Integrand size = 25, antiderivative size = 112 \[ \int \frac {A+B x+C x^2+D x^3}{(a+b x)^{3/2}} \, dx=-\frac {2 \left (A b^3-a \left (b^2 B-a b C+a^2 D\right )\right )}{b^4 \sqrt {a+b x}}+\frac {2 \left (b^2 B-2 a b C+3 a^2 D\right ) \sqrt {a+b x}}{b^4}+\frac {2 (b C-3 a D) (a+b x)^{3/2}}{3 b^4}+\frac {2 D (a+b x)^{5/2}}{5 b^4} \] Output:
(-2*A*b^3+2*a*(B*b^2-C*a*b+D*a^2))/b^4/(b*x+a)^(1/2)+2*(B*b^2-2*C*a*b+3*D* a^2)*(b*x+a)^(1/2)/b^4+2/3*(C*b-3*D*a)*(b*x+a)^(3/2)/b^4+2/5*D*(b*x+a)^(5/ 2)/b^4
Time = 0.08 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.73 \[ \int \frac {A+B x+C x^2+D x^3}{(a+b x)^{3/2}} \, dx=\frac {2 \left (-15 A b^3+48 a^3 D-8 a^2 b (5 C-3 D x)+2 a b^2 \left (15 B-10 C x-3 D x^2\right )+b^3 x \left (15 B+5 C x+3 D x^2\right )\right )}{15 b^4 \sqrt {a+b x}} \] Input:
Integrate[(A + B*x + C*x^2 + D*x^3)/(a + b*x)^(3/2),x]
Output:
(2*(-15*A*b^3 + 48*a^3*D - 8*a^2*b*(5*C - 3*D*x) + 2*a*b^2*(15*B - 10*C*x - 3*D*x^2) + b^3*x*(15*B + 5*C*x + 3*D*x^2)))/(15*b^4*Sqrt[a + b*x])
Time = 0.29 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {2389, 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.
\(\displaystyle \int \frac {A+B x+C x^2+D x^3}{(a+b x)^{3/2}} \, dx\) |
\(\Big \downarrow \) 2389 |
\(\displaystyle \int \left (\frac {A b^3-a \left (a^2 D-a b C+b^2 B\right )}{b^3 (a+b x)^{3/2}}+\frac {3 a^2 D-2 a b C+b^2 B}{b^3 \sqrt {a+b x}}+\frac {\sqrt {a+b x} (b C-3 a D)}{b^3}+\frac {D (a+b x)^{3/2}}{b^3}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {2 \left (A b^3-a \left (a^2 D-a b C+b^2 B\right )\right )}{b^4 \sqrt {a+b x}}+\frac {2 \sqrt {a+b x} \left (3 a^2 D-2 a b C+b^2 B\right )}{b^4}+\frac {2 (a+b x)^{3/2} (b C-3 a D)}{3 b^4}+\frac {2 D (a+b x)^{5/2}}{5 b^4}\) |
Input:
Int[(A + B*x + C*x^2 + D*x^3)/(a + b*x)^(3/2),x]
Output:
(-2*(A*b^3 - a*(b^2*B - a*b*C + a^2*D)))/(b^4*Sqrt[a + b*x]) + (2*(b^2*B - 2*a*b*C + 3*a^2*D)*Sqrt[a + b*x])/b^4 + (2*(b*C - 3*a*D)*(a + b*x)^(3/2)) /(3*b^4) + (2*D*(a + b*x)^(5/2))/(5*b^4)
Int[(Pq_)*((a_) + (b_.)*(x_)^(n_.))^(p_.), x_Symbol] :> Int[ExpandIntegrand [Pq*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, n}, x] && PolyQ[Pq, x] && (IGtQ[p , 0] || EqQ[n, 1])
Time = 0.37 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.66
method | result | size |
pseudoelliptic | \(\frac {\left (6 D x^{3}+10 C \,x^{2}+30 B x -30 A \right ) b^{3}+60 \left (-\frac {1}{5} D x^{2}-\frac {2}{3} C x +B \right ) a \,b^{2}-80 \left (-\frac {3 D x}{5}+C \right ) a^{2} b +96 a^{3} D}{15 \sqrt {b x +a}\, b^{4}}\) | \(74\) |
gosper | \(-\frac {2 \left (-3 D x^{3} b^{3}-5 C \,x^{2} b^{3}+6 D x^{2} a \,b^{2}-15 b^{3} B x +20 C x a \,b^{2}-24 D x \,a^{2} b +15 b^{3} A -30 a \,b^{2} B +40 a^{2} b C -48 a^{3} D\right )}{15 \sqrt {b x +a}\, b^{4}}\) | \(91\) |
