Integrand size = 25, antiderivative size = 113 \[ \int \frac {A+B x+C x^2+D x^3}{(c+d x)^{3/2}} \, dx=-\frac {2 \left (c^2 C d-B c d^2+A d^3-c^3 D\right )}{d^4 \sqrt {c+d x}}-\frac {2 \left (2 c C d-B d^2-3 c^2 D\right ) \sqrt {c+d x}}{d^4}+\frac {2 (C d-3 c D) (c+d x)^{3/2}}{3 d^4}+\frac {2 D (c+d x)^{5/2}}{5 d^4} \] Output:
(-2*A*d^3+2*B*c*d^2-2*C*c^2*d+2*D*c^3)/d^4/(d*x+c)^(1/2)-2*(-B*d^2+2*C*c*d -3*D*c^2)*(d*x+c)^(1/2)/d^4+2/3*(C*d-3*D*c)*(d*x+c)^(3/2)/d^4+2/5*D*(d*x+c )^(5/2)/d^4
Time = 0.06 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.73 \[ \int \frac {A+B x+C x^2+D x^3}{(c+d x)^{3/2}} \, dx=\frac {2 \left (48 c^3 D-8 c^2 d (5 C-3 D x)+2 c d^2 (15 B-x (10 C+3 D x))+d^3 \left (-15 A+x \left (15 B+5 C x+3 D x^2\right )\right )\right )}{15 d^4 \sqrt {c+d x}} \] Input:
Integrate[(A + B*x + C*x^2 + D*x^3)/(c + d*x)^(3/2),x]
Output:
(2*(48*c^3*D - 8*c^2*d*(5*C - 3*D*x) + 2*c*d^2*(15*B - x*(10*C + 3*D*x)) + d^3*(-15*A + x*(15*B + 5*C*x + 3*D*x^2))))/(15*d^4*Sqrt[c + d*x])
Time = 0.31 (sec) , antiderivative size = 113, 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.
\(\displaystyle \int \frac {A+B x+C x^2+D x^3}{(c+d x)^{3/2}} \, dx\) |
\(\Big \downarrow \) 2389 |
\(\displaystyle \int \left (\frac {A d^3-B c d^2+c^3 (-D)+c^2 C d}{d^3 (c+d x)^{3/2}}+\frac {B d^2+3 c^2 D-2 c C d}{d^3 \sqrt {c+d x}}+\frac {\sqrt {c+d x} (C d-3 c D)}{d^3}+\frac {D (c+d x)^{3/2}}{d^3}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {2 \left (A d^3-B c d^2+c^3 (-D)+c^2 C d\right )}{d^4 \sqrt {c+d x}}-\frac {2 \sqrt {c+d x} \left (-B d^2-3 c^2 D+2 c C d\right )}{d^4}+\frac {2 (c+d x)^{3/2} (C d-3 c D)}{3 d^4}+\frac {2 D (c+d x)^{5/2}}{5 d^4}\) |
Input:
Int[(A + B*x + C*x^2 + D*x^3)/(c + d*x)^(3/2),x]
Output:
(-2*(c^2*C*d - B*c*d^2 + A*d^3 - c^3*D))/(d^4*Sqrt[c + d*x]) - (2*(2*c*C*d - B*d^2 - 3*c^2*D)*Sqrt[c + d*x])/d^4 + (2*(C*d - 3*c*D)*(c + d*x)^(3/2)) /(3*d^4) + (2*D*(c + d*x)^(5/2))/(5*d^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 = 0.40 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.65
method | result | size |
pseudoelliptic | \(\frac {\left (6 D x^{3}+10 C \,x^{2}+30 B x -30 A \right ) d^{3}+60 \left (-\frac {1}{5} D x^{2}-\frac {2}{3} C x +B \right ) c \,d^{2}-80 \left (-\frac {3 D x}{5}+C \right ) c^{2} d +96 D c^{3}}{15 \sqrt {x d +c}\, d^{4}}\) | \(74\) |
gosper | \(-\frac {2 \left (-3 D x^{3} d^{3}-5 C \,x^{2} d^{3}+6 D x^{2} c \,d^{2}-15 B x \,d^{3}+20 C x c \,d^{2}-24 D x \,c^{2} d +15 A \,d^{3}-30 B c \,d^{2}+40 C \,c^{2} d -48 D c^{3}\right )}{15 \sqrt {x d +c}\, d^{4}}\) | \(91\) |
