Integrand size = 19, antiderivative size = 138 \[ \int \frac {(c+d x)^{5/2}}{(a+b x)^{3/2}} \, dx=\frac {15 d (b c-a d) \sqrt {a+b x} \sqrt {c+d x}}{4 b^3}+\frac {5 d \sqrt {a+b x} (c+d x)^{3/2}}{2 b^2}-\frac {2 (c+d x)^{5/2}}{b \sqrt {a+b x}}+\frac {15 \sqrt {d} (b c-a d)^2 \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{4 b^{7/2}} \]
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Time = 0.04 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {49, 52, 65, 223, 212} \[ \int \frac {(c+d x)^{5/2}}{(a+b x)^{3/2}} \, dx=\frac {15 \sqrt {d} (b c-a d)^2 \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{4 b^{7/2}}+\frac {15 d \sqrt {a+b x} \sqrt {c+d x} (b c-a d)}{4 b^3}+\frac {5 d \sqrt {a+b x} (c+d x)^{3/2}}{2 b^2}-\frac {2 (c+d x)^{5/2}}{b \sqrt {a+b x}} \]
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Rule 49
Rule 52
Rule 65
Rule 212
Rule 223
Rubi steps \begin{align*} \text {integral}& = -\frac {2 (c+d x)^{5/2}}{b \sqrt {a+b x}}+\frac {(5 d) \int \frac {(c+d x)^{3/2}}{\sqrt {a+b x}} \, dx}{b} \\ & = \frac {5 d \sqrt {a+b x} (c+d x)^{3/2}}{2 b^2}-\frac {2 (c+d x)^{5/2}}{b \sqrt {a+b x}}+\frac {(15 d (b c-a d)) \int \frac {\sqrt {c+d x}}{\sqrt {a+b x}} \, dx}{4 b^2} \\ & = \frac {15 d (b c-a d) \sqrt {a+b x} \sqrt {c+d x}}{4 b^3}+\frac {5 d \sqrt {a+b x} (c+d x)^{3/2}}{2 b^2}-\frac {2 (c+d x)^{5/2}}{b \sqrt {a+b x}}+\frac {\left (15 d (b c-a d)^2\right ) \int \frac {1}{\sqrt {a+b x} \sqrt {c+d x}} \, dx}{8 b^3} \\ & = \frac {15 d (b c-a d) \sqrt {a+b x} \sqrt {c+d x}}{4 b^3}+\frac {5 d \sqrt {a+b x} (c+d x)^{3/2}}{2 b^2}-\frac {2 (c+d x)^{5/2}}{b \sqrt {a+b x}}+\frac {\left (15 d (b c-a d)^2\right ) \text {Subst}\left (\int \frac {1}{\sqrt {c-\frac {a d}{b}+\frac {d x^2}{b}}} \, dx,x,\sqrt {a+b x}\right )}{4 b^4} \\ & = \frac {15 d (b c-a d) \sqrt {a+b x} \sqrt {c+d x}}{4 b^3}+\frac {5 d \sqrt {a+b x} (c+d x)^{3/2}}{2 b^2}-\frac {2 (c+d x)^{5/2}}{b \sqrt {a+b x}}+\frac {\left (15 d (b c-a d)^2\right ) \text {Subst}\left (\int \frac {1}{1-\frac {d x^2}{b}} \, dx,x,\frac {\sqrt {a+b x}}{\sqrt {c+d x}}\right )}{4 b^4} \\ & = \frac {15 d (b c-a d) \sqrt {a+b x} \sqrt {c+d x}}{4 b^3}+\frac {5 d \sqrt {a+b x} (c+d x)^{3/2}}{2 b^2}-\frac {2 (c+d x)^{5/2}}{b \sqrt {a+b x}}+\frac {15 \sqrt {d} (b c-a d)^2 \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{4 b^{7/2}} \\ \end{align*}
Time = 0.24 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.90 \[ \int \frac {(c+d x)^{5/2}}{(a+b x)^{3/2}} \, dx=\frac {\sqrt {c+d x} \left (-15 a^2 d^2-5 a b d (-5 c+d x)+b^2 \left (-8 c^2+9 c d x+2 d^2 x^2\right )\right )}{4 b^3 \sqrt {a+b x}}+\frac {15 \sqrt {d} (b c-a d)^2 \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {a+b x}}\right )}{4 b^{7/2}} \]
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\[\int \frac {\left (d x +c \right )^{\frac {5}{2}}}{\left (b x +a \right )^{\frac {3}{2}}}d x\]
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none
