Integrand size = 36, antiderivative size = 279 \[ \int \frac {\sqrt {\tan (c+d x)} (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^2} \, dx=\frac {((-1+3 i) A+(1+3 i) B) \arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{16 \sqrt {2} a^2 d}-\frac {((-1+3 i) A+(1+3 i) B) \arctan \left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{16 \sqrt {2} a^2 d}+\frac {((1+3 i) A+(1-3 i) B) \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{32 \sqrt {2} a^2 d}-\frac {((1+3 i) A+(1-3 i) B) \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{32 \sqrt {2} a^2 d}+\frac {(i A+3 B) \sqrt {\tan (c+d x)}}{8 a^2 d (1+i \tan (c+d x))}+\frac {(i A-B) \sqrt {\tan (c+d x)}}{4 d (a+i a \tan (c+d x))^2} \]
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Time = 0.53 (sec) , antiderivative size = 279, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3676, 3677, 3615, 1182, 1176, 631, 210, 1179, 642} \[ \int \frac {\sqrt {\tan (c+d x)} (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^2} \, dx=\frac {((1+3 i) B-(1-3 i) A) \arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{16 \sqrt {2} a^2 d}-\frac {((1+3 i) B-(1-3 i) A) \arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{16 \sqrt {2} a^2 d}+\frac {(3 B+i A) \sqrt {\tan (c+d x)}}{8 a^2 d (1+i \tan (c+d x))}+\frac {((1+3 i) A+(1-3 i) B) \log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{32 \sqrt {2} a^2 d}-\frac {((1+3 i) A+(1-3 i) B) \log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{32 \sqrt {2} a^2 d}+\frac {(-B+i A) \sqrt {\tan (c+d x)}}{4 d (a+i a \tan (c+d x))^2} \]
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Rule 210
Rule 631
Rule 642
Rule 1176
Rule 1179
Rule 1182
Rule 3615
Rule 3676
Rule 3677
Rubi steps \begin{align*} \text {integral}& = \frac {(i A-B) \sqrt {\tan (c+d x)}}{4 d (a+i a \tan (c+d x))^2}-\frac {\int \frac {\frac {1}{2} a (i A-B)-\frac {1}{2} a (3 A-5 i B) \tan (c+d x)}{\sqrt {\tan (c+d x)} (a+i a \tan (c+d x))} \, dx}{4 a^2} \\ & = \frac {(i A+3 B) \sqrt {\tan (c+d x)}}{8 a^2 d (1+i \tan (c+d x))}+\frac {(i A-B) \sqrt {\tan (c+d x)}}{4 d (a+i a \tan (c+d x))^2}-\frac {\int \frac {\frac {1}{2} a^2 (3 i A+B)-\frac {1}{2} a^2 (A-3 i B) \tan (c+d x)}{\sqrt {\tan (c+d x)}} \, dx}{8 a^4} \\ & = \frac {(i A+3 B) \sqrt {\tan (c+d x)}}{8 a^2 d (1+i \tan (c+d x))}+\frac {(i A-B) \sqrt {\tan (c+d x)}}{4 d (a+i a \tan (c+d x))^2}-\frac {\text {Subst}\left (\int \frac {\frac {1}{2} a^2 (3 i A+B)-\frac {1}{2} a^2 (A-3 i B) x^2}{1+x^4} \, dx,x,\sqrt {\tan (c+d x)}\right )}{4 a^4 d} \\ & = \frac {(i A+3 B) \sqrt {\tan (c+d x)}}{8 a^2 d (1+i \tan (c+d x))}+\frac {(i A-B) \sqrt {\tan (c+d x)}}{4 d (a+i a \tan (c+d x))^2}-\frac {((1+3 i) A+(1-3 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 {((-1+3 i) A+(1+3 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 {(i A+3 B) \sqrt {\tan (c+d x)}}{8 a^2 d (1+i \tan (c+d x))}+\frac {(i A-B) \sqrt {\tan (c+d x)}}{4 d (a+i a \tan (c+d x))^2}+\frac {((1+3 i) A+(1-3 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 {((1+3 i) A+(1-3 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 {((-1+3 i) A+(1+3 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 {((-1+3 i) A+(1+3 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 {((1+3 i) A+(1-3 