Integrand size = 36, antiderivative size = 393 \[ \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^3} \, dx=\frac {\left (\frac {1}{16}-\frac {i}{16}\right ) ((76+i) A+(1+29 i) B) \arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} a^3 d}-\frac {\left (\frac {1}{16}-\frac {i}{16}\right ) ((76+i) A+(1+29 i) B) \arctan \left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} a^3 d}+\frac {((77+75 i) A-(30-28 i) B) \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{32 \sqrt {2} a^3 d}-\frac {\left (\frac {1}{32}-\frac {i}{32}\right ) ((1+76 i) A-(29+i) B) \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{\sqrt {2} a^3 d}-\frac {7 (11 A+4 i B)}{24 a^3 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {15 (5 i A-2 B)}{8 a^3 d \sqrt {\tan (c+d x)}}+\frac {A+i B}{6 d \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^3}+\frac {2 A+i B}{4 a d \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^2}+\frac {3 (5 A+2 i B)}{8 d \tan ^{\frac {3}{2}}(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )} \]
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Time = 1.00 (sec) , antiderivative size = 393, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3677, 3610, 3615, 1182, 1176, 631, 210, 1179, 642} \[ \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^3} \, dx=\frac {\left (\frac {1}{16}-\frac {i}{16}\right ) ((76+i) A+(1+29 i) B) \arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} a^3 d}-\frac {\left (\frac {1}{16}-\frac {i}{16}\right ) ((76+i) A+(1+29 i) B) \arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2} a^3 d}+\frac {3 (5 A+2 i B)}{8 d \tan ^{\frac {3}{2}}(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}-\frac {7 (11 A+4 i B)}{24 a^3 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {15 (-2 B+5 i A)}{8 a^3 d \sqrt {\tan (c+d x)}}+\frac {((77+75 i) A-(30-28 i) B) \log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{32 \sqrt {2} a^3 d}-\frac {\left (\frac {1}{32}-\frac {i}{32}\right ) ((1+76 i) A-(29+i) B) \log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2} a^3 d}+\frac {2 A+i B}{4 a d \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^2}+\frac {A+i B}{6 d \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^3} \]
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Rule 210
Rule 631
Rule 642
Rule 1176
Rule 1179
Rule 1182
Rule 3610
Rule 3615
Rule 3677
Rubi steps \begin{align*} \text {integral}& = \frac {A+i B}{6 d \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^3}+\frac {\int \frac {\frac {3}{2} a (5 A+i B)-\frac {9}{2} a (i A-B) \tan (c+d x)}{\tan ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^2} \, dx}{6 a^2} \\ & = \frac {A+i B}{6 d \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^3}+\frac {2 A+i B}{4 a d \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^2}+\frac {\int \frac {3 a^2 (16 A+5 i B)-21 a^2 (2 i A-B) \tan (c+d x)}{\tan ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))} \, dx}{24 a^4} \\ & = \frac {A+i B}{6 d \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^3}+\frac {2 A+i B}{4 a d \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^2}+\frac {3 (5 A+2 i B)}{8 d \tan ^{\frac {3}{2}}(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}+\frac {\int \frac {21 a^3 (11 A+4 i B)-45 a^3 (5 i A-2 B) \tan (c+d x)}{\tan ^{\frac {5}{2}}(c+d x)} \, dx}{48 a^6} \\ & = -\frac {7 (11 A+4 i B)}{24 a^3 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {A+i B}{6 d \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^3}+\frac {2 A+i B}{4 a d \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^2}+\frac {3 (5 A+2 i B)}{8 d \tan ^{\frac {3}{2}}(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}+\frac {\int \frac {-45 a^3 (5 i A-2 B)-21 a^3 (11 A+4 i B) \tan (c+d x)}{\tan ^{\frac {3}{2}}(c+d x)} \, dx}{48 a^6} \\ & = -\frac {7 (11 A+4 i B)}{24 a^3 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {15 (5 i A-2 B)}{8 a^3 d \sqrt {\tan (c+d x)}}+\frac {A+i B}{6 d \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^3}+\frac {2 A+i B}{4 a d \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^2}+\frac {3 (5 A+2 i B)}{8 d \tan ^{\frac {3}{2}}(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}+\frac {\int \frac {-21 a^3 (11 A+4 i B)+45 a^3 (5 i A-2 B) \tan (c+d x)}{\sqrt {\tan (c+d x)}} \, dx}{48 a^6} \\ & = -\frac {7 (11 A+4 i B)}{24 a^3 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {15 (5 i A-2 B)}{8 a^3 d \sqrt {\tan (c+d x)}}+\frac {A+i B}{6 d \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^3}+\frac {2 A+i B}{4 a d \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^2}+\frac {3 (5 A+2 i B)}{8 d \tan ^{\frac {3}{2}}(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}+\frac {\text {Subst}\left (\int \frac {-21 a^3 (11 A+4 i B)+45 a^3 (5 i A-2 B) x^2}{1+x^4} \, dx,x,\sqrt {\tan (c+d x)}\right )}{24 a^6 d} \\ & = -\frac {7 (11 A+4 i B)}{24 a^3 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {15 (5 i A-2 B)}{8 a^3 d \sqrt {\tan (c+d x)}}+\frac {A+i B}{6 d \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^3}+\frac {2 A+i B}{4 a d \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^2}+\frac {3 (5 A+2 i B)}{8 d \tan ^{\frac {3}{2}}(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}-\frac {((77+75 i) A-(30-28 i) B) \text {Subst}\left (\int \frac {1-x^2}{1+x^4} \, dx,x,\sqrt {\tan (c+d x)}\right )}{16 a^3 d}+-\frac {\left (\left (\frac {1}{16}-\frac {i}{16}\right ) ((76+i) A+(1+29 i) B)\right ) \text {Subst}\left (\int \frac {1+x^2}{1+x^4} \, dx,x,\sqrt {\tan (c+d x)}\right )}{a^3 d} \\ & = -\frac {7 (11 A+4 i B)}{24 a^3 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {15 (5 i A-2 B)}{8 a^3 d \sqrt {\tan (c+d x)}}+\frac {A+i B}{6 d \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^3}+\frac {2 A+i B}{4 a d \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^2}+\frac {3 (5 A+2 i B)}{8 d \tan ^{\frac {3}{2}}(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}+\frac {((77+75 i) A-(30-28 i) B) \text {Subst}\left (\int \frac {\sqrt {2}+2 x}{-1-\sqrt {2} x-x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{32 \sqrt {2} a^3 d}+\frac {((77+75 i) A-(30-28 i) B) \text {Subst}\left (\int \frac {\sqrt {2}-2 x}{-1+\sqrt {2} x-x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{32 \sqrt {2} a^3 d}+-\frac {\left (\left (\frac {1}{32}-\frac {i}{32}\right ) ((76+i) A+(1+29 i) B)\right ) \text {Subst}\left (\int \frac {1}{1-\sqrt {2} x+x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{a^3 d}+-\frac {\left (\left (\frac {1}{32}-\frac {i}{32}\right ) ((76+i) A+(1+29 i) B)\right ) \text {Subst}\left (\int \frac {1}{1+\sqrt {2} x+x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{a^3 d} \\ & = \frac {((77+75 i) A-(30-28 i) B) \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{32 \sqrt {2} a^3 d}-\frac {((77+75 i) A-(30-28 i) B) \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{32 \sqrt {2} a^3 d}-\frac {7 (11 A+4 i B)}{24 a^3 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {15 (5 i A-2 B)}{8 a^3 d \sqrt {\tan (c+d x)}}+\frac {A+i B}{6 d \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^3}+\frac {2 A+i B}{4 a d \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^2}+\frac {3 (5 A+2 i B)}{8 d \tan ^{\frac {3}{2}}(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}+-\frac {\left (\left (\frac {1}{16}-\frac {i}{16}\right ) ((76+i) A+(1+29 i) B)\right ) \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} a^3 d}+\frac {\left (\left (\frac {1}{16}-\frac {i}{16}\right ) ((76+i) A+(1+29 i) B)\right ) \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} a^3 d} \\ & = \frac {\left (\frac {1}{16}-\frac {i}{16}\right ) ((76+i) A+(1+29 i) B) \arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} a^3 d}-\frac {\left (\frac {1}{16}-\frac {i}{16}\right ) ((76+i) A+(1+29 i) B) \arctan \left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} a^3 d}+\frac {((77+75 i) A-(30-28 i) B) \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{32 \sqrt {2} a^3 d}-\frac {((77+75 i) A-(30-28 i) B) \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{32 \sqrt {2} a^3 d}-\frac {7 (11 A+4 i B)}{24 a^3 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {15 (5 i A-2 B)}{8 a^3 d \sqrt {\tan (c+d x)}}+\frac {A+i B}{6 d \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^3}+\frac {2 A+i B}{4 a d \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^2}+\frac {3 (5 A+2 i B)}{8 d \tan ^{\frac {3}{2}}(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )} \\ \end{align*}
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 3.10 (sec) , antiderivative size = 194, normalized size of antiderivative = 0.49 \[ \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^3} \, dx=\frac {\sec ^3(c+d x) \left (2 i \cos (c+d x) (8 A+5 i B+(53 A+23 i B) \cos (2 (c+d x))+(51 i A-21 B) \sin (2 (c+d x)))+2 (-76 i A+29 B) \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},1,-\frac {1}{2},-i \tan (c+d x)\right ) (\cos (3 (c+d x))+i \sin (3 (c+d x)))+2 (A-i B) \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},1,-\frac {1}{2},i \tan (c+d x)\right ) (-i \cos (3 (c+d x))+\sin (3 (c+d x)))\right )}{48 a^3 d \tan ^{\frac {3}{2}}(c+d x) (-i+\tan (c+d x))^3} \]
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Time = 0.07 (sec) , antiderivative size = 196, normalized size of antiderivative = 0.50
method | result | size |
derivativedivides | \(\frac {-\frac {i \left (\frac {i \left (27 i A -14 B \right ) \left (\tan ^{\frac {5}{2}}\left (d x +c \right )\right )+\left (\frac {182 i A}{3}-\frac {98 B}{3}\right ) \left (\tan ^{\frac {3}{2}}\left (d x +c \right )\right )+\left (20 i B +35 A \right ) \left (\sqrt {\tan }\left (d x +c \right )\right )}{\left (\tan \left (d x +c \right )-i\right )^{3}}-\frac {2 \left (29 i B +76 A \right ) \arctan \left (\frac {2 \left (\sqrt {\tan }\left (d x +c \right )\right )}{\sqrt {2}-i \sqrt {2}}\right )}{\sqrt {2}-i \sqrt {2}}\right )}{8}-\frac {2 A}{3 \tan \left (d x +c \right )^{\frac {3}{2}}}-\frac {2 \left (-3 i A +B \right )}{\sqrt {\tan \left (d x +c \right )}}+\frac {4 \left (-\frac {i A}{16}-\frac {B}{16}\right ) \arctan \left (\frac {2 \left (\sqrt {\tan }\left (d x +c \right )\right )}{\sqrt {2}+i \sqrt {2}}\right )}{\sqrt {2}+i \sqrt {2}}}{d \,a^{3}}\) | \(196\) |
default | \(\frac {-\frac {i \left (\frac {i \left (27 i A -14 B \right ) \left (\tan ^{\frac {5}{2}}\left (d x +c \right )\right )+\left (\frac {182 i A}{3}-\frac {98 B}{3}\right ) \left (\tan ^{\frac {3}{2}}\left (d x +c \right )\right )+\left (20 i B +35 A \right ) \left (\sqrt {\tan }\left (d x +c \right )\right )}{\left (\tan \left (d x +c \right )-i\right )^{3}}-\frac {2 \left (29 i B +76 A \right ) \arctan \left (\frac {2 \left (\sqrt {\tan }\left (d x +c \right )\right )}{\sqrt {2}-i \sqrt {2}}\right )}{\sqrt {2}-i \sqrt {2}}\right )}{8}-\frac {2 A}{3 \tan \left (d x +c \right )^{\frac {3}{2}}}-\frac {2 \left (-3 i A +B \right )}{\sqrt {\tan \left (d x +c \right )}}+\frac {4 \left (-\frac {i A}{16}-\frac {B}{16}\right ) \arctan \left (\frac {2 \left (\sqrt {\tan }\left (d x +c \right )\right )}{\sqrt {2}+i \sqrt {2}}\right )}{\sqrt {2}+i \sqrt {2}}}{d \,a^{3}}\) | \(196\) |
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Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 875 vs. \(2 (296) = 592\).
