Integrand size = 31, antiderivative size = 99 \[ \int \frac {A+B x}{\sqrt {x} \sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\frac {2 B \sqrt {x} (a+b x)}{b \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 (A b-a B) (a+b x) \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{\sqrt {a} b^{3/2} \sqrt {a^2+2 a b x+b^2 x^2}} \] Output:
2*B*x^(1/2)*(b*x+a)/b/((b*x+a)^2)^(1/2)+2*(A*b-B*a)*(b*x+a)*arctan(b^(1/2) *x^(1/2)/a^(1/2))/a^(1/2)/b^(3/2)/((b*x+a)^2)^(1/2)
Time = 0.09 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.73 \[ \int \frac {A+B x}{\sqrt {x} \sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\frac {2 (a+b x) \left (\sqrt {a} \sqrt {b} B \sqrt {x}+(A b-a B) \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )\right )}{\sqrt {a} b^{3/2} \sqrt {(a+b x)^2}} \] Input:
Integrate[(A + B*x)/(Sqrt[x]*Sqrt[a^2 + 2*a*b*x + b^2*x^2]),x]
Output:
(2*(a + b*x)*(Sqrt[a]*Sqrt[b]*B*Sqrt[x] + (A*b - a*B)*ArcTan[(Sqrt[b]*Sqrt [x])/Sqrt[a]]))/(Sqrt[a]*b^(3/2)*Sqrt[(a + b*x)^2])
Time = 0.35 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.76, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.161, Rules used = {1187, 27, 90, 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.
\(\displaystyle \int \frac {A+B x}{\sqrt {x} \sqrt {a^2+2 a b x+b^2 x^2}} \, dx\) |
\(\Big \downarrow \) 1187 |
\(\displaystyle \frac {b (a+b x) \int \frac {A+B x}{b \sqrt {x} (a+b x)}dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {(a+b x) \int \frac {A+B x}{\sqrt {x} (a+b x)}dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
\(\Big \downarrow \) 90 |
\(\displaystyle \frac {(a+b x) \left (\frac {(A b-a B) \int \frac {1}{\sqrt {x} (a+b x)}dx}{b}+\frac {2 B \sqrt {x}}{b}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {(a+b x) \left (\frac {2 (A b-a B) \int \frac {1}{a+b x}d\sqrt {x}}{b}+\frac {2 B \sqrt {x}}{b}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {(a+b x) \left (\frac {2 (A b-a B) \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{\sqrt {a} b^{3/2}}+\frac {2 B \sqrt {x}}{b}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
Input:
Int[(A + B*x)/(Sqrt[x]*Sqrt[a^2 + 2*a*b*x + b^2*x^2]),x]
Output:
((a + b*x)*((2*B*Sqrt[x])/b + (2*(A*b - a*B)*ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt [a]])/(Sqrt[a]*b^(3/2))))/Sqrt[a^2 + 2*a*b*x + b^2*x^2]
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] :> 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*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 2))), x] + Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(d*f*(n + p + 2)) Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2, 0]
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[(a + b*x + c*x^2)^FracPart[p]/(c^ IntPart[p]*(b/2 + c*x)^(2*FracPart[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, p}, x] && EqQ[b^2 - 4*a*c, 0] && !IntegerQ[p]
Time = 0.78 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.66
method | result | size |
default | \(\frac {2 \left (b x +a \right ) \left (A \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right ) b +B \sqrt {x}\, \sqrt {a b}-B \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right ) a \right )}{\sqrt {\left (b x +a \right )^{2}}\, b \sqrt {a b}}\) | \(65\) |
risch | \(\frac {2 B \sqrt {x}\, \sqrt {\left (b x +a \right )^{2}}}{b \left (b x +a \right )}+\frac {2 \left (A b -B a \right ) \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right ) \sqrt {\left (b x +a \right )^{2}}}{b \sqrt {a b}\, \left (b x +a \right )}\) | \(72\) |