trager | \(-\frac {2 \left (-3 D x^{3} d^{3}-5 C \,x^{2} d^{3}+6 D x^{2} c \,d^{2}-15 B x \,d^{3}+20 C x c \,d^{2}-24 D x \,c^{2} d +15 A \,d^{3}-30 B c \,d^{2}+40 C \,c^{2} d -48 D c^{3}\right )}{15 \sqrt {x d +c}\, d^{4}}\) | \(91\) |
orering | \(-\frac {2 \left (-3 D x^{3} d^{3}-5 C \,x^{2} d^{3}+6 D x^{2} c \,d^{2}-15 B x \,d^{3}+20 C x c \,d^{2}-24 D x \,c^{2} d +15 A \,d^{3}-30 B c \,d^{2}+40 C \,c^{2} d -48 D c^{3}\right )}{15 \sqrt {x d +c}\, d^{4}}\) | \(91\) |
derivativedivides | \(\frac {\frac {2 D \left (x d +c \right )^{\frac {5}{2}}}{5}+\frac {2 C d \left (x d +c \right )^{\frac {3}{2}}}{3}-2 D c \left (x d +c \right )^{\frac {3}{2}}+2 B \,d^{2} \sqrt {x d +c}-4 C c d \sqrt {x d +c}+6 D c^{2} \sqrt {x d +c}-\frac {2 \left (A \,d^{3}-B c \,d^{2}+C \,c^{2} d -D c^{3}\right )}{\sqrt {x d +c}}}{d^{4}}\) | \(110\) |
default | \(\frac {\frac {2 D \left (x d +c \right )^{\frac {5}{2}}}{5}+\frac {2 C d \left (x d +c \right )^{\frac {3}{2}}}{3}-2 D c \left (x d +c \right )^{\frac {3}{2}}+2 B \,d^{2} \sqrt {x d +c}-4 C c d \sqrt {x d +c}+6 D c^{2} \sqrt {x d +c}-\frac {2 \left (A \,d^{3}-B c \,d^{2}+C \,c^{2} d -D c^{3}\right )}{\sqrt {x d +c}}}{d^{4}}\) | \(110\) |
Input:
int((D*x^3+C*x^2+B*x+A)/(d*x+c)^(3/2),x,method=_RETURNVERBOSE)
Output:
1/15*((6*D*x^3+10*C*x^2+30*B*x-30*A)*d^3+60*(-1/5*D*x^2-2/3*C*x+B)*c*d^2-8 0*(-3/5*D*x+C)*c^2*d+96*D*c^3)/(d*x+c)^(1/2)/d^4
Time = 0.07 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.88 \[ \int \frac {A+B x+C x^2+D x^3}{(c+d x)^{3/2}} \, dx=\frac {2 \, {\left (3 \, D d^{3} x^{3} + 48 \, D c^{3} - 40 \, C c^{2} d + 30 \, B c d^{2} - 15 \, A d^{3} - {\left (6 \, D c d^{2} - 5 \, C d^{3}\right )} x^{2} + {\left (24 \, D c^{2} d - 20 \, C c d^{2} + 15 \, B d^{3}\right )} x\right )} \sqrt {d x + c}}{15 \, {\left (d^{5} x + c d^{4}\right )}} \] Input:
integrate((D*x^3+C*x^2+B*x+A)/(d*x+c)^(3/2),x, algorithm="fricas")
Output:
2/15*(3*D*d^3*x^3 + 48*D*c^3 - 40*C*c^2*d + 30*B*c*d^2 - 15*A*d^3 - (6*D*c *d^2 - 5*C*d^3)*x^2 + (24*D*c^2*d - 20*C*c*d^2 + 15*B*d^3)*x)*sqrt(d*x + c )/(d^5*x + c*d^4)
Time = 2.00 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.27 \[ \int \frac {A+B x+C x^2+D x^3}{(c+d x)^{3/2}} \, dx=\begin {cases} \frac {2 \left (\frac {D \left (c + d x\right )^{\frac {5}{2}}}{5 d^{3}} + \frac {\left (c + d x\right )^{\frac {3}{2}} \left (C d - 3 D c\right )}{3 d^{3}} + \frac {\sqrt {c + d x} \left (B d^{2} - 2 C c d + 3 D c^{2}\right )}{d^{3}} + \frac {- A d^{3} + B c d^{2} - C c^{2} d + D c^{3}}{d^{3} \sqrt {c + d x}}\right )}{d} & \text {for}\: d \neq 0 \\\frac {A x + \frac {B x^{2}}{2} + \frac {C x^{3}}{3} + \frac {D x^{4}}{4}}{c^{\frac {3}{2}}} & \text {otherwise} \end {cases} \] Input:
