Integrand size = 31, antiderivative size = 209 \[ \int \frac {A+B x}{x^{3/2} \left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\frac {5 A b-a B}{4 a^2 b \sqrt {x} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {A b-a B}{2 a b \sqrt {x} (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {3 (5 A b-a B) (a+b x)}{4 a^3 b \sqrt {x} \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {3 (5 A b-a B) (a+b x) \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{4 a^{7/2} \sqrt {b} \sqrt {a^2+2 a b x+b^2 x^2}} \]
-3/4*(5*A*b-B*a)*(b*x+a)*arctan(b^(1/2)*x^(1/2)/a^(1/2))/a^(7/2)/b^(1/2)/( (b*x+a)^2)^(1/2)+1/4*(5*A*b-B*a)/a^2/b/x^(1/2)/((b*x+a)^2)^(1/2)+1/2*(A*b- B*a)/a/b/(b*x+a)/x^(1/2)/((b*x+a)^2)^(1/2)-3/4*(5*A*b-B*a)*(b*x+a)/a^3/b/x ^(1/2)/((b*x+a)^2)^(1/2)
Time = 0.05 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.53 \[ \int \frac {A+B x}{x^{3/2} \left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\frac {(a+b x) \left (\frac {\sqrt {a} \left (-15 A b^2 x^2+a b x (-25 A+3 B x)+a^2 (-8 A+5 B x)\right )}{\sqrt {x}}+\frac {3 (-5 A b+a B) (a+b x)^2 \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{\sqrt {b}}\right )}{4 a^{7/2} \left ((a+b x)^2\right )^{3/2}} \]
((a + b*x)*((Sqrt[a]*(-15*A*b^2*x^2 + a*b*x*(-25*A + 3*B*x) + a^2*(-8*A + 5*B*x)))/Sqrt[x] + (3*(-5*A*b + a*B)*(a + b*x)^2*ArcTan[(Sqrt[b]*Sqrt[x])/ Sqrt[a]])/Sqrt[b]))/(4*a^(7/2)*((a + b*x)^2)^(3/2))
Time = 0.25 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.67, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.226, Rules used = {1187, 27, 87, 52, 61, 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.
| \(\displaystyle \int \frac {A+B x}{x^{3/2} \left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx\) |
| \(\Big \downarrow \) 1187 |
| \(\displaystyle \frac {b^3 (a+b x) \int \frac {A+B x}{b^3 x^{3/2} (a+b x)^3}dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {(a+b x) \int \frac {A+B x}{x^{3/2} (a+b x)^3}dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 87 |
| \(\displaystyle \frac {(a+b x) \left (\frac {(5 A b-a B) \int \frac {1}{x^{3/2} (a+b x)^2}dx}{4 a b}+\frac {A b-a B}{2 a b \sqrt {x} (a+b x)^2}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 52 |
| \(\displaystyle \frac {(a+b x) \left (\frac {(5 A b-a B) \left (\frac {3 \int \frac {1}{x^{3/2} (a+b x)}dx}{2 a}+\frac {1}{a \sqrt {x} (a+b x)}\right )}{4 a b}+\frac {A b-a B}{2 a b \sqrt {x} (a+b x)^2}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 61 |
