Integrand size = 32, antiderivative size = 350 \[ \int \frac {A+B x+C x^2+D x^3}{(a+b x)^3 (c+d x)^{3/2}} \, dx=-\frac {a b^2 B d^3-a^2 b C d^3+a^3 d^3 D-b^3 \left (4 c^2 C d-4 B c d^2+5 A d^3-4 c^3 D\right )}{2 b^3 d (b c-a d)^3 \sqrt {c+d x}}-\frac {A b^3-a \left (b^2 B-a b C+a^2 D\right )}{2 b^3 (b c-a d) (a+b x)^2 \sqrt {c+d x}}-\frac {\left (b^3 (4 B c-5 A d)-a b^2 (8 c C-B d)-7 a^3 d D+3 a^2 b (C d+4 c D)\right ) \sqrt {c+d x}}{4 b^2 (b c-a d)^3 (a+b x)}-\frac {\left (b^3 \left (8 c^2 C-12 B c d+15 A d^2\right )-3 a^3 d^2 D-a^2 b d (C d-12 c D)+a b^2 \left (8 c C d-3 B d^2-24 c^2 D\right )\right ) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{4 b^{5/2} (b c-a d)^{7/2}} \]
-1/4*(b^3*(15*A*d^2-12*B*c*d+8*C*c^2)-3*a^3*d^2*D-a^2*b*d*(C*d-12*D*c)+a*b ^2*(-3*B*d^2+8*C*c*d-24*D*c^2))*arctanh(b^(1/2)*(d*x+c)^(1/2)/(-a*d+b*c)^( 1/2))/b^(5/2)/(-a*d+b*c)^(7/2)+1/2*(-a*b^2*B*d^3+a^2*b*C*d^3-a^3*d^3*D+b^3 *(5*A*d^3-4*B*c*d^2+4*C*c^2*d-4*D*c^3))/b^3/d/(-a*d+b*c)^3/(d*x+c)^(1/2)+1 /2*(-A*b^3+a*(B*b^2-C*a*b+D*a^2))/b^3/(-a*d+b*c)/(b*x+a)^2/(d*x+c)^(1/2)-1 /4*(b^3*(-5*A*d+4*B*c)-a*b^2*(-B*d+8*C*c)-7*a^3*d*D+3*a^2*b*(C*d+4*D*c))*( d*x+c)^(1/2)/b^2/(-a*d+b*c)^3/(b*x+a)
Time = 1.38 (sec) , antiderivative size = 387, normalized size of antiderivative = 1.11 \[ \int \frac {A+B x+C x^2+D x^3}{(a+b x)^3 (c+d x)^{3/2}} \, dx=\frac {\frac {\sqrt {b} \left (-3 a^4 d^2 D (c+d x)-a^3 b d (c+d x) (C d+5 D (-2 c+d x))+4 b^4 c x (2 c (-C d+c D) x+B d (c+3 d x))+a b^3 \left (-8 c x \left (3 c C d-2 c^2 D+C d^2 x\right )+B d \left (2 c^2+21 c d x+3 d^2 x^2\right )\right )-A b^2 d \left (8 a^2 d^2+a b d (9 c+25 d x)+b^2 \left (-2 c^2+5 c d x+15 d^2 x^2\right )\right )+a^2 b^2 \left (8 c^3 D+d^3 x (5 B+C x)-2 c^2 d (7 C-6 D x)+c d^2 \left (13 B-5 C x+12 D x^2\right )\right )\right )}{d (-b c+a d)^3 (a+b x)^2 \sqrt {c+d x}}-\frac {\left (b^3 \left (8 c^2 C-12 B c d+15 A d^2\right )-3 a^3 d^2 D+a^2 b d (-C d+12 c D)+a b^2 \left (8 c C d-3 B d^2-24 c^2 D\right )\right ) \arctan \left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {-b c+a d}}\right )}{(-b c+a d)^{7/2}}}{4 b^{5/2}} \]
