Integrand size = 25, antiderivative size = 113 \[ \int \frac {A+B x+C x^2+D x^3}{(c+d x)^{5/2}} \, dx=-\frac {2 \left (c^2 C d-B c d^2+A d^3-c^3 D\right )}{3 d^4 (c+d x)^{3/2}}+\frac {2 \left (2 c C d-B d^2-3 c^2 D\right )}{d^4 \sqrt {c+d x}}+\frac {2 (C d-3 c D) \sqrt {c+d x}}{d^4}+\frac {2 D (c+d x)^{3/2}}{3 d^4} \]
-2/3*(A*d^3-B*c*d^2+C*c^2*d-D*c^3)/d^4/(d*x+c)^(3/2)+2/3*D*(d*x+c)^(3/2)/d ^4+2*(-B*d^2+2*C*c*d-3*D*c^2)/d^4/(d*x+c)^(1/2)+2*(C*d-3*D*c)*(d*x+c)^(1/2 )/d^4
Time = 0.07 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.66 \[ \int \frac {A+B x+C x^2+D x^3}{(c+d x)^{5/2}} \, dx=-\frac {2 \left (16 c^3 D-8 c^2 d (C-3 D x)+2 c d^2 (B+3 x (-2 C+D x))+d^3 \left (A+3 B x-x^2 (3 C+D x)\right )\right )}{3 d^4 (c+d x)^{3/2}} \]
(-2*(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))))/(3*d^4*(c + d*x)^(3/2))
Time = 0.27 (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.
\(\displaystyle \int \frac {A+B x+C x^2+D x^3}{(c+d x)^{5/2}} \, dx\) |
\(\Big \downarrow \) 2389 |
\(\displaystyle \int \left (\frac {A d^3-B c d^2+c^3 (-D)+c^2 C d}{d^3 (c+d x)^{5/2}}+\frac {B d^2+3 c^2 D-2 c C d}{d^3 (c+d x)^{3/2}}+\frac {C d-3 c D}{d^3 \sqrt {c+d x}}+\frac {D \sqrt {c+d x}}{d^3}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {2 \left (A d^3-B c d^2+c^3 (-D)+c^2 C d\right )}{3 d^4 (c+d x)^{3/2}}+\frac {2 \left (-B d^2-3 c^2 D+2 c C d\right )}{d^4 \sqrt {c+d x}}+\frac {2 \sqrt {c+d x} (C d-3 c D)}{d^4}+\frac {2 D (c+d x)^{3/2}}{3 d^4}\) |
(-2*(c^2*C*d - B*c*d^2 + A*d^3 - c^3*D))/(3*d^4*(c + d*x)^(3/2)) + (2*(2*c *C*d - B*d^2 - 3*c^2*D))/(d^4*Sqrt[c + d*x]) + (2*(C*d - 3*c*D)*Sqrt[c + d *x])/d^4 + (2*D*(c + d*x)^(3/2))/(3*d^4)
3.1.21.3.1 Defintions of rubi rules used
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 = 1.64 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.64
method | result | size |
pseudoelliptic | \(-\frac {2 \left (\left (-D x^{3}-3 C \,x^{2}+3 B x +A \right ) d^{3}+2 c \left (3 D x^{2}-6 C x +B \right ) d^{2}-8 c^{2} \left (-3 D x +C \right ) d +16 D c^{3}\right )}{3 \left (d x +c \right )^{\frac {3}{2}} d^{4}}\) | \(72\) |
gosper | \(-\frac {2 \left (-D x^{3} d^{3}-3 C \,d^{3} x^{2}+6 D c \,d^{2} x^{2}+3 B \,d^{3} x -12 C c \,d^{2} x +24 D c^{2} d x +A \,d^{3}+2 B c \,d^{2}-8 C \,c^{2} d +16 D c^{3}\right )}{3 \left (d x +c \right )^{\frac {3}{2}} d^{4}}\) | \(90\) |
