Integrand size = 19, antiderivative size = 232 \[ \int \sqrt {a+b x} (c+d x)^{3/4} \, dx=\frac {4 (b c-a d) \sqrt {a+b x} (c+d x)^{3/4}}{15 b d}+\frac {4 (a+b x)^{3/2} (c+d x)^{3/4}}{9 b}-\frac {8 (b c-a d)^{11/4} \sqrt {-\frac {d (a+b x)}{b c-a d}} E\left (\left .\arcsin \left (\frac {\sqrt [4]{b} \sqrt [4]{c+d x}}{\sqrt [4]{b c-a d}}\right )\right |-1\right )}{15 b^{7/4} d^2 \sqrt {a+b x}}+\frac {8 (b c-a d)^{11/4} \sqrt {-\frac {d (a+b x)}{b c-a d}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt [4]{b} \sqrt [4]{c+d x}}{\sqrt [4]{b c-a d}}\right ),-1\right )}{15 b^{7/4} d^2 \sqrt {a+b x}} \]
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Time = 0.19 (sec) , antiderivative size = 232, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.421, Rules used = {52, 65, 313, 230, 227, 1214, 1213, 435} \[ \int \sqrt {a+b x} (c+d x)^{3/4} \, dx=\frac {8 (b c-a d)^{11/4} \sqrt {-\frac {d (a+b x)}{b c-a d}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt [4]{b} \sqrt [4]{c+d x}}{\sqrt [4]{b c-a d}}\right ),-1\right )}{15 b^{7/4} d^2 \sqrt {a+b x}}-\frac {8 (b c-a d)^{11/4} \sqrt {-\frac {d (a+b x)}{b c-a d}} E\left (\left .\arcsin \left (\frac {\sqrt [4]{b} \sqrt [4]{c+d x}}{\sqrt [4]{b c-a d}}\right )\right |-1\right )}{15 b^{7/4} d^2 \sqrt {a+b x}}+\frac {4 \sqrt {a+b x} (c+d x)^{3/4} (b c-a d)}{15 b d}+\frac {4 (a+b x)^{3/2} (c+d x)^{3/4}}{9 b} \]
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Rule 52
Rule 65
Rule 227
Rule 230
Rule 313
Rule 435
Rule 1213
Rule 1214
Rubi steps \begin{align*} \text {integral}& = \frac {4 (a+b x)^{3/2} (c+d x)^{3/4}}{9 b}+\frac {(b c-a d) \int \frac {\sqrt {a+b x}}{\sqrt [4]{c+d x}} \, dx}{3 b} \\ & = \frac {4 (b c-a d) \sqrt {a+b x} (c+d x)^{3/4}}{15 b d}+\frac {4 (a+b x)^{3/2} (c+d x)^{3/4}}{9 b}-\frac {\left (2 (b c-a d)^2\right ) \int \frac {1}{\sqrt {a+b x} \sqrt [4]{c+d x}} \, dx}{15 b d} \\ & = \frac {4 (b c-a d) \sqrt {a+b x} (c+d x)^{3/4}}{15 b d}+\frac {4 (a+b x)^{3/2} (c+d x)^{3/4}}{9 b}-\frac {\left (8 (b c-a d)^2\right ) \text {Subst}\left (\int \frac {x^2}{\sqrt {a-\frac {b c}{d}+\frac {b x^4}{d}}} \, dx,x,\sqrt [4]{c+d x}\right )}{15 b d^2} \\ & = \frac {4 (b c-a d) \sqrt {a+b x} (c+d x)^{3/4}}{15 b d}+\frac {4 (a+b x)^{3/2} (c+d x)^{3/4}}{9 b}+\frac {\left (8 (b c-a d)^{5/2}\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 )}{15 b^{3/2} d^2}-\frac {\left (8 (b c-a d)^{5/2}\right ) \text {Subst}\left (\int \frac {1+\frac {\sqrt {b} x^2}{\sqrt {b c-a d}}}{\sqrt {a-\frac {b c}{d}+\frac {b x^4}{d}}} \, dx,x,\sqrt [4]{c+d x}\right )}{15 b^{3/2} d^2} \\ & = \frac {4 (b c-a d) \sqrt {a+b x} (c+d x)^{3/4}}{15 b d}+\frac {4 (a+b x)^{3/2} (c+d x)^{3/4}}{9 b}+\frac {\left (8 (b c-a