Integrand size = 36, antiderivative size = 393 \[ \int \frac {\tan ^{\frac {9}{2}}(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^3} \, dx=\frac {\left (\frac {1}{16}+\frac {i}{16}\right ) ((29+i) A+(1+76 i) B) \arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} a^3 d}-\frac {\left (\frac {1}{16}+\frac {i}{16}\right ) ((29+i) A+(1+76 i) B) \arctan \left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} a^3 d}-\frac {((28-30 i) A+(75+77 i) B) \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{32 \sqrt {2} a^3 d}-\frac {\left (\frac {1}{32}+\frac {i}{32}\right ) ((1+29 i) A-(76+i) B) \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{\sqrt {2} a^3 d}+\frac {15 (2 i A-5 B) \sqrt {\tan (c+d x)}}{8 a^3 d}+\frac {7 (4 A+11 i B) \tan ^{\frac {3}{2}}(c+d x)}{24 a^3 d}+\frac {(i A-B) \tan ^{\frac {9}{2}}(c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac {(A+2 i B) \tan ^{\frac {7}{2}}(c+d x)}{4 a d (a+i a \tan (c+d x))^2}-\frac {3 (2 i A-5 B) \tan ^{\frac {5}{2}}(c+d x)}{8 d \left (a^3+i a^3 \tan (c+d x)\right )} \]
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Time = 0.99 (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 = {3676, 3609, 3615, 1182, 1176, 631, 210, 1179, 642} \[ \int \frac {\tan ^{\frac {9}{2}}(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^3} \, dx=\frac {\left (\frac {1}{16}+\frac {i}{16}\right ) ((29+i) A+(1+76 i) B) \arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} a^3 d}-\frac {\left (\frac {1}{16}+\frac {i}{16}\right ) ((29+i) A+(1+76 i) B) \arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2} a^3 d}-\frac {3 (-5 B+2 i A) \tan ^{\frac {5}{2}}(c+d x)}{8 d \left (a^3+i a^3 \tan (c+d x)\right )}+\frac {7 (4 A+11 i B) \tan ^{\frac {3}{2}}(c+d x)}{24 a^3 d}+\frac {15 (-5 B+2 i A) \sqrt {\tan (c+d x)}}{8 a^3 d}-\frac {((28-30 i) A+(75+77 i) B) \log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{32 \sqrt {2} a^3 d}-\frac {\left (\frac {1}{32}+\frac {i}{32}\right ) ((1+29 i) A-(76+i) B) \log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2} a^3 d}+\frac {(-B+i A) \tan ^{\frac {9}{2}}(c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac {(A+2 i B) \tan ^{\frac {7}{2}}(c+d x)}{4 a d (a+i a \tan (c+d x))^2} \]
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Rule 210
Rule 631
Rule 642
Rule 1176
Rule 1179
Rule 1182
Rule 3609
Rule 3615
Rule 3676
Rubi steps \begin{align*} \text {integral}& = \frac {(i A-B) \tan ^{\frac {9}{2}}(c+d x)}{6 d (a+i a \tan (c+d x))^3}-\frac {\int \frac {\tan ^{\frac {7}{2}}(c+d x) \left (\frac {9}{2} a (i A-B)+\frac {3}{2} a (A+5 i B) \tan (c+d x)\right )}{(a+i a \tan (c+d x))^2} \, dx}{6 a^2} \\ & = \frac {(i A-B) \tan ^{\frac {9}{2}}(c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac {(A+2 i B) \tan ^{\frac {7}{2}}(c+d x)}{4 a d (a+i a \tan (c+d x))^2}+\frac {\int \frac {\tan ^{\frac {5}{2}}(c+d x) \left (-21 a^2 (A+2 i B)+3 a^2 (5 i A-16 B) \tan (c+d x)\right )}{a+i a \tan (c+d x)} \, dx}{24 a^4} \\ & = \frac {(i A-B) \tan ^{\frac {9}{2}}(c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac {(A+2 i B) \tan ^{\frac {7}{2}}(c+d x)}{4 a d (a+i a \tan (c+d x))^2}-\frac {3 (2 i A-5 B) \tan ^{\frac {5}{2}}(c+d x)}{8 d \left (a^3+i a^3 \tan (c+d x)\right )}-\frac {\int \tan ^{\frac {3}{2}}(c+d x) \left (-45 a^3 (2 i A-5 B)-21 a^3 (4 A+11 i B) \tan (c+d x)\right ) \, dx}{48 a^6} \\ & = \frac {7 (4 A+11 i B) \tan ^{\frac {3}{2}}(c+d x)}{24 a^3 d}+\frac {(i A-B) \tan ^{\frac {9}{2}}(c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac {(A+2 