Integrand size = 18, antiderivative size = 69 \[ \int \frac {\sqrt {x} (A+B x)}{a+b x} \, dx=\frac {2 (A b-a B) \sqrt {x}}{b^2}+\frac {2 B x^{3/2}}{3 b}-\frac {2 \sqrt {a} (A b-a B) \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{b^{5/2}} \] Output:
2*(A*b-B*a)*x^(1/2)/b^2+2/3*B*x^(3/2)/b-2*a^(1/2)*(A*b-B*a)*arctan(b^(1/2) *x^(1/2)/a^(1/2))/b^(5/2)
Time = 0.06 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.91 \[ \int \frac {\sqrt {x} (A+B x)}{a+b x} \, dx=\frac {2 \sqrt {x} (3 A b-3 a B+b B x)}{3 b^2}+\frac {2 \sqrt {a} (-A b+a B) \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{b^{5/2}} \] Input:
Integrate[(Sqrt[x]*(A + B*x))/(a + b*x),x]
Output:
(2*Sqrt[x]*(3*A*b - 3*a*B + b*B*x))/(3*b^2) + (2*Sqrt[a]*(-(A*b) + a*B)*Ar cTan[(Sqrt[b]*Sqrt[x])/Sqrt[a]])/b^(5/2)
Time = 0.16 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.96, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {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.
\(\displaystyle \int \frac {\sqrt {x} (A+B x)}{a+b x} \, dx\) |
\(\Big \downarrow \) 90 |
\(\displaystyle \frac {(A b-a B) \int \frac {\sqrt {x}}{a+b x}dx}{b}+\frac {2 B x^{3/2}}{3 b}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {(A b-a B) \left (\frac {2 \sqrt {x}}{b}-\frac {a \int \frac {1}{\sqrt {x} (a+b x)}dx}{b}\right )}{b}+\frac {2 B x^{3/2}}{3 b}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {(A b-a B) \left (\frac {2 \sqrt {x}}{b}-\frac {2 a \int \frac {1}{a+b x}d\sqrt {x}}{b}\right )}{b}+\frac {2 B x^{3/2}}{3 b}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {(A b-a B) \left (\frac {2 \sqrt {x}}{b}-\frac {2 \sqrt {a} \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{b^{3/2}}\right )}{b}+\frac {2 B x^{3/2}}{3 b}\) |
Input:
Int[(Sqrt[x]*(A + B*x))/(a + b*x),x]
Output:
(2*B*x^(3/2))/(3*b) + ((A*b - a*B)*((2*Sqrt[x])/b - (2*Sqrt[a]*ArcTan[(Sqr t[b]*Sqrt[x])/Sqrt[a]])/b^(3/2)))/b
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]
Time = 0.15 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.77
method | result | size |
risch | \(\frac {2 \left (b B x +3 A b -3 B a \right ) \sqrt {x}}{3 b^{2}}-\frac {2 a \left (A b -B a \right ) \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{b^{2} \sqrt {a b}}\) | \(53\) |
derivativedivides | \(\frac {\frac {2 b B \,x^{\frac {3}{2}}}{3}+2 A b \sqrt {x}-2 B a \sqrt {x}}{b^{2}}-\frac {2 a \left (A b -B a \right ) \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{b^{2} \sqrt {a b}}\) | \(58\) |
default | \(\frac {\frac {2 b B \,x^{\frac {3}{2}}}{3}+2 A b \sqrt {x}-2 B a \sqrt {x}}{b^{2}}-\frac {2 a \left (A b -B a \right ) \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{b^{2} \sqrt {a b}}\) | \(58\) |
Input:
int(x^(1/2)*(B*x+A)/(b*x+a),x,method=_RETURNVERBOSE)
Output:
2/3*(B*b*x+3*A*b-3*B*a)*x^(1/2)/b^2-2*a*(A*b-B*a)/b^2/(a*b)^(1/2)*arctan(b *x^(1/2)/(a*b)^(1/2))
Time = 0.09 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.87 \[ \int \frac {\sqrt {x} (A+B x)}{a+b x} \, dx=\left [-\frac {3 \, {\left (B a - A b\right )} \sqrt {-\frac {a}{b}} \log \left (\frac {b x - 2 \, b \sqrt {x} \sqrt {-\frac {a}{b}} - a}{b x + a}\right ) - 2 \, {\left (B b x - 3 \, B a + 3 \, A b\right )} \sqrt {x}}{3 \, b^{2}}, \frac {2 \, {\left (3 \, {\left (B a - A b\right )} \sqrt {\frac {a}{b}} \arctan \left (\frac {b \sqrt {x} \sqrt {\frac {a}{b}}}{a}\right ) + {\left (B b x - 3 \, B a + 3 \, A b\right )} \sqrt {x}\right )}}{3 \, b^{2}}\right ] \] Input:
integrate(x^(1/2)*(B*x+A)/(b*x+a),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]
Leaf count of result is larger than twice the leaf count of optimal. 221 vs. \(2 (65) = 130\).
