Integrand size = 28, antiderivative size = 141 \[ \int (a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx=-\frac {4 \sqrt {2} a^{5/2} (i A+B) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}+\frac {4 a^2 (i A+B) \sqrt {a+i a \tan (c+d x)}}{d}+\frac {2 a (i A+B) (a+i a \tan (c+d x))^{3/2}}{3 d}+\frac {2 B (a+i a \tan (c+d x))^{5/2}}{5 d} \]
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Time = 0.14 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3608, 3559, 3561, 212} \[ \int (a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx=-\frac {4 \sqrt {2} a^{5/2} (B+i A) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}+\frac {4 a^2 (B+i A) \sqrt {a+i a \tan (c+d x)}}{d}+\frac {2 a (B+i A) (a+i a \tan (c+d x))^{3/2}}{3 d}+\frac {2 B (a+i a \tan (c+d x))^{5/2}}{5 d} \]
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Rule 212
Rule 3559
Rule 3561
Rule 3608
Rubi steps \begin{align*} \text {integral}& = \frac {2 B (a+i a \tan (c+d x))^{5/2}}{5 d}-(-A+i B) \int (a+i a \tan (c+d x))^{5/2} \, dx \\ & = \frac {2 a (i A+B) (a+i a \tan (c+d x))^{3/2}}{3 d}+\frac {2 B (a+i a \tan (c+d x))^{5/2}}{5 d}+(2 a (A-i B)) \int (a+i a \tan (c+d x))^{3/2} \, dx \\ & = \frac {4 a^2 (i A+B) \sqrt {a+i a \tan (c+d x)}}{d}+\frac {2 a (i A+B) (a+i a \tan (c+d x))^{3/2}}{3 d}+\frac {2 B (a+i a \tan (c+d x))^{5/2}}{5 d}+\left (4 a^2 (A-i B)\right ) \int \sqrt {a+i a \tan (c+d x)} \, dx \\ & = \frac {4 a^2 (i A+B) \sqrt {a+i a \tan (c+d x)}}{d}+\frac {2 a (i A+B) (a+i a \tan (c+d x))^{3/2}}{3 d}+\frac {2 B (a+i a \tan (c+d x))^{5/2}}{5 d}-\frac {\left (8 a^3 (i A+B)\right ) \text {Subst}\left (\int \frac {1}{2 a-x^2} \, dx,x,\sqrt {a+i a \tan (c+d x)}\right )}{d} \\ & = -\frac {4 \sqrt {2} a^{5/2} (i A+B) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}+\frac {4 a^2 (i A+B) \sqrt {a+i a \tan (c+d x)}}{d}+\frac {2 a (i A+B) (a+i a \tan (c+d x))^{3/2}}{3 d}+\frac {2 B (a+i a \tan (c+d x))^{5/2}}{5 d} \\ \end{align*}
Time = 0.87 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.82 \[ \int (a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx=\frac {-60 i \sqrt {2} a^{5/2} (A-i B) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )+2 a^2 \sqrt {a+i a \tan (c+d x)} \left (35 i A+38 B+(-5 A+11 i B) \tan (c+d x)-3 B \tan ^2(c+d x)\right )}{15 d} \]
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Time = 0.13 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.00
method | result | size |
derivativedivides | \(\frac {2 i \left (-\frac {i B \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}}}{5}-\frac {i B a \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}{3}+\frac {A a \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}{3}-2 i a^{2} B \sqrt {a +i a \tan \left (d x +c \right )}+2 A \,a^{2} \sqrt {a +i a \tan \left (d x +c \right )}-2 a^{\frac {5}{2}} \left (-i B +A \right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )\right )}{d}\) | \(141\) |
default | \(\frac {2 i \left (-\frac {i B \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}}}{5}-\frac {i B a \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}{3}+\frac {A a \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}{3}-2 i a^{2} B \sqrt {a +i a \tan \left (d x +c \right )}+2 A \,a^{2} \sqrt {a +i a \tan \left (d x +c \right )}-2 a^{\frac {5}{2}} \left (-i B +A \right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )\right )}{d}\) | \(141\) |
parts | \(\frac {2 i A a \left (\frac {\left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}{3}+2 a \sqrt {a +i a \tan \left (d x +c \right )}-2 a^{\frac {3}{2}} \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )\right )}{d}+\frac {B \left (\frac {2 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}}}{5}+\frac {2 a \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}{3}+4 a^{2} \sqrt {a +i a \tan \left (d x +c \right )}-4 a^{\frac {5}{2}} \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )\right )}{d}\) | \(164\) |
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Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 421 vs. \(2 (108) = 216\).
