\(\int \frac {x^{3/2}}{(a+b x)^2} \, dx\) [115]

Optimal result

Integrand size = 13, antiderivative size = 50 \[ \int \frac {x^{3/2}}{(a+b x)^2} \, dx=\frac {2 x^{3/2}}{b (a+b x)}+\frac {3 a \sqrt {x} \arctan \left (\frac {b x}{a}\right )}{b^3 \sqrt {a b} (a+b x)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.14, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {43, 52, 65, 211} \[ \int \frac {x^{3/2}}{(a+b x)^2} \, dx=-\frac {3 \sqrt {a} \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{b^{5/2}}-\frac {x^{3/2}}{b (a+b x)}+\frac {3 \sqrt {x}}{b^2} \]

[In]

Int[x^(3/2)/(a + b*x)^2,x]

[Out]

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

Rule 43

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, n
}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, -1] &&  !IntegerQ[n] && GtQ[n, 0]

Rule 52

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^n/(b*(
m + n + 1))), x] + Dist[n*((b*c - a*d)/(b*(m + n + 1))), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 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 211

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/Rt[a/b, 2]], x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

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

Mathematica [A] (verified)

Time = 0.12 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.08 \[ \int \frac {x^{3/2}}{(a+b x)^2} \, dx=\frac {\sqrt {x} (3 a+2 b x)}{b^2 (a+b x)}-\frac {3 \sqrt {a} \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{b^{5/2}} \]

[In]

Integrate[x^(3/2)/(a + b*x)^2,x]

[Out]

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

Maple [A] (verified)

Time = 0.09 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.94

method	result	size
derivativedivides	\(\frac {2 \sqrt {x}}{b^{2}}-\frac {2 a \left (-\frac {\sqrt {x}}{2 \left (b x +a \right )}+\frac {3 \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{2 \sqrt {a b}}\right )}{b^{2}}\)	\(47\)
default	\(\frac {2 \sqrt {x}}{b^{2}}-\frac {2 a \left (-\frac {\sqrt {x}}{2 \left (b x +a \right )}+\frac {3 \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{2 \sqrt {a b}}\right )}{b^{2}}\)	\(47\)
risch	\(\frac {2 \sqrt {x}}{b^{2}}+\frac {a \sqrt {x}}{b^{2} \left (b x +a \right )}-\frac {3 a \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{b^{2} \sqrt {a b}}\)	\(47\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 134, normalized size of antiderivative = 2.68 \[ \int \frac {x^{3/2}}{(a+b x)^2} \, dx=\left [\frac {3 \, {\left (b x + a\right )} \sqrt {-\frac {a}{b}} \log \left (\frac {b x - 2 \, b \sqrt {x} \sqrt {-\frac {a}{b}} - a}{b x + a}\right ) + 2 \, {\left (2 \, b x + 3 \, a\right )} \sqrt {x}}{2 \, {\left (b^{3} x + a b^{2}\right )}}, -\frac {3 \, {\left (b x + a\right )} \sqrt {\frac {a}{b}} \arctan \left (\frac {b \sqrt {x} \sqrt {\frac {a}{b}}}{a}\right ) - {\left (2 \, b x + 3 \, a\right )} \sqrt {x}}{b^{3} x + a b^{2}}\right ] \]

[In]

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

[Out]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 332 vs. \(2 (42) = 84\).

Time = 2.82 (sec) , antiderivative size = 332, normalized size of antiderivative = 6.64 \[ \int \frac {x^{3/2}}{(a+b x)^2} \, dx=\begin {cases} \tilde {\infty } \sqrt {x} & \text {for}\: a = 0 \wedge b = 0 \\\frac {2 x^{\frac {5}{2}}}{5 a^{2}} & \text {for}\: b = 0 \\\frac {2 \sqrt {x}}{b^{2}} & \text {for}\: a = 0 \\- \frac {3 a^{2} \log {\left (\sqrt {x} - \sqrt {- \frac {a}{b}} \right )}}{2 a b^{3} \sqrt {- \frac {a}{b}} + 2 b^{4} x \sqrt {- \frac {a}{b}}} + \frac {3 a^{2} \log {\left (\sqrt {x} + \sqrt {- \frac {a}{b}} \right )}}{2 a b^{3} \sqrt {- \frac {a}{b}} + 2 b^{4} x \sqrt {- \frac {a}{b}}} + \frac {6 a b \sqrt {x} \sqrt {- \frac {a}{b}}}{2 a b^{3} \sqrt {- \frac {a}{b}} + 2 b^{4} x \sqrt {- \frac {a}{b}}} - \frac {3 a b x \log {\left (\sqrt {x} - \sqrt {- \frac {a}{b}} \right )}}{2 a b^{3} \sqrt {- \frac {a}{b}} + 2 b^{4} x \sqrt {- \frac {a}{b}}} + \frac {3 a b x \log {\left (\sqrt {x} + \sqrt {- \frac {a}{b}} \right )}}{2 a b^{3} \sqrt {- \frac {a}{b}} + 2 b^{4} x \sqrt {- \frac {a}{b}}} + \frac {4 b^{2} x^{\frac {3}{2}} \sqrt {- \frac {a}{b}}}{2 a b^{3} \sqrt {- \frac {a}{b}} + 2 b^{4} x \sqrt {- \frac {a}{b}}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((zoo*sqrt(x), Eq(a, 0) & Eq(b, 0)), (2*x**(5/2)/(5*a**2), Eq(b, 0)), (2*sqrt(x)/b**2, Eq(a, 0)), (-3
*a**2*log(sqrt(x) - sqrt(-a/b))/(2*a*b**3*sqrt(-a/b) + 2*b**4*x*sqrt(-a/b)) + 3*a**2*log(sqrt(x) + sqrt(-a/b))
/(2*a*b**3*sqrt(-a/b) + 2*b**4*x*sqrt(-a/b)) + 6*a*b*sqrt(x)*sqrt(-a/b)/(2*a*b**3*sqrt(-a/b) + 2*b**4*x*sqrt(-
a/b)) - 3*a*b*x*log(sqrt(x) - sqrt(-a/b))/(2*a*b**3*sqrt(-a/b) + 2*b**4*x*sqrt(-a/b)) + 3*a*b*x*log(sqrt(x) +
sqrt(-a/b))/(2*a*b**3*sqrt(-a/b) + 2*b**4*x*sqrt(-a/b)) + 4*b**2*x**(3/2)*sqrt(-a/b)/(2*a*b**3*sqrt(-a/b) + 2*
b**4*x*sqrt(-a/b)), True))

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.98 \[ \int \frac {x^{3/2}}{(a+b x)^2} \, dx=\frac {a \sqrt {x}}{b^{3} x + a b^{2}} - \frac {3 \, a \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} b^{2}} + \frac {2 \, \sqrt {x}}{b^{2}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.92 \[ \int \frac {x^{3/2}}{(a+b x)^2} \, dx=-\frac {3 \, a \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} b^{2}} + \frac {a \sqrt {x}}{{\left (b x + a\right )} b^{2}} + \frac {2 \, \sqrt {x}}{b^{2}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 15.30 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.92 \[ \int \frac {x^{3/2}}{(a+b x)^2} \, dx=\frac {2\,\sqrt {x}}{b^2}+\frac {a\,\sqrt {x}}{x\,b^3+a\,b^2}-\frac {3\,\sqrt {a}\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {x}}{\sqrt {a}}\right )}{b^{5/2}} \]

[In]

int(x^(3/2)/(a + b*x)^2,x)

[Out]

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