Optimal. Leaf size=319 \[ -\frac{\left (b^2-4 a c\right )^{5/4} \sqrt{\frac{(b+2 c x)^2}{\left (b^2-4 a c\right ) \left (\frac{2 \sqrt{c} \sqrt{a+b x+c x^2}}{\sqrt{b^2-4 a c}}+1\right )^2}} \left (\frac{2 \sqrt{c} \sqrt{a+b x+c x^2}}{\sqrt{b^2-4 a c}}+1\right ) \left (-4 c e (2 a e+7 b d)+9 b^2 e^2+28 c^2 d^2\right ) \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{c} \sqrt [4]{a+b x+c x^2}}{\sqrt [4]{b^2-4 a c}}\right ),\frac{1}{2}\right )}{168 \sqrt{2} c^{13/4} (b+2 c x)}+\frac{(b+2 c x) \sqrt [4]{a+b x+c x^2} \left (-4 c e (2 a e+7 b d)+9 b^2 e^2+28 c^2 d^2\right )}{84 c^3}+\frac{9 e \left (a+b x+c x^2\right )^{5/4} (2 c d-b e)}{35 c^2}+\frac{2 e (d+e x) \left (a+b x+c x^2\right )^{5/4}}{7 c} \]
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Rubi [A] time = 0.423134, antiderivative size = 319, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.227, Rules used = {742, 640, 612, 623, 220} \[ \frac{(b+2 c x) \sqrt [4]{a+b x+c x^2} \left (-4 c e (2 a e+7 b d)+9 b^2 e^2+28 c^2 d^2\right )}{84 c^3}-\frac{\left (b^2-4 a c\right )^{5/4} \sqrt{\frac{(b+2 c x)^2}{\left (b^2-4 a c\right ) \left (\frac{2 \sqrt{c} \sqrt{a+b x+c x^2}}{\sqrt{b^2-4 a c}}+1\right )^2}} \left (\frac{2 \sqrt{c} \sqrt{a+b x+c x^2}}{\sqrt{b^2-4 a c}}+1\right ) \left (-4 c e (2 a e+7 b d)+9 b^2 e^2+28 c^2 d^2\right ) F\left (2 \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{c} \sqrt [4]{c x^2+b x+a}}{\sqrt [4]{b^2-4 a c}}\right )|\frac{1}{2}\right )}{168 \sqrt{2} c^{13/4} (b+2 c x)}+\frac{9 e \left (a+b x+c x^2\right )^{5/4} (2 c d-b e)}{35 c^2}+\frac{2 e (d+e x) \left (a+b x+c x^2\right )^{5/4}}{7 c} \]
Antiderivative was successfully verified.
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Rule 742
Rule 640
Rule 612
Rule 623
Rule 220
Rubi steps
\begin{align*} \int (d+e x)^2 \sqrt [4]{a+b x+c x^2} \, dx &=\frac{2 e (d+e x) \left (a+b x+c x^2\right )^{5/4}}{7 c}+\frac{2 \int \left (\frac{1}{4} \left (14 c d^2-4 e \left (\frac{5 b d}{4}+a e\right )\right )+\frac{9}{4} e (2 c d-b e) x\right ) \sqrt [4]{a+b x+c x^2} \, dx}{7 c}\\ &=\frac{9 e (2 c d-b e) \left (a+b x+c x^2\right )^{5/4}}{35 c^2}+\frac{2 e (d+e x) \left (a+b x+c x^2\right )^{5/4}}{7 c}+\frac{\left (-\frac{9}{4} b e (2 c d-b e)+\frac{1}{2} c \left (14 c d^2-4 e \left (\frac{5 b d}{4}+a e\right )\right )\right ) \int \sqrt [4]{a+b x+c x^2} \, dx}{7 c^2}\\ &=\frac{\left (28 c^2 d^2+9 b^2 e^2-4 c e (7 b d+2 a e)\right ) (b+2 c x) \sqrt [4]{a+b x+c x^2}}{84 c^3}+\frac{9 e (2 c d-b e) \left (a+b x+c x^2\right )^{5/4}}{35 c^2}+\frac{2 e (d+e x) \left (a+b x+c x^2\right )^{5/4}}{7 c}-\frac{\left (\left (b^2-4 a c\right ) \left (28 c^2 d^2+9 b^2 e^2-4 c e (7 b d+2 a e)\right )\right ) \int \frac{1}{\left (a+b x+c x^2\right )^{3/4}} \, dx}{336 c^3}\\ &=\frac{\left (28 c^2 d^2+9 b^2 e^2-4 c e (7 b d+2 a e)\right ) (b+2 c x) \sqrt [4]{a+b x+c x^2}}{84 c^3}+\frac{9 e (2 c d-b e) \left (a+b x+c x^2\right )^{5/4}}{35 c^2}+\frac{2 e (d+e x) \left (a+b x+c x^2\right )^{5/4}}{7 c}-\frac{\left (\left (b^2-4 a c\right ) \left (28 c^2 d^2+9 b^2 e^2-4 c e (7 b d+2 a e)\right ) \sqrt{(b+2 c x)^2}\right ) \operatorname{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 )}{84 c^3 (b+2 c x)}\\ &=\frac{\left (28 c^2 d^2+9 b^2 e^2-4 c e (7 b d+2 a e)\right ) (b+2 c x) \sqrt [4]{a+b x+c x^2}}{84 c^3}+\frac{9 e (2 c d-b e) \left (a+b x+c x^2\right )^{5/4}}{35 c^2}+\frac{2 e (d+e x) \left (a+b x+c x^2\right )^{5/4}}{7 c}-\frac{\left (b^2-4 a c\right )^{5/4} \left (28 c^2 d^2+9 b^2 e^2-4 c e (7 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 )}{168 \sqrt{2} c^{13/4} (b+2 c x)}\\ \end{align*}
Mathematica [A] time = 0.459999, size = 191, normalized size = 0.6 \[ \frac{\frac{5 \left (-4 c e (2 a e+7 b d)+9 b^2 e^2+28 c^2 d^2\right ) \left (2 c (b+2 c x) (a+x (b+c x))-\sqrt{2} \left (b^2-4 a c\right )^{3/2} \left (\frac{c (a+x (b+c x))}{4 a c-b^2}\right )^{3/4} \text{EllipticF}\left (\frac{1}{2} \sin ^{-1}\left (\frac{b+2 c x}{\sqrt{b^2-4 a c}}\right ),2\right )\right )}{4 c^2}-54 e (a+x (b+c x))^2 (b e-2 c d)+60 c e (d+e x) (a+x (b+c x))^2}{210 c^2 (a+x (b+c x))^{3/4}} \]
Antiderivative was successfully verified.
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Maple [F] time = 0.978, size = 0, normalized size = 0. \begin{align*} \int \left ( ex+d \right ) ^{2}\sqrt [4]{c{x}^{2}+bx+a}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (c x^{2} + b x + a\right )}^{\frac{1}{4}}{\left (e x + d\right )}^{2}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (e^{2} x^{2} + 2 \, d e x + d^{2}\right )}{\left (c x^{2} + b x + a\right )}^{\frac{1}{4}}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (d + e x\right )^{2} \sqrt [4]{a + b x + c x^{2}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (c x^{2} + b x + a\right )}^{\frac{1}{4}}{\left (e x + d\right )}^{2}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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