Integrand size = 32, antiderivative size = 336 \[ \int \frac {A+B x+C x^2+D x^3}{(a+b x)^2 (c+d x)^{5/2}} \, dx=\frac {3 a b^2 B d^3-3 a^2 b C d^3+3 a^3 d^3 D-b^3 \left (2 c^2 C d-2 B c d^2+5 A d^3-2 c^3 D\right )}{3 b^3 d^2 (b c-a d)^2 (c+d x)^{3/2}}-\frac {A-\frac {a \left (b^2 B-a b C+a^2 D\right )}{b^3}}{(b c-a d) (a+b x) (c+d x)^{3/2}}-\frac {a^2 b C d^3-a^3 d^3 D+a b^2 d \left (4 c C d-3 B d^2-6 c^2 D\right )-b^3 \left (2 B c d^2-5 A d^3-2 c^3 D\right )}{b^2 d^2 (b c-a d)^3 \sqrt {c+d x}}-\frac {\left (b^3 (2 B c-5 A d)-a b^2 (4 c C-3 B d)-a^3 d D-a^2 b (C d-6 c D)\right ) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{b^{3/2} (b c-a d)^{7/2}} \]
1/3*(3*a*b^2*B*d^3-3*a^2*b*C*d^3+3*a^3*d^3*D-b^3*(5*A*d^3-2*B*c*d^2+2*C*c^ 2*d-2*D*c^3))/b^3/d^2/(-a*d+b*c)^2/(d*x+c)^(3/2)+(-A+a*(B*b^2-C*a*b+D*a^2) /b^3)/(-a*d+b*c)/(b*x+a)/(d*x+c)^(3/2)-(b^3*(-5*A*d+2*B*c)-a*b^2*(-3*B*d+4 *C*c)-a^3*d*D-a^2*b*(C*d-6*D*c))*arctanh(b^(1/2)*(d*x+c)^(1/2)/(-a*d+b*c)^ (1/2))/b^(3/2)/(-a*d+b*c)^(7/2)+(-a^2*b*C*d^3+a^3*d^3*D-a*b^2*d*(-3*B*d^2+ 4*C*c*d-6*D*c^2)+b^3*(-5*A*d^3+2*B*c*d^2-2*D*c^3))/b^2/d^2/(-a*d+b*c)^3/(d *x+c)^(1/2)
Time = 0.90 (sec) , antiderivative size = 354, normalized size of antiderivative = 1.05 \[ \int \frac {A+B x+C x^2+D x^3}{(a+b x)^2 (c+d x)^{5/2}} \, dx=\frac {-3 a^3 d^2 D (c+d x)^2-a^2 b d \left (16 c^3 D+2 c d^2 (2 B-9 C x)+c^2 (-13 C d+18 d D x)+d^3 \left (2 A+6 B x-3 C x^2\right )\right )+a b^2 \left (4 c^4 D+d^4 x (10 A-9 B x)+2 c^3 d (C-5 D x)+2 c d^3 \left (7 A-8 B x+6 C x^2\right )+c^2 d^2 (-11 B+2 x (5 C-9 D x))\right )+b^3 \left (A d^2 \left (3 c^2+20 c d x+15 d^2 x^2\right )+2 c x \left (-4 B c d^2+2 c^3 D-3 B d^3 x+c^2 d (C+3 D x)\right )\right )}{3 b d^2 (-b c+a d)^3 (a+b x) (c+d x)^{3/2}}-\frac {\left (b^3 (2 B c-5 A d)+a b^2 (-4 c C+3 B d)-a^3 d D+a^2 b (-C d+6 c D)\right ) \arctan \left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {-b c+a d}}\right )}{b^{3/2} (-b c+a d)^{7/2}} \]
(-3*a^3*d^2*D*(c + d*x)^2 - a^2*b*d*(16*c^3*D + 2*c*d^2*(2*B - 9*C*x) + c^ 2*(-13*C*d + 18*d*D*x) + d^3*(2*A + 6*B*x - 3*C*x^2)) + a*b^2*(4*c^4*D + d ^4*x*(10*A - 9*B*x) + 2*c^3*d*(C - 5*D*x) + 2*c*d^3*(7*A - 8*B*x + 6*C*x^2 ) + c^2*d^2*(-11*B + 2*x*(5*C - 9*D*x))) + b^3*(A*d^2*(3*c^2 + 20*c*d*x + 15*d^2*x^2) + 2*c*x*(-4*B*c*d^2 + 2*c^3*D - 3*B*d^3*x + c^2*d*(C + 3*D*x)) ))/(3*b*d^2*(-(b*c) + a*d)^3*(a + b*x)*(c + d*x)^(3/2)) - ((b^3*(2*B*c - 5 *A*d) + a*b^2*(-4*c*C + 3*B*d) - a^3*d*D + a^2*b*(-(C*d) + 6*c*D))*ArcTan[ (Sqrt[b]*Sqrt[c + d*x])/Sqrt[-(b*c) + a*d]])/(b^(3/2)*(-(b*c) + a*d)^(7/2) )
Time = 0.89 (sec) , antiderivative size = 348, normalized size of antiderivative = 1.04, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.156, Rules used = {2124, 27, 1192, 1584, 2009}
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.
