\(\int \frac {1}{\sqrt {x} (a+b x)^{5/2}} \, dx\) [587]

Optimal result

Integrand size = 15, antiderivative size = 43 \[ \int \frac {1}{\sqrt {x} (a+b x)^{5/2}} \, dx=\frac {2 \sqrt {x}}{3 a (a+b x)^{3/2}}+\frac {4 \sqrt {x}}{3 a^2 \sqrt {a+b x}} \]

[Out]

2/3*x^(1/2)/a/(b*x+a)^(3/2)+4/3*x^(1/2)/a^2/(b*x+a)^(1/2)

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {47, 37} \[ \int \frac {1}{\sqrt {x} (a+b x)^{5/2}} \, dx=\frac {4 \sqrt {x}}{3 a^2 \sqrt {a+b x}}+\frac {2 \sqrt {x}}{3 a (a+b x)^{3/2}} \]

[In]

Int[1/(Sqrt[x]*(a + b*x)^(5/2)),x]

[Out]

(2*Sqrt[x])/(3*a*(a + b*x)^(3/2)) + (4*Sqrt[x])/(3*a^2*Sqrt[a + b*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 {2 \sqrt {x}}{3 a (a+b x)^{3/2}}+\frac {2 \int \frac {1}{\sqrt {x} (a+b x)^{3/2}} \, dx}{3 a} \\ & = \frac {2 \sqrt {x}}{3 a (a+b x)^{3/2}}+\frac {4 \sqrt {x}}{3 a^2 \sqrt {a+b x}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.67 \[ \int \frac {1}{\sqrt {x} (a+b x)^{5/2}} \, dx=\frac {2 \sqrt {x} (3 a+2 b x)}{3 a^2 (a+b x)^{3/2}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.09 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.56

method	result	size
gosper	\(\frac {2 \sqrt {x}\, \left (2 b x +3 a \right )}{3 \left (b x +a \right )^{\frac {3}{2}} a^{2}}\)	\(24\)
default	\(\frac {2 \sqrt {x}}{3 a \left (b x +a \right )^{\frac {3}{2}}}+\frac {4 \sqrt {x}}{3 a^{2} \sqrt {b x +a}}\)	\(32\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.00 \[ \int \frac {1}{\sqrt {x} (a+b x)^{5/2}} \, dx=\frac {2 \, {\left (2 \, b x + 3 \, a\right )} \sqrt {b x + a} \sqrt {x}}{3 \, {\left (a^{2} b^{2} x^{2} + 2 \, a^{3} b x + a^{4}\right )}} \]

[In]

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

[Out]

2/3*(2*b*x + 3*a)*sqrt(b*x + a)*sqrt(x)/(a^2*b^2*x^2 + 2*a^3*b*x + a^4)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 92 vs. \(2 (37) = 74\).

Time = 1.25 (sec) , antiderivative size = 92, normalized size of antiderivative = 2.14 \[ \int \frac {1}{\sqrt {x} (a+b x)^{5/2}} \, dx=\frac {6 a}{3 a^{3} \sqrt {b} \sqrt {\frac {a}{b x} + 1} + 3 a^{2} b^{\frac {3}{2}} x \sqrt {\frac {a}{b x} + 1}} + \frac {4 b x}{3 a^{3} \sqrt {b} \sqrt {\frac {a}{b x} + 1} + 3 a^{2} b^{\frac {3}{2}} x \sqrt {\frac {a}{b x} + 1}} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.63 \[ \int \frac {1}{\sqrt {x} (a+b x)^{5/2}} \, dx=-\frac {2 \, {\left (b - \frac {3 \, {\left (b x + a\right )}}{x}\right )} x^{\frac {3}{2}}}{3 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{2}} \]

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 81 vs. \(2 (31) = 62\).

Time = 0.31 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.88 \[ \int \frac {1}{\sqrt {x} (a+b x)^{5/2}} \, dx=\frac {8 \, {\left (3 \, {\left (\sqrt {b x + a} \sqrt {b} - \sqrt {{\left (b x + a\right )} b - a b}\right )}^{2} + a b\right )} b^{\frac {5}{2}}}{3 \, {\left ({\left (\sqrt {b x + a} \sqrt {b} - \sqrt {{\left (b x + a\right )} b - a b}\right )}^{2} + a b\right )}^{3} {\left | b \right |}} \]

[In]

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

[Out]

8/3*(3*(sqrt(b*x + a)*sqrt(b) - sqrt((b*x + a)*b - a*b))^2 + a*b)*b^(5/2)/(((sqrt(b*x + a)*sqrt(b) - sqrt((b*x
+ a)*b - a*b))^2 + a*b)^3*abs(b))

Mupad [B] (verification not implemented)

Time = 0.40 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.26 \[ \int \frac {1}{\sqrt {x} (a+b x)^{5/2}} \, dx=\frac {6\,a\,\sqrt {x}\,\sqrt {a+b\,x}+4\,b\,x^{3/2}\,\sqrt {a+b\,x}}{3\,a^4+6\,a^3\,b\,x+3\,a^2\,b^2\,x^2} \]

[In]

int(1/(x^(1/2)*(a + b*x)^(5/2)),x)

[Out]

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