\(\int \frac {1}{x^{9/2} \sqrt {2+b x}} \, dx\) [615]

Optimal result

Integrand size = 15, antiderivative size = 80 \[ \int \frac {1}{x^{9/2} \sqrt {2+b x}} \, dx=-\frac {\sqrt {2+b x}}{7 x^{7/2}}+\frac {3 b \sqrt {2+b x}}{35 x^{5/2}}-\frac {2 b^2 \sqrt {2+b x}}{35 x^{3/2}}+\frac {2 b^3 \sqrt {2+b x}}{35 \sqrt {x}} \]

[Out]

-1/7*(b*x+2)^(1/2)/x^(7/2)+3/35*b*(b*x+2)^(1/2)/x^(5/2)-2/35*b^2*(b*x+2)^(1/2)/x^(3/2)+2/35*b^3*(b*x+2)^(1/2)/
x^(1/2)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {47, 37} \[ \int \frac {1}{x^{9/2} \sqrt {2+b x}} \, dx=\frac {2 b^3 \sqrt {b x+2}}{35 \sqrt {x}}-\frac {2 b^2 \sqrt {b x+2}}{35 x^{3/2}}+\frac {3 b \sqrt {b x+2}}{35 x^{5/2}}-\frac {\sqrt {b x+2}}{7 x^{7/2}} \]

[In]

Int[1/(x^(9/2)*Sqrt[2 + b*x]),x]

[Out]

-1/7*Sqrt[2 + b*x]/x^(7/2) + (3*b*Sqrt[2 + b*x])/(35*x^(5/2)) - (2*b^2*Sqrt[2 + b*x])/(35*x^(3/2)) + (2*b^3*Sq
rt[2 + b*x])/(35*Sqrt[x])

Rule 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n +
1)/((b*c - a*d)*(m + 1))), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

Rule 47

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

Rubi steps \begin{align*} \text {integral}& = -\frac {\sqrt {2+b x}}{7 x^{7/2}}-\frac {1}{7} (3 b) \int \frac {1}{x^{7/2} \sqrt {2+b x}} \, dx \\ & = -\frac {\sqrt {2+b x}}{7 x^{7/2}}+\frac {3 b \sqrt {2+b x}}{35 x^{5/2}}+\frac {1}{35} \left (6 b^2\right ) \int \frac {1}{x^{5/2} \sqrt {2+b x}} \, dx \\ & = -\frac {\sqrt {2+b x}}{7 x^{7/2}}+\frac {3 b \sqrt {2+b x}}{35 x^{5/2}}-\frac {2 b^2 \sqrt {2+b x}}{35 x^{3/2}}-\frac {1}{35} \left (2 b^3\right ) \int \frac {1}{x^{3/2} \sqrt {2+b x}} \, dx \\ & = -\frac {\sqrt {2+b x}}{7 x^{7/2}}+\frac {3 b \sqrt {2+b x}}{35 x^{5/2}}-\frac {2 b^2 \sqrt {2+b x}}{35 x^{3/2}}+\frac {2 b^3 \sqrt {2+b x}}{35 \sqrt {x}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.50 \[ \int \frac {1}{x^{9/2} \sqrt {2+b x}} \, dx=\frac {\sqrt {2+b x} \left (-5+3 b x-2 b^2 x^2+2 b^3 x^3\right )}{35 x^{7/2}} \]

[In]

Integrate[1/(x^(9/2)*Sqrt[2 + b*x]),x]

[Out]

(Sqrt[2 + b*x]*(-5 + 3*b*x - 2*b^2*x^2 + 2*b^3*x^3))/(35*x^(7/2))

Maple [A] (verified)

