Optimal. Leaf size=130 \[ -\frac {2 \sqrt [4]{-1} a (B+i A) \tan ^{-1}\left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )}{d}+\frac {2 a (B+i A) \tan ^{\frac {5}{2}}(c+d x)}{5 d}+\frac {2 a (A-i B) \tan ^{\frac {3}{2}}(c+d x)}{3 d}-\frac {2 a (B+i A) \sqrt {\tan (c+d x)}}{d}+\frac {2 i a B \tan ^{\frac {7}{2}}(c+d x)}{7 d} \]
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Rubi [A] time = 0.20, antiderivative size = 130, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 34, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {3592, 3528, 3533, 205} \[ \frac {2 a (B+i A) \tan ^{\frac {5}{2}}(c+d x)}{5 d}+\frac {2 a (A-i B) \tan ^{\frac {3}{2}}(c+d x)}{3 d}-\frac {2 \sqrt [4]{-1} a (B+i A) \tan ^{-1}\left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )}{d}-\frac {2 a (B+i A) \sqrt {\tan (c+d x)}}{d}+\frac {2 i a B \tan ^{\frac {7}{2}}(c+d x)}{7 d} \]
Antiderivative was successfully verified.
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Rule 205
Rule 3528
Rule 3533
Rule 3592
Rubi steps
\begin {align*} \int \tan ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x)) (A+B \tan (c+d x)) \, dx &=\frac {2 i a B \tan ^{\frac {7}{2}}(c+d x)}{7 d}+\int \tan ^{\frac {5}{2}}(c+d x) (a (A-i B)+a (i A+B) \tan (c+d x)) \, dx\\ &=\frac {2 a (i A+B) \tan ^{\frac {5}{2}}(c+d x)}{5 d}+\frac {2 i a B \tan ^{\frac {7}{2}}(c+d x)}{7 d}+\int \tan ^{\frac {3}{2}}(c+d x) (-a (i A+B)+a (A-i B) \tan (c+d x)) \, dx\\ &=\frac {2 a (A-i B) \tan ^{\frac {3}{2}}(c+d x)}{3 d}+\frac {2 a (i A+B) \tan ^{\frac {5}{2}}(c+d x)}{5 d}+\frac {2 i a B \tan ^{\frac {7}{2}}(c+d x)}{7 d}+\int \sqrt {\tan (c+d x)} (-a (A-i B)-a (i A+B) \tan (c+d x)) \, dx\\ &=-\frac {2 a (i A+B) \sqrt {\tan (c+d x)}}{d}+\frac {2 a (A-i B) \tan ^{\frac {3}{2}}(c+d x)}{3 d}+\frac {2 a (i A+B) \tan ^{\frac {5}{2}}(c+d x)}{5 d}+\frac {2 i a B \tan ^{\frac {7}{2}}(c+d x)}{7 d}+\int \frac {a (i A+B)-a (A-i B) \tan (c+d x)}{\sqrt {\tan (c+d x)}} \, dx\\ &=-\frac {2 a (i A+B) \sqrt {\tan (c+d x)}}{d}+\frac {2 a (A-i B) \tan ^{\frac {3}{2}}(c+d x)}{3 d}+\frac {2 a (i A+B) \tan ^{\frac {5}{2}}(c+d x)}{5 d}+\frac {2 i a B \tan ^{\frac {7}{2}}(c+d x)}{7 d}+\frac {\left (2 a^2 (i A+B)^2\right ) \operatorname {Subst}\left (\int \frac {1}{a (i A+B)+a (A-i B) x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{d}\\ &=-\frac {2 \sqrt [4]{-1} a (i A+B) \tan ^{-1}\left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )}{d}-\frac {2 a (i A+B) \sqrt {\tan (c+d x)}}{d}+\frac {2 a (A-i B) \tan ^{\frac {3}{2}}(c+d x)}{3 d}+\frac {2 a (i A+B) \tan ^{\frac {5}{2}}(c+d x)}{5 d}+\frac {2 i a B \tan ^{\frac {7}{2}}(c+d x)}{7 d}\\ \end {align*}
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Mathematica [B] time = 4.63, size = 280, normalized size = 2.15 \[ \frac {\cos ^2(c+d x) (\cos (d x)-i \sin (d x)) (a+i a \tan (c+d x)) (A+B \tan (c+d x)) \left (\frac {2 e^{-i c} (B+i A) \sqrt {-\frac {i \left (-1+e^{2 i (c+d x)}\right )}{1+e^{2 i (c+d x)}}} \tanh ^{-1}\left (\sqrt {\frac {-1+e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}}\right )}{\sqrt {\frac {-1+e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}}}-\frac {1}{105} \cos (c) (\tan (c)+i) \sqrt {\tan (c+d x)} \sec ^2(c+d x) (5 (4 B+7 i A) \tan (c+d x)+\cos (2 (c+d x)) (5 (10 B+7 i A) \tan (c+d x)+126 (A-i B))+84 (A-i B))\right )}{d (A \cos (c+d x)+B \sin (c+d x))} \]
