\(\int \frac {A+B x+C x^2+D x^3}{(a+b x)^3 \sqrt {c+d x}} \, dx\) [7]

Optimal result

Integrand size = 32, antiderivative size = 279 \[ \int \frac {A+B x+C x^2+D x^3}{(a+b x)^3 \sqrt {c+d x}} \, dx=\frac {2 D \sqrt {c+d x}}{b^3 d}-\frac {\left (A b^3-a \left (b^2 B-a b C+a^2 D\right )\right ) \sqrt {c+d x}}{2 b^3 (b c-a d) (a+b x)^2}-\frac {\left (b^3 (4 B c-3 A d)-a b^2 (8 c C+B d)-9 a^3 d D+a^2 b (5 C d+12 c D)\right ) \sqrt {c+d x}}{4 b^3 (b c-a d)^2 (a+b x)}-\frac {\left (b^3 \left (8 c^2 C-4 B c d+3 A d^2\right )-15 a^3 d^2 D+3 a^2 b d (C d+12 c D)-a b^2 \left (8 c C d-B d^2+24 c^2 D\right )\right ) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{4 b^{7/2} (b c-a d)^{5/2}} \]

[Out]

-1/4*(b^3*(3*A*d^2-4*B*c*d+8*C*c^2)-15*a^3*d^2*D+3*a^2*b*d*(C*d+12*D*c)-a*b^2*(-B*d^2+8*C*c*d+24*D*c^2))*arcta
nh(b^(1/2)*(d*x+c)^(1/2)/(-a*d+b*c)^(1/2))/b^(7/2)/(-a*d+b*c)^(5/2)+2*D*(d*x+c)^(1/2)/b^3/d-1/2*(A*b^3-a*(B*b^
2-C*a*b+D*a^2))*(d*x+c)^(1/2)/b^3/(-a*d+b*c)/(b*x+a)^2-1/4*(b^3*(-3*A*d+4*B*c)-a*b^2*(B*d+8*C*c)-9*a^3*d*D+a^2
*b*(5*C*d+12*D*c))*(d*x+c)^(1/2)/b^3/(-a*d+b*c)^2/(b*x+a)

Rubi [A] (verified)

Time = 0.49 (sec) , antiderivative size = 279, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.156, Rules used = {1635, 911, 1171, 396, 214} \[ \int \frac {A+B x+C x^2+D x^3}{(a+b x)^3 \sqrt {c+d x}} \, dx=-\frac {\sqrt {c+d x} \left (A b^3-a \left (a^2 D-a b C+b^2 B\right )\right )}{2 b^3 (a+b x)^2 (b c-a d)}-\frac {\text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right ) \left (-15 a^3 d^2 D+3 a^2 b d (12 c D+C d)-a b^2 \left (-B d^2+24 c^2 D+8 c C d\right )+b^3 \left (3 A d^2-4 B c d+8 c^2 C\right )\right )}{4 b^{7/2} (b c-a d)^{5/2}}-\frac {\sqrt {c+d x} \left (-9 a^3 d D+a^2 b (12 c D+5 C d)-a b^2 (B d+8 c C)+b^3 (4 B c-3 A d)\right )}{4 b^3 (a+b x) (b c-a d)^2}+\frac {2 D \sqrt {c+d x}}{b^3 d} \]

[In]

Int[(A + B*x + C*x^2 + D*x^3)/((a + b*x)^3*Sqrt[c + d*x]),x]

[Out]

