Integrand size = 17, antiderivative size = 94 \[ \int \frac {(a+b x)^3}{(c+d x)^{3/2}} \, dx=\frac {2 (b c-a d)^3}{d^4 \sqrt {c+d x}}+\frac {6 b (b c-a d)^2 \sqrt {c+d x}}{d^4}-\frac {2 b^2 (b c-a d) (c+d x)^{3/2}}{d^4}+\frac {2 b^3 (c+d x)^{5/2}}{5 d^4} \]
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Time = 0.03 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.059, Rules used = {45} \[ \int \frac {(a+b x)^3}{(c+d x)^{3/2}} \, dx=-\frac {2 b^2 (c+d x)^{3/2} (b c-a d)}{d^4}+\frac {6 b \sqrt {c+d x} (b c-a d)^2}{d^4}+\frac {2 (b c-a d)^3}{d^4 \sqrt {c+d x}}+\frac {2 b^3 (c+d x)^{5/2}}{5 d^4} \]
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Rule 45
Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {(-b c+a d)^3}{d^3 (c+d x)^{3/2}}+\frac {3 b (b c-a d)^2}{d^3 \sqrt {c+d x}}-\frac {3 b^2 (b c-a d) \sqrt {c+d x}}{d^3}+\frac {b^3 (c+d x)^{3/2}}{d^3}\right ) \, dx \\ & = \frac {2 (b c-a d)^3}{d^4 \sqrt {c+d x}}+\frac {6 b (b c-a d)^2 \sqrt {c+d x}}{d^4}-\frac {2 b^2 (b c-a d) (c+d x)^{3/2}}{d^4}+\frac {2 b^3 (c+d x)^{5/2}}{5 d^4} \\ \end{align*}
Time = 0.06 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.05 \[ \int \frac {(a+b x)^3}{(c+d x)^{3/2}} \, dx=\frac {2 \left (-5 a^3 d^3+15 a^2 b d^2 (2 c+d x)+5 a b^2 d \left (-8 c^2-4 c d x+d^2 x^2\right )+b^3 \left (16 c^3+8 c^2 d x-2 c d^2 x^2+d^3 x^3\right )\right )}{5 d^4 \sqrt {c+d x}} \]
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Time = 0.28 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.00
method | result | size |
pseudoelliptic | \(\frac {\frac {2 \left (d^{3} x^{3}-2 c \,d^{2} x^{2}+8 c^{2} d x +16 c^{3}\right ) b^{3}}{5}-16 \left (-\frac {1}{8} d^{2} x^{2}+\frac {1}{2} c d x +c^{2}\right ) d a \,b^{2}+12 \left (\frac {d x}{2}+c \right ) d^{2} a^{2} b -2 a^{3} d^{3}}{\sqrt {d x +c}\, d^{4}}\) | \(94\) |
risch | \(\frac {2 b \left (d^{2} x^{2} b^{2}+5 x a b \,d^{2}-3 x \,b^{2} c d +15 a^{2} d^{2}-25 a b c d +11 b^{2} c^{2}\right ) \sqrt {d x +c}}{5 d^{4}}-\frac {2 \left (a^{3} d^{3}-3 a^{2} b c \,d^{2}+3 a \,b^{2} c^{2} d -b^{3} c^{3}\right )}{d^{4} \sqrt {d x +c}}\) | \(112\) |
gosper | \(-\frac {2 \left (-d^{3} x^{3} b^{3}-5 x^{2} a \,b^{2} d^{3}+2 x^{2} b^{3} c \,d^{2}-15 x \,a^{2} b \,d^{3}+20 x a \,b^{2} c \,d^{2}-8 x \,b^{3} c^{2} d +5 a^{3} d^{3}-30 a^{2} b c \,d^{2}+40 a \,b^{2} c^{2} d -16 b^{3} c^{3}\right )}{5 \sqrt {d x +c}\, d^{4}}\) | \(116\) |
trager | \(-\frac {2 \left (-d^{3} x^{3} b^{3}-5 x^{2} a \,b^{2} d^{3}+2 x^{2} b^{3} c \,d^{2}-15 x \,a^{2} b \,d^{3}+20 x a \,b^{2} c \,d^{2}-8 x \,b^{3} c^{2} d +5 a^{3} d^{3}-30 a^{2} b c \,d^{2}+40 a \,b^{2} c^{2} d -16 b^{3} c^{3}\right )}{5 \sqrt {d x +c}\, d^{4}}\) | \(116\) |
derivativedivides | \(\frac {\frac {2 b^{3} \left (d x +c \right )^{\frac {5}{2}}}{5}+2 a \,b^{2} d \left (d x +c \right )^{\frac {3}{2}}-2 b^{3} c \left (d x +c \right )^{\frac {3}{2}}+6 a^{2} b \,d^{2} \sqrt {d x +c}-12 a \,b^{2} c d \sqrt {d x +c}+6 b^{3} c^{2} \sqrt {d x +c}-\frac {2 \left (a^{3} d^{3}-3 a^{2} b c \,d^{2}+3 a \,b^{2} c^{2} d -b^{3} c^{3}\right )}{\sqrt {d x +c}}}{d^{4}}\) | \(136\) |
