\(\int \frac {A+B x+C x^2+D x^3}{(a+b x)^{5/2} (c+d x)^{5/2}} \, dx\) [150]

Optimal result

Integrand size = 34, antiderivative size = 324 \[ \int \frac {A+B x+C x^2+D x^3}{(a+b x)^{5/2} (c+d x)^{5/2}} \, dx=\frac {2 \left (c^2 C d-B c d^2+A d^3-c^3 D\right ) \sqrt {a+b x}}{3 d (b c-a d)^3 (c+d x)^{3/2}}-\frac {2 \left (A b^3-a \left (b^2 B-a b C+a^2 D\right )\right )}{3 b^2 (b c-a d)^2 (a+b x)^{3/2} \sqrt {c+d x}}-\frac {2 \left (b^3 (3 B c-7 A d)-a b^2 (6 c C-4 B d)-2 a^3 d D-a^2 b (C d-9 c D)\right )}{3 b^2 (b c-a d)^3 \sqrt {a+b x} \sqrt {c+d x}}+\frac {2 \left (a^3 d^3 D+a^2 b d^2 (2 C d-9 c D)+a b^2 d \left (12 c C d-8 B d^2-9 c^2 D\right )+b^3 \left (2 c^2 C d-8 B c d^2+16 A d^3+c^3 D\right )\right ) \sqrt {a+b x}}{3 b^2 d (b c-a d)^4 \sqrt {c+d x}} \] Output:

2/3*(A*d^3-B*c*d^2+C*c^2*d-D*c^3)*(b*x+a)^(1/2)/d/(-a*d+b*c)^3/(d*x+c)^(3/ 
2)-2/3*(A*b^3-a*(B*b^2-C*a*b+D*a^2))/b^2/(-a*d+b*c)^2/(b*x+a)^(3/2)/(d*x+c 
)^(1/2)-2/3*(b^3*(-7*A*d+3*B*c)-a*b^2*(-4*B*d+6*C*c)-2*a^3*d*D-a^2*b*(C*d- 
9*D*c))/b^2/(-a*d+b*c)^3/(b*x+a)^(1/2)/(d*x+c)^(1/2)+2/3*(a^3*d^3*D+a^2*b* 
d^2*(2*C*d-9*D*c)+a*b^2*d*(-8*B*d^2+12*C*c*d-9*D*c^2)+b^3*(16*A*d^3-8*B*c* 
d^2+2*C*c^2*d+D*c^3))*(b*x+a)^(1/2)/b^2/d/(-a*d+b*c)^4/(d*x+c)^(1/2)
 

Mathematica [A] (verified)

Time = 0.48 (sec) , antiderivative size = 361, normalized size of antiderivative = 1.11 \[ \int \frac {A+B x+C x^2+D x^3}{(a+b x)^{5/2} (c+d x)^{5/2}} \, dx=-\frac {2 \left (c^2 C d (a+b x)^3-B c d^2 (a+b x)^3+A d^3 (a+b x)^3-c^3 D (a+b x)^3-3 b c^2 C (a+b x)^2 (c+d x)+6 b B c d (a+b x)^2 (c+d x)-6 a c C d (a+b x)^2 (c+d x)-9 A b d^2 (a+b x)^2 (c+d x)+3 a B d^2 (a+b x)^2 (c+d x)+9 a c^2 D (a+b x)^2 (c+d x)+3 b^2 B c (a+b x) (c+d x)^2-6 a b c C (a+b x) (c+d x)^2-9 A b^2 d (a+b x) (c+d x)^2+6 a b B d (a+b x) (c+d x)^2-3 a^2 C d (a+b x) (c+d x)^2+9 a^2 c D (a+b x) (c+d x)^2+A b^3 (c+d x)^3-a b^2 B (c+d x)^3+a^2 b C (c+d x)^3-a^3 D (c+d x)^3\right )}{3 (b c-a d)^4 (a+b x)^{3/2} (c+d x)^{3/2}} \] Input:

Integrate[(A + B*x + C*x^2 + D*x^3)/((a + b*x)^(5/2)*(c + d*x)^(5/2)),x]
 

Output:

