Integrand size = 36, antiderivative size = 318 \[ \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^2} \, dx=\frac {((25+21 i) A-(9-5 i) B) \arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{16 \sqrt {2} a^2 d}-\frac {\left (\frac {1}{16}-\frac {i}{16}\right ) ((2+23 i) A-(7+2 i) B) \arctan \left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} a^2 d}-\frac {\left (\frac {1}{32}-\frac {i}{32}\right ) ((23+2 i) A+(2+7 i) B) \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{\sqrt {2} a^2 d}+\frac {\left (\frac {1}{32}-\frac {i}{32}\right ) ((23+2 i) A+(2+7 i) B) \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{\sqrt {2} a^2 d}-\frac {5 (5 A+i B)}{8 a^2 d \sqrt {\tan (c+d x)}}+\frac {7 A+3 i B}{8 a^2 d (1+i \tan (c+d x)) \sqrt {\tan (c+d x)}}+\frac {A+i B}{4 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^2} \]
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Time = 0.89 (sec) , antiderivative size = 318, 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 = {3677, 3610, 3615, 1182, 1176, 631, 210, 1179, 642} \[ \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^2} \, dx=\frac {((25+21 i) A-(9-5 i) B) \arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{16 \sqrt {2} a^2 d}-\frac {\left (\frac {1}{16}-\frac {i}{16}\right ) ((2+23 i) A-(7+2 i) B) \arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2} a^2 d}-\frac {5 (5 A+i B)}{8 a^2 d \sqrt {\tan (c+d x)}}+\frac {7 A+3 i B}{8 a^2 d (1+i \tan (c+d x)) \sqrt {\tan (c+d x)}}-\frac {\left (\frac {1}{32}-\frac {i}{32}\right ) ((23+2 i) A+(2+7 i) B) \log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2} a^2 d}+\frac {\left (\frac {1}{32}-\frac {i}{32}\right ) ((23+2 i) A+(2+7 i) B) \log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2} a^2 d}+\frac {A+i B}{4 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^2} \]
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Rule 210
Rule 631
Rule 642
Rule 1176
Rule 1179
Rule 1182
Rule 3610
Rule 3615
Rule 3677
Rubi steps \begin{align*} \text {integral}& = \frac {A+i B}{4 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^2}+\frac {\int \frac {\frac {1}{2} a (9 A+i B)-\frac {5}{2} a (i A-B) \tan (c+d x)}{\tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))} \, dx}{4 a^2} \\ & = \frac {7 A+3 i B}{8 a^2 d (1+i \tan (c+d x)) \sqrt {\tan (c+d x)}}+\frac {A+i B}{4 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^2}+\frac {\int \frac {\frac {5}{2} a^2 (5 A+i B)-\frac {3}{2} a^2 (7 i A-3 B) \tan (c+d x)}{\tan ^{\frac {3}{2}}(c+d x)} \, dx}{8 a^4} \\ & = -\frac {5 (5 A+i B)}{8 a^2 d \sqrt {\tan (c+d x)}}+\frac {7 A+3 i B}{8 a^2 d (1+i \tan (c+d x)) \sqrt {\tan (c+d x)}}+\frac {A+i B}{4 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^2}+\frac {\int \frac {-\frac {3}{2} a^2 (7 i A-3 B)-\frac {5}{2} a^2 (5 A+i B) \tan (c+d x)}{\sqrt {\tan (c+d x)}} \, dx}{8 a^4} \\ & = -\frac {5 (5 A+i B)}{8 a^2 d \sqrt {\tan (c+d x)}}+\frac {7 A+3 i B}{8 a^2 d (1+i \tan (c+d x)) \sqrt {\tan (c+d x)}}+\frac {A+i B}{4 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^2}+\frac {\text {Subst}\left (\int \frac {-\frac {3}{2} a^2 (7 i A-3 B)-\frac {5}{2} a^2 (5 A+i B) x^2}{1+x^4} \, dx,x,\sqrt {\tan (c+d x)}\right )}{4 a^4 d} \\ & = -\frac {5 (5 A+i B)}{8 a^2 d \sqrt {\tan (c+d x)}}+\frac {7 A+3 i B}{8 a^2 d (1+i \tan (c+d x)) \sqrt {\tan (c+d x)}}+\frac {A+i B}{4 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^2}-\frac {((25+21 i) A-(9-5 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 {((25-21 i) A+(9+5 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 (5 A+i B)}{8 a^2 d \sqrt {\tan (c+d x)}}+\frac {7 A+3 i B}{8 a^2 d (1+i \tan (c+d x)) \sqrt {\tan (c+d x)}}+\frac {A+i B}{4 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^2}-\frac {((25+21 i) A-(9-5 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 {((25+21 i) A-(9-5 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 {((25-21 i) A+(9+5 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 {((25-21 i) A+(9+5 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 {((25-21 i) A+(9+5 i) B) \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{32 \sqrt {2} a^2 d}+\frac {((25-21 i) A+(9+5 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 (5 A+i B)}{8 a^2 d \sqrt {\tan (c+d x)}}+\frac {7 A+3 i B}{8 a^2 d (1+i \tan (c+d x)) \sqrt {\tan (c+d x)}}+\frac {A+i B}{4 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^2}-\frac {((25+21 i) A-(9-5 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 {((25+21 i) A-(9-5 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 {((25+21 i) A-(9-5 i) B) \arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{16 \sqrt {2} a^2 d}-\frac {((25+21 i) A-(9-5 i) B) \arctan \left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{16 \sqrt {2} a^2 d}-\frac {((25-21 i) A+(9+5 i) B) \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{32 \sqrt {2} a^2 d}+\frac {((25-21 i) A+(9+5 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 (5 A+i B)}{8 a^2 d \sqrt {\tan (c+d x)}}+\frac {7 A+3 i B}{8 a^2 d (1+i \tan (c+d x)) \sqrt {\tan (c+d x)}}+\frac {A+i B}{4 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^2} \\ \end{align*}
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 2.04 (sec) , antiderivative size = 180, normalized size of antiderivative = 0.57 \[ \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^2} \, dx=\frac {\sec ^2(c+d x) \left (-2 \cos (c+d x) ((9 A+5 i B) \cos (c+d x)+(7 i A-3 B) \sin (c+d x))+2 (23 A+7 i B) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},-i \tan (c+d x)\right ) (\cos (2 (c+d x))+i \sin (2 (c+d x)))+4 (A-i B) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},i \tan (c+d x)\right ) (\cos (2 (c+d x))+i \sin (2 (c+d x)))\right )}{16 a^2 d \sqrt {\tan (c+d x)} (-i+\tan (c+d x))^2} \]
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Time = 0.06 (sec) , antiderivative size = 157, normalized size of antiderivative = 0.49
method | result | size |
derivativedivides | \(\frac {\frac {\left (-\frac {9 A}{2}-\frac {5 i B}{2}\right ) \left (\tan ^{\frac {3}{2}}\left (d x +c \right )\right )+\left (\frac {11 i A}{2}-\frac {7 B}{2}\right ) \left (\sqrt {\tan }\left (d x +c \right )\right )}{4 \left (\tan \left (d x +c \right )-i\right )^{2}}-\frac {\left (7 i B +23 A \right ) \arctan \left (\frac {2 \left (\sqrt {\tan }\left (d x +c \right )\right )}{\sqrt {2}-i \sqrt {2}}\right )}{4 \left (\sqrt {2}-i \sqrt {2}\right )}-\frac {2 A}{\sqrt {\tan \left (d x +c \right )}}+\frac {i \left (i A +B \right ) \arctan \left (\frac {2 \left (\sqrt {\tan }\left (d x +c \right )\right )}{\sqrt {2}+i \sqrt {2}}\right )}{2 \sqrt {2}+2 i \sqrt {2}}}{d \,a^{2}}\) | \(157\) |
default | \(\frac {\frac {\left (-\frac {9 A}{2}-\frac {5 i B}{2}\right ) \left (\tan ^{\frac {3}{2}}\left (d x +c \right )\right )+\left (\frac {11 i A}{2}-\frac {7 B}{2}\right ) \left (\sqrt {\tan }\left (d x +c \right )\right )}{4 \left (\tan \left (d x +c \right )-i\right )^{2}}-\frac {\left (7 i B +23 A \right ) \arctan \left (\frac {2 \left (\sqrt {\tan }\left (d x +c \right )\right )}{\sqrt {2}-i \sqrt {2}}\right )}{4 \left (\sqrt {2}-i \sqrt {2}\right )}-\frac {2 A}{\sqrt {\tan \left (d x +c \right )}}+\frac {i \left (i A +B \right ) \arctan \left (\frac {2 \left (\sqrt {\tan }\left (d x +c \right )\right )}{\sqrt {2}+i \sqrt {2}}\right )}{2 \sqrt {2}+2 i \sqrt {2}}}{d \,a^{2}}\) | \(157\) |
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Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 763 vs. \(2 (235) = 470\).
