Integrand size = 29, antiderivative size = 95 \[ \int \frac {A+B x}{x^{5/2} \left (a^2+2 a b x+b^2 x^2\right )} \, dx=-\frac {2 A}{3 a^2 x^{3/2}}+\frac {2 (2 A b-a B)}{a^3 \sqrt {x}}+\frac {b (A b-a B) \sqrt {x}}{a^3 (a+b x)}+\frac {\sqrt {b} (5 A b-3 a B) \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{a^{7/2}} \] Output:
-2/3*A/a^2/x^(3/2)+2*(2*A*b-B*a)/a^3/x^(1/2)+b*(A*b-B*a)*x^(1/2)/a^3/(b*x+ a)+b^(1/2)*(5*A*b-3*B*a)*arctan(b^(1/2)*x^(1/2)/a^(1/2))/a^(7/2)
Time = 0.18 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.95 \[ \int \frac {A+B x}{x^{5/2} \left (a^2+2 a b x+b^2 x^2\right )} \, dx=\frac {15 A b^2 x^2+a b x (10 A-9 B x)-2 a^2 (A+3 B x)}{3 a^3 x^{3/2} (a+b x)}+\frac {\sqrt {b} (5 A b-3 a B) \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{a^{7/2}} \] Input:
Integrate[(A + B*x)/(x^(5/2)*(a^2 + 2*a*b*x + b^2*x^2)),x]
Output:
(15*A*b^2*x^2 + a*b*x*(10*A - 9*B*x) - 2*a^2*(A + 3*B*x))/(3*a^3*x^(3/2)*( a + b*x)) + (Sqrt[b]*(5*A*b - 3*a*B)*ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[a]])/a^ (7/2)
Time = 0.38 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.12, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.241, Rules used = {1184, 27, 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.
\(\displaystyle \int \frac {A+B x}{x^{5/2} \left (a^2+2 a b x+b^2 x^2\right )} \, dx\) |
\(\Big \downarrow \) 1184 |
\(\displaystyle b^2 \int \frac {A+B x}{b^2 x^{5/2} (a+b x)^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \int \frac {A+B x}{x^{5/2} (a+b x)^2}dx\) |
\(\Big \downarrow \) 87 |
\(\displaystyle \frac {(5 A b-3 a B) \int \frac {1}{x^{5/2} (a+b x)}dx}{2 a b}+\frac {A b-a B}{a b x^{3/2} (a+b x)}\) |
\(\Big \downarrow \) 61 |
\(\displaystyle \frac {(5 A b-3 a B) \left (-\frac {b \int \frac {1}{x^{3/2} (a+b x)}dx}{a}-\frac {2}{3 a x^{3/2}}\right )}{2 a b}+\frac {A b-a B}{a b x^{3/2} (a+b x)}\) |
\(\Big \downarrow \) 61 |
\(\displaystyle \frac {(5 A b-3 a B) \left (-\frac {b \left (-\frac {b \int \frac {1}{\sqrt {x} (a+b x)}dx}{a}-\frac {2}{a \sqrt {x}}\right )}{a}-\frac {2}{3 a x^{3/2}}\right )}{2 a b}+\frac {A b-a B}{a b x^{3/2} (a+b x)}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {(5 A b-3 a B) \left (-\frac {b \left (-\frac {2 b \int \frac {1}{a+b x}d\sqrt {x}}{a}-\frac {2}{a \sqrt {x}}\right )}{a}-\frac {2}{3 a x^{3/2}}\right )}{2 a b}+\frac {A b-a B}{a b x^{3/2} (a+b x)}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {(5 A b-3 a B) \left (-\frac {b \left (-\frac {2 \sqrt {b} \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{a^{3/2}}-\frac {2}{a \sqrt {x}}\right )}{a}-\frac {2}{3 a x^{3/2}}\right )}{2 a b}+\frac {A b-a B}{a b x^{3/2} (a+b x)}\) |
Input:
Int[(A + B*x)/(x^(5/2)*(a^2 + 2*a*b*x + b^2*x^2)),x]
Output:
(A*b - a*B)/(a*b*x^(3/2)*(a + b*x)) + ((5*A*b - 3*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))/(2*a*b)
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[1/c^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}, x] && E qQ[b^2 - 4*a*c, 0] && IntegerQ[p]
