\(\int \frac {1}{x^{3/2} (a+b x)^{5/2}} \, dx\) [588]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [B] (verification not implemented)
   Maxima [A] (verification not implemented)
   Giac [B] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 15, antiderivative size = 64 \[ \int \frac {1}{x^{3/2} (a+b x)^{5/2}} \, dx=\frac {2}{3 a \sqrt {x} (a+b x)^{3/2}}+\frac {8}{3 a^2 \sqrt {x} \sqrt {a+b x}}-\frac {16 \sqrt {a+b x}}{3 a^3 \sqrt {x}} \]

[Out]

2/3/a/(b*x+a)^(3/2)/x^(1/2)+8/3/a^2/x^(1/2)/(b*x+a)^(1/2)-16/3*(b*x+a)^(1/2)/a^3/x^(1/2)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 64, 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} \[ \int \frac {1}{x^{3/2} (a+b x)^{5/2}} \, dx=-\frac {16 \sqrt {a+b x}}{3 a^3 \sqrt {x}}+\frac {8}{3 a^2 \sqrt {x} \sqrt {a+b x}}+\frac {2}{3 a \sqrt {x} (a+b x)^{3/2}} \]

[In]

Int[1/(x^(3/2)*(a + b*x)^(5/2)),x]

[Out]

2/(3*a*Sqrt[x]*(a + b*x)^(3/2)) + 8/(3*a^2*Sqrt[x]*Sqrt[a + b*x]) - (16*Sqrt[a + b*x])/(3*a^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 {2}{3 a \sqrt {x} (a+b x)^{3/2}}+\frac {4 \int \frac {1}{x^{3/2} (a+b x)^{3/2}} \, dx}{3 a} \\ & = \frac {2}{3 a \sqrt {x} (a+b x)^{3/2}}+\frac {8}{3 a^2 \sqrt {x} \sqrt {a+b x}}+\frac {8 \int \frac {1}{x^{3/2} \sqrt {a+b x}} \, dx}{3 a^2} \\ & = \frac {2}{3 a \sqrt {x} (a+b x)^{3/2}}+\frac {8}{3 a^2 \sqrt {x} \sqrt {a+b x}}-\frac {16 \sqrt {a+b x}}{3 a^3 \sqrt {x}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.62 \[ \int \frac {1}{x^{3/2} (a+b x)^{5/2}} \, dx=-\frac {2 \left (3 a^2+12 a b x+8 b^2 x^2\right )}{3 a^3 \sqrt {x} (a+b x)^{3/2}} \]

[In]

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

[Out]

(-2*(3*a^2 + 12*a*b*x + 8*b^2*x^2))/(3*a^3*Sqrt[x]*(a + b*x)^(3/2))

Maple [A] (verified)

Time = 0.10 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.55

method result size
gosper \(-\frac {2 \left (8 b^{2} x^{2}+12 a b x +3 a^{2}\right )}{3 \sqrt {x}\, \left (b x +a \right )^{\frac {3}{2}} a^{3}}\) \(35\)
risch \(-\frac {2 \sqrt {b x +a}}{a^{3} \sqrt {x}}-\frac {2 b \left (5 b x +6 a \right ) \sqrt {x}}{3 \left (b x +a \right )^{\frac {3}{2}} a^{3}}\) \(41\)
default \(-\frac {2}{a \left (b x +a \right )^{\frac {3}{2}} \sqrt {x}}-\frac {4 b \left (\frac {2 \sqrt {x}}{3 a \left (b x +a \right )^{\frac {3}{2}}}+\frac {4 \sqrt {x}}{3 a^{2} \sqrt {b x +a}}\right )}{a}\) \(54\)

[In]

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

[Out]

-2/3*(8*b^2*x^2+12*a*b*x+3*a^2)/x^(1/2)/(b*x+a)^(3/2)/a^3

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.91 \[ \int \frac {1}{x^{3/2} (a+b x)^{5/2}} \, dx=-\frac {2 \, {\left (8 \, b^{2} x^{2} + 12 \, a b x + 3 \, a^{2}\right )} \sqrt {b x + a} \sqrt {x}}{3 \, {\left (a^{3} b^{2} x^{3} + 2 \, a^{4} b x^{2} + a^{5} x\right )}} \]

[In]

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

[Out]

-2/3*(8*b^2*x^2 + 12*a*b*x + 3*a^2)*sqrt(b*x + a)*sqrt(x)/(a^3*b^2*x^3 + 2*a^4*b*x^2 + a^5*x)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 153 vs. \(2 (58) = 116\).