((Sqrt[b]*(-3*a^4*d^2*D*(c + d*x) - a^3*b*d*(c + d*x)*(C*d + 5*D*(-2*c + d *x)) + 4*b^4*c*x*(2*c*(-(C*d) + c*D)*x + B*d*(c + 3*d*x)) + a*b^3*(-8*c*x* (3*c*C*d - 2*c^2*D + C*d^2*x) + B*d*(2*c^2 + 21*c*d*x + 3*d^2*x^2)) - A*b^ 2*d*(8*a^2*d^2 + a*b*d*(9*c + 25*d*x) + b^2*(-2*c^2 + 5*c*d*x + 15*d^2*x^2 )) + a^2*b^2*(8*c^3*D + d^3*x*(5*B + C*x) - 2*c^2*d*(7*C - 6*D*x) + c*d^2* (13*B - 5*C*x + 12*D*x^2))))/(d*(-(b*c) + a*d)^3*(a + b*x)^2*Sqrt[c + d*x] ) - ((b^3*(8*c^2*C - 12*B*c*d + 15*A*d^2) - 3*a^3*d^2*D + a^2*b*d*(-(C*d) + 12*c*D) + a*b^2*(8*c*C*d - 3*B*d^2 - 24*c^2*D))*ArcTan[(Sqrt[b]*Sqrt[c + d*x])/Sqrt[-(b*c) + a*d]])/(-(b*c) + a*d)^(7/2))/(4*b^(5/2))
Time = 0.87 (sec) , antiderivative size = 381, normalized size of antiderivative = 1.09, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {2124, 27, 1192, 25, 1582, 27, 359, 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.
| \(\displaystyle \int \frac {A+B x+C x^2+D x^3}{(a+b x)^3 (c+d x)^{3/2}} \, dx\) |
| \(\Big \downarrow \) 2124 |
| \(\displaystyle -\frac {\int -\frac {4 \left (c-\frac {a d}{b}\right ) D x^2+\frac {4 (b c-a d) (b C-a D) x}{b^2}+\frac {d D a^3-b (C d-4 c D) a^2-b^2 (4 c C-B d) a+b^3 (4 B c-5 A d)}{b^3}}{2 (a+b x)^2 (c+d x)^{3/2}}dx}{2 (b c-a d)}-\frac {A b^3-a \left (a^2 D-a b C+b^2 B\right )}{2 b^3 (a+b x)^2 \sqrt {c+d x} (b c-a d)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {\frac {d D a^3}{b^3}-\frac {(C d-4 c D) a^2}{b^2}-\frac {(4 c C-B d) a}{b}+4 \left (c-\frac {a d}{b}\right ) D x^2+4 B c-5 A d+\frac {4 (b c-a d) (b C-a D) x}{b^2}}{(a+b x)^2 (c+d x)^{3/2}}dx}{4 (b c-a d)}-\frac {A b^3-a \left (a^2 D-a b C+b^2 B\right )}{2 b^3 (a+b x)^2 \sqrt {c+d x} (b c-a d)}\) |
| \(\Big \downarrow \) 1192 |
| \(\displaystyle \frac {\int -\frac {-4 D c^3+4 C d c^2-4 B d^2 c-4 \left (c-\frac {a d}{b}\right ) D (c+d x)^2+d^3 \left (5 A-\frac {a \left (D a^2-b C a+b^2 B\right )}{b^3}\right )-\frac {4 (b c-a d) (b C d-a D d-2 b c D) (c+d x)}{b^2}}{(c+d x) (b c-a d-b (c+d x))^2}d\sqrt {c+d x}}{2 d (b c-a d)}-\frac {A b^3-a \left (a^2 D-a b C+b^2 B\right )}{2 b^3 (a+b x)^2 \sqrt {c+d x} (b c-a d)}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {\int \frac {-4 D c^3+4 C d c^2-4 B d^2 c-4 \left (c-\frac {a d}{b}\right ) D (c+d x)^2+d^3 \left (5 A-\frac {a \left (D a^2-b C a+b^2 B\right )}{b^3}\right )-\frac {4 (b c-a d) (b C d-a D d-2 b c D) (c+d x)}{b^2}}{(c+d x) (b c-a d-b (c+d x))^2}d\sqrt {c+d x}}{2 d (b c-a d)}-\frac {A b^3-a \left (a^2 D-a b C+b^2 B\right )}{2 b^3 (a+b x)^2 \sqrt {c+d x} (b c-a d)}\) |