(-2*(c^2*C*d*(a + b*x)^3 - B*c*d^2*(a + b*x)^3 + A*d^3*(a + b*x)^3 - c^3*D 
*(a + b*x)^3 - 3*b*c^2*C*(a + b*x)^2*(c + d*x) + 6*b*B*c*d*(a + b*x)^2*(c 
+ d*x) - 6*a*c*C*d*(a + b*x)^2*(c + d*x) - 9*A*b*d^2*(a + b*x)^2*(c + d*x) 
 + 3*a*B*d^2*(a + b*x)^2*(c + d*x) + 9*a*c^2*D*(a + b*x)^2*(c + d*x) + 3*b 
^2*B*c*(a + b*x)*(c + d*x)^2 - 6*a*b*c*C*(a + b*x)*(c + d*x)^2 - 9*A*b^2*d 
*(a + b*x)*(c + d*x)^2 + 6*a*b*B*d*(a + b*x)*(c + d*x)^2 - 3*a^2*C*d*(a + 
b*x)*(c + d*x)^2 + 9*a^2*c*D*(a + b*x)*(c + d*x)^2 + A*b^3*(c + d*x)^3 - a 
*b^2*B*(c + d*x)^3 + a^2*b*C*(c + d*x)^3 - a^3*D*(c + d*x)^3))/(3*(b*c - a 
*d)^4*(a + b*x)^(3/2)*(c + d*x)^(3/2))
 

Rubi [A] (verified)

Time = 0.76 (sec) , antiderivative size = 386, normalized size of antiderivative = 1.19, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {2124, 27, 1193, 27, 87, 48}

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.

\(\displaystyle \int \frac {A+B x+C x^2+D x^3}{(a+b x)^{5/2} (c+d x)^{5/2}} \, dx\)

\(\Big \downarrow \) 2124

\(\displaystyle -\frac {2 \int -\frac {3 \left (\left (c-\frac {a d}{b}\right ) D x^2+\frac {(b c-a d) (b C-a D) x}{b^2}+\frac {d D a^3-b (C d-c D) a^2-b^2 (c C-B d) a+b^3 (B c-2 A d)}{b^3}\right )}{2 (a+b x)^{3/2} (c+d x)^{5/2}}dx}{3 (b c-a d)}-\frac {2 \left (A b^3-a \left (a^2 D-a b C+b^2 B\right )\right )}{3 b^3 (a+b x)^{3/2} (c+d x)^{3/2} (b c-a d)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {\frac {d D a^3}{b^3}-\frac {(C d-c D) a^2}{b^2}-\frac {(c C-B d) a}{b}+\left (c-\frac {a d}{b}\right ) D x^2+B c-2 A d+\frac {(b c-a d) (b C-a D) x}{b^2}}{(a+b x)^{3/2} (c+d x)^{5/2}}dx}{b c-a d}-\frac {2 \left (A b^3-a \left (a^2 D-a b C+b^2 B\right )\right )}{3 b^3 (a+b x)^{3/2} (c+d x)^{3/2} (b c-a d)}\)

\(\Big \downarrow \) 1193

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 87

\(\displaystyle \frac {\frac {\frac {b \left (a^3 d^3 D+a^2 b d^2 (2 C d-9 c D)+a b^2 d \left (-8 B d^2-9 c^2 D+12 c C d\right )+b^3 \left (16 A d^3-8 B c d^2+c^3 D+2 c^2 C d\right )\right ) \int \frac {1}{\sqrt {a+b x} (c+d x)^{3/2}}dx}{3 d (b c-a d)}+\frac {2 \sqrt {a+b x} \left (2 a^3 d^3 D+a^2 b d^2 (C d-9 c D)+2 a b^2 d^2 (3 c C-2 B d)+b^3 \left (8 A d^3-4 B c d^2+c^3 (-D)+c^2 C d\right )\right )}{3 d (c+d x)^{3/2} (b c-a d)}}{b^3 (b c-a d)}-\frac {2 \left (-\frac {a^3 d D}{b^3}+\frac {3 a^2 c D}{b^2}-\frac {a (2 c C-B d)}{b}-2 A d+B c\right )}{\sqrt {a+b x} (c+d x)^{3/2} (b c-a d)}}{b c-a d}-\frac {2 \left (A b^3-a \left (a^2 D-a b C+b^2 B\right )\right )}{3 b^3 (a+b x)^{3/2} (c+d x)^{3/2} (b c-a d)}\)

