Integrand size = 36, antiderivative size = 307 \[ \int \frac {\tan ^{\frac {5}{2}}(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^3} \, dx=\frac {(2 A+(5-7 i) B) \arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{16 \sqrt {2} a^3 d}-\frac {(2 A+(5-7 i) B) \arctan \left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{16 \sqrt {2} a^3 d}-\frac {(2 A-(5+7 i) B) \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{32 \sqrt {2} a^3 d}+\frac {(2 A-(5+7 i) B) \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{32 \sqrt {2} a^3 d}+\frac {(i A-B) \tan ^{\frac {5}{2}}(c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac {(A+4 i B) \tan ^{\frac {3}{2}}(c+d x)}{12 a d (a+i a \tan (c+d x))^2}+\frac {5 B \sqrt {\tan (c+d x)}}{8 d \left (a^3+i a^3 \tan (c+d x)\right )} \]
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Time = 0.77 (sec) , antiderivative size = 307, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {3676, 3615, 1182, 1176, 631, 210, 1179, 642} \[ \int \frac {\tan ^{\frac {5}{2}}(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^3} \, dx=\frac {(2 A+(5-7 i) B) \arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{16 \sqrt {2} a^3 d}-\frac {(2 A+(5-7 i) B) \arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{16 \sqrt {2} a^3 d}-\frac {(2 A-(5+7 i) B) \log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{32 \sqrt {2} a^3 d}+\frac {(2 A-(5+7 i) B) \log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{32 \sqrt {2} a^3 d}+\frac {5 B \sqrt {\tan (c+d x)}}{8 d \left (a^3+i a^3 \tan (c+d x)\right )}+\frac {(-B+i A) \tan ^{\frac {5}{2}}(c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac {(A+4 i B) \tan ^{\frac {3}{2}}(c+d x)}{12 a d (a+i a \tan (c+d x))^2} \]
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Rule 210
Rule 631
Rule 642
Rule 1176
Rule 1179
Rule 1182
Rule 3615
Rule 3676
Rubi steps \begin{align*} \text {integral}& = \frac {(i A-B) \tan ^{\frac {5}{2}}(c+d x)}{6 d (a+i a \tan (c+d x))^3}-\frac {\int \frac {\tan ^{\frac {3}{2}}(c+d x) \left (\frac {5}{2} a (i A-B)-\frac {1}{2} a (A-11 i B) \tan (c+d x)\right )}{(a+i a \tan (c+d x))^2} \, dx}{6 a^2} \\ & = \frac {(i A-B) \tan ^{\frac {5}{2}}(c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac {(A+4 i B) \tan ^{\frac {3}{2}}(c+d x)}{12 a d (a+i a \tan (c+d x))^2}+\frac {\int \frac {\sqrt {\tan (c+d x)} \left (-3 a^2 (A+4 i B)-3 a^2 (i A+6 B) \tan (c+d x)\right )}{a+i a \tan (c+d x)} \, dx}{24 a^4} \\ & = \frac {(i A-B) \tan ^{\frac {5}{2}}(c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac {(A+4 i B) \tan ^{\frac {3}{2}}(c+d x)}{12 a d (a+i a \tan (c+d x))^2}+\frac {5 B \sqrt {\tan (c+d x)}}{8 d \left (a^3+i a^3 \tan (c+d x)\right )}-\frac {\int \frac {15 a^3 B+3 a^3 (2 A-7 i B) \tan (c+d x)}{\sqrt {\tan (c+d x)}} \, dx}{48 a^6} \\ & = \frac {(i A-B) \tan ^{\frac {5}{2}}(c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac {(A+4 i B) \tan ^{\frac {3}{2}}(c+d x)}{12 a d (a+i a \tan (c+d x))^2}+\frac {5 B \sqrt {\tan (c+d x)}}{8 d \left (a^3+i a^3 \tan (c+d x)\right )}-\frac {\text {Subst}\left (\int \frac {15 a^3 B+3 a^3 (2 A-7 i B) x^2}{1+x^4} \, dx,x,\sqrt {\tan (c+d x)}\right )}{24 a^6 d} \\ & = \frac {(i A-B) \tan ^{\frac {5}{2}}(c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac {(A+4 i B) \tan ^{\frac {3}{2}}(c+d x)}{12 a d (a+i a \tan (c+d x))^2}+\frac {5 B \sqrt {\tan (c+d x)}}{8 d \left (a^3+i a^3 \tan (c+d x)\right )}+\frac {(2 A-(5+7 i) B) \text {Subst}\left (\int \frac {1-x^2}{1+x^4} \, dx,x,\sqrt {\tan (c+d x)}\right )}{16 a^3 d}-\frac {(2 A+(5-7 i) B) \text {Subst}\left (\int \frac {1+x^2}{1+x^4} \, dx,x,\sqrt {\tan (c+d x)}\right )}{16 a^3 d} \\ & = \frac {(i A-B) \tan ^{\frac {5}{2}}(c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac {(A+4 i B) \tan ^{\frac {3}{2}}(c+d x)}{12 a d (a+i a \tan (c+d x))^2}+\frac {5 B \sqrt {\tan (c+d x)}}{8 d \left (a^3+i a^3 \tan (c+d x)\right )}-\frac {(2 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 )}{32 \sqrt {2} a^3 d}-\frac {(2 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 )}{32 \sqrt {2} a^3 d}-\frac {(2 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 )}{32 a^3 d}-\frac {(2 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 )}{32 a^3 d} \\ & = -\frac {(2 A-(5+7 i) B) \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{32 \sqrt {2} a^3 d}+\frac {(2 A-(5+7 i) B) \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{32 \sqrt {2} a^3 d}+\frac {(i A-B) \tan ^{\frac {5}{2}}(c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac {(A+4 i B) \tan ^{\frac {3}{2}}(c+d x)}{12 a d (a+i a \tan (c+d x))^2}+\frac {5 B \sqrt {\tan (c+d x)}}{8 d \left (a^3+i a^3 \tan (c+d x)\right )}-\frac {(2 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 )}{16 \sqrt {2} a^3 d}+\frac {(2 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 )}{16 \sqrt {2} a^3 d} \\ & = \frac {(2 A+(5-7 i) B) \arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{16 \sqrt {2} a^3 d}-\frac {(2 A+(5-7 i) B) \arctan \left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{16 \sqrt {2} a^3 d}-\frac {(2 A-(5+7 i) B) \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{32 \sqrt {2} a^3 d}+\frac {(2 A-(5+7 i) B) \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{32 \sqrt {2} a^3 d}+\frac {(i A-B) \tan ^{\frac {5}{2}}(c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac {(A+4 i B) \tan ^{\frac {3}{2}}(c+d x)}{12 a d (a+i a \tan (c+d x))^2}+\frac {5 B \sqrt {\tan (c+d x)}}{8 d \left (a^3+i a^3 \tan (c+d x)\right )} \\ \end{align*}
Time = 3.76 (sec) , antiderivative size = 197, normalized size of antiderivative = 0.64 \[ \int \frac {\tan ^{\frac {5}{2}}(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^3} \, dx=-\frac {\sec ^3(c+d x) \left (-3 \sqrt [4]{-1} (A-i B) \arctan \left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right ) (\cos (3 (c+d x))+i \sin (3 (c+d x)))+3 \sqrt [4]{-1} (A-6 i B) \text {arctanh}\left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right ) (\cos (3 (c+d x))+i \sin (3 (c+d x)))+\cos (c+d x) (3 (A+2 i B)-3 (A+7 i B) \cos (2 (c+d x))+(-i A+19 B) \sin (2 (c+d x))) \sqrt {\tan (c+d x)}\right )}{24 a^3 d (-i+\tan (c+d x))^3} \]
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Time = 0.08 (sec) , antiderivative size = 163, normalized size of antiderivative = 0.53
method | result | size |
