Integrand size = 33, antiderivative size = 250 \[ \int \frac {(A+B x) (d+e x)^{5/2}}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=\frac {5 e^2 (6 b B d+A b e-7 a B e) \sqrt {d+e x}}{8 b^4 (b d-a e)}-\frac {5 e (6 b B d+A b e-7 a B e) (d+e x)^{3/2}}{24 b^3 (b d-a e) (a+b x)}-\frac {(6 b B d+A b e-7 a B e) (d+e x)^{5/2}}{12 b^2 (b d-a e) (a+b x)^2}-\frac {(A b-a B) (d+e x)^{7/2}}{3 b (b d-a e) (a+b x)^3}-\frac {5 e^2 (6 b B d+A b e-7 a B e) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{8 b^{9/2} \sqrt {b d-a e}} \]
-5/24*e*(A*b*e-7*B*a*e+6*B*b*d)*(e*x+d)^(3/2)/b^3/(-a*e+b*d)/(b*x+a)-1/12* (A*b*e-7*B*a*e+6*B*b*d)*(e*x+d)^(5/2)/b^2/(-a*e+b*d)/(b*x+a)^2-1/3*(A*b-B* a)*(e*x+d)^(7/2)/b/(-a*e+b*d)/(b*x+a)^3-5/8*e^2*(A*b*e-7*B*a*e+6*B*b*d)*ar ctanh(b^(1/2)*(e*x+d)^(1/2)/(-a*e+b*d)^(1/2))/b^(9/2)/(-a*e+b*d)^(1/2)+5/8 *e^2*(A*b*e-7*B*a*e+6*B*b*d)*(e*x+d)^(1/2)/b^4/(-a*e+b*d)
Time = 1.21 (sec) , antiderivative size = 211, normalized size of antiderivative = 0.84 \[ \int \frac {(A+B x) (d+e x)^{5/2}}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=-\frac {\sqrt {d+e x} \left (B \left (-105 a^3 e^2+20 a^2 b e (d-14 e x)+a b^2 \left (4 d^2+58 d e x-231 e^2 x^2\right )+6 b^3 x \left (2 d^2+9 d e x-8 e^2 x^2\right )\right )+A b \left (15 a^2 e^2+10 a b e (d+4 e x)+b^2 \left (8 d^2+26 d e x+33 e^2 x^2\right )\right )\right )}{24 b^4 (a+b x)^3}+\frac {5 e^2 (6 b B d+A b e-7 a B e) \arctan \left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {-b d+a e}}\right )}{8 b^{9/2} \sqrt {-b d+a e}} \]
-1/24*(Sqrt[d + e*x]*(B*(-105*a^3*e^2 + 20*a^2*b*e*(d - 14*e*x) + a*b^2*(4 *d^2 + 58*d*e*x - 231*e^2*x^2) + 6*b^3*x*(2*d^2 + 9*d*e*x - 8*e^2*x^2)) + A*b*(15*a^2*e^2 + 10*a*b*e*(d + 4*e*x) + b^2*(8*d^2 + 26*d*e*x + 33*e^2*x^ 2))))/(b^4*(a + b*x)^3) + (5*e^2*(6*b*B*d + A*b*e - 7*a*B*e)*ArcTan[(Sqrt[ b]*Sqrt[d + e*x])/Sqrt[-(b*d) + a*e]])/(8*b^(9/2)*Sqrt[-(b*d) + a*e])
Time = 0.33 (sec) , antiderivative size = 198, normalized size of antiderivative = 0.79, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.242, Rules used = {1184, 27, 87, 51, 51, 60, 73, 221}
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.