Time = 0.30 (sec) , antiderivative size = 439, normalized size of antiderivative = 3.18 \[ \int \frac {(c+d x)^{5/2}}{(a+b x)^{3/2}} \, dx=\left [\frac {15 \, {\left (a b^{2} c^{2} - 2 \, a^{2} b c d + a^{3} d^{2} + {\left (b^{3} c^{2} - 2 \, a b^{2} c d + a^{2} b d^{2}\right )} x\right )} \sqrt {\frac {d}{b}} \log \left (8 \, b^{2} d^{2} x^{2} + b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2} + 4 \, {\left (2 \, b^{2} d x + b^{2} c + a b d\right )} \sqrt {b x + a} \sqrt {d x + c} \sqrt {\frac {d}{b}} + 8 \, {\left (b^{2} c d + a b d^{2}\right )} x\right ) + 4 \, {\left (2 \, b^{2} d^{2} x^{2} - 8 \, b^{2} c^{2} + 25 \, a b c d - 15 \, a^{2} d^{2} + {\left (9 \, b^{2} c d - 5 \, a b d^{2}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{16 \, {\left (b^{4} x + a b^{3}\right )}}, -\frac {15 \, {\left (a b^{2} c^{2} - 2 \, a^{2} b c d + a^{3} d^{2} + {\left (b^{3} c^{2} - 2 \, a b^{2} c d + a^{2} b d^{2}\right )} x\right )} \sqrt {-\frac {d}{b}} \arctan \left (\frac {{\left (2 \, b d x + b c + a d\right )} \sqrt {b x + a} \sqrt {d x + c} \sqrt {-\frac {d}{b}}}{2 \, {\left (b d^{2} x^{2} + a c d + {\left (b c d + a d^{2}\right )} x\right )}}\right ) - 2 \, {\left (2 \, b^{2} d^{2} x^{2} - 8 \, b^{2} c^{2} + 25 \, a b c d - 15 \, a^{2} d^{2} + {\left (9 \, b^{2} c d - 5 \, a b d^{2}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{8 \, {\left (b^{4} x + a b^{3}\right )}}\right ] \]
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\[ \int \frac {(c+d x)^{5/2}}{(a+b x)^{3/2}} \, dx=\int \frac {\left (c + d x\right )^{\frac {5}{2}}}{\left (a + b x\right )^{\frac {3}{2}}}\, dx \]
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Exception generated. \[ \int \frac {(c+d x)^{5/2}}{(a+b x)^{3/2}} \, dx=\text {Exception raised: ValueError} \]
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Leaf count of result is larger than twice the leaf count of optimal. 287 vs. \(2 (108) = 216\).
Time = 0.61 (sec) , antiderivative size = 287, normalized size of antiderivative = 2.08 \[ \int \frac {(c+d x)^{5/2}}{(a+b x)^{3/2}} \, dx=\frac {1}{4} \, \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d} \sqrt {b x + a} {\left (\frac {2 \, {\left (b x + a\right )} d^{2} {\left | b \right |}}{b^{5}} + \frac {9 \, {\left (b^{10} c d^{3} {\left | b \right |} - a b^{9} d^{4} {\left | b \right |}\right )}}{b^{14} d^{2}}\right )} - \frac {15 \, {\left (\sqrt {b d} b^{2} c^{2} {\left | b \right |} - 2 \, \sqrt {b d} a b c d {\left | b \right |} + \sqrt {b d} a^{2} d^{2} {\left | b \right |}\right )} \log \left ({\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2}\right )}{8 \, b^{5}} - \frac {4 \, {\left (\sqrt {b d} b^{3} c^{3} {\left | b \right |} - 3 \, \sqrt {b d} a b^{2} c^{2} d {\left | b \right |} + 3 \, \sqrt {b d} a^{2} b c d^{2} {\left | b \right |} - \sqrt {b d} a^{3} d^{3} {\left | b \right |}\right )}}{{\left (b^{2} c - a b d - {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2}\right )} b^{4}} \]
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Timed out. \[ \int \frac {(c+d x)^{5/2}}{(a+b x)^{3/2}} \, dx=\int \frac {{\left (c+d\,x\right )}^{5/2}}{{\left (a+b\,x\right )}^{3/2}} \,d x \]
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