Time = 0.79 (sec) , antiderivative size = 221, normalized size of antiderivative = 3.20 \[ \int \frac {\sqrt {x} (A+B x)}{a+b x} \, dx=\begin {cases} \tilde {\infty } \left (2 A \sqrt {x} + \frac {2 B x^{\frac {3}{2}}}{3}\right ) & \text {for}\: a = 0 \wedge b = 0 \\\frac {\frac {2 A x^{\frac {3}{2}}}{3} + \frac {2 B x^{\frac {5}{2}}}{5}}{a} & \text {for}\: b = 0 \\\frac {2 A \sqrt {x} + \frac {2 B x^{\frac {3}{2}}}{3}}{b} & \text {for}\: a = 0 \\- \frac {A a \log {\left (\sqrt {x} - \sqrt {- \frac {a}{b}} \right )}}{b^{2} \sqrt {- \frac {a}{b}}} + \frac {A a \log {\left (\sqrt {x} + \sqrt {- \frac {a}{b}} \right )}}{b^{2} \sqrt {- \frac {a}{b}}} + \frac {2 A \sqrt {x}}{b} + \frac {B a^{2} \log {\left (\sqrt {x} - \sqrt {- \frac {a}{b}} \right )}}{b^{3} \sqrt {- \frac {a}{b}}} - \frac {B a^{2} \log {\left (\sqrt {x} + \sqrt {- \frac {a}{b}} \right )}}{b^{3} \sqrt {- \frac {a}{b}}} - \frac {2 B a \sqrt {x}}{b^{2}} + \frac {2 B x^{\frac {3}{2}}}{3 b} & \text {otherwise} \end {cases} \] Input:
integrate(x**(1/2)*(B*x+A)/(b*x+a),x)
Output:
Piecewise((zoo*(2*A*sqrt(x) + 2*B*x**(3/2)/3), Eq(a, 0) & Eq(b, 0)), ((2*A *x**(3/2)/3 + 2*B*x**(5/2)/5)/a, Eq(b, 0)), ((2*A*sqrt(x) + 2*B*x**(3/2)/3 )/b, Eq(a, 0)), (-A*a*log(sqrt(x) - sqrt(-a/b))/(b**2*sqrt(-a/b)) + A*a*lo g(sqrt(x) + sqrt(-a/b))/(b**2*sqrt(-a/b)) + 2*A*sqrt(x)/b + B*a**2*log(sqr t(x) - sqrt(-a/b))/(b**3*sqrt(-a/b)) - B*a**2*log(sqrt(x) + sqrt(-a/b))/(b **3*sqrt(-a/b)) - 2*B*a*sqrt(x)/b**2 + 2*B*x**(3/2)/(3*b), True))
Time = 0.15 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.84 \[ \int \frac {\sqrt {x} (A+B x)}{a+b x} \, dx=\frac {2 \, {\left (B a^{2} - A a b\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} b^{2}} + \frac {2 \, {\left (B b x^{\frac {3}{2}} - 3 \, {\left (B a - A b\right )} \sqrt {x}\right )}}{3 \, b^{2}} \] Input:
integrate(x^(1/2)*(B*x+A)/(b*x+a),x, algorithm="maxima")
Output:
2*(B*a^2 - A*a*b)*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*b^2) + 2/3*(B*b*x ^(3/2) - 3*(B*a - A*b)*sqrt(x))/b^2
Time = 0.13 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.93 \[ \int \frac {\sqrt {x} (A+B x)}{a+b x} \, dx=\frac {2 \, {\left (B a^{2} - A a b\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} b^{2}} + \frac {2 \, {\left (B b^{2} x^{\frac {3}{2}} - 3 \, B a b \sqrt {x} + 3 \, A b^{2} \sqrt {x}\right )}}{3 \, b^{3}} \] Input:
integrate(x^(1/2)*(B*x+A)/(b*x+a),x, algorithm="giac")
Output:
2*(B*a^2 - A*a*b)*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*b^2) + 2/3*(B*b^2 *x^(3/2) - 3*B*a*b*sqrt(x) + 3*A*b^2*sqrt(x))/b^3
Time = 0.05 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.10 \[ \int \frac {\sqrt {x} (A+B x)}{a+b x} \, dx=\sqrt {x}\,\left (\frac {2\,A}{b}-\frac {2\,B\,a}{b^2}\right )+\frac {2\,B\,x^{3/2}}{3\,b}+\frac {2\,\sqrt {a}\,\mathrm {atan}\left (\frac {\sqrt {a}\,\sqrt {b}\,\sqrt {x}\,\left (A\,b-B\,a\right )}{B\,a^2-A\,a\,b}\right )\,\left (A\,b-B\,a\right )}{b^{5/2}} \] Input:
int((x^(1/2)*(A + B*x))/(a + b*x),x)
Output:
x^(1/2)*((2*A)/b - (2*B*a)/b^2) + (2*B*x^(3/2))/(3*b) + (2*a^(1/2)*atan((a ^(1/2)*b^(1/2)*x^(1/2)*(A*b - B*a))/(B*a^2 - A*a*b))*(A*b - B*a))/b^(5/2)
Time = 0.15 (sec) , antiderivative size = 5, normalized size of antiderivative = 0.07 \[ \int \frac {\sqrt {x} (A+B x)}{a+b x} \, dx=\frac {2 \sqrt {x}\, x}{3} \] Input:
int(x^(1/2)*(B*x+A)/(b*x+a),x)
Output:
(2*sqrt(x)*x)/3