trager | \(-\frac {2 \left (-3 D x^{3} b^{3}-5 C \,x^{2} b^{3}+6 D x^{2} a \,b^{2}-15 b^{3} B x +20 C x a \,b^{2}-24 D x \,a^{2} b +15 b^{3} A -30 a \,b^{2} B +40 a^{2} b C -48 a^{3} D\right )}{15 \sqrt {b x +a}\, b^{4}}\) | \(91\) |
orering | \(-\frac {2 \left (-3 D x^{3} b^{3}-5 C \,x^{2} b^{3}+6 D x^{2} a \,b^{2}-15 b^{3} B x +20 C x a \,b^{2}-24 D x \,a^{2} b +15 b^{3} A -30 a \,b^{2} B +40 a^{2} b C -48 a^{3} D\right )}{15 \sqrt {b x +a}\, b^{4}}\) | \(91\) |
derivativedivides | \(\frac {\frac {2 D \left (b x +a \right )^{\frac {5}{2}}}{5}+\frac {2 C b \left (b x +a \right )^{\frac {3}{2}}}{3}-2 D a \left (b x +a \right )^{\frac {3}{2}}+2 B \,b^{2} \sqrt {b x +a}-4 C a b \sqrt {b x +a}+6 D a^{2} \sqrt {b x +a}-\frac {2 \left (b^{3} A -a \,b^{2} B +a^{2} b C -a^{3} D\right )}{\sqrt {b x +a}}}{b^{4}}\) | \(110\) |
default | \(\frac {\frac {2 D \left (b x +a \right )^{\frac {5}{2}}}{5}+\frac {2 C b \left (b x +a \right )^{\frac {3}{2}}}{3}-2 D a \left (b x +a \right )^{\frac {3}{2}}+2 B \,b^{2} \sqrt {b x +a}-4 C a b \sqrt {b x +a}+6 D a^{2} \sqrt {b x +a}-\frac {2 \left (b^{3} A -a \,b^{2} B +a^{2} b C -a^{3} D\right )}{\sqrt {b x +a}}}{b^{4}}\) | \(110\) |
Input:
int((D*x^3+C*x^2+B*x+A)/(b*x+a)^(3/2),x,method=_RETURNVERBOSE)
Output:
1/15*((6*D*x^3+10*C*x^2+30*B*x-30*A)*b^3+60*(-1/5*D*x^2-2/3*C*x+B)*a*b^2-8 0*(-3/5*D*x+C)*a^2*b+96*a^3*D)/(b*x+a)^(1/2)/b^4
Time = 0.07 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.89 \[ \int \frac {A+B x+C x^2+D x^3}{(a+b x)^{3/2}} \, dx=\frac {2 \, {\left (3 \, D b^{3} x^{3} + 48 \, D a^{3} - 40 \, C a^{2} b + 30 \, B a b^{2} - 15 \, A b^{3} - {\left (6 \, D a b^{2} - 5 \, C b^{3}\right )} x^{2} + {\left (24 \, D a^{2} b - 20 \, C a b^{2} + 15 \, B b^{3}\right )} x\right )} \sqrt {b x + a}}{15 \, {\left (b^{5} x + a b^{4}\right )}} \] Input:
integrate((D*x^3+C*x^2+B*x+A)/(b*x+a)^(3/2),x, algorithm="fricas")
Output:
2/15*(3*D*b^3*x^3 + 48*D*a^3 - 40*C*a^2*b + 30*B*a*b^2 - 15*A*b^3 - (6*D*a *b^2 - 5*C*b^3)*x^2 + (24*D*a^2*b - 20*C*a*b^2 + 15*B*b^3)*x)*sqrt(b*x + a )/(b^5*x + a*b^4)
Time = 1.81 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.29 \[ \int \frac {A+B x+C x^2+D x^3}{(a+b x)^{3/2}} \, dx=\begin {cases} \frac {2 \left (\frac {D \left (a + b x\right )^{\frac {5}{2}}}{5 b^{3}} + \frac {\left (a + b x\right )^{\frac {3}{2}} \left (C b - 3 D a\right )}{3 b^{3}} + \frac {\sqrt {a + b x} \left (B b^{2} - 2 C a b + 3 D a^{2}\right )}{b^{3}} + \frac {- A b^{3} + B a b^{2} - C a^{2} b + D a^{3}}{b^{3} \sqrt {a + b x}}\right )}{b} & \text {for}\: b \neq 0 \\\frac {A x + \frac {B x^{2}}{2} + \frac {C x^{3}}{3} + \frac {D x^{4}}{4}}{a^{\frac {3}{2}}} & \text {otherwise} \end {cases} \] Input:
integrate((D*x**3+C*x**2+B*x+A)/(b*x+a)**(3/2),x)