| \(\Big \downarrow \) 1582 |
| \(\displaystyle \frac {\frac {\int \frac {2 (b c-a d) \left (-\left (\left (-4 D c^3+4 C d c^2-4 B d^2 c+5 A d^3\right ) b^3\right )+a B d^3 b^2-a^2 C d^3 b+a^3 d^3 D\right )+b \left (\left (-8 D c^3+4 B d^2 c-5 A d^3\right ) b^3-a d \left (-24 D c^2+8 C d c-B d^2\right ) b^2+3 a^2 d^2 (C d-4 c D) b+a^3 d^3 D\right ) (c+d x)}{b (c+d x) (b c-a d-b (c+d x))}d\sqrt {c+d x}}{2 b^2 (b c-a d)^2}+\frac {d^2 \sqrt {c+d x} \left (-7 a^3 d D+3 a^2 b (4 c D+C d)-a b^2 (8 c C-B d)+b^3 (4 B c-5 A d)\right )}{2 b^2 (b c-a d)^2 (-a d-b (c+d x)+b c)}}{2 d (b c-a d)}-\frac {A b^3-a \left (a^2 D-a b C+b^2 B\right )}{2 b^3 (a+b x)^2 \sqrt {c+d x} (b c-a d)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\int \frac {2 (b c-a d) \left (-\left (\left (-4 D c^3+4 C d c^2-4 B d^2 c+5 A d^3\right ) b^3\right )+a B d^3 b^2-a^2 C d^3 b+a^3 d^3 D\right )+b \left (\left (-8 D c^3+4 B d^2 c-5 A d^3\right ) b^3-a d \left (-24 D c^2+8 C d c-B d^2\right ) b^2+3 a^2 d^2 (C d-4 c D) b+a^3 d^3 D\right ) (c+d x)}{(c+d x) (b c-a d-b (c+d x))}d\sqrt {c+d x}}{2 b^3 (b c-a d)^2}+\frac {d^2 \sqrt {c+d x} \left (-7 a^3 d D+3 a^2 b (4 c D+C d)-a b^2 (8 c C-B d)+b^3 (4 B c-5 A d)\right )}{2 b^2 (b c-a d)^2 (-a d-b (c+d x)+b c)}}{2 d (b c-a d)}-\frac {A b^3-a \left (a^2 D-a b C+b^2 B\right )}{2 b^3 (a+b x)^2 \sqrt {c+d x} (b c-a d)}\) |
| \(\Big \downarrow \) 359 |
| \(\displaystyle \frac {\frac {-b d \left (-3 a^3 d^2 D-a^2 b d (C d-12 c D)+a b^2 \left (-3 B d^2-24 c^2 D+8 c C d\right )+b^3 \left (15 A d^2-12 B c d+8 c^2 C\right )\right ) \int \frac {1}{b c-a d-b (c+d x)}d\sqrt {c+d x}-\frac {2 \left (a^3 d^3 D-a^2 b C d^3+a b^2 B d^3-\left (b^3 \left (5 A d^3-4 B c d^2-4 c^3 D+4 c^2 C d\right )\right )\right )}{\sqrt {c+d x}}}{2 b^3 (b c-a d)^2}+\frac {d^2 \sqrt {c+d x} \left (-7 a^3 d D+3 a^2 b (4 c D+C d)-a b^2 (8 c C-B d)+b^3 (4 B c-5 A d)\right )}{2 b^2 (b c-a d)^2 (-a d-b (c+d x)+b c)}}{2 d (b c-a d)}-\frac {A b^3-a \left (a^2 D-a b C+b^2 B\right )}{2 b^3 (a+b x)^2 \sqrt {c+d x} (b c-a d)}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {\frac {-\frac {\sqrt {b} d \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right ) \left (-3 a^3 d^2 D-a^2 b d (C d-12 c D)+a b^2 \left (-3 B d^2-24 