\(\Big \downarrow \) 48

\(\displaystyle \frac {\frac {\frac {2 \sqrt {a+b x} \left (2 a^3 d^3 D+a^2 b d^2 (C d-9 c D)+2 a b^2 d^2 (3 c C-2 B d)+b^3 \left (8 A d^3-4 B c d^2+c^3 (-D)+c^2 C d\right )\right )}{3 d (c+d x)^{3/2} (b c-a d)}+\frac {2 b \sqrt {a+b x} \left (a^3 d^3 D+a^2 b d^2 (2 C d-9 c D)+a b^2 d \left (-8 B d^2-9 c^2 D+12 c C d\right )+b^3 \left (16 A d^3-8 B c d^2+c^3 D+2 c^2 C d\right )\right )}{3 d \sqrt {c+d x} (b c-a d)^2}}{b^3 (b c-a d)}-\frac {2 \left (-\frac {a^3 d D}{b^3}+\frac {3 a^2 c D}{b^2}-\frac {a (2 c C-B d)}{b}-2 A d+B c\right )}{\sqrt {a+b x} (c+d x)^{3/2} (b c-a d)}}{b c-a d}-\frac {2 \left (A b^3-a \left (a^2 D-a b C+b^2 B\right )\right )}{3 b^3 (a+b x)^{3/2} (c+d x)^{3/2} (b c-a d)}\)

Input:

Int[(A + B*x + C*x^2 + D*x^3)/((a + b*x)^(5/2)*(c + d*x)^(5/2)),x]
 

Output:

(-2*(A*b^3 - a*(b^2*B - a*b*C + a^2*D)))/(3*b^3*(b*c - a*d)*(a + b*x)^(3/2 
)*(c + d*x)^(3/2)) + ((-2*(B*c - 2*A*d - (a*(2*c*C - B*d))/b + (3*a^2*c*D) 
/b^2 - (a^3*d*D)/b^3))/((b*c - a*d)*Sqrt[a + b*x]*(c + d*x)^(3/2)) + ((2*( 
2*a*b^2*d^2*(3*c*C - 2*B*d) + 2*a^3*d^3*D + a^2*b*d^2*(C*d - 9*c*D) + b^3* 
(c^2*C*d - 4*B*c*d^2 + 8*A*d^3 - c^3*D))*Sqrt[a + b*x])/(3*d*(b*c - a*d)*( 
c + d*x)^(3/2)) + (2*b*(a^3*d^3*D + a^2*b*d^2*(2*C*d - 9*c*D) + a*b^2*d*(1 
2*c*C*d - 8*B*d^2 - 9*c^2*D) + b^3*(2*c^2*C*d - 8*B*c*d^2 + 16*A*d^3 + c^3 
*D))*Sqrt[a + b*x])/(3*d*(b*c - a*d)^2*Sqrt[c + d*x]))/(b^3*(b*c - a*d)))/ 
(b*c - a*d)
 

Defintions of rubi rules used

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 48

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

rule 87

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]))))
 

rule 1193

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) 
 + (c_.)*(x_)^2)^(p_.), x_Symbol] :> With[{Qx = PolynomialQuotient[(a + b*x 
 + c*x^2)^p, d + e*x, x], R = PolynomialRemainder[(a + b*x + c*x^2)^p, d + 
e*x, x]}, Simp[R*(d + e*x)^(m + 1)*((f + g*x)^(n + 1)/((m + 1)*(e*f - d*g)) 
), x] + Simp[1/((m + 1)*(e*f - d*g))   Int[(d + e*x)^(m + 1)*(f + g*x)^n*Ex 
pandToSum[(m + 1)*(e*f - d*g)*Qx - g*R*(m + n + 2), x], x], x]] /; FreeQ[{a 
, b, c, d, e, f, g, n}, x] && IGtQ[p, 0] && ILtQ[2*m, -2] &&  !IntegerQ[n] 
&&  !(EqQ[m, -2] && EqQ[p, 1] && EqQ[2*c*d - b*e, 0])
 

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 = 0.66 (sec) , antiderivative size = 548, normalized size of antiderivative = 1.69

