Integrand size = 29, antiderivative size = 117 \[ \int \frac {x^{5/2} (A+B x)}{a^2+2 a b x+b^2 x^2} \, dx=-\frac {a (5 A b-7 a B) \sqrt {x}}{b^4}+\frac {(5 A b-7 a B) x^{3/2}}{3 b^3}+\frac {2 B x^{5/2}}{5 b^2}-\frac {(A b-a B) x^{5/2}}{b^2 (a+b x)}+\frac {a^{3/2} (5 A b-7 a B) \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{b^{9/2}} \] Output:
-a*(5*A*b-7*B*a)*x^(1/2)/b^4+1/3*(5*A*b-7*B*a)*x^(3/2)/b^3+2/5*B*x^(5/2)/b ^2-(A*b-B*a)*x^(5/2)/b^2/(b*x+a)+a^(3/2)*(5*A*b-7*B*a)*arctan(b^(1/2)*x^(1 /2)/a^(1/2))/b^(9/2)
Time = 0.22 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.94 \[ \int \frac {x^{5/2} (A+B x)}{a^2+2 a b x+b^2 x^2} \, dx=\frac {\sqrt {x} \left (105 a^3 B+2 b^3 x^2 (5 A+3 B x)-2 a b^2 x (25 A+7 B x)+a^2 (-75 A b+70 b B x)\right )}{15 b^4 (a+b x)}-\frac {a^{3/2} (-5 A b+7 a B) \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{b^{9/2}} \] Input:
Integrate[(x^(5/2)*(A + B*x))/(a^2 + 2*a*b*x + b^2*x^2),x]
Output:
(Sqrt[x]*(105*a^3*B + 2*b^3*x^2*(5*A + 3*B*x) - 2*a*b^2*x*(25*A + 7*B*x) + a^2*(-75*A*b + 70*b*B*x)))/(15*b^4*(a + b*x)) - (a^(3/2)*(-5*A*b + 7*a*B) *ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[a]])/b^(9/2)
Time = 0.41 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.07, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.276, Rules used = {1184, 27, 87, 60, 60, 60, 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.
\(\displaystyle \int \frac {x^{5/2} (A+B x)}{a^2+2 a b x+b^2 x^2} \, dx\) |
\(\Big \downarrow \) 1184 |
\(\displaystyle b^2 \int \frac {x^{5/2} (A+B x)}{b^2 (a+b x)^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \int \frac {x^{5/2} (A+B x)}{(a+b x)^2}dx\) |
\(\Big \downarrow \) 87 |
\(\displaystyle \frac {x^{7/2} (A b-a B)}{a b (a+b x)}-\frac {(5 A b-7 a B) \int \frac {x^{5/2}}{a+b x}dx}{2 a b}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {x^{7/2} (A b-a B)}{a b (a+b x)}-\frac {(5 A b-7 a B) \left (\frac {2 x^{5/2}}{5 b}-\frac {a \int \frac {x^{3/2}}{a+b x}dx}{b}\right )}{2 a b}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {x^{7/2} (A b-a B)}{a b (a+b x)}-\frac {(5 A b-7 a B) \left (\frac {2 x^{5/2}}{5 b}-\frac {a \left (\frac {2 x^{3/2}}{3 b}-\frac {a \int \frac {\sqrt {x}}{a+b x}dx}{b}\right )}{b}\right )}{2 a b}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {x^{7/2} (A b-a B)}{a b (a+b x)}-\frac {(5 A b-7 a B) \left (\frac {2 x^{5/2}}{5 b}-\frac {a \left (\frac {2 x^{3/2}}{3 b}-\frac {a \left (\frac {2 \sqrt {x}}{b}-\frac {a \int \frac {1}{\sqrt {x} (a+b x)}dx}{b}\right )}{b}\right )}{b}\right )}{2 a b}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {x^{7/2} (A b-a B)}{a b (a+b x)}-\frac {(5 A b-7 a B) \left (\frac {2 x^{5/2}}{5 b}-\frac {a \left (\frac {2 x^{3/2}}{3 b}-\frac {a \left (\frac {2 \sqrt {x}}{b}-\frac {2 a \int \frac {1}{a+b x}d\sqrt {x}}{b}\right )}{b}\right )}{b}\right )}{2 a b}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {x^{7/2} (A b-a B)}{a b (a+b x)}-\frac {(5 A b-7 a B) \left (\frac {2 x^{5/2}}{5 b}-\frac {a \left (\frac {2 x^{3/2}}{3 b}-\frac {a \left (\frac {2 \sqrt {x}}{b}-\frac {2 \sqrt {a} \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{b^{3/2}}\right )}{b}\right )}{b}\right )}{2 a b}\) |
Input:
Int[(x^(5/2)*(A + B*x))/(a^2 + 2*a*b*x + b^2*x^2),x]
Output:
((A*b - a*B)*x^(7/2))/(a*b*(a + b*x)) - ((5*A*b - 7*a*B)*((2*x^(5/2))/(5*b ) - (a*((2*x^(3/2))/(3*b) - (a*((2*Sqrt[x])/b - (2*Sqrt[a]*ArcTan[(Sqrt[b] *Sqrt[x])/Sqrt[a]])/b^(3/2)))/b))/b))/(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/(b*(m + n + 1))), x] + Simp[n*((b*c - a*d)/( b*(m + n + 1))) Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] && GtQ[n, 0] && NeQ[m + n + 1, 0] && !(IGtQ[m, 0] && ( !Integer Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) && !ILtQ[m + n + 2, 0] && IntLinear Q[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 = 1.12 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.85
method | result | size |
risch | \(-\frac {2 \left (-3 x^{2} B \,b^{2}-5 x \,b^{2} A +10 x a b B +30 a b A -45 a^{2} B \right ) \sqrt {x}}{15 b^{4}}+\frac {a^{2} \left (\frac {2 \left (-\frac {A b}{2}+\frac {B a}{2}\right ) \sqrt {x}}{b x +a}+\frac {\left (5 A b -7 B a \right ) \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b}}\right )}{b^{4}}\) | \(99\) |
derivativedivides | \(-\frac {2 \left (-\frac {B \,b^{2} x^{\frac {5}{2}}}{5}-\frac {A \,b^{2} x^{\frac {3}{2}}}{3}+\frac {2 B a b \,x^{\frac {3}{2}}}{3}+2 A a b \sqrt {x}-3 B \,a^{2} \sqrt {x}\right )}{b^{4}}+\frac {2 a^{2} \left (\frac {\left (-\frac {A b}{2}+\frac {B a}{2}\right ) \sqrt {x}}{b x +a}+\frac {\left (5 A b -7 B a \right ) \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{2 \sqrt {a b}}\right )}{b^{4}}\) | \(107\) |
