\(\int \frac {(a+b x) (A+B x+C x^2+D x^3)}{(c+d x)^{5/2}} \, dx\) [96]

Optimal result

Integrand size = 30, antiderivative size = 210 \[ \int \frac {(a+b x) \left (A+B x+C x^2+D x^3\right )}{(c+d x)^{5/2}} \, dx=\frac {2 (b c-a d) \left (c^2 C d-B c d^2+A d^3-c^3 D\right )}{3 d^5 (c+d x)^{3/2}}+\frac {2 \left (a d \left (2 c C d-B d^2-3 c^2 D\right )-b \left (3 c^2 C d-2 B c d^2+A d^3-4 c^3 D\right )\right )}{d^5 \sqrt {c+d x}}+\frac {2 \left (a d (C d-3 c D)-b \left (3 c C d-B d^2-6 c^2 D\right )\right ) \sqrt {c+d x}}{d^5}+\frac {2 (b C d-4 b c D+a d D) (c+d x)^{3/2}}{3 d^5}+\frac {2 b D (c+d x)^{5/2}}{5 d^5} \] Output:

2/3*(-a*d+b*c)*(A*d^3-B*c*d^2+C*c^2*d-D*c^3)/d^5/(d*x+c)^(3/2)+2*(a*d*(-B* 
d^2+2*C*c*d-3*D*c^2)-b*(A*d^3-2*B*c*d^2+3*C*c^2*d-4*D*c^3))/d^5/(d*x+c)^(1 
/2)+2*(a*d*(C*d-3*D*c)-b*(-B*d^2+3*C*c*d-6*D*c^2))*(d*x+c)^(1/2)/d^5+2/3*( 
C*b*d+D*a*d-4*D*b*c)*(d*x+c)^(3/2)/d^5+2/5*b*D*(d*x+c)^(5/2)/d^5
 

Mathematica [A] (verified)

Time = 0.18 (sec) , antiderivative size = 177, normalized size of antiderivative = 0.84 \[ \int \frac {(a+b x) \left (A+B x+C x^2+D x^3\right )}{(c+d x)^{5/2}} \, dx=\frac {2 \left (-5 a d \left (16 c^3 D-8 c^2 d (C-3 D x)+2 c d^2 (B+3 x (-2 C+D x))+d^3 \left (A+3 B x-x^2 (3 C+D x)\right )\right )+b \left (128 c^4 D+c^3 (-80 C d+192 d D x)+8 c^2 d^2 (5 B+3 x (-5 C+2 D x))+d^4 x \left (-15 A+x \left (15 B+5 C x+3 D x^2\right )\right )-2 c d^3 \left (5 A+x \left (-30 B+15 C x+4 D x^2\right )\right )\right )\right )}{15 d^5 (c+d x)^{3/2}} \] Input:

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

Output:

(2*(-5*a*d*(16*c^3*D - 8*c^2*d*(C - 3*D*x) + 2*c*d^2*(B + 3*x*(-2*C + D*x) 
) + d^3*(A + 3*B*x - x^2*(3*C + D*x))) + b*(128*c^4*D + c^3*(-80*C*d + 192 
*d*D*x) + 8*c^2*d^2*(5*B + 3*x*(-5*C + 2*D*x)) + d^4*x*(-15*A + x*(15*B + 
5*C*x + 3*D*x^2)) - 2*c*d^3*(5*A + x*(-30*B + 15*C*x + 4*D*x^2)))))/(15*d^ 
5*(c + d*x)^(3/2))
 

Rubi [A] (verified)

Time = 0.44 (sec) , antiderivative size = 210, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {2123, 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.