method	result	size
default	\(-\frac {2 \left (-16 A \,b^{3} d^{3} x^{3}+8 B a \,b^{2} d^{3} x^{3}+8 B \,b^{3} c \,d^{2} x^{3}-2 C \,a^{2} b \,d^{3} x^{3}-12 C a \,b^{2} c \,d^{2} x^{3}-2 C \,b^{3} c^{2} d \,x^{3}-D a^{3} d^{3} x^{3}+9 D a^{2} b c \,d^{2} x^{3}+9 D a \,b^{2} c^{2} d \,x^{3}-D b^{3} c^{3} x^{3}-24 A a \,b^{2} d^{3} x^{2}-24 A \,b^{3} c \,d^{2} x^{2}+12 B \,a^{2} b \,d^{3} x^{2}+24 B a \,b^{2} c \,d^{2} x^{2}+12 B \,b^{3} c^{2} d \,x^{2}-3 C \,a^{3} d^{3} x^{2}-21 C \,a^{2} b c \,d^{2} x^{2}-21 C a \,b^{2} c^{2} d \,x^{2}-3 C \,b^{3} c^{3} x^{2}+6 D a^{3} c \,d^{2} x^{2}+36 D a^{2} b \,c^{2} d \,x^{2}+6 D a \,b^{2} c^{3} x^{2}-6 A \,a^{2} b \,d^{3} x -36 A a \,b^{2} c \,d^{2} x -6 A \,b^{3} c^{2} d x +3 B \,a^{3} d^{3} x +21 B \,a^{2} b c \,d^{2} x +21 B a \,b^{2} c^{2} d x +3 B \,b^{3} c^{3} x -12 C \,a^{3} c \,d^{2} x -24 C \,a^{2} b \,c^{2} d x -12 C a \,b^{2} c^{3} x +24 D a^{3} c^{2} d x +24 D a^{2} b \,c^{3} x +a^{3} A \,d^{3}-9 A \,a^{2} b c \,d^{2}-9 A a \,b^{2} c^{2} d +A \,b^{3} c^{3}+2 B \,a^{3} c \,d^{2}+12 B \,a^{2} b \,c^{2} d +2 B a \,b^{2} c^{3}-8 a^{3} c^{2} C d -8 C \,a^{2} b \,c^{3}+16 D a^{3} c^{3}\right )}{3 \left (a d -b c \right )^{4} \left (b x +a \right )^{\frac {3}{2}} \left (x d +c \right )^{\frac {3}{2}}}\)	\(548\)
gosper	\(-\frac {2 \left (-16 A \,b^{3} d^{3} x^{3}+8 B a \,b^{2} d^{3} x^{3}+8 B \,b^{3} c \,d^{2} x^{3}-2 C \,a^{2} b \,d^{3} x^{3}-12 C a \,b^{2} c \,d^{2} x^{3}-2 C \,b^{3} c^{2} d \,x^{3}-D a^{3} d^{3} x^{3}+9 D a^{2} b c \,d^{2} x^{3}+9 D a \,b^{2} c^{2} d \,x^{3}-D b^{3} c^{3} x^{3}-24 A a \,b^{2} d^{3} x^{2}-24 A \,b^{3} c \,d^{2} x^{2}+12 B \,a^{2} b \,d^{3} x^{2}+24 B a \,b^{2} c \,d^{2} x^{2}+12 B \,b^{3} c^{2} d \,x^{2}-3 C \,a^{3} d^{3} x^{2}-21 C \,a^{2} b c \,d^{2} x^{2}-21 C a \,b^{2} c^{2} d \,x^{2}-3 C \,b^{3} c^{3} x^{2}+6 D a^{3} c \,d^{2} x^{2}+36 D a^{2} b \,c^{2} d \,x^{2}+6 D a \,b^{2} c^{3} x^{2}-6 A \,a^{2} b \,d^{3} x -36 A a \,b^{2} c \,d^{2} x -6 A \,b^{3} c^{2} d x +3 B \,a^{3} d^{3} x +21 B \,a^{2} b c \,d^{2} x +21 B a \,b^{2} c^{2} d x +3 B \,b^{3} c^{3} x -12 C \,a^{3} c \,d^{2} x -24 C \,a^{2} b \,c^{2} d x -12 C a \,b^{2} c^{3} x +24 D a^{3} c^{2} d x +24 D a^{2} b \,c^{3} x +a^{3} A \,d^{3}-9 A \,a^{2} b c \,d^{2}-9 A a \,b^{2} c^{2} d +A \,b^{3} c^{3}+2 B \,a^{3} c \,d^{2}+12 B \,a^{2} b \,c^{2} d +2 B a \,b^{2} c^{3}-8 a^{3} c^{2} C d -8 C \,a^{2} b \,c^{3}+16 D a^{3} c^{3}\right )}{3 \left (b x +a \right )^{\frac {3}{2}} \left (x d +c \right )^{\frac {3}{2}} \left (d^{4} a^{4}-4 a^{3} b c \,d^{3}+6 a^{2} b^{2} c^{2} d^{2}-4 a \,b^{3} c^{3} d +c^{4} b^{4}\right )}\)	\(589\)
orering	\(-\frac {2 \left (-16 A \,b^{3} d^{3} x^{3}+8 B a \,b^{2} d^{3} x^{3}+8 B \,b^{3} c \,d^{2} x^{3}-2 C \,a^{2} b \,d^{3} x^{3}-12 C a \,b^{2} c \,d^{2} x^{3}-2 C \,b^{3} c^{2} d \,x^{3}-D a^{3} d^{3} x^{3}+9 D a^{2} b c \,d^{2} x^{3}+9 D a \,b^{2} c^{2} d \,x^{3}-D b^{3} c^{3} x^{3}-24 A a \,b^{2} d^{3} x^{2}-24 A \,b^{3} c \,d^{2} x^{2}+12 B \,a^{2} b \,d^{3} x^{2}+24 B a \,b^{2} c \,d^{2} x^{2}+12 B \,b^{3} c^{2} d \,x^{2}-3 C \,a^{3} d^{3} x^{2}-21 C \,a^{2} b c \,d^{2} x^{2}-21 C a \,b^{2} c^{2} d \,x^{2}-3 C \,b^{3} c^{3} x^{2}+6 D a^{3} c \,d^{2} x^{2}+36 D a^{2} b \,c^{2} d \,x^{2}+6 D a \,b^{2} c^{3} x^{2}-6 A \,a^{2} b \,d^{3} x -36 A a \,b^{2} c \,d^{2} x -6 A \,b^{3} c^{2} d x +3 B \,a^{3} d^{3} x +21 B \,a^{2} b c \,d^{2} x +21 B a \,b^{2} c^{2} d x +3 B \,b^{3} c^{3} x -12 C \,a^{3} c \,d^{2} x -24 C \,a^{2} b \,c^{2} d x -12 C a \,b^{2} c^{3} x +24 D a^{3} c^{2} d x +24 D a^{2} b \,c^{3} x +a^{3} A \,d^{3}-9 A \,a^{2} b c \,d^{2}-9 A a \,b^{2} c^{2} d +A \,b^{3} c^{3}+2 B \,a^{3} c \,d^{2}+12 B \,a^{2} b \,c^{2} d +2 B a \,b^{2} c^{3}-8 a^{3} c^{2} C d -8 C \,a^{2} b \,c^{3}+16 D a^{3} c^{3}\right )}{3 \left (b x +a \right )^{\frac {3}{2}} \left (x d +c \right )^{\frac {3}{2}} \left (d^{4} a^{4}-4 a^{3} b c \,d^{3}+6 a^{2} b^{2} c^{2} d^{2}-4 a \,b^{3} c^{3} d +c^{4} b^{4}\right )}\)	\(589\)

