Integrand size = 22, antiderivative size = 594 \[ \int \frac {(d+e x)^2}{\left (a+b x+c x^2\right )^{5/4}} \, dx=-\frac {4 (d+e x) (b d-2 a e+(2 c d-b e) x)}{\left (b^2-4 a c\right ) \sqrt [4]{a+b x+c x^2}}+\frac {4 e (2 c d-b e) \left (a+b x+c x^2\right )^{3/4}}{c \left (b^2-4 a c\right )}+\frac {2 \left (4 c^2 d^2+3 b^2 e^2-4 c e (b d+2 a e)\right ) (b+2 c x) \sqrt [4]{a+b x+c x^2}}{c^{3/2} \left (b^2-4 a c\right )^{3/2} \left (1+\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}\right )}-\frac {\sqrt {2} \left (4 c^2 d^2+3 b^2 e^2-4 c e (b d+2 a e)\right ) \sqrt {\frac {(b+2 c x)^2}{\left (b^2-4 a c\right ) \left (1+\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}\right )^2}} \left (1+\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}\right ) E\left (2 \arctan \left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt [4]{a+b x+c x^2}}{\sqrt [4]{b^2-4 a c}}\right )|\frac {1}{2}\right )}{c^{7/4} \sqrt [4]{b^2-4 a c} (b+2 c x)}+\frac {\left (4 c^2 d^2+3 b^2 e^2-4 c e (b d+2 a e)\right ) \sqrt {\frac {(b+2 c x)^2}{\left (b^2-4 a c\right ) \left (1+\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}\right )^2}} \left (1+\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}\right ) \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt [4]{a+b x+c x^2}}{\sqrt [4]{b^2-4 a c}}\right ),\frac {1}{2}\right )}{\sqrt {2} c^{7/4} \sqrt [4]{b^2-4 a c} (b+2 c x)} \]
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Time = 0.45 (sec) , antiderivative size = 594, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {752, 654, 637, 311, 226, 1210} \[ \int \frac {(d+e x)^2}{\left (a+b x+c x^2\right )^{5/4}} \, dx=\frac {\sqrt {\frac {(b+2 c x)^2}{\left (b^2-4 a c\right ) \left (\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}+1\right )^2}} \left (\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}+1\right ) \left (-4 c e (2 a e+b d)+3 b^2 e^2+4 c^2 d^2\right ) \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt [4]{c x^2+b x+a}}{\sqrt [4]{b^2-4 a c}}\right ),\frac {1}{2}\right )}{\sqrt {2} c^{7/4} \sqrt [4]{b^2-4 a c} (b+2 c x)}-\frac {\sqrt {2} \sqrt {\frac {(b+2 c x)^2}{\left (b^2-4 a c\right ) \left (\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}+1\right )^2}} \left (\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}+1\right ) \left (-4 c e (2 a e+b d)+3 b^2 e^2+4 c^2 d^2\right ) E\left (2 \arctan \left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt [4]{c x^2+b x+a}}{\sqrt [4]{b^2-4 a c}}\right )|\frac {1}{2}\right )}{c^{7/4} \sqrt [4]{b^2-4 a c} (b+2 c x)}+\frac {2 (b+2 c x) \sqrt [4]{a+b x+c x^2} \left (-4 c e (2 a e+b d)+3 b^2 e^2+4 c^2 d^2\right )}{c^{3/2} \left (b^2-4 a c\right )^{3/2} \left (\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}+1\right )}+\frac {4 e \left (a+b x+c x^2\right )^{3/4} (2 c d-b e)}{c \left (b^2-4 a c\right )}-\frac {4 (d+e x) (-2 a e+x (2 c d-b e)+b d)}{\left (b^2-4 a c\right ) \sqrt [4]{a+b x+c x^2}} \]
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Rule 226
Rule 311
Rule 637
Rule 654
Rule 752
Rule 1210
Rubi steps \begin{align*} \text {integral}& = -\frac {4 (d+e x) (b d-2 a e+(2 c d-b e) x)}{\left (b^2-4 a c\right ) \sqrt [4]{a+b x+c x^2}}-\frac {4 \int \frac {\frac {1}{2} \left (-2 c d^2-2 e \left (\frac {b d}{2}-2 a e\right )\right )-\frac {3}{2} e (2 c d-b e) x}{\sqrt [4]{a+b x+c x^2}} \, dx}{b^2-4 a c} \\ & = -\frac {4 (d+e x) (b d-2 a e+(2 c d-b e) x)}{\left (b^2-4 a c\right ) \sqrt [4]{a+b x+c x^2}}+\frac {4 e (2 c d-b e) \left (a+b x+c x^2\right )^{3/4}}{c \left (b^2-4 a c\right )}+\frac {\left (4 c^2 d^2+3 b^2 e^2-4 c e (b d+2 a e)\right ) \int \frac {1}{\sqrt [4]{a+b x+c x^2}} \, dx}{c \left (b^2-4 a c\right )} \\ & = -\frac {4 (d+e x) (b d-2 a e+(2 c d-b e) x)}{\left (b^2-4 a c\right ) \sqrt [4]{a+b x+c x^2}}+\frac {4 e (2 c d-b e) \left (a+b x+c x^2\right )^{3/4}}{c \left (b^2-4 a c\right )}+\frac {\left (4 \left (4 c^2 d^2+3 b^2 e^2-4 c e (b d+2 a e)\right ) \sqrt {(b+2 c x)^2}\right ) \text {Subst}\left (\int \frac {x^2}{\sqrt {b^2-4 a c+4 c x^4}} \, dx,x,\sqrt [4]{a+b x+c x^2}\right )}{c \left (b^2-4 a c\right ) (b+2 c x)} \\ & = -\frac {4 (d+e x) (b d-2 a e+(2 c d-b e) x)}{\left (b^2-4 a c\right ) \sqrt [4]{a+b x+c x^2}}+\frac {4 e (2 c d-b e) \left (a+b x+c x^2\right )^{3/4}}{c \left (b^2-4 a c\right )}+\frac {\left (2 \left (4 c^2 d^2+3 b^2 e^2-4 c e (b d+2 a e)\right ) \sqrt {(b+2 c x)^2}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {b^2-4 a c+4 c x^4}} \, dx,x,\sqrt [4]{a+b x+c x^2}\right )}{c^{3/2} \sqrt {b^2-4 a c} (b+2 c x)}-\frac {\left (2 \left (4 c^2 d^2+3 b^2 e^2-4 c e (b d+2 a e)\right ) \sqrt {(b+2 c x)^2}\right ) \text {Subst}\left (\int \frac {1-\frac {2 \sqrt {c} x^2}{\sqrt {b^2-4 a c}}}{\sqrt {b^2-4 a c+4 c x^4}} \, dx,x,\sqrt [4]{a+b x+c x^2}\right )}{c^{3/2} \sqrt {b^2-4 a c} (b+2 c x)} \\ & = -\frac {4 (d+e x) (b d-2 a e+(2 c d-b e) x)}{\left (b^2-4 a c\right ) \sqrt [4]{a+b x+c x^2}}+\frac {4 e (2 c d-b e) \left (a+b x+c x^2\right )^{3/4}}{c \left (b^2-4 a c\right )}+\frac {2 \left (4 c^2 d^2+3 b^2 e^2-4 c e (b d+2 a e)\right ) (b+2 c x) \sqrt [4]{a+b x+c x^2}}{c^{3/2} \left (b^2-4 a c\right )^{3/2} \left (1+\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}\right )}-\frac {\sqrt {2} \left (4 c^2 d^2+3 b^2 e^2-4 c e (b d+2 a e)\right ) \sqrt {\frac {(b+2 c x)^2}{\left (b^2-4 a c\right ) \left (1+\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}\right )^2}} \left (1+\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}\right ) E\left (2 \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt [4]{a+b x+c x^2}}{\sqrt [4]{b^2-4 a c}}\right )|\frac {1}{2}\right )}{c^{7/4} \sqrt [4]{b^2-4 a c} (b+2 c x)}+\frac {\left (4 c^2 d^2+3 b^2 e^2-4 c e (b d+2 a e)\right ) \sqrt {\frac {(b+2 c x)^2}{\left (b^2-4 a c\right ) \left (1+\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}\right )^2}} \left (1+\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}\right ) F\left (2 \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt [4]{a+b x+c x^2}}{\sqrt [4]{b^2-4 a c}}\right )|\frac {1}{2}\right )}{\sqrt {2} c^{7/4} \sqrt [4]{b^2-4 a c} (b+2 c x)} \\ \end{align*}
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 10.29 (sec) , antiderivative size = 177, normalized size of antiderivative = 0.30 \[ \int \frac {(d+e x)^2}{\left (a+b x+c x^2\right )^{5/4}} \, dx=\frac {-8 c \left (a b e^2+2 c^2 d^2 x+b^2 e^2 x+b c d (d-2 e x)-2 a c e (2 d+e x)\right )+\sqrt {2} \left (4 c^2 d^2+3 b^2 e^2-4 c e (b d+2 a e)\right ) (b+2 c x) \sqrt [4]{\frac {c (a+x (b+c x))}{-b^2+4 a c}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {3}{2},\frac {(b+2 c x)^2}{b^2-4 a c}\right )}{2 c^2 \left (b^2-4 a c\right ) \sqrt [4]{a+x (b+c x)}} \]
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\[\int \frac {\left (e x +d \right )^{2}}{\left (c \,x^{2}+b x +a \right )^{\frac {5}{4}}}d x\]
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\[ \int \frac {(d+e x)^2}{\left (a+b x+c x^2\right )^{5/4}} \, dx=\int { \frac {{\left (e x + d\right )}^{2}}{{\left (c x^{2} + b x + a\right )}^{\frac {5}{4}}} \,d x } \]
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\[ \int \frac {(d+e x)^2}{\left (a+b x+c x^2\right )^{5/4}} \, dx=\int \frac {\left (d + e x\right )^{2}}{\left (a + b x + c x^{2}\right )^{\frac {5}{4}}}\, dx \]
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\[ \int \frac {(d+e x)^2}{\left (a+b x+c x^2\right )^{5/4}} \, dx=\int { \frac {{\left (e x + d\right )}^{2}}{{\left (c x^{2} + b x + a\right )}^{\frac {5}{4}}} \,d x } \]
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\[ \int \frac {(d+e x)^2}{\left (a+b x+c x^2\right )^{5/4}} \, dx=\int { \frac {{\left (e x + d\right )}^{2}}{{\left (c x^{2} + b x + a\right )}^{\frac {5}{4}}} \,d x } \]
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Timed out. \[ \int \frac {(d+e x)^2}{\left (a+b x+c x^2\right )^{5/4}} \, dx=\int \frac {{\left (d+e\,x\right )}^2}{{\left (c\,x^2+b\,x+a\right )}^{5/4}} \,d x \]
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