Optimal. Leaf size=748 \[ \frac{4 e \left (\frac{d}{4 e}+x\right ) \left (-256 a e^3-48 d^2 e^2 \left (\frac{d}{4 e}+x\right )^2+13 d^4\right )}{\left (-16384 a^2 e^6-64 a d^4 e^3+5 d^8\right ) \sqrt{8 a e^2-d^3 x+8 d e^2 x^3+8 e^3 x^4}}+\frac{384 d^2 e^2 \left (\frac{d}{4 e}+x\right ) \sqrt{8 a e^2-d^3 x+8 d e^2 x^3+8 e^3 x^4}}{\left (d^4-64 a e^3\right ) \left (256 a e^3+5 d^4\right )^{3/2} \left (\frac{16 e^2 \left (\frac{d}{4 e}+x\right )^2}{\sqrt{256 a e^3+5 d^4}}+1\right )}-\frac{2 \sqrt{2} \left (-3 d^2 \sqrt{256 a e^3+5 d^4}+256 a e^3+5 d^4\right ) \sqrt{\frac{e \left (8 a e^2-d^3 x+8 d e^2 x^3+8 e^3 x^4\right )}{\left (256 a e^3+5 d^4\right ) \left (\frac{16 e^2 \left (\frac{d}{4 e}+x\right )^2}{\sqrt{256 a e^3+5 d^4}}+1\right )^2}} \left (\frac{16 e^2 \left (\frac{d}{4 e}+x\right )^2}{\sqrt{256 a e^3+5 d^4}}+1\right ) F\left (2 \tan ^{-1}\left (\frac{d+4 e x}{\sqrt [4]{5 d^4+256 a e^3}}\right )|\frac{1}{2} \left (\frac{3 d^2}{\sqrt{5 d^4+256 a e^3}}+1\right )\right )}{\left (d^4-64 a e^3\right ) \left (256 a e^3+5 d^4\right )^{3/4} \sqrt{8 a e^2-d^3 x+8 d e^2 x^3+8 e^3 x^4}}-\frac{12 \sqrt{2} d^2 \sqrt{\frac{e \left (8 a e^2-d^3 x+8 d e^2 x^3+8 e^3 x^4\right )}{\left (256 a e^3+5 d^4\right ) \left (\frac{16 e^2 \left (\frac{d}{4 e}+x\right )^2}{\sqrt{256 a e^3+5 d^4}}+1\right )^2}} \left (\frac{16 e^2 \left (\frac{d}{4 e}+x\right )^2}{\sqrt{256 a e^3+5 d^4}}+1\right ) E\left (2 \tan ^{-1}\left (\frac{d+4 e x}{\sqrt [4]{5 d^4+256 a e^3}}\right )|\frac{1}{2} \left (\frac{3 d^2}{\sqrt{5 d^4+256 a e^3}}+1\right )\right )}{\left (d^4-64 a e^3\right ) \sqrt [4]{256 a e^3+5 d^4} \sqrt{8 a e^2-d^3 x+8 d e^2 x^3+8 e^3 x^4}} \]
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Rubi [A] time = 0.786038, antiderivative size = 748, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 34, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.147, Rules used = {1106, 1092, 1197, 1103, 1195} \[ \frac{4 e \left (\frac{d}{4 e}+x\right ) \left (-256 a e^3-48 d^2 e^2 \left (\frac{d}{4 e}+x\right )^2+13 d^4\right )}{\left (-16384 a^2 e^6-64 a d^4 e^3+5 d^8\right ) \sqrt{8 a e^2-d^3 x+8 d e^2 x^3+8 e^3 x^4}}+\frac{384 d^2 e^2 \left (\frac{d}{4 e}+x\right ) \sqrt{8 a e^2-d^3 x+8 d e^2 x^3+8 e^3 x^4}}{\left (d^4-64 a e^3\right ) \left (256 a e^3+5 d^4\right )^{3/2} \left (\frac{16 e^2 \left (\frac{d}{4 e}+x\right )^2}{\sqrt{256 a e^3+5 d^4}}+1\right )}-\frac{2 \sqrt{2} \left (-3 d^2 \sqrt{256 a e^3+5 d^4}+256 a e^3+5 d^4\right ) \sqrt{\frac{e \left (8 a e^2-d^3 x+8 d e^2 x^3+8 e^3 x^4\right )}{\left (256 a e^3+5 d^4\right ) \left (\frac{16 e^2 \left (\frac{d}{4 