Time = 2.29 (sec) , antiderivative size = 153, normalized size of antiderivative = 2.39 \[ \int \frac {1}{x^{3/2} (a+b x)^{5/2}} \, dx=- \frac {6 a^{2} b^{\frac {9}{2}} \sqrt {\frac {a}{b x} + 1}}{3 a^{5} b^{4} + 6 a^{4} b^{5} x + 3 a^{3} b^{6} x^{2}} - \frac {24 a b^{\frac {11}{2}} x \sqrt {\frac {a}{b x} + 1}}{3 a^{5} b^{4} + 6 a^{4} b^{5} x + 3 a^{3} b^{6} x^{2}} - \frac {16 b^{\frac {13}{2}} x^{2} \sqrt {\frac {a}{b x} + 1}}{3 a^{5} b^{4} + 6 a^{4} b^{5} x + 3 a^{3} b^{6} x^{2}} \]

[In]

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

[Out]

-6*a**2*b**(9/2)*sqrt(a/(b*x) + 1)/(3*a**5*b**4 + 6*a**4*b**5*x + 3*a**3*b**6*x**2) - 24*a*b**(11/2)*x*sqrt(a/
(b*x) + 1)/(3*a**5*b**4 + 6*a**4*b**5*x + 3*a**3*b**6*x**2) - 16*b**(13/2)*x**2*sqrt(a/(b*x) + 1)/(3*a**5*b**4
 + 6*a**4*b**5*x + 3*a**3*b**6*x**2)

Maxima [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.72 \[ \int \frac {1}{x^{3/2} (a+b x)^{5/2}} \, dx=\frac {2 \, {\left (b^{2} - \frac {6 \, {\left (b x + a\right )} b}{x}\right )} x^{\frac {3}{2}}}{3 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{3}} - \frac {2 \, \sqrt {b x + a}}{a^{3} \sqrt {x}} \]

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 159 vs. \(2 (46) = 92\).

Time = 0.34 (sec) , antiderivative size = 159, normalized size of antiderivative = 2.48 \[ \int \frac {1}{x^{3/2} (a+b x)^{5/2}} \, dx=-\frac {2 \, \sqrt {b x + a} b^{2}}{\sqrt {{\left (b x + a\right )} b - a b} a^{3} {\left | b \right |}} - \frac {4 \, {\left (3 \, {\left (\sqrt {b x + a} \sqrt {b} - \sqrt {{\left (b x + a\right )} b - a b}\right )}^{4} b^{\frac {5}{2}} + 12 \, a {\left (\sqrt {b x + a} \sqrt {b} - \sqrt {{\left (b x + a\right )} b - a b}\right )}^{2} b^{\frac {7}{2}} + 5 \, a^{2} b^{\frac {9}{2}}\right )}}{3 \, {\left ({\left (\sqrt {b x + a} \sqrt {b} - \sqrt {{\left (b x + a\right )} b - a b}\right )}^{2} + a b\right )}^{3} a^{2} {\left | b \right |}} \]

[In]

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

[Out]

-2*sqrt(b*x + a)*b^2/(sqrt((b*x + a)*b - a*b)*a^3*abs(b)) - 4/3*(3*(sqrt(b*x + a)*sqrt(b) - sqrt((b*x + a)*b -
 a*b))^4*b^(5/2) + 12*a*(sqrt(b*x + a)*sqrt(b) - sqrt((b*x + a)*b - a*b))^2*b^(7/2) + 5*a^2*b^(9/2))/(((sqrt(b
*x + a)*sqrt(b) - sqrt((b*x + a)*b - a*b))^2 + a*b)^3*a^2*abs(b))

Mupad [B] (verification not implemented)

Time = 0.46 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.11 \[ \int \frac {1}{x^{3/2} (a+b x)^{5/2}} \, dx=-\frac {6\,a^2\,\sqrt {a+b\,x}+16\,b^2\,x^2\,\sqrt {a+b\,x}+24\,a\,b\,x\,\sqrt {a+b\,x}}{\sqrt {x}\,\left (x\,\left (6\,a^4\,b+3\,x\,a^3\,b^2\right )+3\,a^5\right )} \]

[In]

int(1/(x^(3/2)*(a + b*x)^(5/2)),x)

[Out]

-(6*a^2*(a + b*x)^(1/2) + 16*b^2*x^2*(a + b*x)^(1/2) + 24*a*b*x*(a + b*x)^(1/2))/(x^(1/2)*(x*(6*a^4*b + 3*a^3*
b^2*x) + 3*a^5))