trager | \(-\frac {2 \left (-D x^{3} d^{3}-3 C \,d^{3} x^{2}+6 D c \,d^{2} x^{2}+3 B \,d^{3} x -12 C c \,d^{2} x +24 D c^{2} d x +A \,d^{3}+2 B c \,d^{2}-8 C \,c^{2} d +16 D c^{3}\right )}{3 \left (d x +c \right )^{\frac {3}{2}} d^{4}}\) | \(90\) |
derivativedivides | \(\frac {\frac {2 D \left (d x +c \right )^{\frac {3}{2}}}{3}+2 d C \sqrt {d x +c}-6 D c \sqrt {d x +c}-\frac {2 \left (B \,d^{2}-2 C c d +3 D c^{2}\right )}{\sqrt {d x +c}}-\frac {2 \left (A \,d^{3}-B c \,d^{2}+C \,c^{2} d -D c^{3}\right )}{3 \left (d x +c \right )^{\frac {3}{2}}}}{d^{4}}\) | \(98\) |
default | \(\frac {\frac {2 D \left (d x +c \right )^{\frac {3}{2}}}{3}+2 d C \sqrt {d x +c}-6 D c \sqrt {d x +c}-\frac {2 \left (B \,d^{2}-2 C c d +3 D c^{2}\right )}{\sqrt {d x +c}}-\frac {2 \left (A \,d^{3}-B c \,d^{2}+C \,c^{2} d -D c^{3}\right )}{3 \left (d x +c \right )^{\frac {3}{2}}}}{d^{4}}\) | \(98\) |
-2/3*((-D*x^3-3*C*x^2+3*B*x+A)*d^3+2*c*(3*D*x^2-6*C*x+B)*d^2-8*c^2*(-3*D*x +C)*d+16*D*c^3)/(d*x+c)^(3/2)/d^4
Time = 0.28 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.97 \[ \int \frac {A+B x+C x^2+D x^3}{(c+d x)^{5/2}} \, dx=\frac {2 \, {\left (D d^{3} x^{3} - 16 \, D c^{3} + 8 \, C c^{2} d - 2 \, B c d^{2} - A d^{3} - 3 \, {\left (2 \, D c d^{2} - C d^{3}\right )} x^{2} - 3 \, {\left (8 \, D c^{2} d - 4 \, C c d^{2} + B d^{3}\right )} x\right )} \sqrt {d x + c}}{3 \, {\left (d^{6} x^{2} + 2 \, c d^{5} x + c^{2} d^{4}\right )}} \]
2/3*(D*d^3*x^3 - 16*D*c^3 + 8*C*c^2*d - 2*B*c*d^2 - A*d^3 - 3*(2*D*c*d^2 - C*d^3)*x^2 - 3*(8*D*c^2*d - 4*C*c*d^2 + B*d^3)*x)*sqrt(d*x + c)/(d^6*x^2 + 2*c*d^5*x + c^2*d^4)
Leaf count of result is larger than twice the leaf count of optimal. 425 vs. \(2 (119) = 238\).
Time = 0.36 (sec) , antiderivative size = 425, normalized size of antiderivative = 3.76 \[ \int \frac {A+B x+C x^2+D x^3}{(c+d x)^{5/2}} \, dx=\begin {cases} - \frac {2 A d^{3}}{3 c d^{4} \sqrt {c + d x} + 3 d^{5} x \sqrt {c + d x}} - \frac {4 B c d^{2}}{3 c d^{4} \sqrt {c + d x} + 3 d^{5} x \sqrt {c + d x}} - \frac {6 B d^{3} x}{3 c d^{4} \sqrt {c + d x} + 3 d^{5} x \sqrt {c + d x}} + \frac {16 C c^{2} d}{3 c d^{4} \sqrt {c + d x} + 3 d^{5} x \sqrt {c + d x}} + \frac {24 C c d^{2} x}{3 c d^{4} \sqrt {c + d x} + 3 d^{5} x \sqrt {c + d x}} + \frac {6 C d^{3} x^{2}}{3 c d^{4} \sqrt {c + d x} + 3 d^{5} x \sqrt {c + d x}} - \frac {32 D c^{3}}{3 c d^{4} \sqrt {c + d x} + 3 d^{5} x \sqrt {c + d x}} - \frac {48 D c^{2} d x}{3 c d^{4} \sqrt {c + d x} + 3 d^{5} x \sqrt {c + d x}} - \frac {12 D c d^{2} x^{2}}{3 c d^{4} \sqrt {c + d x} + 3 d^{5} x \sqrt {c + d x}} + \frac {2 D d^{3} x^{3}}{3 c d^{4} \sqrt {c + d x} + 3 d^{5} x \sqrt {c + d x}} & \text {for}\: d \neq 0 \\\frac {A x + \frac {B x^{2}}{2} + \frac {C x^{3}}{3} + \frac {D x^{4}}{4}}{c^{\frac {5}{2}}} & \text {otherwise} \end {cases} \]
