} + \frac {6 \, {\left (b x + 2\right )}^{2} b^{4}}{x^{2}} - \frac {4 \, {\left (b x + 2\right )}^{3} b^{3}}{x^{3}} + \frac {{\left (b x + 2\right )}^{4} b^{2}}{x^{4}}\right )}} - \frac {3 \, \log \left (-\frac {\sqrt {b} - \frac {\sqrt {b x + 2}}{\sqrt {x}}}{\sqrt {b} + \frac {\sqrt {b x + 2}}{\sqrt {x}}}\right )}{8 \, b^{\frac {5}{2}}} \]

[In]

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

[Out]

-1/4*(3*sqrt(b*x + 2)*b^3/sqrt(x) - 11*(b*x + 2)^(3/2)*b^2/x^(3/2) - 11*(b*x + 2)^(5/2)*b/x^(5/2) + 3*(b*x + 2
)^(7/2)/x^(7/2))/(b^6 - 4*(b*x + 2)*b^5/x + 6*(b*x + 2)^2*b^4/x^2 - 4*(b*x + 2)^3*b^3/x^3 + (b*x + 2)^4*b^2/x^
4) - 3/8*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. 262 vs. \(2 (72) = 144\).

Time = 16.64 (sec) , antiderivative size = 262, normalized size of antiderivative = 2.50 \[ \int x^{3/2} (2+b x)^{3/2} \, dx=\frac {{\left ({\left ({\left (b x + 2\right )} {\left (2 \, {\left (b x + 2\right )} {\left (\frac {3 \, {\left (b x + 2\right )}}{b^{3}} - \frac {25}{b^{3}}\right )} + \frac {163}{b^{3}}\right )} - \frac {279}{b^{3}}\right )} \sqrt {{\left (b x + 2\right )} b - 2 \, b} \sqrt {b x + 2} - \frac {210 \, \log \left ({\left | -\sqrt {b x + 2} \sqrt {b} + \sqrt {{\left (b x + 2\right )} b - 2 \, b} \right |}\right )}{b^{\frac {5}{2}}}\right )} {\left | b \right |} + \frac {16 \, {\left (\sqrt {{\left (b x + 2\right )} b - 2 \, b} \sqrt {b x + 2} {\left ({\left (b x + 2\right )} {\left (\frac {2 \, {\left (b x + 2\right )}}{b^{2}} - \frac {13}{b^{2}}\right )} + \frac {33}{b^{2}}\right )} + \frac {30 \, \log \left ({\left | -\sqrt {b x + 2} \sqrt {b} + \sqrt {{\left (b x + 2\right )} b - 2 \, b} \right |}\right )}{b^{\frac {3}{2}}}\right )} {\left | b \right |}}{b} + \frac {48 \, {\left (\sqrt {{\left (b x + 2\right )} b - 2 \, b} \sqrt {b x + 2} {\left (b x - 3\right )} - 6 \, \sqrt {b} \log \left ({\left | -\sqrt {b x + 2} \sqrt {b} + \sqrt {{\left (b x + 2\right )} b - 2 \, b} \right |}\right )\right )} {\left | b \right |}}{b^{3}}}{24 \, b} \]

[In]

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

[Out]

1/24*((((b*x + 2)*(2*(b*x + 2)*(3*(b*x + 2)/b^3 - 25/b^3) + 163/b^3) - 279/b^3)*sqrt((b*x + 2)*b - 2*b)*sqrt(b
*x + 2) - 210*log(abs(-sqrt(b*x + 2)*sqrt(b) + sqrt((b*x + 2)*b - 2*b)))/b^(5/2))*abs(b) + 16*(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) + s
qrt((b*x + 2)*b - 2*b)))/b^(3/2))*abs(b)/b + 48*(sqrt((b*x + 2)*b - 2*b)*sqrt(b*x + 2)*(b*x - 3) - 6*sqrt(b)*l
og(abs(-sqrt(b*x + 2)*sqrt(b) + sqrt((b*x + 2)*b - 2*b))))*abs(b)/b^3)/b

Mupad [F(-1)]

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

[In]

int(x^(3/2)*(b*x + 2)^(3/2),x)

[Out]

int(x^(3/2)*(b*x + 2)^(3/2), x)