\(\displaystyle \int \frac {(A+B x) (d+e x)^{5/2}}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx\) |
\(\Big \downarrow \) 1184 |
\(\displaystyle b^4 \int \frac {(A+B x) (d+e x)^{5/2}}{b^4 (a+b x)^4}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \int \frac {(A+B x) (d+e x)^{5/2}}{(a+b x)^4}dx\) |
\(\Big \downarrow \) 87 |
\(\displaystyle \frac {(-7 a B e+A b e+6 b B d) \int \frac {(d+e x)^{5/2}}{(a+b x)^3}dx}{6 b (b d-a e)}-\frac {(d+e x)^{7/2} (A b-a B)}{3 b (a+b x)^3 (b d-a e)}\) |
\(\Big \downarrow \) 51 |
\(\displaystyle \frac {(-7 a B e+A b e+6 b B d) \left (\frac {5 e \int \frac {(d+e x)^{3/2}}{(a+b x)^2}dx}{4 b}-\frac {(d+e x)^{5/2}}{2 b (a+b x)^2}\right )}{6 b (b d-a e)}-\frac {(d+e x)^{7/2} (A b-a B)}{3 b (a+b x)^3 (b d-a e)}\) |
\(\Big \downarrow \) 51 |
\(\displaystyle \frac {(-7 a B e+A b e+6 b B d) \left (\frac {5 e \left (\frac {3 e \int \frac {\sqrt {d+e x}}{a+b x}dx}{2 b}-\frac {(d+e x)^{3/2}}{b (a+b x)}\right )}{4 b}-\frac {(d+e x)^{5/2}}{2 b (a+b x)^2}\right )}{6 b (b d-a e)}-\frac {(d+e x)^{7/2} (A b-a B)}{3 b (a+b x)^3 (b d-a e)}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {(-7 a B e+A b e+6 b B d) \left (\frac {5 e \left (\frac {3 e \left (\frac {(b d-a e) \int \frac {1}{(a+b x) \sqrt {d+e x}}dx}{b}+\frac {2 \sqrt {d+e x}}{b}\right )}{2 b}-\frac {(d+e x)^{3/2}}{b (a+b x)}\right )}{4 b}-\frac {(d+e x)^{5/2}}{2 b (a+b x)^2}\right )}{6 b (b d-a e)}-\frac {(d+e x)^{7/2} (A b-a B)}{3 b (a+b x)^3 (b d-a e)}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {(-7 a B e+A b e+6 b B d) \left (\frac {5 e \left (\frac {3 e \left (\frac {2 (b d-a e) \int \frac {1}{a+\frac {b (d+e x)}{e}-\frac {b d}{e}}d\sqrt {d+e x}}{b e}+\frac {2 \sqrt {d+e x}}{b}\right )}{2 b}-\frac {(d+e x)^{3/2}}{b (a+b x)}\right )}{4 b}-\frac {(d+e x)^{5/2}}{2 b (a+b x)^2}\right )}{6 b (b d-a e)}-\frac {(d+e x)^{7/2} (A b-a B)}{3 b (a+b x)^3 (b d-a e)}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {(-7 a B e+A b e+6 b B d) \left (\frac {5 e \left (\frac {3 e \left (\frac {2 \sqrt {d+e x}}{b}-\frac {2 \sqrt {b d-a e} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{b^{3/2}}\right )}{2 b}-\frac {(d+e x)^{3/2}}{b (a+b x)}\right )}{4 b}-\frac {(d+e x)^{5/2}}{2 b (a+b x)^2}\right )}{6 b (b d-a e)}-\frac {(d+e x)^{7/2} (A b-a B)}{3 b (a+b x)^3 (b d-a e)}\) |
