Integrand size = 31, antiderivative size = 238 \[ \int \frac {A+B x}{x^{9/2} \sqrt {a^2+2 a b x+b^2 x^2}} \, dx=-\frac {2 A (a+b x)}{7 a x^{7/2} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 (A b-a B) (a+b x)}{5 a^2 x^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {2 b (A b-a B) (a+b x)}{3 a^3 x^{3/2} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 b^2 (A b-a B) (a+b x)}{a^4 \sqrt {x} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 b^{5/2} (A b-a B) (a+b x) \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{a^{9/2} \sqrt {a^2+2 a b x+b^2 x^2}} \]
-2/7*A*(b*x+a)/a/x^(7/2)/((b*x+a)^2)^(1/2)+2/5*(A*b-B*a)*(b*x+a)/a^2/x^(5/ 2)/((b*x+a)^2)^(1/2)-2/3*b*(A*b-B*a)*(b*x+a)/a^3/x^(3/2)/((b*x+a)^2)^(1/2) +2*b^(5/2)*(A*b-B*a)*(b*x+a)*arctan(b^(1/2)*x^(1/2)/a^(1/2))/a^(9/2)/((b*x +a)^2)^(1/2)+2*b^2*(A*b-B*a)*(b*x+a)/a^4/x^(1/2)/((b*x+a)^2)^(1/2)
Time = 0.04 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.53 \[ \int \frac {A+B x}{x^{9/2} \sqrt {a^2+2 a b x+b^2 x^2}} \, dx=-\frac {2 (a+b x) \left (\sqrt {a} \left (-105 A b^3 x^3+35 a b^2 x^2 (A+3 B x)-7 a^2 b x (3 A+5 B x)+3 a^3 (5 A+7 B x)\right )-105 b^{5/2} (A b-a B) x^{7/2} \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )\right )}{105 a^{9/2} x^{7/2} \sqrt {(a+b x)^2}} \]
(-2*(a + b*x)*(Sqrt[a]*(-105*A*b^3*x^3 + 35*a*b^2*x^2*(A + 3*B*x) - 7*a^2* b*x*(3*A + 5*B*x) + 3*a^3*(5*A + 7*B*x)) - 105*b^(5/2)*(A*b - a*B)*x^(7/2) *ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[a]]))/(105*a^(9/2)*x^(7/2)*Sqrt[(a + b*x)^2 ])
Time = 0.24 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.55, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.258, Rules used = {1187, 27, 87, 61, 61, 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.
| \(\displaystyle \int \frac {A+B x}{x^{9/2} \sqrt {a^2+2 a b x+b^2 x^2}} \, dx\) |
| \(\Big \downarrow \) 1187 |
| \(\displaystyle \frac {b (a+b x) \int \frac {A+B x}{b x^{9/2} (a+b x)}dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {(a+b x) \int \frac {A+B x}{x^{9/2} (a+b x)}dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 87 |
| \(\displaystyle \frac {(a+b x) \left (-\frac {(A b-a B) \int \frac {1}{x^{7/2} (a+b x)}dx}{a}-\frac {2 A}{7 a x^{7/2}}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 61 |
| \(\displaystyle \frac {(a+b x) \left (-\frac {(A b-a B) \left (-\frac {b \int \frac {1}{x^{5/2} (a+b x)}dx}{a}-\frac {2}{5 a x^{5/2}}\right )}{a}-\frac {2 A}{7 a x^{7/2}}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 61 |