integrate((D*x**3+C*x**2+B*x+A)/(d*x+c)**(3/2),x)
Output:
Piecewise((2*(D*(c + d*x)**(5/2)/(5*d**3) + (c + d*x)**(3/2)*(C*d - 3*D*c) /(3*d**3) + sqrt(c + d*x)*(B*d**2 - 2*C*c*d + 3*D*c**2)/d**3 + (-A*d**3 + B*c*d**2 - C*c**2*d + D*c**3)/(d**3*sqrt(c + d*x)))/d, Ne(d, 0)), ((A*x + B*x**2/2 + C*x**3/3 + D*x**4/4)/c**(3/2), True))
Time = 0.03 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.90 \[ \int \frac {A+B x+C x^2+D x^3}{(c+d x)^{3/2}} \, dx=\frac {2 \, {\left (\frac {3 \, {\left (d x + c\right )}^{\frac {5}{2}} D - 5 \, {\left (3 \, D c - C d\right )} {\left (d x + c\right )}^{\frac {3}{2}} + 15 \, {\left (3 \, D c^{2} - 2 \, C c d + B d^{2}\right )} \sqrt {d x + c}}{d^{3}} + \frac {15 \, {\left (D c^{3} - C c^{2} d + B c d^{2} - A d^{3}\right )}}{\sqrt {d x + c} d^{3}}\right )}}{15 \, d} \] Input:
integrate((D*x^3+C*x^2+B*x+A)/(d*x+c)^(3/2),x, algorithm="maxima")
Output:
2/15*((3*(d*x + c)^(5/2)*D - 5*(3*D*c - C*d)*(d*x + c)^(3/2) + 15*(3*D*c^2 - 2*C*c*d + B*d^2)*sqrt(d*x + c))/d^3 + 15*(D*c^3 - C*c^2*d + B*c*d^2 - A *d^3)/(sqrt(d*x + c)*d^3))/d
Time = 0.13 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.12 \[ \int \frac {A+B x+C x^2+D x^3}{(c+d x)^{3/2}} \, dx=\frac {2 \, {\left (D c^{3} - C c^{2} d + B c d^{2} - A d^{3}\right )}}{\sqrt {d x + c} d^{4}} + \frac {2 \, {\left (3 \, {\left (d x + c\right )}^{\frac {5}{2}} D d^{16} - 15 \, {\left (d x + c\right )}^{\frac {3}{2}} D c d^{16} + 45 \, \sqrt {d x + c} D c^{2} d^{16} + 5 \, {\left (d x + c\right )}^{\frac {3}{2}} C d^{17} - 30 \, \sqrt {d x + c} C c d^{17} + 15 \, \sqrt {d x + c} B d^{18}\right )}}{15 \, d^{20}} \] Input:
integrate((D*x^3+C*x^2+B*x+A)/(d*x+c)^(3/2),x, algorithm="giac")
Output:
2*(D*c^3 - C*c^2*d + B*c*d^2 - A*d^3)/(sqrt(d*x + c)*d^4) + 2/15*(3*(d*x + c)^(5/2)*D*d^16 - 15*(d*x + c)^(3/2)*D*c*d^16 + 45*sqrt(d*x + c)*D*c^2*d^ 16 + 5*(d*x + c)^(3/2)*C*d^17 - 30*sqrt(d*x + c)*C*c*d^17 + 15*sqrt(d*x + c)*B*d^18)/d^20
Timed out. \[ \int \frac {A+B x+C x^2+D x^3}{(c+d x)^{3/2}} \, dx=\int \frac {A+B\,x+C\,x^2+x^3\,D}{{\left (c+d\,x\right )}^{3/2}} \,d x \] Input:
int((A + B*x + C*x^2 + x^3*D)/(c + d*x)^(3/2),x)
Output:
int((A + B*x + C*x^2 + x^3*D)/(c + d*x)^(3/2), x)
Time = 0.14 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.54 \[ \int \frac {A+B x+C x^2+D x^3}{(c+d x)^{3/2}} \, dx=\frac {\frac {2}{5} d^{3} x^{3}-\frac {2}{15} c \,d^{2} x^{2}+2 b \,d^{2} x +\frac {8}{15} c^{2} d x -2 a \,d^{2}+4 b c d +\frac {16}{15} c^{3}}{\sqrt {d x +c}\, d^{3}} \] Input:
int((D*x^3+C*x^2+B*x+A)/(d*x+c)^(3/2),x)
Output:
(2*( - 15*a*d**2 + 30*b*c*d + 15*b*d**2*x + 8*c**3 + 4*c**2*d*x - c*d**2*x **2 + 3*d**3*x**3))/(15*sqrt(c + d*x)*d**3)