i) B) \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{32 \sqrt {2} a^2 d}-\frac {((1+3 i) A+(1-3 i) B) \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{32 \sqrt {2} a^2 d}+\frac {(i A+3 B) \sqrt {\tan (c+d x)}}{8 a^2 d (1+i \tan (c+d x))}+\frac {(i A-B) \sqrt {\tan (c+d x)}}{4 d (a+i a \tan (c+d x))^2}-\frac {((-1+3 i) A+(1+3 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 {((-1+3 i) A+(1+3 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 {((-1+3 i) A+(1+3 i) B) \arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{16 \sqrt {2} a^2 d}-\frac {((-1+3 i) A+(1+3 i) B) \arctan \left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{16 \sqrt {2} a^2 d}+\frac {((1+3 i) A+(1-3 i) B) \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{32 \sqrt {2} a^2 d}-\frac {((1+3 i) A+(1-3 i) B) \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{32 \sqrt {2} a^2 d}+\frac {(i A+3 B) \sqrt {\tan (c+d x)}}{8 a^2 d (1+i \tan (c+d x))}+\frac {(i A-B) \sqrt {\tan (c+d x)}}{4 d (a+i a \tan (c+d x))^2} \\ \end{align*}
Time = 2.81 (sec) , antiderivative size = 176, normalized size of antiderivative = 0.63 \[ \int \frac {\sqrt {\tan (c+d x)} (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^2} \, dx=\frac {-2 \sqrt [4]{-1} (i A+B) \arctan \left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right ) \sec ^2(c+d x) (\cos (2 (c+d x))+i \sin (2 (c+d x)))+\sqrt [4]{-1} (-i A+B) \text {arctanh}\left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right ) \sec ^2(c+d x) (\cos (2 (c+d x))+i \sin (2 (c+d x)))+\sqrt {\tan (c+d x)} (-3 i A-B+(A-3 i B) \tan (c+d x))}{8 a^2 d (-i+\tan (c+d x))^2} \]
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Time = 0.10 (sec) , antiderivative size = 144, normalized size of antiderivative = 0.52
method | result | size |
derivativedivides | \(\frac {\frac {\left (-\frac {3 i B}{2}+\frac {A}{2}\right ) \left (\tan ^{\frac {3}{2}}\left (d x +c \right )\right )+\left (-\frac {B}{2}-\frac {3 i A}{2}\right ) \left (\sqrt {\tan }\left (d x +c \right )\right )}{4 \left (\tan \left (d x +c \right )-i\right )^{2}}-\frac {\left (i B +A \right ) \arctan \left (\frac {2 \left (\sqrt {\tan }\left (d x +c \right )\right )}{\sqrt {2}-i \sqrt {2}}\right )}{4 \left (\sqrt {2}-i \sqrt {2}\right )}-\frac {i \left (i A +B \right ) \arctan \left (\frac {2 \left (\sqrt {\tan }\left (d x +c \right )\right )}{\sqrt {2}+i \sqrt {2}}\right )}{2 \left (\sqrt {2}+i \sqrt {2}\right )}}{d \,a^{2}}\) | \(144\) |
default | \(\frac {\frac {\left (-\frac {3 i B}{2}+\frac {A}{2}\right ) \left (\tan ^{\frac {3}{2}}\left (d x +c \right )\right )+\left (-\frac {B}{2}-\frac {3 i A}{2}\right ) \left (\sqrt {\tan }\left (d x +c \right )\right )}{4 \left (\tan \left (d x +c \right )-i\right )^{2}}-\frac {\left (i B +A \right ) \arctan \left (\frac {2 \left (\sqrt {\tan }\left (d x +c \right )\right )}{\sqrt {2}-i \sqrt {2}}\right )}{4 \left (\sqrt {2}-i \sqrt {2}\right )}-\frac {i \left (i A +B \right ) \arctan \left (\frac {2 \left (\sqrt {\tan }\left (d x +c \right )\right )}{\sqrt {2}+i \sqrt {2}}\right )}{2 \left (\sqrt {2}+i \sqrt {2}\right )}}{d \,a^{2}}\) | \(144\) |
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Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 658 vs. \(2 (212) = 424\).