| \(\displaystyle \frac {(a+b x) \left (-\frac {(A b-a B) \left (-\frac {b \left (-\frac {b \int \frac {1}{x^{3/2} (a+b x)}dx}{a}-\frac {2}{3 a x^{3/2}}\right )}{a}-\frac {2}{5 a x^{5/2}}\right )}{a}-\frac {2 A}{7 a x^{7/2}}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 61 |
| \(\displaystyle \frac {(a+b x) \left (-\frac {(A b-a B) \left (-\frac {b \left (-\frac {b \left (-\frac {b \int \frac {1}{\sqrt {x} (a+b x)}dx}{a}-\frac {2}{a \sqrt {x}}\right )}{a}-\frac {2}{3 a x^{3/2}}\right )}{a}-\frac {2}{5 a x^{5/2}}\right )}{a}-\frac {2 A}{7 a x^{7/2}}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {(a+b x) \left (-\frac {(A b-a B) \left (-\frac {b \left (-\frac {b \left (-\frac {2 b \int \frac {1}{a+b x}d\sqrt {x}}{a}-\frac {2}{a \sqrt {x}}\right )}{a}-\frac {2}{3 a x^{3/2}}\right )}{a}-\frac {2}{5 a x^{5/2}}\right )}{a}-\frac {2 A}{7 a x^{7/2}}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {(a+b x) \left (-\frac {(A b-a B) \left (-\frac {b \left (-\frac {b \left (-\frac {2 \sqrt {b} \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{a^{3/2}}-\frac {2}{a \sqrt {x}}\right )}{a}-\frac {2}{3 a x^{3/2}}\right )}{a}-\frac {2}{5 a x^{5/2}}\right )}{a}-\frac {2 A}{7 a x^{7/2}}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
((a + b*x)*((-2*A)/(7*a*x^(7/2)) - ((A*b - a*B)*(-2/(5*a*x^(5/2)) - (b*(-2 /(3*a*x^(3/2)) - (b*(-2/(a*Sqrt[x]) - (2*Sqrt[b]*ArcTan[(Sqrt[b]*Sqrt[x])/ Sqrt[a]])/a^(3/2)))/a))/a))/a))/Sqrt[a^2 + 2*a*b*x + b^2*x^2]
3.9.17.3.1 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] && 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 = 0.14 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.57
| method | result | size |
| risch | \(-\frac {2 \left (-105 A \,b^{3} x^{3}+105 B a \,b^{2} x^{3}+35 A a \,b^{2} x^{2}-35 B \,a^{2} b \,x^{2}-21 A \,a^{2} b x +21 a^{3} B x +15 A \,a^{3}\right ) \sqrt {\left (b x +a \right )^{2}}}{105 a^{4} x^{\frac {7}{2}} \left (b x +a \right )}+\frac {2 \left (A b -B a \right ) b^{3} \arctan \left (\frac {b \sqrt {x}}{\sqrt {b a}}\right ) \sqrt {\left (b x +a \right )^{2}}}{a^{4} \sqrt {b a}\, \left (b x +a \right )}\) | \(135\) |
| default | \(\frac {2 \left (b x +a \right ) \left (105 A \,x^{\frac {7}{2}} \arctan \left (\frac {b \sqrt {x}}{\sqrt {b a}}\right ) b^{4}-105 B \,x^{\frac {7}{2}} \arctan \left (\frac {b \sqrt {x}}{\sqrt {b a}}\right ) a \,b^{3}+105 A \,x^{3} \sqrt {b a}\, b^{3}-105 B \,x^{3} \sqrt {b a}\, a \,b^{2}-35 A \,x^{2} \sqrt {b a}\, a \,b^{2}+35 B \,x^{2} \sqrt {b a}\, a^{2} b +21 A x \sqrt {b a}\, a^{2} b -21 B x \sqrt {b a}\, a^{3}-15 A \,a^{3} \sqrt {b a}\right )}{105 \sqrt {\left (b x +a \right )^{2}}\, a^{4} \sqrt {b a}\, x^{\frac {7}{2}}}\) | \(165\) |