-1/3*((A*b - a*B)*(d + e*x)^(7/2))/(b*(b*d - a*e)*(a + b*x)^3) + ((6*b*B*d + A*b*e - 7*a*B*e)*(-1/2*(d + e*x)^(5/2)/(b*(a + b*x)^2) + (5*e*(-((d + e *x)^(3/2)/(b*(a + b*x))) + (3*e*((2*Sqrt[d + e*x])/b - (2*Sqrt[b*d - a*e]* ArcTanh[(Sqrt[b]*Sqrt[d + e*x])/Sqrt[b*d - a*e]])/b^(3/2)))/(2*b)))/(4*b)) )/(6*b*(b*d - a*e))
3.19.18.3.1 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/(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)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[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 = 0.52 (sec) , antiderivative size = 214, normalized size of antiderivative = 0.86
method | result | size |
pseudoelliptic | \(\frac {-\frac {5 \left (\left (\left (-7 a^{3}-\frac {16}{5} x^{3} b^{3}-\frac {56}{3} b \,a^{2} x -\frac {77}{5} a \,b^{2} x^{2}\right ) e^{2}+\frac {4 b \left (\frac {27}{10} b^{2} x^{2}+\frac {29}{10} a b x +a^{2}\right ) d e}{3}+\frac {4 b^{2} d^{2} \left (3 b x +a \right )}{15}\right ) B +b A \left (\left (\frac {11}{5} b^{2} x^{2}+a^{2}+\frac {8}{3} a b x \right ) e^{2}+\frac {2 b \left (\frac {13 b x}{5}+a \right ) d e}{3}+\frac {8 b^{2} d^{2}}{15}\right )\right ) \sqrt {\left (a e -b d \right ) b}\, \sqrt {e x +d}}{8}+\frac {5 \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right ) \left (\left (-7 a e +6 b d \right ) B +A b e \right ) \left (b x +a \right )^{3} e^{2}}{8}}{b^{4} \left (b x +a \right )^{3} \sqrt {\left (a e -b d \right ) b}}\) | \(214\) |
derivativedivides | \(2 e^{2} \left (\frac {B \sqrt {e x +d}}{b^{4}}+\frac {\frac {\left (-\frac {11}{16} A \,b^{3} e +\frac {29}{16} B e \,b^{2} a -\frac {9}{8} B \,b^{3} d \right ) \left (e x +d \right )^{\frac {5}{2}}-\frac {b \left (5 A a b \,e^{2}-5 A \,b^{2} d e -17 a^{2} B \,e^{2}+29 B a b d e -12 B \,b^{2} d^{2}\right ) \left (e x +d \right )^{\frac {3}{2}}}{6}+\left (-\frac {5}{16} A \,a^{2} b \,e^{3}+\frac {5}{8} A a \,b^{2} d \,e^{2}-\frac {5}{16} A \,b^{3} d^{2} e +\frac {19}{16} B \,e^{3} a^{3}-\frac {13}{4} B \,a^{2} b d \,e^{2}+\frac {47}{16} B a \,b^{2} d^{2} e -\frac {7}{8} B \,b^{3} d^{3}\right ) \sqrt {e x +d}}{\left (b \left (e x +d \right )+a e -b d \right )^{3}}+\frac {5 \left (A b e -7 B a e +6 B b d \right ) \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right )}{16 \sqrt {\left (a e -b d \right ) b}}}{b^{4}}\right )\) | \(257\) |
