3.1.14 \(\int \frac {A+B x+C x^2+D x^3}{(a+b x) (c+d x)^{3/2}} \, dx\) [14]

3.1.14.1 Optimal result
3.1.14.2 Mathematica [A] (verified)
3.1.14.3 Rubi [A] (verified)
3.1.14.4 Maple [A] (verified)
3.1.14.5 Fricas [B] (verification not implemented)
3.1.14.6 Sympy [A] (verification not implemented)
3.1.14.7 Maxima [F(-2)]
3.1.14.8 Giac [A] (verification not implemented)
3.1.14.9 Mupad [F(-1)]

3.1.14.1 Optimal result

Integrand size = 32, antiderivative size = 193 \[ \int \frac {A+B x+C x^2+D x^3}{(a+b x) (c+d x)^{3/2}} \, dx=\frac {2 \left (c^2 C d-B c d^2+A d^3-c^3 D\right )}{d^3 (b c-a d) \sqrt {c+d x}}-\frac {2 c D \sqrt {c+d x}}{b d^3}+\frac {2 (b C d-b c D-a d D) \sqrt {c+d x}}{b^2 d^3}+\frac {2 D (c+d x)^{3/2}}{3 b d^3}-\frac {2 \left (A b^3-a \left (b^2 B-a b C+a^2 D\right )\right ) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{b^{5/2} (b c-a d)^{3/2}} \]

output
2/3*D*(d*x+c)^(3/2)/b/d^3-2*(A*b^3-a*(B*b^2-C*a*b+D*a^2))*arctanh(b^(1/2)* 
(d*x+c)^(1/2)/(-a*d+b*c)^(1/2))/b^(5/2)/(-a*d+b*c)^(3/2)+2*(A*d^3-B*c*d^2+ 
C*c^2*d-D*c^3)/d^3/(-a*d+b*c)/(d*x+c)^(1/2)-2*c*D*(d*x+c)^(1/2)/b/d^3+2*(C 
*b*d-D*a*d-D*b*c)*(d*x+c)^(1/2)/b^2/d^3
 
3.1.14.2 Mathematica [A] (verified)

Time = 0.35 (sec) , antiderivative size = 191, normalized size of antiderivative = 0.99 \[ \int \frac {A+B x+C x^2+D x^3}{(a+b x) (c+d x)^{3/2}} \, dx=\frac {-6 a^2 d^2 D (c+d x)+2 a b d (c+d x) (3 C d-2 c D+d D x)+2 b^2 \left (-3 A d^3+8 c^3 D+c^2 (-6 C d+4 d D x)+c d^2 (3 B-x (3 C+D x))\right )}{3 b^2 d^3 (-b c+a d) \sqrt {c+d x}}-\frac {2 \left (A b^3-a \left (b^2 B-a b C+a^2 D\right )\right ) \arctan \left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {-b c+a d}}\right )}{b^{5/2} (-b c+a d)^{3/2}} \]

input
Integrate[(A + B*x + C*x^2 + D*x^3)/((a + b*x)*(c + d*x)^(3/2)),x]
 
output
(-6*a^2*d^2*D*(c + d*x) + 2*a*b*d*(c + d*x)*(3*C*d - 2*c*D + d*D*x) + 2*b^ 
2*(-3*A*d^3 + 8*c^3*D + c^2*(-6*C*d + 4*d*D*x) + c*d^2*(3*B - x*(3*C + D*x 
))))/(3*b^2*d^3*(-(b*c) + a*d)*Sqrt[c + d*x]) - (2*(A*b^3 - a*(b^2*B - a*b 
*C + a^2*D))*ArcTan[(Sqrt[b]*Sqrt[c + d*x])/Sqrt[-(b*c) + a*d]])/(b^(5/2)* 
(-(b*c) + a*d)^(3/2))
 
3.1.14.3 Rubi [A] (verified)

Time = 0.45 (sec) , antiderivative size = 193, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {2122, 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.