Time = 0.09 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.44

method	result	size
gosper	\(\frac {\sqrt {b x +2}\, \left (2 b^{3} x^{3}-2 b^{2} x^{2}+3 b x -5\right )}{35 x^{\frac {7}{2}}}\)	\(35\)
meijerg	\(-\frac {\sqrt {2}\, \left (-\frac {2}{5} b^{3} x^{3}+\frac {2}{5} b^{2} x^{2}-\frac {3}{5} b x +1\right ) \sqrt {\frac {b x}{2}+1}}{7 x^{\frac {7}{2}}}\)	\(39\)
risch	\(\frac {2 b^{4} x^{4}+2 b^{3} x^{3}-b^{2} x^{2}+b x -10}{35 x^{\frac {7}{2}} \sqrt {b x +2}}\)	\(42\)
default	\(-\frac {\sqrt {b x +2}}{7 x^{\frac {7}{2}}}-\frac {3 b \left (-\frac {\sqrt {b x +2}}{5 x^{\frac {5}{2}}}-\frac {2 b \left (-\frac {\sqrt {b x +2}}{3 x^{\frac {3}{2}}}+\frac {b \sqrt {b x +2}}{3 \sqrt {x}}\right )}{5}\right )}{7}\)	\(59\)

[In]

int(1/x^(9/2)/(b*x+2)^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/35*(b*x+2)^(1/2)*(2*b^3*x^3-2*b^2*x^2+3*b*x-5)/x^(7/2)

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.42 \[ \int \frac {1}{x^{9/2} \sqrt {2+b x}} \, dx=\frac {{\left (2 \, b^{3} x^{3} - 2 \, b^{2} x^{2} + 3 \, b x - 5\right )} \sqrt {b x + 2}}{35 \, x^{\frac {7}{2}}} \]

[In]

integrate(1/x^(9/2)/(b*x+2)^(1/2),x, algorithm="fricas")

[Out]

1/35*(2*b^3*x^3 - 2*b^2*x^2 + 3*b*x - 5)*sqrt(b*x + 2)/x^(7/2)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 374 vs. \(2 (73) = 146\).

Time = 10.02 (sec) , antiderivative size = 374, normalized size of antiderivative = 4.68 \[ \int \frac {1}{x^{9/2} \sqrt {2+b x}} \, dx=\frac {2 b^{\frac {31}{2}} x^{6} \sqrt {1 + \frac {2}{b x}}}{35 b^{12} x^{6} + 210 b^{11} x^{5} + 420 b^{10} x^{4} + 280 b^{9} x^{3}} + \frac {10 b^{\frac {29}{2}} x^{5} \sqrt {1 + \frac {2}{b x}}}{35 b^{12} x^{6} + 210 b^{11} x^{5} + 420 b^{10} x^{4} + 280 b^{9} x^{3}} + \frac {15 b^{\frac {27}{2}} x^{4} \sqrt {1 + \frac {2}{b x}}}{35 b^{12} x^{6} + 210 b^{11} x^{5} + 420 b^{10} x^{4} + 280 b^{9} x^{3}} + \frac {5 b^{\frac {25}{2}} x^{3} \sqrt {1 + \frac {2}{b x}}}{35 b^{12} x^{6} + 210 b^{11} x^{5} + 420 b^{10} x^{4} + 280 b^{9} x^{3}} - \frac {10 b^{\frac {23}{2}} x^{2} \sqrt {1 + \frac {2}{b x}}}{35 b^{12} x^{6} + 210 b^{11} x^{5} + 420 b^{10} x^{4} + 280 b^{9} x^{3}} - \frac {36 b^{\frac {21}{2}} x \sqrt {1 + \frac {2}{b x}}}{35 b^{12} x^{6} + 210 b^{11} x^{5} + 420 b^{10} x^{4} + 280 b^{9} x^{3}} - \frac {40 b^{\frac {19}{2}} \sqrt {1 + \frac {2}{b x}}}{35 b^{12} x^{6} + 210 b^{11} x^{5} + 420 b^{10} x^{4} + 280 b^{9} x^{3}} \]

[In]

integrate(1/x**(9/2)/(b*x+2)**(1/2),x)

[Out]

