Integrand size = 26, antiderivative size = 59 \[ \int \frac {\sec ^4(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx=-\frac {4 i (a+i a \tan (c+d x))^{3/2}}{3 a^2 d}+\frac {2 i (a+i a \tan (c+d x))^{5/2}}{5 a^3 d} \]
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Time = 0.09 (sec) , antiderivative size = 59, 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} \[ \int \frac {\sec ^4(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx=\frac {2 i (a+i a \tan (c+d x))^{5/2}}{5 a^3 d}-\frac {4 i (a+i a \tan (c+d x))^{3/2}}{3 a^2 d} \]
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Rule 45
Rule 3568
Rubi steps \begin{align*} \text {integral}& = -\frac {i \text {Subst}\left (\int (a-x) \sqrt {a+x} \, dx,x,i a \tan (c+d x)\right )}{a^3 d} \\ & = -\frac {i \text {Subst}\left (\int \left (2 a \sqrt {a+x}-(a+x)^{3/2}\right ) \, dx,x,i a \tan (c+d x)\right )}{a^3 d} \\ & = -\frac {4 i (a+i a \tan (c+d x))^{3/2}}{3 a^2 d}+\frac {2 i (a+i a \tan (c+d x))^{5/2}}{5 a^3 d} \\ \end{align*}
Time = 0.12 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.83 \[ \int \frac {\sec ^4(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx=\frac {2 (7-3 i \tan (c+d x)) (-i+\tan (c+d x)) \sqrt {a+i a \tan (c+d x)}}{15 a d} \]
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Time = 1.12 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.75
method | result | size |
derivativedivides | \(\frac {2 i \left (\frac {\left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}}}{5}-\frac {2 a \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}{3}\right )}{d \,a^{3}}\) | \(44\) |
default | \(\frac {2 i \left (\frac {\left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}}}{5}-\frac {2 a \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}{3}\right )}{d \,a^{3}}\) | \(44\) |
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none
Time = 0.24 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.29 \[ \int \frac {\sec ^4(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx=-\frac {8 \, \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (2 i \, e^{\left (5 i \, d x + 5 i \, c\right )} + 5 i \, e^{\left (3 i \, d x + 3 i \, c\right )}\right )}}{15 \, {\left (a d e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, a d e^{\left (2 i \, d x + 2 i \, c\right )} + a d\right )}} \]
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\[ \int \frac {\sec ^4(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx=\int \frac {\sec ^{4}{\left (c + d x \right )}}{\sqrt {i a \left (\tan {\left (c + d x \right )} - i\right )}}\, dx \]
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none
Time = 0.23 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.34 \[ \int \frac {\sec ^4(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx=-\frac {2 i \, {\left (15 \, \sqrt {i \, a \tan \left (d x + c\right ) + a} - \frac {3 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}} - 10 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}} a + 15 \, \sqrt {i \, a \tan \left (d x + c\right ) + a} a^{2}}{a^{2}}\right )}}{15 \, a d} \]
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\[ \int \frac {\sec ^4(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx=\int { \frac {\sec \left (d x + c\right )^{4}}{\sqrt {i \, a \tan \left (d x + c\right ) + a}} \,d x } \]
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Time = 1.43 (sec) , antiderivative size = 155, normalized size of antiderivative = 2.63 \[ \int \frac {\sec ^4(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx=-\frac {8\,\sqrt {\frac {a\,\left (\cos \left (2\,c+2\,d\,x\right )+1+\sin \left (2\,c+2\,d\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,c+2\,d\,x\right )+1}}\,\left (\cos \left (2\,c+2\,d\,x\right )\,27{}\mathrm {i}+\cos \left (4\,c+4\,d\,x\right )\,9{}\mathrm {i}+\cos \left (6\,c+6\,d\,x\right )\,1{}\mathrm {i}-5\,\sin \left (2\,c+2\,d\,x\right )-4\,\sin \left (4\,c+4\,d\,x\right )-\sin \left (6\,c+6\,d\,x\right )+19{}\mathrm {i}\right )}{15\,a\,d\,\left (15\,\cos \left (2\,c+2\,d\,x\right )+6\,\cos \left (4\,c+4\,d\,x\right )+\cos \left (6\,c+6\,d\,x\right )+10\right )} \]
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