Integrand size = 25, antiderivative size = 112 \[ \int \frac {A+B x+C x^2+D x^3}{(a+b x)^{5/2}} \, dx=-\frac {2 \left (A b^3-a \left (b^2 B-a b C+a^2 D\right )\right )}{3 b^4 (a+b x)^{3/2}}-\frac {2 \left (b^2 B-2 a b C+3 a^2 D\right )}{b^4 \sqrt {a+b x}}+\frac {2 (b C-3 a D) \sqrt {a+b x}}{b^4}+\frac {2 D (a+b x)^{3/2}}{3 b^4} \] Output:
1/3*(-2*A*b^3+2*a*(B*b^2-C*a*b+D*a^2))/b^4/(b*x+a)^(3/2)-2*(B*b^2-2*C*a*b+ 3*D*a^2)/b^4/(b*x+a)^(1/2)+2*(C*b-3*D*a)*(b*x+a)^(1/2)/b^4+2/3*D*(b*x+a)^( 3/2)/b^4
Time = 0.08 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.69 \[ \int \frac {A+B x+C x^2+D x^3}{(a+b x)^{5/2}} \, dx=-\frac {2 \left (A b^3+16 a^3 D-8 a^2 b (C-3 D x)-b^3 x \left (-3 B+3 C x+D x^2\right )+2 a b^2 \left (B-6 C x+3 D x^2\right )\right )}{3 b^4 (a+b x)^{3/2}} \] Input:
Integrate[(A + B*x + C*x^2 + D*x^3)/(a + b*x)^(5/2),x]
Output:
(-2*(A*b^3 + 16*a^3*D - 8*a^2*b*(C - 3*D*x) - b^3*x*(-3*B + 3*C*x + D*x^2) + 2*a*b^2*(B - 6*C*x + 3*D*x^2)))/(3*b^4*(a + b*x)^(3/2))
Time = 0.27 (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.
\(\displaystyle \int \frac {A+B x+C x^2+D x^3}{(a+b x)^{5/2}} \, dx\) |
\(\Big \downarrow \) 2389 |
\(\displaystyle \int \left (\frac {A b^3-a \left (a^2 D-a b C+b^2 B\right )}{b^3 (a+b x)^{5/2}}+\frac {3 a^2 D-2 a b C+b^2 B}{b^3 (a+b x)^{3/2}}+\frac {b C-3 a D}{b^3 \sqrt {a+b x}}+\frac {D \sqrt {a+b x}}{b^3}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {2 \left (A b^3-a \left (a^2 D-a b C+b^2 B\right )\right )}{3 b^4 (a+b x)^{3/2}}-\frac {2 \left (3 a^2 D-2 a b C+b^2 B\right )}{b^4 \sqrt {a+b x}}+\frac {2 \sqrt {a+b x} (b C-3 a D)}{b^4}+\frac {2 D (a+b x)^{3/2}}{3 b^4}\) |
Input:
Int[(A + B*x + C*x^2 + D*x^3)/(a + b*x)^(5/2),x]
Output:
(-2*(A*b^3 - a*(b^2*B - a*b*C + a^2*D)))/(3*b^4*(a + b*x)^(3/2)) - (2*(b^2 *B - 2*a*b*C + 3*a^2*D))/(b^4*Sqrt[a + b*x]) + (2*(b*C - 3*a*D)*Sqrt[a + b *x])/b^4 + (2*D*(a + b*x)^(3/2))/(3*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 = 72, normalized size of antiderivative = 0.64
method | result | size |
pseudoelliptic | \(-\frac {2 \left (\left (-D x^{3}-3 C \,x^{2}+3 B x +A \right ) b^{3}+2 a \left (3 D x^{2}-6 C x +B \right ) b^{2}-8 a^{2} \left (-3 D x +C \right ) b +16 a^{3} D\right )}{3 \left (b x +a \right )^{\frac {3}{2}} b^{4}}\) | \(72\) |