-2/105*(-105*A*b^3*x^3+105*B*a*b^2*x^3+35*A*a*b^2*x^2-35*B*a^2*b*x^2-21*A* a^2*b*x+21*B*a^3*x+15*A*a^3)/a^4/x^(7/2)*((b*x+a)^2)^(1/2)/(b*x+a)+2*(A*b- B*a)/a^4*b^3/(b*a)^(1/2)*arctan(b*x^(1/2)/(b*a)^(1/2))*((b*x+a)^2)^(1/2)/( b*x+a)
Time = 0.33 (sec) , antiderivative size = 246, normalized size of antiderivative = 1.03 \[ \int \frac {A+B x}{x^{9/2} \sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\left [-\frac {105 \, {\left (B a b^{2} - A b^{3}\right )} x^{4} \sqrt {-\frac {b}{a}} \log \left (\frac {b x + 2 \, a \sqrt {x} \sqrt {-\frac {b}{a}} - a}{b x + a}\right ) + 2 \, {\left (15 \, A a^{3} + 105 \, {\left (B a b^{2} - A b^{3}\right )} x^{3} - 35 \, {\left (B a^{2} b - A a b^{2}\right )} x^{2} + 21 \, {\left (B a^{3} - A a^{2} b\right )} x\right )} \sqrt {x}}{105 \, a^{4} x^{4}}, \frac {2 \, {\left (105 \, {\left (B a b^{2} - A b^{3}\right )} x^{4} \sqrt {\frac {b}{a}} \arctan \left (\frac {a \sqrt {\frac {b}{a}}}{b \sqrt {x}}\right ) - {\left (15 \, A a^{3} + 105 \, {\left (B a b^{2} - A b^{3}\right )} x^{3} - 35 \, {\left (B a^{2} b - A a b^{2}\right )} x^{2} + 21 \, {\left (B a^{3} - A a^{2} b\right )} x\right )} \sqrt {x}\right )}}{105 \, a^{4} x^{4}}\right ] \]
[-1/105*(105*(B*a*b^2 - A*b^3)*x^4*sqrt(-b/a)*log((b*x + 2*a*sqrt(x)*sqrt( -b/a) - a)/(b*x + a)) + 2*(15*A*a^3 + 105*(B*a*b^2 - A*b^3)*x^3 - 35*(B*a^ 2*b - A*a*b^2)*x^2 + 21*(B*a^3 - A*a^2*b)*x)*sqrt(x))/(a^4*x^4), 2/105*(10 5*(B*a*b^2 - A*b^3)*x^4*sqrt(b/a)*arctan(a*sqrt(b/a)/(b*sqrt(x))) - (15*A* a^3 + 105*(B*a*b^2 - A*b^3)*x^3 - 35*(B*a^2*b - A*a*b^2)*x^2 + 21*(B*a^3 - A*a^2*b)*x)*sqrt(x))/(a^4*x^4)]
Timed out. \[ \int \frac {A+B x}{x^{9/2} \sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\text {Timed out} \]
Leaf count of result is larger than twice the leaf count of optimal. 357 vs. \(2 (159) = 318\).
Time = 0.34 (sec) , antiderivative size = 357, normalized size of antiderivative = 1.50 \[ \int \frac {A+B x}{x^{9/2} \sqrt {a^2+2 a b x+b^2 x^2}} \, dx=-\frac {35 \, {\left ({\left (5 \, B a b^{5} - 7 \, A b^{6}\right )} x^{2} + 3 \, {\left (7 \, B a^{2} b^{4} - 9 \, A a b^{5}\right )} x\right )} \sqrt {x} - \frac {70 \, {\left ({\left (5 \, B a^{2} b^{4} - 7 \, A a b^{5}\right )} x^{2} - 3 \, {\left (7 \, B a^{3} b^{3} - 9 \, A a^{2} b^{4}\right )} x\right )}}{\sqrt {x}} - \frac {70 \, {\left (3 \, {\left (5 \, B a^{3} b^{3} - 7 \, A a^{2} b^{4}\right )} x^{2} - {\left (7 \, B a^{4} b^{2} - 9 \, A a^{3} b^{3}\right )} x\right )}}{x^{\frac {3}{2}}} - \frac {14 \, {\left (5 \, {\left (5 \, B