Integrand size = 36, antiderivative size = 306 \[ \int \frac {\tan ^{\frac {5}{2}}(c+d x) (A+B \tan (c+d x))}{a+i a \tan (c+d x)} \, dx=-\frac {\left (\frac {1}{4}+\frac {i}{4}\right ) ((4+i) A+(1+6 i) B) \arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} a d}+\frac {\left (\frac {1}{4}+\frac {i}{4}\right ) ((4+i) A+(1+6 i) B) \arctan \left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} a d}-\frac {\left (\frac {1}{8}+\frac {i}{8}\right ) ((1+4 i) A-(6+i) B) \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{\sqrt {2} a d}-\frac {((3-5 i) A+(5+7 i) B) \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{8 \sqrt {2} a d}-\frac {5 (i A-B) \sqrt {\tan (c+d x)}}{2 a d}-\frac {(3 A+7 i B) \tan ^{\frac {3}{2}}(c+d x)}{6 a d}+\frac {(i A-B) \tan ^{\frac {5}{2}}(c+d x)}{2 d (a+i a \tan (c+d x))} \]
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Time = 0.50 (sec) , antiderivative size = 306, 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} \[ \int \frac {\tan ^{\frac {5}{2}}(c+d x) (A+B \tan (c+d x))}{a+i a \tan (c+d x)} \, dx=-\frac {\left (\frac {1}{4}+\frac {i}{4}\right ) ((4+i) A+(1+6 i) B) \arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} a d}+\frac {\left (\frac {1}{4}+\frac {i}{4}\right ) ((4+i) A+(1+6 i) B) \arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2} a d}+\frac {(-B+i A) \tan ^{\frac {5}{2}}(c+d x)}{2 d (a+i a \tan (c+d x))}-\frac {(3 A+7 i B) \tan ^{\frac {3}{2}}(c+d x)}{6 a d}-\frac {5 (-B+i A) \sqrt {\tan (c+d x)}}{2 a d}-\frac {\left (\frac {1}{8}+\frac {i}{8}\right ) ((1+4 i) A-(6+i) B) \log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2} a d}-\frac {((3-5 i) A+(5+7 i) B) \log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{8 \sqrt {2} a d} \]
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Rule 210
Rule 631
Rule 642
Rule 1176
Rule 1179
Rule 1182
Rule 3609
Rule 3615
Rule 3676
Rubi steps \begin{align*} \text {integral}& = \frac {(i A-B) \tan ^{\frac {5}{2}}(c+d x)}{2 d (a+i a \tan (c+d x))}-\frac {\int \tan ^{\frac {3}{2}}(c+d x) \left (\frac {5}{2} a (i A-B)+\frac {1}{2} a (3 A+7 i B) \tan (c+d x)\right ) \, dx}{2 a^2} \\ & = -\frac {(3 A+7 i B) \tan ^{\frac {3}{2}}(c+d x)}{6 a d}+\frac {(i A-B) \tan ^{\frac {5}{2}}(c+d x)}{2 d (a+i a \tan (c+d x))}-\frac {\int \sqrt {\tan (c+d x)} \left (-\frac {1}{2} a (3 A+7 i B)+\frac {5}{2} a (i A-B) \tan (c+d x)\right ) \, dx}{2 a^2} \\ & = -\frac {5 (i A-B) \sqrt {\tan (c+d x)}}{2 a d}-\frac {(3 A+7 i B) \tan ^{\frac {3}{2}}(c+d x)}{6 a d}+\frac {(i A-B) \tan ^{\frac {5}{2}}(c+d x)}{2 d (a+i a \tan (c+d x))}-\frac {\int \frac {-\frac {5}{2} a (i A-B)-\frac {1}{2} a (3 A+7 i B) \tan (c+d x)}{\sqrt {\tan (c+d x)}} \, dx}{2 a^2} \\ & = -\frac {5 (i A-B) \sqrt {\tan (c+d x)}}{2 a d}-\frac {(3 A+7 i B) \tan ^{\frac {3}{2}}(c+d x)}{6 a d}+\frac {(i A-B) \tan ^{\frac {5}{2}}(c+d x)}{2 d (a+i a \tan (c+d x))}-\frac {\text {Subst}\left (\int \frac {-\frac {5}{2} a (i A-B)-\frac {1}{2} a (3 A+7 i B) x^2}{1+x^4} \, dx,x,\sqrt {\tan (c+d x)}\right )}{a^2 d} \\ & = -\frac {5 (i A-B) \sqrt {\tan (c+d x)}}{2 a d}-\frac {(3 A+7 i B) \tan ^{\frac {3}{2}}(c+d x)}{6 a d}+\frac {(i A-B) \tan ^{\frac {5}{2}}(c+d x)}{2 d (a+i a \tan (c+d x))}+\frac {((3+5 i) A-(5-7 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 {((3-5 i) A+(5+7 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 (i A-B) \sqrt {\tan (c+d x)}}{2 a d}-\frac {(3 A+7 i B) \tan ^{\frac {3}{2}}(c+d x)}{6 a d}+\frac {(i A-B) \tan ^{\frac {5}{2}}(c+d x)}{2 d (a+i a \tan (c+d x))}+\frac {((3+5 i) A-(5-7 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 {((3+5 i) A-(5-7 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 {((3-5 i) A+(5+7 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 {((3-5 i) A+(5+7 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 {((3-5 i) A+(5+7 i) B) \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{8 \sqrt {2} a d}-\frac {((3-5 i) A+(5+7 i) B) \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{8 \sqrt {2} a d}-\frac {5 (i A-B) \sqrt {\tan (c+d x)}}{2 a d}-\frac {(3 A+7 i B) \tan ^{\frac {3}{2}}(c+d x)}{6 a d}+\frac {(i A-B) \tan ^{\frac {5}{2}}(c+d x)}{2 d (a+i a \tan (c+d x))}+\frac {((3+5 i) A-(5-7 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 {((3+5 i) A-(5-7 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 {((3+5 i) A-(5-7 i) B) \arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{4 \sqrt {2} a d}+\frac {((3+5 i) A-(5-7 i) B) \arctan \left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{4 \sqrt {2} a d}+\frac {((3-5 i) A+(5+7 i) B) \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{8 \sqrt {2} a d}-\frac {((3-5 i) A+(5+7 i) B) \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{8 \sqrt {2} a d}-\frac {5 (i A-B) \sqrt {\tan (c+d x)}}{2 a d}-\frac {(3 A+7 i B) \tan ^{\frac {3}{2}}(c+d x)}{6 a d}+\frac {(i A-B) \tan ^{\frac {5}{2}}(c+d x)}{2 d (a+i a \tan (c+d x))} \\ \end{align*}
Time = 2.47 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.45 \[ \int \frac {\tan ^{\frac {5}{2}}(c+d x) (A+B \tan (c+d x))}{a+i a \tan (c+d x)} \, dx=\frac {-3 (-1)^{3/4} (A-i B) \arctan \left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )-6 (-1)^{3/4} (2 A+3 i B) \text {arctanh}\left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )+\frac {\sqrt {\tan (c+d x)} \left (-15 (A+i B)+4 (-3 i A+2 B) \tan (c+d x)-4 i B \tan ^2(c+d x)\right )}{-i+\tan (c+d x)}}{6 a d} \]
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Time = 0.08 (sec) , antiderivative size = 166, normalized size of antiderivative = 0.54
method | result | size |
derivativedivides | \(\frac {-\frac {2 i B \left (\tan ^{\frac {3}{2}}\left (d x +c \right )\right )}{3}+2 B \left (\sqrt {\tan }\left (d x +c \right )\right )-2 i A \left (\sqrt {\tan }\left (d x +c \right )\right )-\frac {i \left (-\frac {i \left (i B +A \right ) \left (\sqrt {\tan }\left (d x +c \right )\right )}{\tan \left (d x +c \right )-i}+\frac {4 \left (2 i A -3 B \right ) \arctan \left (\frac {2 \left (\sqrt {\tan }\left (d x +c \right )\right )}{\sqrt {2}-i \sqrt {2}}\right )}{\sqrt {2}-i \sqrt {2}}\right )}{2}+\frac {4 \left (-\frac {A}{4}+\frac {i B}{4}\right ) \arctan \left (\frac {2 \left (\sqrt {\tan }\left (d x +c \right )\right )}{\sqrt {2}+i \sqrt {2}}\right )}{\sqrt {2}+i \sqrt {2}}}{d a}\) | \(166\) |
default | \(\frac {-\frac {2 i B \left (\tan ^{\frac {3}{2}}\left (d x +c \right )\right )}{3}+2 B \left (\sqrt {\tan }\left (d x +c \right )\right )-2 i A \left (\sqrt {\tan }\left (d x +c \right )\right )-\frac {i \left (-\frac {i \left (i B +A \right ) \left (\sqrt {\tan }\left (d x +c \right )\right )}{\tan \left (d x +c \right )-i}+\frac {4 \left (2 i A -3 B \right ) \arctan \left (\frac {2 \left (\sqrt {\tan }\left (d x +c \right )\right )}{\sqrt {2}-i \sqrt {2}}\right )}{\sqrt {2}-i \sqrt {2}}\right )}{2}+\frac {4 \left (-\frac {A}{4}+\frac {i B}{4}\right ) \arctan \left (\frac {2 \left (\sqrt {\tan }\left (d x +c \right )\right )}{\sqrt {2}+i \sqrt {2}}\right )}{\sqrt {2}+i \sqrt {2}}}{d a}\) | \(166\) |
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Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 707 vs. \(2 (221) = 442\).