e}+x\right )^2}{\sqrt{256 a e^3+5 d^4}}+1\right )^2}} \left (\frac{16 e^2 \left (\frac{d}{4 e}+x\right )^2}{\sqrt{256 a e^3+5 d^4}}+1\right ) F\left (2 \tan ^{-1}\left (\frac{d+4 e x}{\sqrt [4]{5 d^4+256 a e^3}}\right )|\frac{1}{2} \left (\frac{3 d^2}{\sqrt{5 d^4+256 a e^3}}+1\right )\right )}{\left (d^4-64 a e^3\right ) \left (256 a e^3+5 d^4\right )^{3/4} \sqrt{8 a e^2-d^3 x+8 d e^2 x^3+8 e^3 x^4}}-\frac{12 \sqrt{2} d^2 \sqrt{\frac{e \left (8 a e^2-d^3 x+8 d e^2 x^3+8 e^3 x^4\right )}{\left (256 a e^3+5 d^4\right ) \left (\frac{16 e^2 \left (\frac{d}{4 e}+x\right )^2}{\sqrt{256 a e^3+5 d^4}}+1\right )^2}} \left (\frac{16 e^2 \left (\frac{d}{4 e}+x\right )^2}{\sqrt{256 a e^3+5 d^4}}+1\right ) E\left (2 \tan ^{-1}\left (\frac{d+4 e x}{\sqrt [4]{5 d^4+256 a e^3}}\right )|\frac{1}{2} \left (\frac{3 d^2}{\sqrt{5 d^4+256 a e^3}}+1\right )\right )}{\left (d^4-64 a e^3\right ) \sqrt [4]{256 a e^3+5 d^4} \sqrt{8 a e^2-d^3 x+8 d e^2 x^3+8 e^3 x^4}} \]
Antiderivative was successfully verified.
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Rule 1106
Rule 1092
Rule 1197
Rule 1103
Rule 1195
Rubi steps
\begin{align*} \int \frac{1}{\left (8 a e^2-d^3 x+8 d e^2 x^3+8 e^3 x^4\right )^{3/2}} \, dx &=\operatorname{Subst}\left (\int \frac{1}{\left (\frac{1}{32} \left (\frac{5 d^4}{e}+256 a e^2\right )-3 d^2 e x^2+8 e^3 x^4\right )^{3/2}} \, dx,x,\frac{d}{4 e}+x\right )\\ &=\frac{4 e \left (\frac{d}{4 e}+x\right ) \left (13 d^4-256 a e^3-48 d^2 e^2 \left (\frac{d}{4 e}+x\right )^2\right )}{\left (5 d^8-64 a d^4 e^3-16384 a^2 e^6\right ) \sqrt{8 a e^2-d^3 x+8 d e^2 x^3+8 e^3 x^4}}-\frac{8 \operatorname{Subst}\left (\int \frac{\frac{1}{2} e^3 \left (\frac{5 d^4}{e}+256 a e^2\right )-24 d^2 e^4 x^2}{\sqrt{\frac{1}{32} \left (\frac{5 d^4}{e}+256 a e^2\right )-3 d^2 e x^2+8 e^3 x^4}} \, dx,x,\frac{d}{4 e}+x\right )}{e \left (5 d^8-64 a d^4 e^3-16384 a^2 e^6\right )}\\ &=\frac{4 e \left (\frac{d}{4 e}+x\right ) \left (13 d^4-256 a e^3-48 d^2 e^2 \left (\frac{d}{4 e}+x\right )^2\right )}{\left (5 d^8-64 a d^4 e^3-16384 a^2 e^6\right ) \sqrt{8 a e^2-d^3 x+8 d e^2 x^3+8 e^3 x^4}}-\frac{\left (12 d^2 e\right ) \operatorname{Subst}\left (\int \frac{1-\frac{16 e^2 x^2}{\sqrt{5 d^4+256 a e^3}}}{\sqrt{\frac{1}{32} \left (\frac{5 d^4}{e}+256 a e^2\right )-3 d^2 e x^2+8 e^3 x^4}} \, dx,x,\frac{d}{4 e}+x\right )}{\left (d^4-64 a e^3\right ) \sqrt{5 d^4+256 a e^3}}-\frac{\left (4 e \left (5 d^4+256 a e^3-3 d^2 \sqrt{5 d^4+256 a e^3}\right )\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{\frac{1}{32} \left (\frac{5 d^4}{e}+256 a e^2\right )-3 d^2 e x^2+8 e^3 x^4}} \, dx,x,\frac{d}{4 