i B) \tan ^{\frac {7}{2}}(c+d x)}{4 a d (a+i a \tan (c+d x))^2}-\frac {3 (2 i A-5 B) \tan ^{\frac {5}{2}}(c+d x)}{8 d \left (a^3+i a^3 \tan (c+d x)\right )}-\frac {\int \sqrt {\tan (c+d x)} \left (21 a^3 (4 A+11 i B)-45 a^3 (2 i A-5 B) \tan (c+d x)\right ) \, dx}{48 a^6} \\ & = \frac {15 (2 i A-5 B) \sqrt {\tan (c+d x)}}{8 a^3 d}+\frac {7 (4 A+11 i B) \tan ^{\frac {3}{2}}(c+d x)}{24 a^3 d}+\frac {(i A-B) \tan ^{\frac {9}{2}}(c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac {(A+2 i B) \tan ^{\frac {7}{2}}(c+d x)}{4 a d (a+i a \tan (c+d x))^2}-\frac {3 (2 i A-5 B) \tan ^{\frac {5}{2}}(c+d x)}{8 d \left (a^3+i a^3 \tan (c+d x)\right )}-\frac {\int \frac {45 a^3 (2 i A-5 B)+21 a^3 (4 A+11 i B) \tan (c+d x)}{\sqrt {\tan (c+d x)}} \, dx}{48 a^6} \\ & = \frac {15 (2 i A-5 B) \sqrt {\tan (c+d x)}}{8 a^3 d}+\frac {7 (4 A+11 i B) \tan ^{\frac {3}{2}}(c+d x)}{24 a^3 d}+\frac {(i A-B) \tan ^{\frac {9}{2}}(c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac {(A+2 i B) \tan ^{\frac {7}{2}}(c+d x)}{4 a d (a+i a \tan (c+d x))^2}-\frac {3 (2 i A-5 B) \tan ^{\frac {5}{2}}(c+d x)}{8 d \left (a^3+i a^3 \tan (c+d x)\right )}-\frac {\text {Subst}\left (\int \frac {45 a^3 (2 i A-5 B)+21 a^3 (4 A+11 i B) x^2}{1+x^4} \, dx,x,\sqrt {\tan (c+d x)}\right )}{24 a^6 d} \\ & = \frac {15 (2 i A-5 B) \sqrt {\tan (c+d x)}}{8 a^3 d}+\frac {7 (4 A+11 i B) \tan ^{\frac {3}{2}}(c+d x)}{24 a^3 d}+\frac {(i A-B) \tan ^{\frac {9}{2}}(c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac {(A+2 i B) \tan ^{\frac {7}{2}}(c+d x)}{4 a d (a+i a \tan (c+d x))^2}-\frac {3 (2 i A-5 B) \tan ^{\frac {5}{2}}(c+d x)}{8 d \left (a^3+i a^3 \tan (c+d x)\right )}-\frac {\left (\left (\frac {1}{16}+\frac {i}{16}\right ) ((29+i) A+(1+76 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 {((28-30 i) A+(75+77 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 {15 (2 i A-5 B) \sqrt {\tan (c+d x)}}{8 a^3 d}+\frac {7 (4 A+11 i B) \tan ^{\frac {3}{2}}(c+d x)}{24 a^3 d}+\frac {(i A-B) \tan ^{\frac {9}{2}}(c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac {(A+2 i B) \tan ^{\frac {7}{2}}(c+d x)}{4 a d (a+i a \tan (c+d x))^2}-\frac {3 (2 i A-5 B) \tan ^{\frac {5}{2}}(c+d x)}{8 d \left (a^3+i a^3 \tan (c+d x)\right )}-\frac {\left (\left (\frac {1}{32}+\frac {i}{32}\right ) ((29+i) A+(1+76 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 ) ((29+i) A+(1+76 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 {((28-30 i) A+(75+77 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 {((28-30 i) A+(75+77 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 {((28-30 i) A+(75+77 i) B) \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{32 \sqrt {2} a^3 d}+\frac {((28-30 i) A+(75+77 i) B) \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{32 \sqrt {2} a^3 d}+\frac {15 (2 i A-5 B) \sqrt {\tan (c+d x)}}{8 a^3 d}+\frac {7 (4 A+11 i B) \tan ^{\frac {3}{2}}(c+d x)}{24 a^3 d}+\frac {(i A-B) \tan ^{\frac {9}{2}}(c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac {(A+2 i B) \tan ^{\frac {7}{2}}(c+d x)}{4 a d (a+i a \tan (c+d x))^2}-\frac {3 (2 i A-5 B) \tan ^{\frac {5}{2}}(c+d x)}{8 d \left (a^3+i a^3 \tan (c+d x)\right )}--\frac {\left (\left (\frac {1}{16}+\frac {i}{16}\right ) ((29+i) A+(1+76 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 ) ((29+i) A+(1+76 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 ) ((29+i) A+(1+76 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 ) ((29+i) A+(1+76 i) B) \arctan \left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} a^3 d}-\frac {((28-30 i) A+(75+77 i) B) \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{32 \sqrt {2} a^3 d}+\frac {((28-30 i) A+(75+77 i) B) \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{32 \sqrt {2} a^3 d}+\frac {15 (2 i A-5 B) \sqrt {\tan (c+d x)}}{8 a^3 d}+\frac {7 (4 A+11 i B) \tan ^{\frac {3}{2}}(c+d x)}{24 a^3 d}+\frac {(i A-B) \tan ^{\frac {9}{2}}(c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac {(A+2 i B) \tan ^{\frac {7}{2}}(c+d x)}{4 a d (a+i a \tan (c+d x))^2}-\frac {3 (2 i A-5 B) \tan ^{\frac {5}{2}}(c+d x)}{8 d \left (a^3+i a^3 \tan (c+d x)\right )} \\ \end{align*}
Time = 6.17 (sec) , antiderivative size = 235, normalized size of antiderivative = 0.60 \[ \int \frac {\tan ^{\frac {9}{2}}(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^3} \, dx=\frac {-3 \sqrt [4]{-1} (A-i B) \arctan \left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right ) \sec ^3(c+d x) (\cos (3 (c+d x))+i \sin (3 (c+d x)))-3 \sqrt [4]{-1} (29 A+76 i B) \text {arctanh}\left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right ) \sec ^3(c+d x) (\cos (3 (c+d x))+i \sin (3 (c+d x)))+\sqrt {\tan (c+d x)} \left (-45 (2 A+5 i B)+(-242 i A+598 B) \tan (c+d x)+3 (68 A+163 i B) \tan ^2(c+d x)+48 i (A+2 i B) \tan ^3(c+d x)+16 i B \tan ^4(c+d x)\right )}{24 a^3 d (-i+\tan (c+d x))^3} \]
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Time = 0.07 (sec) , antiderivative size = 204, normalized size of antiderivative = 0.52
method | result | size |
derivativedivides | \(\frac {\frac {2 i B \left (\tan ^{\frac {3}{2}}\left (d x +c \right )\right )}{3}-6 B \left (\sqrt {\tan }\left (d x +c \right )\right )+2 i A \left (\sqrt {\tan }\left (d x +c \right )\right )+\frac {i \left (\frac {-5 i \left (7 i B +4 A \right ) \left (\tan ^{\frac {5}{2}}\left (d x +c \right )\right )+\left (-\frac {182 i B}{3}-\frac {98 A}{3}\right ) \left (\tan ^{\frac {3}{2}}\left (d x +c \right )\right )+\left (14 i A -27 B \right ) \left (\sqrt {\tan }\left (d x +c \right )\right )}{\left (\tan \left (d x +c \right )-i\right )^{3}}+\frac {2 \left (29 i A -76 B \right ) \arctan \left (\frac {2 \left (\sqrt {\tan }\left (d x +c \right )\right )}{\sqrt {2}-i \sqrt {2}}\right )}{\sqrt {2}-i \sqrt {2}}\right )}{8}+\frac {4 \left (\frac {A}{16}-\frac {i B}{16}\right ) \arctan \left (\frac {2 \left (\sqrt {\tan }\left (d x +c \right )\right )}{\sqrt {2}+i \sqrt {2}}\right )}{\sqrt {2}+i \sqrt {2}}}{d \,a^{3}}\) | \(204\) |
default | \(\frac {\frac {2 i B \left (\tan ^{\frac {3}{2}}\left (d x +c \right )\right )}{3}-6 B \left (\sqrt {\tan }\left (d x +c \right )\right )+2 i A \left (\sqrt {\tan }\left (d x +c \right )\right )+\frac {i \left (\frac {-5 i \left (7 i B +4 A \right ) \left (\tan ^{\frac {5}{2}}\left (d x +c \right )\right )+\left (-\frac {182 i B}{3}-\frac {98 A}{3}\right ) \left (\tan ^{\frac {3}{2}}\left (d x +c \right )\right )+\left (14 i A -27 B \right ) \left (\sqrt {\tan }\left (d x +c \right )\right )}{\left (\tan \left (d x +c \right )-i\right )^{3}}+\frac {2 \left (29 i A -76 B \right ) \arctan \left (\frac {2 \left (\sqrt {\tan }\left (d x +c \right )\right )}{\sqrt {2}-i \sqrt {2}}\right )}{\sqrt {2}-i \sqrt {2}}\right )}{8}+\frac {4 \left (\frac {A}{16}-\frac {i B}{16}\right ) \arctan \left (\frac {2 \left (\sqrt {\tan }\left (d x +c \right )\right )}{\sqrt {2}+i \sqrt {2}}\right )}{\sqrt {2}+i \sqrt {2}}}{d \,a^{3}}\) | \(204\) |
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Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 779 vs. \(2 (294) = 588\).