a^{4} b^{2} - 7 \, A a^{3} b^{3}\right )} x^{2} + {\left (7 \, B a^{5} b - 9 \, A a^{4} b^{2}\right )} x\right )}}{x^{\frac {5}{2}}} + \frac {2 \, {\left (7 \, {\left (5 \, B a^{5} b - 7 \, A a^{4} b^{2}\right )} x^{2} + 3 \, {\left (7 \, B a^{6} - 9 \, A a^{5} b\right )} x\right )}}{x^{\frac {7}{2}}} + \frac {6 \, {\left (7 \, A a^{5} b x^{2} + 5 \, A a^{6} x\right )}}{x^{\frac {9}{2}}}}{105 \, {\left (a^{6} b x + a^{7}\right )}} - \frac {2 \, {\left (B a b^{3} - A b^{4}\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} a^{4}} + \frac {{\left (5 \, B a b^{4} - 7 \, A b^{5}\right )} x^{\frac {3}{2}} + 6 \, {\left (B a^{2} b^{3} - A a b^{4}\right )} \sqrt {x}}{3 \, a^{6}} \]
-1/105*(35*((5*B*a*b^5 - 7*A*b^6)*x^2 + 3*(7*B*a^2*b^4 - 9*A*a*b^5)*x)*sqr t(x) - 70*((5*B*a^2*b^4 - 7*A*a*b^5)*x^2 - 3*(7*B*a^3*b^3 - 9*A*a^2*b^4)*x )/sqrt(x) - 70*(3*(5*B*a^3*b^3 - 7*A*a^2*b^4)*x^2 - (7*B*a^4*b^2 - 9*A*a^3 *b^3)*x)/x^(3/2) - 14*(5*(5*B*a^4*b^2 - 7*A*a^3*b^3)*x^2 + (7*B*a^5*b - 9* A*a^4*b^2)*x)/x^(5/2) + 2*(7*(5*B*a^5*b - 7*A*a^4*b^2)*x^2 + 3*(7*B*a^6 - 9*A*a^5*b)*x)/x^(7/2) + 6*(7*A*a^5*b*x^2 + 5*A*a^6*x)/x^(9/2))/(a^6*b*x + a^7) - 2*(B*a*b^3 - A*b^4)*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*a^4) + 1 /3*((5*B*a*b^4 - 7*A*b^5)*x^(3/2) + 6*(B*a^2*b^3 - A*a*b^4)*sqrt(x))/a^6
Time = 0.27 (sec) , antiderivative size = 158, normalized size of antiderivative = 0.66 \[ \int \frac {A+B x}{x^{9/2} \sqrt {a^2+2 a b x+b^2 x^2}} \, dx=-\frac {2 \, {\left (B a b^{3} \mathrm {sgn}\left (b x + a\right ) - A b^{4} \mathrm {sgn}\left (b x + a\right )\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} a^{4}} - \frac {2 \, {\left (105 \, B a b^{2} x^{3} \mathrm {sgn}\left (b x + a\right ) - 105 \, A b^{3} x^{3} \mathrm {sgn}\left (b x + a\right ) - 35 \, B a^{2} b x^{2} \mathrm {sgn}\left (b x + a\right ) + 35 \, A a b^{2} x^{2} \mathrm {sgn}\left (b x + a\right ) + 21 \, B a^{3} x \mathrm {sgn}\left (b x + a\right ) - 21 \, A a^{2} b x \mathrm {sgn}\left (b x + a\right ) + 15 \, A a^{3} \mathrm {sgn}\left (b x + a\right )\right )}}{105 \, a^{4} x^{\frac {7}{2}}} \]
-2*(B*a*b^3*sgn(b*x + a) - A*b^4*sgn(b*x + a))*arctan(b*sqrt(x)/sqrt(a*b)) /(sqrt(a*b)*a^4) - 2/105*(105*B*a*b^2*x^3*sgn(b*x + a) - 105*A*b^3*x^3*sgn (b*x + a) - 35*B*a^2*b*x^2*sgn(b*x + a) + 35*A*a*b^2*x^2*sgn(b*x + a) + 21 *B*a^3*x*sgn(b*x + a) - 21*A*a^2*b*x*sgn(b*x + a) + 15*A*a^3*sgn(b*x + a)) /(a^4*x^(7/2))
Timed out. \[ \int \frac {A+B x}{x^{9/2} \sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\int \frac {A+B\,x}{x^{9/2}\,\sqrt {{\left (a+b\,x\right )}^2}} \,d x \]