\(\int \frac {x^{5/2}}{(2+b x)^{5/2}} \, dx\) [623]

Optimal result

Integrand size = 15, antiderivative size = 86 \[ \int \frac {x^{5/2}}{(2+b x)^{5/2}} \, dx=-\frac {2 x^{5/2}}{3 b (2+b x)^{3/2}}-\frac {10 x^{3/2}}{3 b^2 \sqrt {2+b x}}+\frac {5 \sqrt {x} \sqrt {2+b x}}{b^3}-\frac {10 \text {arcsinh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right )}{b^{7/2}} \]

[Out]

-2/3*x^(5/2)/b/(b*x+2)^(3/2)-10*arcsinh(1/2*b^(1/2)*x^(1/2)*2^(1/2))/b^(7/2)-10/3*x^(3/2)/b^2/(b*x+2)^(1/2)+5*
x^(1/2)*(b*x+2)^(1/2)/b^3

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {49, 52, 56, 221} \[ \int \frac {x^{5/2}}{(2+b x)^{5/2}} \, dx=-\frac {10 \text {arcsinh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right )}{b^{7/2}}+\frac {5 \sqrt {x} \sqrt {b x+2}}{b^3}-\frac {10 x^{3/2}}{3 b^2 \sqrt {b x+2}}-\frac {2 x^{5/2}}{3 b (b x+2)^{3/2}} \]

[In]

Int[x^(5/2)/(2 + b*x)^(5/2),x]

[Out]

(-2*x^(5/2))/(3*b*(2 + b*x)^(3/2)) - (10*x^(3/2))/(3*b^2*Sqrt[2 + b*x]) + (5*Sqrt[x]*Sqrt[2 + b*x])/b^3 - (10*
ArcSinh[(Sqrt[b]*Sqrt[x])/Sqrt[2]])/b^(7/2)

Rule 49

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

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 {2 x^{5/2}}{3 b (2+b x)^{3/2}}+\frac {5 \int \frac {x^{3/2}}{(2+b x)^{3/2}} \, dx}{3 b} \\ & = -\frac {2 x^{5/2}}{3 b (2+b x)^{3/2}}-\frac {10 x^{3/2}}{3 b^2 \sqrt {2+b x}}+\frac {5 \int \frac {\sqrt {x}}{\sqrt {2+b x}} \, dx}{b^2} \\ & = -\frac {2 x^{5/2}}{3 b (2+b x)^{3/2}}-\frac {10 x^{3/2}}{3 b^2 \sqrt {2+b x}}+\frac {5 \sqrt {x} \sqrt {2+b x}}{b^3}-\frac {5 \int \frac {1}{\sqrt {x} \sqrt {2+b x}} \, dx}{b^3} \\ & = -\frac {2 x^{5/2}}{3 b (2+b x)^{3/2}}-\frac {10 x^{3/2}}{3 b^2 \sqrt {2+b x}}+\frac {5 \sqrt {x} \sqrt {2+b x}}{b^3}-\frac {10 \text {Subst}\left (\int \frac {1}{\sqrt {2+b x^2}} \, dx,x,\sqrt {x}\right )}{b^3} \\ & = -\frac {2 x^{5/2}}{3 b (2+b x)^{3/2}}-\frac {10 x^{3/2}}{3 b^2 \sqrt {2+b x}}+\frac {5 \sqrt {x} \sqrt {2+b x}}{b^3}-\frac {10 \sinh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right )}{b^{7/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.21 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.86 \[ \int \frac {x^{5/2}}{(2+b x)^{5/2}} \, dx=\frac {\sqrt {x} \left (60+40 b x+3 b^2 x^2\right )}{3 b^3 (2+b x)^{3/2}}+\frac {20 \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}-\sqrt {2+b x}}\right )}{b^{7/2}} \]

[In]

Integrate[x^(5/2)/(2 + b*x)^(5/2),x]

[Out]

(Sqrt[x]*(60 + 40*b*x + 3*b^2*x^2))/(3*b^3*(2 + b*x)^(3/2)) + (20*ArcTanh[(Sqrt[b]*Sqrt[x])/(Sqrt[2] - Sqrt[2
+ b*x])])/b^(7/2)

Maple [A] (verified)