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

\(\Big \downarrow \) 2122

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

\(\Big \downarrow \) 2009

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

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

3.1.14.3.1 Defintions of rubi rules used

rule 2009
Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2122
Int[((Px_)*((c_.) + (d_.)*(x_))^(n_))/((a_.) + (b_.)*(x_)), x_Symbol] :> In 
t[ExpandIntegrand[1/Sqrt[c + d*x], Px*((c + d*x)^(n + 1/2)/(a + b*x)), x], 
x] /; FreeQ[{a, b, c, d, n}, x] && PolyQ[Px, x] && ILtQ[n + 1/2, 0]
 
3.1.14.4 Maple [A] (verified)

Time = 1.72 (sec) , antiderivative size = 161, normalized size of antiderivative = 0.83

method result size
pseudoelliptic \(\frac {\frac {2 \sqrt {d x +c}\, \left (D b d x +3 C b d -3 D a d -5 D b c \right )}{3 b^{2}}-\frac {2 \left (A \,d^{3}-B c \,d^{2}+C \,c^{2} d -D c^{3}\right )}{\left (a d -b c \right ) \sqrt {d x +c}}-\frac {2 d^{3} \left (b^{3} A -a \,b^{2} B +C \,a^{2} b -D a^{3}\right ) \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{\left (a d -b c \right ) b^{2} \sqrt {\left (a d -b c \right ) b}}}{d^{3}}\) \(161\)
derivativedivides \(\frac {\frac {2 \left (\frac {D \left (d x +c \right )^{\frac {3}{2}} b}{3}+d b C \sqrt {d x +c}-D a d \sqrt {d x +c}-2 D c b \sqrt {d x +c}\right )}{b^{2}}-\frac {2 \left (A \,d^{3}-B c \,d^{2}+C \,c^{2} d -D c^{3}\right )}{\left (a d -b c \right ) \sqrt {d x +c}}-\frac {2 d^{3} \left (b^{3} A -a \,b^{2} B +C \,a^{2} b -D a^{3}\right ) \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{\left (a d -b c \right ) b^{2} \sqrt {\left (a d -b c \right ) b}}}{d^{3}}\) \(179\)
default \(\frac {\frac {2 \left (\frac {D \left (d x +c \right )^{\frac {3}{2}} b}{3}+d b C \sqrt {d x +c}-D a d \sqrt {d x +c}-2 D c b \sqrt {d x +c}\right )}{b^{2}}-\frac {2 \left (A \,d^{3}-B c \,d^{2}+C \,c^{2} d -D c^{3}\right )}{\left (a d -b c \right ) \sqrt {d x +c}}-\frac {2 d^{3} \left (b^{3} A -a \,b^{2} B +C \,a^{2} b -D a^{3}\right ) \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{\left (a d -b c \right ) b^{2} \sqrt {\left (a d -b c \right ) b}}}{d^{3}}\) \(179\)

input
int((D*x^3+C*x^2+B*x+A)/(b*x+a)/(d*x+c)^(3/2),x,method=_RETURNVERBOSE)
 
output
2*(1/3*(d*x+c)^(1/2)*(D*b*d*x+3*C*b*d-3*D*a*d-5*D*b*c)/b^2-(A*d^3-B*c*d^2+ 
C*c^2*d-D*c^3)/(a*d-b*c)/(d*x+c)^(1/2)-d^3*(A*b^3-B*a*b^2+C*a^2*b-D*a^3)/( 
a*d-b*c)/b^2/((a*d-b*c)*b)^(1/2)*arctan(b*(d*x+c)^(1/2)/((a*d-b*c)*b)^(1/2 
)))/d^3
 
3.1.14.5 Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 426 vs. \(2 (172) = 344\).

