\(\int x^{5/2} \sqrt {2+b x} \, dx\) [505]

Optimal result

Integrand size = 15, antiderivative size = 108 \[ \int x^{5/2} \sqrt {2+b x} \, dx=\frac {5 \sqrt {x} \sqrt {2+b x}}{8 b^3}-\frac {5 x^{3/2} \sqrt {2+b x}}{24 b^2}+\frac {x^{5/2} \sqrt {2+b x}}{12 b}+\frac {1}{4} x^{7/2} \sqrt {2+b x}-\frac {5 \text {arcsinh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right )}{4 b^{7/2}} \]

[Out]

-5/4*arcsinh(1/2*b^(1/2)*x^(1/2)*2^(1/2))/b^(7/2)-5/24*x^(3/2)*(b*x+2)^(1/2)/b^2+1/12*x^(5/2)*(b*x+2)^(1/2)/b+
1/4*x^(7/2)*(b*x+2)^(1/2)+5/8*x^(1/2)*(b*x+2)^(1/2)/b^3

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 108, 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} \[ \int x^{5/2} \sqrt {2+b x} \, dx=-\frac {5 \text {arcsinh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right )}{4 b^{7/2}}+\frac {5 \sqrt {x} \sqrt {b x+2}}{8 b^3}-\frac {5 x^{3/2} \sqrt {b x+2}}{24 b^2}+\frac {1}{4} x^{7/2} \sqrt {b x+2}+\frac {x^{5/2} \sqrt {b x+2}}{12 b} \]

[In]

Int[x^(5/2)*Sqrt[2 + b*x],x]

[Out]

(5*Sqrt[x]*Sqrt[2 + b*x])/(8*b^3) - (5*x^(3/2)*Sqrt[2 + b*x])/(24*b^2) + (x^(5/2)*Sqrt[2 + b*x])/(12*b) + (x^(
7/2)*Sqrt[2 + b*x])/4 - (5*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}{4} x^{7/2} \sqrt {2+b x}+\frac {1}{4} \int \frac {x^{5/2}}{\sqrt {2+b x}} \, dx \\ & = \frac {x^{5/2} \sqrt {2+b x}}{12 b}+\frac {1}{4} x^{7/2} \sqrt {2+b x}-\frac {5 \int \frac {x^{3/2}}{\sqrt {2+b x}} \, dx}{12 b} \\ & = -\frac {5 x^{3/2} \sqrt {2+b x}}{24 b^2}+\frac {x^{5/2} \sqrt {2+b x}}{12 b}+\frac {1}{4} x^{7/2} \sqrt {2+b x}+\frac {5 \int \frac {\sqrt {x}}{\sqrt {2+b x}} \, dx}{8 b^2} \\ & = \frac {5 \sqrt {x} \sqrt {2+b x}}{8 b^3}-\frac {5 x^{3/2} \sqrt {2+b x}}{24 b^2}+\frac {x^{5/2} \sqrt {2+b x}}{12 b}+\frac {1}{4} x^{7/2} \sqrt {2+b x}-\frac {5 \int \frac {1}{\sqrt {x} \sqrt {2+b x}} \, dx}{8 b^3} \\ & = \frac {5 \sqrt {x} \sqrt {2+b x}}{8 b^3}-\frac {5 x^{3/2} \sqrt {2+b x}}{24 b^2}+\frac {x^{5/2} \sqrt {2+b x}}{12 b}+\frac {1}{4} x^{7/2} \sqrt {2+b x}-\frac {5 \text {Subst}\left (\int \frac {1}{\sqrt {2+b x^2}} \, dx,x,\sqrt {x}\right )}{4 b^3} \\ & = \frac {5 \sqrt {x} \sqrt {2+b x}}{8 b^3}-\frac {5 x^{3/2} \sqrt {2+b x}}{24 b^2}+\frac {x^{5/2} \sqrt {2+b x}}{12 b}+\frac {1}{4} x^{7/2} \sqrt {2+b x}-\frac {5 \sinh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right )}{4 b^{7/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.28 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.78 \[ \int x^{5/2} \sqrt {2+b x} \, dx=\frac {\sqrt {x} \sqrt {2+b x} \left (15-5 b x+2 b^2 x^2+6 b^3 x^3\right )}{24 b^3}+\frac {5 \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}-\sqrt {2+b x}}\right )}{2 b^{7/2}} \]

[In]

Integrate[x^(5/2)*Sqrt[2 + b*x],x]

[Out]

(Sqrt[x]*Sqrt[2 + b*x]*(15 - 5*b*x + 2*b^2*x^2 + 6*b^3*x^3))/(24*b^3) + (5*ArcTanh[(Sqrt[b]*Sqrt[x])/(Sqrt[2]
- Sqrt[2 + b*x])])/(2*b^(7/2))