| \(\displaystyle \int \frac {A+B x+C x^2+D x^3}{(a+b x)^2 (c+d x)^{5/2}} \, dx\) |
| \(\Big \downarrow \) 2124 |
| \(\displaystyle -\frac {\int -\frac {2 \left (c-\frac {a d}{b}\right ) D x^2+\frac {2 (b c-a d) (b C-a D) x}{b^2}+\frac {3 d D a^3-b (3 C d-2 c D) a^2-b^2 (2 c C-3 B d) a+b^3 (2 B c-5 A d)}{b^3}}{2 (a+b x) (c+d x)^{5/2}}dx}{b c-a d}-\frac {A-\frac {a \left (a^2 D-a b C+b^2 B\right )}{b^3}}{(a+b x) (c+d x)^{3/2} (b c-a d)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {\frac {3 d D a^3}{b^3}-\frac {(3 C d-2 c D) a^2}{b^2}-\frac {(2 c C-3 B d) a}{b}+2 \left (c-\frac {a d}{b}\right ) D x^2+2 B c-5 A d+\frac {2 (b c-a d) (b C-a D) x}{b^2}}{(a+b x) (c+d x)^{5/2}}dx}{2 (b c-a d)}-\frac {A-\frac {a \left (a^2 D-a b C+b^2 B\right )}{b^3}}{(a+b x) (c+d x)^{3/2} (b c-a d)}\) |
| \(\Big \downarrow \) 1192 |
| \(\displaystyle \frac {\int \frac {-2 D c^3+2 C d c^2-2 B d^2 c-2 \left (c-\frac {a d}{b}\right ) D (c+d x)^2+d^3 \left (5 A-\frac {3 a \left (D a^2-b C a+b^2 B\right )}{b^3}\right )-\frac {2 (b c-a d) (b C d-a D d-2 b c D) (c+d x)}{b^2}}{(c+d x)^2 (b c-a d-b (c+d x))}d\sqrt {c+d x}}{d^2 (b c-a d)}-\frac {A-\frac {a \left (a^2 D-a b C+b^2 B\right )}{b^3}}{(a+b x) (c+d x)^{3/2} (b c-a d)}\) |
| \(\Big \downarrow \) 1584 |
| \(\displaystyle \frac {\int \left (\frac {\left (d D a^3+b (C d-6 c D) a^2+b^2 (4 c C-3 B d) a-b^3 (2 B c-5 A d)\right ) d^2}{b (b c-a d)^2 (b c-a d-b (c+d x))}+\frac {-\left (\left (-2 D c^3+2 B d^2 c-5 A d^3\right ) b^3\right )+a d \left (-6 D c^2+4 C d c-3 B d^2\right ) b^2+a^2 C d^3 b-a^3 d^3 D}{b^2 (b c-a d)^2 (c+d x)}+\frac {\left (-2 D c^3+2 C d c^2-2 B d^2 c+5 A d^3\right ) b^3-3 a B d^3 b^2+3 a^2 C d^3 b-3 a^3 d^3 D}{b^3 (b c-a d) (c+d x)^2}\right )d\sqrt {c+d x}}{d^2 (b c-a d)}-\frac {A-\frac {a \left (a^2 D-a b C+b^2 B\right )}{b^3}}{(a+b x) (c+d x)^{3/2} (b c-a d)}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {-\frac {d^2 \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right ) \left (a^3 (-d) D-a^2 b (C d-6 c D)-a b^2 (4 c C-3 B d)+b^3 (2 B c-5 A d)\right )}{b^{3/2} (b c-a d)^{5/2}}-\frac {-a^3 d^3 D+a^2 b C d^3+a b^2 d \left (-3 B d^2-6 c^2 D+4 c C d\right )-\left (b^3 \left (-5 A d^3+2 B c d^2-2 c^3 D\right )\right )}{b^2 \sqrt {c+d x} (b c-a d)^2}+\frac {3 a^3 d^3 D-3 a^2 b C d^3+3 a b^2 B d^3-\left (b^3 \left (5 A d^3-2 B c d^2-2 c^3 D+2 c^2 C d\right )\right )}{3 b^3 (c+d x)^{3/2} (b c-a d)}}{d^2 (b c-a d)}-\frac {A-\frac {a \left (a^2 D-a b C+b^2 B\right )}{b^3}}{(a+b x) (c+d x)^{3/2} (b c-a d)}\) |