c^2 D+8 c C d\right )+b^3 \left (15 A d^2-12 B c d+8 c^2 C\right )\right )}{\sqrt {b c-a d}}-\frac {2 \left (a^3 d^3 D-a^2 b C d^3+a b^2 B d^3-\left (b^3 \left (5 A d^3-4 B c d^2-4 c^3 D+4 c^2 C d\right )\right )\right )}{\sqrt {c+d x}}}{2 b^3 (b c-a d)^2}+\frac {d^2 \sqrt {c+d x} \left (-7 a^3 d D+3 a^2 b (4 c D+C d)-a b^2 (8 c C-B d)+b^3 (4 B c-5 A d)\right )}{2 b^2 (b c-a d)^2 (-a d-b (c+d x)+b c)}}{2 d (b c-a d)}-\frac {A b^3-a \left (a^2 D-a b C+b^2 B\right )}{2 b^3 (a+b x)^2 \sqrt {c+d x} (b c-a d)}\) |
-1/2*(A*b^3 - a*(b^2*B - a*b*C + a^2*D))/(b^3*(b*c - a*d)*(a + b*x)^2*Sqrt [c + d*x]) + ((d^2*(b^3*(4*B*c - 5*A*d) - a*b^2*(8*c*C - B*d) - 7*a^3*d*D + 3*a^2*b*(C*d + 4*c*D))*Sqrt[c + d*x])/(2*b^2*(b*c - a*d)^2*(b*c - a*d - b*(c + d*x))) + ((-2*(a*b^2*B*d^3 - a^2*b*C*d^3 + a^3*d^3*D - b^3*(4*c^2*C *d - 4*B*c*d^2 + 5*A*d^3 - 4*c^3*D)))/Sqrt[c + d*x] - (Sqrt[b]*d*(b^3*(8*c ^2*C - 12*B*c*d + 15*A*d^2) - 3*a^3*d^2*D - a^2*b*d*(C*d - 12*c*D) + a*b^2 *(8*c*C*d - 3*B*d^2 - 24*c^2*D))*ArcTanh[(Sqrt[b]*Sqrt[c + d*x])/Sqrt[b*c - a*d]])/Sqrt[b*c - a*d])/(2*b^3*(b*c - a*d)^2))/(2*d*(b*c - a*d))
3.1.16.3.1 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_)^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[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2), x _Symbol] :> Simp[c*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(a*e*(m + 1))), x] + Simp[(a*d*(m + 1) - b*c*(m + 2*p + 3))/(a*e^2*(m + 1)) Int[(e*x)^(m + 2)* (a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] && !ILtQ[p, -1]
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[(x_)^(m_)*((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^ 4)^(p_.), x_Symbol] :> Simp[(-d)^(m/2 - 1)*(c*d^2 - b*d*e + a*e^2)^p*x*((d + e*x^2)^(q + 1)/(2*e^(2*p + m/2)*(q + 1))), x] + Simp[(-d)^(m/2 - 1)/(2*e^ (2*p)*(q + 1)) Int[x^m*(d + e*x^2)^(q + 1)*ExpandToSum[Together[(1/(d + e *x^2))*(2*(-d)^(-m/2 + 1)*e^(2*p)*(q + 1)*(a + b*x^2 + c*x^4)^p - ((c*d^2 - b*d*e + a*e^2)^p/(e^(m/2)*x^m))*(d + e*(2*q + 3)*x^2))], x], x], x] /; Fre eQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && IGtQ[p, 0] && ILtQ[q, -1] && ILtQ[m/2, 0]
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 = 1.96 (sec) , antiderivative size = 379, normalized size of antiderivative = 1.08
| method | result | size |
