Integrand size = 26, antiderivative size = 57 \[ \int \frac {\sec ^4(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=-\frac {4 i \sqrt {a+i a \tan (c+d x)}}{a^2 d}+\frac {2 i (a+i a \tan (c+d x))^{3/2}}{3 a^3 d} \]
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Time = 0.10 (sec) , antiderivative size = 57, 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} \[ \int \frac {\sec ^4(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=\frac {2 i (a+i a \tan (c+d x))^{3/2}}{3 a^3 d}-\frac {4 i \sqrt {a+i a \tan (c+d x)}}{a^2 d} \]
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Rule 45
Rule 3568
Rubi steps \begin{align*} \text {integral}& = -\frac {i \text {Subst}\left (\int \frac {a-x}{\sqrt {a+x}} \, dx,x,i a \tan (c+d x)\right )}{a^3 d} \\ & = -\frac {i \text {Subst}\left (\int \left (\frac {2 a}{\sqrt {a+x}}-\sqrt {a+x}\right ) \, dx,x,i a \tan (c+d x)\right )}{a^3 d} \\ & = -\frac {4 i \sqrt {a+i a \tan (c+d x)}}{a^2 d}+\frac {2 i (a+i a \tan (c+d x))^{3/2}}{3 a^3 d} \\ \end{align*}
Time = 0.11 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.65 \[ \int \frac {\sec ^4(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=-\frac {2 (5 i+\tan (c+d x)) \sqrt {a+i a \tan (c+d x)}}{3 a^2 d} \]
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Time = 1.00 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.77
method | result | size |
derivativedivides | \(\frac {2 i \left (\frac {\left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}{3}-2 a \sqrt {a +i a \tan \left (d x +c \right )}\right )}{d \,a^{3}}\) | \(44\) |
default | \(\frac {2 i \left (\frac {\left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}{3}-2 a \sqrt {a +i a \tan \left (d x +c \right )}\right )}{d \,a^{3}}\) | \(44\) |
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Time = 0.24 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.18 \[ \int \frac {\sec ^4(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=-\frac {4 \, \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (2 i \, e^{\left (3 i \, d x + 3 i \, c\right )} + 3 i \, e^{\left (i \, d x + i \, c\right )}\right )}}{3 \, {\left (a^{2} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2} d\right )}} \]
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\[ \int \frac {\sec ^4(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=\int \frac {\sec ^{4}{\left (c + d x \right )}}{\left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{\frac {3}{2}}}\, dx \]
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Time = 0.23 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.67 \[ \int \frac {\sec ^4(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=\frac {2 i \, {\left ({\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}} - 6 \, \sqrt {i \, a \tan \left (d x + c\right ) + a} a\right )}}{3 \, a^{3} d} \]
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\[ \int \frac {\sec ^4(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=\int { \frac {\sec \left (d x + c\right )^{4}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}}} \,d x } \]
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Time = 4.71 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.49 \[ \int \frac {\sec ^4(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=-\frac {2\,\left (\cos \left (2\,c+2\,d\,x\right )\,5{}\mathrm {i}+\sin \left (2\,c+2\,d\,x\right )+5{}\mathrm {i}\right )\,\sqrt {\frac {a\,\left (\cos \left (2\,c+2\,d\,x\right )+1+\sin \left (2\,c+2\,d\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,c+2\,d\,x\right )+1}}}{3\,a^2\,d\,\left (\cos \left (2\,c+2\,d\,x\right )+1\right )} \]
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