Integrand size = 22, antiderivative size = 573 \[ \int \frac {(d+e x)^2}{\sqrt [4]{a+b x+c x^2}} \, dx=\frac {7 e (2 c d-b e) \left (a+b x+c x^2\right )^{3/4}}{15 c^2}+\frac {2 e (d+e x) \left (a+b x+c x^2\right )^{3/4}}{5 c}+\frac {\left (20 c^2 d^2+7 b^2 e^2-4 c e (5 b d+2 a e)\right ) (b+2 c x) \sqrt [4]{a+b x+c x^2}}{10 c^{5/2} \sqrt {b^2-4 a c} \left (1+\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}\right )}-\frac {\left (b^2-4 a c\right )^{3/4} \left (20 c^2 d^2+7 b^2 e^2-4 c e (5 b d+2 a e)\right ) \sqrt {\frac {(b+2 c x)^2}{\left (b^2-4 a c\right ) \left (1+\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}\right )^2}} \left (1+\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}\right ) E\left (2 \arctan \left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt [4]{a+b x+c x^2}}{\sqrt [4]{b^2-4 a c}}\right )|\frac {1}{2}\right )}{10 \sqrt {2} c^{11/4} (b+2 c x)}+\frac {\left (b^2-4 a c\right )^{3/4} \left (20 c^2 d^2+7 b^2 e^2-4 c e (5 b d+2 a e)\right ) \sqrt {\frac {(b+2 c x)^2}{\left (b^2-4 a c\right ) \left (1+\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}\right )^2}} \left (1+\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}\right ) \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt [4]{a+b x+c x^2}}{\sqrt [4]{b^2-4 a c}}\right ),\frac {1}{2}\right )}{20 \sqrt {2} c^{11/4} (b+2 c x)} \]
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Time = 0.47 (sec) , antiderivative size = 573, 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 = {756, 654, 637, 311, 226, 1210} \[ \int \frac {(d+e x)^2}{\sqrt [4]{a+b x+c x^2}} \, dx=\frac {\left (b^2-4 a c\right )^{3/4} \sqrt {\frac {(b+2 c x)^2}{\left (b^2-4 a c\right ) \left (\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}+1\right )^2}} \left (\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}+1\right ) \left (-4 c e (2 a e+5 b d)+7 b^2 e^2+20 c^2 d^2\right ) \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt [4]{c x^2+b x+a}}{\sqrt [4]{b^2-4 a c}}\right ),\frac {1}{2}\right )}{20 \sqrt {2} c^{11/4} (b+2 c x)}-\frac {\left (b^2-4 a c\right )^{3/4} \sqrt {\frac {(b+2 c x)^2}{\left (b^2-4 a c\right ) \left (\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}+1\right )^2}} \left (\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}+1\right ) \left (-4 c e (2 a e+5 b d)+7 b^2 e^2+20 c^2 d^2\right ) E\left (2 \arctan \left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt [4]{c x^2+b x+a}}{\sqrt [4]{b^2-4 a c}}\right )|\frac {1}{2}\right )}{10 \sqrt {2} c^{11/4} (b+2 c x)}+\frac {(b+2 c x) \sqrt [4]{a+b x+c x^2} \left (-4 c e (2 a e+5 b d)+7 b^2 e^2+20 c^2 d^2\right )}{10 c^{5/2} \sqrt {b^2-4 a c} \left (\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}+1\right )}+\frac {7 e \left (a+b x+c x^2\right )^{3/4} (2 c d-b e)}{15 c^2}+\frac {2 e (d+e x) \left (a+b x+c x^2\right )^{3/4}}{5 c} \]
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Rule 226
Rule 311
Rule 637
Rule 654
Rule 756
Rule 1210