default | \(2 e^{2} \left (\frac {B \sqrt {e x +d}}{b^{4}}+\frac {\frac {\left (-\frac {11}{16} A \,b^{3} e +\frac {29}{16} B e \,b^{2} a -\frac {9}{8} B \,b^{3} d \right ) \left (e x +d \right )^{\frac {5}{2}}-\frac {b \left (5 A a b \,e^{2}-5 A \,b^{2} d e -17 a^{2} B \,e^{2}+29 B a b d e -12 B \,b^{2} d^{2}\right ) \left (e x +d \right )^{\frac {3}{2}}}{6}+\left (-\frac {5}{16} A \,a^{2} b \,e^{3}+\frac {5}{8} A a \,b^{2} d \,e^{2}-\frac {5}{16} A \,b^{3} d^{2} e +\frac {19}{16} B \,e^{3} a^{3}-\frac {13}{4} B \,a^{2} b d \,e^{2}+\frac {47}{16} B a \,b^{2} d^{2} e -\frac {7}{8} B \,b^{3} d^{3}\right ) \sqrt {e x +d}}{\left (b \left (e x +d \right )+a e -b d \right )^{3}}+\frac {5 \left (A b e -7 B a e +6 B b d \right ) \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right )}{16 \sqrt {\left (a e -b d \right ) b}}}{b^{4}}\right )\) | \(257\) |
risch | \(\frac {2 B \,e^{2} \sqrt {e x +d}}{b^{4}}+\frac {2 e^{2} \left (\frac {\left (-\frac {11}{16} A \,b^{3} e +\frac {29}{16} B e \,b^{2} a -\frac {9}{8} B \,b^{3} d \right ) \left (e x +d \right )^{\frac {5}{2}}-\frac {b \left (5 A a b \,e^{2}-5 A \,b^{2} d e -17 a^{2} B \,e^{2}+29 B a b d e -12 B \,b^{2} d^{2}\right ) \left (e x +d \right )^{\frac {3}{2}}}{6}+\left (-\frac {5}{16} A \,a^{2} b \,e^{3}+\frac {5}{8} A a \,b^{2} d \,e^{2}-\frac {5}{16} A \,b^{3} d^{2} e +\frac {19}{16} B \,e^{3} a^{3}-\frac {13}{4} B \,a^{2} b d \,e^{2}+\frac {47}{16} B a \,b^{2} d^{2} e -\frac {7}{8} B \,b^{3} d^{3}\right ) \sqrt {e x +d}}{\left (b \left (e x +d \right )+a e -b d \right )^{3}}+\frac {5 \left (A b e -7 B a e +6 B b d \right ) \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right )}{16 \sqrt {\left (a e -b d \right ) b}}\right )}{b^{4}}\) | \(260\) |
5/8/((a*e-b*d)*b)^(1/2)*(-(((-7*a^3-16/5*x^3*b^3-56/3*b*a^2*x-77/5*a*b^2*x ^2)*e^2+4/3*b*(27/10*b^2*x^2+29/10*a*b*x+a^2)*d*e+4/15*b^2*d^2*(3*b*x+a))* B+b*A*((11/5*b^2*x^2+a^2+8/3*a*b*x)*e^2+2/3*b*(13/5*b*x+a)*d*e+8/15*b^2*d^ 2))*((a*e-b*d)*b)^(1/2)*(e*x+d)^(1/2)+arctan(b*(e*x+d)^(1/2)/((a*e-b*d)*b) ^(1/2))*((-7*a*e+6*b*d)*B+A*b*e)*(b*x+a)^3*e^2)/b^4/(b*x+a)^3
Leaf count of result is larger than twice the leaf count of optimal. 531 vs. \(2 (222) = 444\).
