\(\int x^{5/2} (2+b x)^{3/2} \, dx\) [533]

Optimal result

Integrand size = 15, antiderivative size = 126 \[ \int x^{5/2} (2+b x)^{3/2} \, dx=\frac {3 \sqrt {x} \sqrt {2+b x}}{8 b^3}-\frac {x^{3/2} \sqrt {2+b x}}{8 b^2}+\frac {x^{5/2} \sqrt {2+b x}}{20 b}+\frac {3}{20} x^{7/2} \sqrt {2+b x}+\frac {1}{5} x^{7/2} (2+b x)^{3/2}-\frac {3 \text {arcsinh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right )}{4 b^{7/2}} \]

[Out]

1/5*x^(7/2)*(b*x+2)^(3/2)-3/4*arcsinh(1/2*b^(1/2)*x^(1/2)*2^(1/2))/b^(7/2)-1/8*x^(3/2)*(b*x+2)^(1/2)/b^2+1/20*
x^(5/2)*(b*x+2)^(1/2)/b+3/20*x^(7/2)*(b*x+2)^(1/2)+3/8*x^(1/2)*(b*x+2)^(1/2)/b^3

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {52, 56, 221} \[ \int x^{5/2} (2+b x)^{3/2} \, dx=-\frac {3 \text {arcsinh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right )}{4 b^{7/2}}+\frac {3 \sqrt {x} \sqrt {b x+2}}{8 b^3}-\frac {x^{3/2} \sqrt {b x+2}}{8 b^2}+\frac {1}{5} x^{7/2} (b x+2)^{3/2}+\frac {3}{20} x^{7/2} \sqrt {b x+2}+\frac {x^{5/2} \sqrt {b x+2}}{20 b} \]

[In]

Int[x^(5/2)*(2 + b*x)^(3/2),x]

[Out]

(3*Sqrt[x]*Sqrt[2 + b*x])/(8*b^3) - (x^(3/2)*Sqrt[2 + b*x])/(8*b^2) + (x^(5/2)*Sqrt[2 + b*x])/(20*b) + (3*x^(7
/2)*Sqrt[2 + b*x])/20 + (x^(7/2)*(2 + b*x)^(3/2))/5 - (3*ArcSinh[(Sqrt[b]*Sqrt[x])/Sqrt[2]])/(4*b^(7/2))

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}{5} x^{7/2} (2+b x)^{3/2}+\frac {3}{5} \int x^{5/2} \sqrt {2+b x} \, dx \\ & = \frac {3}{20} x^{7/2} \sqrt {2+b x}+\frac {1}{5} x^{7/2} (2+b x)^{3/2}+\frac {3}{20} \int \frac {x^{5/2}}{\sqrt {2+b x}} \, dx \\ & = \frac {x^{5/2} \sqrt {2+b x}}{20 b}+\frac {3}{20} x^{7/2} \sqrt {2+b x}+\frac {1}{5} x^{7/2} (2+b x)^{3/2}-\frac {\int \frac {x^{3/2}}{\sqrt {2+b x}} \, dx}{4 b} \\ & = -\frac {x^{3/2} \sqrt {2+b x}}{8 b^2}+\frac {x^{5/2} \sqrt {2+b x}}{20 b}+\frac {3}{20} x^{7/2} \sqrt {2+b x}+\frac {1}{5} x^{7/2} (2+b x)^{3/2}+\frac {3 \int \frac {\sqrt {x}}{\sqrt {2+b x}} \, dx}{8 b^2} \\ & = \frac {3 \sqrt {x} \sqrt {2+b x}}{8 b^3}-\frac {x^{3/2} \sqrt {2+b x}}{8 b^2}+\frac {x^{5/2} \sqrt {2+b x}}{20 b}+\frac {3}{20} x^{7/2} \sqrt {2+b x}+\frac {1}{5} x^{7/2} (2+b x)^{3/2}-\frac {3 \int \frac {1}{\sqrt {x} \sqrt {2+b x}} \, dx}{8 b^3} \\ & = \frac {3 \sqrt {x} \sqrt {2+b x}}{8 b^3}-\frac {x^{3/2} \sqrt {2+b x}}{8 b^2}+\frac {x^{5/2} \sqrt {2+b x}}{20 b}+\frac {3}{20} x^{7/2} \sqrt {2+b x}+\frac {1}{5} x^{7/2} (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^3} \\ & = \frac {3 \sqrt {x} \sqrt {2+b x}}{8 b^3}-\frac {x^{3/2} \sqrt {2+b x}}{8 b^2}+\frac {x^{5/2} \sqrt {2+b x}}{20 b}+\frac {3}{20} x^{7/2} \sqrt {2+b x}+\frac {1}{5} x^{7/2} (2+b x)^{3/2}-\frac {3 \sinh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right )}{4 b^{7/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.38 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.73 \[ \int x^{5/2} (2+b x)^{3/2} \, dx=\frac {\sqrt {x} \sqrt {2+b x} \left (15-5 b x+2 b^2 x^2+22 b^3 x^3+8 b^4 x^4\right )}{40 b^3}+\frac {3 \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}-\sqrt {2+b x}}\right )}{2 b^{7/2}} \]