derivativedivides | \(\frac {-\frac {i \left (\frac {-i \left (9 i B +2 A \right ) \left (\tan ^{\frac {5}{2}}\left (d x +c \right )\right )+\left (-\frac {38 i B}{3}-\frac {2 A}{3}\right ) \left (\tan ^{\frac {3}{2}}\left (d x +c \right )\right )-5 B \left (\sqrt {\tan }\left (d x +c \right )\right )}{\left (\tan \left (d x +c \right )-i\right )^{3}}-\frac {2 \left (i A +6 B \right ) \arctan \left (\frac {2 \left (\sqrt {\tan }\left (d x +c \right )\right )}{\sqrt {2}-i \sqrt {2}}\right )}{\sqrt {2}-i \sqrt {2}}\right )}{8}+\frac {4 \left (-\frac {A}{16}+\frac {i B}{16}\right ) \arctan \left (\frac {2 \left (\sqrt {\tan }\left (d x +c \right )\right )}{\sqrt {2}+i \sqrt {2}}\right )}{\sqrt {2}+i \sqrt {2}}}{d \,a^{3}}\) | \(163\) |
default | \(\frac {-\frac {i \left (\frac {-i \left (9 i B +2 A \right ) \left (\tan ^{\frac {5}{2}}\left (d x +c \right )\right )+\left (-\frac {38 i B}{3}-\frac {2 A}{3}\right ) \left (\tan ^{\frac {3}{2}}\left (d x +c \right )\right )-5 B \left (\sqrt {\tan }\left (d x +c \right )\right )}{\left (\tan \left (d x +c \right )-i\right )^{3}}-\frac {2 \left (i A +6 B \right ) \arctan \left (\frac {2 \left (\sqrt {\tan }\left (d x +c \right )\right )}{\sqrt {2}-i \sqrt {2}}\right )}{\sqrt {2}-i \sqrt {2}}\right )}{8}+\frac {4 \left (-\frac {A}{16}+\frac {i B}{16}\right ) \arctan \left (\frac {2 \left (\sqrt {\tan }\left (d x +c \right )\right )}{\sqrt {2}+i \sqrt {2}}\right )}{\sqrt {2}+i \sqrt {2}}}{d \,a^{3}}\) | \(163\) |
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Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 681 vs. \(2 (242) = 484\).
Time = 0.27 (sec) , antiderivative size = 681, normalized size of antiderivative = 2.22 \[ \int \frac {\tan ^{\frac {5}{2}}(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^3} \, dx=\frac {{\left (3 \, a^{3} d \sqrt {\frac {i \, A^{2} + 2 \, A B - i \, B^{2}}{a^{6} d^{2}}} e^{\left (6 i \, d x + 6 i \, c\right )} \log \left (\frac {2 \, {\left ({\left (a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{3} 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^{6} 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 \, a^{3} d \sqrt {\frac {i \, A^{2} + 2 \, A B - i \, B^{2}}{a^{6} d^{2}}} e^{\left (6 i \, d x + 6 i \, c\right )} \log \left (-\frac {2 \, {\left ({\left (a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{3} 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^{6} 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 \, a^{3} d \sqrt {\frac {-i \, A^{2} - 12 \, A B + 36 i \, B^{2}}{a^{6} d^{2}}} e^{\left (6 i \, d x + 6 i \, c\right )} \log \left (\frac {{\left ({\left (a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{3} 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} - 12 \, A B + 36 i \, B^{2}}{a^{6} d^{2}}} + A - 6 i \, B\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{8 \, a^{3} d}\right ) - 3 \, a^{3} d \sqrt {\frac {-i \, A^{2} - 12 \, A B + 36 i \, B^{2}}{a^{6} d^{2}}} e^{\left (6 i \, d x + 6 i \, c\right )} \log \left (-\frac {{\left ({\left (a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{3} 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} - 12 \, A B + 36 i \, B^{2}}{a^{6} d^{2}}} - A + 6 i \, B\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{8 \, a^{3} d}\right ) - 2 \, {\left (2 \, {\left (i \, A - 10 \, B\right )} e^{\left (6 i \, d x + 6 i \, c\right )} - {\left (i \, A + 14 \, B\right )} e^{\left (4 i \, d x + 4 i \, c\right )} - {\left (2 i \, A - 5 \, 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 (-6 i \, d x - 6 i \, c\right )}}{96 \, a^{3} d} \]