Time = 0.36 (sec) , antiderivative size = 1075, normalized size of antiderivative = 4.30 \[ \int \frac {(A+B x) (d+e x)^{5/2}}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=\left [\frac {15 \, {\left (6 \, B a^{3} b d e^{2} - {\left (7 \, B a^{4} - A a^{3} b\right )} e^{3} + {\left (6 \, B b^{4} d e^{2} - {\left (7 \, B a b^{3} - A b^{4}\right )} e^{3}\right )} x^{3} + 3 \, {\left (6 \, B a b^{3} d e^{2} - {\left (7 \, B a^{2} b^{2} - A a b^{3}\right )} e^{3}\right )} x^{2} + 3 \, {\left (6 \, B a^{2} b^{2} d e^{2} - {\left (7 \, B a^{3} b - A a^{2} b^{2}\right )} e^{3}\right )} x\right )} \sqrt {b^{2} d - a b e} \log \left (\frac {b e x + 2 \, b d - a e - 2 \, \sqrt {b^{2} d - a b e} \sqrt {e x + d}}{b x + a}\right ) - 2 \, {\left (4 \, {\left (B a b^{4} + 2 \, A b^{5}\right )} d^{3} + 2 \, {\left (8 \, B a^{2} b^{3} + A a b^{4}\right )} d^{2} e - 5 \, {\left (25 \, B a^{3} b^{2} - A a^{2} b^{3}\right )} d e^{2} + 15 \, {\left (7 \, B a^{4} b - A a^{3} b^{2}\right )} e^{3} - 48 \, {\left (B b^{5} d e^{2} - B a b^{4} e^{3}\right )} x^{3} + 3 \, {\left (18 \, B b^{5} d^{2} e - {\left (95 \, B a b^{4} - 11 \, A b^{5}\right )} d e^{2} + 11 \, {\left (7 \, B a^{2} b^{3} - A a b^{4}\right )} e^{3}\right )} x^{2} + 2 \, {\left (6 \, B b^{5} d^{3} + {\left (23 \, B a b^{4} + 13 \, A b^{5}\right )} d^{2} e - {\left (169 \, B a^{2} b^{3} - 7 \, A a b^{4}\right )} d e^{2} + 20 \, {\left (7 \, B a^{3} b^{2} - A a^{2} b^{3}\right )} e^{3}\right )} x\right )} \sqrt {e x + d}}{48 \, {\left (a^{3} b^{6} d - a^{4} b^{5} e + {\left (b^{9} d - a b^{8} e\right )} x^{3} + 3 \, {\left (a b^{8} d - a^{2} b^{7} e\right )} x^{2} + 3 \, {\left (a^{2} b^{7} d - a^{3} b^{6} e\right )} x\right )}}, \frac {15 \, {\left (6 \, B a^{3} b d e^{2} - {\left (7 \, B a^{4} - A a^{3} b\right )} e^{3} + {\left (6 \, B b^{4} d e^{2} - {\left (7 \, B a b^{3} - A b^{4}\right )} e^{3}\right )} x^{3} + 3 \, {\left (6 \, B a b^{3} d e^{2} - {\left (7 \, B a^{2} b^{2} - A a b^{3}\right )} e^{3}\right )} x^{2} + 3 \, {\left (6 \, B a^{2} b^{2} d e^{2} - {\left (7 \, B a^{3} b - A a^{2} b^{2}\right )} e^{3}\right )} x\right )} \sqrt {-b^{2} d + a b e} \arctan \left (\frac {\sqrt {-b^{2} d + a b e} \sqrt {e x + d}}{b e x + b d}\right ) - {\left (4 \, {\left (B a b^{4} + 2 \, A b^{5}\right )} d^{3} + 2 \, {\left (8 \, B a^{2} b^{3} + A a b^{4}\right )} d^{2} e - 5 \, {\left (25 \, B a^{3} b^{2} - A