gosper | \(-\frac {2 \left (-D x^{3} b^{3}-3 C \,x^{2} b^{3}+6 D x^{2} a \,b^{2}+3 b^{3} B x -12 C x a \,b^{2}+24 D x \,a^{2} b +b^{3} A +2 a \,b^{2} B -8 a^{2} b C +16 a^{3} D\right )}{3 \left (b x +a \right )^{\frac {3}{2}} b^{4}}\) | \(90\) |
trager | \(-\frac {2 \left (-D x^{3} b^{3}-3 C \,x^{2} b^{3}+6 D x^{2} a \,b^{2}+3 b^{3} B x -12 C x a \,b^{2}+24 D x \,a^{2} b +b^{3} A +2 a \,b^{2} B -8 a^{2} b C +16 a^{3} D\right )}{3 \left (b x +a \right )^{\frac {3}{2}} b^{4}}\) | \(90\) |
orering | \(-\frac {2 \left (-D x^{3} b^{3}-3 C \,x^{2} b^{3}+6 D x^{2} a \,b^{2}+3 b^{3} B x -12 C x a \,b^{2}+24 D x \,a^{2} b +b^{3} A +2 a \,b^{2} B -8 a^{2} b C +16 a^{3} D\right )}{3 \left (b x +a \right )^{\frac {3}{2}} b^{4}}\) | \(90\) |
derivativedivides | \(\frac {\frac {2 D \left (b x +a \right )^{\frac {3}{2}}}{3}+2 b C \sqrt {b x +a}-6 D a \sqrt {b x +a}-\frac {2 \left (B \,b^{2}-2 C a b +3 D a^{2}\right )}{\sqrt {b x +a}}-\frac {2 \left (b^{3} A -a \,b^{2} B +a^{2} b C -a^{3} D\right )}{3 \left (b x +a \right )^{\frac {3}{2}}}}{b^{4}}\) | \(98\) |
default | \(\frac {\frac {2 D \left (b x +a \right )^{\frac {3}{2}}}{3}+2 b C \sqrt {b x +a}-6 D a \sqrt {b x +a}-\frac {2 \left (B \,b^{2}-2 C a b +3 D a^{2}\right )}{\sqrt {b x +a}}-\frac {2 \left (b^{3} A -a \,b^{2} B +a^{2} b C -a^{3} D\right )}{3 \left (b x +a \right )^{\frac {3}{2}}}}{b^{4}}\) | \(98\) |
Input:
int((D*x^3+C*x^2+B*x+A)/(b*x+a)^(5/2),x,method=_RETURNVERBOSE)
Output:
-2/3*((-D*x^3-3*C*x^2+3*B*x+A)*b^3+2*a*(3*D*x^2-6*C*x+B)*b^2-8*a^2*(-3*D*x +C)*b+16*a^3*D)/(b*x+a)^(3/2)/b^4
Time = 0.08 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.98 \[ \int \frac {A+B x+C x^2+D x^3}{(a+b x)^{5/2}} \, dx=\frac {2 \, {\left (D b^{3} x^{3} - 16 \, D a^{3} + 8 \, C a^{2} b - 2 \, B a b^{2} - A b^{3} - 3 \, {\left (2 \, D a b^{2} - C b^{3}\right )} x^{2} - 3 \, {\left (8 \, D a^{2} b - 4 \, C a b^{2} + B b^{3}\right )} x\right )} \sqrt {b x + a}}{3 \, {\left (b^{6} x^{2} + 2 \, a b^{5} x + a^{2} b^{4}\right )}} \] Input:
integrate((D*x^3+C*x^2+B*x+A)/(b*x+a)^(5/2),x, algorithm="fricas")
Output:
2/3*(D*b^3*x^3 - 16*D*a^3 + 8*C*a^2*b - 2*B*a*b^2 - A*b^3 - 3*(2*D*a*b^2 - C*b^3)*x^2 - 3*(8*D*a^2*b - 4*C*a*b^2 + B*b^3)*x)*sqrt(b*x + a)/(b^6*x^2 + 2*a*b^5*x + a^2*b^4)
Leaf count of result is larger than twice the leaf count of optimal. 425 vs. \(2 (110) = 220\).