Input:

int((D*x^3+C*x^2+B*x+A)/(b*x+a)^(5/2)/(d*x+c)^(5/2),x,method=_RETURNVERBOS 
E)
 

Output:

-2/3*(-16*A*b^3*d^3*x^3+8*B*a*b^2*d^3*x^3+8*B*b^3*c*d^2*x^3-2*C*a^2*b*d^3* 
x^3-12*C*a*b^2*c*d^2*x^3-2*C*b^3*c^2*d*x^3-D*a^3*d^3*x^3+9*D*a^2*b*c*d^2*x 
^3+9*D*a*b^2*c^2*d*x^3-D*b^3*c^3*x^3-24*A*a*b^2*d^3*x^2-24*A*b^3*c*d^2*x^2 
+12*B*a^2*b*d^3*x^2+24*B*a*b^2*c*d^2*x^2+12*B*b^3*c^2*d*x^2-3*C*a^3*d^3*x^ 
2-21*C*a^2*b*c*d^2*x^2-21*C*a*b^2*c^2*d*x^2-3*C*b^3*c^3*x^2+6*D*a^3*c*d^2* 
x^2+36*D*a^2*b*c^2*d*x^2+6*D*a*b^2*c^3*x^2-6*A*a^2*b*d^3*x-36*A*a*b^2*c*d^ 
2*x-6*A*b^3*c^2*d*x+3*B*a^3*d^3*x+21*B*a^2*b*c*d^2*x+21*B*a*b^2*c^2*d*x+3* 
B*b^3*c^3*x-12*C*a^3*c*d^2*x-24*C*a^2*b*c^2*d*x-12*C*a*b^2*c^3*x+24*D*a^3* 
c^2*d*x+24*D*a^2*b*c^3*x+A*a^3*d^3-9*A*a^2*b*c*d^2-9*A*a*b^2*c^2*d+A*b^3*c 
^3+2*B*a^3*c*d^2+12*B*a^2*b*c^2*d+2*B*a*b^2*c^3-8*C*a^3*c^2*d-8*C*a^2*b*c^ 
3+16*D*a^3*c^3)/(a*d-b*c)^4/(b*x+a)^(3/2)/(d*x+c)^(3/2)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 734 vs. \(2 (304) = 608\).

