Integrand size = 29, antiderivative size = 85 \[ \int \frac {\sqrt {x} (A+B x)}{a^2+2 a b x+b^2 x^2} \, dx=-\frac {(A b-3 a B) \sqrt {x}}{a b^2}+\frac {(A b-a B) x^{3/2}}{a b (a+b x)}+\frac {(A b-3 a B) \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{\sqrt {a} b^{5/2}} \]
(A*b-B*a)*x^(3/2)/a/b/(b*x+a)+(A*b-3*B*a)*arctan(b^(1/2)*x^(1/2)/a^(1/2))/ b^(5/2)/a^(1/2)-(A*b-3*B*a)*x^(1/2)/a/b^2
Time = 0.02 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.79 \[ \int \frac {\sqrt {x} (A+B x)}{a^2+2 a b x+b^2 x^2} \, dx=\frac {\sqrt {x} (-A b+3 a B+2 b B x)}{b^2 (a+b x)}+\frac {(A b-3 a B) \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{\sqrt {a} b^{5/2}} \]
(Sqrt[x]*(-(A*b) + 3*a*B + 2*b*B*x))/(b^2*(a + b*x)) + ((A*b - 3*a*B)*ArcT an[(Sqrt[b]*Sqrt[x])/Sqrt[a]])/(Sqrt[a]*b^(5/2))
Time = 0.22 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.01, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.207, Rules used = {1184, 27, 87, 60, 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.
\(\displaystyle \int \frac {\sqrt {x} (A+B x)}{a^2+2 a b x+b^2 x^2} \, dx\) |
\(\Big \downarrow \) 1184 |
\(\displaystyle b^2 \int \frac {\sqrt {x} (A+B x)}{b^2 (a+b x)^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \int \frac {\sqrt {x} (A+B x)}{(a+b x)^2}dx\) |
\(\Big \downarrow \) 87 |
\(\displaystyle \frac {x^{3/2} (A b-a B)}{a b (a+b x)}-\frac {(A b-3 a B) \int \frac {\sqrt {x}}{a+b x}dx}{2 a b}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {x^{3/2} (A b-a B)}{a b (a+b x)}-\frac {(A b-3 a B) \left (\frac {2 \sqrt {x}}{b}-\frac {a \int \frac {1}{\sqrt {x} (a+b x)}dx}{b}\right )}{2 a b}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {x^{3/2} (A b-a B)}{a b (a+b x)}-\frac {(A b-3 a B) \left (\frac {2 \sqrt {x}}{b}-\frac {2 a \int \frac {1}{a+b x}d\sqrt {x}}{b}\right )}{2 a b}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {x^{3/2} (A b-a B)}{a b (a+b x)}-\frac {(A b-3 a B) \left (\frac {2 \sqrt {x}}{b}-\frac {2 \sqrt {a} \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{b^{3/2}}\right )}{2 a b}\) |
((A*b - a*B)*x^(3/2))/(a*b*(a + b*x)) - ((A*b - 3*a*B)*((2*Sqrt[x])/b - (2 *Sqrt[a]*ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[a]])/b^(3/2)))/(2*a*b)
3.8.60.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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] + Simp[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] && GtQ[n, 0] && NeQ[m + n + 1, 0] && !(IGtQ[m, 0] && ( !Integer Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) && !ILtQ[m + n + 2, 0] && IntLinear Q[a, b, c, d, m, n, x]
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_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p _.), x_] :> Simp[(-(b*e - a*f))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p + 1)*(c*f - d*e))), x] - Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)) Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || Intege rQ[p] || !(IntegerQ[n] || !(EqQ[e, 0] || !(EqQ[c, 0] || LtQ[p, n]))))
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]
Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.)*((a_) + (b_.)*(x_ ) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[1/c^p Int[(d + e*x)^m*(f + g*x )^n*(b/2 + c*x)^(2*p), x], x] /; FreeQ[{a, b, c, d, e, f, g, m, n}, x] && E qQ[b^2 - 4*a*c, 0] && IntegerQ[p]