e}+x\right )}{5 d^8-64 a d^4 e^3-16384 a^2 e^6}\\ &=\frac{4 e \left (\frac{d}{4 e}+x\right ) \left (13 d^4-256 a e^3-48 d^2 e^2 \left (\frac{d}{4 e}+x\right )^2\right )}{\left (5 d^8-64 a d^4 e^3-16384 a^2 e^6\right ) \sqrt{8 a e^2-d^3 x+8 d e^2 x^3+8 e^3 x^4}}+\frac{96 d^2 e (d+4 e x) \sqrt{8 a e^2-d^3 x+8 d e^2 x^3+8 e^3 x^4}}{\left (d^4-64 a e^3\right ) \left (5 d^4+256 a e^3\right )^{3/2} \left (1+\frac{(d+4 e x)^2}{\sqrt{5 d^4+256 a e^3}}\right )}-\frac{12 \sqrt{2} d^2 \sqrt{\frac{e \left (8 a e^2-d^3 x+8 d e^2 x^3+8 e^3 x^4\right )}{\left (5 d^4+256 a e^3\right ) \left (1+\frac{(d+4 e x)^2}{\sqrt{5 d^4+256 a e^3}}\right )^2}} \left (1+\frac{(d+4 e x)^2}{\sqrt{5 d^4+256 a e^3}}\right ) E\left (2 \tan ^{-1}\left (\frac{d+4 e x}{\sqrt [4]{5 d^4+256 a e^3}}\right )|\frac{1}{2} \left (1+\frac{3 d^2}{\sqrt{5 d^4+256 a e^3}}\right )\right )}{\left (d^4-64 a e^3\right ) \sqrt [4]{5 d^4+256 a e^3} \sqrt{8 a e^2-d^3 x+8 d e^2 x^3+8 e^3 x^4}}-\frac{2 \sqrt{2} \left (5 d^4+256 a e^3-3 d^2 \sqrt{5 d^4+256 a e^3}\right ) \sqrt{\frac{e \left (8 a e^2-d^3 x+8 d e^2 x^3+8 e^3 x^4\right )}{\left (5 d^4+256 a e^3\right ) \left (1+\frac{(d+4 e x)^2}{\sqrt{5 d^4+256 a e^3}}\right )^2}} \left (1+\frac{(d+4 e x)^2}{\sqrt{5 d^4+256 a e^3}}\right ) F\left (2 \tan ^{-1}\left (\frac{d+4 e x}{\sqrt [4]{5 d^4+256 a e^3}}\right )|\frac{1}{2} \left (1+\frac{3 d^2}{\sqrt{5 d^4+256 a e^3}}\right )\right )}{\left (d^4-64 a e^3\right ) \left (5 d^4+256 a e^3\right )^{3/4} \sqrt{8 a e^2-d^3 x+8 d e^2 x^3+8 e^3 x^4}}\\ \end{align*}
Mathematica [B] time = 6.15134, size = 7629, normalized size = 10.2 \[ \text{Result too large to show} \]
Warning: Unable to verify antiderivative.
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Maple [B] time = 0.04, size = 8103, normalized size = 10.8 \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{1}{{\left (8 \, e^{3} x^{4} + 8 \, d e^{2} x^{3} - d^{3} x + 8 \, a e^{2}\right )}^{\frac{3}{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{\sqrt{8 \, e^{3} x^{4} + 8 \, d e^{2} x^{3} - d^{3} x + 8 \, a e^{2}}}{64 \, e^{6} x^{8} + 128 \, d e^{5} x^{7} + 64 \, d^{2} e^{4} x^{6} - 16 \, d^{3} e^{3} x^{5} + 128 \, a d e^{4} x^{3} + d^{6} x^{2} - 16 \, a d^{3} e^{2} x + 64 \, a^{2} e^{4} - 16 \,{\left (d^{4} e^{2} - 8 \, a e^{5}\right )} x^{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{1}{\left (8 a e^{2} - d^{3} x + 8 d e^{2} x^{3} + 8 e^{3} x^{4}\right )^{\frac{3}{2}}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: RuntimeError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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