Maple [A] (verified)

Time = 0.10 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.66

method	result	size
meijerg	\(-\frac {8 \left (-\frac {\sqrt {\pi }\, \sqrt {x}\, \sqrt {2}\, \sqrt {b}\, \left (42 b^{3} x^{3}+14 b^{2} x^{2}-35 b x +105\right ) \sqrt {\frac {b x}{2}+1}}{1344}+\frac {5 \sqrt {\pi }\, \operatorname {arcsinh}\left (\frac {\sqrt {b}\, \sqrt {x}\, \sqrt {2}}{2}\right )}{32}\right )}{b^{\frac {7}{2}} \sqrt {\pi }}\)	\(71\)
risch	\(\frac {\left (6 b^{3} x^{3}+2 b^{2} x^{2}-5 b x +15\right ) \sqrt {x}\, \sqrt {b x +2}}{24 b^{3}}-\frac {5 \ln \left (\frac {b x +1}{\sqrt {b}}+\sqrt {b \,x^{2}+2 x}\right ) \sqrt {x \left (b x +2\right )}}{8 b^{\frac {7}{2}} \sqrt {x}\, \sqrt {b x +2}}\)	\(85\)
default	\(\frac {x^{\frac {5}{2}} \left (b x +2\right )^{\frac {3}{2}}}{4 b}-\frac {5 \left (\frac {x^{\frac {3}{2}} \left (b x +2\right )^{\frac {3}{2}}}{3 b}-\frac {\frac {\sqrt {x}\, \left (b x +2\right )^{\frac {3}{2}}}{2 b}-\frac {\sqrt {x}\, \sqrt {b x +2}+\frac {\sqrt {x \left (b x +2\right )}\, \ln \left (\frac {b x +1}{\sqrt {b}}+\sqrt {b \,x^{2}+2 x}\right )}{\sqrt {b x +2}\, \sqrt {x}\, \sqrt {b}}}{2 b}}{b}\right )}{4 b}\)	\(121\)

[In]

int(x^(5/2)*(b*x+2)^(1/2),x,method=_RETURNVERBOSE)

[Out]

-8/b^(7/2)/Pi^(1/2)*(-1/1344*Pi^(1/2)*x^(1/2)*2^(1/2)*b^(1/2)*(42*b^3*x^3+14*b^2*x^2-35*b*x+105)*(1/2*b*x+1)^(
1/2)+5/32*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 = 140, normalized size of antiderivative = 1.30 \[ \int x^{5/2} \sqrt {2+b x} \, dx=\left [\frac {{\left (6 \, b^{4} x^{3} + 2 \, b^{3} x^{2} - 5 \, b^{2} x + 15 \, b\right )} \sqrt {b x + 2} \sqrt {x} + 15 \, \sqrt {b} \log \left (b x - \sqrt {b x + 2} \sqrt {b} \sqrt {x} + 1\right )}{24 \, b^{4}}, \frac {{\left (6 \, b^{4} x^{3} + 2 \, b^{3} x^{2} - 5 \, b^{2} x + 15 \, b\right )} \sqrt {b x + 2} \sqrt {x} + 30 \, \sqrt {-b} \arctan \left (\frac {\sqrt {b x + 2} \sqrt {-b}}{b \sqrt {x}}\right )}{24 \, b^{4}}\right ] \]

[In]

integrate(x^(5/2)*(b*x+2)^(1/2),x, algorithm="fricas")

[Out]

[1/24*((6*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)*sqr
t(b)*sqrt(x) + 1))/b^4, 1/24*((6*b^4*x^3 + 2*b^3*x^2 - 5*b^2*x + 15*b)*sqrt(b*x + 2)*sqrt(x) + 30*sqrt(-b)*arc
tan(sqrt(b*x + 2)*sqrt(-b)/(b*sqrt(x))))/b^4]

Sympy [A] (verification not implemented)

Time = 26.37 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.08 \[ \int x^{5/2} \sqrt {2+b x} \, dx=\frac {b x^{\frac {9}{2}}}{4 \sqrt {b x + 2}} + \frac {7 x^{\frac {7}{2}}}{12 \sqrt {b x + 2}} - \frac {x^{\frac {5}{2}}}{24 b \sqrt {b x + 2}} + \frac {5 x^{\frac {3}{2}}}{24 b^{2} \sqrt {b x + 2}} + \frac {5 \sqrt {x}}{4 b^{3} \sqrt {b x + 2}} - \frac {5 \operatorname {asinh}{\left (\frac {\sqrt {2} \sqrt {b} \sqrt {x}}{2} \right )}}{4 b^{\frac {7}{2}}} \]

[In]

integrate(x**(5/2)*(b*x+2)**(1/2),x)

[Out]

b*x**(9/2)/(4*sqrt(b*x + 2)) + 7*x**(7/2)/(12*sqrt(b*x + 2)) - x**(5/2)/(24*b*sqrt(b*x + 2)) + 5*x**(3/2)/(24*
b**2*sqrt(b*x + 2)) + 5*sqrt(x)/(4*b**3*sqrt(b*x + 2)) - 5*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. 163 vs. \(2 (75) = 150\).