(2*D*Sqrt[c + d*x])/(b^3*d) - ((A*b^3 - a*(b^2*B - a*b*C + a^2*D))*Sqrt[c + d*x])/(2*b^3*(b*c - a*d)*(a + b*x)
^2) - ((b^3*(4*B*c - 3*A*d) - a*b^2*(8*c*C + B*d) - 9*a^3*d*D + a^2*b*(5*C*d + 12*c*D))*Sqrt[c + d*x])/(4*b^3*
(b*c - a*d)^2*(a + b*x)) - ((b^3*(8*c^2*C - 4*B*c*d + 3*A*d^2) - 15*a^3*d^2*D + 3*a^2*b*d*(C*d + 12*c*D) - a*b
^2*(8*c*C*d - B*d^2 + 24*c^2*D))*ArcTanh[(Sqrt[b]*Sqrt[c + d*x])/Sqrt[b*c - a*d]])/(4*b^(7/2)*(b*c - a*d)^(5/2
))

Rule 214

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]

Rule 396

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[d*x*((a + b*x^n)^(p + 1)/(b*(n*(
p + 1) + 1))), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(b*(n*(p + 1) + 1)), Int[(a + b*x^n)^p, x], x] /; FreeQ[{
a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && NeQ[n*(p + 1) + 1, 0]

Rule 911

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :
> With[{q = Denominator[m]}, Dist[q/e, Subst[Int[x^(q*(m + 1) - 1)*((e*f - d*g)/e + g*(x^q/e))^n*((c*d^2 - b*d
*e + a*e^2)/e^2 - (2*c*d - b*e)*(x^q/e^2) + c*(x^(2*q)/e^2))^p, x], x, (d + e*x)^(1/q)], x]] /; FreeQ[{a, b, c
, d, e, f, g}, x] && NeQ[e*f - d*g, 0] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegersQ[n,
p] && FractionQ[m]

Rule 1171

Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> With[{Qx = PolynomialQ
uotient[(a + b*x^2 + c*x^4)^p, d + e*x^2, x], R = Coeff[PolynomialRemainder[(a + b*x^2 + c*x^4)^p, d + e*x^2,
x], x, 0]}, Simp[(-R)*x*((d + e*x^2)^(q + 1)/(2*d*(q + 1))), x] + Dist[1/(2*d*(q + 1)), Int[(d + e*x^2)^(q + 1
)*ExpandToSum[2*d*(q + 1)*Qx + R*(2*q + 3), x], x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] &&
NeQ[c*d^2 - b*d*e + a*e^2, 0] && IGtQ[p, 0] && LtQ[q, -1]

Rule 1635

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] + Dist[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]] /; FreeQ[{a, b, c, d, n}, x] && PolyQ[Px, x] && ILtQ[m, -1] && GtQ[Expo
n[Px, x], 2]

