Optimal. Leaf size=182 \[ \frac {2 e^2 (10 d+11 e x)}{5 d^7 \sqrt {d^2-e^2 x^2}}-\frac {2 e \sqrt {d^2-e^2 x^2}}{d^7 x}-\frac {9 e^2 \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{2 d^7}-\frac {\sqrt {d^2-e^2 x^2}}{2 d^6 x^2}+\frac {e^2 (5 d+6 e x)}{5 d^5 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {2 e^2 (d+e x)}{5 d^3 \left (d^2-e^2 x^2\right )^{5/2}} \]
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Rubi [A] time = 0.36, antiderivative size = 182, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {1805, 1807, 807, 266, 63, 208} \[ \frac {2 e^2 (10 d+11 e x)}{5 d^7 \sqrt {d^2-e^2 x^2}}+\frac {e^2 (5 d+6 e x)}{5 d^5 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {2 e^2 (d+e x)}{5 d^3 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {2 e \sqrt {d^2-e^2 x^2}}{d^7 x}-\frac {\sqrt {d^2-e^2 x^2}}{2 d^6 x^2}-\frac {9 e^2 \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{2 d^7} \]
Antiderivative was successfully verified.
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Rule 63
Rule 208
Rule 266
Rule 807
Rule 1805
Rule 1807
Rubi steps
\begin {align*} \int \frac {(d+e x)^2}{x^3 \left (d^2-e^2 x^2\right )^{7/2}} \, dx &=\frac {2 e^2 (d+e x)}{5 d^3 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {\int \frac {-5 d^2-10 d e x-10 e^2 x^2-\frac {8 e^3 x^3}{d}}{x^3 \left (d^2-e^2 x^2\right )^{5/2}} \, dx}{5 d^2}\\ &=\frac {2 e^2 (d+e x)}{5 d^3 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {e^2 (5 d+6 e x)}{5 d^5 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {\int \frac {15 d^2+30 d e x+45 e^2 x^2+\frac {36 e^3 x^3}{d}}{x^3 \left (d^2-e^2 x^2\right )^{3/2}} \, dx}{15 d^4}\\ &=\frac {2 e^2 (d+e x)}{5 d^3 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {e^2 (5 d+6 e x)}{5 d^5 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {2 e^2 (10 d+11 e x)}{5 d^7 \sqrt {d^2-e^2 x^2}}-\frac {\int \frac {-15 d^2-30 d e x-60 e^2 x^2}{x^3 \sqrt {d^2-e^2 x^2}} \, dx}{15 d^6}\\ &=\frac {2 e^2 (d+e x)}{5 d^3 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {e^2 (5 d+6 e x)}{5 d^5 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {2 e^2 (10 d+11 e x)}{5 d^7 \sqrt {d^2-e^2 x^2}}-\frac {\sqrt {d^2-e^2 x^2}}{2 d^6 x^2}+\frac {\int \frac {60 d^3 e+135 d^2 e^2 x}{x^2 \sqrt {d^2-e^2 x^2}} \, dx}{30 d^8}\\ &=\frac {2 e^2 (d+e x)}{5 d^3 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {e^2 (5 d+6 e x)}{5 d^5 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {2 e^2 (10 d+11 e x)}{5 d^7 \sqrt {d^2-e^2 x^2}}-\frac {\sqrt {d^2-e^2 x^2}}{2 d^6 x^2}-\frac {2 e \sqrt {d^2-e^2 x^2}}{d^7 x}+\frac {\left (9 e^2\right ) \int \frac {1}{x \sqrt {d^2-e^2 x^2}} \, dx}{2 d^6}\\ &=\frac {2 e^2 (d+e x)}{5 d^3 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {e^2 (5 d+6 e x)}{5 d^5 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {2 e^2 (10 d+11 e x)}{5 d^7 \sqrt {d^2-e^2 x^2}}-\frac {\sqrt {d^2-e^2 x^2}}{2 d^6 x^2}-\frac {2 e \sqrt {d^2-e^2 x^2}}{d^7 x}+\frac {\left (9 e^2\right ) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {d^2-e^2 x}} \, dx,x,x^2\right )}{4 d^6}\\ &=\frac {2 e^2 (d+e x)}{5 d^3 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {e^2 (5 d+6 e x)}{5 d^5 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {2 e^2 (10 d+11 e x)}{5 d^7 \sqrt {d^2-e^2 x^2}}-\frac {\sqrt {d^2-e^2 x^2}}{2 d^6 x^2}-\frac {2 e \sqrt {d^2-e^2 x^2}}{d^7 x}-\frac {9 \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^6}\\ &=\frac {2 e^2 (d+e x)}{5 d^3 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {e^2 (5 d+6 e x)}{5 d^5 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {2 e^2 (10 d+11 e x)}{5 d^7 \sqrt {d^2-e^2 x^2}}-\frac {\sqrt {d^2-e^2 x^2}}{2 d^6 x^2}-\frac {2 e \sqrt {d^2-e^2 x^2}}{d^7 x}-\frac {9 e^2 \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{2 d^7}\\ \end {align*}