[In]

Integrate[x^(5/2)*(2 + b*x)^(3/2),x]

[Out]

(Sqrt[x]*Sqrt[2 + b*x]*(15 - 5*b*x + 2*b^2*x^2 + 22*b^3*x^3 + 8*b^4*x^4))/(40*b^3) + (3*ArcTanh[(Sqrt[b]*Sqrt[
x])/(Sqrt[2] - Sqrt[2 + b*x])])/(2*b^(7/2))

Maple [A] (verified)

Time = 0.20 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.63

method	result	size
meijerg	\(\frac {\frac {\sqrt {\pi }\, \sqrt {x}\, \sqrt {2}\, \sqrt {b}\, \left (56 b^{4} x^{4}+154 b^{3} x^{3}+14 b^{2} x^{2}-35 b x +105\right ) \sqrt {\frac {b x}{2}+1}}{280}-\frac {3 \sqrt {\pi }\, \operatorname {arcsinh}\left (\frac {\sqrt {b}\, \sqrt {x}\, \sqrt {2}}{2}\right )}{4}}{b^{\frac {7}{2}} \sqrt {\pi }}\)	\(79\)
risch	\(\frac {\left (8 b^{4} x^{4}+22 b^{3} x^{3}+2 b^{2} x^{2}-5 b x +15\right ) \sqrt {x}\, \sqrt {b x +2}}{40 b^{3}}-\frac {3 \ln \left (\frac {b x +1}{\sqrt {b}}+\sqrt {b \,x^{2}+2 x}\right ) \sqrt {x \left (b x +2\right )}}{8 b^{\frac {7}{2}} \sqrt {x}\, \sqrt {b x +2}}\)	\(93\)
default	\(\frac {x^{\frac {5}{2}} \left (b x +2\right )^{\frac {5}{2}}}{5 b}-\frac {\frac {x^{\frac {3}{2}} \left (b x +2\right )^{\frac {5}{2}}}{4 b}-\frac {3 \left (\frac {\sqrt {x}\, \left (b x +2\right )^{\frac {5}{2}}}{3 b}-\frac {\frac {\left (b x +2\right )^{\frac {3}{2}} \sqrt {x}}{2}+\frac {3 \sqrt {x}\, \sqrt {b x +2}}{2}+\frac {3 \sqrt {x \left (b x +2\right )}\, \ln \left (\frac {b x +1}{\sqrt {b}}+\sqrt {b \,x^{2}+2 x}\right )}{2 \sqrt {b x +2}\, \sqrt {x}\, \sqrt {b}}}{3 b}\right )}{4 b}}{b}\)	\(135\)

[In]

int(x^(5/2)*(b*x+2)^(3/2),x,method=_RETURNVERBOSE)

[Out]

24/b^(7/2)/Pi^(1/2)*(1/6720*Pi^(1/2)*x^(1/2)*2^(1/2)*b^(1/2)*(56*b^4*x^4+154*b^3*x^3+14*b^2*x^2-35*b*x+105)*(1
/2*b*x+1)^(1/2)-1/32*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 = 156, normalized size of antiderivative = 1.24 \[ \int x^{5/2} (2+b x)^{3/2} \, dx=\left [\frac {{\left (8 \, b^{5} x^{4} + 22 \, b^{4} x^{3} + 2 \, b^{3} x^{2} - 5 \, b^{2} x + 15 \, b\right )} \sqrt {b x + 2} \sqrt {x} + 15 \, \sqrt {b} \log \left (b x - \sqrt {b x + 2} \sqrt {b} \sqrt {x} + 1\right )}{40 \, b^{4}}, \frac {{\left (8 \, b^{5} x^{4} + 22 \, b^{4} x^{3} + 2 \, b^{3} x^{2} - 5 \, b^{2} x + 15 \, b\right )} \sqrt {b x + 2} \sqrt {x} + 30 \, \sqrt {-b} \arctan \left (\frac {\sqrt {b x + 2} \sqrt {-b}}{b \sqrt {x}}\right )}{40 \, b^{4}}\right ] \]