Rubi steps \begin{align*} \text {integral}& = -\frac {\left (A b^3-a \left (b^2 B-a b C+a^2 D\right )\right ) \sqrt {c+d x}}{2 b^3 (b c-a d) (a+b x)^2}-\frac {\int \frac {-\frac {b^3 (4 B c-3 A d)-a b^2 (4 c C+B d)-a^3 d D+a^2 b (C d+4 c D)}{2 b^3}-\frac {2 (b c-a d) (b C-a D) x}{b^2}-2 \left (c-\frac {a d}{b}\right ) D x^2}{(a+b x)^2 \sqrt {c+d x}} \, dx}{2 (b c-a d)} \\ & = -\frac {\left (A b^3-a \left (b^2 B-a b C+a^2 D\right )\right ) \sqrt {c+d x}}{2 b^3 (b c-a d) (a+b x)^2}-\frac {\text {Subst}\left (\int \frac {\frac {-2 c^2 \left (c-\frac {a d}{b}\right ) D+\frac {2 c d (b c-a d) (b C-a D)}{b^2}-\frac {d^2 \left (b^3 (4 B c-3 A d)-a b^2 (4 c C+B d)-a^3 d D+a^2 b (C d+4 c D)\right )}{2 b^3}}{d^2}-\frac {\left (-4 c \left (c-\frac {a d}{b}\right ) D+\frac {2 d (b c-a d) (b C-a D)}{b^2}\right ) x^2}{d^2}-\frac {2 \left (c-\frac {a d}{b}\right ) D x^4}{d^2}}{\left (\frac {-b c+a d}{d}+\frac {b x^2}{d}\right )^2} \, dx,x,\sqrt {c+d x}\right )}{d (b c-a d)} \\ & = -\frac {\left (A b^3-a \left (b^2 B-a b C+a^2 D\right )\right ) \sqrt {c+d x}}{2 b^3 (b c-a d) (a+b x)^2}-\frac {\left (b^3 (4 B c-3 A d)-a b^2 (8 c C+B d)-9 a^3 d D+a^2 b (5 C d+12 c D)\right ) \sqrt {c+d x}}{4 b^3 (b c-a d)^2 (a+b x)}-\frac {\text {Subst}\left (\int \frac {\frac {1}{2} \left (4 B c-\frac {8 c^2 C}{d}-3 A d+\frac {a (8 c C-B d)}{b}+\frac {8 c^3 D}{d^2}+\frac {7 a^3 d D}{b^3}-\frac {3 a^2 (C d+4 c D)}{b^2}\right )-\frac {4 (b c-a d)^2 D x^2}{b^2 d^2}}{\frac {-b c+a d}{d}+\frac {b x^2}{d}} \, dx,x,\sqrt {c+d x}\right )}{2 (b c-a d)^2} \\ & = \frac {2 D \sqrt {c+d x}}{b^3 d}-\frac {\left (A b^3-a \left (b^2 B-a b C+a^2 D\right )\right ) \sqrt {c+d x}}{2 b^3 (b c-a d) (a+b x)^2}-\frac {\left (b^3 (4 B c-3 A d)-a b^2 (8 c C+B d)-9 a^3 d D+a^2 b (5 C d+12 c D)\right ) \sqrt {c+d x}}{4 b^3 (b c-a d)^2 (a+b x)}+\frac {\left (b^3 \left (8 c^2 C-4 B c d+3 A d^2\right )-15 a^3 d^2 D+3 a^2 b d (C d+12 c D)-a b^2 \left (8 c C d-B d^2+24 c^2 D\right )\right ) \text {Subst}\left (\int \frac {1}{\frac {-b c+a d}{d}+\frac {b x^2}{d}} \, dx,x,\sqrt {c+d x}\right )}{4 b^3 d (b c-a d)^2} \\ & = \frac {2 D \sqrt {c+d x}}{b^3 d}-\frac {\left (A b^3-a \left (b^2 B-a b C+a^2 D\right )\right ) \sqrt {c+d x}}{2 b^3 (b c-a d) (a+b x)^2}-\frac {\left (b^3 (4 B c-3 A d)-a b^2 (8 c C+B d)-9 a^3 d D+a^2 b (5 C d+12 c D)\right ) \sqrt {c+d x}}{4 b^3 (b c-a d)^2 (a+b x)}-\frac {\left (b^3 \left (8 c^2 C-4 B c d+3 A d^2\right )-15 a^3 d^2 D+3 a^2 b d (C d+12 c D)-a b^2 \left (8 c C d-B d^2+24 c^2 D\right )\right ) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{4 b^{7/2} (b c-a d)^{5/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.86 (sec) , antiderivative size = 298, normalized size of antiderivative = 1.07 \[ \int \frac {A+B x+C x^2+D x^3}{(a+b x)^3 \sqrt {c+d x}} \, dx=\frac {\frac {\sqrt {b} \sqrt {c+d x} \left (15 a^4 d^2 D+A b^3 d (-2 b c+5 a d+3 b d x)+4 b^4 c x (-B d+2 c D x)+a^3 b d (-3 C d-26 c D+25 d D x)+a b^3 (B d (-2 c+d x)+8 c x (C d+2 c D-2 d D x))+a^2 b^2 \left (8 c^2 D+c (6 C d-44 d D x)-d^2 \left (B+5 C x-8 D x^2\right )\right )\right )}{d (b c-a d)^2 (a+b x)^2}+\frac {\left (b^3 \left (8 c^2 C-4 B c d+3 A d^2\right )-15 a^3 d^2 D+3 a^2 b d (C d+12 c D)+a b^2 \left (-8 c C d+B d^2-24 c^2 D\right )\right ) \arctan \left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {-b c+a d}}\right )}{(-b c+a d)^{5/2}}}{4 b^{7/2}} \]