Time = 0.33 (sec) , antiderivative size = 866, normalized size of antiderivative = 4.49 \[ \int \frac {A+B x+C x^2+D x^3}{(a+b x) (c+d x)^{3/2}} \, dx=\left [-\frac {3 \, {\left ({\left (D a^{3} - C a^{2} b + B a b^{2} - A b^{3}\right )} d^{4} x + {\left (D a^{3} c - {\left (C a^{2} b - B a b^{2} + A b^{3}\right )} c\right )} d^{3}\right )} \sqrt {b^{2} c - a b d} \log \left (\frac {b d x + 2 \, b c - a d - 2 \, \sqrt {b^{2} c - a b d} \sqrt {d x + c}}{b x + a}\right ) + 2 \, {\left (8 \, D b^{4} c^{4} + 3 \, A a b^{3} d^{4} + 3 \, {\left (D a^{3} b c - {\left (C a^{2} b^{2} + B a b^{3} + A b^{4}\right )} c\right )} d^{3} - {\left (D a^{2} b^{2} c^{2} - 3 \, {\left (3 \, C a b^{3} + B b^{4}\right )} c^{2}\right )} d^{2} - {\left (D b^{4} c^{2} d^{2} - 2 \, D a b^{3} c d^{3} + D a^{2} b^{2} d^{4}\right )} x^{2} - 2 \, {\left (5 \, D a b^{3} c^{3} + 3 \, C b^{4} c^{3}\right )} d + {\left (4 \, D b^{4} c^{3} d + 3 \, {\left (D a^{3} b - C a^{2} b^{2}\right )} d^{4} - 2 \, {\left (D a^{2} b^{2} c - 3 \, C a b^{3} c\right )} d^{3} - {\left (5 \, D a b^{3} c^{2} + 3 \, C b^{4} c^{2}\right )} d^{2}\right )} x\right )} \sqrt {d x + c}}{3 \, {\left (b^{5} c^{3} d^{3} - 2 \, a b^{4} c^{2} d^{4} + a^{2} b^{3} c d^{5} + {\left (b^{5} c^{2} d^{4} - 2 \, a b^{4} c d^{5} + a^{2} b^{3} d^{6}\right )} x\right )}}, -\frac {2 \, {\left (3 \, {\left ({\left (D a^{3} - C a^{2} b + B a b^{2} - A b^{3}\right )} d^{4} x + {\left (D a^{3} c - {\left (C a^{2} b - B a b^{2} + A b^{3}\right )} c\right )} d^{3}\right )} \sqrt {-b^{2} c + a b d} \arctan \left (\frac {\sqrt {-b^{2} c + a b d} \sqrt {d x + c}}{b d x + b c}\right ) + {\left (8 \, D b^{4} c^{4} + 3 \, A a b^{3} d^{4} + 3 \, {\left (D a^{3} b c - {\left (C a^{2} b^{2} + B a b^{3} + A b^{4}\right )} c\right )} d^{3} - {\left (D a^{2} b^{2} c^{2} - 3 \, {\left (3 \, C a b^{3} + B b^{4}\right )} c^{2}\right )} d^{2} - {\left (D b^{4} c^{2} d^{2} - 2 \, D a b^{3} c d^{3} + D a^{2} b^{2} d^{4}\right )} x^{2} - 2 \, {\left (5 \, D a b^{3} c^{3} + 3 \, C b^{4} c^{3}\right )} d + {\left (4 \, D b^{4} c^{3} d + 3 \, {\left (D a^{3} b - C a^{2} b^{2}\right )} d^{4} - 2 \, {\left (D a^{2} b^{2} c - 3 \, C a b^{3} c\right )} d^{3} - {\left (5 \, D a b^{3} c^{2} + 3 \, C b^{4} c^{2}\right )} d^{2}\right )} x\right )} \sqrt {d x + c}\right )}}{3 \, {\left (b^{5} c^{3} d^{3} - 2 \, a b^{4} c^{2} d^{4} + a^{2} b^{3} c d^{5} + {\left (b^{5} c^{2} d^{4} - 2 \, a b^{4} c d^{5} + a^{2} b^{3} d^{6}\right )} x\right )}}\right ] \]

