Integrand size = 36, antiderivative size = 347 \[ \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^2} \, dx=\frac {\left (\frac {1}{16}-\frac {i}{16}\right ) ((47+2 i) A+(2+23 i) B) \arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} a^2 d}-\frac {\left (\frac {1}{16}-\frac {i}{16}\right ) ((47+2 i) A+(2+23 i) B) \arctan \left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} a^2 d}+\frac {((49+45 i) A-(25-21 i) B) \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{32 \sqrt {2} a^2 d}-\frac {\left (\frac {1}{32}-\frac {i}{32}\right ) ((2+47 i) A-(23+2 i) B) \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{\sqrt {2} a^2 d}-\frac {7 (7 A+3 i B)}{24 a^2 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {9 A+5 i B}{8 a^2 d (1+i \tan (c+d x)) \tan ^{\frac {3}{2}}(c+d x)}+\frac {5 (9 i A-5 B)}{8 a^2 d \sqrt {\tan (c+d x)}}+\frac {A+i B}{4 d \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^2} \]
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Time = 0.90 (sec) , antiderivative size = 347, normalized size of antiderivative = 1.00, number of steps used = 14, 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} \[ \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^2} \, dx=\frac {\left (\frac {1}{16}-\frac {i}{16}\right ) ((47+2 i) A+(2+23 i) B) \arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} a^2 d}-\frac {\left (\frac {1}{16}-\frac {i}{16}\right ) ((47+2 i) A+(2+23 i) B) \arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2} a^2 d}-\frac {7 (7 A+3 i B)}{24 a^2 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {9 A+5 i B}{8 a^2 d (1+i \tan (c+d x)) \tan ^{\frac {3}{2}}(c+d x)}+\frac {5 (-5 B+9 i A)}{8 a^2 d \sqrt {\tan (c+d x)}}+\frac {((49+45 i) A-(25-21 i) B) \log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{32 \sqrt {2} a^2 d}-\frac {\left (\frac {1}{32}-\frac {i}{32}\right ) ((2+47 i) A-(23+2 i) B) \log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2} a^2 d}+\frac {A+i B}{4 d \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^2} \]
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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 \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^2}+\frac {\int \frac {\frac {1}{2} a (11 A+3 i B)-\frac {7}{2} a (i A-B) \tan (c+d x)}{\tan ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))} \, dx}{4 a^2} \\ & = \frac {9 A+5 i B}{8 a^2 d (1+i \tan (c+d x)) \tan ^{\frac {3}{2}}(c+d x)}+\frac {A+i B}{4 d \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^2}+\frac {\int \frac {\frac {7}{2} a^2 (7 A+3 i B)-\frac {5}{2} a^2 (9 i A-5 B) \tan (c+d x)}{\tan ^{\frac {5}{2}}(c+d x)} \, dx}{8 a^4} \\ & = -\frac {7 (7 A+3 i B)}{24 a^2 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {9 A+5 i B}{8 a^2 d (1+i \tan (c+d x)) \tan ^{\frac {3}{2}}(c+d x)}+\frac {A+i B}{4 d \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^2}+\frac {\int \frac {-\frac {5}{2} a^2 (9 i A-5 B)-\frac {7}{2} a^2 (7 A+3 i B) \tan (c+d x)}{\tan ^{\frac {3}{2}}(c+d x)} \, dx}{8 a^4} \\ & = -\frac {7 (7 A+3 i B)}{24 