Integrand size = 31, antiderivative size = 158 \[ \int \frac {\sqrt {x} (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=-\frac {(A b+3 a B) \sqrt {x}}{4 a b^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {(A b-a B) x^{3/2}}{2 a b (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {(A b+3 a B) (a+b x) \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{4 a^{3/2} b^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}} \]
1/2*(A*b-B*a)*x^(3/2)/a/b/(b*x+a)/((b*x+a)^2)^(1/2)+1/4*(A*b+3*B*a)*(b*x+a )*arctan(b^(1/2)*x^(1/2)/a^(1/2))/a^(3/2)/b^(5/2)/((b*x+a)^2)^(1/2)-1/4*(A *b+3*B*a)*x^(1/2)/a/b^2/((b*x+a)^2)^(1/2)
Time = 0.04 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.65 \[ \int \frac {\sqrt {x} (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\frac {(a+b x) \left (\sqrt {a} \sqrt {b} \sqrt {x} \left (-3 a^2 B+A b^2 x-a b (A+5 B x)\right )+(A b+3 a B) (a+b x)^2 \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )\right )}{4 a^{3/2} b^{5/2} \left ((a+b x)^2\right )^{3/2}} \]
((a + b*x)*(Sqrt[a]*Sqrt[b]*Sqrt[x]*(-3*a^2*B + A*b^2*x - a*b*(A + 5*B*x)) + (A*b + 3*a*B)*(a + b*x)^2*ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[a]]))/(4*a^(3/2 )*b^(5/2)*((a + b*x)^2)^(3/2))
Time = 0.24 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.77, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.194, Rules used = {1187, 27, 87, 51, 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.
\(\displaystyle \int \frac {\sqrt {x} (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 1187 |
\(\displaystyle \frac {b^3 (a+b x) \int \frac {\sqrt {x} (A+B x)}{b^3 (a+b x)^3}dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {(a+b x) \int \frac {\sqrt {x} (A+B x)}{(a+b x)^3}dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
\(\Big \downarrow \) 87 |
\(\displaystyle \frac {(a+b x) \left (\frac {(3 a B+A b) \int \frac {\sqrt {x}}{(a+b x)^2}dx}{4 a b}+\frac {x^{3/2} (A b-a B)}{2 a b (a+b x)^2}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
\(\Big \downarrow \) 51 |
\(\displaystyle \frac {(a+b x) \left (\frac {(3 a B+A b) \left (\frac {\int \frac {1}{\sqrt {x} (a+b x)}dx}{2 b}-\frac {\sqrt {x}}{b (a+b x)}\right )}{4 a b}+\frac {x^{3/2} (A b-a B)}{2 a b (a+b x)^2}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {(a+b x) \left (\frac {(3 a B+A b) \left (\frac {\int \frac {1}{a+b x}d\sqrt {x}}{b}-\frac {\sqrt {x}}{b (a+b x)}\right )}{4 a b}+\frac {x^{3/2} (A b-a B)}{2 a b (a+b x)^2}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {(a+b x) \left (\frac {(3 a B+A b) \left (\frac {\arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{\sqrt {a} b^{3/2}}-\frac {\sqrt {x}}{b (a+b x)}\right )}{4 a b}+\frac {x^{3/2} (A b-a B)}{2 a b (a+b x)^2}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
((a + b*x)*(((A*b - a*B)*x^(3/2))/(2*a*b*(a + b*x)^2) + ((A*b + 3*a*B)*(-( Sqrt[x]/(b*(a + b*x))) + ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[a]]/(Sqrt[a]*b^(3/2 ))))/(4*a*b)))/Sqrt[a^2 + 2*a*b*x + b^2*x^2]