input
integrate((D*x^3+C*x^2+B*x+A)/(b*x+a)/(d*x+c)^(3/2),x, algorithm="fricas")
 
output
[-1/3*(3*((D*a^3 - C*a^2*b + B*a*b^2 - A*b^3)*d^4*x + (D*a^3*c - (C*a^2*b 
- B*a*b^2 + A*b^3)*c)*d^3)*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*b^4*c^4 + 3*A*a*b 
^3*d^4 + 3*(D*a^3*b*c - (C*a^2*b^2 + B*a*b^3 + A*b^4)*c)*d^3 - (D*a^2*b^2* 
c^2 - 3*(3*C*a*b^3 + B*b^4)*c^2)*d^2 - (D*b^4*c^2*d^2 - 2*D*a*b^3*c*d^3 + 
D*a^2*b^2*d^4)*x^2 - 2*(5*D*a*b^3*c^3 + 3*C*b^4*c^3)*d + (4*D*b^4*c^3*d + 
3*(D*a^3*b - C*a^2*b^2)*d^4 - 2*(D*a^2*b^2*c - 3*C*a*b^3*c)*d^3 - (5*D*a*b 
^3*c^2 + 3*C*b^4*c^2)*d^2)*x)*sqrt(d*x + c))/(b^5*c^3*d^3 - 2*a*b^4*c^2*d^ 
4 + a^2*b^3*c*d^5 + (b^5*c^2*d^4 - 2*a*b^4*c*d^5 + a^2*b^3*d^6)*x), -2/3*( 
3*((D*a^3 - C*a^2*b + B*a*b^2 - A*b^3)*d^4*x + (D*a^3*c - (C*a^2*b - B*a*b 
^2 + A*b^3)*c)*d^3)*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*b^4*c^4 + 3*A*a*b^3*d^4 + 3*(D*a^3*b*c - (C 
*a^2*b^2 + B*a*b^3 + A*b^4)*c)*d^3 - (D*a^2*b^2*c^2 - 3*(3*C*a*b^3 + B*b^4 
)*c^2)*d^2 - (D*b^4*c^2*d^2 - 2*D*a*b^3*c*d^3 + D*a^2*b^2*d^4)*x^2 - 2*(5* 
D*a*b^3*c^3 + 3*C*b^4*c^3)*d + (4*D*b^4*c^3*d + 3*(D*a^3*b - C*a^2*b^2)*d^ 
4 - 2*(D*a^2*b^2*c - 3*C*a*b^3*c)*d^3 - (5*D*a*b^3*c^2 + 3*C*b^4*c^2)*d^2) 
*x)*sqrt(d*x + c))/(b^5*c^3*d^3 - 2*a*b^4*c^2*d^4 + a^2*b^3*c*d^5 + (b^5*c 
^2*d^4 - 2*a*b^4*c*d^5 + a^2*b^3*d^6)*x)]
 
3.1.14.6 Sympy [A] (verification not implemented)

Time = 8.48 (sec) , antiderivative size = 262, normalized size of antiderivative = 1.36 \[ \int \frac {A+B x+C x^2+D x^3}{(a+b x) (c+d x)^{3/2}} \, dx=\begin {cases} \frac {2 \left (\frac {D \left (c + d x\right )^{\frac {3}{2}}}{3 b d^{2}} + \frac {- A d^{3} + B c d^{2} - C c^{2} d + D c^{3}}{d^{2} \sqrt {c + d x} \left (a d - b c\right )} + \frac {\sqrt {c + d x} \left (C b d - D a d - 2 D b c\right )}{b^{2} d^{2}} + \frac {d \left (- A b^{3} + B a b^{2} - C a^{2} b + D a^{3}\right ) \operatorname {atan}{\left (\frac {\sqrt {c + d x}}{\sqrt {\frac {a d - b c}{b}}} \right )}}{b^{3} \sqrt {\frac {a d - b c}{b}} \left (a d - b c\right )}\right )}{d} & \text {for}\: d \neq 0 \\\frac {\frac {D x^{3}}{3 b} + \frac {x^{2} \left (C b - D a\right )}{2 b^{2}} + \frac {x \left (B b^{2} - C a b + D a^{2}\right )}{b^{3}} - \frac {\left (- A b^{3} + B a b^{2} - C a^{2} b + D a^{3}\right ) \left (\begin {cases} \frac {x}{a} & \text {for}\: b = 0 \\\frac {\log {\left (a + b x \right )}}{b} & \text {otherwise} \end {cases}\right )}{b^{3}}}{c^{\frac {3}{2}}} & \text {otherwise} \end {cases} \]