a^2 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {9 A+5 i B}{8 a^2 d (1+i \tan (c+d x)) \tan ^{\frac {3}{2}}(c+d x)}+\frac {5 (9 i A-5 B)}{8 a^2 d \sqrt {\tan (c+d x)}}+\frac {A+i B}{4 d \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^2}+\frac {\int \frac {-\frac {7}{2} a^2 (7 A+3 i B)+\frac {5}{2} a^2 (9 i A-5 B) \tan (c+d x)}{\sqrt {\tan (c+d x)}} \, dx}{8 a^4} \\ & = -\frac {7 (7 A+3 i B)}{24 a^2 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {9 A+5 i B}{8 a^2 d (1+i \tan (c+d x)) \tan ^{\frac {3}{2}}(c+d x)}+\frac {5 (9 i A-5 B)}{8 a^2 d \sqrt {\tan (c+d x)}}+\frac {A+i B}{4 d \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^2}+\frac {\text {Subst}\left (\int \frac {-\frac {7}{2} a^2 (7 A+3 i B)+\frac {5}{2} a^2 (9 i A-5 B) x^2}{1+x^4} \, dx,x,\sqrt {\tan (c+d x)}\right )}{4 a^4 d} \\ & = -\frac {7 (7 A+3 i B)}{24 a^2 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {9 A+5 i B}{8 a^2 d (1+i \tan (c+d x)) \tan ^{\frac {3}{2}}(c+d x)}+\frac {5 (9 i A-5 B)}{8 a^2 d \sqrt {\tan (c+d x)}}+\frac {A+i B}{4 d \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^2}-\frac {((49+45 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 {\left (\left (\frac {1}{16}-\frac {i}{16}\right ) ((47+2 i) A+(2+23 i) B)\right ) \text {Subst}\left (\int \frac {1+x^2}{1+x^4} \, dx,x,\sqrt {\tan (c+d x)}\right )}{a^2 d} \\ & = -\frac {7 (7 A+3 i B)}{24 a^2 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {9 A+5 i B}{8 a^2 d (1+i \tan (c+d x)) \tan ^{\frac {3}{2}}(c+d x)}+\frac {5 (9 i A-5 B)}{8 a^2 d \sqrt {\tan (c+d x)}}+\frac {A+i B}{4 d \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^2}+\frac {((49+45 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 {((49+45 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 {\left (\left (\frac {1}{32}-\frac {i}{32}\right ) ((47+2 i) A+(2+23 i) B)\right ) \text {Subst}\left (\int \frac {1}{1-\sqrt {2} x+x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{a^2 d}+-\frac {\left (\left (\frac {1}{32}-\frac {i}{32}\right ) ((47+2 i) A+(2+23 i) B)\right ) \text {Subst}\left (\int \frac {1}{1+\sqrt {2} x+x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{a^2 d} \\ & = \frac {((49+45 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 {((49+45 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 {7 (7 A+3 i B)}{24 a^2 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {9 A+5 i B}{8 a^2 d (1+i \tan (c+d x)) \tan ^{\frac {3}{2}}(c+d x)}+\frac {5 (9 i A-5 B)}{8 a^2 d \sqrt {\tan (c+d x)}}+\frac {A+i B}{4 d \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^2}+-\frac {\left (\left (\frac {1}{16}-\frac {i}{16}\right ) ((47+2 i) A+(2+23 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^2 d}+\frac {\left (\left (\frac {1}{16}-\frac {i}{16}\right ) ((47+2 i) A+(2+23 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^2 d} \\ & = \frac {\left (\frac {1}{16}-\frac {i}{16}\right ) ((47+2 i) A+(2+23 i) B) \arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} a^2 d}-\frac {\left (\frac {1}{16}-\frac {i}{16}\right ) ((47+2 i) A+(2+23 