Time = 0.28 (sec) , antiderivative size = 763, normalized size of antiderivative = 2.40 \[ \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^2} \, dx=\frac {2 \, {\left (a^{2} d e^{\left (6 i \, d x + 6 i \, c\right )} - a^{2} d e^{\left (4 i \, d x + 4 i \, c\right )}\right )} \sqrt {\frac {i \, A^{2} + 2 \, A B - i \, B^{2}}{a^{4} d^{2}}} \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 \, {\left (a^{2} d e^{\left (6 i \, d x + 6 i \, c\right )} - a^{2} d e^{\left (4 i \, d x + 4 i \, c\right )}\right )} \sqrt {\frac {i \, A^{2} + 2 \, A B - i \, B^{2}}{a^{4} d^{2}}} \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 ) + {\left (a^{2} d e^{\left (6 i \, d x + 6 i \, c\right )} - a^{2} d e^{\left (4 i \, d x + 4 i \, c\right )}\right )} \sqrt {\frac {-529 i \, A^{2} + 322 \, A B + 49 i \, B^{2}}{a^{4} d^{2}}} \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 {-529 i \, A^{2} + 322 \, A B + 49 i \, B^{2}}{a^{4} d^{2}}} + 23 \, A + 7 i \, B\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{8 \, a^{2} d}\right ) - {\left (a^{2} d e^{\left (6 i \, d x + 6 i \, c\right )} - a^{2} d e^{\left (4 i \, d x + 4 i \, c\right )}\right )} \sqrt {\frac {-529 i \, A^{2} + 322 \, A B + 49 i \, B^{2}}{a^{4} d^{2}}} \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 {-529 i \, A^{2} + 322 \, A B + 49 i \, B^{2}}{a^{4} d^{2}}} - 23 \, A - 7 i \, B\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{8 \, a^{2} d}\right ) - 2 \, {\left (6 \, {\left (7 i \, A - B\right )} e^{\left (6 i \, d x + 6 i \, c\right )} - {\left (-33 i \, A + B\right )} e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, {\left (-5 i \, A + 3 \, 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}}}{32 \, {\left (a^{2} d e^{\left (6 i \, d x + 6 i \, c\right )} - a^{2} d e^{\left (4 i \, d x + 4 i \, c\right )}\right )}} \]
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Exception generated. \[ \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^2} \, dx=\text {Exception raised: TypeError} \]
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Exception generated. \[ \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^2} \, dx=\text {Exception raised: RuntimeError} \]
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none
Time = 1.22 (sec) , antiderivative size = 143, normalized size of antiderivative = 0.45 \[ \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^2} \, dx=-\frac {\left (i + 1\right ) \, \sqrt {2} {\left (23 \, A + 7 i \, B\right )} \arctan \left (\left (\frac {1}{2} i + \frac {1}{2}\right ) \, \sqrt {2} \sqrt {\tan \left (d x + c\right )}\right )}{16 \, a^{2} d} + \frac {\left (i - 1\right ) \, \sqrt {2} {\left (A - i \, B\right )} \arctan \left (-\left (\frac {1}{2} i - \frac {1}{2}\right ) \, \sqrt {2} \sqrt {\tan \left (d x + c\right )}\right )}{8 \, a^{2} d} - \frac {2 \, A}{a^{2} d \sqrt {\tan \left (d x + c\right )}} - \frac {9 \, A \tan \left (d x + c\right )^{\frac {3}{2}} + 5 i \, B \tan \left (d x + c\right )^{\frac {3}{2}} - 11 i \, A \sqrt {\tan \left (d x + c\right )} + 7 \, B \sqrt {\tan \left (d x + c\right )}}{8 \, a^{2} d {\left (\tan \left (d x + c\right ) - i\right )}^{2}} \]
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Time = 11.93 (sec) , antiderivative size = 338, normalized size of antiderivative = 1.06 \[ \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^2} \, dx=-\frac {-\frac {5\,B\,{\mathrm {tan}\left (c+d\,x\right )}^{3/2}}{8\,a^2\,d}+\frac {B\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,7{}\mathrm {i}}{8\,a^2\,d}}{{\mathrm {tan}\left (c+d\,x\right )}^2\,1{}\mathrm {i}+2\,\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}}+2\,\mathrm {atanh}\left (\frac {8\,a^2\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {\frac {A^2\,1{}\mathrm {i}}{64\,a^4\,d^2}}}{A}\right )\,\sqrt {\frac {A^2\,1{}\mathrm {i}}{64\,a^4\,d^2}}+2\,\mathrm {atanh}\left (\frac {16\,a^2\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {-\frac {A^2\,529{}\mathrm {i}}{256\,a^4\,d^2}}}{23\,A}\right )\,\sqrt {-\frac {A^2\,529{}\mathrm {i}}{256\,a^4\,d^2}}+\mathrm {atan}\left (\frac {8\,a^2\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {-\frac {B^2\,1{}\mathrm {i}}{64\,a^4\,d^2}}}{B}\right )\,\sqrt {-\frac {B^2\,1{}\mathrm {i}}{64\,a^4\,d^2}}\,2{}\mathrm {i}-\mathrm {atan}\left (\frac {16\,a^2\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {\frac {B^2\,49{}\mathrm {i}}{256\,a^4\,d^2}}}{7\,B}\right )\,\sqrt {\frac {B^2\,49{}\mathrm {i}}{256\,a^4\,d^2}}\,2{}\mathrm {i}-\frac {\frac {43\,A\,\mathrm {tan}\left (c+d\,x\right )}{8\,a^2\,d}-\frac {A\,2{}\mathrm {i}}{a^2\,d}+\frac {A\,{\mathrm {tan}\left (c+d\,x\right )}^2\,25{}\mathrm {i}}{8\,a^2\,d}}{2\,{\mathrm {tan}\left (c+d\,x\right )}^{3/2}-\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,1{}\mathrm {i}+{\mathrm {tan}\left (c+d\,x\right )}^{5/2}\,1{}\mathrm {i}} \]
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