a^{2} b^{3}\right )} d e^{2} + 15 \, {\left (7 \, B a^{4} b - A a^{3} b^{2}\right )} e^{3} - 48 \, {\left (B b^{5} d e^{2} - B a b^{4} e^{3}\right )} x^{3} + 3 \, {\left (18 \, B b^{5} d^{2} e - {\left (95 \, B a b^{4} - 11 \, A b^{5}\right )} d e^{2} + 11 \, {\left (7 \, B a^{2} b^{3} - A a b^{4}\right )} e^{3}\right )} x^{2} + 2 \, {\left (6 \, B b^{5} d^{3} + {\left (23 \, B a b^{4} + 13 \, A b^{5}\right )} d^{2} e - {\left (169 \, B a^{2} b^{3} - 7 \, A a b^{4}\right )} d e^{2} + 20 \, {\left (7 \, B a^{3} b^{2} - A a^{2} b^{3}\right )} e^{3}\right )} x\right )} \sqrt {e x + d}}{24 \, {\left (a^{3} b^{6} d - a^{4} b^{5} e + {\left (b^{9} d - a b^{8} e\right )} x^{3} + 3 \, {\left (a b^{8} d - a^{2} b^{7} e\right )} x^{2} + 3 \, {\left (a^{2} b^{7} d - a^{3} b^{6} e\right )} x\right )}}\right ] \]
[1/48*(15*(6*B*a^3*b*d*e^2 - (7*B*a^4 - A*a^3*b)*e^3 + (6*B*b^4*d*e^2 - (7 *B*a*b^3 - A*b^4)*e^3)*x^3 + 3*(6*B*a*b^3*d*e^2 - (7*B*a^2*b^2 - A*a*b^3)* e^3)*x^2 + 3*(6*B*a^2*b^2*d*e^2 - (7*B*a^3*b - A*a^2*b^2)*e^3)*x)*sqrt(b^2 *d - a*b*e)*log((b*e*x + 2*b*d - a*e - 2*sqrt(b^2*d - a*b*e)*sqrt(e*x + d) )/(b*x + a)) - 2*(4*(B*a*b^4 + 2*A*b^5)*d^3 + 2*(8*B*a^2*b^3 + A*a*b^4)*d^ 2*e - 5*(25*B*a^3*b^2 - A*a^2*b^3)*d*e^2 + 15*(7*B*a^4*b - A*a^3*b^2)*e^3 - 48*(B*b^5*d*e^2 - B*a*b^4*e^3)*x^3 + 3*(18*B*b^5*d^2*e - (95*B*a*b^4 - 1 1*A*b^5)*d*e^2 + 11*(7*B*a^2*b^3 - A*a*b^4)*e^3)*x^2 + 2*(6*B*b^5*d^3 + (2 3*B*a*b^4 + 13*A*b^5)*d^2*e - (169*B*a^2*b^3 - 7*A*a*b^4)*d*e^2 + 20*(7*B* a^3*b^2 - A*a^2*b^3)*e^3)*x)*sqrt(e*x + d))/(a^3*b^6*d - a^4*b^5*e + (b^9* d - a*b^8*e)*x^3 + 3*(a*b^8*d - a^2*b^7*e)*x^2 + 3*(a^2*b^7*d - a^3*b^6*e) *x), 1/24*(15*(6*B*a^3*b*d*e^2 - (7*B*a^4 - A*a^3*b)*e^3 + (6*B*b^4*d*e^2 - (7*B*a*b^3 - A*b^4)*e^3)*x^3 + 3*(6*B*a*b^3*d*e^2 - (7*B*a^2*b^2 - A*a*b ^3)*e^3)*x^2 + 3*(6*B*a^2*b^2*d*e^2 - (7*B*a^3*b - A*a^2*b^2)*e^3)*x)*sqrt (-b^2*d + a*b*e)*arctan(sqrt(-b^2*d + a*b*e)*sqrt(e*x + d)/(b*e*x + b*d)) - (4*(B*a*b^4 + 2*A*b^5)*d^3 + 2*(8*B*a^2*b^3 + A*a*b^4)*d^2*e - 5*(25*B*a ^3*b^2 - A*a^2*b^3)*d*e^2 + 15*(7*B*a^4*b - A*a^3*b^2)*e^3 - 48*(B*b^5*d*e ^2 - B*a*b^4*e^3)*x^3 + 3*(18*B*b^5*d^2*e - (95*B*a*b^4 - 11*A*b^5)*d*e^2 + 11*(7*B*a^2*b^3 - A*a*b^4)*e^3)*x^2 + 2*(6*B*b^5*d^3 + (23*B*a*b^4 + 13* A*b^5)*d^2*e - (169*B*a^2*b^3 - 7*A*a*b^4)*d*e^2 + 20*(7*B*a^3*b^2 - A*...