i) B) \arctan \left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} a^2 d}+\frac {((49+45 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 {((49+45 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 {7 (7 A+3 i B)}{24 a^2 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {9 A+5 i B}{8 a^2 d (1+i \tan (c+d x)) \tan ^{\frac {3}{2}}(c+d x)}+\frac {5 (9 i A-5 B)}{8 a^2 d \sqrt {\tan (c+d x)}}+\frac {A+i B}{4 d \tan ^{\frac {3}{2}}(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 = 1.97 (sec) , antiderivative size = 180, normalized size of antiderivative = 0.52 \[ \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^2} \, dx=\frac {\sec ^2(c+d x) \left (-6 \cos (c+d x) ((11 A+7 i B) \cos (c+d x)+(9 i A-5 B) \sin (c+d x))+2 (47 A+23 i B) \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},1,-\frac {1}{2},-i \tan (c+d x)\right ) (\cos (2 (c+d x))+i \sin (2 (c+d x)))+4 (A-i B) \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},1,-\frac {1}{2},i \tan (c+d x)\right ) (\cos (2 (c+d x))+i \sin (2 (c+d x)))\right )}{48 a^2 d \tan ^{\frac {3}{2}}(c+d x) (-i+\tan (c+d x))^2} \]
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Time = 0.06 (sec) , antiderivative size = 176, normalized size of antiderivative = 0.51
method | result | size |
derivativedivides | \(\frac {-\frac {i \left (\frac {\left (-\frac {13 A}{2}-\frac {9 i B}{2}\right ) \left (\tan ^{\frac {3}{2}}\left (d x +c \right )\right )+\left (\frac {15 i A}{2}-\frac {11 B}{2}\right ) \left (\sqrt {\tan }\left (d x +c \right )\right )}{\left (\tan \left (d x +c \right )-i\right )^{2}}-\frac {\left (23 i B +47 A \right ) \arctan \left (\frac {2 \left (\sqrt {\tan }\left (d x +c \right )\right )}{\sqrt {2}-i \sqrt {2}}\right )}{\sqrt {2}-i \sqrt {2}}\right )}{4}-\frac {2 A}{3 \tan \left (d x +c \right )^{\frac {3}{2}}}-\frac {2 \left (-2 i A +B \right )}{\sqrt {\tan \left (d x +c \right )}}-\frac {i \left (-i B +A \right ) \arctan \left (\frac {2 \left (\sqrt {\tan }\left (d x +c \right )\right )}{\sqrt {2}+i \sqrt {2}}\right )}{2 \left (\sqrt {2}+i \sqrt {2}\right )}}{d \,a^{2}}\) | \(176\) |
default | \(\frac {-\frac {i \left (\frac {\left (-\frac {13 A}{2}-\frac {9 i B}{2}\right ) \left (\tan ^{\frac {3}{2}}\left (d x +c \right )\right )+\left (\frac {15 i A}{2}-\frac {11 B}{2}\right ) \left (\sqrt {\tan }\left (d x +c \right )\right )}{\left (\tan \left (d x +c \right )-i\right )^{2}}-\frac {\left (23 i B +47 A \right ) \arctan \left (\frac {2 \left (\sqrt {\tan }\left (d x +c \right )\right )}{\sqrt {2}-i \sqrt {2}}\right )}{\sqrt {2}-i \sqrt {2}}\right )}{4}-\frac {2 A}{3 \tan \left (d x +c \right )^{\frac {3}{2}}}-\frac {2 \left (-2 i A +B \right )}{\sqrt {\tan \left (d x +c \right )}}-\frac {i \left (-i B +A \right ) \arctan \left (\frac {2 \left (\sqrt {\tan }\left (d x +c \right )\right )}{\sqrt {2}+i \sqrt {2}}\right )}{2 \left (\sqrt {2}+i \sqrt {2}\right )}}{d \,a^{2}}\) | \(176\) |
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Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 861 vs. \(2 (258) = 516\).