[In]

Integrate[(A + B*x + C*x^2 + D*x^3)/((a + b*x)^3*Sqrt[c + d*x]),x]

[Out]

((Sqrt[b]*Sqrt[c + d*x]*(15*a^4*d^2*D + A*b^3*d*(-2*b*c + 5*a*d + 3*b*d*x) + 4*b^4*c*x*(-(B*d) + 2*c*D*x) + a^
3*b*d*(-3*C*d - 26*c*D + 25*d*D*x) + a*b^3*(B*d*(-2*c + d*x) + 8*c*x*(C*d + 2*c*D - 2*d*D*x)) + a^2*b^2*(8*c^2
*D + c*(6*C*d - 44*d*D*x) - d^2*(B + 5*C*x - 8*D*x^2))))/(d*(b*c - a*d)^2*(a + b*x)^2) + ((b^3*(8*c^2*C - 4*B*
c*d + 3*A*d^2) - 15*a^3*d^2*D + 3*a^2*b*d*(C*d + 12*c*D) + a*b^2*(-8*c*C*d + B*d^2 - 24*c^2*D))*ArcTan[(Sqrt[b
]*Sqrt[c + d*x])/Sqrt[-(b*c) + a*d]])/(-(b*c) + a*d)^(5/2))/(4*b^(7/2))

Maple [A] (verified)

Time = 1.84 (sec) , antiderivative size = 289, normalized size of antiderivative = 1.04

