Integrand size = 36, antiderivative size = 154 \[ \int \frac {(a+i a \tan (c+d x))^2 (A+B \tan (c+d x))}{\tan ^{\frac {9}{2}}(c+d x)} \, dx=-\frac {4 \sqrt [4]{-1} a^2 (A-i B) \arctan \left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )}{d}-\frac {2 a^2 (9 i A+7 B)}{35 d \tan ^{\frac {5}{2}}(c+d x)}+\frac {4 a^2 (A-i B)}{3 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {4 a^2 (i A+B)}{d \sqrt {\tan (c+d x)}}-\frac {2 A \left (a^2+i a^2 \tan (c+d x)\right )}{7 d \tan ^{\frac {7}{2}}(c+d x)} \]
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Time = 0.35 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.139, Rules used = {3674, 3672, 3610, 3614, 211} \[ \int \frac {(a+i a \tan (c+d x))^2 (A+B \tan (c+d x))}{\tan ^{\frac {9}{2}}(c+d x)} \, dx=-\frac {4 \sqrt [4]{-1} a^2 (A-i B) \arctan \left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )}{d}+\frac {4 a^2 (A-i B)}{3 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {2 a^2 (7 B+9 i A)}{35 d \tan ^{\frac {5}{2}}(c+d x)}+\frac {4 a^2 (B+i A)}{d \sqrt {\tan (c+d x)}}-\frac {2 A \left (a^2+i a^2 \tan (c+d x)\right )}{7 d \tan ^{\frac {7}{2}}(c+d x)} \]
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Rule 211
Rule 3610
Rule 3614
Rule 3672
Rule 3674
Rubi steps \begin{align*} \text {integral}& = -\frac {2 A \left (a^2+i a^2 \tan (c+d x)\right )}{7 d \tan ^{\frac {7}{2}}(c+d x)}+\frac {2}{7} \int \frac {(a+i a \tan (c+d x)) \left (\frac {1}{2} a (9 i A+7 B)-\frac {1}{2} a (5 A-7 i B) \tan (c+d x)\right )}{\tan ^{\frac {7}{2}}(c+d x)} \, dx \\ & = -\frac {2 a^2 (9 i A+7 B)}{35 d \tan ^{\frac {5}{2}}(c+d x)}-\frac {2 A \left (a^2+i a^2 \tan (c+d x)\right )}{7 d \tan ^{\frac {7}{2}}(c+d x)}+\frac {2}{7} \int \frac {-7 a^2 (A-i B)-7 a^2 (i A+B) \tan (c+d x)}{\tan ^{\frac {5}{2}}(c+d x)} \, dx \\ & = -\frac {2 a^2 (9 i A+7 B)}{35 d \tan ^{\frac {5}{2}}(c+d x)}+\frac {4 a^2 (A-i B)}{3 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {2 A \left (a^2+i a^2 \tan (c+d x)\right )}{7 d \tan ^{\frac {7}{2}}(c+d x)}+\frac {2}{7} \int \frac {-7 a^2 (i A+B)+7 a^2 (A-i B) \tan (c+d x)}{\tan ^{\frac {3}{2}}(c+d x)} \, dx \\ & = -\frac {2 a^2 (9 i A+7 B)}{35 d \tan ^{\frac {5}{2}}(c+d x)}+\frac {4 a^2 (A-i B)}{3 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {4 a^2 (i A+B)}{d \sqrt {\tan (c+d x)}}-\frac {2 A \left (a^2+i a^2 \tan (c+d x)\right )}{7 d \tan ^{\frac {7}{2}}(c+d x)}+\frac {2}{7} \int \frac {7 a^2 (A-i B)+7 a^2 (i A+B) \tan (c+d x)}{\sqrt {\tan (c+d x)}} \, dx \\ & = -\frac {2 a^2 (9 i A+7 B)}{35 d \tan ^{\frac {5}{2}}(c+d x)}+\frac {4 a^2 (A-i B)}{3 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {4 a^2 (i A+B)}{d \sqrt {\tan (c+d x)}}-\frac {2 A \left (a^2+i a^2 \tan (c+d x)\right )}{7 d \tan ^{\frac {7}{2}}(c+d x)}+\frac {\left (28 a^4 (A-i B)^2\right ) \text {Subst}\left (\int \frac {1}{7 a^2 (A-i B)-7 a^2 (i A+B) x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{d} \\ & = -\frac {4 \sqrt [4]{-1} a^2 (A-i B) \arctan \left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )}{d}-\frac {2 a^2 (9 i A+7 B)}{35 d \tan ^{\frac {5}{2}}(c+d x)}+\frac {4 a^2 (A-i B)}{3 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {4 a^2 (i A+B)}{d \sqrt {\tan (c+d x)}}-\frac {2 A \left (a^2+i a^2 \tan (c+d x)\right )}{7 d \tan ^{\frac {7}{2}}(c+d x)} \\ \end{align*}