-((A - (a*(b^2*B - a*b*C + a^2*D))/b^3)/((b*c - a*d)*(a + b*x)*(c + d*x)^( 3/2))) + ((3*a*b^2*B*d^3 - 3*a^2*b*C*d^3 + 3*a^3*d^3*D - b^3*(2*c^2*C*d - 2*B*c*d^2 + 5*A*d^3 - 2*c^3*D))/(3*b^3*(b*c - a*d)*(c + d*x)^(3/2)) - (a^2 *b*C*d^3 - a^3*d^3*D + a*b^2*d*(4*c*C*d - 3*B*d^2 - 6*c^2*D) - b^3*(2*B*c* d^2 - 5*A*d^3 - 2*c^3*D))/(b^2*(b*c - a*d)^2*Sqrt[c + d*x]) - (d^2*(b^3*(2 *B*c - 5*A*d) - a*b^2*(4*c*C - 3*B*d) - a^3*d*D - a^2*b*(C*d - 6*c*D))*Arc Tanh[(Sqrt[b]*Sqrt[c + d*x])/Sqrt[b*c - a*d]])/(b^(3/2)*(b*c - a*d)^(5/2)) )/(d^2*(b*c - a*d))
3.1.23.3.1 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[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[2/e^(n + 2*p + 1) Subst[Int[x^( 2*m + 1)*(e*f - d*g + g*x^2)^n*(c*d^2 - b*d*e + a*e^2 - (2*c*d - b*e)*x^2 + c*x^4)^p, x], x, Sqrt[d + e*x]], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && IGtQ[p, 0] && ILtQ[n, 0] && IntegerQ[m + 1/2]
Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + ( c_.)*(x_)^4)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(f*x)^m*(d + e*x^2)^q* (a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, q}, x] && NeQ[ b^2 - 4*a*c, 0] && IGtQ[p, 0] && IGtQ[q, -2]
Int[(Px_)*((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] : > With[{Qx = PolynomialQuotient[Px, a + b*x, x], R = PolynomialRemainder[Px , a + b*x, x]}, Simp[R*(a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((m + 1)*(b*c - a*d))), x] + Simp[1/((m + 1)*(b*c - a*d)) Int[(a + b*x)^(m + 1)*(c + d*x )^n*ExpandToSum[(m + 1)*(b*c - a*d)*Qx - d*R*(m + n + 2), x], x], x]] /; Fr eeQ[{a, b, c, d, n}, x] && PolyQ[Px, x] && LtQ[m, -1] && (IntegerQ[m] || ! ILtQ[n, -1])
Time = 2.00 (sec) , antiderivative size = 275, normalized size of antiderivative = 0.82
| method | result | size |
| derivativedivides | \(\frac {\frac {2 d^{2} \left (\frac {d \left (b^{3} A -a \,b^{2} B +C \,a^{2} b -D a^{3}\right ) \sqrt {d x +c}}{2 b \left (\left (d x +c \right ) b +a d -b c \right )}+\frac {\left (5 A \,b^{3} d -3 B a \,b^{2} d -2 B \,b^{3} c +C \,a^{2} b d +4 C a \,b^{2} c +a^{3} d D-6 D a^{2} b c \right ) \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{2 b \sqrt {\left (a d -b c \right ) b}}\right )}{\left (a d -b c \right )^{3}}-\frac {2 \left (A \,d^{3}-B c \,d^{2}+C \,c^{2} d -D c^{3}\right )}{3 \left (a d -b c \right )^{2} \left (d x +c \right )^{\frac {3}{2}}}-\frac {2 \left (-2 A b \,d^{3}+B a \,d^{3}+B b c \,d^{2}-2 C a c \,d^{2}+3 D a \,c^{2} d -D b \,c^{3}\right )}{\left (a d -b c \right )^{3} \sqrt {d x +c}}}{d^{2}}\) | \(275\) |