Time = 0.28 (sec) , antiderivative size = 658, normalized size of antiderivative = 2.36 \[ \int \frac {\sqrt {\tan (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 {-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 {{\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}}} + A + i \, B\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{8 \, a^{2} d}\right ) + 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 {{\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}}} - A - i \, B\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{8 \, a^{2} d}\right ) + 2 \, {\left (2 \, {\left (-i \, A - B\right )} e^{\left (4 i \, d x + 4 i \, c\right )} - {\left (3 i \, A + 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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\[ \int \frac {\sqrt {\tan (c+d x)} (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^2} \, dx=- \frac {\int \frac {A \sqrt {\tan {\left (c + d x \right )}}}{\tan ^{2}{\left (c + d x \right )} - 2 i \tan {\left (c + d x \right )} - 1}\, dx + \int \frac {B \tan ^{\frac {3}{2}}{\left (c + d x \right )}}{\tan ^{2}{\left (c + d x \right )} - 2 i \tan {\left (c + d x \right )} - 1}\, dx}{a^{2}} \]
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Exception generated. \[ \int \frac {\sqrt {\tan (c+d x)} (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^2} \, dx=\text {Exception raised: RuntimeError} \]
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none
Time = 0.56 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.44 \[ \int \frac {\sqrt {\tan (c+d x)} (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^2} \, dx=-\frac {\left (i + 1\right ) \, \sqrt {2} {\left (A + i \, B\right )} \arctan \left (\left (\frac {1}{2} i + \frac {1}{2}\right ) \, \sqrt {2} \sqrt {\tan \left (d x + c\right )}\right )}{16 \, a^{2} d} - \frac {\left (i - 1\right ) \, \sqrt {2} {\left (A - i \, B\right )} \arctan \left (-\left (\frac {1}{2} i - \frac {1}{2}\right ) \, \sqrt {2} \sqrt {\tan \left (d x + c\right )}\right )}{8 \, a^{2} d} + \frac {A \tan \left (d x + c\right )^{\frac {3}{2}} - 3 i \, B \tan \left (d x + c\right )^{\frac {3}{2}} - 3 i \, A \sqrt {\tan \left (d x + c\right )} - B \sqrt {\tan \left (d x + c\right )}}{8 \, a^{2} d {\left (\tan \left (d x + c\right ) - i\right )}^{2}} \]
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Time = 11.29 (sec) , antiderivative size = 318, normalized size of antiderivative = 1.14 \[ \int \frac {\sqrt {\tan (c+d x)} (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^2} \, dx=\frac {\frac {3\,A\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}}{8\,a^2\,d}+\frac {A\,{\mathrm {tan}\left (c+d\,x\right )}^{3/2}\,1{}\mathrm {i}}{8\,a^2\,d}}{{\mathrm {tan}\left (c+d\,x\right )}^2\,1{}\mathrm {i}+2\,\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}}-\frac {-\frac {3\,B\,{\mathrm {tan}\left (c+d\,x\right )}^{3/2}}{8\,a^2\,d}+\frac {B\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,1{}\mathrm {i}}{8\,a^2\,d}}{{\mathrm {tan}\left (c+d\,x\right )}^2\,1{}\mathrm {i}+2\,\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}}-2\,\mathrm {atanh}\left (\frac {8\,a^2\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {\frac {A^2\,1{}\mathrm {i}}{64\,a^4\,d^2}}}{A}\right )\,\sqrt {\frac {A^2\,1{}\mathrm {i}}{64\,a^4\,d^2}}+2\,\mathrm {atanh}\left (\frac {16\,a^2\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {-\frac {A^2\,1{}\mathrm {i}}{256\,a^4\,d^2}}}{A}\right )\,\sqrt {-\frac {A^2\,1{}\mathrm {i}}{256\,a^4\,d^2}}-\mathrm {atan}\left (\frac {8\,a^2\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {-\frac {B^2\,1{}\mathrm {i}}{64\,a^4\,d^2}}}{B}\right )\,\sqrt {-\frac {B^2\,1{}\mathrm {i}}{64\,a^4\,d^2}}\,2{}\mathrm {i}-\mathrm {atan}\left (\frac {16\,a^2\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {\frac {B^2\,1{}\mathrm {i}}{256\,a^4\,d^2}}}{B}\right )\,\sqrt {\frac {B^2\,1{}\mathrm {i}}{256\,a^4\,d^2}}\,2{}\mathrm {i} \]
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