Integrand size = 36, antiderivative size = 316 \[ \int \frac {\tan ^{\frac {5}{2}}(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^2} \, dx=\frac {((9+5 i) A-(25-21 i) B) \arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{16 \sqrt {2} a^2 d}-\frac {((9+5 i) A-(25-21 i) B) \arctan \left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{16 \sqrt {2} a^2 d}-\frac {\left (\frac {1}{32}-\frac {i}{32}\right ) ((7+2 i) A+(2+23 i) B) \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{\sqrt {2} a^2 d}+\frac {\left (\frac {1}{32}-\frac {i}{32}\right ) ((7+2 i) A+(2+23 i) B) \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{\sqrt {2} a^2 d}+\frac {5 (i A-5 B) \sqrt {\tan (c+d x)}}{8 a^2 d}+\frac {(3 A+7 i B) \tan ^{\frac {3}{2}}(c+d x)}{8 a^2 d (1+i \tan (c+d x))}+\frac {(i A-B) \tan ^{\frac {5}{2}}(c+d x)}{4 d (a+i a \tan (c+d x))^2} \]
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Time = 0.67 (sec) , antiderivative size = 316, normalized size of antiderivative = 1.00, number of steps used = 13, 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} \[ \int \frac {\tan ^{\frac {5}{2}}(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^2} \, dx=\frac {((9+5 i) A-(25-21 i) B) \arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{16 \sqrt {2} a^2 d}-\frac {((9+5 i) A-(25-21 i) B) \arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{16 \sqrt {2} a^2 d}+\frac {(3 A+7 i B) \tan ^{\frac {3}{2}}(c+d x)}{8 a^2 d (1+i \tan (c+d x))}+\frac {5 (-5 B+i A) \sqrt {\tan (c+d x)}}{8 a^2 d}-\frac {\left (\frac {1}{32}-\frac {i}{32}\right ) ((7+2 i) A+(2+23 i) B) \log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2} a^2 d}+\frac {\left (\frac {1}{32}-\frac {i}{32}\right ) ((7+2 i) A+(2+23 i) B) \log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2} a^2 d}+\frac {(-B+i A) \tan ^{\frac {5}{2}}(c+d x)}{4 d (a+i a \tan (c+d x))^2} \]
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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 {5}{2}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}-\frac {\int \frac {\tan ^{\frac {3}{2}}(c+d x) \left (\frac {5}{2} a (i A-B)+\frac {1}{2} a (A+9 i B) \tan (c+d x)\right )}{a+i a \tan (c+d x)} \, dx}{4 a^2} \\ & = \frac {(3 A+7 i B) \tan ^{\frac {3}{2}}(c+d x)}{8 a^2 d (1+i \tan (c+d x))}+\frac {(i A-B) \tan ^{\frac {5}{2}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}+\frac {\int \sqrt {\tan (c+d x)} \left (-\frac {3}{2} a^2 (3 A+7 i B)+\frac {5}{2} a^2 (i A-5 B) \tan (c+d x)\right ) \, dx}{8 a^4} \\ & = \frac {5 (i A-5 B) \sqrt {\tan (c+d x)}}{8 a^2 d}+\frac {(3 A+7 i B) \tan ^{\frac {3}{2}}(c+d x)}{8 a^2 d (1+i \tan (c+d x))}+\frac {(i A-B) \tan ^{\frac {5}{2}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}+\frac {\int \frac {-\frac {5}{2} a^2 (i A-5 B)-\frac {3}{2} a^2 (3 A+7 i B) \tan (c+d x)}{\sqrt {\tan (c+d x)}} \, dx}{8 a^4} \\ & = \frac {5 (i A-5 B) \sqrt {\tan (c+d x)}}{8 a^2 d}+\frac {(3 A+7 i B) \tan ^{\frac {3}{2}}(c+d x)}{8 a^2 d (1+i \tan (c+d x))}+\frac {(i A-B) \tan ^{\frac {5}{2}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}+\frac {\text {Subst}\left (\int \frac {-\frac {5}{2} a^2 (i A-5 B)-\frac {3}{2} a^2 (3 A+7 i B) x^2}{1+x^4} \, dx,x,\sqrt {\tan (c+d x)}\right )}{4 a^4 d} \\ & = \frac {5 (i A-5 B) \sqrt {\tan (c+d x)}}{8 a^2 d}+\frac {(3 A+7 i B) \tan ^{\frac {3}{2}}(c+d x)}{8 a^2 d (1+i \tan (c+d x))}+\frac {(i