Integrand size = 19, antiderivative size = 135 \[ \int \frac {(c+d x)^{5/4}}{(a+b x)^{5/2}} \, dx=-\frac {5 d \sqrt [4]{c+d x}}{3 b^2 \sqrt {a+b x}}-\frac {2 (c+d x)^{5/4}}{3 b (a+b x)^{3/2}}+\frac {5 d \sqrt [4]{b c-a d} \sqrt {-\frac {d (a+b x)}{b c-a d}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt [4]{b} \sqrt [4]{c+d x}}{\sqrt [4]{b c-a d}}\right ),-1\right )}{3 b^{9/4} \sqrt {a+b x}} \]
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Time = 0.06 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {49, 65, 230, 227} \[ \int \frac {(c+d x)^{5/4}}{(a+b x)^{5/2}} \, dx=\frac {5 d \sqrt [4]{b c-a d} \sqrt {-\frac {d (a+b x)}{b c-a d}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt [4]{b} \sqrt [4]{c+d x}}{\sqrt [4]{b c-a d}}\right ),-1\right )}{3 b^{9/4} \sqrt {a+b x}}-\frac {5 d \sqrt [4]{c+d x}}{3 b^2 \sqrt {a+b x}}-\frac {2 (c+d x)^{5/4}}{3 b (a+b x)^{3/2}} \]
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Rule 49
Rule 65
Rule 227
Rule 230
Rubi steps \begin{align*} \text {integral}& = -\frac {2 (c+d x)^{5/4}}{3 b (a+b x)^{3/2}}+\frac {(5 d) \int \frac {\sqrt [4]{c+d x}}{(a+b x)^{3/2}} \, dx}{6 b} \\ & = -\frac {5 d \sqrt [4]{c+d x}}{3 b^2 \sqrt {a+b x}}-\frac {2 (c+d x)^{5/4}}{3 b (a+b x)^{3/2}}+\frac {\left (5 d^2\right ) \int \frac {1}{\sqrt {a+b x} (c+d x)^{3/4}} \, dx}{12 b^2} \\ & = -\frac {5 d \sqrt [4]{c+d x}}{3 b^2 \sqrt {a+b x}}-\frac {2 (c+d x)^{5/4}}{3 b (a+b x)^{3/2}}+\frac {(5 d) \text {Subst}\left (\int \frac {1}{\sqrt {a-\frac {b c}{d}+\frac {b x^4}{d}}} \, dx,x,\sqrt [4]{c+d x}\right )}{3 b^2} \\ & = -\frac {5 d \sqrt [4]{c+d x}}{3 b^2 \sqrt {a+b x}}-\frac {2 (c+d x)^{5/4}}{3 b (a+b x)^{3/2}}+\frac {\left (5 d \sqrt {\frac {d (a+b x)}{-b c+a d}}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {1+\frac {b x^4}{\left (a-\frac {b c}{d}\right ) d}}} \, dx,x,\sqrt [4]{c+d x}\right )}{3 b^2 \sqrt {a+b x}} \\ & = -\frac {5 d \sqrt [4]{c+d x}}{3 b^2 \sqrt {a+b x}}-\frac {2 (c+d x)^{5/4}}{3 b (a+b x)^{3/2}}+\frac {5 d \sqrt [4]{b c-a d} \sqrt {-\frac {d (a+b x)}{b c-a d}} F\left (\left .\sin ^{-1}\left (\frac {\sqrt [4]{b} \sqrt [4]{c+d x}}{\sqrt [4]{b c-a d}}\right )\right |-1\right )}{3 b^{9/4} \sqrt {a+b x}} \\ \end{align*}
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 0.03 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.54 \[ \int \frac {(c+d x)^{5/4}}{(a+b x)^{5/2}} \, dx=-\frac {2 (c+d x)^{5/4} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},-\frac {5}{4},-\frac {1}{2},\frac {d (a+b x)}{-b c+a d}\right )}{3 b (a+b x)^{3/2} \left (\frac {b (c+d x)}{b c-a d}\right )^{5/4}} \]
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\[\int \frac {\left (d x +c \right )^{\frac {5}{4}}}{\left (b x +a \right )^{\frac {5}{2}}}d x\]
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\[ \int \frac {(c+d x)^{5/4}}{(a+b x)^{5/2}} \, dx=\int { \frac {{\left (d x + c\right )}^{\frac {5}{4}}}{{\left (b x + a\right )}^{\frac {5}{2}}} \,d x } \]
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\[ \int \frac {(c+d x)^{5/4}}{(a+b x)^{5/2}} \, dx=\int \frac {\left (c + d x\right )^{\frac {5}{4}}}{\left (a + b x\right )^{\frac {5}{2}}}\, dx \]
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\[ \int \frac {(c+d x)^{5/4}}{(a+b x)^{5/2}} \, dx=\int { \frac {{\left (d x + c\right )}^{\frac {5}{4}}}{{\left (b x + a\right )}^{\frac {5}{2}}} \,d x } \]
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\[ \int \frac {(c+d x)^{5/4}}{(a+b x)^{5/2}} \, dx=\int { \frac {{\left (d x + c\right )}^{\frac {5}{4}}}{{\left (b x + a\right )}^{\frac {5}{2}}} \,d x } \]
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Timed out. \[ \int \frac {(c+d x)^{5/4}}{(a+b x)^{5/2}} \, dx=\int \frac {{\left (c+d\,x\right )}^{5/4}}{{\left (a+b\,x\right )}^{5/2}} \,d x \]
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