[In]

integrate(x^(5/2)*(b*x+2)^(3/2),x, algorithm="fricas")

[Out]

[1/40*((8*b^5*x^4 + 22*b^4*x^3 + 2*b^3*x^2 - 5*b^2*x + 15*b)*sqrt(b*x + 2)*sqrt(x) + 15*sqrt(b)*log(b*x - sqrt
(b*x + 2)*sqrt(b)*sqrt(x) + 1))/b^4, 1/40*((8*b^5*x^4 + 22*b^4*x^3 + 2*b^3*x^2 - 5*b^2*x + 15*b)*sqrt(b*x + 2)
*sqrt(x) + 30*sqrt(-b)*arctan(sqrt(b*x + 2)*sqrt(-b)/(b*sqrt(x))))/b^4]

Sympy [A] (verification not implemented)

Time = 81.39 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.08 \[ \int x^{5/2} (2+b x)^{3/2} \, dx=\frac {b^{2} x^{\frac {11}{2}}}{5 \sqrt {b x + 2}} + \frac {19 b x^{\frac {9}{2}}}{20 \sqrt {b x + 2}} + \frac {23 x^{\frac {7}{2}}}{20 \sqrt {b x + 2}} - \frac {x^{\frac {5}{2}}}{40 b \sqrt {b x + 2}} + \frac {x^{\frac {3}{2}}}{8 b^{2} \sqrt {b x + 2}} + \frac {3 \sqrt {x}}{4 b^{3} \sqrt {b x + 2}} - \frac {3 \operatorname {asinh}{\left (\frac {\sqrt {2} \sqrt {b} \sqrt {x}}{2} \right )}}{4 b^{\frac {7}{2}}} \]

[In]

integrate(x**(5/2)*(b*x+2)**(3/2),x)

[Out]

b**2*x**(11/2)/(5*sqrt(b*x + 2)) + 19*b*x**(9/2)/(20*sqrt(b*x + 2)) + 23*x**(7/2)/(20*sqrt(b*x + 2)) - x**(5/2
)/(40*b*sqrt(b*x + 2)) + x**(3/2)/(8*b**2*sqrt(b*x + 2)) + 3*sqrt(x)/(4*b**3*sqrt(b*x + 2)) - 3*asinh(sqrt(2)*
sqrt(b)*sqrt(x)/2)/(4*b**(7/2))

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 194 vs. \(2 (87) = 174\).

Time = 0.30 (sec) , antiderivative size = 194, normalized size of antiderivative = 1.54 \[ \int x^{5/2} (2+b x)^{3/2} \, dx=\frac {\frac {15 \, \sqrt {b x + 2} b^{4}}{\sqrt {x}} - \frac {70 \, {\left (b x + 2\right )}^{\frac {3}{2}} b^{3}}{x^{\frac {3}{2}}} - \frac {128 \, {\left (b x + 2\right )}^{\frac {5}{2}} b^{2}}{x^{\frac {5}{2}}} + \frac {70 \, {\left (b x + 2\right )}^{\frac {7}{2}} b}{x^{\frac {7}{2}}} - \frac {15 \, {\left (b x + 2\right )}^{\frac {9}{2}}}{x^{\frac {9}{2}}}}{20 \, {\left (b^{8} - \frac {5 \, {\left (b x + 2\right )} b^{7}}{x} + \frac {10 \, {\left (b x + 2\right )}^{2} b^{6}}{x^{2}} - \frac {10 \, {\left (b x + 2\right )}^{3} b^{5}}{x^{3}} + \frac {5 \, {\left (b x + 2\right )}^{4} b^{4}}{x^{4}} - \frac {{\left (b x + 2\right )}^{5} b^{3}}{x^{5}}\right )}} + \frac {3 \, \log \left (-\frac {\sqrt {b} - \frac {\sqrt {b x + 2}}{\sqrt {x}}}{\sqrt {b} + \frac {\sqrt {b x + 2}}{\sqrt {x}}}\right )}{8 \, b^{\frac {7}{2}}} \]

[In]

integrate(x^(5/2)*(b*x+2)^(3/2),x, algorithm="maxima")

[Out]