Time = 0.11 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.73

method	result	size
meijerg	\(\frac {\frac {\sqrt {\pi }\, \sqrt {x}\, \sqrt {2}\, \sqrt {b}\, \left (\frac {21}{4} b^{2} x^{2}+70 b x +105\right )}{21 \left (\frac {b x}{2}+1\right )^{\frac {3}{2}}}-10 \sqrt {\pi }\, \operatorname {arcsinh}\left (\frac {\sqrt {b}\, \sqrt {x}\, \sqrt {2}}{2}\right )}{b^{\frac {7}{2}} \sqrt {\pi }}\)	\(63\)
risch	\(\frac {\sqrt {x}\, \sqrt {b x +2}}{b^{3}}+\frac {\left (-\frac {5 \ln \left (\frac {b x +1}{\sqrt {b}}+\sqrt {b \,x^{2}+2 x}\right )}{b^{\frac {7}{2}}}-\frac {8 \sqrt {b \left (x +\frac {2}{b}\right )^{2}-2 x -\frac {4}{b}}}{3 b^{5} \left (x +\frac {2}{b}\right )^{2}}+\frac {28 \sqrt {b \left (x +\frac {2}{b}\right )^{2}-2 x -\frac {4}{b}}}{3 b^{4} \left (x +\frac {2}{b}\right )}\right ) \sqrt {x \left (b x +2\right )}}{\sqrt {x}\, \sqrt {b x +2}}\)	\(136\)

[In]

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

[Out]

8/3/b^(7/2)/Pi^(1/2)*(1/56*Pi^(1/2)*x^(1/2)*2^(1/2)*b^(1/2)*(21/4*b^2*x^2+70*b*x+105)/(1/2*b*x+1)^(3/2)-15/4*P
i^(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 = 186, normalized size of antiderivative = 2.16 \[ \int \frac {x^{5/2}}{(2+b x)^{5/2}} \, dx=\left [\frac {15 \, {\left (b^{2} x^{2} + 4 \, b x + 4\right )} \sqrt {b} \log \left (b x - \sqrt {b x + 2} \sqrt {b} \sqrt {x} + 1\right ) + {\left (3 \, b^{3} x^{2} + 40 \, b^{2} x + 60 \, b\right )} \sqrt {b x + 2} \sqrt {x}}{3 \, {\left (b^{6} x^{2} + 4 \, b^{5} x + 4 \, b^{4}\right )}}, \frac {30 \, {\left (b^{2} x^{2} + 4 \, b x + 4\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {b x + 2} \sqrt {-b}}{b \sqrt {x}}\right ) + {\left (3 \, b^{3} x^{2} + 40 \, b^{2} x + 60 \, b\right )} \sqrt {b x + 2} \sqrt {x}}{3 \, {\left (b^{6} x^{2} + 4 \, b^{5} x + 4 \, b^{4}\right )}}\right ] \]

[In]

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

[Out]

[1/3*(15*(b^2*x^2 + 4*b*x + 4)*sqrt(b)*log(b*x - sqrt(b*x + 2)*sqrt(b)*sqrt(x) + 1) + (3*b^3*x^2 + 40*b^2*x +
60*b)*sqrt(b*x + 2)*sqrt(x))/(b^6*x^2 + 4*b^5*x + 4*b^4), 1/3*(30*(b^2*x^2 + 4*b*x + 4)*sqrt(-b)*arctan(sqrt(b
*x + 2)*sqrt(-b)/(b*sqrt(x))) + (3*b^3*x^2 + 40*b^2*x + 60*b)*sqrt(b*x + 2)*sqrt(x))/(b^6*x^2 + 4*b^5*x + 4*b^
4)]

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 308 vs. \(2 (82) = 164\).

Time = 5.45 (sec) , antiderivative size = 308, normalized size of antiderivative = 3.58 \[ \int \frac {x^{5/2}}{(2+b x)^{5/2}} \, dx=\frac {3 b^{\frac {23}{2}} x^{15}}{3 b^{\frac {27}{2}} x^{\frac {27}{2}} \sqrt {b x + 2} + 6 b^{\frac {25}{2}} x^{\frac {25}{2}} \sqrt {b x + 2}} + \frac {40 b^{\frac {21}{2}} x^{14}}{3 b^{\frac {27}{2}} x^{\frac {27}{2}} \sqrt {b x + 2} + 6 b^{\frac {25}{2}} x^{\frac {25}{2}} \sqrt {b x + 2}} + \frac {60 b^{\frac {19}{2}} x^{13}}{3 b^{\frac {27}{2}} x^{\frac {27}{2}} \sqrt {b x + 2} + 6 b^{\frac {25}{2}} x^{\frac {25}{2}} \sqrt {b x + 2}} - \frac {30 b^{10} x^{\frac {27}{2}} \sqrt {b x + 2} \operatorname {asinh}{\left (\frac {\sqrt {2} \sqrt {b} \sqrt {x}}{2} \right )}}{3 b^{\frac {27}{2}} x^{\frac {27}{2}} \sqrt {b x + 2} + 6 b^{\frac {25}{2}} x^{\frac {25}{2}} \sqrt {b x + 2}} - \frac {60 b^{9} x^{\frac {25}{2}} \sqrt {b x + 2} \operatorname {asinh}{\left (\frac {\sqrt {2} \sqrt {b} \sqrt {x}}{2} \right )}}{3 b^{\frac {27}{2}} x^{\frac {27}{2}} \sqrt {b x + 2} + 6 b^{\frac {25}{2}} x^{\frac {25}{2}} \sqrt {b x + 2}} \]