Timed out. \[ \int \frac {(A+B x) (d+e x)^{5/2}}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=\text {Timed out} \]
Exception generated. \[ \int \frac {(A+B x) (d+e x)^{5/2}}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=\text {Exception raised: ValueError} \]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(a*e-b*d>0)', see `assume?` for m ore detail
Time = 0.29 (sec) , antiderivative size = 368, normalized size of antiderivative = 1.47 \[ \int \frac {(A+B x) (d+e x)^{5/2}}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=\frac {2 \, \sqrt {e x + d} B e^{2}}{b^{4}} + \frac {5 \, {\left (6 \, B b d e^{2} - 7 \, B a e^{3} + A b e^{3}\right )} \arctan \left (\frac {\sqrt {e x + d} b}{\sqrt {-b^{2} d + a b e}}\right )}{8 \, \sqrt {-b^{2} d + a b e} b^{4}} - \frac {54 \, {\left (e x + d\right )}^{\frac {5}{2}} B b^{3} d e^{2} - 96 \, {\left (e x + d\right )}^{\frac {3}{2}} B b^{3} d^{2} e^{2} + 42 \, \sqrt {e x + d} B b^{3} d^{3} e^{2} - 87 \, {\left (e x + d\right )}^{\frac {5}{2}} B a b^{2} e^{3} + 33 \, {\left (e x + d\right )}^{\frac {5}{2}} A b^{3} e^{3} + 232 \, {\left (e x + d\right )}^{\frac {3}{2}} B a b^{2} d e^{3} - 40 \, {\left (e x + d\right )}^{\frac {3}{2}} A b^{3} d e^{3} - 141 \, \sqrt {e x + d} B a b^{2} d^{2} e^{3} + 15 \, \sqrt {e x + d} A b^{3} d^{2} e^{3} - 136 \, {\left (e x + d\right )}^{\frac {3}{2}} B a^{2} b e^{4} + 40 \, {\left (e x + d\right )}^{\frac {3}{2}} A a b^{2} e^{4} + 156 \, \sqrt {e x + d} B a^{2} b d e^{4} - 30 \, \sqrt {e x + d} A a b^{2} d e^{4} - 57 \, \sqrt {e x + d} B a^{3} e^{5} + 15 \, \sqrt {e x + d} A a^{2} b e^{5}}{24 \, {\left ({\left (e x + d\right )} b - b d + a e\right )}^{3} b^{4}} \]
2*sqrt(e*x + d)*B*e^2/b^4 + 5/8*(6*B*b*d*e^2 - 7*B*a*e^3 + A*b*e^3)*arctan (sqrt(e*x + d)*b/sqrt(-b^2*d + a*b*e))/(sqrt(-b^2*d + a*b*e)*b^4) - 1/24*( 54*(e*x + d)^(5/2)*B*b^3*d*e^2 - 96*(e*x + d)^(3/2)*B*b^3*d^2*e^2 + 42*sqr t(e*x + d)*B*b^3*d^3*e^2 - 87*(e*x + d)^(5/2)*B*a*b^2*e^3 + 33*(e*x + d)^( 5/2)*A*b^3*e^3 + 232*(e*x + d)^(3/2)*B*a*b^2*d*e^3 - 40*(e*x + d)^(3/2)*A* b^3*d*e^3 - 141*sqrt(e*x + d)*B*a*b^2*d^2*e^3 + 15*sqrt(e*x + d)*A*b^3*d^2 *e^3 - 136*(e*x + d)^(3/2)*B*a^2*b*e^4 + 40*(e*x + d)^(3/2)*A*a*b^2*e^4 + 156*sqrt(e*x + d)*B*a^2*b*d*e^4 - 30*sqrt(e*x + d)*A*a*b^2*d*e^4 - 57*sqrt (e*x + d)*B*a^3*e^5 + 15*sqrt(e*x + d)*A*a^2*b*e^5)/(((e*x + d)*b - b*d + a*e)^3*b^4)