Time = 1.17 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.81
method | result | size |
risch | \(-\frac {2 \left (-6 A b x +3 B a x +A a \right )}{3 a^{3} x^{\frac {3}{2}}}+\frac {b \left (\frac {2 \left (\frac {A b}{2}-\frac {B a}{2}\right ) \sqrt {x}}{b x +a}+\frac {\left (5 A b -3 B a \right ) \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b}}\right )}{a^{3}}\) | \(77\) |
derivativedivides | \(-\frac {2 A}{3 a^{2} x^{\frac {3}{2}}}-\frac {2 \left (-2 A b +B a \right )}{a^{3} \sqrt {x}}+\frac {2 b \left (\frac {\left (\frac {A b}{2}-\frac {B a}{2}\right ) \sqrt {x}}{b x +a}+\frac {\left (5 A b -3 B a \right ) \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{2 \sqrt {a b}}\right )}{a^{3}}\) | \(81\) |
default | \(-\frac {2 A}{3 a^{2} x^{\frac {3}{2}}}-\frac {2 \left (-2 A b +B a \right )}{a^{3} \sqrt {x}}+\frac {2 b \left (\frac {\left (\frac {A b}{2}-\frac {B a}{2}\right ) \sqrt {x}}{b x +a}+\frac {\left (5 A b -3 B a \right ) \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{2 \sqrt {a b}}\right )}{a^{3}}\) | \(81\) |
Input:
int((B*x+A)/x^(5/2)/(b^2*x^2+2*a*b*x+a^2),x,method=_RETURNVERBOSE)
Output:
-2/3*(-6*A*b*x+3*B*a*x+A*a)/a^3/x^(3/2)+b/a^3*(2*(1/2*A*b-1/2*B*a)*x^(1/2) /(b*x+a)+(5*A*b-3*B*a)/(a*b)^(1/2)*arctan(b*x^(1/2)/(a*b)^(1/2)))
Time = 0.09 (sec) , antiderivative size = 257, normalized size of antiderivative = 2.71 \[ \int \frac {A+B x}{x^{5/2} \left (a^2+2 a b x+b^2 x^2\right )} \, dx=\left [-\frac {3 \, {\left ({\left (3 \, B a b - 5 \, A b^{2}\right )} x^{3} + {\left (3 \, B a^{2} - 5 \, A a b\right )} x^{2}\right )} \sqrt {-\frac {b}{a}} \log \left (\frac {b x + 2 \, a \sqrt {x} \sqrt {-\frac {b}{a}} - a}{b x + a}\right ) + 2 \, {\left (2 \, A a^{2} + 3 \, {\left (3 \, B a b - 5 \, A b^{2}\right )} x^{2} + 2 \, {\left (3 \, B a^{2} - 5 \, A a b\right )} x\right )} \sqrt {x}}{6 \, {\left (a^{3} b x^{3} + a^{4} x^{2}\right )}}, -\frac {3 \, {\left ({\left (3 \, B a b - 5 \, A b^{2}\right )} x^{3} + {\left (3 \, B a^{2} - 5 \, A a b\right )} x^{2}\right )} \sqrt {\frac {b}{a}} \arctan \left (\sqrt {x} \sqrt {\frac {b}{a}}\right ) + {\left (2 \, A a^{2} + 3 \, {\left (3 \, B a b - 5 \, A b^{2}\right )} x^{2} + 2 \, {\left (3 \, B a^{2} - 5 \, A a b\right )} x\right )} \sqrt {x}}{3 \, {\left (a^{3} b x^{3} + a^{4} x^{2}\right )}}\right ] \] Input:
integrate((B*x+A)/x^(5/2)/(b^2*x^2+2*a*b*x+a^2),x, algorithm="fricas")
Output:
[-1/6*(3*((3*B*a*b - 5*A*b^2)*x^3 + (3*B*a^2 - 5*A*a*b)*x^2)*sqrt(-b/a)*lo g((b*x + 2*a*sqrt(x)*sqrt(-b/a) - a)/(b*x + a)) + 2*(2*A*a^2 + 3*(3*B*a*b - 5*A*b^2)*x^2 + 2*(3*B*a^2 - 5*A*a*b)*x)*sqrt(x))/(a^3*b*x^3 + a^4*x^2), -1/3*(3*((3*B*a*b - 5*A*b^2)*x^3 + (3*B*a^2 - 5*A*a*b)*x^2)*sqrt(b/a)*arct an(sqrt(x)*sqrt(b/a)) + (2*A*a^2 + 3*(3*B*a*b - 5*A*b^2)*x^2 + 2*(3*B*a^2 - 5*A*a*b)*x)*sqrt(x))/(a^3*b*x^3 + a^4*x^2)]
Leaf count of result is larger than twice the leaf count of optimal. 882 vs. \(2 (90) = 180\).