3.9.21.3.1 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 + 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] :> 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[(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 = 0.40 (sec) , antiderivative size = 194, normalized size of antiderivative = 1.23
method | result | size |
default | \(\frac {\left (A \,x^{\frac {3}{2}} \sqrt {b a}\, b^{2}+A \arctan \left (\frac {b \sqrt {x}}{\sqrt {b a}}\right ) b^{3} x^{2}-5 B \,x^{\frac {3}{2}} \sqrt {b a}\, a b +3 B \arctan \left (\frac {b \sqrt {x}}{\sqrt {b a}}\right ) a \,b^{2} x^{2}+2 A \arctan \left (\frac {b \sqrt {x}}{\sqrt {b a}}\right ) a \,b^{2} x +6 B \arctan \left (\frac {b \sqrt {x}}{\sqrt {b a}}\right ) a^{2} b x -A \sqrt {x}\, \sqrt {b a}\, a b +A \arctan \left (\frac {b \sqrt {x}}{\sqrt {b a}}\right ) a^{2} b -3 B \sqrt {x}\, \sqrt {b a}\, a^{2}+3 B \arctan \left (\frac {b \sqrt {x}}{\sqrt {b a}}\right ) a^{3}\right ) \left (b x +a \right )}{4 \sqrt {b a}\, a \,b^{2} \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}\) | \(194\) |
1/4*(A*x^(3/2)*(b*a)^(1/2)*b^2+A*arctan(b*x^(1/2)/(b*a)^(1/2))*b^3*x^2-5*B *x^(3/2)*(b*a)^(1/2)*a*b+3*B*arctan(b*x^(1/2)/(b*a)^(1/2))*a*b^2*x^2+2*A*a rctan(b*x^(1/2)/(b*a)^(1/2))*a*b^2*x+6*B*arctan(b*x^(1/2)/(b*a)^(1/2))*a^2 *b*x-A*x^(1/2)*(b*a)^(1/2)*a*b+A*arctan(b*x^(1/2)/(b*a)^(1/2))*a^2*b-3*B*x ^(1/2)*(b*a)^(1/2)*a^2+3*B*arctan(b*x^(1/2)/(b*a)^(1/2))*a^3)*(b*x+a)/(b*a )^(1/2)/a/b^2/((b*x+a)^2)^(3/2)
Time = 0.29 (sec) , antiderivative size = 291, normalized size of antiderivative = 1.84 \[ \int \frac {\sqrt {x} (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\left [-\frac {{\left (3 \, B a^{3} + A a^{2} b + {\left (3 \, B a b^{2} + A b^{3}\right )} x^{2} + 2 \, {\left (3 \, B a^{2} b + A a b^{2}\right )} x\right )} \sqrt {-a b} \log \left (\frac {b x - a - 2 \, \sqrt {-a b} \sqrt {x}}{b x + a}\right ) + 2 \, {\left (3 \, B a^{3} b + A a^{2} b^{2} + {\left (5 \, B a^{2} b^{2} - A a b^{3}\right )} x\right )} \sqrt {x}}{8 \, {\left (a^{2} b^{5} x^{2} + 2 \, a^{3} b^{4} x + a^{4} b^{3}\right )}}, -\frac {{\left (3 \, B a^{3} + A a^{2} b + {\left (3 \, B a b^{2} + A b^{3}\right )} x^{2} + 2 \, {\left (3 \, B a^{2} b + A a b^{2}\right )} x\right )} \sqrt {a b} \arctan \left (\frac {\sqrt {a b}}{b \sqrt {x}}\right ) + {\left (3 \, B a^{3} b + A a^{2} b^{2} + {\left (5 \, B a^{2} b^{2} - A a b^{3}\right )} x\right )} \sqrt {x}}{4 \, {\left (a^{2} b^{5} x^{2} + 2 \, a^{3} b^{4} x + a^{4} b^{3}\right )}}\right ] \]
[-1/8*((3*B*a^3 + A*a^2*b + (3*B*a*b^2 + A*b^3)*x^2 + 2*(3*B*a^2*b + A*a*b ^2)*x)*sqrt(-a*b)*log((b*x - a - 2*sqrt(-a*b)*sqrt(x))/(b*x + a)) + 2*(3*B *a^3*b + A*a^2*b^2 + (5*B*a^2*b^2 - A*a*b^3)*x)*sqrt(x))/(a^2*b^5*x^2 + 2* a^3*b^4*x + a^4*b^3), -1/4*((3*B*a^3 + A*a^2*b + (3*B*a*b^2 + A*b^3)*x^2 + 2*(3*B*a^2*b + A*a*b^2)*x)*sqrt(a*b)*arctan(sqrt(a*b)/(b*sqrt(x))) + (3*B *a^3*b + A*a^2*b^2 + (5*B*a^2*b^2 - A*a*b^3)*x)*sqrt(x))/(a^2*b^5*x^2 + 2* a^3*b^4*x + a^4*b^3)]
\[ \int \frac {\sqrt {x} (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\int \frac {\sqrt {x} \left (A + B x\right )}{\left (\left (a + b x\right )^{2}\right )^{\frac {3}{2}}}\, dx \]
Leaf count of result is larger than twice the leaf count of optimal. 218 vs. \(2 (105) = 210\).