method	result	size
pseudoelliptic	\(\frac {5 \left (\left (\frac {3 A \,b^{4} x}{5}+a \left (\frac {B x}{5}+A \right ) b^{3}-\frac {a^{2} \left (-8 D x^{2}+5 C x +B \right ) b^{2}}{5}-\frac {3 a^{3} \left (-\frac {25 D x}{3}+C \right ) b}{5}+3 D a^{4}\right ) d^{2}-\frac {2 b c \left (\left (2 B x +A \right ) b^{3}+a \left (8 D x^{2}-4 C x +B \right ) b^{2}-3 a^{2} \left (-\frac {22 D x}{3}+C \right ) b +13 D a^{3}\right ) d}{5}+\frac {8 D b^{2} c^{2} \left (b x +a \right )^{2}}{5}\right ) \sqrt {\left (a d -b c \right ) b}\, \sqrt {d x +c}+3 \left (b x +a \right )^{2} \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (a d -b c \right ) b}}\right ) \left (\left (b^{3} A +\frac {1}{3} a \,b^{2} B +C \,a^{2} b -5 D a^{3}\right ) d^{2}-\frac {4 b c \left (B \,b^{2}+2 C a b -9 D a^{2}\right ) d}{3}+\frac {8 b^{2} c^{2} \left (C b -3 D a \right )}{3}\right ) d}{4 \sqrt {\left (a d -b c \right ) b}\, \left (a d -b c \right )^{2} b^{3} \left (b x +a \right )^{2} d}\)	\(289\)
derivativedivides	\(\frac {\frac {2 D \sqrt {d x +c}}{b^{3}}+\frac {2 d \left (\frac {\frac {b d \left (3 A \,b^{3} d +B a \,b^{2} d -4 B \,b^{3} c -5 C \,a^{2} b d +8 C a \,b^{2} c +9 a^{3} d D-12 D a^{2} b c \right ) \left (d x +c \right )^{\frac {3}{2}}}{8 a^{2} d^{2}-16 a b c d +8 b^{2} c^{2}}+\frac {\left (5 A \,b^{3} d -B a \,b^{2} d -4 B \,b^{3} c -3 C \,a^{2} b d +8 C a \,b^{2} c +7 a^{3} d D-12 D a^{2} b c \right ) d \sqrt {d x +c}}{8 a d -8 b c}}{\left (\left (d x +c \right ) b +a d -b c \right )^{2}}+\frac {\left (3 A \,b^{3} d^{2}+B a \,b^{2} d^{2}-4 B \,b^{3} c d +3 a^{2} b C \,d^{2}-8 C a \,b^{2} c d +8 C \,b^{3} c^{2}-15 a^{3} d^{2} D+36 D a^{2} b c d -24 D a \,b^{2} c^{2}\right ) \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{8 \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) \sqrt {\left (a d -b c \right ) b}}\right )}{b^{3}}}{d}\)	\(347\)
default	\(\frac {\frac {2 D \sqrt {d x +c}}{b^{3}}+\frac {2 d \left (\frac {\frac {b d \left (3 A \,b^{3} d +B a \,b^{2} d -4 B \,b^{3} c -5 C \,a^{2} b d +8 C a \,b^{2} c +9 a^{3} d D-12 D a^{2} b c \right ) \left (d x +c \right )^{\frac {3}{2}}}{8 a^{2} d^{2}-16 a b c d +8 b^{2} c^{2}}+\frac {\left (5 A \,b^{3} d -B a \,b^{2} d -4 B \,b^{3} c -3 C \,a^{2} b d +8 C a \,b^{2} c +7 a^{3} d D-12 D a^{2} b c \right ) d \sqrt {d x +c}}{8 a d -8 b c}}{\left (\left (d x +c \right ) b +a d -b c \right )^{2}}+\frac {\left (3 A \,b^{3} d^{2}+B a \,b^{2} d^{2}-4 B \,b^{3} c d +3 a^{2} b C \,d^{2}-8 C a \,b^{2} c d +8 C \,b^{3} c^{2}-15 a^{3} d^{2} D+36 D a^{2} b c d -24 D a \,b^{2} c^{2}\right ) \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{8 \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) \sqrt {\left (a d -b c \right ) b}}\right )}{b^{3}}}{d}\)	\(347\)

[In]

int((D*x^3+C*x^2+B*x+A)/(b*x+a)^3/(d*x+c)^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/4*(5*((3/5*A*b^4*x+a*(1/5*B*x+A)*b^3-1/5*a^2*(-8*D*x^2+5*C*x+B)*b^2-3/5*a^3*(-25/3*D*x+C)*b+3*D*a^4)*d^2-2/5
*b*c*((2*B*x+A)*b^3+a*(8*D*x^2-4*C*x+B)*b^2-3*a^2*(-22/3*D*x+C)*b+13*D*a^3)*d+8/5*D*b^2*c^2*(b*x+a)^2)*((a*d-b
*c)*b)^(1/2)*(d*x+c)^(1/2)+3*(b*x+a)^2*arctan(b*(d*x+c)^(1/2)/((a*d-b*c)*b)^(1/2))*((b^3*A+1/3*a*b^2*B+C*a^2*b
-5*D*a^3)*d^2-4/3*b*c*(B*b^2+2*C*a*b-9*D*a^2)*d+8/3*b^2*c^2*(C*b-3*D*a))*d)/((a*d-b*c)*b)^(1/2)/(a*d-b*c)^2/b^
3/(b*x+a)^2/d

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 824 vs. \(2 (258) = 516\).

