Integrand size = 19, antiderivative size = 182 \[ \int \sqrt {a+b x} (c+d x)^{5/4} \, dx=\frac {20 (b c-a d)^2 \sqrt {a+b x} \sqrt [4]{c+d x}}{231 b^2 d}+\frac {20 (b c-a d) (a+b x)^{3/2} \sqrt [4]{c+d x}}{77 b^2}+\frac {4 (a+b x)^{3/2} (c+d x)^{5/4}}{11 b}-\frac {40 (b c-a d)^{13/4} \sqrt {-\frac {d (a+b x)}{b c-a d}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt [4]{b} \sqrt [4]{c+d x}}{\sqrt [4]{b c-a d}}\right ),-1\right )}{231 b^{9/4} d^2 \sqrt {a+b x}} \]
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Time = 0.09 (sec) , antiderivative size = 182, 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, 230, 227} \[ \int \sqrt {a+b x} (c+d x)^{5/4} \, dx=-\frac {40 (b c-a d)^{13/4} \sqrt {-\frac {d (a+b x)}{b c-a d}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt [4]{b} \sqrt [4]{c+d x}}{\sqrt [4]{b c-a d}}\right ),-1\right )}{231 b^{9/4} d^2 \sqrt {a+b x}}+\frac {20 \sqrt {a+b x} \sqrt [4]{c+d x} (b c-a d)^2}{231 b^2 d}+\frac {20 (a+b x)^{3/2} \sqrt [4]{c+d x} (b c-a d)}{77 b^2}+\frac {4 (a+b x)^{3/2} (c+d x)^{5/4}}{11 b} \]
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Rule 52
Rule 65
Rule 227
Rule 230
Rubi steps \begin{align*} \text {integral}& = \frac {4 (a+b x)^{3/2} (c+d x)^{5/4}}{11 b}+\frac {(5 (b c-a d)) \int \sqrt {a+b x} \sqrt [4]{c+d x} \, dx}{11 b} \\ & = \frac {20 (b c-a d) (a+b x)^{3/2} \sqrt [4]{c+d x}}{77 b^2}+\frac {4 (a+b x)^{3/2} (c+d x)^{5/4}}{11 b}+\frac {\left (5 (b c-a d)^2\right ) \int \frac {\sqrt {a+b x}}{(c+d x)^{3/4}} \, dx}{77 b^2} \\ & = \frac {20 (b c-a d)^2 \sqrt {a+b x} \sqrt [4]{c+d x}}{231 b^2 d}+\frac {20 (b c-a d) (a+b x)^{3/2} \sqrt [4]{c+d x}}{77 b^2}+\frac {4 (a+b x)^{3/2} (c+d x)^{5/4}}{11 b}-\frac {\left (10 (b c-a d)^3\right ) \int \frac {1}{\sqrt {a+b x} (c+d x)^{3/4}} \, dx}{231 b^2 d} \\ & = \frac {20 (b c-a d)^2 \sqrt {a+b x} \sqrt [4]{c+d x}}{231 b^2 d}+\frac {20 (b c-a d) (a+b x)^{3/2} \sqrt [4]{c+d x}}{77 b^2}+\frac {4 (a+b x)^{3/2} (c+d x)^{5/4}}{11 b}-\frac {\left (40 (b c-a d)^3\right ) \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 )}{231 b^2 d^2} \\ & = \frac {20 (b c-a d)^2 \sqrt {a+b x} \sqrt [4]{c+d x}}{231 b^2 d}+\frac {20 (b c-a d) (a+b x)^{3/2} \sqrt [4]{c+d x}}{77 b^2}+\frac {4 (a+b x)^{3/2} (c+d x)^{5/4}}{11 b}-\frac {\left (40 (b c-a d)^3 \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 )}{231 b^2 d^2 \sqrt {a+b x}} \\ & = \frac {20 (b c-a d)^2 \sqrt {a+b x} \sqrt [4]{c+d x}}{231 b^2 d}+\frac {20 (b c-a d) (a+b x)^{3/2} \sqrt [4]{c+d x}}{77 b^2}+\frac {4 (a+b x)^{3/2} (c+d x)^{5/4}}{11 b}-\frac {40 (b c-a d)^{13/4} \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 )}{231 b^{9/4} d^2 \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 = 0.40 \[ \int \sqrt {a+b x} (c+d x)^{5/4} \, dx=\frac {2 (a+b x)^{3/2} (c+d x)^{5/4} \operatorname {Hypergeometric2F1}\left (-\frac {5}{4},\frac {3}{2},\frac {5}{2},\frac {d (a+b x)}{-b c+a d}\right )}{3 b \left (\frac {b (c+d x)}{b c-a d}\right )^{5/4}} \]
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\[\int \sqrt {b x +a}\, \left (d x +c \right )^{\frac {5}{4}}d x\]
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\[ \int \sqrt {a+b x} (c+d x)^{5/4} \, dx=\int { \sqrt {b x + a} {\left (d x + c\right )}^{\frac {5}{4}} \,d x } \]
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\[ \int \sqrt {a+b x} (c+d x)^{5/4} \, dx=\int \sqrt {a + b x} \left (c + d x\right )^{\frac {5}{4}}\, dx \]
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\[ \int \sqrt {a+b x} (c+d x)^{5/4} \, dx=\int { \sqrt {b x + a} {\left (d x + c\right )}^{\frac {5}{4}} \,d x } \]
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\[ \int \sqrt {a+b x} (c+d x)^{5/4} \, dx=\int { \sqrt {b x + a} {\left (d x + c\right )}^{\frac {5}{4}} \,d x } \]
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Timed out. \[ \int \sqrt {a+b x} (c+d x)^{5/4} \, dx=\int \sqrt {a+b\,x}\,{\left (c+d\,x\right )}^{5/4} \,d x \]
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