Time = 0.29 (sec) , antiderivative size = 875, normalized size of antiderivative = 2.23 \[ \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^3} \, dx=\text {Too large to display} \]
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Timed out. \[ \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^3} \, dx=\text {Timed out} \]
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Exception generated. \[ \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^3} \, dx=\text {Exception raised: RuntimeError} \]
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Time = 0.72 (sec) , antiderivative size = 182, normalized size of antiderivative = 0.46 \[ \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^3} \, dx=-\frac {\left (i + 1\right ) \, \sqrt {2} {\left (-76 i \, A + 29 \, B\right )} \arctan \left (\left (\frac {1}{2} i + \frac {1}{2}\right ) \, \sqrt {2} \sqrt {\tan \left (d x + c\right )}\right )}{16 \, a^{3} d} + \frac {\left (i - 1\right ) \, \sqrt {2} {\left (i \, A + B\right )} \arctan \left (-\left (\frac {1}{2} i - \frac {1}{2}\right ) \, \sqrt {2} \sqrt {\tan \left (d x + c\right )}\right )}{16 \, a^{3} d} - \frac {225 \, A \tan \left (d x + c\right )^{4} + 90 i \, B \tan \left (d x + c\right )^{4} - 598 i \, A \tan \left (d x + c\right )^{3} + 242 \, B \tan \left (d x + c\right )^{3} - 489 \, A \tan \left (d x + c\right )^{2} - 204 i \, B \tan \left (d x + c\right )^{2} + 96 i \, A \tan \left (d x + c\right ) - 48 \, B \tan \left (d x + c\right ) - 16 \, A}{24 \, {\left (-i \, \tan \left (d x + c\right )^{\frac {3}{2}} - \sqrt {\tan \left (d x + c\right )}\right )}^{3} a^{3} d} \]
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Time = 9.98 (sec) , antiderivative size = 425, normalized size of antiderivative = 1.08 \[ \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^3} \, dx=-\mathrm {atan}\left (\frac {16\,a^3\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {-\frac {A^2\,1{}\mathrm {i}}{256\,a^6\,d^2}}}{A}\right )\,\sqrt {-\frac {A^2\,1{}\mathrm {i}}{256\,a^6\,d^2}}\,2{}\mathrm {i}+\mathrm {atan}\left (\frac {4\,a^3\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {\frac {A^2\,361{}\mathrm {i}}{16\,a^6\,d^2}}}{19\,A}\right )\,\sqrt {\frac {A^2\,361{}\mathrm {i}}{16\,a^6\,d^2}}\,2{}\mathrm {i}-\frac {\frac {A\,2{}\mathrm {i}}{3\,a^3\,d}+\frac {4\,A\,\mathrm {tan}\left (c+d\,x\right )}{a^3\,d}+\frac {A\,{\mathrm {tan}\left (c+d\,x\right )}^2\,163{}\mathrm {i}}{8\,a^3\,d}-\frac {299\,A\,{\mathrm {tan}\left (c+d\,x\right )}^3}{12\,a^3\,d}-\frac {A\,{\mathrm {tan}\left (c+d\,x\right )}^4\,75{}\mathrm {i}}{8\,a^3\,d}}{{\mathrm {tan}\left (c+d\,x\right )}^{9/2}-3\,{\mathrm {tan}\left (c+d\,x\right )}^{5/2}+{\mathrm {tan}\left (c+d\,x\right )}^{3/2}\,1{}\mathrm {i}-{\mathrm {tan}\left (c+d\,x\right )}^{7/2}\,3{}\mathrm {i}}+2\,\mathrm {atanh}\left (\frac {16\,a^3\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {\frac {B^2\,1{}\mathrm {i}}{256\,a^6\,d^2}}}{B}\right )\,\sqrt {\frac {B^2\,1{}\mathrm {i}}{256\,a^6\,d^2}}+2\,\mathrm {atanh}\left (\frac {16\,a^3\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {-\frac {B^2\,841{}\mathrm {i}}{256\,a^6\,d^2}}}{29\,B}\right )\,\sqrt {-\frac {B^2\,841{}\mathrm {i}}{256\,a^6\,d^2}}-\frac {\frac {2\,B}{a^3\,d}+\frac {B\,\mathrm {tan}\left (c+d\,x\right )\,17{}\mathrm {i}}{2\,a^3\,d}-\frac {121\,B\,{\mathrm {tan}\left (c+d\,x\right )}^2}{12\,a^3\,d}-\frac {B\,{\mathrm {tan}\left (c+d\,x\right )}^3\,15{}\mathrm {i}}{4\,a^3\,d}}{\sqrt {\mathrm {tan}\left (c+d\,x\right )}+{\mathrm {tan}\left (c+d\,x\right )}^{3/2}\,3{}\mathrm {i}-3\,{\mathrm {tan}\left (c+d\,x\right )}^{5/2}-{\mathrm {tan}\left (c+d\,x\right )}^{7/2}\,1{}\mathrm {i}} \]
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