\(\int \frac {x^{3/2}}{(a+b x)^{5/2}} \, dx\) [585]

Optimal result

Integrand size = 15, antiderivative size = 69 \[ \int \frac {x^{3/2}}{(a+b x)^{5/2}} \, dx=-\frac {2 x^{3/2}}{3 b (a+b x)^{3/2}}-\frac {2 \sqrt {x}}{b^2 \sqrt {a+b x}}+\frac {2 \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )}{b^{5/2}} \]

[Out]

-2/3*x^(3/2)/b/(b*x+a)^(3/2)+2*arctanh(b^(1/2)*x^(1/2)/(b*x+a)^(1/2))/b^(5/2)-2*x^(1/2)/b^2/(b*x+a)^(1/2)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {49, 65, 223, 212} \[ \int \frac {x^{3/2}}{(a+b x)^{5/2}} \, dx=\frac {2 \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )}{b^{5/2}}-\frac {2 \sqrt {x}}{b^2 \sqrt {a+b x}}-\frac {2 x^{3/2}}{3 b (a+b x)^{3/2}} \]

[In]

Int[x^(3/2)/(a + b*x)^(5/2),x]

[Out]

(-2*x^(3/2))/(3*b*(a + b*x)^(3/2)) - (2*Sqrt[x])/(b^2*Sqrt[a + b*x]) + (2*ArcTanh[(Sqrt[b]*Sqrt[x])/Sqrt[a + b
*x]])/b^(5/2)

Rule 49

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

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 223

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

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

Mathematica [A] (verified)

Time = 0.19 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.99 \[ \int \frac {x^{3/2}}{(a+b x)^{5/2}} \, dx=-\frac {2 \sqrt {x} (3 a+4 b x)}{3 b^2 (a+b x)^{3/2}}+\frac {4 \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{-\sqrt {a}+\sqrt {a+b x}}\right )}{b^{5/2}} \]

[In]

Integrate[x^(3/2)/(a + b*x)^(5/2),x]

[Out]

(-2*Sqrt[x]*(3*a + 4*b*x))/(3*b^2*(a + b*x)^(3/2)) + (4*ArcTanh[(Sqrt[b]*Sqrt[x])/(-Sqrt[a] + Sqrt[a + b*x])])
/b^(5/2)

Maple [F]

\[\int \frac {x^{\frac {3}{2}}}{\left (b x +a \right )^{\frac {5}{2}}}d x\]

[In]

int(x^(3/2)/(b*x+a)^(5/2),x)

[Out]

int(x^(3/2)/(b*x+a)^(5/2),x)

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 186, normalized size of antiderivative = 2.70 \[ \int \frac {x^{3/2}}{(a+b x)^{5/2}} \, dx=\left [\frac {3 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )} \sqrt {b} \log \left (2 \, b x + 2 \, \sqrt {b x + a} \sqrt {b} \sqrt {x} + a\right ) - 2 \, {\left (4 \, b^{2} x + 3 \, a b\right )} \sqrt {b x + a} \sqrt {x}}{3 \, {\left (b^{5} x^{2} + 2 \, a b^{4} x + a^{2} b^{3}\right )}}, -\frac {2 \, {\left (3 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-b}}{b \sqrt {x}}\right ) + {\left (4 \, b^{2} x + 3 \, a b\right )} \sqrt {b x + a} \sqrt {x}\right )}}{3 \, {\left (b^{5} x^{2} + 2 \, a b^{4} x + a^{2} b^{3}\right )}}\right ] \]

[In]

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

[Out]

[1/3*(3*(b^2*x^2 + 2*a*b*x + a^2)*sqrt(b)*log(2*b*x + 2*sqrt(b*x + a)*sqrt(b)*sqrt(x) + a) - 2*(4*b^2*x + 3*a*
b)*sqrt(b*x + a)*sqrt(x))/(b^5*x^2 + 2*a*b^4*x + a^2*b^3), -2/3*(3*(b^2*x^2 + 2*a*b*x + a^2)*sqrt(-b)*arctan(s
qrt(b*x + a)*sqrt(-b)/(b*sqrt(x))) + (4*b^2*x + 3*a*b)*sqrt(b*x + a)*sqrt(x))/(b^5*x^2 + 2*a*b^4*x + a^2*b^3)]

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 328 vs. \(2 (63) = 126\).

Time = 3.01 (sec) , antiderivative size = 328, normalized size of antiderivative = 4.75 \[ \int \frac {x^{3/2}}{(a+b x)^{5/2}} \, dx=\frac {6 a^{\frac {39}{2}} b^{11} x^{\frac {27}{2}} \sqrt {1 + \frac {b x}{a}} \operatorname {asinh}{\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}} \right )}}{3 a^{\frac {39}{2}} b^{\frac {27}{2}} x^{\frac {27}{2}} \sqrt {1 + \frac {b x}{a}} + 3 a^{\frac {37}{2}} b^{\frac {29}{2}} x^{\frac {29}{2}} \sqrt {1 + \frac {b x}{a}}} + \frac {6 a^{\frac {37}{2}} b^{12} x^{\frac {29}{2}} \sqrt {1 + \frac {b x}{a}} \operatorname {asinh}{\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}} \right )}}{3 a^{\frac {39}{2}} b^{\frac {27}{2}} x^{\frac {27}{2}} \sqrt {1 + \frac {b x}{a}} + 3 a^{\frac {37}{2}} b^{\frac {29}{2}} x^{\frac {29}{2}} \sqrt {1 + \frac {b x}{a}}} - \frac {6 a^{19} b^{\frac {23}{2}} x^{14}}{3 a^{\frac {39}{2}} b^{\frac {27}{2}} x^{\frac {27}{2}} \sqrt {1 + \frac {b x}{a}} + 3 a^{\frac {37}{2}} b^{\frac {29}{2}} x^{\frac {29}{2}} \sqrt {1 + \frac {b x}{a}}} - \frac {8 a^{18} b^{\frac {25}{2}} x^{15}}{3 a^{\frac {39}{2}} b^{\frac {27}{2}} x^{\frac {27}{2}} \sqrt {1 + \frac {b x}{a}} + 3 a^{\frac {37}{2}} b^{\frac {29}{2}} x^{\frac {29}{2}} \sqrt {1 + \frac {b x}{a}}} \]