Time = 0.32 (sec) , antiderivative size = 1661, normalized size of antiderivative = 5.95 \[ \int \frac {A+B x+C x^2+D x^3}{(a+b x)^3 \sqrt {c+d x}} \, dx=\text {Too large to display} \]

[In]

integrate((D*x^3+C*x^2+B*x+A)/(b*x+a)^3/(d*x+c)^(1/2),x, algorithm="fricas")

[Out]

[1/8*(((15*D*a^5 - 3*C*a^4*b - B*a^3*b^2 - 3*A*a^2*b^3)*d^3 - 4*(9*D*a^4*b*c - (2*C*a^3*b^2 + B*a^2*b^3)*c)*d^
2 + ((15*D*a^3*b^2 - 3*C*a^2*b^3 - B*a*b^4 - 3*A*b^5)*d^3 - 4*(9*D*a^2*b^3*c - (2*C*a*b^4 + B*b^5)*c)*d^2 + 8*
(3*D*a*b^4*c^2 - C*b^5*c^2)*d)*x^2 + 8*(3*D*a^3*b^2*c^2 - C*a^2*b^3*c^2)*d + 2*((15*D*a^4*b - 3*C*a^3*b^2 - B*
a^2*b^3 - 3*A*a*b^4)*d^3 - 4*(9*D*a^3*b^2*c - (2*C*a^2*b^3 + B*a*b^4)*c)*d^2 + 8*(3*D*a^2*b^3*c^2 - C*a*b^4*c^
2)*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^3 - (15*D*a^5*b - 3*C*a^4*b^2 - B*a^3*b^3 + 5*A*a^2*b^4)*d^3 + (41*D*a^4*b^2*c - (9*C*a^3*b^3 - B
*a^2*b^4 - 7*A*a*b^5)*c)*d^2 + 8*(D*b^6*c^3 - 3*D*a*b^5*c^2*d + 3*D*a^2*b^4*c*d^2 - D*a^3*b^3*d^3)*x^2 - 2*(17
*D*a^3*b^3*c^2 - (3*C*a^2*b^4 - B*a*b^5 - A*b^6)*c^2)*d + (16*D*a*b^5*c^3 - (25*D*a^4*b^2 - 5*C*a^3*b^3 + B*a^
2*b^4 + 3*A*a*b^5)*d^3 + (69*D*a^3*b^3*c - (13*C*a^2*b^4 - 5*B*a*b^5 - 3*A*b^6)*c)*d^2 - 4*(15*D*a^2*b^4*c^2 -
(2*C*a*b^5 - B*b^6)*c^2)*d)*x)*sqrt(d*x + c))/(a^2*b^7*c^3*d - 3*a^3*b^6*c^2*d^2 + 3*a^4*b^5*c*d^3 - a^5*b^4*
