Integrand size = 13, antiderivative size = 73 \[ \int \frac {\sqrt {x}}{(a+b x)^3} \, dx=-\frac {\sqrt {x}}{2 b (a+b x)^2}+\frac {\sqrt {x}}{4 a b (a+b x)}+\frac {\arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{4 a^{3/2} b^{3/2}} \]
1/4*arctan(b^(1/2)*x^(1/2)/a^(1/2))/a^(3/2)/b^(3/2)-1/2*x^(1/2)/b/(b*x+a)^ 2+1/4*x^(1/2)/a/b/(b*x+a)
Time = 0.10 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.82 \[ \int \frac {\sqrt {x}}{(a+b x)^3} \, dx=-\frac {\sqrt {x} (a-b x)}{4 a b (a+b x)^2}+\frac {\arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{4 a^{3/2} b^{3/2}} \]
-1/4*(Sqrt[x]*(a - b*x))/(a*b*(a + b*x)^2) + ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt [a]]/(4*a^(3/2)*b^(3/2))
Time = 0.16 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.99, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {51, 52, 73, 218}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {\sqrt {x}}{(a+b x)^3} \, dx\) |
\(\Big \downarrow \) 51 |
\(\displaystyle \frac {\int \frac {1}{\sqrt {x} (a+b x)^2}dx}{4 b}-\frac {\sqrt {x}}{2 b (a+b x)^2}\) |
\(\Big \downarrow \) 52 |
\(\displaystyle \frac {\frac {\int \frac {1}{\sqrt {x} (a+b x)}dx}{2 a}+\frac {\sqrt {x}}{a (a+b x)}}{4 b}-\frac {\sqrt {x}}{2 b (a+b x)^2}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {\frac {\int \frac {1}{a+b x}d\sqrt {x}}{a}+\frac {\sqrt {x}}{a (a+b x)}}{4 b}-\frac {\sqrt {x}}{2 b (a+b x)^2}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {\frac {\arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{a^{3/2} \sqrt {b}}+\frac {\sqrt {x}}{a (a+b x)}}{4 b}-\frac {\sqrt {x}}{2 b (a+b x)^2}\) |
-1/2*Sqrt[x]/(b*(a + b*x)^2) + (Sqrt[x]/(a*(a + b*x)) + ArcTan[(Sqrt[b]*Sq rt[x])/Sqrt[a]]/(a^(3/2)*Sqrt[b]))/(4*b)
3.5.65.3.1 Defintions of rubi rules used
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] - Simp[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 ] && ILtQ[m, -1] && FractionQ[n] && GtQ[n, 0]
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] - Simp[d*(( m + n + 2)/((b*c - a*d)*(m + 1))) Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && ILtQ[m, -1] && FractionQ[n] && LtQ[n, 0]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[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] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
Time = 0.07 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.71
method | result | size |
derivativedivides | \(\frac {\frac {x^{\frac {3}{2}}}{4 a}-\frac {\sqrt {x}}{4 b}}{\left (b x +a \right )^{2}}+\frac {\arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{4 b a \sqrt {a b}}\) | \(52\) |
default | \(\frac {\frac {x^{\frac {3}{2}}}{4 a}-\frac {\sqrt {x}}{4 b}}{\left (b x +a \right )^{2}}+\frac {\arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{4 b a \sqrt {a b}}\) | \(52\) |
Time = 0.24 (sec) , antiderivative size = 186, normalized size of antiderivative = 2.55 \[ \int \frac {\sqrt {x}}{(a+b x)^3} \, dx=\left [-\frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )} \sqrt {-a b} \log \left (\frac {b x - a - 2 \, \sqrt {-a b} \sqrt {x}}{b x + a}\right ) - 2 \, {\left (a b^{2} x - a^{2} b\right )} \sqrt {x}}{8 \, {\left (a^{2} b^{4} x^{2} + 2 \, a^{3} b^{3} x + a^{4} b^{2}\right )}}, -\frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )} \sqrt {a b} \arctan \left (\frac {\sqrt {a b}}{b \sqrt {x}}\right ) - {\left (a b^{2} x - a^{2} b\right )} \sqrt {x}}{4 \, {\left (a^{2} b^{4} x^{2} + 2 \, a^{3} b^{3} x + a^{4} b^{2}\right )}}\right ] \]
[-1/8*((b^2*x^2 + 2*a*b*x + a^2)*sqrt(-a*b)*log((b*x - a - 2*sqrt(-a*b)*sq rt(x))/(b*x + a)) - 2*(a*b^2*x - a^2*b)*sqrt(x))/(a^2*b^4*x^2 + 2*a^3*b^3* x + a^4*b^2), -1/4*((b^2*x^2 + 2*a*b*x + a^2)*sqrt(a*b)*arctan(sqrt(a*b)/( b*sqrt(x))) - (a*b^2*x - a^2*b)*sqrt(x))/(a^2*b^4*x^2 + 2*a^3*b^3*x + a^4* b^2)]
Leaf count of result is larger than twice the leaf count of optimal. 627 vs. \(2 (58) = 116\).