Time = 27.55 (sec) , antiderivative size = 882, normalized size of antiderivative = 9.28 \[ \int \frac {A+B x}{x^{5/2} \left (a^2+2 a b x+b^2 x^2\right )} \, dx =\text {Too large to display} \] Input:
integrate((B*x+A)/x**(5/2)/(b**2*x**2+2*a*b*x+a**2),x)
Output:
Piecewise((zoo*(-2*A/(7*x**(7/2)) - 2*B/(5*x**(5/2))), Eq(a, 0) & Eq(b, 0) ), ((-2*A/(3*x**(3/2)) - 2*B/sqrt(x))/a**2, Eq(b, 0)), ((-2*A/(7*x**(7/2)) - 2*B/(5*x**(5/2)))/b**2, Eq(a, 0)), (-4*A*a**2*sqrt(-a/b)/(6*a**4*x**(3/ 2)*sqrt(-a/b) + 6*a**3*b*x**(5/2)*sqrt(-a/b)) + 15*A*a*b*x**(3/2)*log(sqrt (x) - sqrt(-a/b))/(6*a**4*x**(3/2)*sqrt(-a/b) + 6*a**3*b*x**(5/2)*sqrt(-a/ b)) - 15*A*a*b*x**(3/2)*log(sqrt(x) + sqrt(-a/b))/(6*a**4*x**(3/2)*sqrt(-a /b) + 6*a**3*b*x**(5/2)*sqrt(-a/b)) + 20*A*a*b*x*sqrt(-a/b)/(6*a**4*x**(3/ 2)*sqrt(-a/b) + 6*a**3*b*x**(5/2)*sqrt(-a/b)) + 15*A*b**2*x**(5/2)*log(sqr t(x) - sqrt(-a/b))/(6*a**4*x**(3/2)*sqrt(-a/b) + 6*a**3*b*x**(5/2)*sqrt(-a /b)) - 15*A*b**2*x**(5/2)*log(sqrt(x) + sqrt(-a/b))/(6*a**4*x**(3/2)*sqrt( -a/b) + 6*a**3*b*x**(5/2)*sqrt(-a/b)) + 30*A*b**2*x**2*sqrt(-a/b)/(6*a**4* x**(3/2)*sqrt(-a/b) + 6*a**3*b*x**(5/2)*sqrt(-a/b)) - 9*B*a**2*x**(3/2)*lo g(sqrt(x) - sqrt(-a/b))/(6*a**4*x**(3/2)*sqrt(-a/b) + 6*a**3*b*x**(5/2)*sq rt(-a/b)) + 9*B*a**2*x**(3/2)*log(sqrt(x) + sqrt(-a/b))/(6*a**4*x**(3/2)*s qrt(-a/b) + 6*a**3*b*x**(5/2)*sqrt(-a/b)) - 12*B*a**2*x*sqrt(-a/b)/(6*a**4 *x**(3/2)*sqrt(-a/b) + 6*a**3*b*x**(5/2)*sqrt(-a/b)) - 9*B*a*b*x**(5/2)*lo g(sqrt(x) - sqrt(-a/b))/(6*a**4*x**(3/2)*sqrt(-a/b) + 6*a**3*b*x**(5/2)*sq rt(-a/b)) + 9*B*a*b*x**(5/2)*log(sqrt(x) + sqrt(-a/b))/(6*a**4*x**(3/2)*sq rt(-a/b) + 6*a**3*b*x**(5/2)*sqrt(-a/b)) - 18*B*a*b*x**2*sqrt(-a/b)/(6*a** 4*x**(3/2)*sqrt(-a/b) + 6*a**3*b*x**(5/2)*sqrt(-a/b)), True))
Time = 0.16 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.98 \[ \int \frac {A+B x}{x^{5/2} \left (a^2+2 a b x+b^2 x^2\right )} \, dx=-\frac {2 \, A a^{2} + 3 \, {\left (3 \, B a b - 5 \, A b^{2}\right )} x^{2} + 2 \, {\left (3 \, B a^{2} - 5 \, A a b\right )} x}{3 \, {\left (a^{3} b x^{\frac {5}{2}} + a^{4} x^{\frac {3}{2}}\right )}} - \frac {{\left (3 \, B a b - 5 \, A b^{2}\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} a^{3}} \] Input:
integrate((B*x+A)/x^(5/2)/(b^2*x^2+2*a*b*x+a^2),x, algorithm="maxima")
Output:
-1/3*(2*A*a^2 + 3*(3*B*a*b - 5*A*b^2)*x^2 + 2*(3*B*a^2 - 5*A*a*b)*x)/(a^3* b*x^(5/2) + a^4*x^(3/2)) - (3*B*a*b - 5*A*b^2)*arctan(b*sqrt(x)/sqrt(a*b)) /(sqrt(a*b)*a^3)
Time = 0.26 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.89 \[ \int \frac {A+B x}{x^{5/2} \left (a^2+2 a b x+b^2 x^2\right )} \, dx=-\frac {{\left (3 \, B a b - 5 \, A b^{2}\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} a^{3}} - \frac {B a b \sqrt {x} - A b^{2} \sqrt {x}}{{\left (b x + a\right )} a^{3}} - \frac {2 \, {\left (3 \, B a x - 6 \, A b x + A a\right )}}{3 \, a^{3} x^{\frac {3}{2}}} \] Input:
integrate((B*x+A)/x^(5/2)/(b^2*x^2+2*a*b*x+a^2),x, algorithm="giac")
Output:
-(3*B*a*b - 5*A*b^2)*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*a^3) - (B*a*b* sqrt(x) - A*b^2*sqrt(x))/((b*x + a)*a^3) - 2/3*(3*B*a*x - 6*A*b*x + A*a)/( a^3*x^(3/2))
Time = 10.79 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.85 \[ \int \frac {A+B x}{x^{5/2} \left (a^2+2 a b x+b^2 x^2\right )} \, dx=\frac {\frac {2\,x\,\left (5\,A\,b-3\,B\,a\right )}{3\,a^2}-\frac {2\,A}{3\,a}+\frac {b\,x^2\,\left (5\,A\,b-3\,B\,a\right )}{a^3}}{a\,x^{3/2}+b\,x^{5/2}}+\frac {\sqrt {b}\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {x}}{\sqrt {a}}\right )\,\left (5\,A\,b-3\,B\,a\right )}{a^{7/2}} \] Input:
int((A + B*x)/(x^(5/2)*(a^2 + b^2*x^2 + 2*a*b*x)),x)
Output:
((2*x*(5*A*b - 3*B*a))/(3*a^2) - (2*A)/(3*a) + (b*x^2*(5*A*b - 3*B*a))/a^3 )/(a*x^(3/2) + b*x^(5/2)) + (b^(1/2)*atan((b^(1/2)*x^(1/2))/a^(1/2))*(5*A* b - 3*B*a))/a^(7/2)
Time = 0.26 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.48 \[ \int \frac {A+B x}{x^{5/2} \left (a^2+2 a b x+b^2 x^2\right )} \, dx=\frac {2 \sqrt {x}\, \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {\sqrt {x}\, b}{\sqrt {b}\, \sqrt {a}}\right ) b x -\frac {2 a^{2}}{3}+2 a b x}{\sqrt {x}\, a^{3} x} \] Input:
int((B*x+A)/x^(5/2)/(b^2*x^2+2*a*b*x+a^2),x)
Output:
(2*(3*sqrt(x)*sqrt(b)*sqrt(a)*atan((sqrt(x)*b)/(sqrt(b)*sqrt(a)))*b*x - a* *2 + 3*a*b*x))/(3*sqrt(x)*a**3*x)