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

\(\Big \downarrow \) 2123

\(\displaystyle \int \left (\frac {b \left (A d^3-2 B c d^2-4 c^3 D+3 c^2 C d\right )-a d \left (-B d^2-3 c^2 D+2 c C d\right )}{d^4 (c+d x)^{3/2}}+\frac {(a d-b c) \left (A d^3-B c d^2+c^3 (-D)+c^2 C d\right )}{d^4 (c+d x)^{5/2}}+\frac {a d (C d-3 c D)-b \left (-B d^2-6 c^2 D+3 c C d\right )}{d^4 \sqrt {c+d x}}+\frac {\sqrt {c+d x} (a d D-4 b c D+b C d)}{d^4}+\frac {b D (c+d x)^{3/2}}{d^4}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {2 \left (a d \left (-B d^2-3 c^2 D+2 c C d\right )-b \left (A d^3-2 B c d^2-4 c^3 D+3 c^2 C d\right )\right )}{d^5 \sqrt {c+d x}}+\frac {2 (b c-a d) \left (A d^3-B c d^2+c^3 (-D)+c^2 C d\right )}{3 d^5 (c+d x)^{3/2}}+\frac {2 \sqrt {c+d x} \left (a d (C d-3 c D)-b \left (-B d^2-6 c^2 D+3 c C d\right )\right )}{d^5}+\frac {2 (c+d x)^{3/2} (a d D-4 b c D+b C d)}{3 d^5}+\frac {2 b D (c+d x)^{5/2}}{5 d^5}\)

Input:

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

Output:

(2*(b*c - a*d)*(c^2*C*d - B*c*d^2 + A*d^3 - c^3*D))/(3*d^5*(c + d*x)^(3/2) 
) + (2*(a*d*(2*c*C*d - B*d^2 - 3*c^2*D) - b*(3*c^2*C*d - 2*B*c*d^2 + A*d^3 
 - 4*c^3*D)))/(d^5*Sqrt[c + d*x]) + (2*(a*d*(C*d - 3*c*D) - b*(3*c*C*d - B 
*d^2 - 6*c^2*D))*Sqrt[c + d*x])/d^5 + (2*(b*C*d - 4*b*c*D + a*d*D)*(c + d* 
x)^(3/2))/(3*d^5) + (2*b*D*(c + d*x)^(5/2))/(5*d^5)
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2123

Int[(Px_)*((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] 
:> Int[ExpandIntegrand[Px*(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c 
, d, m, n}, x] && PolyQ[Px, x] && (IntegersQ[m, n] || IGtQ[m, -2])
 

Maple [A] (verified)

Time = 0.46 (sec) , antiderivative size = 168, normalized size of antiderivative = 0.80

