Integrand size = 26, antiderivative size = 146 \[ \int \frac {\sec ^{10}(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=-\frac {32 i (a+i a \tan (c+d x))^{5/2}}{5 a^5 d}+\frac {64 i (a+i a \tan (c+d x))^{7/2}}{7 a^6 d}-\frac {16 i (a+i a \tan (c+d x))^{9/2}}{3 a^7 d}+\frac {16 i (a+i a \tan (c+d x))^{11/2}}{11 a^8 d}-\frac {2 i (a+i a \tan (c+d x))^{13/2}}{13 a^9 d} \]
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Time = 0.12 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {3568, 45} \[ \int \frac {\sec ^{10}(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=-\frac {2 i (a+i a \tan (c+d x))^{13/2}}{13 a^9 d}+\frac {16 i (a+i a \tan (c+d x))^{11/2}}{11 a^8 d}-\frac {16 i (a+i a \tan (c+d x))^{9/2}}{3 a^7 d}+\frac {64 i (a+i a \tan (c+d x))^{7/2}}{7 a^6 d}-\frac {32 i (a+i a \tan (c+d x))^{5/2}}{5 a^5 d} \]
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Rule 45
Rule 3568
Rubi steps \begin{align*} \text {integral}& = -\frac {i \text {Subst}\left (\int (a-x)^4 (a+x)^{3/2} \, dx,x,i a \tan (c+d x)\right )}{a^9 d} \\ & = -\frac {i \text {Subst}\left (\int \left (16 a^4 (a+x)^{3/2}-32 a^3 (a+x)^{5/2}+24 a^2 (a+x)^{7/2}-8 a (a+x)^{9/2}+(a+x)^{11/2}\right ) \, dx,x,i a \tan (c+d x)\right )}{a^9 d} \\ & = -\frac {32 i (a+i a \tan (c+d x))^{5/2}}{5 a^5 d}+\frac {64 i (a+i a \tan (c+d x))^{7/2}}{7 a^6 d}-\frac {16 i (a+i a \tan (c+d x))^{9/2}}{3 a^7 d}+\frac {16 i (a+i a \tan (c+d x))^{11/2}}{11 a^8 d}-\frac {2 i (a+i a \tan (c+d x))^{13/2}}{13 a^9 d} \\ \end{align*}
Time = 0.40 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.58 \[ \int \frac {\sec ^{10}(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\frac {2 (-i+\tan (c+d x))^2 \sqrt {a+i a \tan (c+d x)} \left (9683 i+16700 \tan (c+d x)-14210 i \tan ^2(c+d x)-6300 \tan ^3(c+d x)+1155 i \tan ^4(c+d x)\right )}{15015 a^3 d} \]
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Time = 1.29 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.69
method | result | size |
derivativedivides | \(\frac {2 i \left (-\frac {\left (a +i a \tan \left (d x +c \right )\right )^{\frac {13}{2}}}{13}+\frac {8 a \left (a +i a \tan \left (d x +c \right )\right )^{\frac {11}{2}}}{11}-\frac {8 a^{2} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {9}{2}}}{3}+\frac {32 a^{3} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {7}{2}}}{7}-\frac {16 a^{4} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}}}{5}\right )}{d \,a^{9}}\) | \(101\) |
default | \(\frac {2 i \left (-\frac {\left (a +i a \tan \left (d x +c \right )\right )^{\frac {13}{2}}}{13}+\frac {8 a \left (a +i a \tan \left (d x +c \right )\right )^{\frac {11}{2}}}{11}-\frac {8 a^{2} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {9}{2}}}{3}+\frac {32 a^{3} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {7}{2}}}{7}-\frac {16 a^{4} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}}}{5}\right )}{d \,a^{9}}\) | \(101\) |
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none
Time = 0.27 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.20 \[ \int \frac {\sec ^{10}(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=-\frac {128 \, \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (128 i \, e^{\left (13 i \, d x + 13 i \, c\right )} + 832 i \, e^{\left (11 i \, d x + 11 i \, c\right )} + 2288 i \, e^{\left (9 i \, d x + 9 i \, c\right )} + 3432 i \, e^{\left (7 i \, d x + 7 i \, c\right )} + 3003 i \, e^{\left (5 i \, d x + 5 i \, c\right )}\right )}}{15015 \, {\left (a^{3} d e^{\left (12 i \, d x + 12 i \, c\right )} + 6 \, a^{3} d e^{\left (10 i \, d x + 10 i \, c\right )} + 15 \, a^{3} d e^{\left (8 i \, d x + 8 i \, c\right )} + 20 \, a^{3} d e^{\left (6 i \, d x + 6 i \, c\right )} + 15 \, a^{3} d e^{\left (4 i \, d x + 4 i \, c\right )} + 6 \, a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{3} d\right )}} \]
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Timed out. \[ \int \frac {\sec ^{10}(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\text {Timed out} \]
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Time = 0.24 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.64 \[ \int \frac {\sec ^{10}(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=-\frac {2 i \, {\left (1155 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {13}{2}} - 10920 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {11}{2}} a + 40040 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {9}{2}} a^{2} - 68640 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {7}{2}} a^{3} + 48048 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}} a^{4}\right )}}{15015 \, a^{9} d} \]
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\[ \int \frac {\sec ^{10}(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\int { \frac {\sec \left (d x + c\right )^{10}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}}} \,d x } \]
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Time = 10.15 (sec) , antiderivative size = 434, normalized size of antiderivative = 2.97 \[ \int \frac {\sec ^{10}(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=-\frac {\sqrt {a-\frac {a\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1}}\,16384{}\mathrm {i}}{15015\,a^3\,d}-\frac {\sqrt {a-\frac {a\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1}}\,8192{}\mathrm {i}}{15015\,a^3\,d\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}-\frac {\sqrt {a-\frac {a\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1}}\,2048{}\mathrm {i}}{5005\,a^3\,d\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^2}-\frac {\sqrt {a-\frac {a\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1}}\,1024{}\mathrm {i}}{3003\,a^3\,d\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^3}-\frac {\sqrt {a-\frac {a\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1}}\,128{}\mathrm {i}}{429\,a^3\,d\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^4}+\frac {\sqrt {a-\frac {a\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1}}\,1792{}\mathrm {i}}{143\,a^3\,d\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^5}-\frac {\sqrt {a-\frac {a\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1}}\,128{}\mathrm {i}}{13\,a^3\,d\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^6} \]
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