\(\int \frac {A+B x+C x^2+D x^3}{(a+b x)^3 \sqrt {c+d x}} \, dx\) [85]

Optimal result

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

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)-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))*arctanh(b^(1/2)*(d*x+c)^(1/2)/(-a*d+b*c)^(1/2))/b^(7/2)/ 
(-a*d+b*c)^(5/2)
 

Mathematica [A] (verified)

Time = 0.88 (sec) , antiderivative size = 298, normalized size of antiderivative = 1.07 \[ \int \frac {A+B x+C x^2+D x^3}{(a+b x)^3 \sqrt {c+d x}} \, dx=\frac {\frac {\sqrt {b} \sqrt {c+d x} \left (15 a^4 d^2 D+A b^3 d (-2 b c+5 a d+3 b d x)+4 b^4 c x (-B d+2 c D x)+a^3 b d (-3 C d-26 c D+25 d D x)+a b^3 (B d (-2 c+d x)+8 c x (C d+2 c D-2 d D x))+a^2 b^2 \left (8 c^2 D+c (6 C d-44 d D x)-d^2 \left (B+5 C x-8 D x^2\right )\right )\right )}{d (b c-a d)^2 (a+b x)^2}+\frac {\left (b^3 \left (8 c^2 C-4 B c d+3 A d^2\right )-15 a^3 d^2 D+3 a^2 b d (C d+12 c D)+a b^2 \left (-8 c C d+B d^2-24 c^2 D\right )\right ) \arctan \left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {-b c+a d}}\right )}{(-b c+a d)^{5/2}}}{4 b^{7/2}} \] Input:

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

Output:

((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]*Sq 
rt[c + d*x])/Sqrt[-(b*c) + a*d]])/(-(b*c) + a*d)^(5/2))/(4*b^(7/2))
 

Rubi [A] (verified)

Time = 0.93 (sec) , antiderivative size = 334, normalized size of antiderivative = 1.20, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.281, Rules used = {2124, 27, 1192, 25, 1471, 25, 27, 299, 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.

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

\(\Big \downarrow \) 2124

\(\displaystyle -\frac {\int -\frac {4 \left (c-\frac {a d}{b}\right ) D x^2+\frac {4 (b c-a d) (b C-a D) x}{b^2}+\frac {-d D a^3+b (C d+4 c D) a^2-b^2 (4 c C+B d) a+b^3 (4 B c-3 A d)}{b^3}}{2 (a+b x)^2 \sqrt {c+d x}}dx}{2 (b c-a d)}-\frac {\sqrt {c+d x} \left (A b^3-a \left (a^2 D-a b C+b^2 B\right )\right )}{2 b^3 (a+b x)^2 (b c-a d)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {-\frac {d D a^3}{b^3}+\frac {(C d+4 c D) a^2}{b^2}-\frac {(4 c C+B d) a}{b}+4 \left (c-\frac {a d}{b}\right ) D x^2+4 B c-3 A d+\frac {4 (b c-a d) (b C-a D) x}{b^2}}{(a+b x)^2 \sqrt {c+d x}}dx}{4 (b c-a d)}-\frac {\sqrt {c+d x} \left (A b^3-a \left (a^2 D-a b C+b^2 B\right )\right )}{2 b^3 (a+b x)^2 (b c-a d)}\)

