Integrand size = 19, antiderivative size = 148 \[ \int \frac {(a+b x)^{5/2}}{\sqrt {c+d x}} \, dx=\frac {5 (b c-a d)^2 \sqrt {a+b x} \sqrt {c+d x}}{8 d^3}-\frac {5 (b c-a d) (a+b x)^{3/2} \sqrt {c+d x}}{12 d^2}+\frac {(a+b x)^{5/2} \sqrt {c+d x}}{3 d}-\frac {5 (b c-a d)^3 \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{8 \sqrt {b} d^{7/2}} \]
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Time = 0.06 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {52, 65, 223, 212} \[ \int \frac {(a+b x)^{5/2}}{\sqrt {c+d x}} \, dx=-\frac {5 (b c-a d)^3 \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{8 \sqrt {b} d^{7/2}}+\frac {5 \sqrt {a+b x} \sqrt {c+d x} (b c-a d)^2}{8 d^3}-\frac {5 (a+b x)^{3/2} \sqrt {c+d x} (b c-a d)}{12 d^2}+\frac {(a+b x)^{5/2} \sqrt {c+d x}}{3 d} \]
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Rule 52
Rule 65
Rule 212
Rule 223
Rubi steps \begin{align*} \text {integral}& = \frac {(a+b x)^{5/2} \sqrt {c+d x}}{3 d}-\frac {(5 (b c-a d)) \int \frac {(a+b x)^{3/2}}{\sqrt {c+d x}} \, dx}{6 d} \\ & = -\frac {5 (b c-a d) (a+b x)^{3/2} \sqrt {c+d x}}{12 d^2}+\frac {(a+b x)^{5/2} \sqrt {c+d x}}{3 d}+\frac {\left (5 (b c-a d)^2\right ) \int \frac {\sqrt {a+b x}}{\sqrt {c+d x}} \, dx}{8 d^2} \\ & = \frac {5 (b c-a d)^2 \sqrt {a+b x} \sqrt {c+d x}}{8 d^3}-\frac {5 (b c-a d) (a+b x)^{3/2} \sqrt {c+d x}}{12 d^2}+\frac {(a+b x)^{5/2} \sqrt {c+d x}}{3 d}-\frac {\left (5 (b c-a d)^3\right ) \int \frac {1}{\sqrt {a+b x} \sqrt {c+d x}} \, dx}{16 d^3} \\ & = \frac {5 (b c-a d)^2 \sqrt {a+b x} \sqrt {c+d x}}{8 d^3}-\frac {5 (b c-a d) (a+b x)^{3/2} \sqrt {c+d x}}{12 d^2}+\frac {(a+b x)^{5/2} \sqrt {c+d x}}{3 d}-\frac {\left (5 (b c-a d)^3\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 )}{8 b d^3} \\ & = \frac {5 (b c-a d)^2 \sqrt {a+b x} \sqrt {c+d x}}{8 d^3}-\frac {5 (b c-a d) (a+b x)^{3/2} \sqrt {c+d x}}{12 d^2}+\frac {(a+b x)^{5/2} \sqrt {c+d x}}{3 d}-\frac {\left (5 (b c-a d)^3\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 )}{8 b d^3} \\ & = \frac {5 (b c-a d)^2 \sqrt {a+b x} \sqrt {c+d x}}{8 d^3}-\frac {5 (b c-a d) (a+b x)^{3/2} \sqrt {c+d x}}{12 d^2}+\frac {(a+b x)^{5/2} \sqrt {c+d x}}{3 d}-\frac {5 (b c-a d)^3 \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{8 \sqrt {b} d^{7/2}} \\ \end{align*}
Time = 0.04 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.84 \[ \int \frac {(a+b x)^{5/2}}{\sqrt {c+d x}} \, dx=\frac {\sqrt {a+b x} \sqrt {c+d x} \left (33 a^2 d^2+2 a b d (-20 c+13 d x)+b^2 \left (15 c^2-10 c d x+8 d^2 x^2\right )\right )}{24 d^3}-\frac {5 (b c-a d)^3 \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {a+b x}}\right )}{8 \sqrt {b} d^{7/2}} \]
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Time = 1.61 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.17
method | result | size |