Time = 40.18 (sec) , antiderivative size = 734, normalized size of antiderivative = 2.27 \[ \int \frac {A+B x+C x^2+D x^3}{(a+b x)^{5/2} (c+d x)^{5/2}} \, dx=-\frac {2 \, {\left (A a^{3} d^{3} + {\left (16 \, D a^{3} - 8 \, C a^{2} b + 2 \, B a b^{2} + A b^{3}\right )} c^{3} - {\left (8 \, C a^{3} - 12 \, B a^{2} b + 9 \, A a b^{2}\right )} c^{2} d + {\left (2 \, B a^{3} - 9 \, A a^{2} b\right )} c d^{2} - {\left (D b^{3} c^{3} - {\left (9 \, D a b^{2} - 2 \, C b^{3}\right )} c^{2} d - {\left (9 \, D a^{2} b - 12 \, C a b^{2} + 8 \, B b^{3}\right )} c d^{2} + {\left (D a^{3} + 2 \, C a^{2} b - 8 \, B a b^{2} + 16 \, A b^{3}\right )} d^{3}\right )} x^{3} + 3 \, {\left ({\left (2 \, D a b^{2} - C b^{3}\right )} c^{3} + {\left (12 \, D a^{2} b - 7 \, C a b^{2} + 4 \, B b^{3}\right )} c^{2} d + {\left (2 \, D a^{3} - 7 \, C a^{2} b + 8 \, B a b^{2} - 8 \, A b^{3}\right )} c d^{2} - {\left (C a^{3} - 4 \, B a^{2} b + 8 \, A a b^{2}\right )} d^{3}\right )} x^{2} + 3 \, {\left ({\left (8 \, D a^{2} b - 4 \, C a b^{2} + B b^{3}\right )} c^{3} + {\left (8 \, D a^{3} - 8 \, C a^{2} b + 7 \, B a b^{2} - 2 \, A b^{3}\right )} c^{2} d - {\left (4 \, C a^{3} - 7 \, B a^{2} b + 12 \, A a b^{2}\right )} c d^{2} + {\left (B a^{3} - 2 \, A a^{2} b\right )} d^{3}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{3 \, {\left (a^{2} b^{4} c^{6} - 4 \, a^{3} b^{3} c^{5} d + 6 \, a^{4} b^{2} c^{4} d^{2} - 4 \, a^{5} b c^{3} d^{3} + a^{6} c^{2} d^{4} + {\left (b^{6} c^{4} d^{2} - 4 \, a b^{5} c^{3} d^{3} + 6 \, a^{2} b^{4} c^{2} d^{4} - 4 \, a^{3} b^{3} c d^{5} + a^{4} b^{2} d^{6}\right )} x^{4} + 2 \, {\left (b^{6} c^{5} d - 3 \, a b^{5} c^{4} d^{2} + 2 \, a^{2} b^{4} c^{3} d^{3} + 2 \, a^{3} b^{3} c^{2} d^{4} - 3 \, a^{4} b^{2} c d^{5} + a^{5} b d^{6}\right )} x^{3} + {\left (b^{6} c^{6} - 9 \, a^{2} b^{4} c^{4} d^{2} + 16 \, a^{3} b^{3} c^{3} d^{3} - 9 \, a^{4} b^{2} c^{2} d^{4} + a^{6} d^{6}\right )} x^{2} + 2 \, {\left (a b^{5} c^{6} - 3 \, a^{2} b^{4} c^{5} d + 2 \, a^{3} b^{3} c^{4} d^{2} + 2 \, a^{4} b^{2} c^{3} d^{3} - 3 \, a^{5} b c^{2} d^{4} + a^{6} c d^{5}\right )} x\right )}} \] Input:

integrate((D*x^3+C*x^2+B*x+A)/(b*x+a)^(5/2)/(d*x+c)^(5/2),x, algorithm="fr 
icas")
 

Output:

-2/3*(A*a^3*d^3 + (16*D*a^3 - 8*C*a^2*b + 2*B*a*b^2 + A*b^3)*c^3 - (8*C*a^ 
3 - 12*B*a^2*b + 9*A*a*b^2)*c^2*d + (2*B*a^3 - 9*A*a^2*b)*c*d^2 - (D*b^3*c 
^3 - (9*D*a*b^2 - 2*C*b^3)*c^2*d - (9*D*a^2*b - 12*C*a*b^2 + 8*B*b^3)*c*d^ 
2 + (D*a^3 + 2*C*a^2*b - 8*B*a*b^2 + 16*A*b^3)*d^3)*x^3 + 3*((2*D*a*b^2 - 
C*b^3)*c^3 + (12*D*a^2*b - 7*C*a*b^2 + 4*B*b^3)*c^2*d + (2*D*a^3 - 7*C*a^2 
*b + 8*B*a*b^2 - 8*A*b^3)*c*d^2 - (C*a^3 - 4*B*a^2*b + 8*A*a*b^2)*d^3)*x^2 
 + 3*((8*D*a^2*b - 4*C*a*b^2 + B*b^3)*c^3 + (8*D*a^3 - 8*C*a^2*b + 7*B*a*b 
^2 - 2*A*b^3)*c^2*d - (4*C*a^3 - 7*B*a^2*b + 12*A*a*b^2)*c*d^2 + (B*a^3 - 
2*A*a^2*b)*d^3)*x)*sqrt(b*x + a)*sqrt(d*x + c)/(a^2*b^4*c^6 - 4*a^3*b^3*c^ 
5*d + 6*a^4*b^2*c^4*d^2 - 4*a^5*b*c^3*d^3 + a^6*c^2*d^4 + (b^6*c^4*d^2 - 4 
*a*b^5*c^3*d^3 + 6*a^2*b^4*c^2*d^4 - 4*a^3*b^3*c*d^5 + a^4*b^2*d^6)*x^4 + 
2*(b^6*c^5*d - 3*a*b^5*c^4*d^2 + 2*a^2*b^4*c^3*d^3 + 2*a^3*b^3*c^2*d^4 - 3 
*a^4*b^2*c*d^5 + a^5*b*d^6)*x^3 + (b^6*c^6 - 9*a^2*b^4*c^4*d^2 + 16*a^3*b^ 
3*c^3*d^3 - 9*a^4*b^2*c^2*d^4 + a^6*d^6)*x^2 + 2*(a*b^5*c^6 - 3*a^2*b^4*c^ 
5*d + 2*a^3*b^3*c^4*d^2 + 2*a^4*b^2*c^3*d^3 - 3*a^5*b*c^2*d^4 + a^6*c*d^5) 
*x)
 

Sympy [F]

\[ \int \frac {A+B x+C x^2+D x^3}{(a+b x)^{5/2} (c+d x)^{5/2}} \, dx=\int \frac {A + B x + C x^{2} + D x^{3}}{\left (a + b x\right )^{\frac {5}{2}} \left (c + d x\right )^{\frac {5}{2}}}\, dx \] Input:

integrate((D*x**3+C*x**2+B*x+A)/(b*x+a)**(5/2)/(d*x+c)**(5/2),x)
 

Output:

Integral((A + B*x + C*x**2 + D*x**3)/((a + b*x)**(5/2)*(c + d*x)**(5/2)), 
x)
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {A+B x+C x^2+D x^3}{(a+b x)^{5/2} (c+d x)^{5/2}} \, dx=\text {Exception raised: ValueError} \] Input:

integrate((D*x^3+C*x^2+B*x+A)/(b*x+a)^(5/2)/(d*x+c)^(5/2),x, algorithm="ma 
xima")
 

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. 2066 vs. \(2 (304) = 608\).

Time = 1.06 (sec) , antiderivative size = 2066, normalized size of antiderivative = 6.38 \[ \int \frac {A+B x+C x^2+D x^3}{(a+b x)^{5/2} (c+d x)^{5/2}} \, dx=\text {Too large to display} \] Input:

integrate((D*x^3+C*x^2+B*x+A)/(b*x+a)^(5/2)/(d*x+c)^(5/2),x, algorithm="gi 
ac")
 

Output:

2/3*sqrt(b*x + a)*((D*b^7*c^6*d*abs(b) - 12*D*a*b^6*c^5*d^2*abs(b) + 2*C*b 
^7*c^5*d^2*abs(b) + 30*D*a^2*b^5*c^4*d^3*abs(b) - 5*B*b^7*c^4*d^3*abs(b) - 
 28*D*a^3*b^4*c^3*d^4*abs(b) - 12*C*a^2*b^5*c^3*d^4*abs(b) + 12*B*a*b^6*c^ 
3*d^4*abs(b) + 8*A*b^7*c^3*d^4*abs(b) + 9*D*a^4*b^3*c^2*d^5*abs(b) + 16*C* 
a^3*b^4*c^2*d^5*abs(b) - 6*B*a^2*b^5*c^2*d^5*abs(b) - 24*A*a*b^6*c^2*d^5*a 
bs(b) - 6*C*a^4*b^3*c*d^6*abs(b) - 4*B*a^3*b^4*c*d^6*abs(b) + 24*A*a^2*b^5 
*c*d^6*abs(b) + 3*B*a^4*b^3*d^7*abs(b) - 8*A*a^3*b^4*d^7*abs(b))*(b*x + a) 
/(b^9*c^7*d - 7*a*b^8*c^6*d^2 + 21*a^2*b^7*c^5*d^3 - 35*a^3*b^6*c^4*d^4 + 
35*a^4*b^5*c^3*d^5 - 21*a^5*b^4*c^2*d^6 + 7*a^6*b^3*c*d^7 - a^7*b^2*d^8) - 
 3*(3*D*a*b^7*c^6*d*abs(b) - C*b^8*c^6*d*abs(b) - 12*D*a^2*b^6*c^5*d^2*abs 
(b) + 2*C*a*b^7*c^5*d^2*abs(b) + 2*B*b^8*c^5*d^2*abs(b) + 18*D*a^3*b^5*c^4 
*d^3*abs(b) + 2*C*a^2*b^6*c^4*d^3*abs(b) - 7*B*a*b^7*c^4*d^3*abs(b) - 3*A* 
b^8*c^4*d^3*abs(b) - 12*D*a^4*b^4*c^3*d^4*abs(b) - 8*C*a^3*b^5*c^3*d^4*abs 
(b) + 8*B*a^2*b^6*c^3*d^4*abs(b) + 12*A*a*b^7*c^3*d^4*abs(b) + 3*D*a^5*b^3 
*c^2*d^5*abs(b) + 7*C*a^4*b^4*c^2*d^5*abs(b) - 2*B*a^3*b^5*c^2*d^5*abs(b) 
- 18*A*a^2*b^6*c^2*d^5*abs(b) - 2*C*a^5*b^3*c*d^6*abs(b) - 2*B*a^4*b^4*c*d 
^6*abs(b) + 12*A*a^3*b^5*c*d^6*abs(b) + B*a^5*b^3*d^7*abs(b) - 3*A*a^4*b^4 
*d^7*abs(b))/(b^9*c^7*d - 7*a*b^8*c^6*d^2 + 21*a^2*b^7*c^5*d^3 - 35*a^3*b^ 
6*c^4*d^4 + 35*a^4*b^5*c^3*d^5 - 21*a^5*b^4*c^2*d^6 + 7*a^6*b^3*c*d^7 - a^ 
7*b^2*d^8))/(b^2*c + (b*x + a)*b*d - a*b*d)^(3/2) - 4/3*(9*sqrt(b*d)*D*...
 

Mupad [F(-1)]

Timed out. \[ \int \frac {A+B x+C x^2+D x^3}{(a+b x)^{5/2} (c+d x)^{5/2}} \, dx=\int \frac {A+B\,x+C\,x^2+x^3\,D}{{\left (a+b\,x\right )}^{5/2}\,{\left (c+d\,x\right )}^{5/2}} \,d x \] Input:

int((A + B*x + C*x^2 + x^3*D)/((a + b*x)^(5/2)*(c + d*x)^(5/2)),x)
 

Output:

int((A + B*x + C*x^2 + x^3*D)/((a + b*x)^(5/2)*(c + d*x)^(5/2)), x)
 

Reduce [F]

\[ \int \frac {A+B x+C x^2+D x^3}{(a+b x)^{5/2} (c+d x)^{5/2}} \, dx=\int \frac {D x^{3}+C \,x^{2}+B x +A}{\left (b x +a \right )^{\frac {5}{2}} \left (d x +c \right )^{\frac {5}{2}}}d x \] Input:

int((D*x^3+C*x^2+B*x+A)/(b*x+a)^(5/2)/(d*x+c)^(5/2),x)
 

Output:

int((D*x^3+C*x^2+B*x+A)/(b*x+a)^(5/2)/(d*x+c)^(5/2),x)