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Mathematica [C] time = 0.06, size = 117, normalized size = 0.64 \[ \frac {e \left (-10 d^6+60 d^4 e^2 x^2-80 d^2 e^4 x^4+d^5 e x \, _2F_1\left (-\frac {5}{2},1;-\frac {3}{2};1-\frac {e^2 x^2}{d^2}\right )+d^5 e x \, _2F_1\left (-\frac {5}{2},2;-\frac {3}{2};1-\frac {e^2 x^2}{d^2}\right )+32 e^6 x^6\right )}{5 d^7 x \left (d^2-e^2 x^2\right )^{5/2}} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.86, size = 216, normalized size = 1.19 \[ \frac {54 \, e^{6} x^{6} - 108 \, d e^{5} x^{5} + 108 \, d^{3} e^{3} x^{3} - 54 \, d^{4} e^{2} x^{2} + 45 \, {\left (e^{6} x^{6} - 2 \, d e^{5} x^{5} + 2 \, d^{3} e^{3} x^{3} - d^{4} e^{2} x^{2}\right )} \log \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{x}\right ) - {\left (64 \, e^{5} x^{5} - 83 \, d e^{4} x^{4} - 58 \, d^{2} e^{3} x^{3} + 94 \, d^{3} e^{2} x^{2} - 10 \, d^{4} e x - 5 \, d^{5}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{10 \, {\left (d^{7} e^{4} x^{6} - 2 \, d^{8} e^{3} x^{5} + 2 \, d^{10} e x^{3} - d^{11} x^{2}\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.30, size = 260, normalized size = 1.43 \[ -\frac {\sqrt {-x^{2} e^{2} + d^{2}} {\left ({\left ({\left (2 \, {\left (x {\left (\frac {11 \, x e^{7}}{d^{7}} + \frac {10 \, e^{6}}{d^{6}}\right )} - \frac {25 \, e^{5}}{d^{5}}\right )} x - \frac {45 \, e^{4}}{d^{4}}\right )} x + \frac {30 \, e^{3}}{d^{3}}\right )} x + \frac {27 \, e^{2}}{d^{2}}\right )}}{5 \, {\left (x^{2} e^{2} - d^{2}\right )}^{3}} - \frac {9 \, e^{2} \log \left (\frac {{\left | -2 \, d e - 2 \, \sqrt {-x^{2} e^{2} + d^{2}} e \right |} e^{\left (-2\right )}}{2 \, {\left | x \right |}}\right )}{2 \, d^{7}} + \frac {x^{2} {\left (\frac {8 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )} e^{4}}{x} + e^{6}\right )}}{8 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{2} d^{7}} - \frac {{\left (\frac {8 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )} d^{7} e^{8}}{x} + \frac {{\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{2} d^{7} e^{6}}{x^{2}}\right )} e^{\left (-8\right )}}{8 \, d^{14}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.02, size = 224, normalized size = 1.23 \[ \frac {12 e^{3} x}{5 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}} d^{3}}+\frac {9 e^{2}}{10 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}} d^{2}}-\frac {2 e}{\left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}} d x}+\frac {16 e^{3} x}{5 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}} d^{5}}+\frac {3 e^{2}}{2 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}} d^{4}}-\frac {1}{2 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}} x^{2}}-\frac {9 e^{2} \ln \left (\frac {2 d^{2}+2 \sqrt {d^{2}}\, \sqrt {-e^{2} x^{2}+d^{2}}}{x}\right )}{2 \sqrt {d^{2}}\, d^{6}}+\frac {32 e^{3} x}{5 \sqrt {-e^{2} x^{2}+d^{2}}\, d^{7}}+\frac {9 e^{2}}{2 \sqrt {-e^{2} x^{2}+d^{2}}\, d^{6}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.47, size = 218, normalized size = 1.20 \[ \frac {12 \, e^{3} x}{5 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} d^{3}} + \frac {9 \, e^{2}}{10 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} d^{2}} + \frac {16 \, e^{3} x}{5 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d^{5}} + \frac {3 \, e^{2}}{2 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d^{4}} - \frac {2 \, e}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} d x} + \frac {32 \, e^{3} x}{5 \, \sqrt {-e^{2} x^{2} + d^{2}} d^{7}} - \frac {9 \, e^{2} \log \left (\frac {2 \, d^{2}}{{\left | x \right |}} + \frac {2 \, \sqrt {-e^{2} x^{2} + d^{2}} d}{{\left | x \right |}}\right )}{2 \, d^{7}} + \frac {9 \, e^{2}}{2 \, \sqrt {-e^{2} x^{2} + d^{2}} d^{6}} - \frac {1}{2 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} x^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (d+e\,x\right )}^2}{x^3\,{\left (d^2-e^2\,x^2\right )}^{7/2}} \,d x \]
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 \[ \int \frac {\left (d + e x\right )^{2}}{x^{3} \left (- \left (- d + e x\right ) \left (d + e x\right )\right )^{\frac {7}{2}}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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