Integrand size = 23, antiderivative size = 115 \[ \int \frac {e^{-\arctan (a x)}}{\left (c+a^2 c x^2\right )^{7/2}} \, dx=-\frac {e^{-\arctan (a x)} (1-5 a x)}{26 a c \left (c+a^2 c x^2\right )^{5/2}}-\frac {e^{-\arctan (a x)} (1-3 a x)}{13 a c^2 \left (c+a^2 c x^2\right )^{3/2}}-\frac {3 e^{-\arctan (a x)} (1-a x)}{13 a c^3 \sqrt {c+a^2 c x^2}} \]
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Time = 0.09 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {5178, 5177} \[ \int \frac {e^{-\arctan (a x)}}{\left (c+a^2 c x^2\right )^{7/2}} \, dx=-\frac {3 (1-a x) e^{-\arctan (a x)}}{13 a c^3 \sqrt {a^2 c x^2+c}}-\frac {(1-3 a x) e^{-\arctan (a x)}}{13 a c^2 \left (a^2 c x^2+c\right )^{3/2}}-\frac {(1-5 a x) e^{-\arctan (a x)}}{26 a c \left (a^2 c x^2+c\right )^{5/2}} \]
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Rule 5177
Rule 5178
Rubi steps \begin{align*} \text {integral}& = -\frac {e^{-\arctan (a x)} (1-5 a x)}{26 a c \left (c+a^2 c x^2\right )^{5/2}}+\frac {10 \int \frac {e^{-\arctan (a x)}}{\left (c+a^2 c x^2\right )^{5/2}} \, dx}{13 c} \\ & = -\frac {e^{-\arctan (a x)} (1-5 a x)}{26 a c \left (c+a^2 c x^2\right )^{5/2}}-\frac {e^{-\arctan (a x)} (1-3 a x)}{13 a c^2 \left (c+a^2 c x^2\right )^{3/2}}+\frac {6 \int \frac {e^{-\arctan (a x)}}{\left (c+a^2 c x^2\right )^{3/2}} \, dx}{13 c^2} \\ & = -\frac {e^{-\arctan (a x)} (1-5 a x)}{26 a c \left (c+a^2 c x^2\right )^{5/2}}-\frac {e^{-\arctan (a x)} (1-3 a x)}{13 a c^2 \left (c+a^2 c x^2\right )^{3/2}}-\frac {3 e^{-\arctan (a x)} (1-a x)}{13 a c^3 \sqrt {c+a^2 c x^2}} \\ \end{align*}
Time = 0.04 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.70 \[ \int \frac {e^{-\arctan (a x)}}{\left (c+a^2 c x^2\right )^{7/2}} \, dx=\frac {e^{-\arctan (a x)} \left (-9+17 a x-14 a^2 x^2+18 a^3 x^3-6 a^4 x^4+6 a^5 x^5\right )}{26 a c^3 \left (1+a^2 x^2\right )^2 \sqrt {c+a^2 c x^2}} \]
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Time = 0.21 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.63
method | result | size |
gosper | \(\frac {\left (a^{2} x^{2}+1\right ) \left (6 a^{5} x^{5}-6 a^{4} x^{4}+18 a^{3} x^{3}-14 a^{2} x^{2}+17 a x -9\right ) {\mathrm e}^{-\arctan \left (a x \right )}}{26 a \left (a^{2} c \,x^{2}+c \right )^{\frac {7}{2}}}\) | \(72\) |
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Time = 0.29 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.86 \[ \int \frac {e^{-\arctan (a x)}}{\left (c+a^2 c x^2\right )^{7/2}} \, dx=\frac {{\left (6 \, a^{5} x^{5} - 6 \, a^{4} x^{4} + 18 \, a^{3} x^{3} - 14 \, a^{2} x^{2} + 17 \, a x - 9\right )} \sqrt {a^{2} c x^{2} + c} e^{\left (-\arctan \left (a x\right )\right )}}{26 \, {\left (a^{7} c^{4} x^{6} + 3 \, a^{5} c^{4} x^{4} + 3 \, a^{3} c^{4} x^{2} + a c^{4}\right )}} \]
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Timed out. \[ \int \frac {e^{-\arctan (a x)}}{\left (c+a^2 c x^2\right )^{7/2}} \, dx=\text {Timed out} \]
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\[ \int \frac {e^{-\arctan (a x)}}{\left (c+a^2 c x^2\right )^{7/2}} \, dx=\int { \frac {e^{\left (-\arctan \left (a x\right )\right )}}{{\left (a^{2} c x^{2} + c\right )}^{\frac {7}{2}}} \,d x } \]
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\[ \int \frac {e^{-\arctan (a x)}}{\left (c+a^2 c x^2\right )^{7/2}} \, dx=\int { \frac {e^{\left (-\arctan \left (a x\right )\right )}}{{\left (a^{2} c x^{2} + c\right )}^{\frac {7}{2}}} \,d x } \]
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Time = 0.76 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.07 \[ \int \frac {e^{-\arctan (a x)}}{\left (c+a^2 c x^2\right )^{7/2}} \, dx=-\frac {{\mathrm {e}}^{-\mathrm {atan}\left (a\,x\right )}\,\left (\frac {9}{26\,a^5\,c^3}-\frac {3\,x^5}{13\,c^3}-\frac {17\,x}{26\,a^4\,c^3}+\frac {3\,x^4}{13\,a\,c^3}-\frac {9\,x^3}{13\,a^2\,c^3}+\frac {7\,x^2}{13\,a^3\,c^3}\right )}{\frac {\sqrt {c\,a^2\,x^2+c}}{a^4}+x^4\,\sqrt {c\,a^2\,x^2+c}+\frac {2\,x^2\,\sqrt {c\,a^2\,x^2+c}}{a^2}} \]
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