Optimal. Leaf size=156 \[ \frac {4 \sqrt [4]{-1} a^2 (A-i B) \tan ^{-1}\left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )}{d}-\frac {2 a^2 (7 A-9 i B) \tan ^{\frac {5}{2}}(c+d x)}{35 d}+\frac {4 a^2 (B+i A) \tan ^{\frac {3}{2}}(c+d x)}{3 d}+\frac {4 a^2 (A-i B) \sqrt {\tan (c+d x)}}{d}+\frac {2 i B \tan ^{\frac {5}{2}}(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{7 d} \]
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Rubi [A] time = 0.31, antiderivative size = 156, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 36, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.139, Rules used = {3594, 3592, 3528, 3533, 205} \[ -\frac {2 a^2 (7 A-9 i B) \tan ^{\frac {5}{2}}(c+d x)}{35 d}+\frac {4 a^2 (B+i A) \tan ^{\frac {3}{2}}(c+d x)}{3 d}+\frac {4 \sqrt [4]{-1} a^2 (A-i B) \tan ^{-1}\left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )}{d}+\frac {4 a^2 (A-i B) \sqrt {\tan (c+d x)}}{d}+\frac {2 i B \tan ^{\frac {5}{2}}(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{7 d} \]
Antiderivative was successfully verified.
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Rule 205
Rule 3528
Rule 3533
Rule 3592
Rule 3594
Rubi steps
\begin {align*} \int \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx &=\frac {2 i B \tan ^{\frac {5}{2}}(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{7 d}+\frac {2}{7} \int \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x)) \left (\frac {1}{2} a (7 A-5 i B)+\frac {1}{2} a (7 i A+9 B) \tan (c+d x)\right ) \, dx\\ &=-\frac {2 a^2 (7 A-9 i B) \tan ^{\frac {5}{2}}(c+d x)}{35 d}+\frac {2 i B \tan ^{\frac {5}{2}}(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{7 d}+\frac {2}{7} \int \tan ^{\frac {3}{2}}(c+d x) \left (7 a^2 (A-i B)+7 a^2 (i A+B) \tan (c+d x)\right ) \, dx\\ &=\frac {4 a^2 (i A+B) \tan ^{\frac {3}{2}}(c+d x)}{3 d}-\frac {2 a^2 (7 A-9 i B) \tan ^{\frac {5}{2}}(c+d x)}{35 d}+\frac {2 i B \tan ^{\frac {5}{2}}(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{7 d}+\frac {2}{7} \int \sqrt {\tan (c+d x)} \left (-7 a^2 (i A+B)+7 a^2 (A-i B) \tan (c+d x)\right ) \, dx\\ &=\frac {4 a^2 (A-i B) \sqrt {\tan (c+d x)}}{d}+\frac {4 a^2 (i A+B) \tan ^{\frac {3}{2}}(c+d x)}{3 d}-\frac {2 a^2 (7 A-9 i B) \tan ^{\frac {5}{2}}(c+d x)}{35 d}+\frac {2 i B \tan ^{\frac {5}{2}}(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{7 d}+\frac {2}{7} \int \frac {-7 a^2 (A-i B)-7 a^2 (i A+B) \tan (c+d x)}{\sqrt {\tan (c+d x)}} \, dx\\ &=\frac {4 a^2 (A-i B) \sqrt {\tan (c+d x)}}{d}+\frac {4 a^2 (i A+B) \tan ^{\frac {3}{2}}(c+d x)}{3 d}-\frac {2 a^2 (7 A-9 i B) \tan ^{\frac {5}{2}}(c+d x)}{35 d}+\frac {2 i B \tan ^{\frac {5}{2}}(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{7 d}+\frac {\left (28 a^4 (A-i B)^2\right ) \operatorname {Subst}\left (\int \frac {1}{-7 a^2 (A-i B)+7 a^2 (i A+B) x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{d}\\ &=\frac {4 \sqrt [4]{-1} a^2 (A-i B) \tan ^{-1}\left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )}{d}+\frac {4 a^2 (A-i B) \sqrt {\tan (c+d x)}}{d}+\frac {4 a^2 (i A+B) \tan ^{\frac {3}{2}}(c+d x)}{3 d}-\frac {2 a^2 (7 A-9 i B) \tan ^{\frac {5}{2}}(c+d x)}{35 d}+\frac {2 i B \tan ^{\frac {5}{2}}(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{7 d}\\ \end {align*}
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Mathematica [A] time = 5.46, size = 307, normalized size = 1.97 \[ \frac {\cos ^3(c+d x) (a+i a \tan (c+d x))^2 (A+B \tan (c+d x)) \left (\frac {4 e^{-2 i c} (B+i A) \sqrt {\frac {-1+e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}} \tanh ^{-1}\left (\sqrt {\frac {-1+e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}}\right )}{\sqrt {-\frac {i \left (-1+e^{2 i (c+d x)}\right )}{1+e^{2 i (c+d x)}}}}+\frac {1}{210} (\cos (2 c)-i \sin (2 c)) \sqrt {\tan (c+d x)} \sec ^3(c+d x) (21 (29 A-28 i B) \cos (c+d x)+21 (11 A-12 i B) \cos (3 (c+d x))+70 i A \sin (c+d x)+70 i A \sin (3 (c+d x))+25 B \sin (c+d x)+85 B \sin (3 (c+d x)))\right )}{d (\cos (d x)+i \sin (d x))^2 (A \cos (c+d x)+B \sin (c+d x))} \]
Antiderivative was successfully verified.
