Integrand size = 19, antiderivative size = 138 \[ \int \frac {(a+b x)^{5/2}}{(c+d x)^{3/2}} \, dx=-\frac {2 (a+b x)^{5/2}}{d \sqrt {c+d x}}-\frac {15 b (b c-a d) \sqrt {a+b x} \sqrt {c+d x}}{4 d^3}+\frac {5 b (a+b x)^{3/2} \sqrt {c+d x}}{2 d^2}+\frac {15 \sqrt {b} (b c-a d)^2 \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{4 d^{7/2}} \]
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Time = 0.05 (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} \[ \int \frac {(a+b x)^{5/2}}{(c+d x)^{3/2}} \, dx=\frac {15 \sqrt {b} (b c-a d)^2 \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{4 d^{7/2}}-\frac {15 b \sqrt {a+b x} \sqrt {c+d x} (b c-a d)}{4 d^3}+\frac {5 b (a+b x)^{3/2} \sqrt {c+d x}}{2 d^2}-\frac {2 (a+b x)^{5/2}}{d \sqrt {c+d x}} \]
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Rule 49
Rule 52
Rule 65
Rule 212
Rule 223
Rubi steps \begin{align*} \text {integral}& = -\frac {2 (a+b x)^{5/2}}{d \sqrt {c+d x}}+\frac {(5 b) \int \frac {(a+b x)^{3/2}}{\sqrt {c+d x}} \, dx}{d} \\ & = -\frac {2 (a+b x)^{5/2}}{d \sqrt {c+d x}}+\frac {5 b (a+b x)^{3/2} \sqrt {c+d x}}{2 d^2}-\frac {(15 b (b c-a d)) \int \frac {\sqrt {a+b x}}{\sqrt {c+d x}} \, dx}{4 d^2} \\ & = -\frac {2 (a+b x)^{5/2}}{d \sqrt {c+d x}}-\frac {15 b (b c-a d) \sqrt {a+b x} \sqrt {c+d x}}{4 d^3}+\frac {5 b (a+b x)^{3/2} \sqrt {c+d x}}{2 d^2}+\frac {\left (15 b (b c-a d)^2\right ) \int \frac {1}{\sqrt {a+b x} \sqrt {c+d x}} \, dx}{8 d^3} \\ & = -\frac {2 (a+b x)^{5/2}}{d \sqrt {c+d x}}-\frac {15 b (b c-a d) \sqrt {a+b x} \sqrt {c+d x}}{4 d^3}+\frac {5 b (a+b x)^{3/2} \sqrt {c+d x}}{2 d^2}+\frac {\left (15 (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 d^3} \\ & = -\frac {2 (a+b x)^{5/2}}{d \sqrt {c+d x}}-\frac {15 b (b c-a d) \sqrt {a+b x} \sqrt {c+d x}}{4 d^3}+\frac {5 b (a+b x)^{3/2} \sqrt {c+d x}}{2 d^2}+\frac {\left (15 (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 d^3} \\ & = -\frac {2 (a+b x)^{5/2}}{d \sqrt {c+d x}}-\frac {15 b (b c-a d) \sqrt {a+b x} \sqrt {c+d x}}{4 d^3}+\frac {5 b (a+b x)^{3/2} \sqrt {c+d x}}{2 d^2}+\frac {15 \sqrt {b} (b c-a d)^2 \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{4 d^{7/2}} \\ \end{align*}
Time = 0.04 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.90 \[ \int \frac {(a+b x)^{5/2}}{(c+d x)^{3/2}} \, dx=\frac {\sqrt {a+b x} \left (-8 a^2 d^2+a b d (25 c+9 d x)+b^2 \left (-15 c^2-5 c d x+2 d^2 x^2\right )\right )}{4 d^3 \sqrt {c+d x}}+\frac {15 \sqrt {b} (b c-a d)^2 \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {a+b x}}\right )}{4 d^{7/2}} \]
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\[\int \frac {\left (b x +a \right )^{\frac {5}{2}}}{\left (d x +c \right )^{\frac {3}{2}}}d x\]
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none