default | \(-\frac {2 \left (-\frac {B \,b^{2} x^{\frac {5}{2}}}{5}-\frac {A \,b^{2} x^{\frac {3}{2}}}{3}+\frac {2 B a b \,x^{\frac {3}{2}}}{3}+2 A a b \sqrt {x}-3 B \,a^{2} \sqrt {x}\right )}{b^{4}}+\frac {2 a^{2} \left (\frac {\left (-\frac {A b}{2}+\frac {B a}{2}\right ) \sqrt {x}}{b x +a}+\frac {\left (5 A b -7 B a \right ) \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{2 \sqrt {a b}}\right )}{b^{4}}\) | \(107\) |
Input:
int(x^(5/2)*(B*x+A)/(b^2*x^2+2*a*b*x+a^2),x,method=_RETURNVERBOSE)
Output:
-2/15*(-3*B*b^2*x^2-5*A*b^2*x+10*B*a*b*x+30*A*a*b-45*B*a^2)*x^(1/2)/b^4+a^ 2/b^4*(2*(-1/2*A*b+1/2*B*a)*x^(1/2)/(b*x+a)+(5*A*b-7*B*a)/(a*b)^(1/2)*arct an(b*x^(1/2)/(a*b)^(1/2)))
Time = 0.09 (sec) , antiderivative size = 290, normalized size of antiderivative = 2.48 \[ \int \frac {x^{5/2} (A+B x)}{a^2+2 a b x+b^2 x^2} \, dx=\left [-\frac {15 \, {\left (7 \, B a^{3} - 5 \, A a^{2} b + {\left (7 \, B a^{2} b - 5 \, A a b^{2}\right )} x\right )} \sqrt {-\frac {a}{b}} \log \left (\frac {b x + 2 \, b \sqrt {x} \sqrt {-\frac {a}{b}} - a}{b x + a}\right ) - 2 \, {\left (6 \, B b^{3} x^{3} + 105 \, B a^{3} - 75 \, A a^{2} b - 2 \, {\left (7 \, B a b^{2} - 5 \, A b^{3}\right )} x^{2} + 10 \, {\left (7 \, B a^{2} b - 5 \, A a b^{2}\right )} x\right )} \sqrt {x}}{30 \, {\left (b^{5} x + a b^{4}\right )}}, -\frac {15 \, {\left (7 \, B a^{3} - 5 \, A a^{2} b + {\left (7 \, B a^{2} b - 5 \, A a b^{2}\right )} x\right )} \sqrt {\frac {a}{b}} \arctan \left (\frac {b \sqrt {x} \sqrt {\frac {a}{b}}}{a}\right ) - {\left (6 \, B b^{3} x^{3} + 105 \, B a^{3} - 75 \, A a^{2} b - 2 \, {\left (7 \, B a b^{2} - 5 \, A b^{3}\right )} x^{2} + 10 \, {\left (7 \, B a^{2} b - 5 \, A a b^{2}\right )} x\right )} \sqrt {x}}{15 \, {\left (b^{5} x + a b^{4}\right )}}\right ] \] Input:
integrate(x^(5/2)*(B*x+A)/(b^2*x^2+2*a*b*x+a^2),x, algorithm="fricas")
Output:
[-1/30*(15*(7*B*a^3 - 5*A*a^2*b + (7*B*a^2*b - 5*A*a*b^2)*x)*sqrt(-a/b)*lo g((b*x + 2*b*sqrt(x)*sqrt(-a/b) - a)/(b*x + a)) - 2*(6*B*b^3*x^3 + 105*B*a ^3 - 75*A*a^2*b - 2*(7*B*a*b^2 - 5*A*b^3)*x^2 + 10*(7*B*a^2*b - 5*A*a*b^2) *x)*sqrt(x))/(b^5*x + a*b^4), -1/15*(15*(7*B*a^3 - 5*A*a^2*b + (7*B*a^2*b - 5*A*a*b^2)*x)*sqrt(a/b)*arctan(b*sqrt(x)*sqrt(a/b)/a) - (6*B*b^3*x^3 + 1 05*B*a^3 - 75*A*a^2*b - 2*(7*B*a*b^2 - 5*A*b^3)*x^2 + 10*(7*B*a^2*b - 5*A* a*b^2)*x)*sqrt(x))/(b^5*x + a*b^4)]
Leaf count of result is larger than twice the leaf count of optimal. 877 vs. \(2 (112) = 224\).