| default | \(\frac {\frac {2 d^{2} \left (\frac {d \left (b^{3} A -a \,b^{2} B +C \,a^{2} b -D a^{3}\right ) \sqrt {d x +c}}{2 b \left (\left (d x +c \right ) b +a d -b c \right )}+\frac {\left (5 A \,b^{3} d -3 B a \,b^{2} d -2 B \,b^{3} c +C \,a^{2} b d +4 C a \,b^{2} c +a^{3} d D-6 D a^{2} b c \right ) \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{2 b \sqrt {\left (a d -b c \right ) b}}\right )}{\left (a d -b c \right )^{3}}-\frac {2 \left (A \,d^{3}-B c \,d^{2}+C \,c^{2} d -D c^{3}\right )}{3 \left (a d -b c \right )^{2} \left (d x +c \right )^{\frac {3}{2}}}-\frac {2 \left (-2 A b \,d^{3}+B a \,d^{3}+B b c \,d^{2}-2 C a c \,d^{2}+3 D a \,c^{2} d -D b \,c^{3}\right )}{\left (a d -b c \right )^{3} \sqrt {d x +c}}}{d^{2}}\) | \(275\) |
| pseudoelliptic | \(\frac {5 \left (d x +c \right )^{\frac {3}{2}} \left (\left (b^{3} A -\frac {3}{5} a \,b^{2} B +\frac {1}{5} C \,a^{2} b +\frac {1}{5} D a^{3}\right ) d -\frac {2 b c \left (B \,b^{2}-2 C a b +3 D a^{2}\right )}{5}\right ) \left (b x +a \right ) d^{2} \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (a d -b c \right ) b}}\right )-\frac {2 \sqrt {\left (a d -b c \right ) b}\, \left (\left (-\frac {15 A \,b^{3} x^{2}}{2}-5 a x \left (-\frac {9 B x}{10}+A \right ) b^{2}+a^{2} \left (-\frac {3}{2} C \,x^{2}+3 B x +A \right ) b +\frac {3 D a^{3} x^{2}}{2}\right ) d^{4}-7 \left (\frac {\left (-3 x^{2} B +10 A x \right ) b^{3}}{7}+a \left (\frac {6}{7} C \,x^{2}-\frac {8}{7} B x +A \right ) b^{2}-\frac {2 a^{2} \left (-\frac {9 C x}{2}+B \right ) b}{7}-\frac {3 D a^{3} x}{7}\right ) c \,d^{3}-\frac {3 \left (\left (-\frac {8 B x}{3}+A \right ) b^{3}-\frac {11 a \left (\frac {18}{11} D x^{2}-\frac {10}{11} C x +B \right ) b^{2}}{3}+\frac {13 a^{2} \left (-\frac {18 D x}{13}+C \right ) b}{3}-D a^{3}\right ) c^{2} d^{2}}{2}-b \,c^{3} \left (b x +a \right ) \left (\left (3 D x +C \right ) b -8 D a \right ) d -2 D b^{2} c^{4} \left (b x +a \right )\right )}{3}}{d^{2} \left (b x +a \right ) \left (a d -b c \right )^{3} b \sqrt {\left (a d -b c \right ) b}\, \left (d x +c \right )^{\frac {3}{2}}}\) | \(355\) |
2/d^2*(d^2/(a*d-b*c)^3*(1/2*d*(A*b^3-B*a*b^2+C*a^2*b-D*a^3)/b*(d*x+c)^(1/2 )/((d*x+c)*b+a*d-b*c)+1/2*(5*A*b^3*d-3*B*a*b^2*d-2*B*b^3*c+C*a^2*b*d+4*C*a *b^2*c+D*a^3*d-6*D*a^2*b*c)/b/((a*d-b*c)*b)^(1/2)*arctan(b*(d*x+c)^(1/2)/( (a*d-b*c)*b)^(1/2)))-1/3*(A*d^3-B*c*d^2+C*c^2*d-D*c^3)/(a*d-b*c)^2/(d*x+c) ^(3/2)-1/(a*d-b*c)^3*(-2*A*b*d^3+B*a*d^3+B*b*c*d^2-2*C*a*c*d^2+3*D*a*c^2*d -D*b*c^3)/(d*x+c)^(1/2))
Leaf count of result is larger than twice the leaf count of optimal. 1215 vs. \(2 (315) = 630\).