method	result	size
pseudoelliptic	\(-\frac {2 \left (\left (-\frac {3 D b \,x^{4}}{5}+\left (-C b -D a \right ) x^{3}+\left (-3 B b -3 C a \right ) x^{2}+\left (3 A b +3 B a \right ) x +A a \right ) d^{4}+2 c \left (\frac {4 D b \,x^{3}}{5}+\left (3 C b +3 D a \right ) x^{2}+\left (-6 B b -6 C a \right ) x +A b +B a \right ) d^{3}-8 \left (\frac {6 D b \,x^{2}}{5}+\left (-3 C b -3 D a \right ) x +B b +C a \right ) c^{2} d^{2}+16 \left (-\frac {12}{5} D b x +C b +D a \right ) c^{3} d -\frac {128 D b \,c^{4}}{5}\right )}{3 \left (x d +c \right )^{\frac {3}{2}} d^{5}}\)	\(168\)
gosper	\(-\frac {2 \left (-3 D b \,x^{4} d^{4}-5 C \,x^{3} b \,d^{4}-5 D x^{3} a \,d^{4}+8 D x^{3} b c \,d^{3}-15 B \,x^{2} b \,d^{4}-15 C \,x^{2} a \,d^{4}+30 C \,x^{2} b c \,d^{3}+30 D x^{2} a c \,d^{3}-48 D x^{2} b \,c^{2} d^{2}+15 A x b \,d^{4}+15 B x a \,d^{4}-60 B x b c \,d^{3}-60 C x a c \,d^{3}+120 C x b \,c^{2} d^{2}+120 D x a \,c^{2} d^{2}-192 D x b \,c^{3} d +5 A a \,d^{4}+10 A c \,d^{3} b +10 B c \,d^{3} a -40 B b \,c^{2} d^{2}-40 C a \,c^{2} d^{2}+80 C b \,c^{3} d +80 D a \,c^{3} d -128 D b \,c^{4}\right )}{15 \left (x d +c \right )^{\frac {3}{2}} d^{5}}\)	\(241\)
trager	\(-\frac {2 \left (-3 D b \,x^{4} d^{4}-5 C \,x^{3} b \,d^{4}-5 D x^{3} a \,d^{4}+8 D x^{3} b c \,d^{3}-15 B \,x^{2} b \,d^{4}-15 C \,x^{2} a \,d^{4}+30 C \,x^{2} b c \,d^{3}+30 D x^{2} a c \,d^{3}-48 D x^{2} b \,c^{2} d^{2}+15 A x b \,d^{4}+15 B x a \,d^{4}-60 B x b c \,d^{3}-60 C x a c \,d^{3}+120 C x b \,c^{2} d^{2}+120 D x a \,c^{2} d^{2}-192 D x b \,c^{3} d +5 A a \,d^{4}+10 A c \,d^{3} b +10 B c \,d^{3} a -40 B b \,c^{2} d^{2}-40 C a \,c^{2} d^{2}+80 C b \,c^{3} d +80 D a \,c^{3} d -128 D b \,c^{4}\right )}{15 \left (x d +c \right )^{\frac {3}{2}} d^{5}}\)	\(241\)
orering	\(-\frac {2 \left (-3 D b \,x^{4} d^{4}-5 C \,x^{3} b \,d^{4}-5 D x^{3} a \,d^{4}+8 D x^{3} b c \,d^{3}-15 B \,x^{2} b \,d^{4}-15 C \,x^{2} a \,d^{4}+30 C \,x^{2} b c \,d^{3}+30 D x^{2} a c \,d^{3}-48 D x^{2} b \,c^{2} d^{2}+15 A x b \,d^{4}+15 B x a \,d^{4}-60 B x b c \,d^{3}-60 C x a c \,d^{3}+120 C x b \,c^{2} d^{2}+120 D x a \,c^{2} d^{2}-192 D x b \,c^{3} d +5 A a \,d^{4}+10 A c \,d^{3} b +10 B c \,d^{3} a -40 B b \,c^{2} d^{2}-40 C a \,c^{2} d^{2}+80 C b \,c^{3} d +80 D a \,c^{3} d -128 D b \,c^{4}\right )}{15 \left (x d +c \right )^{\frac {3}{2}} d^{5}}\)	\(241\)
derivativedivides	\(\frac {\frac {2 b D \left (x d +c \right )^{\frac {5}{2}}}{5}+\frac {2 C b d \left (x d +c \right )^{\frac {3}{2}}}{3}+\frac {2 D a d \left (x d +c \right )^{\frac {3}{2}}}{3}-\frac {8 D b c \left (x d +c \right )^{\frac {3}{2}}}{3}+2 B b \,d^{2} \sqrt {x d +c}+2 C a \,d^{2} \sqrt {x d +c}-6 C b c d \sqrt {x d +c}-6 D a c d \sqrt {x d +c}+12 D b \,c^{2} \sqrt {x d +c}-\frac {2 \left (A b \,d^{3}+B a \,d^{3}-2 B b c \,d^{2}-2 C a c \,d^{2}+3 C b \,c^{2} d +3 D a \,c^{2} d -4 D b \,c^{3}\right )}{\sqrt {x d +c}}-\frac {2 \left (A a \,d^{4}-A c \,d^{3} b -B c \,d^{3} a +B b \,c^{2} d^{2}+C a \,c^{2} d^{2}-C b \,c^{3} d -D a \,c^{3} d +D b \,c^{4}\right )}{3 \left (x d +c \right )^{\frac {3}{2}}}}{d^{5}}\)	\(253\)
default	\(\frac {\frac {2 b D \left (x d +c \right )^{\frac {5}{2}}}{5}+\frac {2 C b d \left (x d +c \right )^{\frac {3}{2}}}{3}+\frac {2 D a d \left (x d +c \right )^{\frac {3}{2}}}{3}-\frac {8 D b c \left (x d +c \right )^{\frac {3}{2}}}{3}+2 B b \,d^{2} \sqrt {x d +c}+2 C a \,d^{2} \sqrt {x d +c}-6 C b c d \sqrt {x d +c}-6 D a c d \sqrt {x d +c}+12 D b \,c^{2} \sqrt {x d +c}-\frac {2 \left (A b \,d^{3}+B a \,d^{3}-2 B b c \,d^{2}-2 C a c \,d^{2}+3 C b \,c^{2} d +3 D a \,c^{2} d -4 D b \,c^{3}\right )}{\sqrt {x d +c}}-\frac {2 \left (A a \,d^{4}-A c \,d^{3} b -B c \,d^{3} a +B b \,c^{2} d^{2}+C a \,c^{2} d^{2}-C b \,c^{3} d -D a \,c^{3} d +D b \,c^{4}\right )}{3 \left (x d +c \right )^{\frac {3}{2}}}}{d^{5}}\)	\(253\)

