Integrand size = 29, antiderivative size = 134 \[ \int \frac {x^{5/2} (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=\frac {2 B \sqrt {x}}{b^4}-\frac {(A b-a B) x^{5/2}}{3 b^2 (a+b x)^3}-\frac {(5 A b-11 a B) x^{3/2}}{12 b^3 (a+b x)^2}-\frac {(5 A b-19 a B) \sqrt {x}}{8 b^4 (a+b x)}+\frac {5 (A b-7 a B) \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{8 \sqrt {a} b^{9/2}} \] Output:
2*B*x^(1/2)/b^4-1/3*(A*b-B*a)*x^(5/2)/b^2/(b*x+a)^3-1/12*(5*A*b-11*B*a)*x^ (3/2)/b^3/(b*x+a)^2-1/8*(5*A*b-19*B*a)*x^(1/2)/b^4/(b*x+a)+5/8*(A*b-7*B*a) *arctan(b^(1/2)*x^(1/2)/a^(1/2))/a^(1/2)/b^(9/2)
Time = 0.31 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.82 \[ \int \frac {x^{5/2} (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=\frac {\sqrt {x} \left (105 a^3 B-5 a^2 b (3 A-56 B x)+3 b^3 x^2 (-11 A+16 B x)+a b^2 x (-40 A+231 B x)\right )}{24 b^4 (a+b x)^3}+\frac {5 (A b-7 a B) \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{8 \sqrt {a} b^{9/2}} \] Input:
Integrate[(x^(5/2)*(A + B*x))/(a^2 + 2*a*b*x + b^2*x^2)^2,x]
Output:
(Sqrt[x]*(105*a^3*B - 5*a^2*b*(3*A - 56*B*x) + 3*b^3*x^2*(-11*A + 16*B*x) + a*b^2*x*(-40*A + 231*B*x)))/(24*b^4*(a + b*x)^3) + (5*(A*b - 7*a*B)*ArcT an[(Sqrt[b]*Sqrt[x])/Sqrt[a]])/(8*Sqrt[a]*b^(9/2))
Time = 0.43 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.05, 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, 51, 51, 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.
\(\displaystyle \int \frac {x^{5/2} (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx\) |
\(\Big \downarrow \) 1184 |
\(\displaystyle b^4 \int \frac {x^{5/2} (A+B x)}{b^4 (a+b x)^4}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \int \frac {x^{5/2} (A+B x)}{(a+b x)^4}dx\) |
\(\Big \downarrow \) 87 |
\(\displaystyle \frac {x^{7/2} (A b-a B)}{3 a b (a+b x)^3}-\frac {(A b-7 a B) \int \frac {x^{5/2}}{(a+b x)^3}dx}{6 a b}\) |
\(\Big \downarrow \) 51 |
\(\displaystyle \frac {x^{7/2} (A b-a B)}{3 a b (a+b x)^3}-\frac {(A b-7 a B) \left (\frac {5 \int \frac {x^{3/2}}{(a+b x)^2}dx}{4 b}-\frac {x^{5/2}}{2 b (a+b x)^2}\right )}{6 a b}\) |
\(\Big \downarrow \) 51 |
\(\displaystyle \frac {x^{7/2} (A b-a B)}{3 a b (a+b x)^3}-\frac {(A b-7 a B) \left (\frac {5 \left (\frac {3 \int \frac {\sqrt {x}}{a+b x}dx}{2 b}-\frac {x^{3/2}}{b (a+b x)}\right )}{4 b}-\frac {x^{5/2}}{2 b (a+b x)^2}\right )}{6 a b}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {x^{7/2} (A b-a B)}{3 a b (a+b x)^3}-\frac {(A b-7 a B) \left (\frac {5 \left (\frac {3 \left (\frac {2 \sqrt {x}}{b}-\frac {a \int \frac {1}{\sqrt {x} (a+b x)}dx}{b}\right )}{2 b}-\frac {x^{3/2}}{b (a+b x)}\right )}{4 b}-\frac {x^{5/2}}{2 b (a+b x)^2}\right )}{6 a b}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {x^{7/2} (A b-a B)}{3 a b (a+b x)^3}-\frac {(A b-7 a B) \left (\frac {5 \left (\frac {3 \left (\frac {2 \sqrt {x}}{b}-\frac {2 a \int \frac {1}{a+b x}d\sqrt {x}}{b}\right )}{2 b}-\frac {x^{3/2}}{b (a+b x)}\right )}{4 b}-\frac {x^{5/2}}{2 b (a+b x)^2}\right )}{6 a b}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {x^{7/2} (A b-a B)}{3 a b (a+b x)^3}-\frac {(A b-7 a B) \left (\frac {5 \left (\frac {3 \left (\frac {2 \sqrt {x}}{b}-\frac {2 \sqrt {a} \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{b^{3/2}}\right )}{2 b}-\frac {x^{3/2}}{b (a+b x)}\right )}{4 b}-\frac {x^{5/2}}{2 b (a+b x)^2}\right )}{6 a b}\) |
Input:
Int[(x^(5/2)*(A + B*x))/(a^2 + 2*a*b*x + b^2*x^2)^2,x]
Output:
((A*b - a*B)*x^(7/2))/(3*a*b*(a + b*x)^3) - ((A*b - 7*a*B)*(-1/2*x^(5/2)/( b*(a + b*x)^2) + (5*(-(x^(3/2)/(b*(a + b*x))) + (3*((2*Sqrt[x])/b - (2*Sqr t[a]*ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[a]])/b^(3/2)))/(2*b)))/(4*b)))/(6*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 + 1))), x] - Simp[d*(n/(b*(m + 1))) Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d, n}, x ] && ILtQ[m, -1] && FractionQ[n] && GtQ[n, 0]
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 = 1.27 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.78
method | result | size |
derivativedivides | \(\frac {2 B \sqrt {x}}{b^{4}}+\frac {\frac {2 \left (\left (\frac {29}{16} B a \,b^{2}-\frac {11}{16} A \,b^{3}\right ) x^{\frac {5}{2}}-\frac {a b \left (5 A b -17 B a \right ) x^{\frac {3}{2}}}{6}+\left (\frac {19}{16} B \,a^{3}-\frac {5}{16} A \,a^{2} b \right ) \sqrt {x}\right )}{\left (b x +a \right )^{3}}+\frac {5 \left (A b -7 B a \right ) \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{8 \sqrt {a b}}}{b^{4}}\) | \(104\) |
default | \(\frac {2 B \sqrt {x}}{b^{4}}+\frac {\frac {2 \left (\left (\frac {29}{16} B a \,b^{2}-\frac {11}{16} A \,b^{3}\right ) x^{\frac {5}{2}}-\frac {a b \left (5 A b -17 B a \right ) x^{\frac {3}{2}}}{6}+\left (\frac {19}{16} B \,a^{3}-\frac {5}{16} A \,a^{2} b \right ) \sqrt {x}\right )}{\left (b x +a \right )^{3}}+\frac {5 \left (A b -7 B a \right ) \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{8 \sqrt {a b}}}{b^{4}}\) | \(104\) |
risch | \(\frac {2 B \sqrt {x}}{b^{4}}+\frac {\frac {2 \left (\frac {29}{16} B a \,b^{2}-\frac {11}{16} A \,b^{3}\right ) x^{\frac {5}{2}}-\frac {a b \left (5 A b -17 B a \right ) x^{\frac {3}{2}}}{3}+2 \left (\frac {19}{16} B \,a^{3}-\frac {5}{16} A \,a^{2} b \right ) \sqrt {x}}{\left (b x +a \right )^{3}}+\frac {5 \left (A b -7 B a \right ) \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{8 \sqrt {a b}}}{b^{4}}\) | \(104\) |