| pseudoelliptic | \(-\frac {15 \left (\left (\left (A \,d^{2}-\frac {4}{5} B c d +\frac {8}{15} C \,c^{2}\right ) b^{3}-\frac {a \left (B \,d^{2}-\frac {8}{3} C c d +8 D c^{2}\right ) b^{2}}{5}-\frac {a^{2} b d \left (C d -12 D c \right )}{15}-\frac {a^{3} d^{2} D}{5}\right ) \sqrt {d x +c}\, \left (b x +a \right )^{2} d \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (a d -b c \right ) b}}\right )+\frac {8 \sqrt {\left (a d -b c \right ) b}\, \left (\left (\frac {15 A \,d^{3} x^{2}}{8}+\frac {5 \left (-\frac {12 B x}{5}+A \right ) x c \,d^{2}}{8}-\frac {c^{2} \left (-4 C \,x^{2}+2 B x +A \right ) d}{4}-D c^{3} x^{2}\right ) b^{4}+\frac {9 a \left (\left (-\frac {1}{3} x^{2} B +\frac {25}{9} A x \right ) d^{3}+c \left (\frac {8}{9} C \,x^{2}-\frac {7}{3} B x +A \right ) d^{2}-\frac {2 c^{2} \left (-12 C x +B \right ) d}{9}-\frac {16 D c^{3} x}{9}\right ) b^{3}}{8}+a^{2} \left (\left (-\frac {5}{8} B x +A -\frac {1}{8} C \,x^{2}\right ) d^{3}-\frac {13 c \left (\frac {12}{13} D x^{2}-\frac {5}{13} C x +B \right ) d^{2}}{8}+\frac {7 c^{2} \left (-\frac {6 D x}{7}+C \right ) d}{4}-D c^{3}\right ) b^{2}+\frac {a^{3} \left (d x +c \right ) \left (\left (5 D x +C \right ) d -10 D c \right ) d b}{8}+\frac {3 D a^{4} d^{2} \left (d x +c \right )}{8}\right )}{15}\right )}{4 \sqrt {\left (a d -b c \right ) b}\, \sqrt {d x +c}\, \left (a d -b c \right )^{3} \left (b x +a \right )^{2} b^{2} d}\) | \(379\) |
| derivativedivides | \(\frac {-\frac {2 d \left (\frac {\frac {d \left (7 A \,b^{3} d -3 B a \,b^{2} d -4 B \,b^{3} c -C \,a^{2} b d +8 C a \,b^{2} c +5 a^{3} d D-12 D a^{2} b c \right ) \left (d x +c \right )^{\frac {3}{2}}}{8 b}+\frac {d \left (9 A a \,b^{3} d^{2}-9 A \,b^{4} c d -5 B \,a^{2} b^{2} d^{2}+B a \,b^{3} c d +4 B \,b^{4} c^{2}+C \,a^{3} b \,d^{2}+7 C \,a^{2} b^{2} c d -8 C a \,b^{3} c^{2}+3 D a^{4} d^{2}-15 D a^{3} b c d +12 D a^{2} b^{2} c^{2}\right ) \sqrt {d x +c}}{8 b^{2}}}{\left (\left (d x +c \right ) b +a d -b c \right )^{2}}+\frac {\left (15 A \,b^{3} d^{2}-3 B a \,b^{2} d^{2}-12 B \,b^{3} c d -a^{2} b C \,d^{2}+8 C a \,b^{2} c d +8 C \,b^{3} c^{2}-3 a^{3} d^{2} D+12 D a^{2} b c d -24 D a \,b^{2} c^{2}\right ) \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{8 b^{2} \sqrt {\left (a d -b c \right ) b}}\right )}{\left (a d -b c \right )^{3}}-\frac {2 \left (A \,d^{3}-B c \,d^{2}+C \,c^{2} d -D c^{3}\right )}{\left (a d -b c \right )^{3} \sqrt {d x +c}}}{d}\) | \(395\) |
