Integrand size = 36, antiderivative size = 267 \[ \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))} \, dx=\frac {((5+3 i) A-(3-i) B) \arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{4 \sqrt {2} a d}+\frac {((-5-3 i) A+(3-i) B) \arctan \left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{4 \sqrt {2} a d}-\frac {\left (\frac {1}{8}-\frac {i}{8}\right ) ((4+i) A+(1+2 i) B) \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{\sqrt {2} a d}+\frac {((5-3 i) A+(3+i) B) \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{8 \sqrt {2} a d}-\frac {5 A+i B}{2 a d \sqrt {\tan (c+d x)}}+\frac {A+i B}{2 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))} \]
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Time = 0.57 (sec) , antiderivative size = 267, 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 = {3677, 3610, 3615, 1182, 1176, 631, 210, 1179, 642} \[ \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))} \, dx=\frac {((5+3 i) A-(3-i) B) \arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{4 \sqrt {2} a d}+\frac {((3-i) B-(5+3 i) A) \arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{4 \sqrt {2} a d}+\frac {A+i B}{2 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))}-\frac {5 A+i B}{2 a d \sqrt {\tan (c+d x)}}-\frac {\left (\frac {1}{8}-\frac {i}{8}\right ) ((4+i) A+(1+2 i) B) \log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2} a d}+\frac {((5-3 i) A+(3+i) B) \log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{8 \sqrt {2} a d} \]
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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}{2 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))}+\frac {\int \frac {\frac {1}{2} a (5 A+i B)-\frac {3}{2} a (i A-B) \tan (c+d x)}{\tan ^{\frac {3}{2}}(c+d x)} \, dx}{2 a^2} \\ & = -\frac {5 A+i B}{2 a d \sqrt {\tan (c+d x)}}+\frac {A+i B}{2 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))}+\frac {\int \frac {-\frac {3}{2} a (i A-B)-\frac {1}{2} a (5 A+i B) \tan (c+d x)}{\sqrt {\tan (c+d x)}} \, dx}{2 a^2} \\ & = -\frac {5 A+i B}{2 a d \sqrt {\tan (c+d x)}}+\frac {A+i B}{2 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))}+\frac {\text {Subst}\left (\int \frac {-\frac {3}{2} a (i A-B)-\frac {1}{2} a (5 A+i B) x^2}{1+x^4} \, dx,x,\sqrt {\tan (c+d x)}\right )}{a^2 d} \\ & = -\frac {5 A+i B}{2 a d \sqrt {\tan (c+d x)}}+\frac {A+i B}{2 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))}-\frac {((5+3 i) A-(3-i) B) \text {Subst}\left (\int \frac {1+x^2}{1+x^4} \, dx,x,\sqrt {\tan (c+d x)}\right )}{4 a d}+\frac {((5-3 i) A+(3+i) B) \text {Subst}\left (\int \frac {1-x^2}{1+x^4} \, dx,x,\sqrt {\tan (c+d x)}\right )}{4 a d} \\ & = -\frac {5 A+i B}{2 a d \sqrt {\tan (c+d x)}}+\frac {A+i B}{2 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))}-\frac {((5+3 i) A-(3-i) B) \text {Subst}\left (\int \frac {1}{1-\sqrt {2} x+x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{8 a d}-\frac {((5+3 i) A-(3-i) B) \text {Subst}\left (\int \frac {1}{1+\sqrt {2} x+x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{8 a d}-\frac {((5-3 i) A+(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 )}{8 \sqrt {2} a d}-\frac {((5-3 i) A+(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 )}{8 \sqrt {2} a d} \\ & = -\frac {((5-3 i) A+(3+i) B) \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{8 \sqrt {2} a d}+\frac {((5-3 i) A+(3+i) B) \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{8 \sqrt {2} a d}-\frac {5 A+i B}{2 a d \sqrt {\tan (c+d x)}}+\frac {A+i B}{2 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))}-\frac {((5+3 