2*b**(31/2)*x**6*sqrt(1 + 2/(b*x))/(35*b**12*x**6 + 210*b**11*x**5 + 420*b**10*x**4 + 280*b**9*x**3) + 10*b**(
29/2)*x**5*sqrt(1 + 2/(b*x))/(35*b**12*x**6 + 210*b**11*x**5 + 420*b**10*x**4 + 280*b**9*x**3) + 15*b**(27/2)*
x**4*sqrt(1 + 2/(b*x))/(35*b**12*x**6 + 210*b**11*x**5 + 420*b**10*x**4 + 280*b**9*x**3) + 5*b**(25/2)*x**3*sq
rt(1 + 2/(b*x))/(35*b**12*x**6 + 210*b**11*x**5 + 420*b**10*x**4 + 280*b**9*x**3) - 10*b**(23/2)*x**2*sqrt(1 +
2/(b*x))/(35*b**12*x**6 + 210*b**11*x**5 + 420*b**10*x**4 + 280*b**9*x**3) - 36*b**(21/2)*x*sqrt(1 + 2/(b*x))
/(35*b**12*x**6 + 210*b**11*x**5 + 420*b**10*x**4 + 280*b**9*x**3) - 40*b**(19/2)*sqrt(1 + 2/(b*x))/(35*b**12*
x**6 + 210*b**11*x**5 + 420*b**10*x**4 + 280*b**9*x**3)

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.70 \[ \int \frac {1}{x^{9/2} \sqrt {2+b x}} \, dx=\frac {\sqrt {b x + 2} b^{3}}{8 \, \sqrt {x}} - \frac {{\left (b x + 2\right )}^{\frac {3}{2}} b^{2}}{8 \, x^{\frac {3}{2}}} + \frac {3 \, {\left (b x + 2\right )}^{\frac {5}{2}} b}{40 \, x^{\frac {5}{2}}} - \frac {{\left (b x + 2\right )}^{\frac {7}{2}}}{56 \, x^{\frac {7}{2}}} \]

[In]

integrate(1/x^(9/2)/(b*x+2)^(1/2),x, algorithm="maxima")

[Out]

1/8*sqrt(b*x + 2)*b^3/sqrt(x) - 1/8*(b*x + 2)^(3/2)*b^2/x^(3/2) + 3/40*(b*x + 2)^(5/2)*b/x^(5/2) - 1/56*(b*x +
2)^(7/2)/x^(7/2)

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.85 \[ \int \frac {1}{x^{9/2} \sqrt {2+b x}} \, dx=-\frac {{\left (35 \, b^{7} - {\left (35 \, b^{7} + 2 \, {\left ({\left (b x + 2\right )} b^{7} - 7 \, b^{7}\right )} {\left (b x + 2\right )}\right )} {\left (b x + 2\right )}\right )} \sqrt {b x + 2} b}{35 \, {\left ({\left (b x + 2\right )} b - 2 \, b\right )}^{\frac {7}{2}} {\left | b \right |}} \]

[In]

integrate(1/x^(9/2)/(b*x+2)^(1/2),x, algorithm="giac")

[Out]

-1/35*(35*b^7 - (35*b^7 + 2*((b*x + 2)*b^7 - 7*b^7)*(b*x + 2))*(b*x + 2))*sqrt(b*x + 2)*b/(((b*x + 2)*b - 2*b)
^(7/2)*abs(b))

Mupad [B] (verification not implemented)

Time = 0.37 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.41 \[ \int \frac {1}{x^{9/2} \sqrt {2+b x}} \, dx=\frac {\sqrt {b\,x+2}\,\left (\frac {2\,b^3\,x^3}{35}-\frac {2\,b^2\,x^2}{35}+\frac {3\,b\,x}{35}-\frac {1}{7}\right )}{x^{7/2}} \]

[In]

int(1/(x^(9/2)*(b*x + 2)^(1/2)),x)

[Out]

((b*x + 2)^(1/2)*((3*b*x)/35 - (2*b^2*x^2)/35 + (2*b^3*x^3)/35 - 1/7))/x^(7/2)