Integrand size = 18, antiderivative size = 90 \[ \int \frac {A+B x}{x^{7/2} (a+b x)} \, dx=-\frac {2 A}{5 a x^{5/2}}+\frac {2 (A b-a B)}{3 a^2 x^{3/2}}-\frac {2 b (A b-a B)}{a^3 \sqrt {x}}-\frac {2 b^{3/2} (A b-a B) \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{a^{7/2}} \]
-2/5*A/a/x^(5/2)+2/3*(A*b-B*a)/a^2/x^(3/2)-2*b^(3/2)*(A*b-B*a)*arctan(b^(1 /2)*x^(1/2)/a^(1/2))/a^(7/2)-2*b*(A*b-B*a)/a^3/x^(1/2)
Time = 0.11 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.92 \[ \int \frac {A+B x}{x^{7/2} (a+b x)} \, dx=-\frac {2 \left (15 A b^2 x^2-5 a b x (A+3 B x)+a^2 (3 A+5 B x)\right )}{15 a^3 x^{5/2}}+\frac {2 b^{3/2} (-A b+a B) \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{a^{7/2}} \]
(-2*(15*A*b^2*x^2 - 5*a*b*x*(A + 3*B*x) + a^2*(3*A + 5*B*x)))/(15*a^3*x^(5 /2)) + (2*b^(3/2)*(-(A*b) + a*B)*ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[a]])/a^(7/2 )
Time = 0.19 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.96, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.278, Rules used = {87, 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.
\(\displaystyle \int \frac {A+B x}{x^{7/2} (a+b x)} \, dx\) |
\(\Big \downarrow \) 87 |
\(\displaystyle -\frac {(A b-a B) \int \frac {1}{x^{5/2} (a+b x)}dx}{a}-\frac {2 A}{5 a x^{5/2}}\) |
\(\Big \downarrow \) 61 |
\(\displaystyle -\frac {(A b-a B) \left (-\frac {b \int \frac {1}{x^{3/2} (a+b x)}dx}{a}-\frac {2}{3 a x^{3/2}}\right )}{a}-\frac {2 A}{5 a x^{5/2}}\) |
\(\Big \downarrow \) 61 |
\(\displaystyle -\frac {(A b-a B) \left (-\frac {b \left (-\frac {b \int \frac {1}{\sqrt {x} (a+b x)}dx}{a}-\frac {2}{a \sqrt {x}}\right )}{a}-\frac {2}{3 a x^{3/2}}\right )}{a}-\frac {2 A}{5 a x^{5/2}}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle -\frac {(A b-a B) \left (-\frac {b \left (-\frac {2 b \int \frac {1}{a+b x}d\sqrt {x}}{a}-\frac {2}{a \sqrt {x}}\right )}{a}-\frac {2}{3 a x^{3/2}}\right )}{a}-\frac {2 A}{5 a x^{5/2}}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle -\frac {(A b-a B) \left (-\frac {b \left (-\frac {2 \sqrt {b} \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{a^{3/2}}-\frac {2}{a \sqrt {x}}\right )}{a}-\frac {2}{3 a x^{3/2}}\right )}{a}-\frac {2 A}{5 a x^{5/2}}\) |
(-2*A)/(5*a*x^(5/2)) - ((A*b - a*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
3.4.51.3.1 Defintions of rubi rules used