[In]

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

[Out]

6*a**(39/2)*b**11*x**(27/2)*sqrt(1 + b*x/a)*asinh(sqrt(b)*sqrt(x)/sqrt(a))/(3*a**(39/2)*b**(27/2)*x**(27/2)*sq
rt(1 + b*x/a) + 3*a**(37/2)*b**(29/2)*x**(29/2)*sqrt(1 + b*x/a)) + 6*a**(37/2)*b**12*x**(29/2)*sqrt(1 + b*x/a)
*asinh(sqrt(b)*sqrt(x)/sqrt(a))/(3*a**(39/2)*b**(27/2)*x**(27/2)*sqrt(1 + b*x/a) + 3*a**(37/2)*b**(29/2)*x**(2
9/2)*sqrt(1 + b*x/a)) - 6*a**19*b**(23/2)*x**14/(3*a**(39/2)*b**(27/2)*x**(27/2)*sqrt(1 + b*x/a) + 3*a**(37/2)
*b**(29/2)*x**(29/2)*sqrt(1 + b*x/a)) - 8*a**18*b**(25/2)*x**15/(3*a**(39/2)*b**(27/2)*x**(27/2)*sqrt(1 + b*x/
a) + 3*a**(37/2)*b**(29/2)*x**(29/2)*sqrt(1 + b*x/a))

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.00 \[ \int \frac {x^{3/2}}{(a+b x)^{5/2}} \, dx=-\frac {2 \, {\left (b + \frac {3 \, {\left (b x + a\right )}}{x}\right )} x^{\frac {3}{2}}}{3 \, {\left (b x + a\right )}^{\frac {3}{2}} b^{2}} - \frac {\log \left (-\frac {\sqrt {b} - \frac {\sqrt {b x + a}}{\sqrt {x}}}{\sqrt {b} + \frac {\sqrt {b x + a}}{\sqrt {x}}}\right )}{b^{\frac {5}{2}}} \]

[In]

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

[Out]

-2/3*(b + 3*(b*x + a)/x)*x^(3/2)/((b*x + a)^(3/2)*b^2) - log(-(sqrt(b) - sqrt(b*x + a)/sqrt(x))/(sqrt(b) + sqr
t(b*x + a)/sqrt(x)))/b^(5/2)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 165 vs. \(2 (51) = 102\).

Time = 16.13 (sec) , antiderivative size = 165, normalized size of antiderivative = 2.39 \[ \int \frac {x^{3/2}}{(a+b x)^{5/2}} \, dx=-\frac {{\left (\frac {3 \, \log \left ({\left (\sqrt {b x + a} \sqrt {b} - \sqrt {{\left (b x + a\right )} b - a b}\right )}^{2}\right )}{\sqrt {b}} + \frac {8 \, {\left (3 \, a {\left (\sqrt {b x + a} \sqrt {b} - \sqrt {{\left (b x + a\right )} b - a b}\right )}^{4} \sqrt {b} + 3 \, a^{2} {\left (\sqrt {b x + a} \sqrt {b} - \sqrt {{\left (b x + a\right )} b - a b}\right )}^{2} b^{\frac {3}{2}} + 2 \, a^{3} b^{\frac {5}{2}}\right )}}{{\left ({\left (\sqrt {b x + a} \sqrt {b} - \sqrt {{\left (b x + a\right )} b - a b}\right )}^{2} + a b\right )}^{3}}\right )} {\left | b \right |}}{3 \, b^{3}} \]

[In]

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

[Out]

-1/3*(3*log((sqrt(b*x + a)*sqrt(b) - sqrt((b*x + a)*b - a*b))^2)/sqrt(b) + 8*(3*a*(sqrt(b*x + a)*sqrt(b) - sqr
t((b*x + a)*b - a*b))^4*sqrt(b) + 3*a^2*(sqrt(b*x + a)*sqrt(b) - sqrt((b*x + a)*b - a*b))^2*b^(3/2) + 2*a^3*b^
(5/2))/((sqrt(b*x + a)*sqrt(b) - sqrt((b*x + a)*b - a*b))^2 + a*b)^3)*abs(b)/b^3

Mupad [F(-1)]

Timed out. \[ \int \frac {x^{3/2}}{(a+b x)^{5/2}} \, dx=\int \frac {x^{3/2}}{{\left (a+b\,x\right )}^{5/2}} \,d x \]

[In]

int(x^(3/2)/(a + b*x)^(5/2),x)

[Out]

int(x^(3/2)/(a + b*x)^(5/2), x)