\(\int \frac {1}{x^{3/2} (2+b x)^{5/2}} \, dx\) [627]

Optimal result

Integrand size = 15, antiderivative size = 55 \[ \int \frac {1}{x^{3/2} (2+b x)^{5/2}} \, dx=\frac {1}{3 \sqrt {x} (2+b x)^{3/2}}+\frac {2}{3 \sqrt {x} \sqrt {2+b x}}-\frac {2 \sqrt {2+b x}}{3 \sqrt {x}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {47, 37} \[ \int \frac {1}{x^{3/2} (2+b x)^{5/2}} \, dx=-\frac {2 \sqrt {b x+2}}{3 \sqrt {x}}+\frac {2}{3 \sqrt {x} \sqrt {b x+2}}+\frac {1}{3 \sqrt {x} (b x+2)^{3/2}} \]

[In]

Int[1/(x^(3/2)*(2 + b*x)^(5/2)),x]

[Out]

1/(3*Sqrt[x]*(2 + b*x)^(3/2)) + 2/(3*Sqrt[x]*Sqrt[2 + b*x]) - (2*Sqrt[2 + b*x])/(3*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 {1}{3 \sqrt {x} (2+b x)^{3/2}}+\frac {2}{3} \int \frac {1}{x^{3/2} (2+b x)^{3/2}} \, dx \\ & = \frac {1}{3 \sqrt {x} (2+b x)^{3/2}}+\frac {2}{3 \sqrt {x} \sqrt {2+b x}}+\frac {2}{3} \int \frac {1}{x^{3/2} \sqrt {2+b x}} \, dx \\ & = \frac {1}{3 \sqrt {x} (2+b x)^{3/2}}+\frac {2}{3 \sqrt {x} \sqrt {2+b x}}-\frac {2 \sqrt {2+b x}}{3 \sqrt {x}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.58 \[ \int \frac {1}{x^{3/2} (2+b x)^{5/2}} \, dx=\frac {-3-6 b x-2 b^2 x^2}{3 \sqrt {x} (2+b x)^{3/2}} \]

[In]

Integrate[1/(x^(3/2)*(2 + b*x)^(5/2)),x]

[Out]

(-3 - 6*b*x - 2*b^2*x^2)/(3*Sqrt[x]*(2 + b*x)^(3/2))

Maple [A] (verified)

Time = 0.10 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.49

method	result	size
gosper	\(-\frac {2 b^{2} x^{2}+6 b x +3}{3 \sqrt {x}\, \left (b x +2\right )^{\frac {3}{2}}}\)	\(27\)
meijerg	\(-\frac {\sqrt {2}\, \left (2 b^{2} x^{2}+6 b x +3\right )}{12 \sqrt {x}\, \left (\frac {b x}{2}+1\right )^{\frac {3}{2}}}\)	\(31\)
risch	\(-\frac {\sqrt {b x +2}}{4 \sqrt {x}}-\frac {b \left (5 b x +12\right ) \sqrt {x}}{12 \left (b x +2\right )^{\frac {3}{2}}}\)	\(33\)
default	\(-\frac {1}{\left (b x +2\right )^{\frac {3}{2}} \sqrt {x}}-2 b \left (\frac {\sqrt {x}}{3 \left (b x +2\right )^{\frac {3}{2}}}+\frac {\sqrt {x}}{3 \sqrt {b x +2}}\right )\)	\(42\)

[In]

int(1/x^(3/2)/(b*x+2)^(5/2),x,method=_RETURNVERBOSE)

[Out]

-1/3*(2*b^2*x^2+6*b*x+3)/x^(1/2)/(b*x+2)^(3/2)

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.82 \[ \int \frac {1}{x^{3/2} (2+b x)^{5/2}} \, dx=-\frac {{\left (2 \, b^{2} x^{2} + 6 \, b x + 3\right )} \sqrt {b x + 2} \sqrt {x}}{3 \, {\left (b^{2} x^{3} + 4 \, b x^{2} + 4 \, x\right )}} \]

[In]

integrate(1/x^(3/2)/(b*x+2)^(5/2),x, algorithm="fricas")

[Out]

-1/3*(2*b^2*x^2 + 6*b*x + 3)*sqrt(b*x + 2)*sqrt(x)/(b^2*x^3 + 4*b*x^2 + 4*x)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 117 vs. \(2 (49) = 98\).