d^4 + (b^9*c^3*d - 3*a*b^8*c^2*d^2 + 3*a^2*b^7*c*d^3 - a^3*b^6*d^4)*x^2 + 2*(a*b^8*c^3*d - 3*a^2*b^7*c^2*d^2 +
3*a^3*b^6*c*d^3 - a^4*b^5*d^4)*x), -1/4*(((15*D*a^5 - 3*C*a^4*b - B*a^3*b^2 - 3*A*a^2*b^3)*d^3 - 4*(9*D*a^4*b
*c - (2*C*a^3*b^2 + B*a^2*b^3)*c)*d^2 + ((15*D*a^3*b^2 - 3*C*a^2*b^3 - B*a*b^4 - 3*A*b^5)*d^3 - 4*(9*D*a^2*b^3
*c - (2*C*a*b^4 + B*b^5)*c)*d^2 + 8*(3*D*a*b^4*c^2 - C*b^5*c^2)*d)*x^2 + 8*(3*D*a^3*b^2*c^2 - C*a^2*b^3*c^2)*d
+ 2*((15*D*a^4*b - 3*C*a^3*b^2 - B*a^2*b^3 - 3*A*a*b^4)*d^3 - 4*(9*D*a^3*b^2*c - (2*C*a^2*b^3 + B*a*b^4)*c)*d
^2 + 8*(3*D*a^2*b^3*c^2 - C*a*b^4*c^2)*d)*x)*sqrt(-b^2*c + a*b*d)*arctan(sqrt(-b^2*c + a*b*d)*sqrt(d*x + c)/(b
*d*x + b*c)) - (8*D*a^2*b^4*c^3 - (15*D*a^5*b - 3*C*a^4*b^2 - B*a^3*b^3 + 5*A*a^2*b^4)*d^3 + (41*D*a^4*b^2*c -
(9*C*a^3*b^3 - B*a^2*b^4 - 7*A*a*b^5)*c)*d^2 + 8*(D*b^6*c^3 - 3*D*a*b^5*c^2*d + 3*D*a^2*b^4*c*d^2 - D*a^3*b^3
*d^3)*x^2 - 2*(17*D*a^3*b^3*c^2 - (3*C*a^2*b^4 - B*a*b^5 - A*b^6)*c^2)*d + (16*D*a*b^5*c^3 - (25*D*a^4*b^2 - 5
*C*a^3*b^3 + B*a^2*b^4 + 3*A*a*b^5)*d^3 + (69*D*a^3*b^3*c - (13*C*a^2*b^4 - 5*B*a*b^5 - 3*A*b^6)*c)*d^2 - 4*(1
5*D*a^2*b^4*c^2 - (2*C*a*b^5 - B*b^6)*c^2)*d)*x)*sqrt(d*x + c))/(a^2*b^7*c^3*d - 3*a^3*b^6*c^2*d^2 + 3*a^4*b^5
*c*d^3 - a^5*b^4*d^4 + (b^9*c^3*d - 3*a*b^8*c^2*d^2 + 3*a^2*b^7*c*d^3 - a^3*b^6*d^4)*x^2 + 2*(a*b^8*c^3*d - 3*
a^2*b^7*c^2*d^2 + 3*a^3*b^6*c*d^3 - a^4*b^5*d^4)*x)]