\(\Big \downarrow \) 1192

\(\displaystyle \frac {\int -\frac {-4 D c^3+4 C d c^2-4 B d^2 c+3 A d^3-4 \left (c-\frac {a d}{b}\right ) D (c+d x)^2+\frac {a d^3 \left (D a^2-b C a+b^2 B\right )}{b^3}-\frac {4 (b c-a d) (b C d-a D d-2 b c D) (c+d x)}{b^2}}{(b c-a d-b (c+d x))^2}d\sqrt {c+d x}}{2 d (b c-a d)}-\frac {\sqrt {c+d x} \left (A b^3-a \left (a^2 D-a b C+b^2 B\right )\right )}{2 b^3 (a+b x)^2 (b c-a d)}\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {\int \frac {-4 D c^3+4 C d c^2-4 B d^2 c-4 \left (c-\frac {a d}{b}\right ) D (c+d x)^2+d^3 \left (3 A+\frac {a \left (D a^2-b C a+b^2 B\right )}{b^3}\right )-\frac {4 (b c-a d) (b C d-a D d-2 b c D) (c+d x)}{b^2}}{(b c-a d-b (c+d x))^2}d\sqrt {c+d x}}{2 d (b c-a d)}-\frac {\sqrt {c+d x} \left (A b^3-a \left (a^2 D-a b C+b^2 B\right )\right )}{2 b^3 (a+b x)^2 (b c-a d)}\)

\(\Big \downarrow \) 1471

\(\displaystyle \frac {\frac {\int -\frac {\left (-8 D c^3+8 C d c^2-\frac {4 d^2 \left (-3 D a^2+2 b C a+b^2 B\right ) c}{b^2}+3 A d^3+\frac {a d^3 \left (-7 D a^2+3 b C a+b^2 B\right )}{b^3}\right ) b^2+8 (b c-a d)^2 D (c+d x)}{b^2 (b c-a d-b (c+d x))}d\sqrt {c+d x}}{2 (b c-a d)}+\frac {d^2 \sqrt {c+d x} \left (-9 a^3 d D+a^2 b (12 c D+5 C d)-a b^2 (B d+8 c C)+b^3 (4 B c-3 A d)\right )}{2 b^3 (b c-a d) (-a d-b (c+d x)+b c)}}{2 d (b c-a d)}-\frac {\sqrt {c+d x} \left (A b^3-a \left (a^2 D-a b C+b^2 B\right )\right )}{2 b^3 (a+b x)^2 (b c-a d)}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {d^2 \sqrt {c+d x} \left (-9 a^3 d D+a^2 b (12 c D+5 C d)-a b^2 (B d+8 c C)+b^3 (4 B c-3 A d)\right )}{2 b^3 (b c-a d) (-a d-b (c+d x)+b c)}-\frac {\int \frac {\left (-8 D c^3+8 C d c^2-\frac {4 d^2 \left (-3 D a^2+2 b C a+b^2 B\right ) c}{b^2}+d^3 \left (3 A+\frac {a \left (-7 D a^2+3 b C a+b^2 B\right )}{b^3}\right )\right ) b^2+8 (b c-a d)^2 D (c+d x)}{b^2 (b c-a d-b (c+d x))}d\sqrt {c+d x}}{2 (b c-a d)}}{2 d (b c-a d)}-\frac {\sqrt {c+d x} \left (A b^3-a \left (a^2 D-a b C+b^2 B\right )\right )}{2 b^3 (a+b x)^2 (b c-a d)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {d^2 \sqrt {c+d x} \left (-9 a^3 d D+a^2 b (12 c D+5 C d)-a b^2 (B d+8 c C)+b^3 (4 B c-3 A d)\right )}{2 b^3 (b c-a d) (-a d-b (c+d x)+b c)}-\frac {\int \frac {\left (-8 D c^3+8 C d c^2-\frac {4 d^2 \left (-3 D a^2+2 b C a+b^2 B\right ) c}{b^2}+d^3 \left (3 A+\frac {a \left (-7 D a^2+3 b C a+b^2 B\right )}{b^3}\right )\right ) b^2+8 (b c-a d)^2 D (c+d x)}{b c-a d-b (c+d x)}d\sqrt {c+d x}}{2 b^2 (b c-a d)}}{2 d (b c-a d)}-\frac {\sqrt {c+d x} \left (A b^3-a \left (a^2 D-a b C+b^2 B\right )\right )}{2 b^3 (a+b x)^2 (b c-a d)}\)