Time = 0.37 (sec) , antiderivative size = 2444, normalized size of antiderivative = 7.27 \[ \int \frac {A+B x+C x^2+D x^3}{(a+b x)^2 (c+d x)^{5/2}} \, dx=\text {Too large to display} \]
[-1/6*(3*((D*a^4*c^2 + (C*a^3*b - 3*B*a^2*b^2 + 5*A*a*b^3)*c^2)*d^3 + ((D* a^3*b + C*a^2*b^2 - 3*B*a*b^3 + 5*A*b^4)*d^5 - 2*(3*D*a^2*b^2*c - (2*C*a*b ^3 - B*b^4)*c)*d^4)*x^3 - 2*(3*D*a^3*b*c^3 - (2*C*a^2*b^2 - B*a*b^3)*c^3)* d^2 + ((D*a^4 + C*a^3*b - 3*B*a^2*b^2 + 5*A*a*b^3)*d^5 - 2*(2*D*a^3*b*c - (3*C*a^2*b^2 - 4*B*a*b^3 + 5*A*b^4)*c)*d^4 - 4*(3*D*a^2*b^2*c^2 - (2*C*a*b ^3 - B*b^4)*c^2)*d^3)*x^2 + (2*(D*a^4*c + (C*a^3*b - 3*B*a^2*b^2 + 5*A*a*b ^3)*c)*d^4 - (11*D*a^3*b*c^2 - (9*C*a^2*b^2 - 7*B*a*b^3 + 5*A*b^4)*c^2)*d^ 3 - 2*(3*D*a^2*b^2*c^3 - (2*C*a*b^3 - B*b^4)*c^3)*d^2)*x)*sqrt(b^2*c - a*b *d)*log((b*d*x + 2*b*c - a*d - 2*sqrt(b^2*c - a*b*d)*sqrt(d*x + c))/(b*x + a)) + 2*(4*D*a*b^4*c^5 + 2*A*a^3*b^2*d^5 + 4*(B*a^3*b^2 - 4*A*a^2*b^3)*c* d^4 + (3*D*a^4*b*c^2 - (13*C*a^3*b^2 - 7*B*a^2*b^3 - 11*A*a*b^4)*c^2)*d^3 + (13*D*a^3*b^2*c^3 + (11*C*a^2*b^3 - 11*B*a*b^4 + 3*A*b^5)*c^3)*d^2 + 3*( 2*D*b^5*c^4*d - 8*D*a*b^4*c^3*d^2 + (D*a^4*b - C*a^3*b^2 + 3*B*a^2*b^3 - 5 *A*a*b^4)*d^5 - (D*a^3*b^2*c + (3*C*a^2*b^3 + B*a*b^4 - 5*A*b^5)*c)*d^4 + 2*(3*D*a^2*b^3*c^2 + (2*C*a*b^4 - B*b^5)*c^2)*d^3)*x^2 - 2*(10*D*a^2*b^3*c ^4 - C*a*b^4*c^4)*d + 2*(2*D*b^5*c^5 + (3*B*a^3*b^2 - 5*A*a^2*b^3)*d^5 + ( 3*D*a^4*b*c - (9*C*a^3*b^2 - 5*B*a^2*b^3 + 5*A*a*b^4)*c)*d^4 + 2*(3*D*a^3* b^2*c^2 + (2*C*a^2*b^3 - 2*B*a*b^4 + 5*A*b^5)*c^2)*d^3 - 4*(D*a^2*b^3*c^3 - (C*a*b^4 - B*b^5)*c^3)*d^2 - (7*D*a*b^4*c^4 - C*b^5*c^4)*d)*x)*sqrt(d*x + c))/(a*b^6*c^6*d^2 - 4*a^2*b^5*c^5*d^3 + 6*a^3*b^4*c^4*d^4 - 4*a^4*b^...