Time = 1.94 (sec) , antiderivative size = 117, normalized size of antiderivative = 2.13 \[ \int \frac {1}{x^{3/2} (2+b x)^{5/2}} \, dx=- \frac {2 b^{\frac {13}{2}} x^{2} \sqrt {1 + \frac {2}{b x}}}{3 b^{6} x^{2} + 12 b^{5} x + 12 b^{4}} - \frac {6 b^{\frac {11}{2}} x \sqrt {1 + \frac {2}{b x}}}{3 b^{6} x^{2} + 12 b^{5} x + 12 b^{4}} - \frac {3 b^{\frac {9}{2}} \sqrt {1 + \frac {2}{b x}}}{3 b^{6} x^{2} + 12 b^{5} x + 12 b^{4}} \]

[In]

integrate(1/x**(3/2)/(b*x+2)**(5/2),x)

[Out]

-2*b**(13/2)*x**2*sqrt(1 + 2/(b*x))/(3*b**6*x**2 + 12*b**5*x + 12*b**4) - 6*b**(11/2)*x*sqrt(1 + 2/(b*x))/(3*b
**6*x**2 + 12*b**5*x + 12*b**4) - 3*b**(9/2)*sqrt(1 + 2/(b*x))/(3*b**6*x**2 + 12*b**5*x + 12*b**4)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.73 \[ \int \frac {1}{x^{3/2} (2+b x)^{5/2}} \, dx=\frac {{\left (b^{2} - \frac {6 \, {\left (b x + 2\right )} b}{x}\right )} x^{\frac {3}{2}}}{12 \, {\left (b x + 2\right )}^{\frac {3}{2}}} - \frac {\sqrt {b x + 2}}{4 \, \sqrt {x}} \]

[In]

integrate(1/x^(3/2)/(b*x+2)^(5/2),x, algorithm="maxima")

[Out]

1/12*(b^2 - 6*(b*x + 2)*b/x)*x^(3/2)/(b*x + 2)^(3/2) - 1/4*sqrt(b*x + 2)/sqrt(x)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 145 vs. \(2 (37) = 74\).

Time = 0.32 (sec) , antiderivative size = 145, normalized size of antiderivative = 2.64 \[ \int \frac {1}{x^{3/2} (2+b x)^{5/2}} \, dx=-\frac {\sqrt {b x + 2} b^{2}}{4 \, \sqrt {{\left (b x + 2\right )} b - 2 \, b} {\left | b \right |}} - \frac {3 \, {\left (\sqrt {b x + 2} \sqrt {b} - \sqrt {{\left (b x + 2\right )} b - 2 \, b}\right )}^{4} b^{\frac {5}{2}} + 24 \, {\left (\sqrt {b x + 2} \sqrt {b} - \sqrt {{\left (b x + 2\right )} b - 2 \, b}\right )}^{2} b^{\frac {7}{2}} + 20 \, b^{\frac {9}{2}}}{3 \, {\left ({\left (\sqrt {b x + 2} \sqrt {b} - \sqrt {{\left (b x + 2\right )} b - 2 \, b}\right )}^{2} + 2 \, b\right )}^{3} {\left | b \right |}} \]

[In]

integrate(1/x^(3/2)/(b*x+2)^(5/2),x, algorithm="giac")

[Out]

-1/4*sqrt(b*x + 2)*b^2/(sqrt((b*x + 2)*b - 2*b)*abs(b)) - 1/3*(3*(sqrt(b*x + 2)*sqrt(b) - sqrt((b*x + 2)*b - 2
*b))^4*b^(5/2) + 24*(sqrt(b*x + 2)*sqrt(b) - sqrt((b*x + 2)*b - 2*b))^2*b^(7/2) + 20*b^(9/2))/(((sqrt(b*x + 2)
*sqrt(b) - sqrt((b*x + 2)*b - 2*b))^2 + 2*b)^3*abs(b))

Mupad [B] (verification not implemented)

Time = 0.38 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.04 \[ \int \frac {1}{x^{3/2} (2+b x)^{5/2}} \, dx=-\frac {3\,\sqrt {b\,x+2}+6\,b\,x\,\sqrt {b\,x+2}+2\,b^2\,x^2\,\sqrt {b\,x+2}}{\sqrt {x}\,\left (x\,\left (3\,x\,b^2+12\,b\right )+12\right )} \]

[In]

int(1/(x^(3/2)*(b*x + 2)^(5/2)),x)

[Out]

-(3*(b*x + 2)^(1/2) + 6*b*x*(b*x + 2)^(1/2) + 2*b^2*x^2*(b*x + 2)^(1/2))/(x^(1/2)*(x*(12*b + 3*b^2*x) + 12))