Time = 0.33 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.51 \[ \int x^{5/2} \sqrt {2+b x} \, dx=\frac {\frac {15 \, \sqrt {b x + 2} b^{3}}{\sqrt {x}} + \frac {73 \, {\left (b x + 2\right )}^{\frac {3}{2}} b^{2}}{x^{\frac {3}{2}}} - \frac {55 \, {\left (b x + 2\right )}^{\frac {5}{2}} b}{x^{\frac {5}{2}}} + \frac {15 \, {\left (b x + 2\right )}^{\frac {7}{2}}}{x^{\frac {7}{2}}}}{12 \, {\left (b^{7} - \frac {4 \, {\left (b x + 2\right )} b^{6}}{x} + \frac {6 \, {\left (b x + 2\right )}^{2} b^{5}}{x^{2}} - \frac {4 \, {\left (b x + 2\right )}^{3} b^{4}}{x^{3}} + \frac {{\left (b x + 2\right )}^{4} b^{3}}{x^{4}}\right )}} + \frac {5 \, \log \left (-\frac {\sqrt {b} - \frac {\sqrt {b x + 2}}{\sqrt {x}}}{\sqrt {b} + \frac {\sqrt {b x + 2}}{\sqrt {x}}}\right )}{8 \, b^{\frac {7}{2}}} \]

[In]

integrate(x^(5/2)*(b*x+2)^(1/2),x, algorithm="maxima")

[Out]

1/12*(15*sqrt(b*x + 2)*b^3/sqrt(x) + 73*(b*x + 2)^(3/2)*b^2/x^(3/2) - 55*(b*x + 2)^(5/2)*b/x^(5/2) + 15*(b*x +
2)^(7/2)/x^(7/2))/(b^7 - 4*(b*x + 2)*b^6/x + 6*(b*x + 2)^2*b^5/x^2 - 4*(b*x + 2)^3*b^4/x^3 + (b*x + 2)^4*b^3/
x^4) + 5/8*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. 198 vs. \(2 (75) = 150\).

Time = 12.34 (sec) , antiderivative size = 198, normalized size of antiderivative = 1.83 \[ \int x^{5/2} \sqrt {2+b x} \, dx=\frac {\frac {{\left ({\left ({\left (b x + 2\right )} {\left (2 \, {\left (b x + 2\right )} {\left (\frac {3 \, {\left (b x + 2\right )}}{b^{3}} - \frac {25}{b^{3}}\right )} + \frac {163}{b^{3}}\right )} - \frac {279}{b^{3}}\right )} \sqrt {{\left (b x + 2\right )} b - 2 \, b} \sqrt {b x + 2} - \frac {210 \, \log \left ({\left | -\sqrt {b x + 2} \sqrt {b} + \sqrt {{\left (b x + 2\right )} b - 2 \, b} \right |}\right )}{b^{\frac {5}{2}}}\right )} {\left | b \right |}}{b} + \frac {8 \, {\left (\sqrt {{\left (b x + 2\right )} b - 2 \, b} \sqrt {b x + 2} {\left ({\left (b x + 2\right )} {\left (\frac {2 \, {\left (b x + 2\right )}}{b^{2}} - \frac {13}{b^{2}}\right )} + \frac {33}{b^{2}}\right )} + \frac {30 \, \log \left ({\left | -\sqrt {b x + 2} \sqrt {b} + \sqrt {{\left (b x + 2\right )} b - 2 \, b} \right |}\right )}{b^{\frac {3}{2}}}\right )} {\left | b \right |}}{b^{2}}}{24 \, b} \]

[In]

integrate(x^(5/2)*(b*x+2)^(1/2),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)/b + 8*(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. \[ \int x^{5/2} \sqrt {2+b x} \, dx=\int x^{5/2}\,\sqrt {b\,x+2} \,d x \]

[In]

int(x^(5/2)*(b*x + 2)^(1/2),x)

[Out]

int(x^(5/2)*(b*x + 2)^(1/2), x)