Sympy [F(-1)]

Timed out. \[ \int \frac {A+B x+C x^2+D x^3}{(a+b x)^3 \sqrt {c+d x}} \, dx=\text {Timed out} \]

[In]

integrate((D*x**3+C*x**2+B*x+A)/(b*x+a)**3/(d*x+c)**(1/2),x)

[Out]

Timed out

Maxima [F(-2)]

Exception generated. \[ \int \frac {A+B x+C x^2+D x^3}{(a+b x)^3 \sqrt {c+d x}} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate((D*x^3+C*x^2+B*x+A)/(b*x+a)^3/(d*x+c)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(a*d-b*c>0)', see `assume?` for
more detail

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 529 vs. \(2 (258) = 516\).

Time = 0.30 (sec) , antiderivative size = 529, normalized size of antiderivative = 1.90 \[ \int \frac {A+B x+C x^2+D x^3}{(a+b x)^3 \sqrt {c+d x}} \, dx=-\frac {{\left (24 \, D a b^{2} c^{2} - 8 \, C b^{3} c^{2} - 36 \, D a^{2} b c d + 8 \, C a b^{2} c d + 4 \, B b^{3} c d + 15 \, D a^{3} d^{2} - 3 \, C a^{2} b d^{2} - B a b^{2} d^{2} - 3 \, A b^{3} d^{2}\right )} \arctan \left (\frac {\sqrt {d x + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{4 \, {\left (b^{5} c^{2} - 2 \, a b^{4} c d + a^{2} b^{3} d^{2}\right )} \sqrt {-b^{2} c + a b d}} - \frac {12 \, {\left (d x + c\right )}^{\frac {3}{2}} D a^{2} b^{2} c d - 8 \, {\left (d x + c\right )}^{\frac {3}{2}} C a b^{3} c d + 4 \, {\left (d x + c\right )}^{\frac {3}{2}} B b^{4} c d - 12 \, \sqrt {d x + c} D a^{2} b^{2} c^{2} d + 8 \, \sqrt {d x + c} C a b^{3} c^{2} d - 4 \, \sqrt {d x + c} B b^{4} c^{2} d - 9 \, {\left (d x + c\right )}^{\frac {3}{2}} D a^{3} b d^{2} + 5 \, {\left (d x + c\right )}^{\frac {3}{2}} C a^{2} b^{2} d^{2} - {\left (d x + c\right )}^{\frac {3}{2}} B a b^{3} d^{2} - 3 \, {\left (d x + c\right )}^{\frac {3}{2}} A b^{4} d^{2} + 19 \, \sqrt {d x + c} D a^{3} b c d^{2} - 11 \, \sqrt {d x + c} C a^{2} b^{2} c d^{2} + 3 \, \sqrt {d x + c} B a b^{3} c d^{2} + 5 \, \sqrt {d x + c} A b^{4} c d^{2} - 7 \, \sqrt {d x + c} D a^{4} d^{3} + 3 \, \sqrt {d x + c} C a^{3} b d^{3} + \sqrt {d x + c} B a^{2} b^{2} d^{3} - 5 \, \sqrt {d x + c} A a b^{3} d^{3}}{4 \, {\left (b^{5} c^{2} - 2 \, a b^{4} c d + a^{2} b^{3} d^{2}\right )} {\left ({\left (d x + c\right )} b - b c + a d\right )}^{2}} + \frac {2 \, \sqrt {d x + c} D}{b^{3} d} \]

[In]

integrate((D*x^3+C*x^2+B*x+A)/(b*x+a)^3/(d*x+c)^(1/2),x, algorithm="giac")

[Out]

-1/4*(24*D*a*b^2*c^2 - 8*C*b^3*c^2 - 36*D*a^2*b*c*d + 8*C*a*b^2*c*d + 4*B*b^3*c*d + 15*D*a^3*d^2 - 3*C*a^2*b*d
^2 - B*a*b^2*d^2 - 3*A*b^3*d^2)*arctan(sqrt(d*x + c)*b/sqrt(-b^2*c + a*b*d))/((b^5*c^2 - 2*a*b^4*c*d + a^2*b^3
*d^2)*sqrt(-b^2*c + a*b*d)) - 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 - 9*(d*x + c)^(3/2)*D*a^3*b*d^2 + 5*(d*x + c)^(3/2)*C*a^2*b^2*d^2 - (d*x + c)^(3/2)*B*a*b^3*d^2 - 3*(d*
x + c)^(3/2)*A*b^4*d^2 + 19*sqrt(d*x + c)*D*a^3*b*c*d^2 - 11*sqrt(d*x + c)*C*a^2*b^2*c*d^2 + 3*sqrt(d*x + c)*B
*a*b^3*c*d^2 + 5*sqrt(d*x + c)*A*b^4*c*d^2 - 7*sqrt(d*x + c)*D*a^4*d^3 + 3*sqrt(d*x + c)*C*a^3*b*d^3 + sqrt(d*
x + c)*B*a^2*b^2*d^3 - 5*sqrt(d*x + c)*A*a*b^3*d^3)/((b^5*c^2 - 2*a*b^4*c*d + a^2*b^3*d^2)*((d*x + c)*b - b*c
+ a*d)^2) + 2*sqrt(d*x + c)*D/(b^3*d)

Mupad [F(-1)]

Timed out. \[ \int \frac {A+B x+C x^2+D x^3}{(a+b x)^3 \sqrt {c+d x}} \, dx=\int \frac {A+B\,x+C\,x^2+x^3\,D}{{\left (a+b\,x\right )}^3\,\sqrt {c+d\,x}} \,d x \]

[In]

int((A + B*x + C*x^2 + x^3*D)/((a + b*x)^3*(c + d*x)^(1/2)),x)

[Out]

int((A + B*x + C*x^2 + x^3*D)/((a + b*x)^3*(c + d*x)^(1/2)), x)