A-B) \tan ^{\frac {5}{2}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}-\frac {((9+5 i) A-(25-21 i) B) \text {Subst}\left (\int \frac {1+x^2}{1+x^4} \, dx,x,\sqrt {\tan (c+d x)}\right )}{16 a^2 d}+\frac {((9-5 i) A+(25+21 i) B) \text {Subst}\left (\int \frac {1-x^2}{1+x^4} \, dx,x,\sqrt {\tan (c+d x)}\right )}{16 a^2 d} \\ & = \frac {5 (i A-5 B) \sqrt {\tan (c+d x)}}{8 a^2 d}+\frac {(3 A+7 i B) \tan ^{\frac {3}{2}}(c+d x)}{8 a^2 d (1+i \tan (c+d x))}+\frac {(i A-B) \tan ^{\frac {5}{2}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}-\frac {((9+5 i) A-(25-21 i) B) \text {Subst}\left (\int \frac {1}{1-\sqrt {2} x+x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{32 a^2 d}-\frac {((9+5 i) A-(25-21 i) B) \text {Subst}\left (\int \frac {1}{1+\sqrt {2} x+x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{32 a^2 d}-\frac {((9-5 i) A+(25+21 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^2 d}-\frac {((9-5 i) A+(25+21 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^2 d} \\ & = -\frac {((9-5 i) A+(25+21 i) B) \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{32 \sqrt {2} a^2 d}+\frac {((9-5 i) A+(25+21 i) B) \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{32 \sqrt {2} a^2 d}+\frac {5 (i A-5 B) \sqrt {\tan (c+d x)}}{8 a^2 d}+\frac {(3 A+7 i B) \tan ^{\frac {3}{2}}(c+d x)}{8 a^2 d (1+i \tan (c+d x))}+\frac {(i A-B) \tan ^{\frac {5}{2}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}-\frac {((9+5 i) A-(25-21 i) B) \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{16 \sqrt {2} a^2 d}+\frac {((9+5 i) A-(25-21 i) B) \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{16 \sqrt {2} a^2 d} \\ & = \frac {((9+5 i) A-(25-21 i) B) \arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{16 \sqrt {2} a^2 d}-\frac {((9+5 i) A-(25-21 i) B) \arctan \left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{16 \sqrt {2} a^2 d}-\frac {((9-5 i) A+(25+21 i) B) \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{32 \sqrt {2} a^2 d}+\frac {((9-5 i) A+(25+21 i) B) \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{32 \sqrt {2} a^2 d}+\frac {5 (i A-5 B) \sqrt {\tan (c+d x)}}{8 a^2 d}+\frac {(3 A+7 i B) \tan ^{\frac {3}{2}}(c+d x)}{8 a^2 d (1+i \tan (c+d x))}+\frac {(i A-B) \tan ^{\frac {5}{2}}(c+d x)}{4 d (a+i a \tan (c+d x))^2} \\ \end{align*}
Time = 3.47 (sec) , antiderivative size = 191, normalized size of antiderivative = 0.60 \[ \int \frac {\tan ^{\frac {5}{2}}(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^2} \, dx=\frac {2 \sqrt [4]{-1} (i A+B) \arctan \left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right ) \sec ^2(c+d x) (\cos (2 (c+d x))+i \sin (2 (c+d x)))+\sqrt [4]{-1} (-7 i A+23 B) \text {arctanh}\left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right ) \sec ^2(c+d x) (\cos (2 (c+d x))+i \sin (2 (c+d x)))+\sqrt {\tan (c+d x)} \left (-5 i A+25 B+(7 A+43 i B) \tan (c+d x)-16 B \tan ^2(c+d x)\right )}{8 a^2 d (-i+\tan (c+d x))^2} \]
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Time = 0.07 (sec) , antiderivative size = 159, normalized size of antiderivative = 0.50
method | result | size |