Time = 0.34 (sec) , antiderivative size = 218, normalized size of antiderivative = 1.38 \[ \int \frac {\sqrt {x} (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\frac {12 \, {\left (B a^{3} b + A a^{2} b^{2}\right )} x^{\frac {5}{2}} - {\left ({\left (5 \, B a b^{3} + A b^{4}\right )} x^{2} - 3 \, {\left (B a^{2} b^{2} + A a b^{3}\right )} x\right )} x^{\frac {5}{2}} - {\left (3 \, {\left (B a^{3} b - 3 \, A a^{2} b^{2}\right )} x^{2} - {\left (B a^{4} + 17 \, A a^{3} b\right )} x\right )} \sqrt {x}}{24 \, {\left (a^{3} b^{4} x^{3} + 3 \, a^{4} b^{3} x^{2} + 3 \, a^{5} b^{2} x + a^{6} b\right )}} + \frac {{\left (3 \, B a + A b\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{4 \, \sqrt {a b} a b^{2}} + \frac {{\left (5 \, B a b + A b^{2}\right )} x^{\frac {3}{2}} - 6 \, {\left (3 \, B a^{2} + A a b\right )} \sqrt {x}}{24 \, a^{3} b^{2}} \]
1/24*(12*(B*a^3*b + A*a^2*b^2)*x^(5/2) - ((5*B*a*b^3 + A*b^4)*x^2 - 3*(B*a ^2*b^2 + A*a*b^3)*x)*x^(5/2) - (3*(B*a^3*b - 3*A*a^2*b^2)*x^2 - (B*a^4 + 1 7*A*a^3*b)*x)*sqrt(x))/(a^3*b^4*x^3 + 3*a^4*b^3*x^2 + 3*a^5*b^2*x + a^6*b) + 1/4*(3*B*a + A*b)*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*a*b^2) + 1/24* ((5*B*a*b + A*b^2)*x^(3/2) - 6*(3*B*a^2 + A*a*b)*sqrt(x))/(a^3*b^2)
Time = 0.27 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.62 \[ \int \frac {\sqrt {x} (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\frac {{\left (3 \, B a + A b\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{4 \, \sqrt {a b} a b^{2} \mathrm {sgn}\left (b x + a\right )} - \frac {5 \, B a b x^{\frac {3}{2}} - A b^{2} x^{\frac {3}{2}} + 3 \, B a^{2} \sqrt {x} + A a b \sqrt {x}}{4 \, {\left (b x + a\right )}^{2} a b^{2} \mathrm {sgn}\left (b x + a\right )} \]
1/4*(3*B*a + A*b)*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*a*b^2*sgn(b*x + a )) - 1/4*(5*B*a*b*x^(3/2) - A*b^2*x^(3/2) + 3*B*a^2*sqrt(x) + A*a*b*sqrt(x ))/((b*x + a)^2*a*b^2*sgn(b*x + a))
Timed out. \[ \int \frac {\sqrt {x} (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\int \frac {\sqrt {x}\,\left (A+B\,x\right )}{{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{3/2}} \,d x \]