[In]

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

[Out]

3*b**(23/2)*x**15/(3*b**(27/2)*x**(27/2)*sqrt(b*x + 2) + 6*b**(25/2)*x**(25/2)*sqrt(b*x + 2)) + 40*b**(21/2)*x
**14/(3*b**(27/2)*x**(27/2)*sqrt(b*x + 2) + 6*b**(25/2)*x**(25/2)*sqrt(b*x + 2)) + 60*b**(19/2)*x**13/(3*b**(2
7/2)*x**(27/2)*sqrt(b*x + 2) + 6*b**(25/2)*x**(25/2)*sqrt(b*x + 2)) - 30*b**10*x**(27/2)*sqrt(b*x + 2)*asinh(s
qrt(2)*sqrt(b)*sqrt(x)/2)/(3*b**(27/2)*x**(27/2)*sqrt(b*x + 2) + 6*b**(25/2)*x**(25/2)*sqrt(b*x + 2)) - 60*b**
9*x**(25/2)*sqrt(b*x + 2)*asinh(sqrt(2)*sqrt(b)*sqrt(x)/2)/(3*b**(27/2)*x**(27/2)*sqrt(b*x + 2) + 6*b**(25/2)*
x**(25/2)*sqrt(b*x + 2))

Maxima [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.22 \[ \int \frac {x^{5/2}}{(2+b x)^{5/2}} \, dx=\frac {2 \, {\left (2 \, b^{2} + \frac {10 \, {\left (b x + 2\right )} b}{x} - \frac {15 \, {\left (b x + 2\right )}^{2}}{x^{2}}\right )}}{3 \, {\left (\frac {{\left (b x + 2\right )}^{\frac {3}{2}} b^{4}}{x^{\frac {3}{2}}} - \frac {{\left (b x + 2\right )}^{\frac {5}{2}} b^{3}}{x^{\frac {5}{2}}}\right )}} + \frac {5 \, \log \left (-\frac {\sqrt {b} - \frac {\sqrt {b x + 2}}{\sqrt {x}}}{\sqrt {b} + \frac {\sqrt {b x + 2}}{\sqrt {x}}}\right )}{b^{\frac {7}{2}}} \]

[In]

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

[Out]

2/3*(2*b^2 + 10*(b*x + 2)*b/x - 15*(b*x + 2)^2/x^2)/((b*x + 2)^(3/2)*b^4/x^(3/2) - (b*x + 2)^(5/2)*b^3/x^(5/2)
) + 5*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. 177 vs. \(2 (63) = 126\).

Time = 1.61 (sec) , antiderivative size = 177, normalized size of antiderivative = 2.06 \[ \int \frac {x^{5/2}}{(2+b x)^{5/2}} \, dx=\frac {{\left (\frac {15 \, \log \left ({\left (\sqrt {b x + 2} \sqrt {b} - \sqrt {{\left (b x + 2\right )} b - 2 \, b}\right )}^{2}\right )}{b^{\frac {5}{2}}} + \frac {3 \, \sqrt {{\left (b x + 2\right )} b - 2 \, b} \sqrt {b x + 2}}{b^{3}} + \frac {16 \, {\left (9 \, {\left (\sqrt {b x + 2} \sqrt {b} - \sqrt {{\left (b x + 2\right )} b - 2 \, b}\right )}^{4} + 24 \, {\left (\sqrt {b x + 2} \sqrt {b} - \sqrt {{\left (b x + 2\right )} b - 2 \, b}\right )}^{2} b + 28 \, b^{2}\right )}}{{\left ({\left (\sqrt {b x + 2} \sqrt {b} - \sqrt {{\left (b x + 2\right )} b - 2 \, b}\right )}^{2} + 2 \, b\right )}^{3} b^{\frac {3}{2}}}\right )} {\left | b \right |}}{3 \, b^{2}} \]

[In]

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

[Out]

1/3*(15*log((sqrt(b*x + 2)*sqrt(b) - sqrt((b*x + 2)*b - 2*b))^2)/b^(5/2) + 3*sqrt((b*x + 2)*b - 2*b)*sqrt(b*x
+ 2)/b^3 + 16*(9*(sqrt(b*x + 2)*sqrt(b) - sqrt((b*x + 2)*b - 2*b))^4 + 24*(sqrt(b*x + 2)*sqrt(b) - sqrt((b*x +
2)*b - 2*b))^2*b + 28*b^2)/(((sqrt(b*x + 2)*sqrt(b) - sqrt((b*x + 2)*b - 2*b))^2 + 2*b)^3*b^(3/2)))*abs(b)/b^
2

Mupad [F(-1)]

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

[In]

int(x^(5/2)/(b*x + 2)^(5/2),x)

[Out]

int(x^(5/2)/(b*x + 2)^(5/2), x)