Integrand size = 25, antiderivative size = 113 \[ \int \frac {A+B x+C x^2+D x^3}{(c+d x)^{5/2}} \, dx=-\frac {2 \left (c^2 C d-B c d^2+A d^3-c^3 D\right )}{3 d^4 (c+d x)^{3/2}}+\frac {2 \left (2 c C d-B d^2-3 c^2 D\right )}{d^4 \sqrt {c+d x}}+\frac {2 (C d-3 c D) \sqrt {c+d x}}{d^4}+\frac {2 D (c+d x)^{3/2}}{3 d^4} \]
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Time = 0.04 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.040, Rules used = {1864} \[ \int \frac {A+B x+C x^2+D x^3}{(c+d x)^{5/2}} \, dx=-\frac {2 \left (A d^3-B c d^2+c^3 (-D)+c^2 C d\right )}{3 d^4 (c+d x)^{3/2}}+\frac {2 \left (-B d^2-3 c^2 D+2 c C d\right )}{d^4 \sqrt {c+d x}}+\frac {2 \sqrt {c+d x} (C d-3 c D)}{d^4}+\frac {2 D (c+d x)^{3/2}}{3 d^4} \]
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Rule 1864
Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {c^2 C d-B c d^2+A d^3-c^3 D}{d^3 (c+d x)^{5/2}}+\frac {-2 c C d+B d^2+3 c^2 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 \\ & = -\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} \\ \end{align*}
Time = 0.07 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.66 \[ \int \frac {A+B x+C x^2+D x^3}{(c+d x)^{5/2}} \, dx=-\frac {2 \left (16 c^3 D-8 c^2 d (C-3 D x)+2 c d^2 (B+3 x (-2 C+D x))+d^3 \left (A+3 B x-x^2 (3 C+D x)\right )\right )}{3 d^4 (c+d x)^{3/2}} \]
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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 | \(\frac {\frac {2 D \left (d x +c \right )^{\frac {3}{2}}}{3}+2 d C \sqrt {d x +c}-6 D c \sqrt {d x +c}-\frac {2 \left (B \,d^{2}-2 C c d +3 D c^{2}\right )}{\sqrt {d x +c}}-\frac {2 \left (A \,d^{3}-B c \,d^{2}+C \,c^{2} d -D c^{3}\right )}{3 \left (d x +c \right )^{\frac {3}{2}}}}{d^{4}}\) | \(98\) |
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Time = 0.28 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.97 \[ \int \frac {A+B x+C x^2+D x^3}{(c+d x)^{5/2}} \, dx=\frac {2 \, {\left (D d^{3} x^{3} - 16 \, D c^{3} + 8 \, C c^{2} d - 2 \, B c d^{2} - A d^{3} - 3 \, {\left (2 \, D c d^{2} - C d^{3}\right )} x^{2} - 3 \, {\left (8 \, D c^{2} d - 4 \, C c d^{2} + B d^{3}\right )} x\right )} \sqrt {d x + c}}{3 \, {\left (d^{6} x^{2} + 2 \, c d^{5} x + c^{2} d^{4}\right )}} \]
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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 \[ \int \frac {A+B x+C x^2+D x^3}{(c+d x)^{5/2}} \, dx=\begin {cases} - \frac {2 A d^{3}}{3 c d^{4} \sqrt {c + d x} + 3 d^{5} x \sqrt {c + d x}} - \frac {4 B c d^{2}}{3 c d^{4} \sqrt {c + d x} + 3 d^{5} x \sqrt {c + d x}} - \frac {6 B d^{3} x}{3 c d^{4} \sqrt {c + d x} + 3 d^{5} x \sqrt {c + d x}} + \frac {16 C c^{2} d}{3 c d^{4} \sqrt {c + d x} + 3 d^{5} x \sqrt {c + d x}} + \frac {24 C c d^{2} x}{3 c d^{4} \sqrt {c + d x} + 3 d^{5} x \sqrt {c + d x}} + \frac {6 C d^{3} x^{2}}{3 c d^{4} \sqrt {c + d x} + 3 d^{5} x \sqrt {c + d x}} - \frac {32 D c^{3}}{3 c d^{4} \sqrt {c + d x} + 3 d^{5} x \sqrt {c + d x}} - \frac {48 D c^{2} d x}{3 c d^{4} \sqrt {c + d x} + 3 d^{5} x \sqrt {c + d x}} - \frac {12 D c d^{2} x^{2}}{3 c d^{4} \sqrt {c + d x} + 3 d^{5} x \sqrt {c + d x}} + \frac {2 D d^{3} x^{3}}{3 c d^{4} \sqrt {c + d x} + 3 d^{5} x \sqrt {c + d x}} & \text {for}\: d \neq 0 \\\frac {A x + \frac {B x^{2}}{2} + \frac {C x^{3}}{3} + \frac {D x^{4}}{4}}{c^{\frac {5}{2}}} & \text {otherwise} \end {cases} \]
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Time = 0.19 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.87 \[ \int \frac {A+B x+C x^2+D x^3}{(c+d x)^{5/2}} \, dx=\frac {2 \, {\left (\frac {{\left (d x + c\right )}^{\frac {3}{2}} D - 3 \, {\left (3 \, D c - C d\right )} \sqrt {d x + c}}{d^{3}} + \frac {D c^{3} - C c^{2} d + B c d^{2} - A d^{3} - 3 \, {\left (3 \, D c^{2} - 2 \, C c d + B d^{2}\right )} {\left (d x + c\right )}}{{\left (d x + c\right )}^{\frac {3}{2}} d^{3}}\right )}}{3 \, d} \]
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Time = 0.29 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.02 \[ \int \frac {A+B x+C x^2+D x^3}{(c+d x)^{5/2}} \, dx=-\frac {2 \, {\left (9 \, {\left (d x + c\right )} D c^{2} - D c^{3} - 6 \, {\left (d x + c\right )} C c d + C c^{2} d + 3 \, {\left (d x + c\right )} B d^{2} - B c d^{2} + A d^{3}\right )}}{3 \, {\left (d x + c\right )}^{\frac {3}{2}} d^{4}} + \frac {2 \, {\left ({\left (d x + c\right )}^{\frac {3}{2}} D d^{8} - 9 \, \sqrt {d x + c} D c d^{8} + 3 \, \sqrt {d x + c} C d^{9}\right )}}{3 \, d^{12}} \]
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Timed out. \[ \int \frac {A+B x+C x^2+D x^3}{(c+d x)^{5/2}} \, dx=\int \frac {A+B\,x+C\,x^2+x^3\,D}{{\left (c+d\,x\right )}^{5/2}} \,d x \]
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