Integrand size = 20, antiderivative size = 490 \[ \int \frac {d+e x}{\left (a+b x+c x^2\right )^{5/4}} \, dx=-\frac {4 (b d-2 a e+(2 c d-b e) x)}{\left (b^2-4 a c\right ) \sqrt [4]{a+b x+c x^2}}+\frac {4 (2 c d-b e) (b+2 c x) \sqrt [4]{a+b x+c x^2}}{\sqrt {c} \left (b^2-4 a c\right )^{3/2} \left (1+\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}\right )}-\frac {2 \sqrt {2} (2 c d-b e) \sqrt {\frac {(b+2 c x)^2}{\left (b^2-4 a c\right ) \left (1+\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}\right )^2}} \left (1+\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}\right ) E\left (2 \arctan \left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt [4]{a+b x+c x^2}}{\sqrt [4]{b^2-4 a c}}\right )|\frac {1}{2}\right )}{c^{3/4} \sqrt [4]{b^2-4 a c} (b+2 c x)}+\frac {\sqrt {2} (2 c d-b e) \sqrt {\frac {(b+2 c x)^2}{\left (b^2-4 a c\right ) \left (1+\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}\right )^2}} \left (1+\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}\right ) \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt [4]{a+b x+c x^2}}{\sqrt [4]{b^2-4 a c}}\right ),\frac {1}{2}\right )}{c^{3/4} \sqrt [4]{b^2-4 a c} (b+2 c x)} \]
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Time = 0.26 (sec) , antiderivative size = 490, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {652, 637, 311, 226, 1210} \[ \int \frac {d+e x}{\left (a+b x+c x^2\right )^{5/4}} \, dx=\frac {\sqrt {2} \sqrt {\frac {(b+2 c x)^2}{\left (b^2-4 a c\right ) \left (\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}+1\right )^2}} \left (\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}+1\right ) (2 c d-b e) \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt [4]{c x^2+b x+a}}{\sqrt [4]{b^2-4 a c}}\right ),\frac {1}{2}\right )}{c^{3/4} \sqrt [4]{b^2-4 a c} (b+2 c x)}-\frac {2 \sqrt {2} \sqrt {\frac {(b+2 c x)^2}{\left (b^2-4 a c\right ) \left (\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}+1\right )^2}} \left (\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}+1\right ) (2 c d-b e) E\left (2 \arctan \left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt [4]{c x^2+b x+a}}{\sqrt [4]{b^2-4 a c}}\right )|\frac {1}{2}\right )}{c^{3/4} \sqrt [4]{b^2-4 a c} (b+2 c x)}+\frac {4 (b+2 c x) \sqrt [4]{a+b x+c x^2} (2 c d-b e)}{\sqrt {c} \left (b^2-4 a c\right )^{3/2} \left (\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}+1\right )}-\frac {4 (-2 a e+x (2 c d-b e)+b d)}{\left (b^2-4 a c\right ) \sqrt [4]{a+b x+c x^2}} \]
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Rule 226
Rule 311
Rule 637
Rule 652
Rule 1210
Rubi steps \begin{align*} \text {integral}& = -\frac {4 (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 {(2 (2 c d-b e)) \int \frac {1}{\sqrt [4]{a+b x+c x^2}} \, dx}{b^2-4 a c} \\ & = -\frac {4 (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 {\left (8 (2 c d-b e) \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 )}{\left (b^2-4 a c\right ) (b+2 c x)} \\ & = -\frac {4 (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 {\left (4 (2 c d-b e) \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 )}{\sqrt {c} \sqrt {b^2-4 a c} (b+2 c x)}-\frac {\left (4 (2 c d-b e) \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 )}{\sqrt {c} \sqrt {b^2-4 a c} (b+2 c x)} \\ & = -\frac {4 (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 (2 c d-b e) (b+2 c x) \sqrt [4]{a+b x+c x^2}}{\sqrt {c} \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 {2 \sqrt {2} (2 c d-b e) \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^{3/4} \sqrt [4]{b^2-4 a c} (b+2 c x)}+\frac {\sqrt {2} (2 c d-b e) \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 )}{c^{3/4} \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 = 11.19 (sec) , antiderivative size = 167, normalized size of antiderivative = 0.34 \[ \int \frac {d+e x}{\left (a+b x+c x^2\right )^{5/4}} \, dx=-\frac {2 \left (6 c (-2 a e+2 c d x+b (d-e x))+2^{3/4} (-2 c d+b e) \left (b-\sqrt {b^2-4 a c}+2 c x\right ) \sqrt [4]{\frac {b+\sqrt {b^2-4 a c}+2 c x}{\sqrt {b^2-4 a c}}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {3}{4},\frac {7}{4},\frac {-b+\sqrt {b^2-4 a c}-2 c x}{2 \sqrt {b^2-4 a c}}\right )\right )}{3 c \left (b^2-4 a c\right ) \sqrt [4]{a+x (b+c x)}} \]
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\[\int \frac {e x +d}{\left (c \,x^{2}+b x +a \right )^{\frac {5}{4}}}d x\]
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\[ \int \frac {d+e x}{\left (a+b x+c x^2\right )^{5/4}} \, dx=\int { \frac {e x + d}{{\left (c x^{2} + b x + a\right )}^{\frac {5}{4}}} \,d x } \]
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\[ \int \frac {d+e x}{\left (a+b x+c x^2\right )^{5/4}} \, dx=\int \frac {d + e x}{\left (a + b x + c x^{2}\right )^{\frac {5}{4}}}\, dx \]
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\[ \int \frac {d+e x}{\left (a+b x+c x^2\right )^{5/4}} \, dx=\int { \frac {e x + d}{{\left (c x^{2} + b x + a\right )}^{\frac {5}{4}}} \,d x } \]
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\[ \int \frac {d+e x}{\left (a+b x+c x^2\right )^{5/4}} \, dx=\int { \frac {e x + d}{{\left (c x^{2} + b x + a\right )}^{\frac {5}{4}}} \,d x } \]
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Timed out. \[ \int \frac {d+e x}{\left (a+b x+c x^2\right )^{5/4}} \, dx=\int \frac {d+e\,x}{{\left (c\,x^2+b\,x+a\right )}^{5/4}} \,d x \]
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