| default | \(\frac {-\frac {2 d \left (\frac {\frac {d \left (7 A \,b^{3} d -3 B a \,b^{2} d -4 B \,b^{3} c -C \,a^{2} b d +8 C a \,b^{2} c +5 a^{3} d D-12 D a^{2} b c \right ) \left (d x +c \right )^{\frac {3}{2}}}{8 b}+\frac {d \left (9 A a \,b^{3} d^{2}-9 A \,b^{4} c d -5 B \,a^{2} b^{2} d^{2}+B a \,b^{3} c d +4 B \,b^{4} c^{2}+C \,a^{3} b \,d^{2}+7 C \,a^{2} b^{2} c d -8 C a \,b^{3} c^{2}+3 D a^{4} d^{2}-15 D a^{3} b c d +12 D a^{2} b^{2} c^{2}\right ) \sqrt {d x +c}}{8 b^{2}}}{\left (\left (d x +c \right ) b +a d -b c \right )^{2}}+\frac {\left (15 A \,b^{3} d^{2}-3 B a \,b^{2} d^{2}-12 B \,b^{3} c d -a^{2} b C \,d^{2}+8 C a \,b^{2} c d +8 C \,b^{3} c^{2}-3 a^{3} d^{2} D+12 D a^{2} b c d -24 D a \,b^{2} c^{2}\right ) \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{8 b^{2} \sqrt {\left (a d -b c \right ) b}}\right )}{\left (a d -b c \right )^{3}}-\frac {2 \left (A \,d^{3}-B c \,d^{2}+C \,c^{2} d -D c^{3}\right )}{\left (a d -b c \right )^{3} \sqrt {d x +c}}}{d}\) | \(395\) |
-15/4/((a*d-b*c)*b)^(1/2)*(((A*d^2-4/5*B*c*d+8/15*C*c^2)*b^3-1/5*a*(B*d^2- 8/3*C*c*d+8*D*c^2)*b^2-1/15*a^2*b*d*(C*d-12*D*c)-1/5*a^3*d^2*D)*(d*x+c)^(1 /2)*(b*x+a)^2*d*arctan(b*(d*x+c)^(1/2)/((a*d-b*c)*b)^(1/2))+8/15*((a*d-b*c )*b)^(1/2)*((15/8*A*d^3*x^2+5/8*(-12/5*B*x+A)*x*c*d^2-1/4*c^2*(-4*C*x^2+2* B*x+A)*d-D*c^3*x^2)*b^4+9/8*a*((-1/3*x^2*B+25/9*A*x)*d^3+c*(8/9*C*x^2-7/3* B*x+A)*d^2-2/9*c^2*(-12*C*x+B)*d-16/9*D*c^3*x)*b^3+a^2*((-5/8*B*x+A-1/8*C* x^2)*d^3-13/8*c*(12/13*D*x^2-5/13*C*x+B)*d^2+7/4*c^2*(-6/7*D*x+C)*d-D*c^3) *b^2+1/8*a^3*(d*x+c)*((5*D*x+C)*d-10*D*c)*d*b+3/8*D*a^4*d^2*(d*x+c)))/(d*x +c)^(1/2)/(a*d-b*c)^3/(b*x+a)^2/b^2/d
Leaf count of result is larger than twice the leaf count of optimal. 1290 vs. \(2 (327) = 654\).
Time = 0.39 (sec) , antiderivative size = 2594, normalized size of antiderivative = 7.41 \[ \int \frac {A+B x+C x^2+D x^3}{(a+b x)^3 (c+d x)^{3/2}} \, dx=\text {Too large to display} \]