i) A-(3-i) B) \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{4 \sqrt {2} a d}+\frac {((5+3 i) A-(3-i) B) \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{4 \sqrt {2} a d} \\ & = \frac {((5+3 i) A-(3-i) B) \arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{4 \sqrt {2} a d}-\frac {((5+3 i) A-(3-i) B) \arctan \left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{4 \sqrt {2} a d}-\frac {((5-3 i) A+(3+i) B) \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{8 \sqrt {2} a d}+\frac {((5-3 i) A+(3+i) B) \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{8 \sqrt {2} a d}-\frac {5 A+i B}{2 a d \sqrt {\tan (c+d x)}}+\frac {A+i B}{2 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))} \\ \end{align*}
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 1.38 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.43 \[ \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))} \, dx=\frac {-i A+B-2 (2 A+i B) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},-i \tan (c+d x)\right ) (-i+\tan (c+d x))-(A-i B) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},i \tan (c+d x)\right ) (-i+\tan (c+d x))}{2 a d \sqrt {\tan (c+d x)} (-i+\tan (c+d x))} \]
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Time = 0.05 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.52
method | result | size |
derivativedivides | \(\frac {\frac {i \left (i A -B \right ) \left (\sqrt {\tan }\left (d x +c \right )\right )}{2 \tan \left (d x +c \right )-2 i}-\frac {2 \left (i B +2 A \right ) \arctan \left (\frac {2 \left (\sqrt {\tan }\left (d x +c \right )\right )}{\sqrt {2}-i \sqrt {2}}\right )}{\sqrt {2}-i \sqrt {2}}-\frac {2 A}{\sqrt {\tan \left (d x +c \right )}}+\frac {4 \left (-\frac {A}{4}+\frac {i B}{4}\right ) \arctan \left (\frac {2 \left (\sqrt {\tan }\left (d x +c \right )\right )}{\sqrt {2}+i \sqrt {2}}\right )}{\sqrt {2}+i \sqrt {2}}}{d a}\) | \(140\) |
default | \(\frac {\frac {i \left (i A -B \right ) \left (\sqrt {\tan }\left (d x +c \right )\right )}{2 \tan \left (d x +c \right )-2 i}-\frac {2 \left (i B +2 A \right ) \arctan \left (\frac {2 \left (\sqrt {\tan }\left (d x +c \right )\right )}{\sqrt {2}-i \sqrt {2}}\right )}{\sqrt {2}-i \sqrt {2}}-\frac {2 A}{\sqrt {\tan \left (d x +c \right )}}+\frac {4 \left (-\frac {A}{4}+\frac {i B}{4}\right ) \arctan \left (\frac {2 \left (\sqrt {\tan }\left (d x +c \right )\right )}{\sqrt {2}+i \sqrt {2}}\right )}{\sqrt {2}+i \sqrt {2}}}{d a}\) | \(140\) |
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Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 703 vs. \(2 (200) = 400\).
Time = 0.27 (sec) , antiderivative size = 703, normalized size of antiderivative = 2.63 \[ \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))} \, dx=\frac {{\left (a d e^{\left (4 i \, d x + 4 i \, c\right )} - a d e^{\left (2 i \, d x + 2 i \, c\right )}\right )} \sqrt {\frac {i \, A^{2} + 2 \, A B - i \, B^{2}}{a^{2} d^{2}}} \log \left (\frac {2 \, {\left ({\left (a d e^{\left (2 i \, d x + 2 i \, c\right )} + a 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^{2} 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 d e^{\left (4 i \, d x + 4 i \, c\right )} - a d e^{\left (2 i \, d x + 2 i \, c\right )}\right )} \sqrt {\frac {i \, A^{2} + 2 \, A B - i \, B^{2}}{a^{2} d^{2}}} \log \left (-\frac {2 \, {\left ({\left (a d e^{\left (2 i \, d x + 2 i \, c\right )} + a 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^{2} 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 d e^{\left (4 i \, d x + 4 i \, c\right )} - a d e^{\left (2 i \, d x + 2 i \, c\right )}\right )} \sqrt {\frac {-4 i \, A^{2} + 4 \, A B + i \, B^{2}}{a^{2} d^{2}}} \log \left (\frac {{\left ({\left (a d e^{\left (2 i \, d x + 2 i \, c\right )} + a 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 {-4 i \, A^{2} + 4 \, A B + i \, B^{2}}{a^{2} d^{2}}} + 2 \, A + i \, B\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{a d}\right ) - 2 \, {\left (a d e^{\left (4 i \, d x + 4 i \, c\right )} - a d e^{\left (2 i \, d x + 2 i \, c\right )}\right )} \sqrt {\frac {-4 i \, A^{2} + 4 \, A B + i \, B^{2}}{a^{2} d^{2}}} \log \left (-\frac {{\left ({\left (a d e^{\left (2 i \, d x + 2 i \, c\right )} + a 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 {-4 i \, A^{2} + 4 \, A B + i \, B^{2}}{a^{2} d^{2}}} - 2 \, A - i \, B\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{a d}\right ) + 2 \, {\left ({\left (-9 i \, A + B\right )} e^{\left (4 i \, d x + 4 i \, c\right )} - 8 i \, A 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}}}{8 \, {\left (a d e^{\left (4 i \, d x + 4 i \, c\right )} - a d e^{\left (2 i \, d x + 2 i \, c\right )}\right )}} \]
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Exception generated. \[ \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))} \, dx=\text {Exception raised: TypeError} \]
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Exception generated. \[ \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))} \, dx=\text {Exception raised: RuntimeError} \]
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none
Time = 0.77 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.42 \[ \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))} \, dx=-\frac {\left (i + 1\right ) \, \sqrt {2} {\left (2 \, A + i \, B\right )} \arctan \left (\left (\frac {1}{2} i + \frac {1}{2}\right ) \, \sqrt {2} \sqrt {\tan \left (d x + c\right )}\right )}{2 \, a d} + \frac {\left (i - 1\right ) \, \sqrt {2} {\left (A - i \, B\right )} \arctan \left (-\left (\frac {1}{2} i - \frac {1}{2}\right ) \, \sqrt {2} \sqrt {\tan \left (d x + c\right )}\right )}{4 \, a d} + \frac {-5 i \, A \tan \left (d x + c\right ) + B \tan \left (d x + c\right ) - 4 \, A}{2 \, {\left (i \, \tan \left (d x + c\right )^{\frac {3}{2}} + \sqrt {\tan \left (d x + c\right )}\right )} a d} \]
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Time = 9.16 (sec) , antiderivative size = 266, normalized size of antiderivative = 1.00 \[ \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))} \, dx=2\,\mathrm {atanh}\left (\frac {a\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {-\frac {A^2\,1{}\mathrm {i}}{a^2\,d^2}}}{A}\right )\,\sqrt {-\frac {A^2\,1{}\mathrm {i}}{a^2\,d^2}}+2\,\mathrm {atanh}\left (\frac {4\,a\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {\frac {A^2\,1{}\mathrm {i}}{16\,a^2\,d^2}}}{A}\right )\,\sqrt {\frac {A^2\,1{}\mathrm {i}}{16\,a^2\,d^2}}-\mathrm {atan}\left (\frac {2\,a\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {\frac {B^2\,1{}\mathrm {i}}{4\,a^2\,d^2}}}{B}\right )\,\sqrt {\frac {B^2\,1{}\mathrm {i}}{4\,a^2\,d^2}}\,2{}\mathrm {i}+\mathrm {atan}\left (\frac {4\,a\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {-\frac {B^2\,1{}\mathrm {i}}{16\,a^2\,d^2}}}{B}\right )\,\sqrt {-\frac {B^2\,1{}\mathrm {i}}{16\,a^2\,d^2}}\,2{}\mathrm {i}-\frac {\frac {2\,A}{a\,d}+\frac {A\,\mathrm {tan}\left (c+d\,x\right )\,5{}\mathrm {i}}{2\,a\,d}}{\sqrt {\mathrm {tan}\left (c+d\,x\right )}+{\mathrm {tan}\left (c+d\,x\right )}^{3/2}\,1{}\mathrm {i}}+\frac {B\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}}{2\,a\,d\,\left (1+\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )} \]
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