Optimal. Leaf size=603 \[ -\frac{2 \sqrt{2} \sqrt{b^2-4 a c} \sqrt{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}} (2 c d-b e) \left (-4 c e (32 b d-17 a e)+15 b^2 e^2+128 c^2 d^2\right ) \sqrt{\frac{c (d+e x)}{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}} \text{EllipticF}\left (\sin ^{-1}\left (\frac{\sqrt{\frac{\sqrt{b^2-4 a c}+b+2 c x}{\sqrt{b^2-4 a c}}}}{\sqrt{2}}\right ),-\frac{2 e \sqrt{b^2-4 a c}}{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}\right )}{15 c e^6 \sqrt{d+e x} \sqrt{a+b x+c x^2}}-\frac{2 \sqrt{a+b x+c x^2} \left (-4 c e (28 b d-9 a e)+15 b^2 e^2+16 c e x (2 c d-b e)+128 c^2 d^2\right )}{15 e^5 \sqrt{d+e x}}+\frac{2 \sqrt{2} \sqrt{b^2-4 a c} \sqrt{d+e x} \sqrt{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}} \left (-4 c e (32 b d-9 a e)+23 b^2 e^2+128 c^2 d^2\right ) E\left (\sin ^{-1}\left (\frac{\sqrt{\frac{b+2 c x+\sqrt{b^2-4 a c}}{\sqrt{b^2-4 a c}}}}{\sqrt{2}}\right )|-\frac{2 \sqrt{b^2-4 a c} e}{2 c d-\left (b+\sqrt{b^2-4 a c}\right ) e}\right )}{15 e^6 \sqrt{a+b x+c x^2} \sqrt{\frac{c (d+e x)}{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}}}+\frac{2 \left (a+b x+c x^2\right )^{3/2} (-5 b e+16 c d+6 c e x)}{15 e^3 (d+e x)^{3/2}}-\frac{2 \left (a+b x+c x^2\right )^{5/2}}{5 e (d+e x)^{5/2}} \]
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Rubi [A] time = 0.689455, antiderivative size = 603, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {732, 812, 843, 718, 424, 419} \[ -\frac{2 \sqrt{a+b x+c x^2} \left (-4 c e (28 b d-9 a e)+15 b^2 e^2+16 c e x (2 c d-b e)+128 c^2 d^2\right )}{15 e^5 \sqrt{d+e x}}-\frac{2 \sqrt{2} \sqrt{b^2-4 a c} \sqrt{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}} (2 c d-b e) \left (-4 c e (32 b d-17 a e)+15 b^2 e^2+128 c^2 d^2\right ) \sqrt{\frac{c (d+e x)}{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}} F\left (\sin ^{-1}\left (\frac{\sqrt{\frac{b+2 c x+\sqrt{b^2-4 a c}}{\sqrt{b^2-4 a c}}}}{\sqrt{2}}\right )|-\frac{2 \sqrt{b^2-4 a c} e}{2 c d-\left (b+\sqrt{b^2-4 a c}\right ) e}\right )}{15 c e^6 \sqrt{d+e x} \sqrt{a+b x+c x^2}}+\frac{2 \sqrt{2} \sqrt{b^2-4 a c} \sqrt{d+e x} \sqrt{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}} \left (-4 c e (32 b d-9 a e)+23 b^2 e^2+128 c^2 d^2\right ) E\left (\sin ^{-1}\left (\frac{\sqrt{\frac{b+2 c x+\sqrt{b^2-4 a c}}{\sqrt{b^2-4 a c}}}}{\sqrt{2}}\right )|-\frac{2 \sqrt{b^2-4 a c} e}{2 c d-\left (b+\sqrt{b^2-4 a c}\right ) e}\right )}{15 e^6 \sqrt{a+b x+c x^2} \sqrt{\frac{c (d+e x)}{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}}}+\frac{2 \left (a+b x+c x^2\right )^{3/2} (-5 b e+16 c d+6 c e x)}{15 e^3 (d+e x)^{3/2}}-\frac{2 \left (a+b x+c x^2\right )^{5/2}}{5 e (d+e x)^{5/2}} \]
Antiderivative was successfully verified.