Time = 0.15 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.73
method | result | size |
risch | \(\frac {2 B \sqrt {x}}{b^{2}}+\frac {\frac {2 \left (-\frac {A b}{2}+\frac {B a}{2}\right ) \sqrt {x}}{b x +a}+\frac {\left (A b -3 B a \right ) \arctan \left (\frac {b \sqrt {x}}{\sqrt {b a}}\right )}{\sqrt {b a}}}{b^{2}}\) | \(62\) |
derivativedivides | \(\frac {2 B \sqrt {x}}{b^{2}}+\frac {\frac {2 \left (-\frac {A b}{2}+\frac {B a}{2}\right ) \sqrt {x}}{b x +a}+\frac {\left (A b -3 B a \right ) \arctan \left (\frac {b \sqrt {x}}{\sqrt {b a}}\right )}{\sqrt {b a}}}{b^{2}}\) | \(63\) |
default | \(\frac {2 B \sqrt {x}}{b^{2}}+\frac {\frac {2 \left (-\frac {A b}{2}+\frac {B a}{2}\right ) \sqrt {x}}{b x +a}+\frac {\left (A b -3 B a \right ) \arctan \left (\frac {b \sqrt {x}}{\sqrt {b a}}\right )}{\sqrt {b a}}}{b^{2}}\) | \(63\) |
2*B/b^2*x^(1/2)+1/b^2*(2*(-1/2*A*b+1/2*B*a)*x^(1/2)/(b*x+a)+(A*b-3*B*a)/(b *a)^(1/2)*arctan(b*x^(1/2)/(b*a)^(1/2)))
Time = 0.30 (sec) , antiderivative size = 198, normalized size of antiderivative = 2.33 \[ \int \frac {\sqrt {x} (A+B x)}{a^2+2 a b x+b^2 x^2} \, dx=\left [\frac {{\left (3 \, B a^{2} - A a b + {\left (3 \, B a b - A b^{2}\right )} x\right )} \sqrt {-a b} \log \left (\frac {b x - a - 2 \, \sqrt {-a b} \sqrt {x}}{b x + a}\right ) + 2 \, {\left (2 \, B a b^{2} x + 3 \, B a^{2} b - A a b^{2}\right )} \sqrt {x}}{2 \, {\left (a b^{4} x + a^{2} b^{3}\right )}}, \frac {{\left (3 \, B a^{2} - A a b + {\left (3 \, B a b - A b^{2}\right )} x\right )} \sqrt {a b} \arctan \left (\frac {\sqrt {a b}}{b \sqrt {x}}\right ) + {\left (2 \, B a b^{2} x + 3 \, B a^{2} b - A a b^{2}\right )} \sqrt {x}}{a b^{4} x + a^{2} b^{3}}\right ] \]
[1/2*((3*B*a^2 - A*a*b + (3*B*a*b - A*b^2)*x)*sqrt(-a*b)*log((b*x - a - 2* sqrt(-a*b)*sqrt(x))/(b*x + a)) + 2*(2*B*a*b^2*x + 3*B*a^2*b - A*a*b^2)*sqr t(x))/(a*b^4*x + a^2*b^3), ((3*B*a^2 - A*a*b + (3*B*a*b - A*b^2)*x)*sqrt(a *b)*arctan(sqrt(a*b)/(b*sqrt(x))) + (2*B*a*b^2*x + 3*B*a^2*b - A*a*b^2)*sq rt(x))/(a*b^4*x + a^2*b^3)]
Leaf count of result is larger than twice the leaf count of optimal. 634 vs. \(2 (73) = 146\).
Time = 2.30 (sec) , antiderivative size = 634, normalized size of antiderivative = 7.46 \[ \int \frac {\sqrt {x} (A+B x)}{a^2+2 a b x+b^2 x^2} \, dx=\begin {cases} \tilde {\infty } \left (- \frac {2 A}{\sqrt {x}} + 2 B \sqrt {x}\right ) & \text {for}\: a = 0 \wedge b = 0 \\\frac {\frac {2 A x^{\frac {3}{2}}}{3} + \frac {2 B x^{\frac {5}{2}}}{5}}{a^{2}} & \text {for}\: b = 0 \\\frac {- \frac {2 A}{\sqrt {x}} + 2 B \sqrt {x}}{b^{2}} & \text {for}\: a = 0 \\\frac {A a b \log {\left (\sqrt {x} - \sqrt {- \frac {a}{b}} \right )}}{2 a b^{3} \sqrt {- \frac {a}{b}} + 2 b^{4} x \sqrt {- \frac {a}{b}}} - \frac {A a b \log {\left (\sqrt {x} + \sqrt {- \frac {a}{b}} \right )}}{2 a b^{3} \sqrt {- \frac {a}{b}} + 2 b^{4} x \sqrt {- \frac {a}{b}}} - \frac {2 A b^{2} \sqrt {x} \sqrt {- \frac {a}{b}}}{2 a b^{3} \sqrt {- \frac {a}{b}} + 2 b^{4} x \sqrt {- \frac {a}{b}}} + \frac {A b^{2} x \log {\left (\sqrt {x} - \sqrt {- \frac {a}{b}} \right )}}{2 a b^{3} \sqrt {- \frac {a}{b}} + 2 b^{4} x \sqrt {- \frac {a}{b}}} - \frac {A b^{2} x \log {\left (\sqrt {x} + \sqrt {- \frac {a}{b}} \right )}}{2 a b^{3} \sqrt {- \frac {a}{b}} + 2 b^{4} x \sqrt {- \frac {a}{b}}} - \frac {3 B a^{2} \log {\left (\sqrt {x} - \sqrt {- \frac {a}{b}} \right )}}{2 a b^{3} \sqrt {- \frac {a}{b}} + 2 b^{4} x \sqrt {- \frac {a}{b}}} + \frac {3 B a^{2} \log {\left (\sqrt {x} + \sqrt {- \frac {a}{b}} \right )}}{2 a b^{3} \sqrt {- \frac {a}{b}} + 2 b^{4} x \sqrt {- \frac {a}{b}}} + \frac {6 B a b \sqrt {x} \sqrt {- \frac {a}{b}}}{2 a b^{3} \sqrt {- \frac {a}{b}} + 2 b^{4} x \sqrt {- \frac {a}{b}}} - \frac {3 B a b x \log {\left (\sqrt {x} - \sqrt {- \frac {a}{b}} \right )}}{2 a b^{3} \sqrt {- \frac {a}{b}} + 2 b^{4} x \sqrt {- \frac {a}{b}}} + \frac {3 B a b x \log {\left (\sqrt {x} + \sqrt {- \frac {a}{b}} \right )}}{2 a b^{3} \sqrt {- \frac {a}{b}} + 2 b^{4} x \sqrt {- \frac {a}{b}}} + \frac {4 B b^{2} x^{\frac {3}{2}} \sqrt {- \frac {a}{b}}}{2 a b^{3} \sqrt {- \frac {a}{b}} + 2 b^{4} x \sqrt {- \frac {a}{b}}} & \text {otherwise} \end {cases} \]
Piecewise((zoo*(-2*A/sqrt(x) + 2*B*sqrt(x)), Eq(a, 0) & Eq(b, 0)), ((2*A*x **(3/2)/3 + 2*B*x**(5/2)/5)/a**2, Eq(b, 0)), ((-2*A/sqrt(x) + 2*B*sqrt(x)) /b**2, Eq(a, 0)), (A*a*b*log(sqrt(x) - sqrt(-a/b))/(2*a*b**3*sqrt(-a/b) + 2*b**4*x*sqrt(-a/b)) - A*a*b*log(sqrt(x) + sqrt(-a/b))/(2*a*b**3*sqrt(-a/b ) + 2*b**4*x*sqrt(-a/b)) - 2*A*b**2*sqrt(x)*sqrt(-a/b)/(2*a*b**3*sqrt(-a/b ) + 2*b**4*x*sqrt(-a/b)) + A*b**2*x*log(sqrt(x) - sqrt(-a/b))/(2*a*b**3*sq rt(-a/b) + 2*b**4*x*sqrt(-a/b)) - A*b**2*x*log(sqrt(x) + sqrt(-a/b))/(2*a* b**3*sqrt(-a/b) + 2*b**4*x*sqrt(-a/b)) - 3*B*a**2*log(sqrt(x) - sqrt(-a/b) )/(2*a*b**3*sqrt(-a/b) + 2*b**4*x*sqrt(-a/b)) + 3*B*a**2*log(sqrt(x) + sqr t(-a/b))/(2*a*b**3*sqrt(-a/b) + 2*b**4*x*sqrt(-a/b)) + 6*B*a*b*sqrt(x)*sqr t(-a/b)/(2*a*b**3*sqrt(-a/b) + 2*b**4*x*sqrt(-a/b)) - 3*B*a*b*x*log(sqrt(x ) - sqrt(-a/b))/(2*a*b**3*sqrt(-a/b) + 2*b**4*x*sqrt(-a/b)) + 3*B*a*b*x*lo g(sqrt(x) + sqrt(-a/b))/(2*a*b**3*sqrt(-a/b) + 2*b**4*x*sqrt(-a/b)) + 4*B* b**2*x**(3/2)*sqrt(-a/b)/(2*a*b**3*sqrt(-a/b) + 2*b**4*x*sqrt(-a/b)), True ))
Time = 0.28 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.76 \[ \int \frac {\sqrt {x} (A+B x)}{a^2+2 a b x+b^2 x^2} \, dx=\frac {{\left (B a - A b\right )} \sqrt {x}}{b^{3} x + a b^{2}} + \frac {2 \, B \sqrt {x}}{b^{2}} - \frac {{\left (3 \, B a - A b\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} b^{2}} \]
(B*a - A*b)*sqrt(x)/(b^3*x + a*b^2) + 2*B*sqrt(x)/b^2 - (3*B*a - A*b)*arct an(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*b^2)
Time = 0.28 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.76 \[ \int \frac {\sqrt {x} (A+B x)}{a^2+2 a b x+b^2 x^2} \, dx=\frac {2 \, B \sqrt {x}}{b^{2}} - \frac {{\left (3 \, B a - A b\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} b^{2}} + \frac {B a \sqrt {x} - A b \sqrt {x}}{{\left (b x + a\right )} b^{2}} \]
2*B*sqrt(x)/b^2 - (3*B*a - A*b)*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*b^2 ) + (B*a*sqrt(x) - A*b*sqrt(x))/((b*x + a)*b^2)
Time = 0.10 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.73 \[ \int \frac {\sqrt {x} (A+B x)}{a^2+2 a b x+b^2 x^2} \, dx=\frac {2\,B\,\sqrt {x}}{b^2}-\frac {\sqrt {x}\,\left (A\,b-B\,a\right )}{x\,b^3+a\,b^2}+\frac {\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {x}}{\sqrt {a}}\right )\,\left (A\,b-3\,B\,a\right )}{\sqrt {a}\,b^{5/2}} \]