| \(\displaystyle \frac {(a+b x) \left (\frac {(5 A b-a B) \left (\frac {3 \left (-\frac {b \int \frac {1}{\sqrt {x} (a+b x)}dx}{a}-\frac {2}{a \sqrt {x}}\right )}{2 a}+\frac {1}{a \sqrt {x} (a+b x)}\right )}{4 a b}+\frac {A b-a B}{2 a b \sqrt {x} (a+b x)^2}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {(a+b x) \left (\frac {(5 A b-a B) \left (\frac {3 \left (-\frac {2 b \int \frac {1}{a+b x}d\sqrt {x}}{a}-\frac {2}{a \sqrt {x}}\right )}{2 a}+\frac {1}{a \sqrt {x} (a+b x)}\right )}{4 a b}+\frac {A b-a B}{2 a b \sqrt {x} (a+b x)^2}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {(a+b x) \left (\frac {(5 A b-a B) \left (\frac {3 \left (-\frac {2 \sqrt {b} \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{a^{3/2}}-\frac {2}{a \sqrt {x}}\right )}{2 a}+\frac {1}{a \sqrt {x} (a+b x)}\right )}{4 a b}+\frac {A b-a B}{2 a b \sqrt {x} (a+b x)^2}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
((a + b*x)*((A*b - a*B)/(2*a*b*Sqrt[x]*(a + b*x)^2) + ((5*A*b - a*B)*(1/(a *Sqrt[x]*(a + b*x)) + (3*(-2/(a*Sqrt[x]) - (2*Sqrt[b]*ArcTan[(Sqrt[b]*Sqrt [x])/Sqrt[a]])/a^(3/2)))/(2*a)))/(4*a*b)))/Sqrt[a^2 + 2*a*b*x + b^2*x^2]
3.9.23.3.1 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 + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(( m + n + 2)/((b*c - a*d)*(m + 1))) Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && ILtQ[m, -1] && FractionQ[n] && LtQ[n, 0]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(( m + n + 2)/((b*c - a*d)*(m + 1))) Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && LtQ[m, -1] && !(LtQ[n, -1] && (EqQ[a, 0 ] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[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[(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 = 0.19 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.56
| method | result | size |
| risch | \(-\frac {2 A \sqrt {\left (b x +a \right )^{2}}}{a^{3} \sqrt {x}\, \left (b x +a \right )}-\frac {\left (\frac {2 \left (\frac {7}{8} A \,b^{2}-\frac {3}{8} a b B \right ) x^{\frac {3}{2}}+\frac {a \left (9 A b -5 B a \right ) \sqrt {x}}{4}}{\left (b x +a \right )^{2}}+\frac {3 \left (5 A b -B a \right ) \arctan \left (\frac {b \sqrt {x}}{\sqrt {b a}}\right )}{4 \sqrt {b a}}\right ) \sqrt {\left (b x +a \right )^{2}}}{a^{3} \left (b x +a \right )}\) | \(117\) |
| default | \(-\frac {\left (15 A \,x^{2} \sqrt {b a}\, b^{2}+15 A \,x^{\frac {5}{2}} \arctan \left (\frac {b \sqrt {x}}{\sqrt {b a}}\right ) b^{3}-3 B \,x^{2} \sqrt {b a}\, a b -3 B \,x^{\frac {5}{2}} \arctan \left (\frac {b \sqrt {x}}{\sqrt {b a}}\right ) a \,b^{2}+30 A \,x^{\frac {3}{2}} \arctan \left (\frac {b \sqrt {x}}{\sqrt {b a}}\right ) a \,b^{2}-6 B \,x^{\frac {3}{2}} \arctan \left (\frac {b \sqrt {x}}{\sqrt {b a}}\right ) a^{2} b +25 A x \sqrt {b a}\, a b +15 A \sqrt {x}\, \arctan \left (\frac {b \sqrt {x}}{\sqrt {b a}}\right ) a^{2} b -5 B x \sqrt {b a}\, a^{2}-3 B \sqrt {x}\, \arctan \left (\frac {b \sqrt {x}}{\sqrt {b a}}\right ) a^{3}+8 A \,a^{2} \sqrt {b a}\right ) \left (b x +a \right )}{4 \sqrt {x}\, \sqrt {b a}\, a^{3} \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}\) | \(214\) |