\(\Big \downarrow \) 299

\(\displaystyle \frac {\frac {d^2 \sqrt {c+d x} \left (-9 a^3 d D+a^2 b (12 c D+5 C d)-a b^2 (B d+8 c C)+b^3 (4 B c-3 A d)\right )}{2 b^3 (b c-a d) (-a d-b (c+d x)+b c)}-\frac {\frac {d \left (-15 a^3 d^2 D+3 a^2 b d (12 c D+C d)-a b^2 \left (-B d^2+24 c^2 D+8 c C d\right )+b^3 \left (3 A d^2-4 B c d+8 c^2 C\right )\right ) \int \frac {1}{b c-a d-b (c+d x)}d\sqrt {c+d x}}{b}-\frac {8 D \sqrt {c+d x} (b c-a d)^2}{b}}{2 b^2 (b c-a d)}}{2 d (b c-a d)}-\frac {\sqrt {c+d x} \left (A b^3-a \left (a^2 D-a b C+b^2 B\right )\right )}{2 b^3 (a+b x)^2 (b c-a d)}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {\frac {d^2 \sqrt {c+d x} \left (-9 a^3 d D+a^2 b (12 c D+5 C d)-a b^2 (B d+8 c C)+b^3 (4 B c-3 A d)\right )}{2 b^3 (b c-a d) (-a d-b (c+d x)+b c)}-\frac {\frac {d \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right ) \left (-15 a^3 d^2 D+3 a^2 b d (12 c D+C d)-a b^2 \left (-B d^2+24 c^2 D+8 c C d\right )+b^3 \left (3 A d^2-4 B c d+8 c^2 C\right )\right )}{b^{3/2} \sqrt {b c-a d}}-\frac {8 D \sqrt {c+d x} (b c-a d)^2}{b}}{2 b^2 (b c-a d)}}{2 d (b c-a d)}-\frac {\sqrt {c+d x} \left (A b^3-a \left (a^2 D-a b C+b^2 B\right )\right )}{2 b^3 (a+b x)^2 (b c-a d)}\)

Input:

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

Output:

-1/2*((A*b^3 - a*(b^2*B - a*b*C + a^2*D))*Sqrt[c + d*x])/(b^3*(b*c - a*d)* 
(a + b*x)^2) + ((d^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])/(2*b^3*(b*c - a*d)*(b*c - a*d - 
 b*(c + d*x))) - ((-8*(b*c - a*d)^2*D*Sqrt[c + d*x])/b + (d*(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] 
])/(b^(3/2)*Sqrt[b*c - a*d]))/(2*b^2*(b*c - a*d)))/(2*d*(b*c - a*d))
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 221

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 299

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[d*x 
*((a + b*x^2)^(p + 1)/(b*(2*p + 3))), x] - Simp[(a*d - b*c*(2*p + 3))/(b*(2 
*p + 3))   Int[(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - 
 a*d, 0] && NeQ[2*p + 3, 0]
 

rule 1192

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]
 

rule 1471

Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), 
x_Symbol] :> With[{Qx = PolynomialQuotient[(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] + Simp[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 2124

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])
 

Maple [A] (verified)

Time = 0.90 (sec) , antiderivative size = 301, normalized size of antiderivative = 1.08