Time = 0.30 (sec) , antiderivative size = 779, normalized size of antiderivative = 1.98 \[ \int \frac {\tan ^{\frac {9}{2}}(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^3} \, dx=-\frac {3 \, {\left (a^{3} d e^{\left (8 i \, d x + 8 i \, c\right )} + a^{3} d e^{\left (6 i \, d x + 6 i \, c\right )}\right )} \sqrt {\frac {i \, A^{2} + 2 \, A B - i \, B^{2}}{a^{6} d^{2}}} \log \left (\frac {2 \, {\left ({\left (a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{3} 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}} \sqrt {\frac {i \, A^{2} + 2 \, A B - i \, B^{2}}{a^{6} d^{2}}} + {\left (A - i \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{i \, A + B}\right ) - 3 \, {\left (a^{3} d e^{\left (8 i \, d x + 8 i \, c\right )} + a^{3} d e^{\left (6 i \, d x + 6 i \, c\right )}\right )} \sqrt {\frac {i \, A^{2} + 2 \, A B - i \, B^{2}}{a^{6} d^{2}}} \log \left (-\frac {2 \, {\left ({\left (a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{3} 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}} \sqrt {\frac {i \, A^{2} + 2 \, A B - i \, B^{2}}{a^{6} d^{2}}} - {\left (A - i \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{i \, A + B}\right ) - 3 \, {\left (a^{3} d e^{\left (8 i \, d x + 8 i \, c\right )} + a^{3} d e^{\left (6 i \, d x + 6 i \, c\right )}\right )} \sqrt {\frac {-841 i \, A^{2} + 4408 \, A B + 5776 i \, B^{2}}{a^{6} d^{2}}} \log \left (\frac {{\left ({\left (a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{3} 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}} \sqrt {\frac {-841 i \, A^{2} + 4408 \, A B + 5776 i \, B^{2}}{a^{6} d^{2}}} + 29 \, A + 76 i \, B\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{8 \, a^{3} d}\right ) + 3 \, {\left (a^{3} d e^{\left (8 i \, d x + 8 i \, c\right )} + a^{3} d e^{\left (6 i \, d x + 6 i \, c\right )}\right )} \sqrt {\frac {-841 i \, A^{2} + 4408 \, A B + 5776 i \, B^{2}}{a^{6} d^{2}}} \log \left (-\frac {{\left ({\left (a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{3} 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}} \sqrt {\frac {-841 i \, A^{2} + 4408 \, A B + 5776 i \, B^{2}}{a^{6} d^{2}}} - 29 \, A - 76 i \, B\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{8 \, a^{3} d}\right ) + 2 \, {\left (2 \, {\left (-73 i \, A + 174 \, B\right )} e^{\left (8 i \, d x + 8 i \, c\right )} - {\left (187 i \, A - 492 \, B\right )} e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, {\left (-11 i \, A + 23 \, B\right )} e^{\left (4 i \, d x + 4 i \, c\right )} - {\left (-7 i \, A + 10 \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} - i \, A + B\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}}}{96 \, {\left (a^{3} d e^{\left (8 i \, d x + 8 i \, c\right )} + a^{3} d e^{\left (6 i \, d x + 6 i \, c\right )}\right )}} \]
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Timed out. \[ \int \frac {\tan ^{\frac {9}{2}}(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^3} \, dx=\text {Timed out} \]
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Exception generated. \[ \int \frac {\tan ^{\frac {9}{2}}(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^3} \, dx=\text {Exception raised: RuntimeError} \]
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none