Antiderivative was successfully verified.
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fricas [B] time = 0.80, size = 475, normalized size = 3.65 \[ \frac {105 \, {\left (d e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {\frac {{\left (4 i \, A^{2} + 8 \, A B - 4 i \, B^{2}\right )} a^{2}}{d^{2}}} \log \left (\frac {{\left (2 \, {\left (A - i \, B\right )} a e^{\left (2 i \, d x + 2 i \, c\right )} + {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {\frac {{\left (4 i \, A^{2} + 8 \, A B - 4 i \, B^{2}\right )} a^{2}}{d^{2}}} \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 (-2 i \, d x - 2 i \, c\right )}}{{\left (i \, A + B\right )} a}\right ) - 105 \, {\left (d e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {\frac {{\left (4 i \, A^{2} + 8 \, A B - 4 i \, B^{2}\right )} a^{2}}{d^{2}}} \log \left (\frac {{\left (2 \, {\left (A - i \, B\right )} a e^{\left (2 i \, d x + 2 i \, c\right )} - {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {\frac {{\left (4 i \, A^{2} + 8 \, A B - 4 i \, B^{2}\right )} a^{2}}{d^{2}}} \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 (-2 i \, d x - 2 i \, c\right )}}{{\left (i \, A + B\right )} a}\right ) + {\left ({\left (-1288 i \, A - 1408 \, B\right )} a e^{\left (6 i \, d x + 6 i \, c\right )} + {\left (-2632 i \, A - 2272 \, B\right )} a e^{\left (4 i \, d x + 4 i \, c\right )} + {\left (-2072 i \, A - 2432 \, B\right )} a e^{\left (2 i \, d x + 2 i \, c\right )} + {\left (-728 i \, A - 608 \, B\right )} a\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}}}{420 \, {\left (d e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, 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 \[ \int {\left (B \tan \left (d x + c\right ) + A\right )} {\left (i \, a \tan \left (d x + c\right ) + a\right )} \tan \left (d x + c\right )^{\frac {5}{2}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.10, size = 537, normalized size = 4.13 \[ \frac {i a A \sqrt {2}\, \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )}{2 d}+\frac {2 i a A \left (\tan ^{\frac {5}{2}}\left (d x +c \right )\right )}{5 d}+\frac {2 a B \left (\tan ^{\frac {5}{2}}\left (d x +c \right )\right )}{5 d}+\frac {i a A \sqrt {2}\, \ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )}{4 d}+\frac {2 a A \left (\tan ^{\frac {3}{2}}\left (d x +c \right )\right )}{3 d}+\frac {2 i a B \left (\tan ^{\frac {7}{2}}\left (d x +c \right )\right )}{7 d}-\frac {2 a B \left (\sqrt {\tan }\left (d x +c \right )\right )}{d}+\frac {i a B \sqrt {2}\, \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )}{2 d}+\frac {i a A \sqrt {2}\, \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )}{2 d}-\frac {2 i a A \left (\sqrt {\tan }\left (d x +c \right )\right )}{d}+\frac {a B \sqrt {2}\, \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )}{2 d}+\frac {a B \sqrt {2}\, \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )}{2 