Piecewise((-2*A*d**3/(3*c*d**4*sqrt(c + d*x) + 3*d**5*x*sqrt(c + d*x)) - 4 *B*c*d**2/(3*c*d**4*sqrt(c + d*x) + 3*d**5*x*sqrt(c + d*x)) - 6*B*d**3*x/( 3*c*d**4*sqrt(c + d*x) + 3*d**5*x*sqrt(c + d*x)) + 16*C*c**2*d/(3*c*d**4*s qrt(c + d*x) + 3*d**5*x*sqrt(c + d*x)) + 24*C*c*d**2*x/(3*c*d**4*sqrt(c + d*x) + 3*d**5*x*sqrt(c + d*x)) + 6*C*d**3*x**2/(3*c*d**4*sqrt(c + d*x) + 3 *d**5*x*sqrt(c + d*x)) - 32*D*c**3/(3*c*d**4*sqrt(c + d*x) + 3*d**5*x*sqrt (c + d*x)) - 48*D*c**2*d*x/(3*c*d**4*sqrt(c + d*x) + 3*d**5*x*sqrt(c + d*x )) - 12*D*c*d**2*x**2/(3*c*d**4*sqrt(c + d*x) + 3*d**5*x*sqrt(c + d*x)) + 2*D*d**3*x**3/(3*c*d**4*sqrt(c + d*x) + 3*d**5*x*sqrt(c + d*x)), Ne(d, 0)) , ((A*x + B*x**2/2 + C*x**3/3 + D*x**4/4)/c**(5/2), True))
Time = 0.19 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.87 \[ \int \frac {A+B x+C x^2+D x^3}{(c+d x)^{5/2}} \, dx=\frac {2 \, {\left (\frac {{\left (d x + c\right )}^{\frac {3}{2}} D - 3 \, {\left (3 \, D c - C d\right )} \sqrt {d x + c}}{d^{3}} + \frac {D c^{3} - C c^{2} d + B c d^{2} - A d^{3} - 3 \, {\left (3 \, D c^{2} - 2 \, C c d + B d^{2}\right )} {\left (d x + c\right )}}{{\left (d x + c\right )}^{\frac {3}{2}} d^{3}}\right )}}{3 \, d} \]
2/3*(((d*x + c)^(3/2)*D - 3*(3*D*c - C*d)*sqrt(d*x + c))/d^3 + (D*c^3 - C* c^2*d + B*c*d^2 - A*d^3 - 3*(3*D*c^2 - 2*C*c*d + B*d^2)*(d*x + c))/((d*x + c)^(3/2)*d^3))/d
Time = 0.29 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.02 \[ \int \frac {A+B x+C x^2+D x^3}{(c+d x)^{5/2}} \, dx=-\frac {2 \, {\left (9 \, {\left (d x + c\right )} D c^{2} - D c^{3} - 6 \, {\left (d x + c\right )} C c d + C c^{2} d + 3 \, {\left (d x + c\right )} B d^{2} - B c d^{2} + A d^{3}\right )}}{3 \, {\left (d x + c\right )}^{\frac {3}{2}} d^{4}} + \frac {2 \, {\left ({\left (d x + c\right )}^{\frac {3}{2}} D d^{8} - 9 \, \sqrt {d x + c} D c d^{8} + 3 \, \sqrt {d x + c} C d^{9}\right )}}{3 \, d^{12}} \]
-2/3*(9*(d*x + c)*D*c^2 - D*c^3 - 6*(d*x + c)*C*c*d + C*c^2*d + 3*(d*x + c )*B*d^2 - B*c*d^2 + A*d^3)/((d*x + c)^(3/2)*d^4) + 2/3*((d*x + c)^(3/2)*D* d^8 - 9*sqrt(d*x + c)*D*c*d^8 + 3*sqrt(d*x + c)*C*d^9)/d^12
Timed out. \[ \int \frac {A+B x+C x^2+D x^3}{(c+d x)^{5/2}} \, dx=\int \frac {A+B\,x+C\,x^2+x^3\,D}{{\left (c+d\,x\right )}^{5/2}} \,d x \]