Integrand size = 31, antiderivative size = 144 \[ \int \frac {A+B x}{x^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}} \, dx=-\frac {2 A (a+b x)}{3 a x^{3/2} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 (A b-a B) (a+b x)}{a^2 \sqrt {x} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 \sqrt {b} (A b-a B) (a+b x) \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{a^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}} \]
-2/3*A*(b*x+a)/a/x^(3/2)/((b*x+a)^2)^(1/2)+2*(A*b-B*a)*(b*x+a)*arctan(b^(1 /2)*x^(1/2)/a^(1/2))*b^(1/2)/a^(5/2)/((b*x+a)^2)^(1/2)+2*(A*b-B*a)*(b*x+a) /a^2/x^(1/2)/((b*x+a)^2)^(1/2)
Time = 0.03 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.61 \[ \int \frac {A+B x}{x^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}} \, dx=-\frac {2 (a+b x) \left (\sqrt {a} (-3 A b x+a (A+3 B x))-3 \sqrt {b} (A b-a B) x^{3/2} \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )\right )}{3 a^{5/2} x^{3/2} \sqrt {(a+b x)^2}} \]
(-2*(a + b*x)*(Sqrt[a]*(-3*A*b*x + a*(A + 3*B*x)) - 3*Sqrt[b]*(A*b - a*B)* x^(3/2)*ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[a]]))/(3*a^(5/2)*x^(3/2)*Sqrt[(a + b *x)^2])
Time = 0.22 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.65, 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, 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.
\(\displaystyle \int \frac {A+B x}{x^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}} \, dx\) |
\(\Big \downarrow \) 1187 |
\(\displaystyle \frac {b (a+b x) \int \frac {A+B x}{b x^{5/2} (a+b x)}dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {(a+b x) \int \frac {A+B x}{x^{5/2} (a+b x)}dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
\(\Big \downarrow \) 87 |
\(\displaystyle \frac {(a+b x) \left (-\frac {(A b-a B) \int \frac {1}{x^{3/2} (a+b x)}dx}{a}-\frac {2 A}{3 a x^{3/2}}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
\(\Big \downarrow \) 61 |
\(\displaystyle \frac {(a+b x) \left (-\frac {(A b-a B) \left (-\frac {b \int \frac {1}{\sqrt {x} (a+b x)}dx}{a}-\frac {2}{a \sqrt {x}}\right )}{a}-\frac {2 A}{3 a x^{3/2}}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {(a+b x) \left (-\frac {(A b-a B) \left (-\frac {2 b \int \frac {1}{a+b x}d\sqrt {x}}{a}-\frac {2}{a \sqrt {x}}\right )}{a}-\frac {2 A}{3 a x^{3/2}}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {(a+b x) \left (-\frac {(A b-a B) \left (-\frac {2 \sqrt {b} \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{a^{3/2}}-\frac {2}{a \sqrt {x}}\right )}{a}-\frac {2 A}{3 a x^{3/2}}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
((a + b*x)*((-2*A)/(3*a*x^(3/2)) - ((A*b - a*B)*(-2/(a*Sqrt[x]) - (2*Sqrt[ b]*ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[a]])/a^(3/2)))/a))/Sqrt[a^2 + 2*a*b*x + b ^2*x^2]
3.9.15.3.1 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 = 0.13 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.60
method | result | size |
risch | \(-\frac {2 \left (-3 A b x +3 a B x +a A \right ) \sqrt {\left (b x +a \right )^{2}}}{3 a^{2} x^{\frac {3}{2}} \left (b x +a \right )}+\frac {2 \left (A b -B a \right ) b \arctan \left (\frac {b \sqrt {x}}{\sqrt {b a}}\right ) \sqrt {\left (b x +a \right )^{2}}}{a^{2} \sqrt {b a}\, \left (b x +a \right )}\) | \(86\) |
default | \(\frac {2 \left (b x +a \right ) \left (3 A \,x^{\frac {3}{2}} \arctan \left (\frac {b \sqrt {x}}{\sqrt {b a}}\right ) b^{2}-3 B \,x^{\frac {3}{2}} \arctan \left (\frac {b \sqrt {x}}{\sqrt {b a}}\right ) a b +3 A x \sqrt {b a}\, b -3 B x \sqrt {b a}\, a -A a \sqrt {b a}\right )}{3 \sqrt {\left (b x +a \right )^{2}}\, a^{2} \sqrt {b a}\, x^{\frac {3}{2}}}\) | \(97\) |
-2/3*(-3*A*b*x+3*B*a*x+A*a)/a^2/x^(3/2)*((b*x+a)^2)^(1/2)/(b*x+a)+2*(A*b-B *a)/a^2*b/(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 = 146, normalized size of antiderivative = 1.01 \[ \int \frac {A+B x}{x^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\left [-\frac {3 \, {\left (B a - A b\right )} x^{2} \sqrt {-\frac {b}{a}} \log \left (\frac {b x + 2 \, a \sqrt {x} \sqrt {-\frac {b}{a}} - a}{b x + a}\right ) + 2 \, {\left (A a + 3 \, {\left (B a - A b\right )} x\right )} \sqrt {x}}{3 \, a^{2} x^{2}}, \frac {2 \, {\left (3 \, {\left (B a - A b\right )} x^{2} \sqrt {\frac {b}{a}} \arctan \left (\frac {a \sqrt {\frac {b}{a}}}{b \sqrt {x}}\right ) - {\left (A a + 3 \, {\left (B a - A b\right )} x\right )} \sqrt {x}\right )}}{3 \, a^{2} x^{2}}\right ] \]
[-1/3*(3*(B*a - A*b)*x^2*sqrt(-b/a)*log((b*x + 2*a*sqrt(x)*sqrt(-b/a) - a) /(b*x + a)) + 2*(A*a + 3*(B*a - A*b)*x)*sqrt(x))/(a^2*x^2), 2/3*(3*(B*a - A*b)*x^2*sqrt(b/a)*arctan(a*sqrt(b/a)/(b*sqrt(x))) - (A*a + 3*(B*a - A*b)* x)*sqrt(x))/(a^2*x^2)]
Timed out. \[ \int \frac {A+B x}{x^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\text {Timed out} \]
Leaf count of result is larger than twice the leaf count of optimal. 244 vs. \(2 (95) = 190\).