Input:
int(x^(5/2)*(B*x+A)/(b^2*x^2+2*a*b*x+a^2)^2,x,method=_RETURNVERBOSE)
Output:
2*B*x^(1/2)/b^4+2/b^4*(((29/16*B*a*b^2-11/16*A*b^3)*x^(5/2)-1/6*a*b*(5*A*b -17*B*a)*x^(3/2)+(19/16*B*a^3-5/16*A*a^2*b)*x^(1/2))/(b*x+a)^3+5/16*(A*b-7 *B*a)/(a*b)^(1/2)*arctan(b*x^(1/2)/(a*b)^(1/2)))
Time = 0.09 (sec) , antiderivative size = 437, normalized size of antiderivative = 3.26 \[ \int \frac {x^{5/2} (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=\left [\frac {15 \, {\left (7 \, B a^{4} - A a^{3} b + {\left (7 \, B a b^{3} - A b^{4}\right )} x^{3} + 3 \, {\left (7 \, B a^{2} b^{2} - A a b^{3}\right )} x^{2} + 3 \, {\left (7 \, B a^{3} b - A a^{2} b^{2}\right )} x\right )} \sqrt {-a b} \log \left (\frac {b x - a - 2 \, \sqrt {-a b} \sqrt {x}}{b x + a}\right ) + 2 \, {\left (48 \, B a b^{4} x^{3} + 105 \, B a^{4} b - 15 \, A a^{3} b^{2} + 33 \, {\left (7 \, B a^{2} b^{3} - A a b^{4}\right )} x^{2} + 40 \, {\left (7 \, B a^{3} b^{2} - A a^{2} b^{3}\right )} x\right )} \sqrt {x}}{48 \, {\left (a b^{8} x^{3} + 3 \, a^{2} b^{7} x^{2} + 3 \, a^{3} b^{6} x + a^{4} b^{5}\right )}}, \frac {15 \, {\left (7 \, B a^{4} - A a^{3} b + {\left (7 \, B a b^{3} - A b^{4}\right )} x^{3} + 3 \, {\left (7 \, B a^{2} b^{2} - A a b^{3}\right )} x^{2} + 3 \, {\left (7 \, B a^{3} b - A a^{2} b^{2}\right )} x\right )} \sqrt {a b} \arctan \left (\frac {\sqrt {a b}}{b \sqrt {x}}\right ) + {\left (48 \, B a b^{4} x^{3} + 105 \, B a^{4} b - 15 \, A a^{3} b^{2} + 33 \, {\left (7 \, B a^{2} b^{3} - A a b^{4}\right )} x^{2} + 40 \, {\left (7 \, B a^{3} b^{2} - A a^{2} b^{3}\right )} x\right )} \sqrt {x}}{24 \, {\left (a b^{8} x^{3} + 3 \, a^{2} b^{7} x^{2} + 3 \, a^{3} b^{6} x + a^{4} b^{5}\right )}}\right ] \] Input:
integrate(x^(5/2)*(B*x+A)/(b^2*x^2+2*a*b*x+a^2)^2,x, algorithm="fricas")
Output:
[1/48*(15*(7*B*a^4 - A*a^3*b + (7*B*a*b^3 - A*b^4)*x^3 + 3*(7*B*a^2*b^2 - A*a*b^3)*x^2 + 3*(7*B*a^3*b - A*a^2*b^2)*x)*sqrt(-a*b)*log((b*x - a - 2*sq rt(-a*b)*sqrt(x))/(b*x + a)) + 2*(48*B*a*b^4*x^3 + 105*B*a^4*b - 15*A*a^3* b^2 + 33*(7*B*a^2*b^3 - A*a*b^4)*x^2 + 40*(7*B*a^3*b^2 - A*a^2*b^3)*x)*sqr t(x))/(a*b^8*x^3 + 3*a^2*b^7*x^2 + 3*a^3*b^6*x + a^4*b^5), 1/24*(15*(7*B*a ^4 - A*a^3*b + (7*B*a*b^3 - A*b^4)*x^3 + 3*(7*B*a^2*b^2 - A*a*b^3)*x^2 + 3 *(7*B*a^3*b - A*a^2*b^2)*x)*sqrt(a*b)*arctan(sqrt(a*b)/(b*sqrt(x))) + (48* B*a*b^4*x^3 + 105*B*a^4*b - 15*A*a^3*b^2 + 33*(7*B*a^2*b^3 - A*a*b^4)*x^2 + 40*(7*B*a^3*b^2 - A*a^2*b^3)*x)*sqrt(x))/(a*b^8*x^3 + 3*a^2*b^7*x^2 + 3* a^3*b^6*x + a^4*b^5)]
Leaf count of result is larger than twice the leaf count of optimal. 2271 vs. \(2 (133) = 266\).