Input:

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

Output:

-2/3*((-3/5*D*b*x^4+(-C*b-D*a)*x^3+(-3*B*b-3*C*a)*x^2+(3*A*b+3*B*a)*x+A*a) 
*d^4+2*c*(4/5*D*b*x^3+(3*C*b+3*D*a)*x^2+(-6*B*b-6*C*a)*x+A*b+B*a)*d^3-8*(6 
/5*D*b*x^2+(-3*C*b-3*D*a)*x+B*b+C*a)*c^2*d^2+16*(-12/5*D*b*x+C*b+D*a)*c^3* 
d-128/5*D*b*c^4)/(d*x+c)^(3/2)/d^5
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 218, normalized size of antiderivative = 1.04 \[ \int \frac {(a+b x) \left (A+B x+C x^2+D x^3\right )}{(c+d x)^{5/2}} \, dx=\frac {2 \, {\left (3 \, D b d^{4} x^{4} + 128 \, D b c^{4} - 5 \, A a d^{4} - 80 \, {\left (D a + C b\right )} c^{3} d + 40 \, {\left (C a + B b\right )} c^{2} d^{2} - 10 \, {\left (B a + A b\right )} c d^{3} - {\left (8 \, D b c d^{3} - 5 \, {\left (D a + C b\right )} d^{4}\right )} x^{3} + 3 \, {\left (16 \, D b c^{2} d^{2} - 10 \, {\left (D a + C b\right )} c d^{3} + 5 \, {\left (C a + B b\right )} d^{4}\right )} x^{2} + 3 \, {\left (64 \, D b c^{3} d - 40 \, {\left (D a + C b\right )} c^{2} d^{2} + 20 \, {\left (C a + B b\right )} c d^{3} - 5 \, {\left (B a + A b\right )} d^{4}\right )} x\right )} \sqrt {d x + c}}{15 \, {\left (d^{7} x^{2} + 2 \, c d^{6} x + c^{2} d^{5}\right )}} \] Input:

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

Output:

2/15*(3*D*b*d^4*x^4 + 128*D*b*c^4 - 5*A*a*d^4 - 80*(D*a + C*b)*c^3*d + 40* 
(C*a + B*b)*c^2*d^2 - 10*(B*a + A*b)*c*d^3 - (8*D*b*c*d^3 - 5*(D*a + C*b)* 
d^4)*x^3 + 3*(16*D*b*c^2*d^2 - 10*(D*a + C*b)*c*d^3 + 5*(C*a + B*b)*d^4)*x 
^2 + 3*(64*D*b*c^3*d - 40*(D*a + C*b)*c^2*d^2 + 20*(C*a + B*b)*c*d^3 - 5*( 
B*a + A*b)*d^4)*x)*sqrt(d*x + c)/(d^7*x^2 + 2*c*d^6*x + c^2*d^5)
 

Sympy [A] (verification not implemented)

Time = 9.16 (sec) , antiderivative size = 282, normalized size of antiderivative = 1.34 \[ \int \frac {(a+b x) \left (A+B x+C x^2+D x^3\right )}{(c+d x)^{5/2}} \, dx=\begin {cases} \frac {2 \left (\frac {D b \left (c + d x\right )^{\frac {5}{2}}}{5 d^{4}} + \frac {\left (c + d x\right )^{\frac {3}{2}} \left (C b d + D a d - 4 D b c\right )}{3 d^{4}} + \frac {\sqrt {c + d x} \left (B b d^{2} + C a d^{2} - 3 C b c d - 3 D a c d + 6 D b c^{2}\right )}{d^{4}} - \frac {A b d^{3} + B a d^{3} - 2 B b c d^{2} - 2 C a c d^{2} + 3 C b c^{2} d + 3 D a c^{2} d - 4 D b c^{3}}{d^{4} \sqrt {c + d x}} + \frac {\left (a d - b c\right ) \left (- A d^{3} + B c d^{2} - C c^{2} d + D c^{3}\right )}{3 d^{4} \left (c + d x\right )^{\frac {3}{2}}}\right )}{d} & \text {for}\: d \neq 0 \\\frac {A a x + \frac {D b x^{5}}{5} + \frac {x^{4} \left (C b + D a\right )}{4} + \frac {x^{3} \left (B b + C a\right )}{3} + \frac {x^{2} \left (A b + B a\right )}{2}}{c^{\frac {5}{2}}} & \text {otherwise} \end {cases} \] Input:

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

Output:

Piecewise((2*(D*b*(c + d*x)**(5/2)/(5*d**4) + (c + d*x)**(3/2)*(C*b*d + D* 
a*d - 4*D*b*c)/(3*d**4) + sqrt(c + d*x)*(B*b*d**2 + C*a*d**2 - 3*C*b*c*d - 
 3*D*a*c*d + 6*D*b*c**2)/d**4 - (A*b*d**3 + B*a*d**3 - 2*B*b*c*d**2 - 2*C* 
a*c*d**2 + 3*C*b*c**2*d + 3*D*a*c**2*d - 4*D*b*c**3)/(d**4*sqrt(c + d*x)) 
+ (a*d - b*c)*(-A*d**3 + B*c*d**2 - C*c**2*d + D*c**3)/(3*d**4*(c + d*x)** 
(3/2)))/d, Ne(d, 0)), ((A*a*x + D*b*x**5/5 + x**4*(C*b + D*a)/4 + x**3*(B* 
b + C*a)/3 + x**2*(A*b + B*a)/2)/c**(5/2), True))
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 204, normalized size of antiderivative = 0.97 \[ \int \frac {(a+b x) \left (A+B x+C x^2+D x^3\right )}{(c+d x)^{5/2}} \, dx=\frac {2 \, {\left (\frac {3 \, {\left (d x + c\right )}^{\frac {5}{2}} D b - 5 \, {\left (4 \, D b c - {\left (D a + C b\right )} d\right )} {\left (d x + c\right )}^{\frac {3}{2}} + 15 \, {\left (6 \, D b c^{2} - 3 \, {\left (D a + C b\right )} c d + {\left (C a + B b\right )} d^{2}\right )} \sqrt {d x + c}}{d^{4}} - \frac {5 \, {\left (D b c^{4} + A a d^{4} - {\left (D a + C b\right )} c^{3} d + {\left (C a + B b\right )} c^{2} d^{2} - {\left (B a + A b\right )} c d^{3} - 3 \, {\left (4 \, D b c^{3} - 3 \, {\left (D a + C b\right )} c^{2} d + 2 \, {\left (C a + B b\right )} c d^{2} - {\left (B a + A b\right )} d^{3}\right )} {\left (d x + c\right )}\right )}}{{\left (d x + c\right )}^{\frac {3}{2}} d^{4}}\right )}}{15 \, d} \] Input:

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

Output:

2/15*((3*(d*x + c)^(5/2)*D*b - 5*(4*D*b*c - (D*a + C*b)*d)*(d*x + c)^(3/2) 
 + 15*(6*D*b*c^2 - 3*(D*a + C*b)*c*d + (C*a + B*b)*d^2)*sqrt(d*x + c))/d^4 
 - 5*(D*b*c^4 + A*a*d^4 - (D*a + C*b)*c^3*d + (C*a + B*b)*c^2*d^2 - (B*a + 
 A*b)*c*d^3 - 3*(4*D*b*c^3 - 3*(D*a + C*b)*c^2*d + 2*(C*a + B*b)*c*d^2 - ( 
B*a + A*b)*d^3)*(d*x + c))/((d*x + c)^(3/2)*d^4))/d
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 302, normalized size of antiderivative = 1.44 \[ \int \frac {(a+b x) \left (A+B x+C x^2+D x^3\right )}{(c+d x)^{5/2}} \, dx=\frac {2 \, {\left (12 \, {\left (d x + c\right )} D b c^{3} - D b c^{4} - 9 \, {\left (d x + c\right )} D a c^{2} d - 9 \, {\left (d x + c\right )} C b c^{2} d + D a c^{3} d + C b c^{3} d + 6 \, {\left (d x + c\right )} C a c d^{2} + 6 \, {\left (d x + c\right )} B b c d^{2} - C a c^{2} d^{2} - B b c^{2} d^{2} - 3 \, {\left (d x + c\right )} B a d^{3} - 3 \, {\left (d x + c\right )} A b d^{3} + B a c d^{3} + A b c d^{3} - A a d^{4}\right )}}{3 \, {\left (d x + c\right )}^{\frac {3}{2}} d^{5}} + \frac {2 \, {\left (3 \, {\left (d x + c\right )}^{\frac {5}{2}} D b d^{20} - 20 \, {\left (d x + c\right )}^{\frac {3}{2}} D b c d^{20} + 90 \, \sqrt {d x + c} D b c^{2} d^{20} + 5 \, {\left (d x + c\right )}^{\frac {3}{2}} D a d^{21} + 5 \, {\left (d x + c\right )}^{\frac {3}{2}} C b d^{21} - 45 \, \sqrt {d x + c} D a c d^{21} - 45 \, \sqrt {d x + c} C b c d^{21} + 15 \, \sqrt {d x + c} C a d^{22} + 15 \, \sqrt {d x + c} B b d^{22}\right )}}{15 \, d^{25}} \] Input:

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

Output:

2/3*(12*(d*x + c)*D*b*c^3 - D*b*c^4 - 9*(d*x + c)*D*a*c^2*d - 9*(d*x + c)* 
C*b*c^2*d + D*a*c^3*d + C*b*c^3*d + 6*(d*x + c)*C*a*c*d^2 + 6*(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 - 3*(d*x + c)*A* 
b*d^3 + B*a*c*d^3 + A*b*c*d^3 - A*a*d^4)/((d*x + c)^(3/2)*d^5) + 2/15*(3*( 
d*x + c)^(5/2)*D*b*d^20 - 20*(d*x + c)^(3/2)*D*b*c*d^20 + 90*sqrt(d*x + c) 
*D*b*c^2*d^20 + 5*(d*x + c)^(3/2)*D*a*d^21 + 5*(d*x + c)^(3/2)*C*b*d^21 - 
45*sqrt(d*x + c)*D*a*c*d^21 - 45*sqrt(d*x + c)*C*b*c*d^21 + 15*sqrt(d*x + 
c)*C*a*d^22 + 15*sqrt(d*x + c)*B*b*d^22)/d^25
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 156, normalized size of antiderivative = 0.74 \[ \int \frac {(a+b x) \left (A+B x+C x^2+D x^3\right )}{(c+d x)^{5/2}} \, dx=\frac {\frac {2}{5} b \,d^{4} x^{4}+\frac {2}{3} a \,d^{4} x^{3}-\frac {2}{5} b c \,d^{3} x^{3}-2 a c \,d^{3} x^{2}+2 b^{2} d^{3} x^{2}+\frac {12}{5} b \,c^{2} d^{2} x^{2}-4 a b \,d^{3} x -8 a \,c^{2} d^{2} x +8 b^{2} c \,d^{2} x +\frac {48}{5} b \,c^{3} d x -\frac {2}{3} a^{2} d^{3}-\frac {8}{3} a b c \,d^{2}-\frac {16}{3} a \,c^{3} d +\frac {16}{3} b^{2} c^{2} d +\frac {32}{5} b \,c^{4}}{\sqrt {d x +c}\, d^{4} \left (d x +c \right )} \] Input:

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

Output:

(2*( - 5*a**2*d**3 - 20*a*b*c*d**2 - 30*a*b*d**3*x - 40*a*c**3*d - 60*a*c* 
*2*d**2*x - 15*a*c*d**3*x**2 + 5*a*d**4*x**3 + 40*b**2*c**2*d + 60*b**2*c* 
d**2*x + 15*b**2*d**3*x**2 + 48*b*c**4 + 72*b*c**3*d*x + 18*b*c**2*d**2*x* 
*2 - 3*b*c*d**3*x**3 + 3*b*d**4*x**4))/(15*sqrt(c + d*x)*d**4*(c + d*x))