Time = 28.79 (sec) , antiderivative size = 877, normalized size of antiderivative = 7.50 \[ \int \frac {x^{5/2} (A+B x)}{a^2+2 a b x+b^2 x^2} \, dx =\text {Too large to display} \] Input:
integrate(x**(5/2)*(B*x+A)/(b**2*x**2+2*a*b*x+a**2),x)
Output:
Piecewise((zoo*(2*A*x**(3/2)/3 + 2*B*x**(5/2)/5), Eq(a, 0) & Eq(b, 0)), (( 2*A*x**(7/2)/7 + 2*B*x**(9/2)/9)/a**2, Eq(b, 0)), ((2*A*x**(3/2)/3 + 2*B*x **(5/2)/5)/b**2, Eq(a, 0)), (75*A*a**3*b*log(sqrt(x) - sqrt(-a/b))/(30*a*b **5*sqrt(-a/b) + 30*b**6*x*sqrt(-a/b)) - 75*A*a**3*b*log(sqrt(x) + sqrt(-a /b))/(30*a*b**5*sqrt(-a/b) + 30*b**6*x*sqrt(-a/b)) - 150*A*a**2*b**2*sqrt( x)*sqrt(-a/b)/(30*a*b**5*sqrt(-a/b) + 30*b**6*x*sqrt(-a/b)) + 75*A*a**2*b* *2*x*log(sqrt(x) - sqrt(-a/b))/(30*a*b**5*sqrt(-a/b) + 30*b**6*x*sqrt(-a/b )) - 75*A*a**2*b**2*x*log(sqrt(x) + sqrt(-a/b))/(30*a*b**5*sqrt(-a/b) + 30 *b**6*x*sqrt(-a/b)) - 100*A*a*b**3*x**(3/2)*sqrt(-a/b)/(30*a*b**5*sqrt(-a/ b) + 30*b**6*x*sqrt(-a/b)) + 20*A*b**4*x**(5/2)*sqrt(-a/b)/(30*a*b**5*sqrt (-a/b) + 30*b**6*x*sqrt(-a/b)) - 105*B*a**4*log(sqrt(x) - sqrt(-a/b))/(30* a*b**5*sqrt(-a/b) + 30*b**6*x*sqrt(-a/b)) + 105*B*a**4*log(sqrt(x) + sqrt( -a/b))/(30*a*b**5*sqrt(-a/b) + 30*b**6*x*sqrt(-a/b)) + 210*B*a**3*b*sqrt(x )*sqrt(-a/b)/(30*a*b**5*sqrt(-a/b) + 30*b**6*x*sqrt(-a/b)) - 105*B*a**3*b* x*log(sqrt(x) - sqrt(-a/b))/(30*a*b**5*sqrt(-a/b) + 30*b**6*x*sqrt(-a/b)) + 105*B*a**3*b*x*log(sqrt(x) + sqrt(-a/b))/(30*a*b**5*sqrt(-a/b) + 30*b**6 *x*sqrt(-a/b)) + 140*B*a**2*b**2*x**(3/2)*sqrt(-a/b)/(30*a*b**5*sqrt(-a/b) + 30*b**6*x*sqrt(-a/b)) - 28*B*a*b**3*x**(5/2)*sqrt(-a/b)/(30*a*b**5*sqrt (-a/b) + 30*b**6*x*sqrt(-a/b)) + 12*B*b**4*x**(7/2)*sqrt(-a/b)/(30*a*b**5* sqrt(-a/b) + 30*b**6*x*sqrt(-a/b)), True))
Time = 0.16 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.98 \[ \int \frac {x^{5/2} (A+B x)}{a^2+2 a b x+b^2 x^2} \, dx=\frac {{\left (B a^{3} - A a^{2} b\right )} \sqrt {x}}{b^{5} x + a b^{4}} - \frac {{\left (7 \, B a^{3} - 5 \, A a^{2} b\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} b^{4}} + \frac {2 \, {\left (3 \, B b^{2} x^{\frac {5}{2}} - 5 \, {\left (2 \, B a b - A b^{2}\right )} x^{\frac {3}{2}} + 15 \, {\left (3 \, B a^{2} - 2 \, A a b\right )} \sqrt {x}\right )}}{15 \, b^{4}} \] Input:
integrate(x^(5/2)*(B*x+A)/(b^2*x^2+2*a*b*x+a^2),x, algorithm="maxima")
Output:
(B*a^3 - A*a^2*b)*sqrt(x)/(b^5*x + a*b^4) - (7*B*a^3 - 5*A*a^2*b)*arctan(b *sqrt(x)/sqrt(a*b))/(sqrt(a*b)*b^4) + 2/15*(3*B*b^2*x^(5/2) - 5*(2*B*a*b - A*b^2)*x^(3/2) + 15*(3*B*a^2 - 2*A*a*b)*sqrt(x))/b^4