Time = 87.83 (sec) , antiderivative size = 2271, normalized size of antiderivative = 16.95 \[ \int \frac {x^{5/2} (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=\text {Too large to display} \] Input:
integrate(x**(5/2)*(B*x+A)/(b**2*x**2+2*a*b*x+a**2)**2,x)
Output:
Piecewise((zoo*(-2*A/sqrt(x) + 2*B*sqrt(x)), Eq(a, 0) & Eq(b, 0)), ((2*A*x **(7/2)/7 + 2*B*x**(9/2)/9)/a**4, Eq(b, 0)), ((-2*A/sqrt(x) + 2*B*sqrt(x)) /b**4, Eq(a, 0)), (15*A*a**3*b*log(sqrt(x) - sqrt(-a/b))/(48*a**3*b**5*sqr t(-a/b) + 144*a**2*b**6*x*sqrt(-a/b) + 144*a*b**7*x**2*sqrt(-a/b) + 48*b** 8*x**3*sqrt(-a/b)) - 15*A*a**3*b*log(sqrt(x) + sqrt(-a/b))/(48*a**3*b**5*s qrt(-a/b) + 144*a**2*b**6*x*sqrt(-a/b) + 144*a*b**7*x**2*sqrt(-a/b) + 48*b **8*x**3*sqrt(-a/b)) - 30*A*a**2*b**2*sqrt(x)*sqrt(-a/b)/(48*a**3*b**5*sqr t(-a/b) + 144*a**2*b**6*x*sqrt(-a/b) + 144*a*b**7*x**2*sqrt(-a/b) + 48*b** 8*x**3*sqrt(-a/b)) + 45*A*a**2*b**2*x*log(sqrt(x) - sqrt(-a/b))/(48*a**3*b **5*sqrt(-a/b) + 144*a**2*b**6*x*sqrt(-a/b) + 144*a*b**7*x**2*sqrt(-a/b) + 48*b**8*x**3*sqrt(-a/b)) - 45*A*a**2*b**2*x*log(sqrt(x) + sqrt(-a/b))/(48 *a**3*b**5*sqrt(-a/b) + 144*a**2*b**6*x*sqrt(-a/b) + 144*a*b**7*x**2*sqrt( -a/b) + 48*b**8*x**3*sqrt(-a/b)) - 80*A*a*b**3*x**(3/2)*sqrt(-a/b)/(48*a** 3*b**5*sqrt(-a/b) + 144*a**2*b**6*x*sqrt(-a/b) + 144*a*b**7*x**2*sqrt(-a/b ) + 48*b**8*x**3*sqrt(-a/b)) + 45*A*a*b**3*x**2*log(sqrt(x) - sqrt(-a/b))/ (48*a**3*b**5*sqrt(-a/b) + 144*a**2*b**6*x*sqrt(-a/b) + 144*a*b**7*x**2*sq rt(-a/b) + 48*b**8*x**3*sqrt(-a/b)) - 45*A*a*b**3*x**2*log(sqrt(x) + sqrt( -a/b))/(48*a**3*b**5*sqrt(-a/b) + 144*a**2*b**6*x*sqrt(-a/b) + 144*a*b**7* x**2*sqrt(-a/b) + 48*b**8*x**3*sqrt(-a/b)) - 66*A*b**4*x**(5/2)*sqrt(-a/b) /(48*a**3*b**5*sqrt(-a/b) + 144*a**2*b**6*x*sqrt(-a/b) + 144*a*b**7*x**...
Time = 0.13 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.01 \[ \int \frac {x^{5/2} (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=\frac {3 \, {\left (29 \, B a b^{2} - 11 \, A b^{3}\right )} x^{\frac {5}{2}} + 8 \, {\left (17 \, B a^{2} b - 5 \, A a b^{2}\right )} x^{\frac {3}{2}} + 3 \, {\left (19 \, B a^{3} - 5 \, A a^{2} b\right )} \sqrt {x}}{24 \, {\left (b^{7} x^{3} + 3 \, a b^{6} x^{2} + 3 \, a^{2} b^{5} x + a^{3} b^{4}\right )}} + \frac {2 \, B \sqrt {x}}{b^{4}} - \frac {5 \, {\left (7 \, B a - A b\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} b^{4}} \] Input:
integrate(x^(5/2)*(B*x+A)/(b^2*x^2+2*a*b*x+a^2)^2,x, algorithm="maxima")
Output:
1/24*(3*(29*B*a*b^2 - 11*A*b^3)*x^(5/2) + 8*(17*B*a^2*b - 5*A*a*b^2)*x^(3/ 2) + 3*(19*B*a^3 - 5*A*a^2*b)*sqrt(x))/(b^7*x^3 + 3*a*b^6*x^2 + 3*a^2*b^5* x + a^3*b^4) + 2*B*sqrt(x)/b^4 - 5/8*(7*B*a - A*b)*arctan(b*sqrt(x)/sqrt(a *b))/(sqrt(a*b)*b^4)