Output:
Piecewise((2*(D*(a + b*x)**(5/2)/(5*b**3) + (a + b*x)**(3/2)*(C*b - 3*D*a) /(3*b**3) + sqrt(a + b*x)*(B*b**2 - 2*C*a*b + 3*D*a**2)/b**3 + (-A*b**3 + B*a*b**2 - C*a**2*b + D*a**3)/(b**3*sqrt(a + b*x)))/b, Ne(b, 0)), ((A*x + B*x**2/2 + C*x**3/3 + D*x**4/4)/a**(3/2), True))
Time = 0.03 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.91 \[ \int \frac {A+B x+C x^2+D x^3}{(a+b x)^{3/2}} \, dx=\frac {2 \, {\left (\frac {3 \, {\left (b x + a\right )}^{\frac {5}{2}} D - 5 \, {\left (3 \, D a - C b\right )} {\left (b x + a\right )}^{\frac {3}{2}} + 15 \, {\left (3 \, D a^{2} - 2 \, C a b + B b^{2}\right )} \sqrt {b x + a}}{b^{3}} + \frac {15 \, {\left (D a^{3} - C a^{2} b + B a b^{2} - A b^{3}\right )}}{\sqrt {b x + a} b^{3}}\right )}}{15 \, b} \] Input:
integrate((D*x^3+C*x^2+B*x+A)/(b*x+a)^(3/2),x, algorithm="maxima")
Output:
2/15*((3*(b*x + a)^(5/2)*D - 5*(3*D*a - C*b)*(b*x + a)^(3/2) + 15*(3*D*a^2 - 2*C*a*b + B*b^2)*sqrt(b*x + a))/b^3 + 15*(D*a^3 - C*a^2*b + B*a*b^2 - A *b^3)/(sqrt(b*x + a)*b^3))/b
Time = 0.13 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.13 \[ \int \frac {A+B x+C x^2+D x^3}{(a+b x)^{3/2}} \, dx=\frac {2 \, {\left (D a^{3} - C a^{2} b + B a b^{2} - A b^{3}\right )}}{\sqrt {b x + a} b^{4}} + \frac {2 \, {\left (3 \, {\left (b x + a\right )}^{\frac {5}{2}} D b^{16} - 15 \, {\left (b x + a\right )}^{\frac {3}{2}} D a b^{16} + 45 \, \sqrt {b x + a} D a^{2} b^{16} + 5 \, {\left (b x + a\right )}^{\frac {3}{2}} C b^{17} - 30 \, \sqrt {b x + a} C a b^{17} + 15 \, \sqrt {b x + a} B b^{18}\right )}}{15 \, b^{20}} \] Input:
integrate((D*x^3+C*x^2+B*x+A)/(b*x+a)^(3/2),x, algorithm="giac")
Output:
2*(D*a^3 - C*a^2*b + B*a*b^2 - A*b^3)/(sqrt(b*x + a)*b^4) + 2/15*(3*(b*x + a)^(5/2)*D*b^16 - 15*(b*x + a)^(3/2)*D*a*b^16 + 45*sqrt(b*x + a)*D*a^2*b^ 16 + 5*(b*x + a)^(3/2)*C*b^17 - 30*sqrt(b*x + a)*C*a*b^17 + 15*sqrt(b*x + a)*B*b^18)/b^20
Timed out. \[ \int \frac {A+B x+C x^2+D x^3}{(a+b x)^{3/2}} \, dx=\int \frac {A+B\,x+C\,x^2+x^3\,D}{{\left (a+b\,x\right )}^{3/2}} \,d x \] Input:
int((A + B*x + C*x^2 + x^3*D)/(a + b*x)^(3/2),x)
Output:
int((A + B*x + C*x^2 + x^3*D)/(a + b*x)^(3/2), x)
Time = 0.14 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.74 \[ \int \frac {A+B x+C x^2+D x^3}{(a+b x)^{3/2}} \, dx=\frac {\frac {2}{5} b^{3} d \,x^{3}-\frac {4}{5} a \,b^{2} d \,x^{2}+\frac {2}{3} b^{3} c \,x^{2}+\frac {16}{5} a^{2} b d x -\frac {8}{3} a \,b^{2} c x +2 b^{4} x +\frac {32}{5} a^{3} d -\frac {16}{3} a^{2} b c +2 a \,b^{3}}{\sqrt {b x +a}\, b^{4}} \] Input:
int((D*x^3+C*x^2+B*x+A)/(b*x+a)^(3/2),x)
Output:
(2*(48*a**3*d - 40*a**2*b*c + 24*a**2*b*d*x + 15*a*b**3 - 20*a*b**2*c*x - 6*a*b**2*d*x**2 + 15*b**4*x + 5*b**3*c*x**2 + 3*b**3*d*x**3))/(15*sqrt(a + b*x)*b**4)