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]
Time = 0.47 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.84
method | result | size |
derivativedivides | \(-\frac {2 \left (A b -B a \right ) b^{2} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{a^{3} \sqrt {a b}}-\frac {2 A}{5 a \,x^{\frac {5}{2}}}-\frac {2 \left (-A b +B a \right )}{3 a^{2} x^{\frac {3}{2}}}-\frac {2 b \left (A b -B a \right )}{a^{3} \sqrt {x}}\) | \(76\) |
default | \(-\frac {2 \left (A b -B a \right ) b^{2} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{a^{3} \sqrt {a b}}-\frac {2 A}{5 a \,x^{\frac {5}{2}}}-\frac {2 \left (-A b +B a \right )}{3 a^{2} x^{\frac {3}{2}}}-\frac {2 b \left (A b -B a \right )}{a^{3} \sqrt {x}}\) | \(76\) |
risch | \(-\frac {2 \left (15 A \,b^{2} x^{2}-15 B a b \,x^{2}-5 a A b x +5 a^{2} B x +3 a^{2} A \right )}{15 a^{3} x^{\frac {5}{2}}}-\frac {2 \left (A b -B a \right ) b^{2} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{a^{3} \sqrt {a b}}\) | \(79\) |
-2*(A*b-B*a)/a^3*b^2/(a*b)^(1/2)*arctan(b*x^(1/2)/(a*b)^(1/2))-2/5*A/a/x^( 5/2)-2/3*(-A*b+B*a)/a^2/x^(3/2)-2*b*(A*b-B*a)/a^3/x^(1/2)
Time = 0.24 (sec) , antiderivative size = 195, normalized size of antiderivative = 2.17 \[ \int \frac {A+B x}{x^{7/2} (a+b x)} \, dx=\left [-\frac {15 \, {\left (B a b - A b^{2}\right )} x^{3} \sqrt {-\frac {b}{a}} \log \left (\frac {b x - 2 \, a \sqrt {x} \sqrt {-\frac {b}{a}} - a}{b x + a}\right ) + 2 \, {\left (3 \, A a^{2} - 15 \, {\left (B a b - A b^{2}\right )} x^{2} + 5 \, {\left (B a^{2} - A a b\right )} x\right )} \sqrt {x}}{15 \, a^{3} x^{3}}, -\frac {2 \, {\left (15 \, {\left (B a b - A b^{2}\right )} x^{3} \sqrt {\frac {b}{a}} \arctan \left (\frac {a \sqrt {\frac {b}{a}}}{b \sqrt {x}}\right ) + {\left (3 \, A a^{2} - 15 \, {\left (B a b - A b^{2}\right )} x^{2} + 5 \, {\left (B a^{2} - A a b\right )} x\right )} \sqrt {x}\right )}}{15 \, a^{3} x^{3}}\right ] \]
[-1/15*(15*(B*a*b - A*b^2)*x^3*sqrt(-b/a)*log((b*x - 2*a*sqrt(x)*sqrt(-b/a ) - a)/(b*x + a)) + 2*(3*A*a^2 - 15*(B*a*b - A*b^2)*x^2 + 5*(B*a^2 - A*a*b )*x)*sqrt(x))/(a^3*x^3), -2/15*(15*(B*a*b - A*b^2)*x^3*sqrt(b/a)*arctan(a* sqrt(b/a)/(b*sqrt(x))) + (3*A*a^2 - 15*(B*a*b - A*b^2)*x^2 + 5*(B*a^2 - A* a*b)*x)*sqrt(x))/(a^3*x^3)]
Leaf count of result is larger than twice the leaf count of optimal. 262 vs. \(2 (87) = 174\).