[-1/8*(((3*D*a^5*c + (C*a^4*b + 3*B*a^3*b^2 - 15*A*a^2*b^3)*c)*d^3 + ((3*D *a^3*b^2 + C*a^2*b^3 + 3*B*a*b^4 - 15*A*b^5)*d^4 - 4*(3*D*a^2*b^3*c + (2*C *a*b^4 - 3*B*b^5)*c)*d^3 + 8*(3*D*a*b^4*c^2 - C*b^5*c^2)*d^2)*x^3 - 4*(3*D *a^4*b*c^2 + (2*C*a^3*b^2 - 3*B*a^2*b^3)*c^2)*d^2 + (2*(3*D*a^4*b + C*a^3* b^2 + 3*B*a^2*b^3 - 15*A*a*b^4)*d^4 - 3*(7*D*a^3*b^2*c + (5*C*a^2*b^3 - 9* B*a*b^4 + 5*A*b^5)*c)*d^3 + 12*(3*D*a^2*b^3*c^2 - (2*C*a*b^4 - B*b^5)*c^2) *d^2 + 8*(3*D*a*b^4*c^3 - C*b^5*c^3)*d)*x^2 + 8*(3*D*a^3*b^2*c^3 - C*a^2*b ^3*c^3)*d - (24*(C*a^2*b^3 - B*a*b^4)*c^2*d^2 - (3*D*a^5 + C*a^4*b + 3*B*a ^3*b^2 - 15*A*a^2*b^3)*d^4 + 6*(D*a^4*b*c + (C*a^3*b^2 - 3*B*a^2*b^3 + 5*A *a*b^4)*c)*d^3 - 16*(3*D*a^2*b^3*c^3 - C*a*b^4*c^3)*d)*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*(8*D*a^2*b^4*c^4 + 8*A*a^3*b^3*d^4 + (3*D*a^5*b*c + (C*a^4*b^2 - 13*B*a^3*b^3 + A*a^2*b^4)*c)*d^3 - (13*D*a^4*b^2*c^2 - (13*C*a^3*b^3 + 11* B*a^2*b^4 - 11*A*a*b^5)*c^2)*d^2 + (8*D*b^6*c^4 + (5*D*a^4*b^2 - C*a^3*b^3 - 3*B*a^2*b^4 + 15*A*a*b^5)*d^4 - (17*D*a^3*b^3*c - 3*(3*C*a^2*b^4 - 3*B* a*b^5 - 5*A*b^6)*c)*d^3 + 12*(D*a^2*b^4*c^2 + B*b^6*c^2)*d^2 - 8*(D*a*b^5* c^3 + C*b^6*c^3)*d)*x^2 + 2*(D*a^3*b^3*c^3 - (7*C*a^2*b^4 - B*a*b^5 - A*b^ 6)*c^3)*d + (16*D*a*b^5*c^4 + (3*D*a^5*b + C*a^4*b^2 - 5*B*a^3*b^3 + 25*A* a^2*b^4)*d^4 - 4*(2*D*a^4*b^2*c - (C*a^3*b^3 - 4*B*a^2*b^4 - 5*A*a*b^5)*c) *d^3 - (7*D*a^3*b^3*c^2 - (19*C*a^2*b^4 + 17*B*a*b^5 - 5*A*b^6)*c^2)*d^...
Timed out. \[ \int \frac {A+B x+C x^2+D x^3}{(a+b x)^3 (c+d x)^{3/2}} \, dx=\text {Timed out} \]
Exception generated. \[ \int \frac {A+B x+C x^2+D x^3}{(a+b x)^3 (c+d x)^{3/2}} \, dx=\text {Exception raised: ValueError} \]
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 = 0.29 (sec) , antiderivative size = 617, normalized size of antiderivative = 1.76 \[ \int \frac {A+B x+C x^2+D x^3}{(a+b x)^3 (c+d x)^{3/2}} \, dx=-\frac {{\left (24 \, D a b^{2} c^{2} - 8 \, C b^{3} c^{2} - 12 \, D a^{2} b c d - 8 \, C a b^{2} c d + 12 \, B b^{3} c d + 3 \, D a^{3} d^{2} + C a^{2} b d^{2} + 3 \, B a b^{2} d^{2} - 15 \, A b^{3} d^{2}\right )} \arctan \left (\frac {\sqrt {d x + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{4 \, {\left (b^{5} c^{3} - 3 \, a b^{4} c^{2} d + 3 \, a^{2} b^{3} c d^{2} - a^{3} b^{2} d^{3}\right )} \sqrt {-b^{2} c + a b d}} - \frac {2 \, {\left (D c^{3} - C c^{2} d + B c d^{2} - A d^{3}\right )}}{{\left (b^{3} c^{3} d - 3 \, a b^{2} c^{2} d^{2} + 3 \, a^{2} b c d^{3} - a^{3} d^{4}\right )} \sqrt {d x + c}} - \frac {12 \, {\left (d x + c\right )}^{\frac {3}{2}} D a^{2} b^{2} c d - 8 \, {\left (d x + c\right )}^{\frac {3}{2}} C a b^{3} c d + 4 \, {\left (d x + c\right )}^{\frac {3}{2}} B b^{4} c d - 12 \, \sqrt {d x + c} D a^{2} b^{2} c^{2} d + 8 \, \sqrt {d x + c} C a b^{3} c^{2} d - 4 \, \sqrt {d x + c} B b^{4} c^{2} d - 5 \, {\left (d x + c\right )}^{\frac {3}{2}} D a^{3} b d^{2} + {\left (d x + c\right )}^{\frac {3}{2}} C a^{2} b^{2} d^{2} + 3 \, {\left (d x + c\right )}^{\frac {3}{2}} B a b^{3} d^{2} - 7 \, {\left (d x + c\right )}^{\frac {3}{2}} A b^{4} d^{2} + 15 \, \sqrt {d x + c} D a^{3} b c d^{2} - 7 \, \sqrt {d x + c} C a^{2} b^{2} c d^{2} - \sqrt {d x + c} B a b^{3} c d^{2} + 9 \, \sqrt {d x + c} A b^{4} c d^{2} - 3 \, \sqrt {d x + c} D a^{4} d^{3} - \sqrt {d x + c} C a^{3} b d^{3} + 5 \, \sqrt {d x + c} B a^{2} b^{2} d^{3} - 9 \, \sqrt {d x + c} A a b^{3} d^{3}}{4 \, {\left (b^{5} c^{3} - 3 \, a b^{4} c^{2} d + 3 \, a^{2} b^{3} c d^{2} - a^{3} b^{2} d^{3}\right )} {\left ({\left (d x + c\right )} b - b c + a d\right )}^{2}} \]
-1/4*(24*D*a*b^2*c^2 - 8*C*b^3*c^2 - 12*D*a^2*b*c*d - 8*C*a*b^2*c*d + 12*B *b^3*c*d + 3*D*a^3*d^2 + C*a^2*b*d^2 + 3*B*a*b^2*d^2 - 15*A*b^3*d^2)*arcta n(sqrt(d*x + c)*b/sqrt(-b^2*c + a*b*d))/((b^5*c^3 - 3*a*b^4*c^2*d + 3*a^2* b^3*c*d^2 - a^3*b^2*d^3)*sqrt(-b^2*c + a*b*d)) - 2*(D*c^3 - C*c^2*d + B*c* d^2 - A*d^3)/((b^3*c^3*d - 3*a*b^2*c^2*d^2 + 3*a^2*b*c*d^3 - a^3*d^4)*sqrt (d*x + c)) - 1/4*(12*(d*x + c)^(3/2)*D*a^2*b^2*c*d - 8*(d*x + c)^(3/2)*C*a *b^3*c*d + 4*(d*x + c)^(3/2)*B*b^4*c*d - 12*sqrt(d*x + c)*D*a^2*b^2*c^2*d + 8*sqrt(d*x + c)*C*a*b^3*c^2*d - 4*sqrt(d*x + c)*B*b^4*c^2*d - 5*(d*x + c )^(3/2)*D*a^3*b*d^2 + (d*x + c)^(3/2)*C*a^2*b^2*d^2 + 3*(d*x + c)^(3/2)*B* a*b^3*d^2 - 7*(d*x + c)^(3/2)*A*b^4*d^2 + 15*sqrt(d*x + c)*D*a^3*b*c*d^2 - 7*sqrt(d*x + c)*C*a^2*b^2*c*d^2 - sqrt(d*x + c)*B*a*b^3*c*d^2 + 9*sqrt(d* x + c)*A*b^4*c*d^2 - 3*sqrt(d*x + c)*D*a^4*d^3 - sqrt(d*x + c)*C*a^3*b*d^3 + 5*sqrt(d*x + c)*B*a^2*b^2*d^3 - 9*sqrt(d*x + c)*A*a*b^3*d^3)/((b^5*c^3 - 3*a*b^4*c^2*d + 3*a^2*b^3*c*d^2 - a^3*b^2*d^3)*((d*x + c)*b - b*c + a*d) ^2)
Timed out. \[ \int \frac {A+B x+C x^2+D x^3}{(a+b x)^3 (c+d x)^{3/2}} \, dx=\int \frac {A+B\,x+C\,x^2+x^3\,D}{{\left (a+b\,x\right )}^3\,{\left (c+d\,x\right )}^{3/2}} \,d x \]