Integrand size = 31, antiderivative size = 144 \[ \int \frac {\sqrt {x} (A+B x)}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\frac {2 (A b-a B) \sqrt {x} (a+b x)}{b^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 B x^{3/2} (a+b x)}{3 b \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {2 \sqrt {a} (A b-a B) (a+b x) \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{b^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}} \] Output:
2*(A*b-B*a)*x^(1/2)*(b*x+a)/b^2/((b*x+a)^2)^(1/2)+2/3*B*x^(3/2)*(b*x+a)/b/ ((b*x+a)^2)^(1/2)-2*a^(1/2)*(A*b-B*a)*(b*x+a)*arctan(b^(1/2)*x^(1/2)/a^(1/ 2))/b^(5/2)/((b*x+a)^2)^(1/2)
Time = 0.10 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.57 \[ \int \frac {\sqrt {x} (A+B x)}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\frac {2 (a+b x) \left (\sqrt {b} \sqrt {x} (3 A b-3 a B+b B x)+3 \sqrt {a} (-A b+a B) \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )\right )}{3 b^{5/2} \sqrt {(a+b x)^2}} \] Input:
Integrate[(Sqrt[x]*(A + B*x))/Sqrt[a^2 + 2*a*b*x + b^2*x^2],x]
Output:
(2*(a + b*x)*(Sqrt[b]*Sqrt[x]*(3*A*b - 3*a*B + b*B*x) + 3*Sqrt[a]*(-(A*b) + a*B)*ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[a]]))/(3*b^(5/2)*Sqrt[(a + b*x)^2])
Time = 0.35 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.64, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.194, Rules used = {1187, 27, 90, 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.
\(\displaystyle \int \frac {\sqrt {x} (A+B x)}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx\) |
\(\Big \downarrow \) 1187 |
\(\displaystyle \frac {b (a+b x) \int \frac {\sqrt {x} (A+B x)}{b (a+b x)}dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {(a+b x) \int \frac {\sqrt {x} (A+B x)}{a+b x}dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
\(\Big \downarrow \) 90 |
\(\displaystyle \frac {(a+b x) \left (\frac {(A b-a B) \int \frac {\sqrt {x}}{a+b x}dx}{b}+\frac {2 B x^{3/2}}{3 b}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {(a+b x) \left (\frac {(A b-a B) \left (\frac {2 \sqrt {x}}{b}-\frac {a \int \frac {1}{\sqrt {x} (a+b x)}dx}{b}\right )}{b}+\frac {2 B x^{3/2}}{3 b}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {(a+b x) \left (\frac {(A b-a B) \left (\frac {2 \sqrt {x}}{b}-\frac {2 a \int \frac {1}{a+b x}d\sqrt {x}}{b}\right )}{b}+\frac {2 B x^{3/2}}{3 b}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {(a+b x) \left (\frac {(A b-a B) \left (\frac {2 \sqrt {x}}{b}-\frac {2 \sqrt {a} \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{b^{3/2}}\right )}{b}+\frac {2 B x^{3/2}}{3 b}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
Input:
Int[(Sqrt[x]*(A + B*x))/Sqrt[a^2 + 2*a*b*x + b^2*x^2],x]
Output:
((a + b*x)*((2*B*x^(3/2))/(3*b) + ((A*b - a*B)*((2*Sqrt[x])/b - (2*Sqrt[a] *ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[a]])/b^(3/2)))/b))/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] :> 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*(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 = 0.72 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.59
method | result | size |
risch | \(\frac {2 \left (B b x +3 A b -3 B a \right ) \sqrt {x}\, \sqrt {\left (b x +a \right )^{2}}}{3 b^{2} \left (b x +a \right )}-\frac {2 a \left (A b -B a \right ) \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right ) \sqrt {\left (b x +a \right )^{2}}}{b^{2} \sqrt {a b}\, \left (b x +a \right )}\) | \(85\) |
default | \(\frac {2 \left (b x +a \right ) \left (B \sqrt {a b}\, x^{\frac {3}{2}} b +3 A \sqrt {a b}\, \sqrt {x}\, b -3 A \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right ) a b -3 B \sqrt {a b}\, \sqrt {x}\, a +3 B \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right ) a^{2}\right )}{3 \sqrt {\left (b x +a \right )^{2}}\, b^{2} \sqrt {a b}}\) | \(94\) |