Time = 0.31 (sec) , antiderivative size = 441, normalized size of antiderivative = 3.20 \[ \int \frac {(a+b x)^{5/2}}{(c+d x)^{3/2}} \, dx=\left [\frac {15 \, {\left (b^{2} c^{3} - 2 \, a b c^{2} d + a^{2} c d^{2} + {\left (b^{2} c^{2} d - 2 \, a b c d^{2} + a^{2} d^{3}\right )} x\right )} \sqrt {\frac {b}{d}} \log \left (8 \, b^{2} d^{2} x^{2} + b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2} + 4 \, {\left (2 \, b d^{2} x + b c d + a d^{2}\right )} \sqrt {b x + a} \sqrt {d x + c} \sqrt {\frac {b}{d}} + 8 \, {\left (b^{2} c d + a b d^{2}\right )} x\right ) + 4 \, {\left (2 \, b^{2} d^{2} x^{2} - 15 \, b^{2} c^{2} + 25 \, a b c d - 8 \, a^{2} d^{2} - {\left (5 \, b^{2} c d - 9 \, a b d^{2}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{16 \, {\left (d^{4} x + c d^{3}\right )}}, -\frac {15 \, {\left (b^{2} c^{3} - 2 \, a b c^{2} d + a^{2} c d^{2} + {\left (b^{2} c^{2} d - 2 \, a b c d^{2} + a^{2} d^{3}\right )} x\right )} \sqrt {-\frac {b}{d}} \arctan \left (\frac {{\left (2 \, b d x + b c + a d\right )} \sqrt {b x + a} \sqrt {d x + c} \sqrt {-\frac {b}{d}}}{2 \, {\left (b^{2} d x^{2} + a b c + {\left (b^{2} c + a b d\right )} x\right )}}\right ) - 2 \, {\left (2 \, b^{2} d^{2} x^{2} - 15 \, b^{2} c^{2} + 25 \, a b c d - 8 \, a^{2} d^{2} - {\left (5 \, b^{2} c d - 9 \, a b d^{2}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{8 \, {\left (d^{4} x + c d^{3}\right )}}\right ] \]
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\[ \int \frac {(a+b x)^{5/2}}{(c+d x)^{3/2}} \, dx=\int \frac {\left (a + b x\right )^{\frac {5}{2}}}{\left (c + d x\right )^{\frac {3}{2}}}\, dx \]
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Exception generated. \[ \int \frac {(a+b x)^{5/2}}{(c+d x)^{3/2}} \, dx=\text {Exception raised: ValueError} \]
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Time = 0.34 (sec) , antiderivative size = 201, normalized size of antiderivative = 1.46 \[ \int \frac {(a+b x)^{5/2}}{(c+d x)^{3/2}} \, dx=\frac {\sqrt {b x + a} {\left ({\left (b x + a\right )} {\left (\frac {2 \, {\left (b x + a\right )} b^{2}}{d {\left | b \right |}} - \frac {5 \, {\left (b^{3} c d^{3} - a b^{2} d^{4}\right )}}{d^{5} {\left | b \right |}}\right )} - \frac {15 \, {\left (b^{4} c^{2} d^{2} - 2 \, a b^{3} c d^{3} + a^{2} b^{2} d^{4}\right )}}{d^{5} {\left | b \right |}}\right )}}{4 \, \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}} - \frac {15 \, {\left (b^{4} c^{2} - 2 \, a b^{3} c d + a^{2} b^{2} d^{2}\right )} \log \left ({\left | -\sqrt {b d} \sqrt {b x + a} + \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d} \right |}\right )}{4 \, \sqrt {b d} d^{3} {\left | b \right |}} \]
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Timed out. \[ \int \frac {(a+b x)^{5/2}}{(c+d x)^{3/2}} \, dx=\int \frac {{\left (a+b\,x\right )}^{5/2}}{{\left (c+d\,x\right )}^{3/2}} \,d x \]
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