d)^{5/2} \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 )}{15 b^{3/2} d^2 \sqrt {a+b x}}-\frac {\left (8 (b c-a d)^{5/2} \sqrt {\frac {d (a+b x)}{-b c+a d}}\right ) \text {Subst}\left (\int \frac {1+\frac {\sqrt {b} x^2}{\sqrt {b c-a d}}}{\sqrt {1+\frac {b x^4}{\left (a-\frac {b c}{d}\right ) d}}} \, dx,x,\sqrt [4]{c+d x}\right )}{15 b^{3/2} d^2 \sqrt {a+b x}} \\ & = \frac {4 (b c-a d) \sqrt {a+b x} (c+d x)^{3/4}}{15 b d}+\frac {4 (a+b x)^{3/2} (c+d x)^{3/4}}{9 b}+\frac {8 (b c-a d)^{11/4} \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 )}{15 b^{7/4} d^2 \sqrt {a+b x}}-\frac {\left (8 (b c-a d)^{5/2} \sqrt {\frac {d (a+b x)}{-b c+a d}}\right ) \text {Subst}\left (\int \frac {\sqrt {1+\frac {\sqrt {b} x^2}{\sqrt {b c-a d}}}}{\sqrt {1-\frac {\sqrt {b} x^2}{\sqrt {b c-a d}}}} \, dx,x,\sqrt [4]{c+d x}\right )}{15 b^{3/2} d^2 \sqrt {a+b x}} \\ & = \frac {4 (b c-a d) \sqrt {a+b x} (c+d x)^{3/4}}{15 b d}+\frac {4 (a+b x)^{3/2} (c+d x)^{3/4}}{9 b}-\frac {8 (b c-a d)^{11/4} \sqrt {-\frac {d (a+b x)}{b c-a d}} E\left (\left .\sin ^{-1}\left (\frac {\sqrt [4]{b} \sqrt [4]{c+d x}}{\sqrt [4]{b c-a d}}\right )\right |-1\right )}{15 b^{7/4} d^2 \sqrt {a+b x}}+\frac {8 (b c-a d)^{11/4} \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 )}{15 b^{7/4} d^2 \sqrt {a+b x}} \\ \end{align*}
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 0.02 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.31 \[ \int \sqrt {a+b x} (c+d x)^{3/4} \, dx=\frac {2 (a+b x)^{3/2} (c+d x)^{3/4} \operatorname {Hypergeometric2F1}\left (-\frac {3}{4},\frac {3}{2},\frac {5}{2},\frac {d (a+b x)}{-b c+a d}\right )}{3 b \left (\frac {b (c+d x)}{b c-a d}\right )^{3/4}} \]
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\[\int \sqrt {b x +a}\, \left (d x +c \right )^{\frac {3}{4}}d x\]
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\[ \int \sqrt {a+b x} (c+d x)^{3/4} \, dx=\int { \sqrt {b x + a} {\left (d x + c\right )}^{\frac {3}{4}} \,d x } \]
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\[ \int \sqrt {a+b x} (c+d x)^{3/4} \, dx=\int \sqrt {a + b x} \left (c + d x\right )^{\frac {3}{4}}\, dx \]
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\[ \int \sqrt {a+b x} (c+d x)^{3/4} \, dx=\int { \sqrt {b x + a} {\left (d x + c\right )}^{\frac {3}{4}} \,d x } \]
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\[ \int \sqrt {a+b x} (c+d x)^{3/4} \, dx=\int { \sqrt {b x + a} {\left (d x + c\right )}^{\frac {3}{4}} \,d x } \]
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Timed out. \[ \int \sqrt {a+b x} (c+d x)^{3/4} \, dx=\int \sqrt {a+b\,x}\,{\left (c+d\,x\right )}^{3/4} \,d x \]
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