Time = 3.47 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.78 \[ \int \frac {(a+i a \tan (c+d x))^2 (A+B \tan (c+d x))}{\tan ^{\frac {9}{2}}(c+d x)} \, dx=\frac {2 a^2 \left (-15 A+(-42 i A-21 B) \tan (c+d x)+70 (A-i B) \tan ^2(c+d x)+210 (i A+B) \tan ^3(c+d x)+(105-105 i) \sqrt {2} (A-i B) \text {arctanh}\left (\frac {(1+i) \sqrt {\tan (c+d x)}}{\sqrt {2}}\right ) \tan ^{\frac {7}{2}}(c+d x)\right )}{105 d \tan ^{\frac {7}{2}}(c+d x)} \]
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Time = 0.04 (sec) , antiderivative size = 258, normalized size of antiderivative = 1.68
method | result | size |
derivativedivides | \(\frac {a^{2} \left (-\frac {2 A}{7 \tan \left (d x +c \right )^{\frac {7}{2}}}-\frac {2 \left (2 i B -2 A \right )}{3 \tan \left (d x +c \right )^{\frac {3}{2}}}-\frac {2 \left (-2 i A -2 B \right )}{\sqrt {\tan \left (d x +c \right )}}-\frac {2 \left (2 i A +B \right )}{5 \tan \left (d x +c \right )^{\frac {5}{2}}}+\frac {\left (-2 i B +2 A \right ) \sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}+\frac {\left (2 i A +2 B \right ) \sqrt {2}\, \left (\ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}\right )}{d}\) | \(258\) |
default | \(\frac {a^{2} \left (-\frac {2 A}{7 \tan \left (d x +c \right )^{\frac {7}{2}}}-\frac {2 \left (2 i B -2 A \right )}{3 \tan \left (d x +c \right )^{\frac {3}{2}}}-\frac {2 \left (-2 i A -2 B \right )}{\sqrt {\tan \left (d x +c \right )}}-\frac {2 \left (2 i A +B \right )}{5 \tan \left (d x +c \right )^{\frac {5}{2}}}+\frac {\left (-2 i B +2 A \right ) \sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}+\frac {\left (2 i A +2 B \right ) \sqrt {2}\, \left (\ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}\right )}{d}\) | \(258\) |
parts | \(\frac {\left (2 i A \,a^{2}+B \,a^{2}\right ) \left (\frac {\sqrt {2}\, \left (\ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}-\frac {2}{5 \tan \left (d x +c \right )^{\frac {5}{2}}}+\frac {2}{\sqrt {\tan \left (d x +c \right )}}\right )}{d}+\frac {\left (2 i B \,a^{2}-A \,a^{2}\right ) \left (-\frac {2}{3 \tan \left (d x +c \right )^{\frac {3}{2}}}-\frac {\sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}\right )}{d}+\frac {A \,a^{2} \left (\frac {\sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}-\frac {2}{7 \tan \left (d x +c \right )^{\frac {7}{2}}}+\frac {2}{3 \tan \left (d x +c \right )^{\frac {3}{2}}}\right )}{d}-\frac {B \,a^{2} \left (-\frac {\sqrt {2}\, \left (\ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}-\frac {2}{\sqrt {\tan \left (d x +c \right )}}\right )}{d}\) | \(458\) |
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Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 561 vs. \(2 (124) = 248\).