Time = 8.31 (sec) , antiderivative size = 627, normalized size of antiderivative = 8.59 \[ \int \frac {\sqrt {x}}{(a+b x)^3} \, dx=\begin {cases} \frac {\tilde {\infty }}{x^{\frac {3}{2}}} & \text {for}\: a = 0 \wedge b = 0 \\\frac {2 x^{\frac {3}{2}}}{3 a^{3}} & \text {for}\: b = 0 \\- \frac {2}{3 b^{3} x^{\frac {3}{2}}} & \text {for}\: a = 0 \\\frac {a^{2} \log {\left (\sqrt {x} - \sqrt {- \frac {a}{b}} \right )}}{8 a^{3} b^{2} \sqrt {- \frac {a}{b}} + 16 a^{2} b^{3} x \sqrt {- \frac {a}{b}} + 8 a b^{4} x^{2} \sqrt {- \frac {a}{b}}} - \frac {a^{2} \log {\left (\sqrt {x} + \sqrt {- \frac {a}{b}} \right )}}{8 a^{3} b^{2} \sqrt {- \frac {a}{b}} + 16 a^{2} b^{3} x \sqrt {- \frac {a}{b}} + 8 a b^{4} x^{2} \sqrt {- \frac {a}{b}}} - \frac {2 a b \sqrt {x} \sqrt {- \frac {a}{b}}}{8 a^{3} b^{2} \sqrt {- \frac {a}{b}} + 16 a^{2} b^{3} x \sqrt {- \frac {a}{b}} + 8 a b^{4} x^{2} \sqrt {- \frac {a}{b}}} + \frac {2 a b x \log {\left (\sqrt {x} - \sqrt {- \frac {a}{b}} \right )}}{8 a^{3} b^{2} \sqrt {- \frac {a}{b}} + 16 a^{2} b^{3} x \sqrt {- \frac {a}{b}} + 8 a b^{4} x^{2} \sqrt {- \frac {a}{b}}} - \frac {2 a b x \log {\left (\sqrt {x} + \sqrt {- \frac {a}{b}} \right )}}{8 a^{3} b^{2} \sqrt {- \frac {a}{b}} + 16 a^{2} b^{3} x \sqrt {- \frac {a}{b}} + 8 a b^{4} x^{2} \sqrt {- \frac {a}{b}}} + \frac {2 b^{2} x^{\frac {3}{2}} \sqrt {- \frac {a}{b}}}{8 a^{3} b^{2} \sqrt {- \frac {a}{b}} + 16 a^{2} b^{3} x \sqrt {- \frac {a}{b}} + 8 a b^{4} x^{2} \sqrt {- \frac {a}{b}}} + \frac {b^{2} x^{2} \log {\left (\sqrt {x} - \sqrt {- \frac {a}{b}} \right )}}{8 a^{3} b^{2} \sqrt {- \frac {a}{b}} + 16 a^{2} b^{3} x \sqrt {- \frac {a}{b}} + 8 a b^{4} x^{2} \sqrt {- \frac {a}{b}}} - \frac {b^{2} x^{2} \log {\left (\sqrt {x} + \sqrt {- \frac {a}{b}} \right )}}{8 a^{3} b^{2} \sqrt {- \frac {a}{b}} + 16 a^{2} b^{3} x \sqrt {- \frac {a}{b}} + 8 a b^{4} x^{2} \sqrt {- \frac {a}{b}}} & \text {otherwise} \end {cases} \]
