Optimal. Leaf size=302 \[ \frac{f^2 \left (4 a e^2-b^2 f^2\right ) \sqrt{f \sqrt{a+b x+\frac{e^2 x^2}{f^2}}+d+e x}}{4 e^3}-\frac{f^2 \left (4 a e^2-b^2 f^2\right ) \left (2 d e-b f^2\right ) \sqrt{f \sqrt{a+b x+\frac{e^2 x^2}{f^2}}+d+e x}}{8 e^3 \left (2 e \left (f \sqrt{a+\frac{x \left (b f^2+e^2 x\right )}{f^2}}+e x\right )+b f^2\right )}-\frac{3 f^2 \left (4 a e^2-b^2 f^2\right ) \sqrt{2 d e-b f^2} \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{e} \sqrt{f \sqrt{a+b x+\frac{e^2 x^2}{f^2}}+d+e x}}{\sqrt{2 d e-b f^2}}\right )}{8 \sqrt{2} e^{7/2}}+\frac{\left (f \sqrt{a+b x+\frac{e^2 x^2}{f^2}}+d+e x\right )^{5/2}}{5 e} \]
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Rubi [A] time = 0.41462, antiderivative size = 302, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {2116, 897, 1257, 1810, 208} \[ \frac{f^2 \left (4 a e^2-b^2 f^2\right ) \sqrt{f \sqrt{a+b x+\frac{e^2 x^2}{f^2}}+d+e x}}{4 e^3}-\frac{f^2 \left (4 a e^2-b^2 f^2\right ) \left (2 d e-b f^2\right ) \sqrt{f \sqrt{a+b x+\frac{e^2 x^2}{f^2}}+d+e x}}{8 e^3 \left (2 e \left (f \sqrt{a+\frac{x \left (b f^2+e^2 x\right )}{f^2}}+e x\right )+b f^2\right )}-\frac{3 f^2 \left (4 a e^2-b^2 f^2\right ) \sqrt{2 d e-b f^2} \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{e} \sqrt{f \sqrt{a+b x+\frac{e^2 x^2}{f^2}}+d+e x}}{\sqrt{2 d e-b f^2}}\right )}{8 \sqrt{2} e^{7/2}}+\frac{\left (f \sqrt{a+b x+\frac{e^2 x^2}{f^2}}+d+e x\right )^{5/2}}{5 e} \]
Antiderivative was successfully verified.
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Rule 2116
Rule 897
Rule 1257
Rule 1810
Rule 208
Rubi steps
\begin{align*} \int \left (d+e x+f \sqrt{a+b x+\frac{e^2 x^2}{f^2}}\right )^{3/2} \, dx &=2 \operatorname{Subst}\left (\int \frac{x^{3/2} \left (d^2 e-(b d-a e) f^2-\left (2 d e-b f^2\right ) x+e x^2\right )}{\left (-2 d e+b f^2+2 e x\right )^2} \, dx,x,d+e x+f \sqrt{a+b x+\frac{e^2 x^2}{f^2}}\right )\\ &=4 \operatorname{Subst}\left (\int \frac{x^4 \left (d^2 e-(b d-a e) f^2+\left (-2 d e+b f^2\right ) x^2+e x^4\right )}{\left (-2 d e+b f^2+2 e x^2\right )^2} \, dx,x,\sqrt{d+e x+f \sqrt{a+b x+\frac{e^2 x^2}{f^2}}}\right )\\ &=-\frac{f^2 \left (2 d e-b f^2\right ) \left (4 a e^2-b^2 f^2\right ) \sqrt{d+e x+f \sqrt{a+b x+\frac{e^2 x^2}{f^2}}}}{8 e^3 \left (b f^2+2 e \left (e x+f \sqrt{a+\frac{x \left (b f^2+e^2 x\right )}{f^2}}\right )\right )}-\frac{\operatorname{Subst}\left (\int \frac{-e f^2 \left (2 d e-b f^2\right ) \left (4 a e^2-b^2 f^2\right )-4 e^2 f^2 \left (4 a e^2-b^2 f^2\right ) x^2+8 e^3 \left (2 d e-b f^2\right ) x^4-16 e^4 x^6}{-2 d e+b f^2+2 e