Time = 1.02 (sec) , antiderivative size = 208, normalized size of antiderivative = 0.53 \[ \int \frac {\tan ^{\frac {9}{2}}(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^3} \, dx=-\frac {\left (i + 1\right ) \, \sqrt {2} {\left (29 \, A + 76 i \, B\right )} \arctan \left (\left (\frac {1}{2} i + \frac {1}{2}\right ) \, \sqrt {2} \sqrt {\tan \left (d x + c\right )}\right )}{16 \, a^{3} d} - \frac {\left (i - 1\right ) \, \sqrt {2} {\left (A - i \, B\right )} \arctan \left (-\left (\frac {1}{2} i - \frac {1}{2}\right ) \, \sqrt {2} \sqrt {\tan \left (d x + c\right )}\right )}{16 \, a^{3} d} + \frac {60 \, A \tan \left (d x + c\right )^{\frac {5}{2}} + 105 i \, B \tan \left (d x + c\right )^{\frac {5}{2}} - 98 i \, A \tan \left (d x + c\right )^{\frac {3}{2}} + 182 \, B \tan \left (d x + c\right )^{\frac {3}{2}} - 42 \, A \sqrt {\tan \left (d x + c\right )} - 81 i \, B \sqrt {\tan \left (d x + c\right )}}{24 \, a^{3} d {\left (\tan \left (d x + c\right ) - i\right )}^{3}} - \frac {2 \, {\left (-i \, B a^{6} d^{2} \tan \left (d x + c\right )^{\frac {3}{2}} - 3 i \, A a^{6} d^{2} \sqrt {\tan \left (d x + c\right )} + 9 \, B a^{6} d^{2} \sqrt {\tan \left (d x + c\right )}\right )}}{3 \, a^{9} d^{3}} \]
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Time = 11.60 (sec) , antiderivative size = 431, normalized size of antiderivative = 1.10 \[ \int \frac {\tan ^{\frac {9}{2}}(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^3} \, dx=\mathrm {atan}\left (\frac {a^3\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {\frac {A^2\,1{}\mathrm {i}}{256\,a^6\,d^2}}\,16{}\mathrm {i}}{A}\right )\,\sqrt {\frac {A^2\,1{}\mathrm {i}}{256\,a^6\,d^2}}\,2{}\mathrm {i}-\mathrm {atan}\left (\frac {a^3\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {-\frac {A^2\,841{}\mathrm {i}}{256\,a^6\,d^2}}\,16{}\mathrm {i}}{29\,A}\right )\,\sqrt {-\frac {A^2\,841{}\mathrm {i}}{256\,a^6\,d^2}}\,2{}\mathrm {i}-\mathrm {atan}\left (\frac {16\,a^3\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {-\frac {B^2\,1{}\mathrm {i}}{256\,a^6\,d^2}}}{B}\right )\,\sqrt {-\frac {B^2\,1{}\mathrm {i}}{256\,a^6\,d^2}}\,2{}\mathrm {i}-\mathrm {atan}\left (\frac {4\,a^3\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {\frac {B^2\,361{}\mathrm {i}}{16\,a^6\,d^2}}}{19\,B}\right )\,\sqrt {\frac {B^2\,361{}\mathrm {i}}{16\,a^6\,d^2}}\,2{}\mathrm {i}-\frac {\frac {49\,A\,{\mathrm {tan}\left (c+d\,x\right )}^{3/2}}{12\,a^3\,d}-\frac {A\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,7{}\mathrm {i}}{4\,a^3\,d}+\frac {A\,{\mathrm {tan}\left (c+d\,x\right )}^{5/2}\,5{}\mathrm {i}}{2\,a^3\,d}}{-{\mathrm {tan}\left (c+d\,x\right )}^3\,1{}\mathrm {i}-3\,{\mathrm {tan}\left (c+d\,x\right )}^2+\mathrm {tan}\left (c+d\,x\right )\,3{}\mathrm {i}+1}-\frac {\frac {27\,B\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}}{8\,a^3\,d}-\frac {35\,B\,{\mathrm {tan}\left (c+d\,x\right )}^{5/2}}{8\,a^3\,d}+\frac {B\,{\mathrm {tan}\left (c+d\,x\right )}^{3/2}\,91{}\mathrm {i}}{12\,a^3\,d}}{-{\mathrm {tan}\left (c+d\,x\right )}^3\,1{}\mathrm {i}-3\,{\mathrm {tan}\left (c+d\,x\right )}^2+\mathrm {tan}\left (c+d\,x\right )\,3{}\mathrm {i}+1}+\frac {A\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,2{}\mathrm {i}}{a^3\,d}-\frac {6\,B\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}}{a^3\,d}+\frac {B\,{\mathrm {tan}\left (c+d\,x\right )}^{3/2}\,2{}\mathrm {i}}{3\,a^3\,d} \]
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