Time = 0.28 (sec) , antiderivative size = 561, normalized size of antiderivative = 3.64 \[ \int \frac {(a+i a \tan (c+d x))^2 (A+B \tan (c+d x))}{\tan ^{\frac {9}{2}}(c+d x)} \, dx=\frac {105 \, \sqrt {-\frac {{\left (i \, A^{2} + 2 \, A B - i \, B^{2}\right )} a^{4}}{d^{2}}} {\left (d e^{\left (8 i \, d x + 8 i \, c\right )} - 4 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 6 \, d e^{\left (4 i \, d x + 4 i \, c\right )} - 4 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \log \left (-\frac {2 \, {\left ({\left (A - i \, B\right )} a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + \sqrt {-\frac {{\left (i \, A^{2} + 2 \, A B - i \, B^{2}\right )} a^{4}}{d^{2}}} {\left (i \, d e^{\left (2 i \, d x + 2 i \, c\right )} + i \, d\right )} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{{\left (-i \, A - B\right )} a^{2}}\right ) - 105 \, \sqrt {-\frac {{\left (i \, A^{2} + 2 \, A B - i \, B^{2}\right )} a^{4}}{d^{2}}} {\left (d e^{\left (8 i \, d x + 8 i \, c\right )} - 4 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 6 \, d e^{\left (4 i \, d x + 4 i \, c\right )} - 4 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \log \left (-\frac {2 \, {\left ({\left (A - i \, B\right )} a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + \sqrt {-\frac {{\left (i \, A^{2} + 2 \, A B - i \, B^{2}\right )} a^{4}}{d^{2}}} {\left (-i \, d e^{\left (2 i \, d x + 2 i \, c\right )} - i \, d\right )} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{{\left (-i \, A - B\right )} a^{2}}\right ) - 2 \, {\left ({\left (337 \, A - 301 i \, B\right )} a^{2} e^{\left (8 i \, d x + 8 i \, c\right )} - 6 \, {\left (46 \, A - 63 i \, B\right )} a^{2} e^{\left (6 i \, d x + 6 i \, c\right )} - 10 \, {\left (5 \, A - 14 i \, B\right )} a^{2} e^{\left (4 i \, d x + 4 i \, c\right )} + 18 \, {\left (22 \, A - 21 i \, B\right )} a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} - {\left (167 \, A - 161 i \, B\right )} a^{2}\right )} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}}{105 \, {\left (d e^{\left (8 i \, d x + 8 i \, c\right )} - 4 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 6 \, d e^{\left (4 i \, d x + 4 i \, c\right )} - 4 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]
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\[ \int \frac {(a+i a \tan (c+d x))^2 (A+B \tan (c+d x))}{\tan ^{\frac {9}{2}}(c+d x)} \, dx=- a^{2} \left (\int \left (- \frac {A}{\tan ^{\frac {9}{2}}{\left (c + d x \right )}}\right )\, dx + \int \frac {A}{\tan ^{\frac {5}{2}}{\left (c + d x \right )}}\, dx + \int \left (- \frac {B}{\tan ^{\frac {7}{2}}{\left (c + d x \right )}}\right )\, dx + \int \frac {B}{\tan ^{\frac {3}{2}}{\left (c + d x \right )}}\, dx + \int \left (- \frac {2 i A}{\tan ^{\frac {7}{2}}{\left (c + d x \right )}}\right )\, dx + \int \left (- \frac {2 i B}{\tan ^{\frac {5}{2}}{\left (c + d x \right )}}\right )\, dx\right ) \]