x^2} \, dx,x,\sqrt{d+e x+f \sqrt{a+b x+\frac{e^2 x^2}{f^2}}}\right )}{8 e^4}\\ &=-\frac{f^2 \left (2 d e-b f^2\right ) \left (4 a e^2-b^2 f^2\right ) \sqrt{d+e x+f \sqrt{a+b x+\frac{e^2 x^2}{f^2}}}}{8 e^3 \left (b f^2+2 e \left (e x+f \sqrt{a+\frac{x \left (b f^2+e^2 x\right )}{f^2}}\right )\right )}-\frac{\operatorname{Subst}\left (\int \left (-2 e f^2 \left (4 a e^2-b^2 f^2\right )-8 e^3 x^4-\frac{3 \left (8 a d e^4 f^2-2 b^2 d e^2 f^4-4 a b e^3 f^4+b^3 e f^6\right )}{-2 d e+b f^2+2 e x^2}\right ) \, dx,x,\sqrt{d+e x+f \sqrt{a+b x+\frac{e^2 x^2}{f^2}}}\right )}{8 e^4}\\ &=\frac{f^2 \left (4 a e^2-b^2 f^2\right ) \sqrt{d+e x+f \sqrt{a+b x+\frac{e^2 x^2}{f^2}}}}{4 e^3}+\frac{\left (d+e x+f \sqrt{a+b x+\frac{e^2 x^2}{f^2}}\right )^{5/2}}{5 e}-\frac{f^2 \left (2 d e-b f^2\right ) \left (4 a e^2-b^2 f^2\right ) \sqrt{d+e x+f \sqrt{a+b x+\frac{e^2 x^2}{f^2}}}}{8 e^3 \left (b f^2+2 e \left (e x+f \sqrt{a+\frac{x \left (b f^2+e^2 x\right )}{f^2}}\right )\right )}+\frac{\left (3 f^2 \left (2 d e-b f^2\right ) \left (4 a e^2-b^2 f^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{-2 d e+b f^2+2 e x^2} \, dx,x,\sqrt{d+e x+f \sqrt{a+b x+\frac{e^2 x^2}{f^2}}}\right )}{8 e^3}\\ &=\frac{f^2 \left (4 a e^2-b^2 f^2\right ) \sqrt{d+e x+f \sqrt{a+b x+\frac{e^2 x^2}{f^2}}}}{4 e^3}+\frac{\left (d+e x+f \sqrt{a+b x+\frac{e^2 x^2}{f^2}}\right )^{5/2}}{5 e}-\frac{f^2 \left (2 d e-b f^2\right ) \left (4 a e^2-b^2 f^2\right ) \sqrt{d+e x+f \sqrt{a+b x+\frac{e^2 x^2}{f^2}}}}{8 e^3 \left (b f^2+2 e \left (e x+f \sqrt{a+\frac{x \left (b f^2+e^2 x\right )}{f^2}}\right )\right )}-\frac{3 f^2 \sqrt{2 d e-b f^2} \left (4 a e^2-b^2 f^2\right ) \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{e} \sqrt{d+e x+f \sqrt{a+b x+\frac{e^2 x^2}{f^2}}}}{\sqrt{2 d e-b f^2}}\right )}{8 \sqrt{2} e^{7/2}}\\ \end{align*}
Mathematica [A] time = 0.577791, size = 291, normalized size = 0.96 \[ \frac{2 e f^2 \left (4 a e^2-b^2 f^2\right ) \sqrt{f \sqrt{a+x \left (b+\frac{e^2 x}{f^2}\right )}+d+e x}-\frac{\left (4 a e^3 f^2-b^2 e f^4\right ) \left (2 d e-b f^2\right ) \sqrt{f \sqrt{a+x \left (b+\frac{e^2 x}{f^2}\right )}+d+e x}}{2 e \left (f \sqrt{a+x \left (b+\frac{e^2 x}{f^2}\right )}+e x\right )+b f^2}-\frac{3 \sqrt{e} f^2 \left (4 a e^2-b^2 f^2\right ) \sqrt{2 d e-b f^2} \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{e} \sqrt{f \sqrt{a+x \left (b+\frac{e^2 x}{f^2}\right )}+d+e x}}{\sqrt{2 d e-b f^2}}\right )}{\sqrt{2}}+\frac{8}{5} e^3 \left (f \sqrt{a+x \left (b+\frac{e^2 x}{f^2}\right )}+d+e x\right )^{5/2}}{8 e^4} \]
Antiderivative was successfully verified.