Time = 0.33 (sec) , antiderivative size = 244, normalized size of antiderivative = 1.69 \[ \int \frac {A+B x}{x^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}} \, dx=-\frac {{\left ({\left (B a b^{3} - 3 \, A b^{4}\right )} x^{2} + 3 \, {\left (3 \, B a^{2} b^{2} - 5 \, A a b^{3}\right )} x\right )} \sqrt {x} - \frac {2 \, {\left ({\left (B a^{2} b^{2} - 3 \, A a b^{3}\right )} x^{2} - 3 \, {\left (3 \, B a^{3} b - 5 \, A a^{2} b^{2}\right )} x\right )}}{\sqrt {x}} - \frac {2 \, {\left (3 \, {\left (B a^{3} b - 3 \, A a^{2} b^{2}\right )} x^{2} - {\left (3 \, B a^{4} - 5 \, A a^{3} b\right )} x\right )}}{x^{\frac {3}{2}}} + \frac {2 \, {\left (3 \, A a^{3} b x^{2} + A a^{4} x\right )}}{x^{\frac {5}{2}}}}{3 \, {\left (a^{4} b x + a^{5}\right )}} - \frac {2 \, {\left (B a b - A b^{2}\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} a^{2}} + \frac {{\left (B a b^{2} - 3 \, A b^{3}\right )} x^{\frac {3}{2}} + 6 \, {\left (B a^{2} b - A a b^{2}\right )} \sqrt {x}}{3 \, a^{4}} \]
-1/3*(((B*a*b^3 - 3*A*b^4)*x^2 + 3*(3*B*a^2*b^2 - 5*A*a*b^3)*x)*sqrt(x) - 2*((B*a^2*b^2 - 3*A*a*b^3)*x^2 - 3*(3*B*a^3*b - 5*A*a^2*b^2)*x)/sqrt(x) - 2*(3*(B*a^3*b - 3*A*a^2*b^2)*x^2 - (3*B*a^4 - 5*A*a^3*b)*x)/x^(3/2) + 2*(3 *A*a^3*b*x^2 + A*a^4*x)/x^(5/2))/(a^4*b*x + a^5) - 2*(B*a*b - A*b^2)*arcta n(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*a^2) + 1/3*((B*a*b^2 - 3*A*b^3)*x^(3/2) + 6*(B*a^2*b - A*a*b^2)*sqrt(x))/a^4
Time = 0.28 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.59 \[ \int \frac {A+B x}{x^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}} \, dx=-\frac {2 \, {\left (B a b \mathrm {sgn}\left (b x + a\right ) - A b^{2} \mathrm {sgn}\left (b x + a\right )\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} a^{2}} - \frac {2 \, {\left (3 \, B a x \mathrm {sgn}\left (b x + a\right ) - 3 \, A b x \mathrm {sgn}\left (b x + a\right ) + A a \mathrm {sgn}\left (b x + a\right )\right )}}{3 \, a^{2} x^{\frac {3}{2}}} \]
-2*(B*a*b*sgn(b*x + a) - A*b^2*sgn(b*x + a))*arctan(b*sqrt(x)/sqrt(a*b))/( sqrt(a*b)*a^2) - 2/3*(3*B*a*x*sgn(b*x + a) - 3*A*b*x*sgn(b*x + a) + A*a*sg n(b*x + a))/(a^2*x^(3/2))
Timed out. \[ \int \frac {A+B x}{x^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\int \frac {A+B\,x}{x^{5/2}\,\sqrt {{\left (a+b\,x\right )}^2}} \,d x \]