Time = 0.34 (sec) , antiderivative size = 425, normalized size of antiderivative = 3.79 \[ \int \frac {A+B x+C x^2+D x^3}{(a+b x)^{5/2}} \, dx=\begin {cases} - \frac {2 A b^{3}}{3 a b^{4} \sqrt {a + b x} + 3 b^{5} x \sqrt {a + b x}} - \frac {4 B a b^{2}}{3 a b^{4} \sqrt {a + b x} + 3 b^{5} x \sqrt {a + b x}} - \frac {6 B b^{3} x}{3 a b^{4} \sqrt {a + b x} + 3 b^{5} x \sqrt {a + b x}} + \frac {16 C a^{2} b}{3 a b^{4} \sqrt {a + b x} + 3 b^{5} x \sqrt {a + b x}} + \frac {24 C a b^{2} x}{3 a b^{4} \sqrt {a + b x} + 3 b^{5} x \sqrt {a + b x}} + \frac {6 C b^{3} x^{2}}{3 a b^{4} \sqrt {a + b x} + 3 b^{5} x \sqrt {a + b x}} - \frac {32 D a^{3}}{3 a b^{4} \sqrt {a + b x} + 3 b^{5} x \sqrt {a + b x}} - \frac {48 D a^{2} b x}{3 a b^{4} \sqrt {a + b x} + 3 b^{5} x \sqrt {a + b x}} - \frac {12 D a b^{2} x^{2}}{3 a b^{4} \sqrt {a + b x} + 3 b^{5} x \sqrt {a + b x}} + \frac {2 D b^{3} x^{3}}{3 a b^{4} \sqrt {a + b x} + 3 b^{5} x \sqrt {a + b x}} & \text {for}\: b \neq 0 \\\frac {A x + \frac {B x^{2}}{2} + \frac {C x^{3}}{3} + \frac {D x^{4}}{4}}{a^{\frac {5}{2}}} & \text {otherwise} \end {cases} \] Input:
integrate((D*x**3+C*x**2+B*x+A)/(b*x+a)**(5/2),x)
Output:
Piecewise((-2*A*b**3/(3*a*b**4*sqrt(a + b*x) + 3*b**5*x*sqrt(a + b*x)) - 4 *B*a*b**2/(3*a*b**4*sqrt(a + b*x) + 3*b**5*x*sqrt(a + b*x)) - 6*B*b**3*x/( 3*a*b**4*sqrt(a + b*x) + 3*b**5*x*sqrt(a + b*x)) + 16*C*a**2*b/(3*a*b**4*s qrt(a + b*x) + 3*b**5*x*sqrt(a + b*x)) + 24*C*a*b**2*x/(3*a*b**4*sqrt(a + b*x) + 3*b**5*x*sqrt(a + b*x)) + 6*C*b**3*x**2/(3*a*b**4*sqrt(a + b*x) + 3 *b**5*x*sqrt(a + b*x)) - 32*D*a**3/(3*a*b**4*sqrt(a + b*x) + 3*b**5*x*sqrt (a + b*x)) - 48*D*a**2*b*x/(3*a*b**4*sqrt(a + b*x) + 3*b**5*x*sqrt(a + b*x )) - 12*D*a*b**2*x**2/(3*a*b**4*sqrt(a + b*x) + 3*b**5*x*sqrt(a + b*x)) + 2*D*b**3*x**3/(3*a*b**4*sqrt(a + b*x) + 3*b**5*x*sqrt(a + b*x)), Ne(b, 0)) , ((A*x + B*x**2/2 + C*x**3/3 + D*x**4/4)/a**(5/2), True))