1/20*(15*sqrt(b*x + 2)*b^4/sqrt(x) - 70*(b*x + 2)^(3/2)*b^3/x^(3/2) - 128*(b*x + 2)^(5/2)*b^2/x^(5/2) + 70*(b*
x + 2)^(7/2)*b/x^(7/2) - 15*(b*x + 2)^(9/2)/x^(9/2))/(b^8 - 5*(b*x + 2)*b^7/x + 10*(b*x + 2)^2*b^6/x^2 - 10*(b
*x + 2)^3*b^5/x^3 + 5*(b*x + 2)^4*b^4/x^4 - (b*x + 2)^5*b^3/x^5) + 3/8*log(-(sqrt(b) - sqrt(b*x + 2)/sqrt(x))/
(sqrt(b) + sqrt(b*x + 2)/sqrt(x)))/b^(7/2)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 311 vs. \(2 (87) = 174\).

Time = 16.78 (sec) , antiderivative size = 311, normalized size of antiderivative = 2.47 \[ \int x^{5/2} (2+b x)^{3/2} \, dx=\frac {3 \, {\left ({\left ({\left (2 \, {\left (b x + 2\right )} {\left ({\left (b x + 2\right )} {\left (\frac {4 \, {\left (b x + 2\right )}}{b^{4}} - \frac {41}{b^{4}}\right )} + \frac {171}{b^{4}}\right )} - \frac {745}{b^{4}}\right )} {\left (b x + 2\right )} + \frac {965}{b^{4}}\right )} \sqrt {{\left (b x + 2\right )} b - 2 \, b} \sqrt {b x + 2} + \frac {630 \, \log \left ({\left | -\sqrt {b x + 2} \sqrt {b} + \sqrt {{\left (b x + 2\right )} b - 2 \, b} \right |}\right )}{b^{\frac {7}{2}}}\right )} {\left | b \right |} + \frac {20 \, {\left ({\left ({\left (b x + 2\right )} {\left (2 \, {\left (b x + 2\right )} {\left (\frac {3 \, {\left (b x + 2\right )}}{b^{3}} - \frac {25}{b^{3}}\right )} + \frac {163}{b^{3}}\right )} - \frac {279}{b^{3}}\right )} \sqrt {{\left (b x + 2\right )} b - 2 \, b} \sqrt {b x + 2} - \frac {210 \, \log \left ({\left | -\sqrt {b x + 2} \sqrt {b} + \sqrt {{\left (b x + 2\right )} b - 2 \, b} \right |}\right )}{b^{\frac {5}{2}}}\right )} {\left | b \right |}}{b} + \frac {80 \, {\left (\sqrt {{\left (b x + 2\right )} b - 2 \, b} \sqrt {b x + 2} {\left ({\left (b x + 2\right )} {\left (\frac {2 \, {\left (b x + 2\right )}}{b^{2}} - \frac {13}{b^{2}}\right )} + \frac {33}{b^{2}}\right )} + \frac {30 \, \log \left ({\left | -\sqrt {b x + 2} \sqrt {b} + \sqrt {{\left (b x + 2\right )} b - 2 \, b} \right |}\right )}{b^{\frac {3}{2}}}\right )} {\left | b \right |}}{b^{2}}}{120 \, b} \]

[In]

integrate(x^(5/2)*(b*x+2)^(3/2),x, algorithm="giac")

[Out]

1/120*(3*(((2*(b*x + 2)*((b*x + 2)*(4*(b*x + 2)/b^4 - 41/b^4) + 171/b^4) - 745/b^4)*(b*x + 2) + 965/b^4)*sqrt(
(b*x + 2)*b - 2*b)*sqrt(b*x + 2) + 630*log(abs(-sqrt(b*x + 2)*sqrt(b) + sqrt((b*x + 2)*b - 2*b)))/b^(7/2))*abs
(b) + 20*(((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)*sqr
t(b*x + 2) - 210*log(abs(-sqrt(b*x + 2)*sqrt(b) + sqrt((b*x + 2)*b - 2*b)))/b^(5/2))*abs(b)/b + 80*(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^2)/b

Mupad [F(-1)]

Timed out. \[ \int x^{5/2} (2+b x)^{3/2} \, dx=\int x^{5/2}\,{\left (b\,x+2\right )}^{3/2} \,d x \]

[In]

int(x^(5/2)*(b*x + 2)^(3/2),x)

[Out]

int(x^(5/2)*(b*x + 2)^(3/2), x)