Optimal. Leaf size=277 \[ \frac {(4+4 i) a^{5/2} (A-i B) \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}+\frac {8 a^2 (60 B+59 i A) \sqrt {a+i a \tan (c+d x)}}{315 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {2 a^2 (46 A-45 i B) \sqrt {a+i a \tan (c+d x)}}{105 d \tan ^{\frac {5}{2}}(c+d x)}-\frac {2 a^2 (3 B+4 i A) \sqrt {a+i a \tan (c+d x)}}{21 d \tan ^{\frac {7}{2}}(c+d x)}-\frac {8 a^2 (197 A-195 i B) \sqrt {a+i a \tan (c+d x)}}{315 d \sqrt {\tan (c+d x)}}-\frac {2 a A (a+i a \tan (c+d x))^{3/2}}{9 d \tan ^{\frac {9}{2}}(c+d x)} \]
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Rubi [A] time = 0.95, antiderivative size = 277, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 5, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.132, Rules used = {3593, 3598, 12, 3544, 205} \[ \frac {8 a^2 (60 B+59 i A) \sqrt {a+i a \tan (c+d x)}}{315 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {2 a^2 (46 A-45 i B) \sqrt {a+i a \tan (c+d x)}}{105 d \tan ^{\frac {5}{2}}(c+d x)}-\frac {2 a^2 (3 B+4 i A) \sqrt {a+i a \tan (c+d x)}}{21 d \tan ^{\frac {7}{2}}(c+d x)}-\frac {8 a^2 (197 A-195 i B) \sqrt {a+i a \tan (c+d x)}}{315 d \sqrt {\tan (c+d x)}}+\frac {(4+4 i) a^{5/2} (A-i B) \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-\frac {2 a A (a+i a \tan (c+d x))^{3/2}}{9 d \tan ^{\frac {9}{2}}(c+d x)} \]
Antiderivative was successfully verified.
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Rule 12
Rule 205
Rule 3544
Rule 3593
Rule 3598
Rubi steps
\begin {align*} \int \frac {(a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\tan ^{\frac {11}{2}}(c+d x)} \, dx &=-\frac {2 a A (a+i a \tan (c+d x))^{3/2}}{9 d \tan ^{\frac {9}{2}}(c+d x)}+\frac {2}{9} \int \frac {(a+i a \tan (c+d x))^{3/2} \left (\frac {3}{2} a (4 i A+3 B)-\frac {3}{2} a (2 A-3 i B) \tan (c+d x)\right )}{\tan ^{\frac {9}{2}}(c+d x)} \, dx\\ &=-\frac {2 a^2 (4 i A+3 B) \sqrt {a+i a \tan (c+d x)}}{21 d \tan ^{\frac {7}{2}}(c+d x)}-\frac {2 a A (a+i a \tan (c+d x))^{3/2}}{9 d \tan ^{\frac {9}{2}}(c+d x)}+\frac {4}{63} \int \frac {\sqrt {a+i a \tan (c+d x)} \left (-\frac {3}{4} a^2 (46 A-45 i B)-\frac {3}{4} a^2 (38 i A+39 B) \tan (c+d x)\right )}{\tan ^{\frac {7}{2}}(c+d x)} \, dx\\ &=-\frac {2 a^2 (4 i A+3 B) \sqrt {a+i a \tan (c+d x)}}{21 d \tan ^{\frac {7}{2}}(c+d x)}+\frac {2 a^2 (46 A-45 i B) \sqrt {a+i a \tan (c+d x)}}{105 d \tan ^{\frac {5}{2}}(c+d x)}-\frac {2 a A (a+i a \tan (c+d x))^{3/2}}{9 d \tan ^{\frac {9}{2}}(c+d x)}+\frac {8 \int \frac {\sqrt {a+i a \tan (c+d x)} \left (-\frac {3}{2} a^3 (59 i A+60 B)+\frac {3}{2} a^3 (46 A-45 i B) \tan (c+d x)\right )}{\tan ^{\frac {5}{2}}(c+d x)} \, dx}{315 a}\\ &=-\frac {2 a^2 (4 i A+3 B) \sqrt {a+i a \tan (c+d x)}}{21 d \tan ^{\frac {7}{2}}(c+d x)}+\frac {2 a^2 (46 A-45 i B) \sqrt {a+i a \tan (c+d x)}}{105 d \tan ^{\frac {5}{2}}(c+d x)}+\frac {8 a^2 (59 i A+60 B) \sqrt {a+i a \tan (c+d x)}}{315 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {2 a A (a+i a \tan (c+d x))^{3/2}}{9 d \tan ^{\frac {9}{2}}(c+d x)}+\frac {16 \int \frac {\sqrt {a+i a \tan (c+d x)} \left (\frac {3}{4} a^4 (197 A-195 i B)+\frac {3}{2} a^4 (59 i A+60 B) \tan (c+d x)\right )}{\tan ^{\frac {3}{2}}(c+d x)} \, dx}{945 a^2}\\ &=-\frac {2 a^2 (4 i A+3 B) \sqrt {a+i a \tan (c+d x)}}{21 d \tan ^{\frac {7}{2}}(c+d x)}+\frac {2 a^2 (46 A-45 i B) \sqrt {a+i a \tan (c+d x)}}{105 d \tan ^{\frac {5}{2}}(c+d x)}+\frac {8 a^2 (59 i A+60 B) \sqrt {a+i a \tan (c+d x)}}{315 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {8 a^2 (197 A-195 i B) \sqrt {a+i a \tan (c+d x)}}{315 d \sqrt {\tan (c+d x)}}-\frac {2 a A (a+i a \tan (c+d x))^{3/2}}{9 d \tan ^{\frac {9}{2}}(c+d x)}+\frac {32 \int \frac {945 a^5 (i A+B) \sqrt {a+i a \tan (c+d x)}}{8 \sqrt {\tan (c+d x)}} \, dx}{945 a^3}\\ &=-\frac {2 a^2 (4 i A+3 B) \sqrt {a+i a \tan (c+d x)}}{21 d \tan ^{\frac {7}{2}}(c+d x)}+\frac {2 a^2 (46 A-45 i B) \sqrt {a+i a \tan (c+d x)}}{105 d \tan ^{\frac {5}{2}}(c+d x)}+\frac {8 a^2 (59 i A+60 B) \sqrt {a+i a \tan (c+d x)}}{315 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {8 a^2 (197 A-195 i B) \sqrt {a+i a \tan (c+d x)}}{315 d \sqrt {\tan (c+d x)}}-\frac {2 a A (a+i a \tan (c+d x))^{3/2}}{9 d \tan ^{\frac {9}{2}}(c+d x)}+\left (4 a^2 (i A+B)\right ) \int \frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {\tan (c+d x)}} \, dx\\ &=-\frac {2 a^2 (4 i A+3 B) \sqrt {a+i a \tan (c+d x)}}{21 d \tan ^{\frac {7}{2}}(c+d x)}+\frac {2 a^2 (46 A-45 i B) \sqrt {a+i a \tan (c+d x)}}{105 d \tan ^{\frac {5}{2}}(c+d x)}+\frac {8 a^2 (59 i A+60 B) \sqrt {a+i a \tan (c+d x)}}{315 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {8 a^2 (197 A-195 i B) \sqrt {a+i a \tan (c+d x)}}{315 d \sqrt {\tan (c+d x)}}-\frac {2 a A (a+i a \tan (c+d x))^{3/2}}{9 d \tan ^{\frac {9}{2}}(c+d x)}+\frac {\left (8 a^4 (A-i B)\right ) \operatorname {Subst}\left (\int \frac {1}{-i a-2 a^2 x^2} \, dx,x,\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}\\ &=\frac {(4+4 i) a^{5/2} (A-i B) \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-\frac {2 a^2 (4 i A+3 B) \sqrt {a+i a \tan (c+d x)}}{21 d \tan ^{\frac {7}{2}}(c+d x)}+\frac {2 a^2 (46 A-45 i B) \sqrt {a+i a \tan (c+d x)}}{105 d \tan ^{\frac {5}{2}}(c+d x)}+\frac {8 a^2 (59 i A+60 B) \sqrt {a+i a \tan (c+d x)}}{315 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {8 a^2 (197 A-195 i B) \sqrt {a+i a \tan (c+d x)}}{315 d \sqrt {\tan (c+d x)}}-\frac {2 a A (a+i a \tan (c+d x))^{3/2}}{9 d \tan ^{\frac {9}{2}}(c+d x)}\\ \end {align*}
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Mathematica [A] time = 16.57, size = 246, normalized size = 0.89 \[ \frac {a^2 \sqrt {a+i a \tan (c+d x)} \left (\frac {1260 (A-i B) e^{-i (c+d x)} \sqrt {-1+e^{2 i (c+d x)}} \tanh ^{-1}\left (\frac {e^{i (c+d x)}}{\sqrt {-1+e^{2 i (c+d x)}}}\right )}{\sqrt {-\frac {i \left (-1+e^{2 i (c+d x)}\right )}{1+e^{2 i (c+d x)}}}}+\frac {\csc ^2(2 (c+d x)) (12 (251 A-260 i B) \cos (2 (c+d x))+(-961 A+915 i B) \cos (4 (c+d x))+282 i A \sin (2 (c+d x))-331 i A \sin (4 (c+d x))-2331 A+390 B \sin (2 (c+d x))-285 B \sin (4 (c+d x))+2205 i B)}{\tan ^{\frac {5}{2}}(c+d x)}\right )}{315 d} \]
Antiderivative was successfully verified.