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Rule 732
Rule 812
Rule 843
Rule 718
Rule 424
Rule 419
Rubi steps
\begin{align*} \int \frac{\left (a+b x+c x^2\right )^{5/2}}{(d+e x)^{7/2}} \, dx &=-\frac{2 \left (a+b x+c x^2\right )^{5/2}}{5 e (d+e x)^{5/2}}+\frac{\int \frac{(b+2 c x) \left (a+b x+c x^2\right )^{3/2}}{(d+e x)^{5/2}} \, dx}{e}\\ &=\frac{2 (16 c d-5 b e+6 c e x) \left (a+b x+c x^2\right )^{3/2}}{15 e^3 (d+e x)^{3/2}}-\frac{2 \left (a+b x+c x^2\right )^{5/2}}{5 e (d+e x)^{5/2}}-\frac{2 \int \frac{\left (\frac{1}{2} \left (16 b c d-5 b^2 e-12 a c e\right )+8 c (2 c d-b e) x\right ) \sqrt{a+b x+c x^2}}{(d+e x)^{3/2}} \, dx}{5 e^3}\\ &=-\frac{2 \left (128 c^2 d^2+15 b^2 e^2-4 c e (28 b d-9 a e)+16 c e (2 c d-b e) x\right ) \sqrt{a+b x+c x^2}}{15 e^5 \sqrt{d+e x}}+\frac{2 (16 c d-5 b e+6 c e x) \left (a+b x+c x^2\right )^{3/2}}{15 e^3 (d+e x)^{3/2}}-\frac{2 \left (a+b x+c x^2\right )^{5/2}}{5 e (d+e x)^{5/2}}+\frac{4 \int \frac{\frac{1}{4} \left (-112 b^2 c d e-64 a c^2 d e+15 b^3 e^2+4 b c \left (32 c d^2+17 a e^2\right )\right )+\frac{1}{2} c \left (128 c^2 d^2+23 b^2 e^2-4 c e (32 b d-9 a e)\right ) x}{\sqrt{d+e x} \sqrt{a+b x+c x^2}} \, dx}{15 e^5}\\ &=-\frac{2 \left (128 c^2 d^2+15 b^2 e^2-4 c e (28 b d-9 a e)+16 c e (2 c d-b e) x\right ) \sqrt{a+b x+c x^2}}{15 e^5 \sqrt{d+e x}}+\frac{2 (16 c d-5 b e+6 c e x) \left (a+b x+c x^2\right )^{3/2}}{15 e^3 (d+e x)^{3/2}}-\frac{2 \left (a+b x+c x^2\right )^{5/2}}{5 e (d+e x)^{5/2}}-\frac{\left ((2 c d-b e) \left (128 c^2 d^2+15 b^2 e^2-4 c e (32 b d-17 a e)\right )\right ) \int \frac{1}{\sqrt{d+e x} \sqrt{a+b x+c x^2}} \, dx}{15 e^6}+\frac{\left (2 c \left (128 c^2 d^2+23 b^2 e^2-4 c e (32 b d-9 a e)\right )\right ) \int \frac{\sqrt{d+e x}}{\sqrt{a+b x+c x^2}} \, dx}{15 e^6}\\ &=-\frac{2 \left (128 c^2 d^2+15 b^2 e^2-4 c e (28 b d-9 a e)+16 c e (2 c d-b e) x\right ) \sqrt{a+b x+c x^2}}{15 e^5 \sqrt{d+e x}}+\frac{2 (16 c d-5 b e+6 c e x) \left (a+b x+c x^2\right )^{3/2}}{15 e^3 (d+e x)^{3/2}}-\frac{2 \left (a+b x+c x^2\right )^{5/2}}{5 e (d+e x)^{5/2}}+\frac{\left (2 \sqrt{2} \sqrt{b^2-4 a c} \left (128 c^2 d^2+23 b^2 e^2-4 c e (32 b d-9 a e)\right ) \sqrt{d+e x} \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{2 \sqrt{b^2-4 a c} e x^2}{2 c d-b e-\sqrt{b^2-4 a c} e}}}{\sqrt{1-x^2}} \, dx,x,\frac{\sqrt{\frac{b+\sqrt{b^2-4 a c}+2 c x}{\sqrt{b^2-4 a c}}}}{\sqrt{2}}\right )}{15 e^6 \sqrt{\frac{c (d+e x)}{2 c d-b e-\sqrt{b^2-4 a c} e}} \sqrt{a+b x+c x^2}}-\frac{\left (2 \sqrt{2} \sqrt{b^2-4 a c} (2 c d-b e) \left (128 c^2 d^2+15 b^2 e^2-4 c e (32 b d-17 a e)\right ) \sqrt{\frac{c (d+e x)}{2 c d-b e-\sqrt{b^2-4 a c} e}} \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-x^2} \sqrt{1+\frac{2 \sqrt{b^2-4 a c} e x^2}{2 c d-b e-\sqrt{b^2-4 a c} e}}} \, dx,x,\frac{\sqrt{\frac{b+\sqrt{b^2-4 a c}+2 c x}{\sqrt{b^2-4 a c}}}}{\sqrt{2}}\right )}{15 c e^6 \sqrt{d+e x} \sqrt{a+b x+c x^2}}\\ &=-\frac{2 \left (128 c^2 d^2+15 b^2 e^2-4 c e (28 b d-9 a e)+16 c e (2 c d-b e) x\right ) \sqrt{a+b x+c x^2}}{15 e^5 \sqrt{d+e x}}+\frac{2 (16 c d-5 b e+6 c e x) \left (a+b x+c x^2\right )^{3/2}}{15 e^3 (d+e x)^{3/2}}-\frac{2 \left (a+b x+c x^2\right )^{5/2}}{5 e (d+e x)^{5/2}}+\frac{2 \sqrt{2} \sqrt{b^2-4 a c} \left (128 c^2 d^2+23 b^2 e^2-4 c e (32 b d-9 a e)\right ) \sqrt{d+e x} \sqrt{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}} E\left (\sin ^{-1}\left (\frac{\sqrt{\frac{b+\sqrt{b^2-4 a c}+2 c x}{\sqrt{b^2-4 a c}}}}{\sqrt{2}}\right )|-\frac{2 \sqrt{b^2-4 a c} e}{2 c d-\left (b+\sqrt{b^2-4 a c}\right ) e}\right )}{15 e^6 \sqrt{\frac{c (d+e x)}{2 c d-\left (b+\sqrt{b^2-4 a c}\right ) e}} \sqrt{a+b x+c x^2}}-\frac{2 \sqrt{2} \sqrt{b^2-4 a c} (2 c d-b e) \left (128 c^2 d^2+15 b^2 e^2-4 c e (32 b d-17 a e)\right ) \sqrt{\frac{c (d+e x)}{2 c d-\left (b+\sqrt{b^2-4 a c}\right ) e}} \sqrt{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}} F\left (\sin ^{-1}\left (\frac{\sqrt{\frac{b+\sqrt{b^2-4 a c}+2 c x}{\sqrt{b^2-4 a c}}}}{\sqrt{2}}\right )|-\frac{2 \sqrt{b^2-4 a c} e}{2 c d-\left (b+\sqrt{b^2-4 a c}\right ) e}\right )}{15 c e^6 \sqrt{d+e x} \sqrt{a+b x+c x^2}}\\ \end{align*}
Mathematica [C] time = 13.2677, size = 8961, normalized size = 14.86 \[ \text{Result too large to show} \]
Warning: Unable to verify antiderivative.
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Maple [B] time = 0.378, size = 14453, normalized size = 24. \begin{align*} \text{output 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 (e x + 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{c x^{2} + b x + a} \sqrt{e x + d}}{e^{4} x^{4} + 4 \, d e^{3} x^{3} + 6 \, d^{2} e^{2} x^{2} + 4 \, d^{3} e x + d^{4}}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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