Input:
int(x^(1/2)*(B*x+A)/((b*x+a)^2)^(1/2),x,method=_RETURNVERBOSE)
Output:
2/3*(B*b*x+3*A*b-3*B*a)*x^(1/2)/b^2*((b*x+a)^2)^(1/2)/(b*x+a)-2*a*(A*b-B*a )/b^2/(a*b)^(1/2)*arctan(b*x^(1/2)/(a*b)^(1/2))*((b*x+a)^2)^(1/2)/(b*x+a)
Time = 0.08 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.90 \[ \int \frac {\sqrt {x} (A+B x)}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\left [-\frac {3 \, {\left (B a - A b\right )} \sqrt {-\frac {a}{b}} \log \left (\frac {b x - 2 \, b \sqrt {x} \sqrt {-\frac {a}{b}} - a}{b x + a}\right ) - 2 \, {\left (B b x - 3 \, B a + 3 \, A b\right )} \sqrt {x}}{3 \, b^{2}}, \frac {2 \, {\left (3 \, {\left (B a - A b\right )} \sqrt {\frac {a}{b}} \arctan \left (\frac {b \sqrt {x} \sqrt {\frac {a}{b}}}{a}\right ) + {\left (B b x - 3 \, B a + 3 \, A b\right )} \sqrt {x}\right )}}{3 \, b^{2}}\right ] \] Input:
integrate(x^(1/2)*(B*x+A)/((b*x+a)^2)^(1/2),x, algorithm="fricas")
Output:
[-1/3*(3*(B*a - A*b)*sqrt(-a/b)*log((b*x - 2*b*sqrt(x)*sqrt(-a/b) - a)/(b* x + a)) - 2*(B*b*x - 3*B*a + 3*A*b)*sqrt(x))/b^2, 2/3*(3*(B*a - A*b)*sqrt( a/b)*arctan(b*sqrt(x)*sqrt(a/b)/a) + (B*b*x - 3*B*a + 3*A*b)*sqrt(x))/b^2]
\[ \int \frac {\sqrt {x} (A+B x)}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\int \frac {\sqrt {x} \left (A + B x\right )}{\sqrt {\left (a + b x\right )^{2}}}\, dx \] Input:
integrate(x**(1/2)*(B*x+A)/((b*x+a)**2)**(1/2),x)
Output:
Integral(sqrt(x)*(A + B*x)/sqrt((a + b*x)**2), x)
Time = 0.17 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.85 \[ \int \frac {\sqrt {x} (A+B x)}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx=-\frac {{\left ({\left (B a b - A b^{2}\right )} x^{2} + {\left (B a^{2} - A a b\right )} x\right )} \sqrt {x}}{a b^{2} x + a^{2} b} + \frac {2 \, {\left (B a^{2} - A a b\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} b^{2}} + \frac {{\left (5 \, B a b - 3 \, A b^{2}\right )} x^{\frac {3}{2}} - 6 \, {\left (B a^{2} - A a b\right )} \sqrt {x}}{3 \, a b^{2}} \] Input:
integrate(x^(1/2)*(B*x+A)/((b*x+a)^2)^(1/2),x, algorithm="maxima")
Output:
-((B*a*b - A*b^2)*x^2 + (B*a^2 - A*a*b)*x)*sqrt(x)/(a*b^2*x + a^2*b) + 2*( B*a^2 - A*a*b)*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*b^2) + 1/3*((5*B*a*b - 3*A*b^2)*x^(3/2) - 6*(B*a^2 - A*a*b)*sqrt(x))/(a*b^2)
Time = 0.23 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.65 \[ \int \frac {\sqrt {x} (A+B x)}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\frac {2 \, {\left (B a^{2} \mathrm {sgn}\left (b x + a\right ) - A a b \mathrm {sgn}\left (b x + a\right )\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} b^{2}} + \frac {2 \, {\left (B b^{2} x^{\frac {3}{2}} \mathrm {sgn}\left (b x + a\right ) - 3 \, B a b \sqrt {x} \mathrm {sgn}\left (b x + a\right ) + 3 \, A b^{2} \sqrt {x} \mathrm {sgn}\left (b x + a\right )\right )}}{3 \, b^{3}} \] Input:
integrate(x^(1/2)*(B*x+A)/((b*x+a)^2)^(1/2),x, algorithm="giac")
Output:
2*(B*a^2*sgn(b*x + a) - A*a*b*sgn(b*x + a))*arctan(b*sqrt(x)/sqrt(a*b))/(s qrt(a*b)*b^2) + 2/3*(B*b^2*x^(3/2)*sgn(b*x + a) - 3*B*a*b*sqrt(x)*sgn(b*x + a) + 3*A*b^2*sqrt(x)*sgn(b*x + a))/b^3
Timed out. \[ \int \frac {\sqrt {x} (A+B x)}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\int \frac {\sqrt {x}\,\left (A+B\,x\right )}{\sqrt {{\left (a+b\,x\right )}^2}} \,d x \] Input:
int((x^(1/2)*(A + B*x))/((a + b*x)^2)^(1/2),x)
Output:
int((x^(1/2)*(A + B*x))/((a + b*x)^2)^(1/2), x)
Time = 0.27 (sec) , antiderivative size = 5, normalized size of antiderivative = 0.03 \[ \int \frac {\sqrt {x} (A+B x)}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\frac {2 \sqrt {x}\, x}{3} \] Input:
int(x^(1/2)*(B*x+A)/((b*x+a)^2)^(1/2),x)
Output:
(2*sqrt(x)*x)/3