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Maple [F] time = 0.012, size = 0, normalized size = 0. \begin{align*} \int \left ( d+ex+f\sqrt{a+bx+{\frac{{e}^{2}{x}^{2}}{{f}^{2}}}} \right ) ^{{\frac{3}{2}}}\, 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 (e x + \sqrt{b x + \frac{e^{2} x^{2}}{f^{2}} + a} f + d\right )}^{\frac{3}{2}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 2.59244, size = 1439, normalized size = 4.76 \begin{align*} \left [-\frac{15 \, \sqrt{\frac{1}{2}}{\left (b^{2} f^{4} - 4 \, a e^{2} f^{2}\right )} \sqrt{-\frac{b f^{2} - 2 \, d e}{e}} \log \left (-b^{2} f^{4} + 4 \,{\left (b d e - a e^{2}\right )} f^{2} - 4 \,{\left (b e^{2} f^{2} - 2 \, d e^{3}\right )} x + 4 \,{\left (2 \, \sqrt{\frac{1}{2}} e^{2} f \sqrt{-\frac{b f^{2} - 2 \, d e}{e}} \sqrt{\frac{b f^{2} x + e^{2} x^{2} + a f^{2}}{f^{2}}} - \sqrt{\frac{1}{2}}{\left (b e f^{2} + 2 \, e^{3} x\right )} \sqrt{-\frac{b f^{2} - 2 \, d e}{e}}\right )} \sqrt{e x + f \sqrt{\frac{b f^{2} x + e^{2} x^{2} + a f^{2}}{f^{2}}} + d} + 4 \,{\left (b e f^{3} - 2 \, d e^{2} f\right )} \sqrt{\frac{b f^{2} x + e^{2} x^{2} + a f^{2}}{f^{2}}}\right ) + 2 \,{\left (15 \, b^{2} f^{4} - 16 \, e^{4} x^{2} - 8 \, d^{2} e^{2} - 2 \,{\left (5 \, b d e + 24 \, a e^{2}\right )} f^{2} + 2 \,{\left (b e^{2} f^{2} - 18 \, d e^{3}\right )} x - 2 \,{\left (5 \, b e f^{3} + 8 \, e^{3} f x - 2 \, d e^{2} f\right )} \sqrt{\frac{b f^{2} x + e^{2} x^{2} + a f^{2}}{f^{2}}}\right )} \sqrt{e x + f \sqrt{\frac{b f^{2} x + e^{2} x^{2} + a f^{2}}{f^{2}}} + d}}{80 \, e^{3}}, \frac{15 \, \sqrt{\frac{1}{2}}{\left (b^{2} f^{4} - 4 \, a e^{2} f^{2}\right )} \sqrt{\frac{b f^{2} - 2 \, d e}{e}} \arctan \left (\frac{2 \, \sqrt{\frac{1}{2}} \sqrt{e x + f \sqrt{\frac{b f^{2} x + e^{2} x^{2} + a f^{2}}{f^{2}}} + d} e \sqrt{\frac{b f^{2} - 2 \, d e}{e}}}{b f^{2} - 2 \, d e}\right ) -{\left (15 \, b^{2} f^{4} - 16 \, e^{4} x^{2} - 8 \, d^{2} e^{2} - 2 \,{\left (5 \, b d e + 24 \, a e^{2}\right )} f^{2} + 2 \,{\left (b e^{2} f^{2} - 18 \, d e^{3}\right )} x - 2 \,{\left (5 \, b e f^{3} + 8 \, e^{3} f x - 2 \, d e^{2} f\right )} \sqrt{\frac{b f^{2} x + e^{2} x^{2} + a f^{2}}{f^{2}}}\right )} \sqrt{e x + f \sqrt{\frac{b f^{2} x + e^{2} x^{2} + a f^{2}}{f^{2}}} + d}}{40 \, e^{3}}\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 + f \sqrt{a + b x + \frac{e^{2} x^{2}}{f^{2}}}\right )^{\frac{3}{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 (e x + \sqrt{b x + \frac{e^{2} x^{2}}{f^{2}} + a} f + d\right )}^{\frac{3}{2}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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