Integrand size = 19, antiderivative size = 332 \[ \int \frac {(c+d x)^{5/4}}{(a+b x)^{3/4}} \, dx=\frac {5 (b c-a d) \sqrt [4]{a+b x} \sqrt [4]{c+d x}}{3 b^2}+\frac {2 \sqrt [4]{a+b x} (c+d x)^{5/4}}{3 b}+\frac {5 (b c-a d)^{5/2} ((a+b x) (c+d x))^{3/4} \sqrt {(b c+a d+2 b d x)^2} \left (1+\frac {2 \sqrt {b} \sqrt {d} \sqrt {(a+b x) (c+d x)}}{b c-a d}\right ) \sqrt {\frac {(a d+b (c+2 d x))^2}{(b c-a d)^2 \left (1+\frac {2 \sqrt {b} \sqrt {d} \sqrt {(a+b x) (c+d x)}}{b c-a d}\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{d} \sqrt [4]{(a+b x) (c+d x)}}{\sqrt {b c-a d}}\right ),\frac {1}{2}\right )}{6 \sqrt {2} b^{9/4} \sqrt [4]{d} (a+b x)^{3/4} (c+d x)^{3/4} (b c+a d+2 b d x) \sqrt {(a d+b (c+2 d x))^2}} \]
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Time = 0.24 (sec) , antiderivative size = 332, 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 = {52, 64, 637, 226} \[ \int \frac {(c+d x)^{5/4}}{(a+b x)^{3/4}} \, dx=\frac {5 (b c-a d)^{5/2} ((a+b x) (c+d x))^{3/4} \sqrt {(a d+b c+2 b d x)^2} \left (\frac {2 \sqrt {b} \sqrt {d} \sqrt {(a+b x) (c+d x)}}{b c-a d}+1\right ) \sqrt {\frac {(a d+b (c+2 d x))^2}{(b c-a d)^2 \left (\frac {2 \sqrt {b} \sqrt {d} \sqrt {(a+b x) (c+d x)}}{b c-a d}+1\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{d} \sqrt [4]{(a+b x) (c+d x)}}{\sqrt {b c-a d}}\right ),\frac {1}{2}\right )}{6 \sqrt {2} b^{9/4} \sqrt [4]{d} (a+b x)^{3/4} (c+d x)^{3/4} (a d+b c+2 b d x) \sqrt {(a d+b (c+2 d x))^2}}+\frac {5 \sqrt [4]{a+b x} \sqrt [4]{c+d x} (b c-a d)}{3 b^2}+\frac {2 \sqrt [4]{a+b x} (c+d x)^{5/4}}{3 b} \]
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Rule 52
Rule 64
Rule 226
Rule 637
Rubi steps \begin{align*} \text {integral}& = \frac {2 \sqrt [4]{a+b x} (c+d x)^{5/4}}{3 b}+\frac {(5 (b c-a d)) \int \frac {\sqrt [4]{c+d x}}{(a+b x)^{3/4}} \, dx}{6 b} \\ & = \frac {5 (b c-a d) \sqrt [4]{a+b x} \sqrt [4]{c+d x}}{3 b^2}+\frac {2 \sqrt [4]{a+b x} (c+d x)^{5/4}}{3 b}+\frac {\left (5 (b c-a d)^2\right ) \int \frac {1}{(a+b x)^{3/4} (c+d x)^{3/4}} \, dx}{12 b^2} \\ & = \frac {5 (b c-a d) \sqrt [4]{a+b x} \sqrt [4]{c+d x}}{3 b^2}+\frac {2 \sqrt [4]{a+b x} (c+d x)^{5/4}}{3 b}+\frac {\left (5 (b c-a d)^2 ((a+b x) (c+d x))^{3/4}\right ) \int \frac {1}{\left (a c+(b c+a d) x+b d x^2\right )^{3/4}} \, dx}{12 b^2 (a+b x)^{3/4} (c+d x)^{3/4}} \\ & = \frac {5 (b c-a d) \sqrt [4]{a+b x} \sqrt [4]{c+d x}}{3 b^2}+\frac {2 \sqrt [4]{a+b x} (c+d x)^{5/4}}{3 b}+\frac {\left (5 (b c-a d)^2 ((a+b x) (c+d x))^{3/4} \sqrt {(b c+a d+2 b d x)^2}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {-4 a b c d+(b c+a d)^2+4 b d x^4}} \, dx,x,\sqrt [4]{(a+b x) (c+d x)}\right )}{3 b^2 (a+b x)^{3/4} (c+d x)^{3/4} (b c+a d+2 b d x)} \\ & = \frac {5 (b c-a d) \sqrt [4]{a+b x} \sqrt [4]{c+d x}}{3 b^2}+\frac {2 \sqrt [4]{a+b x} (c+d x)^{5/4}}{3 b}+\frac {5 (b c-a d)^{5/2} ((a+b x) (c+d x))^{3/4} \sqrt {(b c+a d+2 b d x)^2} \left (1+\frac {2 \sqrt {b} \sqrt {d} \sqrt {(a+b x) (c+d x)}}{b c-a d}\right ) \sqrt {\frac {(a d+b (c+2 d x))^2}{(b c-a d)^2 \left (1+\frac {2 \sqrt {b} \sqrt {d} \sqrt {(a+b x) (c+d x)}}{b c-a d}\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{d} \sqrt [4]{(a+b x) (c+d x)}}{\sqrt {b c-a d}}\right )|\frac {1}{2}\right )}{6 \sqrt {2} b^{9/4} \sqrt [4]{d} (a+b x)^{3/4} (c+d x)^{3/4} (b c+a d+2 b d x) \sqrt {(a d+b (c+2 d x))^2}} \\ \end{align*}
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 0.03 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.21 \[ \int \frac {(c+d x)^{5/4}}{(a+b x)^{3/4}} \, dx=\frac {4 \sqrt [4]{a+b x} (c+d x)^{5/4} \operatorname {Hypergeometric2F1}\left (-\frac {5}{4},\frac {1}{4},\frac {5}{4},\frac {d (a+b x)}{-b c+a d}\right )}{b \left (\frac {b (c+d x)}{b c-a d}\right )^{5/4}} \]
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\[\int \frac {\left (d x +c \right )^{\frac {5}{4}}}{\left (b x +a \right )^{\frac {3}{4}}}d x\]
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\[ \int \frac {(c+d x)^{5/4}}{(a+b x)^{3/4}} \, dx=\int { \frac {{\left (d x + c\right )}^{\frac {5}{4}}}{{\left (b x + a\right )}^{\frac {3}{4}}} \,d x } \]
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\[ \int \frac {(c+d x)^{5/4}}{(a+b x)^{3/4}} \, dx=\int \frac {\left (c + d x\right )^{\frac {5}{4}}}{\left (a + b x\right )^{\frac {3}{4}}}\, dx \]
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\[ \int \frac {(c+d x)^{5/4}}{(a+b x)^{3/4}} \, dx=\int { \frac {{\left (d x + c\right )}^{\frac {5}{4}}}{{\left (b x + a\right )}^{\frac {3}{4}}} \,d x } \]
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\[ \int \frac {(c+d x)^{5/4}}{(a+b x)^{3/4}} \, dx=\int { \frac {{\left (d x + c\right )}^{\frac {5}{4}}}{{\left (b x + a\right )}^{\frac {3}{4}}} \,d x } \]
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Timed out. \[ \int \frac {(c+d x)^{5/4}}{(a+b x)^{3/4}} \, dx=\int \frac {{\left (c+d\,x\right )}^{5/4}}{{\left (a+b\,x\right )}^{3/4}} \,d x \]
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