Time = 0.29 (sec) , antiderivative size = 861, normalized size of antiderivative = 2.48 \[ \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^2} \, dx=\text {Too large to display} \]
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Timed out. \[ \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^2} \, dx=\text {Timed out} \]
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Exception generated. \[ \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^2} \, dx=\text {Exception raised: RuntimeError} \]
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none
Time = 1.49 (sec) , antiderivative size = 162, normalized size of antiderivative = 0.47 \[ \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^2} \, dx=-\frac {\left (i + 1\right ) \, \sqrt {2} {\left (-47 i \, A + 23 \, B\right )} \arctan \left (\left (\frac {1}{2} i + \frac {1}{2}\right ) \, \sqrt {2} \sqrt {\tan \left (d x + c\right )}\right )}{16 \, a^{2} d} + \frac {\left (i - 1\right ) \, \sqrt {2} {\left (i \, A + B\right )} \arctan \left (-\left (\frac {1}{2} i - \frac {1}{2}\right ) \, \sqrt {2} \sqrt {\tan \left (d x + c\right )}\right )}{8 \, a^{2} d} - \frac {2 \, {\left (-6 i \, A \tan \left (d x + c\right ) + 3 \, B \tan \left (d x + c\right ) + A\right )}}{3 \, a^{2} d \tan \left (d x + c\right )^{\frac {3}{2}}} - \frac {-13 i \, A \tan \left (d x + c\right )^{\frac {3}{2}} + 9 \, B \tan \left (d x + c\right )^{\frac {3}{2}} - 15 \, A \sqrt {\tan \left (d x + c\right )} - 11 i \, B \sqrt {\tan \left (d x + c\right )}}{8 \, a^{2} d {\left (\tan \left (d x + c\right ) - i\right )}^{2}} \]
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Time = 12.22 (sec) , antiderivative size = 373, normalized size of antiderivative = 1.07 \[ \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^2} \, dx=-\mathrm {atan}\left (\frac {8\,a^2\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {-\frac {A^2\,1{}\mathrm {i}}{64\,a^4\,d^2}}}{A}\right )\,\sqrt {-\frac {A^2\,1{}\mathrm {i}}{64\,a^4\,d^2}}\,2{}\mathrm {i}+\mathrm {atan}\left (\frac {16\,a^2\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {\frac {A^2\,2209{}\mathrm {i}}{256\,a^4\,d^2}}}{47\,A}\right )\,\sqrt {\frac {A^2\,2209{}\mathrm {i}}{256\,a^4\,d^2}}\,2{}\mathrm {i}+2\,\mathrm {atanh}\left (\frac {8\,a^2\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {\frac {B^2\,1{}\mathrm {i}}{64\,a^4\,d^2}}}{B}\right )\,\sqrt {\frac {B^2\,1{}\mathrm {i}}{64\,a^4\,d^2}}+2\,\mathrm {atanh}\left (\frac {16\,a^2\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {-\frac {B^2\,529{}\mathrm {i}}{256\,a^4\,d^2}}}{23\,B}\right )\,\sqrt {-\frac {B^2\,529{}\mathrm {i}}{256\,a^4\,d^2}}+\frac {\frac {8\,A\,\mathrm {tan}\left (c+d\,x\right )}{3\,a^2\,d}-\frac {45\,A\,{\mathrm {tan}\left (c+d\,x\right )}^3}{8\,a^2\,d}+\frac {A\,2{}\mathrm {i}}{3\,a^2\,d}+\frac {A\,{\mathrm {tan}\left (c+d\,x\right )}^2\,221{}\mathrm {i}}{24\,a^2\,d}}{2\,{\mathrm {tan}\left (c+d\,x\right )}^{5/2}-{\mathrm {tan}\left (c+d\,x\right )}^{3/2}\,1{}\mathrm {i}+{\mathrm {tan}\left (c+d\,x\right )}^{7/2}\,1{}\mathrm {i}}-\frac {\frac {43\,B\,\mathrm {tan}\left (c+d\,x\right )}{8\,a^2\,d}-\frac {B\,2{}\mathrm {i}}{a^2\,d}+\frac {B\,{\mathrm {tan}\left (c+d\,x\right )}^2\,25{}\mathrm {i}}{8\,a^2\,d}}{2\,{\mathrm {tan}\left (c+d\,x\right )}^{3/2}-\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,1{}\mathrm {i}+{\mathrm {tan}\left (c+d\,x\right )}^{5/2}\,1{}\mathrm {i}} \]
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