Timed out. \[ \int \frac {A+B x+C x^2+D x^3}{(a+b x)^2 (c+d x)^{5/2}} \, dx=\text {Timed out} \]
Exception generated. \[ \int \frac {A+B x+C x^2+D x^3}{(a+b x)^2 (c+d x)^{5/2}} \, dx=\text {Exception raised: ValueError} \]
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*d-b*c>0)', see `assume?` for m ore detail
Time = 0.30 (sec) , antiderivative size = 439, normalized size of antiderivative = 1.31 \[ \int \frac {A+B x+C x^2+D x^3}{(a+b x)^2 (c+d x)^{5/2}} \, dx=\frac {{\left (6 \, D a^{2} b c - 4 \, C a b^{2} c + 2 \, B b^{3} c - D a^{3} d - C a^{2} b d + 3 \, B a b^{2} d - 5 \, A b^{3} d\right )} \arctan \left (\frac {\sqrt {d x + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{{\left (b^{4} c^{3} - 3 \, a b^{3} c^{2} d + 3 \, a^{2} b^{2} c d^{2} - a^{3} b d^{3}\right )} \sqrt {-b^{2} c + a b d}} + \frac {\sqrt {d x + c} D a^{3} d - \sqrt {d x + c} C a^{2} b d + \sqrt {d x + c} B a b^{2} d - \sqrt {d x + c} A b^{3} d}{{\left (b^{4} c^{3} - 3 \, a b^{3} c^{2} d + 3 \, a^{2} b^{2} c d^{2} - a^{3} b d^{3}\right )} {\left ({\left (d x + c\right )} b - b c + a d\right )}} - \frac {2 \, {\left (3 \, {\left (d x + c\right )} D b c^{3} - D b c^{4} - 9 \, {\left (d x + c\right )} D a c^{2} d + D a c^{3} d + C b c^{3} d + 6 \, {\left (d x + c\right )} C a c d^{2} - 3 \, {\left (d x + c\right )} B b c d^{2} - C a c^{2} d^{2} - B b c^{2} d^{2} - 3 \, {\left (d x + c\right )} B a d^{3} + 6 \, {\left (d x + c\right )} A b d^{3} + B a c d^{3} + A b c d^{3} - A a d^{4}\right )}}{3 \, {\left (b^{3} c^{3} d^{2} - 3 \, a b^{2} c^{2} d^{3} + 3 \, a^{2} b c d^{4} - a^{3} d^{5}\right )} {\left (d x + c\right )}^{\frac {3}{2}}} \]
(6*D*a^2*b*c - 4*C*a*b^2*c + 2*B*b^3*c - D*a^3*d - C*a^2*b*d + 3*B*a*b^2*d - 5*A*b^3*d)*arctan(sqrt(d*x + c)*b/sqrt(-b^2*c + a*b*d))/((b^4*c^3 - 3*a *b^3*c^2*d + 3*a^2*b^2*c*d^2 - a^3*b*d^3)*sqrt(-b^2*c + a*b*d)) + (sqrt(d* x + c)*D*a^3*d - sqrt(d*x + c)*C*a^2*b*d + sqrt(d*x + c)*B*a*b^2*d - sqrt( d*x + c)*A*b^3*d)/((b^4*c^3 - 3*a*b^3*c^2*d + 3*a^2*b^2*c*d^2 - a^3*b*d^3) *((d*x + c)*b - b*c + a*d)) - 2/3*(3*(d*x + c)*D*b*c^3 - D*b*c^4 - 9*(d*x + c)*D*a*c^2*d + D*a*c^3*d + C*b*c^3*d + 6*(d*x + c)*C*a*c*d^2 - 3*(d*x + c)*B*b*c*d^2 - C*a*c^2*d^2 - B*b*c^2*d^2 - 3*(d*x + c)*B*a*d^3 + 6*(d*x + c)*A*b*d^3 + B*a*c*d^3 + A*b*c*d^3 - A*a*d^4)/((b^3*c^3*d^2 - 3*a*b^2*c^2* d^3 + 3*a^2*b*c*d^4 - a^3*d^5)*(d*x + c)^(3/2))
Timed out. \[ \int \frac {A+B x+C x^2+D x^3}{(a+b x)^2 (c+d x)^{5/2}} \, dx=\int \frac {A+B\,x+C\,x^2+x^3\,D}{{\left (a+b\,x\right )}^2\,{\left (c+d\,x\right )}^{5/2}} \,d x \]