Optimal. Leaf size=310 \[ \frac{3 \left (b^2-4 a c\right )^{7/4} \sqrt{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}} \text{EllipticF}\left (\sin ^{-1}\left (\frac{\sqrt{b d+2 c d x}}{\sqrt{d} \sqrt [4]{b^2-4 a c}}\right ),-1\right )}{20 c^4 d^{7/2} \sqrt{a+b x+c x^2}}-\frac{3 \left (b^2-4 a c\right )^{7/4} \sqrt{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}} E\left (\left .\sin ^{-1}\left (\frac{\sqrt{b d+2 c x d}}{\sqrt [4]{b^2-4 a c} \sqrt{d}}\right )\right |-1\right )}{20 c^4 d^{7/2} \sqrt{a+b x+c x^2}}-\frac{\left (a+b x+c x^2\right )^{3/2}}{2 c^2 d^3 \sqrt{b d+2 c d x}}+\frac{3 \sqrt{a+b x+c x^2} (b d+2 c d x)^{3/2}}{20 c^3 d^5}-\frac{\left (a+b x+c x^2\right )^{5/2}}{5 c d (b d+2 c d x)^{5/2}} \]
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Rubi [A] time = 0.285641, antiderivative size = 310, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 8, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {684, 685, 691, 690, 307, 221, 1199, 424} \[ \frac{3 \left (b^2-4 a c\right )^{7/4} \sqrt{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}} F\left (\left .\sin ^{-1}\left (\frac{\sqrt{b d+2 c x d}}{\sqrt [4]{b^2-4 a c} \sqrt{d}}\right )\right |-1\right )}{20 c^4 d^{7/2} \sqrt{a+b x+c x^2}}-\frac{3 \left (b^2-4 a c\right )^{7/4} \sqrt{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}} E\left (\left .\sin ^{-1}\left (\frac{\sqrt{b d+2 c x d}}{\sqrt [4]{b^2-4 a c} \sqrt{d}}\right )\right |-1\right )}{20 c^4 d^{7/2} \sqrt{a+b x+c x^2}}-\frac{\left (a+b x+c x^2\right )^{3/2}}{2 c^2 d^3 \sqrt{b d+2 c d x}}+\frac{3 \sqrt{a+b x+c x^2} (b d+2 c d x)^{3/2}}{20 c^3 d^5}-\frac{\left (a+b x+c x^2\right )^{5/2}}{5 c d (b d+2 c d x)^{5/2}} \]
Antiderivative was successfully verified.
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Rule 684
Rule 685
Rule 691
Rule 690
Rule 307
Rule 221
Rule 1199
Rule 424
Rubi steps
\begin{align*} \int \frac{\left (a+b x+c x^2\right )^{5/2}}{(b d+2 c d x)^{7/2}} \, dx &=-\frac{\left (a+b x+c x^2\right )^{5/2}}{5 c d (b d+2 c d x)^{5/2}}+\frac{\int \frac{\left (a+b x+c x^2\right )^{3/2}}{(b d+2 c d x)^{3/2}} \, dx}{2 c d^2}\\ &=-\frac{\left (a+b x+c x^2\right )^{3/2}}{2 c^2 d^3 \sqrt{b d+2 c d x}}-\frac{\left (a+b x+c x^2\right )^{5/2}}{5 c d (b d+2 c d x)^{5/2}}+\frac{3 \int \sqrt{b d+2 c d x} \sqrt{a+b x+c x^2} \, dx}{4 c^2 d^4}\\ &=\frac{3 (b d+2 c d x)^{3/2} \sqrt{a+b x+c x^2}}{20 c^3 d^5}-\frac{\left (a+b x+c x^2\right )^{3/2}}{2 c^2 d^3 \sqrt{b d+2 c d x}}-\frac{\left (a+b x+c x^2\right )^{5/2}}{5 c d (b d+2 c d x)^{5/2}}-\frac{\left (3 \left (b^2-4 a c\right )\right ) \int \frac{\sqrt{b d+2 c d x}}{\sqrt{a+b x+c x^2}} \, dx}{40 c^3 d^4}\\ &=\frac{3 (b d+2 c d x)^{3/2} \sqrt{a+b x+c x^2}}{20 c^3 d^5}-\frac{\left (a+b x+c x^2\right )^{3/2}}{2 c^2 d^3 \sqrt{b d+2 c d x}}-\frac{\left (a+b x+c x^2\right )^{5/2}}{5 c d (b d+2 c d x)^{5/2}}-\frac{\left (3 \left (b^2-4 a c\right ) \sqrt{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}}\right ) \int \frac{\sqrt{b d+2 c d x}}{\sqrt{-\frac{a c}{b^2-4 a c}-\frac{b c x}{b^2-4 a c}-\frac{c^2 x^2}{b^2-4 a c}}} \, dx}{40 c^3 d^4 \sqrt{a+b x+c x^2}}\\ &=\frac{3 (b d+2 c d x)^{3/2} \sqrt{a+b x+c x^2}}{20 c^3 d^5}-\frac{\left (a+b x+c x^2\right )^{3/2}}{2 c^2 d^3 \sqrt{b d+2 c d x}}-\frac{\left (a+b x+c x^2\right )^{5/2}}{5 c d (b d+2 c d x)^{5/2}}-\frac{\left (3 \left (b^2-4 a c\right ) \sqrt{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}}\right ) \operatorname{Subst}\left (\int \frac{x^2}{\sqrt{1-\frac{x^4}{\left (b^2-4 a c\right ) d^2}}} \, dx,x,\sqrt{b d+2 c d x}\right )}{20 c^4 d^5 \sqrt{a+b x+c x^2}}\\ &=\frac{3 (b d+2 c d x)^{3/2} \sqrt{a+b x+c x^2}}{20 c^3 d^5}-\frac{\left (a+b x+c x^2\right )^{3/2}}{2 c^2 d^3 \sqrt{b d+2 c d x}}-\frac{\left (a+b x+c x^2\right )^{5/2}}{5 c d (b d+2 c d x)^{5/2}}+\frac{\left (3 \left (b^2-4 a c\right )^{3/2} \sqrt{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{1-\frac{x^4}{\left (b^2-4 a c\right ) d^2}}} \, dx,x,\sqrt{b d+2 c d x}\right )}{20 c^4 d^4 \sqrt{a+b x+c x^2}}-\frac{\left (3 \left (b^2-4 a c\right )^{3/2} \sqrt{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}}\right ) \operatorname{Subst}\left (\int \frac{1+\frac{x^2}{\sqrt{b^2-4 a c} d}}{\sqrt{1-\frac{x^4}{\left (b^2-4 a c\right ) d^2}}} \, dx,x,\sqrt{b d+2 c d x}\right )}{20 c^4 d^4 \sqrt{a+b x+c x^2}}\\ &=\frac{3 (b d+2 c d x)^{3/2} \sqrt{a+b x+c x^2}}{20 c^3 d^5}-\frac{\left (a+b x+c x^2\right )^{3/2}}{2 c^2 d^3 \sqrt{b d+2 c d x}}-\frac{\left (a+b x+c x^2\right )^{5/2}}{5 c d (b d+2 c d x)^{5/2}}+\frac{3 \left (b^2-4 a c\right )^{7/4} \sqrt{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}} F\left (\left .\sin ^{-1}\left (\frac{\sqrt{b d+2 c d x}}{\sqrt [4]{b^2-4 a c} \sqrt{d}}\right )\right |-1\right )}{20 c^4 d^{7/2} \sqrt{a+b x+c x^2}}-\frac{\left (3 \left (b^2-4 a c\right )^{3/2} \sqrt{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}}\right ) \operatorname{Subst}\left (\int \frac{\sqrt{1+\frac{x^2}{\sqrt{b^2-4 a c} d}}}{\sqrt{1-\frac{x^2}{\sqrt{b^2-4 a c} d}}} \, dx,x,\sqrt{b d+2 c d x}\right )}{20 c^4 d^4 \sqrt{a+b x+c x^2}}\\ &=\frac{3 (b d+2 c d x)^{3/2} \sqrt{a+b x+c x^2}}{20 c^3 d^5}-\frac{\left (a+b x+c x^2\right )^{3/2}}{2 c^2 d^3 \sqrt{b d+2 c d x}}-\frac{\left (a+b x+c x^2\right )^{5/2}}{5 c d (b d+2 c d x)^{5/2}}-\frac{3 \left (b^2-4 a c\right )^{7/4} \sqrt{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}} E\left (\left .\sin ^{-1}\left (\frac{\sqrt{b d+2 c d x}}{\sqrt [4]{b^2-4 a c} \sqrt{d}}\right )\right |-1\right )}{20 c^4 d^{7/2} \sqrt{a+b x+c x^2}}+\frac{3 \left (b^2-4 a c\right )^{7/4} \sqrt{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}} F\left (\left .\sin ^{-1}\left (\frac{\sqrt{b d+2 c d x}}{\sqrt [4]{b^2-4 a c} \sqrt{d}}\right )\right |-1\right )}{20 c^4 d^{7/2} \sqrt{a+b x+c x^2}}\\ \end{align*}
Mathematica [C] time = 0.0647175, size = 101, normalized size = 0.33 \[ -\frac{\left (b^2-4 a c\right )^2 \sqrt{a+x (b+c x)} \, _2F_1\left (-\frac{5}{2},-\frac{5}{4};-\frac{1}{4};\frac{(b+2 c x)^2}{b^2-4 a c}\right )}{160 c^3 d \sqrt{\frac{c (a+x (b+c x))}{4 a c-b^2}} (d (b+2 c x))^{5/2}} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.228, size = 1362, normalized size = 4.4 \begin{align*} \text{result too large to display} \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 \frac{{\left (c x^{2} + b x + a\right )}^{\frac{5}{2}}}{{\left (2 \, c d x + b d\right )}^{\frac{7}{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 (\frac{{\left (c^{2} x^{4} + 2 \, b c x^{3} + 2 \, a b x +{\left (b^{2} + 2 \, a c\right )} x^{2} + a^{2}\right )} \sqrt{2 \, c d x + b d} \sqrt{c x^{2} + b x + a}}{16 \, c^{4} d^{4} x^{4} + 32 \, b c^{3} d^{4} x^{3} + 24 \, b^{2} c^{2} d^{4} x^{2} + 8 \, b^{3} c d^{4} x + b^{4} d^{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 \frac{\left (a + b x + c x^{2}\right )^{\frac{5}{2}}}{\left (d \left (b + 2 c x\right )\right )^{\frac{7}{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 \frac{{\left (c x^{2} + b x + a\right )}^{\frac{5}{2}}}{{\left (2 \, c d x + b d\right )}^{\frac{7}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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