Time = 0.26 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.04 \[ \int \frac {x^{5/2} (A+B x)}{a^2+2 a b x+b^2 x^2} \, dx=-\frac {{\left (7 \, B a^{3} - 5 \, A a^{2} b\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} b^{4}} + \frac {B a^{3} \sqrt {x} - A a^{2} b \sqrt {x}}{{\left (b x + a\right )} b^{4}} + \frac {2 \, {\left (3 \, B b^{8} x^{\frac {5}{2}} - 10 \, B a b^{7} x^{\frac {3}{2}} + 5 \, A b^{8} x^{\frac {3}{2}} + 45 \, B a^{2} b^{6} \sqrt {x} - 30 \, A a b^{7} \sqrt {x}\right )}}{15 \, b^{10}} \] Input:
integrate(x^(5/2)*(B*x+A)/(b^2*x^2+2*a*b*x+a^2),x, algorithm="giac")
Output:
-(7*B*a^3 - 5*A*a^2*b)*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*b^4) + (B*a^ 3*sqrt(x) - A*a^2*b*sqrt(x))/((b*x + a)*b^4) + 2/15*(3*B*b^8*x^(5/2) - 10* B*a*b^7*x^(3/2) + 5*A*b^8*x^(3/2) + 45*B*a^2*b^6*sqrt(x) - 30*A*a*b^7*sqrt (x))/b^10
Time = 0.06 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.25 \[ \int \frac {x^{5/2} (A+B x)}{a^2+2 a b x+b^2 x^2} \, dx=x^{3/2}\,\left (\frac {2\,A}{3\,b^2}-\frac {4\,B\,a}{3\,b^3}\right )-\sqrt {x}\,\left (\frac {2\,a\,\left (\frac {2\,A}{b^2}-\frac {4\,B\,a}{b^3}\right )}{b}+\frac {2\,B\,a^2}{b^4}\right )+\frac {2\,B\,x^{5/2}}{5\,b^2}+\frac {\sqrt {x}\,\left (B\,a^3-A\,a^2\,b\right )}{x\,b^5+a\,b^4}-\frac {a^{3/2}\,\mathrm {atan}\left (\frac {a^{3/2}\,\sqrt {b}\,\sqrt {x}\,\left (5\,A\,b-7\,B\,a\right )}{7\,B\,a^3-5\,A\,a^2\,b}\right )\,\left (5\,A\,b-7\,B\,a\right )}{b^{9/2}} \] Input:
int((x^(5/2)*(A + B*x))/(a^2 + b^2*x^2 + 2*a*b*x),x)
Output:
x^(3/2)*((2*A)/(3*b^2) - (4*B*a)/(3*b^3)) - x^(1/2)*((2*a*((2*A)/b^2 - (4* B*a)/b^3))/b + (2*B*a^2)/b^4) + (2*B*x^(5/2))/(5*b^2) + (x^(1/2)*(B*a^3 - A*a^2*b))/(a*b^4 + b^5*x) - (a^(3/2)*atan((a^(3/2)*b^(1/2)*x^(1/2)*(5*A*b - 7*B*a))/(7*B*a^3 - 5*A*a^2*b))*(5*A*b - 7*B*a))/b^(9/2)
Time = 0.28 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.47 \[ \int \frac {x^{5/2} (A+B x)}{a^2+2 a b x+b^2 x^2} \, dx=\frac {-2 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {\sqrt {x}\, b}{\sqrt {b}\, \sqrt {a}}\right ) a^{2}+2 \sqrt {x}\, a^{2} b -\frac {2 \sqrt {x}\, a \,b^{2} x}{3}+\frac {2 \sqrt {x}\, b^{3} x^{2}}{5}}{b^{4}} \] Input:
int(x^(5/2)*(B*x+A)/(b^2*x^2+2*a*b*x+a^2),x)
Output:
(2*( - 15*sqrt(b)*sqrt(a)*atan((sqrt(x)*b)/(sqrt(b)*sqrt(a)))*a**2 + 15*sq rt(x)*a**2*b - 5*sqrt(x)*a*b**2*x + 3*sqrt(x)*b**3*x**2))/(15*b**4)