method	result	size
pseudoelliptic	\(\frac {\frac {3 \left (\left (d^{2} A -\frac {4}{3} c d B +\frac {8}{3} C \,c^{2}\right ) b^{3}+\frac {a \left (B \,d^{2}-8 C c d -24 D c^{2}\right ) b^{2}}{3}+a^{2} b d \left (C d +12 D c \right )-5 a^{3} d^{2} D\right ) \left (b x +a \right )^{2} d \arctan \left (\frac {b \sqrt {x d +c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{4}+\frac {5 \left (\frac {\left (3 A \,d^{2} x -2 c \left (2 B x +A \right ) d +8 D c^{2} x^{2}\right ) b^{4}}{5}+\left (\left (\frac {B x}{5}+A \right ) d^{2}-\frac {2 c \left (8 D x^{2}-4 C x +B \right ) d}{5}+\frac {16 D c^{2} x}{5}\right ) a \,b^{3}-\frac {\left (\left (-8 D x^{2}+5 C x +B \right ) d^{2}-6 \left (-\frac {22 D x}{3}+C \right ) c d -8 D c^{2}\right ) a^{2} b^{2}}{5}-\frac {3 d \left (\left (-\frac {25 D x}{3}+C \right ) d +\frac {26 D c}{3}\right ) a^{3} b}{5}+3 D a^{4} d^{2}\right ) \sqrt {\left (a d -b c \right ) b}\, \sqrt {x d +c}}{4}}{d \sqrt {\left (a d -b c \right ) b}\, \left (a d -b c \right )^{2} \left (b x +a \right )^{2} b^{3}}\)	\(301\)
derivativedivides	\(\frac {\frac {2 D \sqrt {x d +c}}{b^{3}}+\frac {2 d \left (\frac {\frac {b d \left (3 b^{3} d A +B a \,b^{2} d -4 B \,b^{3} c -5 C \,a^{2} b d +8 a \,b^{2} c C +9 a^{3} d D-12 a^{2} b c D\right ) \left (x d +c \right )^{\frac {3}{2}}}{8 a^{2} d^{2}-16 a b c d +8 b^{2} c^{2}}+\frac {\left (5 b^{3} d A -B a \,b^{2} d -4 B \,b^{3} c -3 C \,a^{2} b d +8 a \,b^{2} c C +7 a^{3} d D-12 a^{2} b c D\right ) d \sqrt {x d +c}}{8 a d -8 b c}}{\left (\left (x d +c \right ) b +a d -b c \right )^{2}}+\frac {\left (3 b^{3} d^{2} A +B a \,b^{2} d^{2}-4 B \,b^{3} c d +3 C \,a^{2} b \,d^{2}-8 C a \,b^{2} c d +8 C \,b^{3} c^{2}-15 a^{3} d^{2} D+36 a^{2} b c d D-24 a \,b^{2} c^{2} D\right ) \arctan \left (\frac {b \sqrt {x d +c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{8 \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) \sqrt {\left (a d -b c \right ) b}}\right )}{b^{3}}}{d}\)	\(347\)
default	\(\frac {\frac {2 D \sqrt {x d +c}}{b^{3}}+\frac {2 d \left (\frac {\frac {b d \left (3 b^{3} d A +B a \,b^{2} d -4 B \,b^{3} c -5 C \,a^{2} b d +8 a \,b^{2} c C +9 a^{3} d D-12 a^{2} b c D\right ) \left (x d +c \right )^{\frac {3}{2}}}{8 a^{2} d^{2}-16 a b c d +8 b^{2} c^{2}}+\frac {\left (5 b^{3} d A -B a \,b^{2} d -4 B \,b^{3} c -3 C \,a^{2} b d +8 a \,b^{2} c C +7 a^{3} d D-12 a^{2} b c D\right ) d \sqrt {x d +c}}{8 a d -8 b c}}{\left (\left (x d +c \right ) b +a d -b c \right )^{2}}+\frac {\left (3 b^{3} d^{2} A +B a \,b^{2} d^{2}-4 B \,b^{3} c d +3 C \,a^{2} b \,d^{2}-8 C a \,b^{2} c d +8 C \,b^{3} c^{2}-15 a^{3} d^{2} D+36 a^{2} b c d D-24 a \,b^{2} c^{2} D\right ) \arctan \left (\frac {b \sqrt {x d +c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{8 \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) \sqrt {\left (a d -b c \right ) b}}\right )}{b^{3}}}{d}\)	\(347\)

Input:

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

Output:

5/4*(3/5*((d^2*A-4/3*c*d*B+8/3*C*c^2)*b^3+1/3*a*(B*d^2-8*C*c*d-24*D*c^2)*b 
^2+a^2*b*d*(C*d+12*D*c)-5*a^3*d^2*D)*(b*x+a)^2*d*arctan(b*(d*x+c)^(1/2)/(( 
a*d-b*c)*b)^(1/2))+(1/5*(3*A*d^2*x-2*c*(2*B*x+A)*d+8*D*c^2*x^2)*b^4+((1/5* 
B*x+A)*d^2-2/5*c*(8*D*x^2-4*C*x+B)*d+16/5*D*c^2*x)*a*b^3-1/5*((-8*D*x^2+5* 
C*x+B)*d^2-6*(-22/3*D*x+C)*c*d-8*D*c^2)*a^2*b^2-3/5*d*((-25/3*D*x+C)*d+26/ 
3*D*c)*a^3*b+3*D*a^4*d^2)*((a*d-b*c)*b)^(1/2)*(d*x+c)^(1/2))/(a*d-b*c)^2/( 
(a*d-b*c)*b)^(1/2)/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. 782 vs. \(2 (258) = 516\).

Time = 0.13 (sec) , antiderivative size = 1577, normalized size of antiderivative = 5.65 \[ \int \frac {A+B x+C x^2+D x^3}{(a+b x)^3 \sqrt {c+d x}} \, dx=\text {Too large to display} \] Input:

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

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F(-2)]

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

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

Output:

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
 

Giac [B] (verification not implemented)

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

Time = 0.14 (sec) , antiderivative size = 529, normalized size of antiderivative = 1.90 \[ \int \frac {A+B x+C x^2+D x^3}{(a+b x)^3 \sqrt {c+d x}} \, dx=-\frac {{\left (24 \, D a b^{2} c^{2} - 8 \, C b^{3} c^{2} - 36 \, D a^{2} b c d + 8 \, C a b^{2} c d + 4 \, B b^{3} c d + 15 \, D a^{3} d^{2} - 3 \, C a^{2} b d^{2} - B a b^{2} d^{2} - 3 \, A b^{3} d^{2}\right )} \arctan \left (\frac {\sqrt {d x + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{4 \, {\left (b^{5} c^{2} - 2 \, a b^{4} c d + a^{2} b^{3} d^{2}\right )} \sqrt {-b^{2} c + a b d}} - \frac {12 \, {\left (d x + c\right )}^{\frac {3}{2}} D a^{2} b^{2} c d - 8 \, {\left (d x + c\right )}^{\frac {3}{2}} C a b^{3} c d + 4 \, {\left (d x + c\right )}^{\frac {3}{2}} B b^{4} c d - 12 \, \sqrt {d x + c} D a^{2} b^{2} c^{2} d + 8 \, \sqrt {d x + c} C a b^{3} c^{2} d - 4 \, \sqrt {d x + c} B b^{4} c^{2} d - 9 \, {\left (d x + c\right )}^{\frac {3}{2}} D a^{3} b d^{2} + 5 \, {\left (d x + c\right )}^{\frac {3}{2}} C a^{2} b^{2} d^{2} - {\left (d x + c\right )}^{\frac {3}{2}} B a b^{3} d^{2} - 3 \, {\left (d x + c\right )}^{\frac {3}{2}} A b^{4} d^{2} + 19 \, \sqrt {d x + c} D a^{3} b c d^{2} - 11 \, \sqrt {d x + c} C a^{2} b^{2} c d^{2} + 3 \, \sqrt {d x + c} B a b^{3} c d^{2} + 5 \, \sqrt {d x + c} A b^{4} c d^{2} - 7 \, \sqrt {d x + c} D a^{4} d^{3} + 3 \, \sqrt {d x + c} C a^{3} b d^{3} + \sqrt {d x + c} B a^{2} b^{2} d^{3} - 5 \, \sqrt {d x + c} A a b^{3} d^{3}}{4 \, {\left (b^{5} c^{2} - 2 \, a b^{4} c d + a^{2} b^{3} d^{2}\right )} {\left ({\left (d x + c\right )} b - b c + a d\right )}^{2}} + \frac {2 \, \sqrt {d x + c} D}{b^{3} d} \] Input:

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

Output:

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 847, normalized size of antiderivative = 3.04 \[ \int \frac {A+B x+C x^2+D x^3}{(a+b x)^3 \sqrt {c+d x}} \, dx =\text {Too large to display} \] Input:

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

Output:

( - 15*sqrt(b)*sqrt(a*d - b*c)*atan((sqrt(c + d*x)*b)/(sqrt(b)*sqrt(a*d - 
b*c)))*a**4*d**2 + 24*sqrt(b)*sqrt(a*d - b*c)*atan((sqrt(c + d*x)*b)/(sqrt 
(b)*sqrt(a*d - b*c)))*a**3*b*c*d - 30*sqrt(b)*sqrt(a*d - b*c)*atan((sqrt(c 
 + d*x)*b)/(sqrt(b)*sqrt(a*d - b*c)))*a**3*b*d**2*x + 4*sqrt(b)*sqrt(a*d - 
 b*c)*atan((sqrt(c + d*x)*b)/(sqrt(b)*sqrt(a*d - b*c)))*a**2*b**3*d - 8*sq 
rt(b)*sqrt(a*d - b*c)*atan((sqrt(c + d*x)*b)/(sqrt(b)*sqrt(a*d - b*c)))*a* 
*2*b**2*c**2 + 48*sqrt(b)*sqrt(a*d - b*c)*atan((sqrt(c + d*x)*b)/(sqrt(b)* 
sqrt(a*d - b*c)))*a**2*b**2*c*d*x - 15*sqrt(b)*sqrt(a*d - b*c)*atan((sqrt( 
c + d*x)*b)/(sqrt(b)*sqrt(a*d - b*c)))*a**2*b**2*d**2*x**2 + 8*sqrt(b)*sqr 
t(a*d - b*c)*atan((sqrt(c + d*x)*b)/(sqrt(b)*sqrt(a*d - b*c)))*a*b**4*d*x 
- 16*sqrt(b)*sqrt(a*d - b*c)*atan((sqrt(c + d*x)*b)/(sqrt(b)*sqrt(a*d - b* 
c)))*a*b**3*c**2*x + 24*sqrt(b)*sqrt(a*d - b*c)*atan((sqrt(c + d*x)*b)/(sq 
rt(b)*sqrt(a*d - b*c)))*a*b**3*c*d*x**2 + 4*sqrt(b)*sqrt(a*d - b*c)*atan(( 
sqrt(c + d*x)*b)/(sqrt(b)*sqrt(a*d - b*c)))*b**5*d*x**2 - 8*sqrt(b)*sqrt(a 
*d - b*c)*atan((sqrt(c + d*x)*b)/(sqrt(b)*sqrt(a*d - b*c)))*b**4*c**2*x**2 
 + 15*sqrt(c + d*x)*a**4*b*d**2 - 29*sqrt(c + d*x)*a**3*b**2*c*d + 25*sqrt 
(c + d*x)*a**3*b**2*d**2*x + 4*sqrt(c + d*x)*a**2*b**4*d + 14*sqrt(c + d*x 
)*a**2*b**3*c**2 - 49*sqrt(c + d*x)*a**2*b**3*c*d*x + 8*sqrt(c + d*x)*a**2 
*b**3*d**2*x**2 - 4*sqrt(c + d*x)*a*b**5*c + 4*sqrt(c + d*x)*a*b**5*d*x + 
24*sqrt(c + d*x)*a*b**4*c**2*x - 16*sqrt(c + d*x)*a*b**4*c*d*x**2 - 4*s...