derivativedivides | \(\frac {-2 B \left (\sqrt {\tan }\left (d x +c \right )\right )+\frac {i \left (\frac {\left (-\frac {7 i A}{2}+\frac {11 B}{2}\right ) \left (\tan ^{\frac {3}{2}}\left (d x +c \right )\right )+\left (-\frac {5 A}{2}-\frac {9 i B}{2}\right ) \left (\sqrt {\tan }\left (d x +c \right )\right )}{\left (\tan \left (d x +c \right )-i\right )^{2}}+\frac {\left (7 i A -23 B \right ) \arctan \left (\frac {2 \left (\sqrt {\tan }\left (d x +c \right )\right )}{\sqrt {2}-i \sqrt {2}}\right )}{\sqrt {2}-i \sqrt {2}}\right )}{4}+\frac {i \left (i A +B \right ) \arctan \left (\frac {2 \left (\sqrt {\tan }\left (d x +c \right )\right )}{\sqrt {2}+i \sqrt {2}}\right )}{2 \sqrt {2}+2 i \sqrt {2}}}{d \,a^{2}}\) | \(159\) |
default | \(\frac {-2 B \left (\sqrt {\tan }\left (d x +c \right )\right )+\frac {i \left (\frac {\left (-\frac {7 i A}{2}+\frac {11 B}{2}\right ) \left (\tan ^{\frac {3}{2}}\left (d x +c \right )\right )+\left (-\frac {5 A}{2}-\frac {9 i B}{2}\right ) \left (\sqrt {\tan }\left (d x +c \right )\right )}{\left (\tan \left (d x +c \right )-i\right )^{2}}+\frac {\left (7 i A -23 B \right ) \arctan \left (\frac {2 \left (\sqrt {\tan }\left (d x +c \right )\right )}{\sqrt {2}-i \sqrt {2}}\right )}{\sqrt {2}-i \sqrt {2}}\right )}{4}+\frac {i \left (i A +B \right ) \arctan \left (\frac {2 \left (\sqrt {\tan }\left (d x +c \right )\right )}{\sqrt {2}+i \sqrt {2}}\right )}{2 \sqrt {2}+2 i \sqrt {2}}}{d \,a^{2}}\) | \(159\) |
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Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 664 vs. \(2 (235) = 470\).
Time = 0.27 (sec) , antiderivative size = 664, normalized size of antiderivative = 2.10 \[ \int \frac {\tan ^{\frac {5}{2}}(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^2} \, dx=\frac {{\left (2 \, a^{2} d \sqrt {\frac {i \, A^{2} + 2 \, A B - i \, B^{2}}{a^{4} d^{2}}} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (\frac {2 \, {\left ({\left (a^{2} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2} 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^{4} 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 ) - 2 \, a^{2} d \sqrt {\frac {i \, A^{2} + 2 \, A B - i \, B^{2}}{a^{4} d^{2}}} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (-\frac {2 \, {\left ({\left (a^{2} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2} 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^{4} 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 ) + a^{2} d \sqrt {\frac {-49 i \, A^{2} + 322 \, A B + 529 i \, B^{2}}{a^{4} d^{2}}} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (\frac {{\left ({\left (a^{2} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2} 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 {-49 i \, A^{2} + 322 \, A B + 529 i \, B^{2}}{a^{4} d^{2}}} + 7 \, A + 23 i \, B\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{8 \, a^{2} d}\right ) - a^{2} d \sqrt {\frac {-49 i \, A^{2} + 322 \, A B + 529 i \, B^{2}}{a^{4} d^{2}}} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (-\frac {{\left ({\left (a^{2} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2} 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 {-49 i \, A^{2} + 322 \, A B + 529 i \, B^{2}}{a^{4} d^{2}}} - 7 \, A - 23 i \, B\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{8 \, a^{2} d}\right ) - 2 \, {\left (6 \, {\left (-i \, A + 7 \, B\right )} e^{\left (4 i \, d x + 4 i \, c\right )} - {\left (5 i \, A - 9 \, 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}}\right )} e^{\left (-4 i \, d x - 4 i \, c\right )}}{32 \, a^{2} d} \]