Time = 0.21 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.83 \[ \int \frac {x^{5/2} (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=\frac {2 \, B \sqrt {x}}{b^{4}} - \frac {5 \, {\left (7 \, B a - A b\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} b^{4}} + \frac {87 \, B a b^{2} x^{\frac {5}{2}} - 33 \, A b^{3} x^{\frac {5}{2}} + 136 \, B a^{2} b x^{\frac {3}{2}} - 40 \, A a b^{2} x^{\frac {3}{2}} + 57 \, B a^{3} \sqrt {x} - 15 \, A a^{2} b \sqrt {x}}{24 \, {\left (b x + a\right )}^{3} b^{4}} \] Input:
integrate(x^(5/2)*(B*x+A)/(b^2*x^2+2*a*b*x+a^2)^2,x, algorithm="giac")
Output:
2*B*sqrt(x)/b^4 - 5/8*(7*B*a - A*b)*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b) *b^4) + 1/24*(87*B*a*b^2*x^(5/2) - 33*A*b^3*x^(5/2) + 136*B*a^2*b*x^(3/2) - 40*A*a*b^2*x^(3/2) + 57*B*a^3*sqrt(x) - 15*A*a^2*b*sqrt(x))/((b*x + a)^3 *b^4)
Time = 10.78 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.98 \[ \int \frac {x^{5/2} (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=\frac {2\,B\,\sqrt {x}}{b^4}-\frac {x^{3/2}\,\left (\frac {5\,A\,a\,b^2}{3}-\frac {17\,B\,a^2\,b}{3}\right )-\sqrt {x}\,\left (\frac {19\,B\,a^3}{8}-\frac {5\,A\,a^2\,b}{8}\right )+x^{5/2}\,\left (\frac {11\,A\,b^3}{8}-\frac {29\,B\,a\,b^2}{8}\right )}{a^3\,b^4+3\,a^2\,b^5\,x+3\,a\,b^6\,x^2+b^7\,x^3}+\frac {5\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {x}}{\sqrt {a}}\right )\,\left (A\,b-7\,B\,a\right )}{8\,\sqrt {a}\,b^{9/2}} \] Input:
int((x^(5/2)*(A + B*x))/(a^2 + b^2*x^2 + 2*a*b*x)^2,x)
Output:
(2*B*x^(1/2))/b^4 - (x^(3/2)*((5*A*a*b^2)/3 - (17*B*a^2*b)/3) - x^(1/2)*(( 19*B*a^3)/8 - (5*A*a^2*b)/8) + x^(5/2)*((11*A*b^3)/8 - (29*B*a*b^2)/8))/(a ^3*b^4 + b^7*x^3 + 3*a^2*b^5*x + 3*a*b^6*x^2) + (5*atan((b^(1/2)*x^(1/2))/ a^(1/2))*(A*b - 7*B*a))/(8*a^(1/2)*b^(9/2))
Time = 0.27 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.90 \[ \int \frac {x^{5/2} (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=\frac {-15 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {\sqrt {x}\, b}{\sqrt {b}\, \sqrt {a}}\right ) a^{2}-30 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {\sqrt {x}\, b}{\sqrt {b}\, \sqrt {a}}\right ) a b x -15 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {\sqrt {x}\, b}{\sqrt {b}\, \sqrt {a}}\right ) b^{2} x^{2}+15 \sqrt {x}\, a^{2} b +25 \sqrt {x}\, a \,b^{2} x +8 \sqrt {x}\, b^{3} x^{2}}{4 b^{4} \left (b^{2} x^{2}+2 a b x +a^{2}\right )} \] Input:
int(x^(5/2)*(B*x+A)/(b^2*x^2+2*a*b*x+a^2)^2,x)
Output:
( - 15*sqrt(b)*sqrt(a)*atan((sqrt(x)*b)/(sqrt(b)*sqrt(a)))*a**2 - 30*sqrt( b)*sqrt(a)*atan((sqrt(x)*b)/(sqrt(b)*sqrt(a)))*a*b*x - 15*sqrt(b)*sqrt(a)* atan((sqrt(x)*b)/(sqrt(b)*sqrt(a)))*b**2*x**2 + 15*sqrt(x)*a**2*b + 25*sqr t(x)*a*b**2*x + 8*sqrt(x)*b**3*x**2)/(4*b**4*(a**2 + 2*a*b*x + b**2*x**2))