Input:
int((B*x+A)/x^(1/2)/((b*x+a)^2)^(1/2),x,method=_RETURNVERBOSE)
Output:
2*(b*x+a)*(A*arctan(b*x^(1/2)/(a*b)^(1/2))*b+B*x^(1/2)*(a*b)^(1/2)-B*arcta n(b*x^(1/2)/(a*b)^(1/2))*a)/((b*x+a)^2)^(1/2)/b/(a*b)^(1/2)
Time = 0.08 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.03 \[ \int \frac {A+B x}{\sqrt {x} \sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\left [\frac {2 \, B a b \sqrt {x} + {\left (B a - A b\right )} \sqrt {-a b} \log \left (\frac {b x - a - 2 \, \sqrt {-a b} \sqrt {x}}{b x + a}\right )}{a b^{2}}, \frac {2 \, {\left (B a b \sqrt {x} + {\left (B a - A b\right )} \sqrt {a b} \arctan \left (\frac {\sqrt {a b}}{b \sqrt {x}}\right )\right )}}{a b^{2}}\right ] \] Input:
integrate((B*x+A)/x^(1/2)/((b*x+a)^2)^(1/2),x, algorithm="fricas")
Output:
[(2*B*a*b*sqrt(x) + (B*a - A*b)*sqrt(-a*b)*log((b*x - a - 2*sqrt(-a*b)*sqr t(x))/(b*x + a)))/(a*b^2), 2*(B*a*b*sqrt(x) + (B*a - A*b)*sqrt(a*b)*arctan (sqrt(a*b)/(b*sqrt(x))))/(a*b^2)]
\[ \int \frac {A+B x}{\sqrt {x} \sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\int \frac {A + B x}{\sqrt {x} \sqrt {\left (a + b x\right )^{2}}}\, dx \] Input:
integrate((B*x+A)/x**(1/2)/((b*x+a)**2)**(1/2),x)
Output:
Integral((A + B*x)/(sqrt(x)*sqrt((a + b*x)**2)), x)
Leaf count of result is larger than twice the leaf count of optimal. 140 vs. \(2 (65) = 130\).
Time = 0.16 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.41 \[ \int \frac {A+B x}{\sqrt {x} \sqrt {a^2+2 a b x+b^2 x^2}} \, dx=-\frac {2 \, {\left (B a - A b\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} b} + \frac {{\left ({\left (3 \, B a b - A b^{2}\right )} x^{2} + 3 \, {\left (B a^{2} + A a b\right )} x\right )} \sqrt {x} + \frac {2 \, {\left (A a b x^{2} + 3 \, A a^{2} x\right )}}{\sqrt {x}}}{3 \, {\left (a^{2} b x + a^{3}\right )}} - \frac {{\left (3 \, B a b - A b^{2}\right )} x^{\frac {3}{2}} - 6 \, {\left (B a^{2} - A a b\right )} \sqrt {x}}{3 \, a^{2} b} \] Input:
integrate((B*x+A)/x^(1/2)/((b*x+a)^2)^(1/2),x, algorithm="maxima")
Output:
-2*(B*a - A*b)*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*b) + 1/3*(((3*B*a*b - A*b^2)*x^2 + 3*(B*a^2 + A*a*b)*x)*sqrt(x) + 2*(A*a*b*x^2 + 3*A*a^2*x)/sq rt(x))/(a^2*b*x + a^3) - 1/3*((3*B*a*b - A*b^2)*x^(3/2) - 6*(B*a^2 - A*a*b )*sqrt(x))/(a^2*b)
Time = 0.27 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.58 \[ \int \frac {A+B x}{\sqrt {x} \sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\frac {2 \, B \sqrt {x} \mathrm {sgn}\left (b x + a\right )}{b} - \frac {2 \, {\left (B a \mathrm {sgn}\left (b x + a\right ) - A b \mathrm {sgn}\left (b x + a\right )\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} b} \] Input:
integrate((B*x+A)/x^(1/2)/((b*x+a)^2)^(1/2),x, algorithm="giac")
Output:
2*B*sqrt(x)*sgn(b*x + a)/b - 2*(B*a*sgn(b*x + a) - A*b*sgn(b*x + a))*arcta n(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*b)
Timed out. \[ \int \frac {A+B x}{\sqrt {x} \sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\int \frac {A+B\,x}{\sqrt {x}\,\sqrt {{\left (a+b\,x\right )}^2}} \,d x \] Input:
int((A + B*x)/(x^(1/2)*((a + b*x)^2)^(1/2)),x)
Output:
int((A + B*x)/(x^(1/2)*((a + b*x)^2)^(1/2)), x)
Time = 0.26 (sec) , antiderivative size = 4, normalized size of antiderivative = 0.04 \[ \int \frac {A+B x}{\sqrt {x} \sqrt {a^2+2 a b x+b^2 x^2}} \, dx=2 \sqrt {x} \] Input:
int((B*x+A)/x^(1/2)/((b*x+a)^2)^(1/2),x)
Output:
2*sqrt(x)