Time = 0.04 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.88 \[ \int \frac {A+B x+C x^2+D x^3}{(a+b x)^{5/2}} \, dx=\frac {2 \, {\left (\frac {{\left (b x + a\right )}^{\frac {3}{2}} D - 3 \, {\left (3 \, D a - C b\right )} \sqrt {b x + a}}{b^{3}} + \frac {D a^{3} - C a^{2} b + B a b^{2} - A b^{3} - 3 \, {\left (3 \, D a^{2} - 2 \, C a b + B b^{2}\right )} {\left (b x + a\right )}}{{\left (b x + a\right )}^{\frac {3}{2}} b^{3}}\right )}}{3 \, b} \] Input:
integrate((D*x^3+C*x^2+B*x+A)/(b*x+a)^(5/2),x, algorithm="maxima")
Output:
2/3*(((b*x + a)^(3/2)*D - 3*(3*D*a - C*b)*sqrt(b*x + a))/b^3 + (D*a^3 - C* a^2*b + B*a*b^2 - A*b^3 - 3*(3*D*a^2 - 2*C*a*b + B*b^2)*(b*x + a))/((b*x + a)^(3/2)*b^3))/b
Time = 0.12 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.03 \[ \int \frac {A+B x+C x^2+D x^3}{(a+b x)^{5/2}} \, dx=-\frac {2 \, {\left (9 \, {\left (b x + a\right )} D a^{2} - D a^{3} - 6 \, {\left (b x + a\right )} C a b + C a^{2} b + 3 \, {\left (b x + a\right )} B b^{2} - B a b^{2} + A b^{3}\right )}}{3 \, {\left (b x + a\right )}^{\frac {3}{2}} b^{4}} + \frac {2 \, {\left ({\left (b x + a\right )}^{\frac {3}{2}} D b^{8} - 9 \, \sqrt {b x + a} D a b^{8} + 3 \, \sqrt {b x + a} C b^{9}\right )}}{3 \, b^{12}} \] Input:
integrate((D*x^3+C*x^2+B*x+A)/(b*x+a)^(5/2),x, algorithm="giac")
Output:
-2/3*(9*(b*x + a)*D*a^2 - D*a^3 - 6*(b*x + a)*C*a*b + C*a^2*b + 3*(b*x + a )*B*b^2 - B*a*b^2 + A*b^3)/((b*x + a)^(3/2)*b^4) + 2/3*((b*x + a)^(3/2)*D* b^8 - 9*sqrt(b*x + a)*D*a*b^8 + 3*sqrt(b*x + a)*C*b^9)/b^12
Timed out. \[ \int \frac {A+B x+C x^2+D x^3}{(a+b x)^{5/2}} \, dx=\int \frac {A+B\,x+C\,x^2+x^3\,D}{{\left (a+b\,x\right )}^{5/2}} \,d x \] Input:
int((A + B*x + C*x^2 + x^3*D)/(a + b*x)^(5/2),x)
Output:
int((A + B*x + C*x^2 + x^3*D)/(a + b*x)^(5/2), x)
Time = 0.14 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.79 \[ \int \frac {A+B x+C x^2+D x^3}{(a+b x)^{5/2}} \, dx=\frac {\frac {2}{3} b^{3} d \,x^{3}-4 a \,b^{2} d \,x^{2}+2 b^{3} c \,x^{2}-16 a^{2} b d x +8 a \,b^{2} c x -2 b^{4} x -\frac {32}{3} a^{3} d +\frac {16}{3} a^{2} b c -2 a \,b^{3}}{\sqrt {b x +a}\, b^{4} \left (b x +a \right )} \] Input:
int((D*x^3+C*x^2+B*x+A)/(b*x+a)^(5/2),x)
Output:
(2*( - 16*a**3*d + 8*a**2*b*c - 24*a**2*b*d*x - 3*a*b**3 + 12*a*b**2*c*x - 6*a*b**2*d*x**2 - 3*b**4*x + 3*b**3*c*x**2 + b**3*d*x**3))/(3*sqrt(a + b* x)*b**4*(a + b*x))