Time = 0.28 (sec) , antiderivative size = 707, normalized size of antiderivative = 2.31 \[ \int \frac {\tan ^{\frac {5}{2}}(c+d x) (A+B \tan (c+d x))}{a+i a \tan (c+d x)} \, dx=\frac {3 \, {\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 ) - 3 \, {\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 ) - 6 \, {\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} + 12 \, A B + 9 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} + 12 \, A B + 9 i \, B^{2}}{a^{2} d^{2}}} + 2 \, A + 3 i \, B\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{a d}\right ) + 6 \, {\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} + 12 \, A B + 9 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} + 12 \, A B + 9 i \, B^{2}}{a^{2} d^{2}}} - 2 \, A - 3 i \, B\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{a d}\right ) + 2 \, {\left ({\left (-27 i \, A + 19 \, B\right )} e^{\left (4 i \, d x + 4 i \, c\right )} - 2 \, {\left (15 i \, A - 19 \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} - 3 i \, A + 3 \, 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}}}{24 \, {\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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\[ \int \frac {\tan ^{\frac {5}{2}}(c+d x) (A+B \tan (c+d x))}{a+i a \tan (c+d x)} \, dx=- \frac {i \left (\int \frac {A \tan ^{\frac {5}{2}}{\left (c + d x \right )}}{\tan {\left (c + d x \right )} - i}\, dx + \int \frac {B \tan ^{\frac {7}{2}}{\left (c + d x \right )}}{\tan {\left (c + d x \right )} - i}\, dx\right )}{a} \]
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Exception generated. \[ \int \frac {\tan ^{\frac {5}{2}}(c+d x) (A+B \tan (c+d x))}{a+i a \tan (c+d x)} \, dx=\text {Exception raised: RuntimeError} \]
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Time = 0.53 (sec) , antiderivative size = 163, normalized size of antiderivative = 0.53 \[ \int \frac {\tan ^{\frac {5}{2}}(c+d x) (A+B \tan (c+d x))}{a+i a \tan (c+d x)} \, dx=\frac {\left (i + 1\right ) \, \sqrt {2} {\left (2 \, A + 3 i \, B\right )} \arctan \left (\left (\frac {1}{2} i + \frac {1}{2}\right ) \, \sqrt {2} \sqrt {\tan \left (d x + c\right )}\right )}{2 \, a d} + \frac {\left (i - 1\right ) \, \sqrt {2} {\left (A - i \, B\right )} \arctan \left (-\left (\frac {1}{2} i - \frac {1}{2}\right ) \, \sqrt {2} \sqrt {\tan \left (d x + c\right )}\right )}{4 \, a d} - \frac {A \sqrt {\tan \left (d x + c\right )} + i \, B \sqrt {\tan \left (d x + c\right )}}{2 \, a d {\left (\tan \left (d x + c\right ) - i\right )}} - \frac {2 \, {\left (i \, B a^{2} d^{2} \tan \left (d x + c\right )^{\frac {3}{2}} + 3 i \, A a^{2} d^{2} \sqrt {\tan \left (d x + c\right )} - 3 \, B a^{2} d^{2} \sqrt {\tan \left (d x + c\right )}\right )}}{3 \, a^{3} d^{3}} \]
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Time = 13.77 (sec) , antiderivative size = 305, normalized size of antiderivative = 1.00 \[ \int \frac {\tan ^{\frac {5}{2}}(c+d x) (A+B \tan (c+d x))}{a+i a \tan (c+d x)} \, dx=\mathrm {atan}\left (\frac {a\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {-\frac {A^2\,1{}\mathrm {i}}{a^2\,d^2}}\,1{}\mathrm {i}}{A}\right )\,\sqrt {-\frac {A^2\,1{}\mathrm {i}}{a^2\,d^2}}\,2{}\mathrm {i}-\mathrm {atan}\left (\frac {a\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {\frac {A^2\,1{}\mathrm {i}}{16\,a^2\,d^2}}\,4{}\mathrm {i}}{A}\right )\,\sqrt {\frac {A^2\,1{}\mathrm {i}}{16\,a^2\,d^2}}\,2{}\mathrm {i}+\mathrm {atan}\left (\frac {2\,a\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {\frac {B^2\,9{}\mathrm {i}}{4\,a^2\,d^2}}}{3\,B}\right )\,\sqrt {\frac {B^2\,9{}\mathrm {i}}{4\,a^2\,d^2}}\,2{}\mathrm {i}+\mathrm {atan}\left (\frac {4\,a\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {-\frac {B^2\,1{}\mathrm {i}}{16\,a^2\,d^2}}}{B}\right )\,\sqrt {-\frac {B^2\,1{}\mathrm {i}}{16\,a^2\,d^2}}\,2{}\mathrm {i}-\frac {A\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,2{}\mathrm {i}}{a\,d}+\frac {2\,B\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}}{a\,d}-\frac {B\,{\mathrm {tan}\left (c+d\,x\right )}^{3/2}\,2{}\mathrm {i}}{3\,a\,d}-\frac {A\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,1{}\mathrm {i}}{2\,a\,d\,\left (1+\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}+\frac {B\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}}{2\,a\,d\,\left (1+\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )} \]
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