Rubi steps \begin{align*} \text {integral}& = \frac {2 e (d+e x) \left (a+b x+c x^2\right )^{3/4}}{5 c}+\frac {2 \int \frac {\frac {1}{4} \left (10 c d^2-4 e \left (\frac {3 b d}{4}+a e\right )\right )+\frac {7}{4} e (2 c d-b e) x}{\sqrt [4]{a+b x+c x^2}} \, dx}{5 c} \\ & = \frac {7 e (2 c d-b e) \left (a+b x+c x^2\right )^{3/4}}{15 c^2}+\frac {2 e (d+e x) \left (a+b x+c x^2\right )^{3/4}}{5 c}+\frac {\left (-\frac {7}{4} b e (2 c d-b e)+\frac {1}{2} c \left (10 c d^2-4 e \left (\frac {3 b d}{4}+a e\right )\right )\right ) \int \frac {1}{\sqrt [4]{a+b x+c x^2}} \, dx}{5 c^2} \\ & = \frac {7 e (2 c d-b e) \left (a+b x+c x^2\right )^{3/4}}{15 c^2}+\frac {2 e (d+e x) \left (a+b x+c x^2\right )^{3/4}}{5 c}+\frac {\left (4 \left (-\frac {7}{4} b e (2 c d-b e)+\frac {1}{2} c \left (10 c d^2-4 e \left (\frac {3 b d}{4}+a e\right )\right )\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 )}{5 c^2 (b+2 c x)} \\ & = \frac {7 e (2 c d-b e) \left (a+b x+c x^2\right )^{3/4}}{15 c^2}+\frac {2 e (d+e x) \left (a+b x+c x^2\right )^{3/4}}{5 c}+\frac {\left (2 \sqrt {b^2-4 a c} \left (-\frac {7}{4} b e (2 c d-b e)+\frac {1}{2} c \left (10 c d^2-4 e \left (\frac {3 b d}{4}+a e\right )\right )\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 )}{5 c^{5/2} (b+2 c x)}-\frac {\left (2 \sqrt {b^2-4 a c} \left (-\frac {7}{4} b e (2 c d-b e)+\frac {1}{2} c \left (10 c d^2-4 e \left (\frac {3 b d}{4}+a e\right )\right )\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 )}{5 c^{5/2} (b+2 c x)} \\ & = \frac {7 e (2 c d-b e) \left (a+b x+c x^2\right )^{3/4}}{15 c^2}+\frac {2 e (d+e x) \left (a+b x+c x^2\right )^{3/4}}{5 c}+\frac {\left (20 c^2 d^2+7 b^2 e^2-4 c e (5 b d+2 a e)\right ) (b+2 c x) \sqrt [4]{a+b x+c x^2}}{10 c^{5/2} \sqrt {b^2-4 a c} \left (1+\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}\right )}-\frac {\left (b^2-4 a c\right )^{3/4} \left (20 c^2 d^2+7 b^2 e^2-4 c e (5 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 )}{10 \sqrt {2} c^{11/4} (b+2 c x)}+\frac {\left (b^2-4 a c\right )^{3/4} \left (20 c^2 d^2+7 b^2 e^2-4 c e (5 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 )}{20 \sqrt {2} c^{11/4} (b+2 c x)} \\ \end{align*}
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 10.15 (sec) , antiderivative size = 165, normalized size of antiderivative = 0.29 \[ \int \frac {(d+e x)^2}{\sqrt [4]{a+b x+c x^2}} \, dx=\frac {-\frac {14 e (-2 c d+b e) (a+x (b+c x))}{c}+12 e (d+e x) (a+x (b+c x))+\frac {3 \left (20 c^2 d^2+7 b^2 e^2-4 c e (5 b d+2 a e)\right ) (b+2 c x) \sqrt [4]{\frac {c (a+x (b+c x))}{-b^2+4 a c}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {3}{2},\frac {(b+2 c x)^2}{b^2-4 a c}\right )}{2 \sqrt {2} c^2}}{30 c \sqrt [4]{a+x (b+c x)}} \]
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\[\int \frac {\left (e x +d \right )^{2}}{\left (c \,x^{2}+b x +a \right )^{\frac {1}{4}}}d x\]
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\[ \int \frac {(d+e x)^2}{\sqrt [4]{a+b x+c x^2}} \, dx=\int { \frac {{\left (e x + d\right )}^{2}}{{\left (c x^{2} + b x + a\right )}^{\frac {1}{4}}} \,d x } \]
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\[ \int \frac {(d+e x)^2}{\sqrt [4]{a+b x+c x^2}} \, dx=\int \frac {\left (d + e x\right )^{2}}{\sqrt [4]{a + b x + c x^{2}}}\, dx \]
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\[ \int \frac {(d+e x)^2}{\sqrt [4]{a+b x+c x^2}} \, dx=\int { \frac {{\left (e x + d\right )}^{2}}{{\left (c x^{2} + b x + a\right )}^{\frac {1}{4}}} \,d x } \]
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\[ \int \frac {(d+e x)^2}{\sqrt [4]{a+b x+c x^2}} \, dx=\int { \frac {{\left (e x + d\right )}^{2}}{{\left (c x^{2} + b x + a\right )}^{\frac {1}{4}}} \,d x } \]
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Timed out. \[ \int \frac {(d+e x)^2}{\sqrt [4]{a+b x+c x^2}} \, dx=\int \frac {{\left (d+e\,x\right )}^2}{{\left (c\,x^2+b\,x+a\right )}^{1/4}} \,d x \]
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