input
integrate((D*x**3+C*x**2+B*x+A)/(b*x+a)/(d*x+c)**(3/2),x)
 
output
Piecewise((2*(D*(c + d*x)**(3/2)/(3*b*d**2) + (-A*d**3 + B*c*d**2 - C*c**2 
*d + D*c**3)/(d**2*sqrt(c + d*x)*(a*d - b*c)) + sqrt(c + d*x)*(C*b*d - D*a 
*d - 2*D*b*c)/(b**2*d**2) + d*(-A*b**3 + B*a*b**2 - C*a**2*b + D*a**3)*ata 
n(sqrt(c + d*x)/sqrt((a*d - b*c)/b))/(b**3*sqrt((a*d - b*c)/b)*(a*d - b*c) 
))/d, Ne(d, 0)), ((D*x**3/(3*b) + x**2*(C*b - D*a)/(2*b**2) + x*(B*b**2 - 
C*a*b + D*a**2)/b**3 - (-A*b**3 + B*a*b**2 - C*a**2*b + D*a**3)*Piecewise( 
(x/a, Eq(b, 0)), (log(a + b*x)/b, True))/b**3)/c**(3/2), True))
 
3.1.14.7 Maxima [F(-2)]

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

input
integrate((D*x^3+C*x^2+B*x+A)/(b*x+a)/(d*x+c)^(3/2),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
 
3.1.14.8 Giac [A] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.04 \[ \int \frac {A+B x+C x^2+D x^3}{(a+b x) (c+d x)^{3/2}} \, dx=-\frac {2 \, {\left (D a^{3} - C a^{2} b + B a b^{2} - A b^{3}\right )} \arctan \left (\frac {\sqrt {d x + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{{\left (b^{3} c - a b^{2} d\right )} \sqrt {-b^{2} c + a b d}} - \frac {2 \, {\left (D c^{3} - C c^{2} d + B c d^{2} - A d^{3}\right )}}{{\left (b c d^{3} - a d^{4}\right )} \sqrt {d x + c}} + \frac {2 \, {\left ({\left (d x + c\right )}^{\frac {3}{2}} D b^{2} d^{6} - 6 \, \sqrt {d x + c} D b^{2} c d^{6} - 3 \, \sqrt {d x + c} D a b d^{7} + 3 \, \sqrt {d x + c} C b^{2} d^{7}\right )}}{3 \, b^{3} d^{9}} \]

input
integrate((D*x^3+C*x^2+B*x+A)/(b*x+a)/(d*x+c)^(3/2),x, algorithm="giac")
 
output
-2*(D*a^3 - C*a^2*b + B*a*b^2 - A*b^3)*arctan(sqrt(d*x + c)*b/sqrt(-b^2*c 
+ a*b*d))/((b^3*c - a*b^2*d)*sqrt(-b^2*c + a*b*d)) - 2*(D*c^3 - C*c^2*d + 
B*c*d^2 - A*d^3)/((b*c*d^3 - a*d^4)*sqrt(d*x + c)) + 2/3*((d*x + c)^(3/2)* 
D*b^2*d^6 - 6*sqrt(d*x + c)*D*b^2*c*d^6 - 3*sqrt(d*x + c)*D*a*b*d^7 + 3*sq 
rt(d*x + c)*C*b^2*d^7)/(b^3*d^9)
 
3.1.14.9 Mupad [F(-1)]

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

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