Optimal. Leaf size=215 \[ \frac {1}{5 d^2 x^3 (d+e x) \left (d^2-e^2 x^2\right )^{3/2}}-\frac {128 e^2 \sqrt {d^2-e^2 x^2}}{15 d^9 x}+\frac {7 e^3 \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{2 d^9}+\frac {7 e \sqrt {d^2-e^2 x^2}}{2 d^8 x^2}-\frac {64 \sqrt {d^2-e^2 x^2}}{15 d^7 x^3}+\frac {48 d-35 e x}{15 d^6 x^3 \sqrt {d^2-e^2 x^2}}+\frac {8 d-7 e x}{15 d^4 x^3 \left (d^2-e^2 x^2\right )^{3/2}} \]
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Rubi [A] time = 0.21, antiderivative size = 215, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.259, Rules used = {857, 823, 835, 807, 266, 63, 208} \begin {gather*} -\frac {128 e^2 \sqrt {d^2-e^2 x^2}}{15 d^9 x}+\frac {7 e \sqrt {d^2-e^2 x^2}}{2 d^8 x^2}-\frac {64 \sqrt {d^2-e^2 x^2}}{15 d^7 x^3}+\frac {48 d-35 e x}{15 d^6 x^3 \sqrt {d^2-e^2 x^2}}+\frac {8 d-7 e x}{15 d^4 x^3 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {1}{5 d^2 x^3 (d+e x) \left (d^2-e^2 x^2\right )^{3/2}}+\frac {7 e^3 \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{2 d^9} \end {gather*}
Antiderivative was successfully verified.
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Rule 63
Rule 208
Rule 266
Rule 807
Rule 823
Rule 835
Rule 857
Rubi steps
\begin {align*} \int \frac {1}{x^4 (d+e x) \left (d^2-e^2 x^2\right )^{5/2}} \, dx &=\frac {1}{5 d^2 x^3 (d+e x) \left (d^2-e^2 x^2\right )^{3/2}}-\frac {\int \frac {-8 d e^2+7 e^3 x}{x^4 \left (d^2-e^2 x^2\right )^{5/2}} \, dx}{5 d^2 e^2}\\ &=\frac {8 d-7 e x}{15 d^4 x^3 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {1}{5 d^2 x^3 (d+e x) \left (d^2-e^2 x^2\right )^{3/2}}-\frac {\int \frac {-48 d^3 e^4+35 d^2 e^5 x}{x^4 \left (d^2-e^2 x^2\right )^{3/2}} \, dx}{15 d^6 e^4}\\ &=\frac {8 d-7 e x}{15 d^4 x^3 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {1}{5 d^2 x^3 (d+e x) \left (d^2-e^2 x^2\right )^{3/2}}+\frac {48 d-35 e x}{15 d^6 x^3 \sqrt {d^2-e^2 x^2}}-\frac {\int \frac {-192 d^5 e^6+105 d^4 e^7 x}{x^4 \sqrt {d^2-e^2 x^2}} \, dx}{15 d^{10} e^6}\\ &=\frac {8 d-7 e x}{15 d^4 x^3 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {1}{5 d^2 x^3 (d+e x) \left (d^2-e^2 x^2\right )^{3/2}}+\frac {48 d-35 e x}{15 d^6 x^3 \sqrt {d^2-e^2 x^2}}-\frac {64 \sqrt {d^2-e^2 x^2}}{15 d^7 x^3}+\frac {\int \frac {-315 d^6 e^7+384 d^5 e^8 x}{x^3 \sqrt {d^2-e^2 x^2}} \, dx}{45 d^{12} e^6}\\ &=\frac {8 d-7 e x}{15 d^4 x^3 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {1}{5 d^2 x^3 (d+e x) \left (d^2-e^2 x^2\right )^{3/2}}+\frac {48 d-35 e x}{15 d^6 x^3 \sqrt {d^2-e^2 x^2}}-\frac {64 \sqrt {d^2-e^2 x^2}}{15 d^7 x^3}+\frac {7 e \sqrt {d^2-e^2 x^2}}{2 d^8 x^2}-\frac {\int \frac {-768 d^7 e^8+315 d^6 e^9 x}{x^2 \sqrt {d^2-e^2 x^2}} \, dx}{90 d^{14} e^6}\\ &=\frac {8 d-7 e x}{15 d^4 x^3 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {1}{5 d^2 x^3 (d+e x) \left (d^2-e^2 x^2\right )^{3/2}}+\frac {48 d-35 e x}{15 d^6 x^3 \sqrt {d^2-e^2 x^2}}-\frac {64 \sqrt {d^2-e^2 x^2}}{15 d^7 x^3}+\frac {7 e \sqrt {d^2-e^2 x^2}}{2 d^8 x^2}-\frac {128 e^2 \sqrt {d^2-e^2 x^2}}{15 d^9 x}-\frac {\left (7 e^3\right ) \int \frac {1}{x \sqrt {d^2-e^2 x^2}} \, dx}{2 d^8}\\ &=\frac {8 d-7 e x}{15 d^4 x^3 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {1}{5 d^2 x^3 (d+e x) \left (d^2-e^2 x^2\right )^{3/2}}+\frac {48 d-35 e x}{15 d^6 x^3 \sqrt {d^2-e^2 x^2}}-\frac {64 \sqrt {d^2-e^2 x^2}}{15 d^7 x^3}+\frac {7 e \sqrt {d^2-e^2 x^2}}{2 d^8 x^2}-\frac {128 e^2 \sqrt {d^2-e^2 x^2}}{15 d^9 x}-\frac {\left (7 e^3\right ) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {d^2-e^2 x}} \, dx,x,x^2\right )}{4 d^8}\\ &=\frac {8 d-7 e x}{15 d^4 x^3 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {1}{5 d^2 x^3 (d+e x) \left (d^2-e^2 x^2\right )^{3/2}}+\frac {48 d-35 e x}{15 d^6 x^3 \sqrt {d^2-e^2 x^2}}-\frac {64 \sqrt {d^2-e^2 x^2}}{15 d^7 x^3}+\frac {7 e \sqrt {d^2-e^2 x^2}}{2 d^8 x^2}-\frac {128 e^2 \sqrt {d^2-e^2 x^2}}{15 d^9 x}+\frac {(7 e) \operatorname {Subst}\left (\int \frac {1}{\frac {d^2}{e^2}-\frac {x^2}{e^2}} \, dx,x,\sqrt {d^2-e^2 x^2}\right )}{2 d^8}\\ &=\frac {8 d-7 e x}{15 d^4 x^3 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {1}{5 d^2 x^3 (d+e x) \left (d^2-e^2 x^2\right )^{3/2}}+\frac {48 d-35 e x}{15 d^6 x^3 \sqrt {d^2-e^2 x^2}}-\frac {64 \sqrt {d^2-e^2 x^2}}{15 d^7 x^3}+\frac {7 e \sqrt {d^2-e^2 x^2}}{2 d^8 x^2}-\frac {128 e^2 \sqrt {d^2-e^2 x^2}}{15 d^9 x}+\frac {7 e^3 \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{2 d^9}\\ \end {align*}