Time = 0.25 (sec) , antiderivative size = 416, normalized size of antiderivative = 1.66 \[ \int \frac {(A+B x) (d+e x)^{5/2}}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=\frac {2\,B\,e^2\,\sqrt {d+e\,x}}{b^4}-\frac {{\left (d+e\,x\right )}^{5/2}\,\left (\frac {11\,A\,b^3\,e^3}{8}+\frac {9\,B\,d\,b^3\,e^2}{4}-\frac {29\,B\,a\,b^2\,e^3}{8}\right )-{\left (d+e\,x\right )}^{3/2}\,\left (\frac {17\,B\,a^2\,b\,e^4}{3}-\frac {29\,B\,a\,b^2\,d\,e^3}{3}-\frac {5\,A\,a\,b^2\,e^4}{3}+4\,B\,b^3\,d^2\,e^2+\frac {5\,A\,b^3\,d\,e^3}{3}\right )+\sqrt {d+e\,x}\,\left (-\frac {19\,B\,a^3\,e^5}{8}+\frac {13\,B\,a^2\,b\,d\,e^4}{2}+\frac {5\,A\,a^2\,b\,e^5}{8}-\frac {47\,B\,a\,b^2\,d^2\,e^3}{8}-\frac {5\,A\,a\,b^2\,d\,e^4}{4}+\frac {7\,B\,b^3\,d^3\,e^2}{4}+\frac {5\,A\,b^3\,d^2\,e^3}{8}\right )}{b^7\,{\left (d+e\,x\right )}^3-\left (3\,b^7\,d-3\,a\,b^6\,e\right )\,{\left (d+e\,x\right )}^2+\left (d+e\,x\right )\,\left (3\,a^2\,b^5\,e^2-6\,a\,b^6\,d\,e+3\,b^7\,d^2\right )-b^7\,d^3+a^3\,b^4\,e^3-3\,a^2\,b^5\,d\,e^2+3\,a\,b^6\,d^2\,e}+\frac {5\,e^2\,\mathrm {atan}\left (\frac {\sqrt {b}\,e^2\,\sqrt {d+e\,x}\,\left (A\,b\,e-7\,B\,a\,e+6\,B\,b\,d\right )}{\sqrt {a\,e-b\,d}\,\left (A\,b\,e^3-7\,B\,a\,e^3+6\,B\,b\,d\,e^2\right )}\right )\,\left (A\,b\,e-7\,B\,a\,e+6\,B\,b\,d\right )}{8\,b^{9/2}\,\sqrt {a\,e-b\,d}} \]
(2*B*e^2*(d + e*x)^(1/2))/b^4 - ((d + e*x)^(5/2)*((11*A*b^3*e^3)/8 - (29*B *a*b^2*e^3)/8 + (9*B*b^3*d*e^2)/4) - (d + e*x)^(3/2)*((17*B*a^2*b*e^4)/3 - (5*A*a*b^2*e^4)/3 + (5*A*b^3*d*e^3)/3 + 4*B*b^3*d^2*e^2 - (29*B*a*b^2*d*e ^3)/3) + (d + e*x)^(1/2)*((5*A*a^2*b*e^5)/8 - (19*B*a^3*e^5)/8 + (5*A*b^3* d^2*e^3)/8 + (7*B*b^3*d^3*e^2)/4 - (47*B*a*b^2*d^2*e^3)/8 - (5*A*a*b^2*d*e ^4)/4 + (13*B*a^2*b*d*e^4)/2))/(b^7*(d + e*x)^3 - (3*b^7*d - 3*a*b^6*e)*(d + e*x)^2 + (d + e*x)*(3*b^7*d^2 + 3*a^2*b^5*e^2 - 6*a*b^6*d*e) - b^7*d^3 + a^3*b^4*e^3 - 3*a^2*b^5*d*e^2 + 3*a*b^6*d^2*e) + (5*e^2*atan((b^(1/2)*e^ 2*(d + e*x)^(1/2)*(A*b*e - 7*B*a*e + 6*B*b*d))/((a*e - b*d)^(1/2)*(A*b*e^3 - 7*B*a*e^3 + 6*B*b*d*e^2)))*(A*b*e - 7*B*a*e + 6*B*b*d))/(8*b^(9/2)*(a*e - b*d)^(1/2))