Time = 0.25 (sec) , antiderivative size = 421, normalized size of antiderivative = 2.99 \[ \int (a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx=\frac {2 \, {\left (15 \, \sqrt {2} \sqrt {-\frac {{\left (A^{2} - 2 i \, A B - B^{2}\right )} a^{5}}{d^{2}}} {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \log \left (\frac {4 \, {\left ({\left (-i \, A - B\right )} a^{3} e^{\left (i \, d x + i \, c\right )} + \sqrt {-\frac {{\left (A^{2} - 2 i \, A B - B^{2}\right )} a^{5}}{d^{2}}} {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-i \, d x - i \, c\right )}}{{\left (-i \, A - B\right )} a^{2}}\right ) - 15 \, \sqrt {2} \sqrt {-\frac {{\left (A^{2} - 2 i \, A B - B^{2}\right )} a^{5}}{d^{2}}} {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \log \left (\frac {4 \, {\left ({\left (-i \, A - B\right )} a^{3} e^{\left (i \, d x + i \, c\right )} - \sqrt {-\frac {{\left (A^{2} - 2 i \, A B - B^{2}\right )} a^{5}}{d^{2}}} {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-i \, d x - i \, c\right )}}{{\left (-i \, A - B\right )} a^{2}}\right ) - 2 \, \sqrt {2} {\left (2 \, {\left (-10 i \, A - 13 \, B\right )} a^{2} e^{\left (5 i \, d x + 5 i \, c\right )} + 35 \, {\left (-i \, A - B\right )} a^{2} e^{\left (3 i \, d x + 3 i \, c\right )} + 15 \, {\left (-i \, A - B\right )} a^{2} e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )}}{15 \, {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]
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\[ \int (a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx=\int \left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{\frac {5}{2}} \left (A + B \tan {\left (c + d x \right )}\right )\, dx \]
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none
Time = 0.30 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.95 \[ \int (a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx=\frac {2 i \, {\left (15 \, \sqrt {2} {\left (A - i \, B\right )} a^{\frac {7}{2}} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {i \, a \tan \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {i \, a \tan \left (d x + c\right ) + a}}\right ) - 3 i \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}} B a + 5 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}} {\left (A - i \, B\right )} a^{2} + 30 \, \sqrt {i \, a \tan \left (d x + c\right ) + a} {\left (A - i \, B\right )} a^{3}\right )}}{15 \, a d} \]
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Timed out. \[ \int (a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx=\text {Timed out} \]
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Time = 1.16 (sec) , antiderivative size = 188, normalized size of antiderivative = 1.33 \[ \int (a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx=\frac {2\,B\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2}}{5\,d}+\frac {A\,a\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{3/2}\,2{}\mathrm {i}}{3\,d}+\frac {2\,B\,a\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{3/2}}{3\,d}+\frac {A\,a^2\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}\,4{}\mathrm {i}}{d}+\frac {4\,B\,a^2\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{d}-\frac {\sqrt {2}\,A\,{\left (-a\right )}^{5/2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {-a}}\right )\,4{}\mathrm {i}}{d}+\frac {\sqrt {2}\,B\,a^{5/2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2\,\sqrt {a}}\right )\,4{}\mathrm {i}}{d} \]
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