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Mathematica [A] time = 0.16, size = 148, normalized size = 0.69 \begin {gather*} -\frac {-105 e^3 \log \left (\sqrt {d^2-e^2 x^2}+d\right )+\frac {\sqrt {d^2-e^2 x^2} \left (10 d^7-5 d^6 e x+75 d^5 e^2 x^2+236 d^4 e^3 x^3-244 d^3 e^4 x^4-489 d^2 e^5 x^5+151 d e^6 x^6+256 e^7 x^7\right )}{x^3 (d-e x)^2 (d+e x)^3}+105 e^3 \log (x)}{30 d^9} \end {gather*}
Antiderivative was successfully verified.
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IntegrateAlgebraic [A] time = 1.42, size = 161, normalized size = 0.75 \begin {gather*} \frac {\sqrt {d^2-e^2 x^2} \left (-10 d^7+5 d^6 e x-75 d^5 e^2 x^2-236 d^4 e^3 x^3+244 d^3 e^4 x^4+489 d^2 e^5 x^5-151 d e^6 x^6-256 e^7 x^7\right )}{30 d^9 x^3 (d-e x)^2 (d+e x)^3}-\frac {7 e^3 \tanh ^{-1}\left (\frac {\sqrt {-e^2} x}{d}-\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{d^9} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.52, size = 297, normalized size = 1.38 \begin {gather*} -\frac {116 \, e^{8} x^{8} + 116 \, d e^{7} x^{7} - 232 \, d^{2} e^{6} x^{6} - 232 \, d^{3} e^{5} x^{5} + 116 \, d^{4} e^{4} x^{4} + 116 \, d^{5} e^{3} x^{3} + 105 \, {\left (e^{8} x^{8} + d e^{7} x^{7} - 2 \, d^{2} e^{6} x^{6} - 2 \, d^{3} e^{5} x^{5} + d^{4} e^{4} x^{4} + d^{5} e^{3} x^{3}\right )} \log \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{x}\right ) + {\left (256 \, e^{7} x^{7} + 151 \, d e^{6} x^{6} - 489 \, d^{2} e^{5} x^{5} - 244 \, d^{3} e^{4} x^{4} + 236 \, d^{4} e^{3} x^{3} + 75 \, d^{5} e^{2} x^{2} - 5 \, d^{6} e x + 10 \, d^{7}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{30 \, {\left (d^{9} e^{5} x^{8} + d^{10} e^{4} x^{7} - 2 \, d^{11} e^{3} x^{6} - 2 \, d^{12} e^{2} x^{5} + d^{13} e x^{4} + d^{14} x^{3}\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.02, size = 326, normalized size = 1.52 \begin {gather*} \frac {4 e^{4} x}{\left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}} d^{7}}+\frac {4 e^{4} x}{15 \left (2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}\right )^{\frac {3}{2}} d^{7}}-\frac {e^{2}}{5 \left (x +\frac {d}{e}\right ) \left (2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}\right )^{\frac {3}{2}} d^{5}}-\frac {7 e^{3}}{6 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}} d^{6}}-\frac {3 e^{2}}{\left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}} d^{5} x}+\frac {7 e^{3} \ln \left (\frac {2 d^{2}+2 \sqrt {d^{2}}\, \sqrt {-e^{2} x^{2}+d^{2}}}{x}\right )}{2 \sqrt {d^{2}}\, d^{8}}+\frac {8 e^{4} x}{\sqrt {-e^{2} x^{2}+d^{2}}\, d^{9}}+\frac {8 e^{4} x}{15 \sqrt {2 \left (x +\frac {d}{e}\right ) d e -\left (x +\frac {d}{e}\right )^{2} e^{2}}\, d^{9}}+\frac {e}{2 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}} d^{4} x^{2}}-\frac {7 e^{3}}{2 \sqrt {-e^{2} x^{2}+d^{2}}\, d^{8}}-\frac {1}{3 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}} d^{3} x^{3}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} {\left (e x + d\right )} x^{4}}\,{d x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {1}{x^4\,{\left (d^2-e^2\,x^2\right )}^{5/2}\,\left (d+e\,x\right )} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{x^{4} \left (- \left (- d + e x\right ) \left (d + e x\right )\right )^{\frac {5}{2}} \left (d + e x\right )}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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