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Time = 0.40 (sec) , antiderivative size = 213, normalized size of antiderivative = 1.38 \[ \int \frac {(a+i a \tan (c+d x))^2 (A+B \tan (c+d x))}{\tan ^{\frac {9}{2}}(c+d x)} \, dx=-\frac {105 \, {\left (2 \, \sqrt {2} {\left (-\left (i + 1\right ) \, A + \left (i - 1\right ) \, B\right )} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) + 2 \, \sqrt {2} {\left (-\left (i + 1\right ) \, A + \left (i - 1\right ) \, B\right )} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) + \sqrt {2} {\left (\left (i - 1\right ) \, A + \left (i + 1\right ) \, B\right )} \log \left (\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right ) - \sqrt {2} {\left (\left (i - 1\right ) \, A + \left (i + 1\right ) \, B\right )} \log \left (-\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right )\right )} a^{2} - \frac {4 \, {\left (210 \, {\left (i \, A + B\right )} a^{2} \tan \left (d x + c\right )^{3} + 70 \, {\left (A - i \, B\right )} a^{2} \tan \left (d x + c\right )^{2} + 21 \, {\left (-2 i \, A - B\right )} a^{2} \tan \left (d x + c\right ) - 15 \, A a^{2}\right )}}{\tan \left (d x + c\right )^{\frac {7}{2}}}}{210 \, d} \]
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Time = 1.14 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.88 \[ \int \frac {(a+i a \tan (c+d x))^2 (A+B \tan (c+d x))}{\tan ^{\frac {9}{2}}(c+d x)} \, dx=\frac {\left (2 i - 2\right ) \, \sqrt {2} {\left (-i \, A a^{2} - B a^{2}\right )} \arctan \left (-\left (\frac {1}{2} i - \frac {1}{2}\right ) \, \sqrt {2} \sqrt {\tan \left (d x + c\right )}\right )}{d} - \frac {2 \, {\left (-210 i \, A a^{2} \tan \left (d x + c\right )^{3} - 210 \, B a^{2} \tan \left (d x + c\right )^{3} - 70 \, A a^{2} \tan \left (d x + c\right )^{2} + 70 i \, B a^{2} \tan \left (d x + c\right )^{2} + 42 i \, A a^{2} \tan \left (d x + c\right ) + 21 \, B a^{2} \tan \left (d x + c\right ) + 15 \, A a^{2}\right )}}{105 \, d \tan \left (d x + c\right )^{\frac {7}{2}}} \]
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Time = 10.92 (sec) , antiderivative size = 293, normalized size of antiderivative = 1.90 \[ \int \frac {(a+i a \tan (c+d x))^2 (A+B \tan (c+d x))}{\tan ^{\frac {9}{2}}(c+d x)} \, dx=-\frac {\frac {2\,A\,a^2}{7\,d}+\frac {A\,a^2\,\mathrm {tan}\left (c+d\,x\right )\,4{}\mathrm {i}}{5\,d}-\frac {4\,A\,a^2\,{\mathrm {tan}\left (c+d\,x\right )}^2}{3\,d}-\frac {A\,a^2\,{\mathrm {tan}\left (c+d\,x\right )}^3\,4{}\mathrm {i}}{d}}{{\mathrm {tan}\left (c+d\,x\right )}^{7/2}}-\frac {\frac {2\,B\,a^2}{5\,d}-\frac {4\,B\,a^2\,{\mathrm {tan}\left (c+d\,x\right )}^2}{d}+\frac {B\,a^2\,\mathrm {tan}\left (c+d\,x\right )\,4{}\mathrm {i}}{3\,d}}{{\mathrm {tan}\left (c+d\,x\right )}^{5/2}}+\frac {\sqrt {2}\,A\,a^2\,\ln \left (A\,a^2\,d\,4{}\mathrm {i}+\sqrt {2}\,A\,a^2\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\left (-2+2{}\mathrm {i}\right )\right )\,\left (1-\mathrm {i}\right )}{d}-\frac {\sqrt {-4{}\mathrm {i}}\,A\,a^2\,\ln \left (A\,a^2\,d\,4{}\mathrm {i}+2\,\sqrt {-4{}\mathrm {i}}\,A\,a^2\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\right )}{d}+\frac {\sqrt {2}\,B\,a^2\,\ln \left (4\,B\,a^2\,d+\sqrt {2}\,B\,a^2\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\left (-2-2{}\mathrm {i}\right )\right )\,\left (1+1{}\mathrm {i}\right )}{d}-\frac {\sqrt {4{}\mathrm {i}}\,B\,a^2\,\ln \left (4\,B\,a^2\,d+2\,\sqrt {4{}\mathrm {i}}\,B\,a^2\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\right )}{d} \]
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