default | \(\frac {\left (b x +a \right )^{\frac {5}{2}} \sqrt {d x +c}}{3 d}-\frac {5 \left (-a d +b c \right ) \left (\frac {\left (b x +a \right )^{\frac {3}{2}} \sqrt {d x +c}}{2 d}-\frac {3 \left (-a d +b c \right ) \left (\frac {\sqrt {b x +a}\, \sqrt {d x +c}}{d}-\frac {\left (-a d +b c \right ) \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \ln \left (\frac {\frac {1}{2} a d +\frac {1}{2} b c +b d x}{\sqrt {b d}}+\sqrt {b d \,x^{2}+\left (a d +b c \right ) x +a c}\right )}{2 d \sqrt {b x +a}\, \sqrt {d x +c}\, \sqrt {b d}}\right )}{4 d}\right )}{6 d}\) | \(173\) |
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Time = 0.25 (sec) , antiderivative size = 412, normalized size of antiderivative = 2.78 \[ \int \frac {(a+b x)^{5/2}}{\sqrt {c+d x}} \, dx=\left [-\frac {15 \, {\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} \sqrt {b d} \log \left (8 \, b^{2} d^{2} x^{2} + b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2} + 4 \, {\left (2 \, b d x + b c + a d\right )} \sqrt {b d} \sqrt {b x + a} \sqrt {d x + c} + 8 \, {\left (b^{2} c d + a b d^{2}\right )} x\right ) - 4 \, {\left (8 \, b^{3} d^{3} x^{2} + 15 \, b^{3} c^{2} d - 40 \, a b^{2} c d^{2} + 33 \, a^{2} b d^{3} - 2 \, {\left (5 \, b^{3} c d^{2} - 13 \, a b^{2} d^{3}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{96 \, b d^{4}}, \frac {15 \, {\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} \sqrt {-b d} \arctan \left (\frac {{\left (2 \, b d x + b c + a d\right )} \sqrt {-b d} \sqrt {b x + a} \sqrt {d x + c}}{2 \, {\left (b^{2} d^{2} x^{2} + a b c d + {\left (b^{2} c d + a b d^{2}\right )} x\right )}}\right ) + 2 \, {\left (8 \, b^{3} d^{3} x^{2} + 15 \, b^{3} c^{2} d - 40 \, a b^{2} c d^{2} + 33 \, a^{2} b d^{3} - 2 \, {\left (5 \, b^{3} c d^{2} - 13 \, a b^{2} d^{3}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{48 \, b d^{4}}\right ] \]
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\[ \int \frac {(a+b x)^{5/2}}{\sqrt {c+d x}} \, dx=\int \frac {\left (a + b x\right )^{\frac {5}{2}}}{\sqrt {c + d x}}\, dx \]
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Exception generated. \[ \int \frac {(a+b x)^{5/2}}{\sqrt {c+d x}} \, dx=\text {Exception raised: ValueError} \]
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Time = 0.31 (sec) , antiderivative size = 198, normalized size of antiderivative = 1.34 \[ \int \frac {(a+b x)^{5/2}}{\sqrt {c+d x}} \, dx=\frac {{\left (\sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d} \sqrt {b x + a} {\left (2 \, {\left (b x + a\right )} {\left (\frac {4 \, {\left (b x + a\right )}}{b d} - \frac {5 \, {\left (b c d^{3} - a d^{4}\right )}}{b d^{5}}\right )} + \frac {15 \, {\left (b^{2} c^{2} d^{2} - 2 \, a b c d^{3} + a^{2} d^{4}\right )}}{b d^{5}}\right )} + \frac {15 \, {\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} \log \left ({\left | -\sqrt {b d} \sqrt {b x + a} + \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d} \right |}\right )}{\sqrt {b d} d^{3}}\right )} b}{24 \, {\left | b \right |}} \]
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Timed out. \[ \int \frac {(a+b x)^{5/2}}{\sqrt {c+d x}} \, dx=\int \frac {{\left (a+b\,x\right )}^{5/2}}{\sqrt {c+d\,x}} \,d x \]
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