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Timed out. \[ \int \frac {\tan ^{\frac {5}{2}}(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^3} \, dx=\text {Timed out} \]
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Exception generated. \[ \int \frac {\tan ^{\frac {5}{2}}(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^3} \, dx=\text {Exception raised: RuntimeError} \]
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none
Time = 0.85 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.44 \[ \int \frac {\tan ^{\frac {5}{2}}(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^3} \, dx=-\frac {\left (i + 1\right ) \, \sqrt {2} {\left (A - 6 i \, B\right )} \arctan \left (\left (\frac {1}{2} i + \frac {1}{2}\right ) \, \sqrt {2} \sqrt {\tan \left (d x + c\right )}\right )}{16 \, a^{3} d} + \frac {\left (i - 1\right ) \, \sqrt {2} {\left (A - i \, B\right )} \arctan \left (-\left (\frac {1}{2} i - \frac {1}{2}\right ) \, \sqrt {2} \sqrt {\tan \left (d x + c\right )}\right )}{16 \, a^{3} d} - \frac {6 \, A \tan \left (d x + c\right )^{\frac {5}{2}} + 27 i \, B \tan \left (d x + c\right )^{\frac {5}{2}} - 2 i \, A \tan \left (d x + c\right )^{\frac {3}{2}} + 38 \, B \tan \left (d x + c\right )^{\frac {3}{2}} - 15 i \, B \sqrt {\tan \left (d x + c\right )}}{24 \, a^{3} d {\left (\tan \left (d x + c\right ) - i\right )}^{3}} \]
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Time = 9.99 (sec) , antiderivative size = 308, normalized size of antiderivative = 1.00 \[ \int \frac {\tan ^{\frac {5}{2}}(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^3} \, dx=\frac {\frac {5\,B\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}}{8\,a^3\,d}-\frac {9\,B\,{\mathrm {tan}\left (c+d\,x\right )}^{5/2}}{8\,a^3\,d}+\frac {B\,{\mathrm {tan}\left (c+d\,x\right )}^{3/2}\,19{}\mathrm {i}}{12\,a^3\,d}}{-{\mathrm {tan}\left (c+d\,x\right )}^3\,1{}\mathrm {i}-3\,{\mathrm {tan}\left (c+d\,x\right )}^2+\mathrm {tan}\left (c+d\,x\right )\,3{}\mathrm {i}+1}+\frac {\frac {A\,{\mathrm {tan}\left (c+d\,x\right )}^{3/2}}{12\,a^3\,d}+\frac {A\,{\mathrm {tan}\left (c+d\,x\right )}^{5/2}\,1{}\mathrm {i}}{4\,a^3\,d}}{-{\mathrm {tan}\left (c+d\,x\right )}^3\,1{}\mathrm {i}-3\,{\mathrm {tan}\left (c+d\,x\right )}^2+\mathrm {tan}\left (c+d\,x\right )\,3{}\mathrm {i}+1}-\frac {{\left (-1\right )}^{1/4}\,A\,\mathrm {atan}\left ({\left (-1\right )}^{1/4}\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\right )}{8\,a^3\,d}+\frac {{\left (-1\right )}^{1/4}\,A\,\mathrm {atanh}\left ({\left (-1\right )}^{1/4}\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\right )}{8\,a^3\,d}+\mathrm {atan}\left (\frac {8\,a^3\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {\frac {B^2\,9{}\mathrm {i}}{64\,a^6\,d^2}}}{3\,B}\right )\,\sqrt {\frac {B^2\,9{}\mathrm {i}}{64\,a^6\,d^2}}\,2{}\mathrm {i}+\mathrm {atan}\left (\frac {16\,a^3\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {-\frac {B^2\,1{}\mathrm {i}}{256\,a^6\,d^2}}}{B}\right )\,\sqrt {-\frac {B^2\,1{}\mathrm {i}}{256\,a^6\,d^2}}\,2{}\mathrm {i} \]
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