Piecewise((zoo/x**(3/2), Eq(a, 0) & Eq(b, 0)), (2*x**(3/2)/(3*a**3), Eq(b, 0)), (-2/(3*b**3*x**(3/2)), Eq(a, 0)), (a**2*log(sqrt(x) - sqrt(-a/b))/(8 *a**3*b**2*sqrt(-a/b) + 16*a**2*b**3*x*sqrt(-a/b) + 8*a*b**4*x**2*sqrt(-a/ b)) - a**2*log(sqrt(x) + sqrt(-a/b))/(8*a**3*b**2*sqrt(-a/b) + 16*a**2*b** 3*x*sqrt(-a/b) + 8*a*b**4*x**2*sqrt(-a/b)) - 2*a*b*sqrt(x)*sqrt(-a/b)/(8*a **3*b**2*sqrt(-a/b) + 16*a**2*b**3*x*sqrt(-a/b) + 8*a*b**4*x**2*sqrt(-a/b) ) + 2*a*b*x*log(sqrt(x) - sqrt(-a/b))/(8*a**3*b**2*sqrt(-a/b) + 16*a**2*b* *3*x*sqrt(-a/b) + 8*a*b**4*x**2*sqrt(-a/b)) - 2*a*b*x*log(sqrt(x) + sqrt(- a/b))/(8*a**3*b**2*sqrt(-a/b) + 16*a**2*b**3*x*sqrt(-a/b) + 8*a*b**4*x**2* sqrt(-a/b)) + 2*b**2*x**(3/2)*sqrt(-a/b)/(8*a**3*b**2*sqrt(-a/b) + 16*a**2 *b**3*x*sqrt(-a/b) + 8*a*b**4*x**2*sqrt(-a/b)) + b**2*x**2*log(sqrt(x) - s qrt(-a/b))/(8*a**3*b**2*sqrt(-a/b) + 16*a**2*b**3*x*sqrt(-a/b) + 8*a*b**4* x**2*sqrt(-a/b)) - b**2*x**2*log(sqrt(x) + sqrt(-a/b))/(8*a**3*b**2*sqrt(- a/b) + 16*a**2*b**3*x*sqrt(-a/b) + 8*a*b**4*x**2*sqrt(-a/b)), True))
Time = 0.39 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.88 \[ \int \frac {\sqrt {x}}{(a+b x)^3} \, dx=\frac {b x^{\frac {3}{2}} - a \sqrt {x}}{4 \, {\left (a b^{3} x^{2} + 2 \, a^{2} b^{2} x + a^{3} b\right )}} + \frac {\arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{4 \, \sqrt {a b} a b} \]
1/4*(b*x^(3/2) - a*sqrt(x))/(a*b^3*x^2 + 2*a^2*b^2*x + a^3*b) + 1/4*arctan (b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*a*b)
Time = 0.30 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.71 \[ \int \frac {\sqrt {x}}{(a+b x)^3} \, dx=\frac {\arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{4 \, \sqrt {a b} a b} + \frac {b x^{\frac {3}{2}} - a \sqrt {x}}{4 \, {\left (b x + a\right )}^{2} a b} \]
1/4*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*a*b) + 1/4*(b*x^(3/2) - a*sqrt( x))/((b*x + a)^2*a*b)
Time = 0.14 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.77 \[ \int \frac {\sqrt {x}}{(a+b x)^3} \, dx=\frac {\frac {x^{3/2}}{4\,a}-\frac {\sqrt {x}}{4\,b}}{a^2+2\,a\,b\,x+b^2\,x^2}+\frac {\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {x}}{\sqrt {a}}\right )}{4\,a^{3/2}\,b^{3/2}} \]