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fricas [B] time = 0.67, size = 700, normalized size = 2.53 \[ \frac {\sqrt {2} {\left ({\left (-5168 i \, A - 4800 \, B\right )} a^{2} e^{\left (11 i \, d x + 11 i \, c\right )} + {\left (8008 i \, A + 9240 \, B\right )} a^{2} e^{\left (9 i \, d x + 9 i \, c\right )} + {\left (-5472 i \, A - 3600 \, B\right )} a^{2} e^{\left (7 i \, d x + 7 i \, c\right )} + {\left (-7728 i \, A - 6720 \, B\right )} a^{2} e^{\left (5 i \, d x + 5 i \, c\right )} + {\left (8400 i \, A + 8400 \, B\right )} a^{2} e^{\left (3 i \, d x + 3 i \, c\right )} + {\left (-2520 i \, A - 2520 \, B\right )} a^{2} e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \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}} + 315 \, \sqrt {\frac {{\left (32 i \, A^{2} + 64 \, A B - 32 i \, B^{2}\right )} a^{5}}{d^{2}}} {\left (d e^{\left (10 i \, d x + 10 i \, c\right )} - 5 \, d e^{\left (8 i \, d x + 8 i \, c\right )} + 10 \, d e^{\left (6 i \, d x + 6 i \, c\right )} - 10 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 5 \, d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )} \log \left (\frac {{\left (\sqrt {2} {\left ({\left (4 i \, A + 4 \, B\right )} a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + {\left (4 i \, A + 4 \, B\right )} a^{2}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \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 {{\left (32 i \, A^{2} + 64 \, A B - 32 i \, B^{2}\right )} a^{5}}{d^{2}}} d e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{{\left (4 i \, A + 4 \, B\right )} a^{2}}\right ) - 315 \, \sqrt {\frac {{\left (32 i \, A^{2} + 64 \, A B - 32 i \, B^{2}\right )} a^{5}}{d^{2}}} {\left (d e^{\left (10 i \, d x + 10 i \, c\right )} - 5 \, d e^{\left (8 i \, d x + 8 i \, c\right )} + 10 \, d e^{\left (6 i \, d x + 6 i \, c\right )} - 10 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 5 \, d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )} \log \left (\frac {{\left (\sqrt {2} {\left ({\left (4 i \, A + 4 \, B\right )} a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + {\left (4 i \, A + 4 \, B\right )} a^{2}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \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 {{\left (32 i \, A^{2} + 64 \, A B - 32 i \, B^{2}\right )} a^{5}}{d^{2}}} d e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{{\left (4 i \, A + 4 \, B\right )} a^{2}}\right )}{630 \, {\left (d e^{\left (10 i \, d x + 10 i \, c\right )} - 5 \, d e^{\left (8 i \, d x + 8 i \, c\right )} + 10 \, d e^{\left (6 i \, d x + 6 i \, c\right )} - 10 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 5 \, d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (B \tan \left (d x + c\right ) + A\right )} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}}}{\tan \left (d x + c\right )^{\frac {11}{2}}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.33, size = 887, normalized size = 3.20 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\left (A+B\,\mathrm {tan}\left (c+d\,x\right )\right )\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2}}{{\mathrm {tan}\left (c+d\,x\right )}^{11/2}} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
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