default | \(\frac {\frac {2 b^{3} \left (d x +c \right )^{\frac {5}{2}}}{5}+2 a \,b^{2} d \left (d x +c \right )^{\frac {3}{2}}-2 b^{3} c \left (d x +c \right )^{\frac {3}{2}}+6 a^{2} b \,d^{2} \sqrt {d x +c}-12 a \,b^{2} c d \sqrt {d x +c}+6 b^{3} c^{2} \sqrt {d x +c}-\frac {2 \left (a^{3} d^{3}-3 a^{2} b c \,d^{2}+3 a \,b^{2} c^{2} d -b^{3} c^{3}\right )}{\sqrt {d x +c}}}{d^{4}}\) | \(136\) |
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Time = 0.22 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.32 \[ \int \frac {(a+b x)^3}{(c+d x)^{3/2}} \, dx=\frac {2 \, {\left (b^{3} d^{3} x^{3} + 16 \, b^{3} c^{3} - 40 \, a b^{2} c^{2} d + 30 \, a^{2} b c d^{2} - 5 \, a^{3} d^{3} - {\left (2 \, b^{3} c d^{2} - 5 \, a b^{2} d^{3}\right )} x^{2} + {\left (8 \, b^{3} c^{2} d - 20 \, a b^{2} c d^{2} + 15 \, a^{2} b d^{3}\right )} x\right )} \sqrt {d x + c}}{5 \, {\left (d^{5} x + c d^{4}\right )}} \]
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Time = 2.41 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.43 \[ \int \frac {(a+b x)^3}{(c+d x)^{3/2}} \, dx=\begin {cases} \frac {2 \left (\frac {b^{3} \left (c + d x\right )^{\frac {5}{2}}}{5 d^{3}} + \frac {\left (c + d x\right )^{\frac {3}{2}} \cdot \left (3 a b^{2} d - 3 b^{3} c\right )}{3 d^{3}} + \frac {\sqrt {c + d x} \left (3 a^{2} b d^{2} - 6 a b^{2} c d + 3 b^{3} c^{2}\right )}{d^{3}} - \frac {\left (a d - b c\right )^{3}}{d^{3} \sqrt {c + d x}}\right )}{d} & \text {for}\: d \neq 0 \\\frac {\begin {cases} a^{3} x & \text {for}\: b = 0 \\\frac {\left (a + b x\right )^{4}}{4 b} & \text {otherwise} \end {cases}}{c^{\frac {3}{2}}} & \text {otherwise} \end {cases} \]
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Time = 0.24 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.33 \[ \int \frac {(a+b x)^3}{(c+d x)^{3/2}} \, dx=\frac {2 \, {\left (\frac {{\left (d x + c\right )}^{\frac {5}{2}} b^{3} - 5 \, {\left (b^{3} c - a b^{2} d\right )} {\left (d x + c\right )}^{\frac {3}{2}} + 15 \, {\left (b^{3} c^{2} - 2 \, a b^{2} c d + a^{2} b d^{2}\right )} \sqrt {d x + c}}{d^{3}} + \frac {5 \, {\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )}}{\sqrt {d x + c} d^{3}}\right )}}{5 \, d} \]
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Time = 0.30 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.62 \[ \int \frac {(a+b x)^3}{(c+d x)^{3/2}} \, dx=\frac {2 \, {\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )}}{\sqrt {d x + c} d^{4}} + \frac {2 \, {\left ({\left (d x + c\right )}^{\frac {5}{2}} b^{3} d^{16} - 5 \, {\left (d x + c\right )}^{\frac {3}{2}} b^{3} c d^{16} + 15 \, \sqrt {d x + c} b^{3} c^{2} d^{16} + 5 \, {\left (d x + c\right )}^{\frac {3}{2}} a b^{2} d^{17} - 30 \, \sqrt {d x + c} a b^{2} c d^{17} + 15 \, \sqrt {d x + c} a^{2} b d^{18}\right )}}{5 \, d^{20}} \]
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Time = 0.08 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.21 \[ \int \frac {(a+b x)^3}{(c+d x)^{3/2}} \, dx=\frac {2\,b^3\,{\left (c+d\,x\right )}^{5/2}}{5\,d^4}-\frac {\left (6\,b^3\,c-6\,a\,b^2\,d\right )\,{\left (c+d\,x\right )}^{3/2}}{3\,d^4}-\frac {2\,a^3\,d^3-6\,a^2\,b\,c\,d^2+6\,a\,b^2\,c^2\,d-2\,b^3\,c^3}{d^4\,\sqrt {c+d\,x}}+\frac {6\,b\,{\left (a\,d-b\,c\right )}^2\,\sqrt {c+d\,x}}{d^4} \]
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