Time = 5.40 (sec) , antiderivative size = 262, normalized size of antiderivative = 2.91 \[ \int \frac {A+B x}{x^{7/2} (a+b x)} \, dx=\begin {cases} \tilde {\infty } \left (- \frac {2 A}{7 x^{\frac {7}{2}}} - \frac {2 B}{5 x^{\frac {5}{2}}}\right ) & \text {for}\: a = 0 \wedge b = 0 \\\frac {- \frac {2 A}{7 x^{\frac {7}{2}}} - \frac {2 B}{5 x^{\frac {5}{2}}}}{b} & \text {for}\: a = 0 \\\frac {- \frac {2 A}{5 x^{\frac {5}{2}}} - \frac {2 B}{3 x^{\frac {3}{2}}}}{a} & \text {for}\: b = 0 \\- \frac {2 A}{5 a x^{\frac {5}{2}}} + \frac {2 A b}{3 a^{2} x^{\frac {3}{2}}} - \frac {A b^{2} \log {\left (\sqrt {x} - \sqrt {- \frac {a}{b}} \right )}}{a^{3} \sqrt {- \frac {a}{b}}} + \frac {A b^{2} \log {\left (\sqrt {x} + \sqrt {- \frac {a}{b}} \right )}}{a^{3} \sqrt {- \frac {a}{b}}} - \frac {2 A b^{2}}{a^{3} \sqrt {x}} - \frac {2 B}{3 a x^{\frac {3}{2}}} + \frac {B b \log {\left (\sqrt {x} - \sqrt {- \frac {a}{b}} \right )}}{a^{2} \sqrt {- \frac {a}{b}}} - \frac {B b \log {\left (\sqrt {x} + \sqrt {- \frac {a}{b}} \right )}}{a^{2} \sqrt {- \frac {a}{b}}} + \frac {2 B b}{a^{2} \sqrt {x}} & \text {otherwise} \end {cases} \]
Piecewise((zoo*(-2*A/(7*x**(7/2)) - 2*B/(5*x**(5/2))), Eq(a, 0) & Eq(b, 0) ), ((-2*A/(7*x**(7/2)) - 2*B/(5*x**(5/2)))/b, Eq(a, 0)), ((-2*A/(5*x**(5/2 )) - 2*B/(3*x**(3/2)))/a, Eq(b, 0)), (-2*A/(5*a*x**(5/2)) + 2*A*b/(3*a**2* x**(3/2)) - A*b**2*log(sqrt(x) - sqrt(-a/b))/(a**3*sqrt(-a/b)) + A*b**2*lo g(sqrt(x) + sqrt(-a/b))/(a**3*sqrt(-a/b)) - 2*A*b**2/(a**3*sqrt(x)) - 2*B/ (3*a*x**(3/2)) + B*b*log(sqrt(x) - sqrt(-a/b))/(a**2*sqrt(-a/b)) - B*b*log (sqrt(x) + sqrt(-a/b))/(a**2*sqrt(-a/b)) + 2*B*b/(a**2*sqrt(x)), True))
Time = 0.29 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.89 \[ \int \frac {A+B x}{x^{7/2} (a+b x)} \, dx=\frac {2 \, {\left (B a b^{2} - A b^{3}\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} a^{3}} - \frac {2 \, {\left (3 \, A a^{2} - 15 \, {\left (B a b - A b^{2}\right )} x^{2} + 5 \, {\left (B a^{2} - A a b\right )} x\right )}}{15 \, a^{3} x^{\frac {5}{2}}} \]
2*(B*a*b^2 - A*b^3)*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*a^3) - 2/15*(3* A*a^2 - 15*(B*a*b - A*b^2)*x^2 + 5*(B*a^2 - A*a*b)*x)/(a^3*x^(5/2))
Time = 0.28 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.89 \[ \int \frac {A+B x}{x^{7/2} (a+b x)} \, dx=\frac {2 \, {\left (B a b^{2} - A b^{3}\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} a^{3}} + \frac {2 \, {\left (15 \, B a b x^{2} - 15 \, A b^{2} x^{2} - 5 \, B a^{2} x + 5 \, A a b x - 3 \, A a^{2}\right )}}{15 \, a^{3} x^{\frac {5}{2}}} \]
2*(B*a*b^2 - A*b^3)*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*a^3) + 2/15*(15 *B*a*b*x^2 - 15*A*b^2*x^2 - 5*B*a^2*x + 5*A*a*b*x - 3*A*a^2)/(a^3*x^(5/2))
Time = 0.43 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.79 \[ \int \frac {A+B x}{x^{7/2} (a+b x)} \, dx=-\frac {\frac {2\,A}{5\,a}-\frac {2\,x\,\left (A\,b-B\,a\right )}{3\,a^2}+\frac {2\,b\,x^2\,\left (A\,b-B\,a\right )}{a^3}}{x^{5/2}}-\frac {2\,b^{3/2}\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {x}}{\sqrt {a}}\right )\,\left (A\,b-B\,a\right )}{a^{7/2}} \]