-2*A/a^3/x^(1/2)*((b*x+a)^2)^(1/2)/(b*x+a)-1/a^3*(2*((7/8*A*b^2-3/8*a*b*B) *x^(3/2)+1/8*a*(9*A*b-5*B*a)*x^(1/2))/(b*x+a)^2+3/4*(5*A*b-B*a)/(b*a)^(1/2 )*arctan(b*x^(1/2)/(b*a)^(1/2)))*((b*x+a)^2)^(1/2)/(b*x+a)
Time = 0.29 (sec) , antiderivative size = 331, normalized size of antiderivative = 1.58 \[ \int \frac {A+B x}{x^{3/2} \left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\left [\frac {3 \, {\left ({\left (B a b^{2} - 5 \, A b^{3}\right )} x^{3} + 2 \, {\left (B a^{2} b - 5 \, A a b^{2}\right )} x^{2} + {\left (B a^{3} - 5 \, A a^{2} b\right )} x\right )} \sqrt {-a b} \log \left (\frac {b x - a + 2 \, \sqrt {-a b} \sqrt {x}}{b x + a}\right ) - 2 \, {\left (8 \, A a^{3} b - 3 \, {\left (B a^{2} b^{2} - 5 \, A a b^{3}\right )} x^{2} - 5 \, {\left (B a^{3} b - 5 \, A a^{2} b^{2}\right )} x\right )} \sqrt {x}}{8 \, {\left (a^{4} b^{3} x^{3} + 2 \, a^{5} b^{2} x^{2} + a^{6} b x\right )}}, -\frac {3 \, {\left ({\left (B a b^{2} - 5 \, A b^{3}\right )} x^{3} + 2 \, {\left (B a^{2} b - 5 \, A a b^{2}\right )} x^{2} + {\left (B a^{3} - 5 \, A a^{2} b\right )} x\right )} \sqrt {a b} \arctan \left (\frac {\sqrt {a b}}{b \sqrt {x}}\right ) + {\left (8 \, A a^{3} b - 3 \, {\left (B a^{2} b^{2} - 5 \, A a b^{3}\right )} x^{2} - 5 \, {\left (B a^{3} b - 5 \, A a^{2} b^{2}\right )} x\right )} \sqrt {x}}{4 \, {\left (a^{4} b^{3} x^{3} + 2 \, a^{5} b^{2} x^{2} + a^{6} b x\right )}}\right ] \]
[1/8*(3*((B*a*b^2 - 5*A*b^3)*x^3 + 2*(B*a^2*b - 5*A*a*b^2)*x^2 + (B*a^3 - 5*A*a^2*b)*x)*sqrt(-a*b)*log((b*x - a + 2*sqrt(-a*b)*sqrt(x))/(b*x + a)) - 2*(8*A*a^3*b - 3*(B*a^2*b^2 - 5*A*a*b^3)*x^2 - 5*(B*a^3*b - 5*A*a^2*b^2)* x)*sqrt(x))/(a^4*b^3*x^3 + 2*a^5*b^2*x^2 + a^6*b*x), -1/4*(3*((B*a*b^2 - 5 *A*b^3)*x^3 + 2*(B*a^2*b - 5*A*a*b^2)*x^2 + (B*a^3 - 5*A*a^2*b)*x)*sqrt(a* b)*arctan(sqrt(a*b)/(b*sqrt(x))) + (8*A*a^3*b - 3*(B*a^2*b^2 - 5*A*a*b^3)* x^2 - 5*(B*a^3*b - 5*A*a^2*b^2)*x)*sqrt(x))/(a^4*b^3*x^3 + 2*a^5*b^2*x^2 + a^6*b*x)]
\[ \int \frac {A+B x}{x^{3/2} \left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\int \frac {A + B x}{x^{\frac {3}{2}} \left (\left (a + b x\right )^{2}\right )^{\frac {3}{2}}}\, dx \]
Leaf count of result is larger than twice the leaf count of optimal. 280 vs. \(2 (138) = 276\).