d}+\frac {a B \sqrt {2}\, \ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )}{4 d}-\frac {2 i a B \left (\tan ^{\frac {3}{2}}\left (d x +c \right )\right )}{3 d}+\frac {i a B \sqrt {2}\, \ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )}{4 d}+\frac {i a B \sqrt {2}\, \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )}{2 d}-\frac {a A \sqrt {2}\, \ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )}{4 d}-\frac {a A \sqrt {2}\, \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )}{2 d}-\frac {a A \sqrt {2}\, \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )}{2 d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.55, size = 204, normalized size = 1.57 \[ -\frac {-120 i \, B a \tan \left (d x + c\right )^{\frac {7}{2}} - 8 \, {\left (21 i \, A + 21 \, B\right )} a \tan \left (d x + c\right )^{\frac {5}{2}} - 280 \, {\left (A - i \, B\right )} a \tan \left (d x + c\right )^{\frac {3}{2}} - 8 \, {\left (-105 i \, A - 105 \, B\right )} a \sqrt {\tan \left (d x + c\right )} - 105 \, {\left (2 \, \sqrt {2} {\left (\left (i - 1\right ) \, A + \left (i + 1\right ) \, B\right )} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) + 2 \, \sqrt {2} {\left (\left (i - 1\right ) \, A + \left (i + 1\right ) \, B\right )} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) - \sqrt {2} {\left (-\left (i + 1\right ) \, A + \left (i - 1\right ) \, B\right )} \log \left (\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right ) + \sqrt {2} {\left (-\left (i + 1\right ) \, A + \left (i - 1\right ) \, B\right )} \log \left (-\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right )\right )} a}{420 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 10.12, size = 161, normalized size = 1.24 \[ \frac {2\,A\,a\,{\mathrm {tan}\left (c+d\,x\right )}^{3/2}}{3\,d}-\frac {A\,a\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,2{}\mathrm {i}}{d}+\frac {A\,a\,{\mathrm {tan}\left (c+d\,x\right )}^{5/2}\,2{}\mathrm {i}}{5\,d}-\frac {2\,B\,a\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}}{d}-\frac {B\,a\,{\mathrm {tan}\left (c+d\,x\right )}^{3/2}\,2{}\mathrm {i}}{3\,d}+\frac {2\,B\,a\,{\mathrm {tan}\left (c+d\,x\right )}^{5/2}}{5\,d}+\frac {B\,a\,{\mathrm {tan}\left (c+d\,x\right )}^{7/2}\,2{}\mathrm {i}}{7\,d}-\frac {{\left (-1\right )}^{1/4}\,A\,a\,\mathrm {atan}\left ({\left (-1\right )}^{1/4}\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,1{}\mathrm {i}\right )\,2{}\mathrm {i}}{d}+\frac {\sqrt {2}\,B\,a\,\mathrm {atan}\left (\sqrt {2}\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\left (\frac {1}{2}-\frac {1}{2}{}\mathrm {i}\right )\right )\,\left (1+1{}\mathrm {i}\right )}{d} \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ i a \left (\int A \tan ^{\frac {7}{2}}{\left (c + d x \right )}\, dx + \int B \tan ^{\frac {9}{2}}{\left (c + d x \right )}\, dx + \int \left (- i A \tan ^{\frac {5}{2}}{\left (c + d x \right )}\right )\, dx + \int \left (- i B \tan ^{\frac {7}{2}}{\left (c + d x \right )}\right )\, dx\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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