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\[ \int \frac {\tan ^{\frac {5}{2}}(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^2} \, dx=- \frac {\int \frac {A \tan ^{\frac {5}{2}}{\left (c + d x \right )}}{\tan ^{2}{\left (c + d x \right )} - 2 i \tan {\left (c + d x \right )} - 1}\, dx + \int \frac {B \tan ^{\frac {7}{2}}{\left (c + d x \right )}}{\tan ^{2}{\left (c + d x \right )} - 2 i \tan {\left (c + d x \right )} - 1}\, dx}{a^{2}} \]
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Exception generated. \[ \int \frac {\tan ^{\frac {5}{2}}(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^2} \, dx=\text {Exception raised: RuntimeError} \]
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Time = 0.77 (sec) , antiderivative size = 143, normalized size of antiderivative = 0.45 \[ \int \frac {\tan ^{\frac {5}{2}}(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^2} \, dx=-\frac {\left (i + 1\right ) \, \sqrt {2} {\left (7 \, A + 23 i \, B\right )} \arctan \left (\left (\frac {1}{2} i + \frac {1}{2}\right ) \, \sqrt {2} \sqrt {\tan \left (d x + c\right )}\right )}{16 \, a^{2} d} + \frac {\left (i - 1\right ) \, \sqrt {2} {\left (A - i \, B\right )} \arctan \left (-\left (\frac {1}{2} i - \frac {1}{2}\right ) \, \sqrt {2} \sqrt {\tan \left (d x + c\right )}\right )}{8 \, a^{2} d} - \frac {2 \, B \sqrt {\tan \left (d x + c\right )}}{a^{2} d} + \frac {7 \, A \tan \left (d x + c\right )^{\frac {3}{2}} + 11 i \, B \tan \left (d x + c\right )^{\frac {3}{2}} - 5 i \, A \sqrt {\tan \left (d x + c\right )} + 9 \, B \sqrt {\tan \left (d x + c\right )}}{8 \, a^{2} d {\left (\tan \left (d x + c\right ) - i\right )}^{2}} \]
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Time = 12.04 (sec) , antiderivative size = 334, normalized size of antiderivative = 1.06 \[ \int \frac {\tan ^{\frac {5}{2}}(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^2} \, dx=\frac {\frac {5\,A\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}}{8\,a^2\,d}+\frac {A\,{\mathrm {tan}\left (c+d\,x\right )}^{3/2}\,7{}\mathrm {i}}{8\,a^2\,d}}{{\mathrm {tan}\left (c+d\,x\right )}^2\,1{}\mathrm {i}+2\,\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}}+\frac {-\frac {11\,B\,{\mathrm {tan}\left (c+d\,x\right )}^{3/2}}{8\,a^2\,d}+\frac {B\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,9{}\mathrm {i}}{8\,a^2\,d}}{{\mathrm {tan}\left (c+d\,x\right )}^2\,1{}\mathrm {i}+2\,\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}}+2\,\mathrm {atanh}\left (\frac {8\,a^2\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {\frac {A^2\,1{}\mathrm {i}}{64\,a^4\,d^2}}}{A}\right )\,\sqrt {\frac {A^2\,1{}\mathrm {i}}{64\,a^4\,d^2}}+2\,\mathrm {atanh}\left (\frac {16\,a^2\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {-\frac {A^2\,49{}\mathrm {i}}{256\,a^4\,d^2}}}{7\,A}\right )\,\sqrt {-\frac {A^2\,49{}\mathrm {i}}{256\,a^4\,d^2}}-\frac {2\,B\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}}{a^2\,d}+\mathrm {atan}\left (\frac {8\,a^2\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {-\frac {B^2\,1{}\mathrm {i}}{64\,a^4\,d^2}}}{B}\right )\,\sqrt {-\frac {B^2\,1{}\mathrm {i}}{64\,a^4\,d^2}}\,2{}\mathrm {i}-\mathrm {atan}\left (\frac {16\,a^2\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {\frac {B^2\,529{}\mathrm {i}}{256\,a^4\,d^2}}}{23\,B}\right )\,\sqrt {\frac {B^2\,529{}\mathrm {i}}{256\,a^4\,d^2}}\,2{}\mathrm {i} \]
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