Time = 0.37 (sec) , antiderivative size = 280, normalized size of antiderivative = 1.34 \[ \int \frac {A+B x}{x^{3/2} \left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\frac {60 \, {\left (B a^{3} b^{2} - 7 \, A a^{2} b^{3}\right )} x^{\frac {5}{2}} - {\left ({\left (B a b^{4} + 5 \, A b^{5}\right )} x^{2} - 15 \, {\left (B a^{2} b^{3} - 7 \, A a b^{4}\right )} x\right )} x^{\frac {5}{2}} + {\left (9 \, {\left (B a^{3} b^{2} + 5 \, A a^{2} b^{3}\right )} x^{2} + 85 \, {\left (B a^{4} b - 7 \, A a^{3} b^{2}\right )} x\right )} \sqrt {x} + \frac {16 \, {\left ({\left (B a^{4} b + 5 \, A a^{3} b^{2}\right )} x^{2} + 3 \, {\left (B a^{5} - 7 \, A a^{4} b\right )} x\right )}}{\sqrt {x}} + \frac {48 \, {\left (A a^{4} b x^{2} - A a^{5} x\right )}}{x^{\frac {3}{2}}}}{24 \, {\left (a^{5} b^{3} x^{3} + 3 \, a^{6} b^{2} x^{2} + 3 \, a^{7} b x + a^{8}\right )}} + \frac {3 \, {\left (B a - 5 \, A b\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{4 \, \sqrt {a b} a^{3}} + \frac {{\left (B a b + 5 \, A b^{2}\right )} x^{\frac {3}{2}} - 18 \, {\left (B a^{2} - 5 \, A a b\right )} \sqrt {x}}{24 \, a^{5}} \]
1/24*(60*(B*a^3*b^2 - 7*A*a^2*b^3)*x^(5/2) - ((B*a*b^4 + 5*A*b^5)*x^2 - 15 *(B*a^2*b^3 - 7*A*a*b^4)*x)*x^(5/2) + (9*(B*a^3*b^2 + 5*A*a^2*b^3)*x^2 + 8 5*(B*a^4*b - 7*A*a^3*b^2)*x)*sqrt(x) + 16*((B*a^4*b + 5*A*a^3*b^2)*x^2 + 3 *(B*a^5 - 7*A*a^4*b)*x)/sqrt(x) + 48*(A*a^4*b*x^2 - A*a^5*x)/x^(3/2))/(a^5 *b^3*x^3 + 3*a^6*b^2*x^2 + 3*a^7*b*x + a^8) + 3/4*(B*a - 5*A*b)*arctan(b*s qrt(x)/sqrt(a*b))/(sqrt(a*b)*a^3) + 1/24*((B*a*b + 5*A*b^2)*x^(3/2) - 18*( B*a^2 - 5*A*a*b)*sqrt(x))/a^5
Time = 0.26 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.53 \[ \int \frac {A+B x}{x^{3/2} \left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\frac {3 \, {\left (B a - 5 \, A b\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{4 \, \sqrt {a b} a^{3} \mathrm {sgn}\left (b x + a\right )} - \frac {2 \, A}{a^{3} \sqrt {x} \mathrm {sgn}\left (b x + a\right )} + \frac {3 \, B a b x^{\frac {3}{2}} - 7 \, A b^{2} x^{\frac {3}{2}} + 5 \, B a^{2} \sqrt {x} - 9 \, A a b \sqrt {x}}{4 \, {\left (b x + a\right )}^{2} a^{3} \mathrm {sgn}\left (b x + a\right )} \]
3/4*(B*a - 5*A*b)*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*a^3*sgn(b*x + a)) - 2*A/(a^3*sqrt(x)*sgn(b*x + a)) + 1/4*(3*B*a*b*x^(3/2) - 7*A*b^2*x^(3/2) + 5*B*a^2*sqrt(x) - 9*A*a*b*sqrt(x))/((b*x + a)^2*a^3*sgn(b*x + a))
Timed out. \[ \int \frac {A+B x}{x^{3/2} \left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\int \frac {A+B\,x}{x^{3/2}\,{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{3/2}} \,d x \]