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fricas [B] time = 1.44, size = 497, normalized size = 3.19 \[ -\frac {105 \, \sqrt {\frac {{\left (-16 i \, A^{2} - 32 \, A B + 16 i \, B^{2}\right )} a^{4}}{d^{2}}} {\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 )} \log \left (\frac {{\left (4 \, {\left (A - i \, B\right )} a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + \sqrt {\frac {{\left (-16 i \, A^{2} - 32 \, A B + 16 i \, B^{2}\right )} a^{4}}{d^{2}}} {\left (i \, d e^{\left (2 i \, d x + 2 i \, c\right )} + i \, 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}}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{{\left (2 i \, A + 2 \, B\right )} a^{2}}\right ) - 105 \, \sqrt {\frac {{\left (-16 i \, A^{2} - 32 \, A B + 16 i \, B^{2}\right )} a^{4}}{d^{2}}} {\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 )} \log \left (\frac {{\left (4 \, {\left (A - i \, B\right )} a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + \sqrt {\frac {{\left (-16 i \, A^{2} - 32 \, A B + 16 i \, B^{2}\right )} a^{4}}{d^{2}}} {\left (-i \, d e^{\left (2 i \, d x + 2 i \, c\right )} - i \, 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}}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{{\left (2 i \, A + 2 \, B\right )} a^{2}}\right ) - 8 \, {\left ({\left (301 \, A - 337 i \, B\right )} a^{2} e^{\left (6 i \, d x + 6 i \, c\right )} + {\left (679 \, A - 613 i \, B\right )} a^{2} e^{\left (4 i \, d x + 4 i \, c\right )} + {\left (539 \, A - 563 i \, B\right )} a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + {\left (161 \, A - 167 i \, B\right )} a^{2}\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 \[ \int {\left (B \tan \left (d x + c\right ) + A\right )} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} \tan \left (d x + c\right )^{\frac {3}{2}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.10, size = 574, normalized size = 3.68 \[ -\frac {2 a^{2} B \left (\tan ^{\frac {7}{2}}\left (d x +c \right )\right )}{7 d}-\frac {4 i a^{2} B \left (\sqrt {\tan }\left (d x +c \right )\right )}{d}-\frac {2 a^{2} A \left (\tan ^{\frac {5}{2}}\left (d x +c \right )\right )}{5 d}+\frac {i a^{2} 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 )}{2 d}+\frac {4 a^{2} B \left (\tan ^{\frac {3}{2}}\left (d x +c \right )\right )}{3 d}-\frac {i a^{2} A \sqrt {2}\, \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )}{d}+\frac {4 a^{2} A \left (\sqrt {\tan }\left (d x +c \right )\right )}{d}+\frac {4 i a^{2} B \left (\tan ^{\frac {5}{2}}\left (d x +c \right )\right )}{5 d}+\frac {4 i a^{2} A \left (\tan ^{\frac {3}{2}}\left (d x +c \right )\right )}{3 d}+\frac {i a^{2} B \sqrt {2}\, \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )}{d}-\frac {a^{2} A \sqrt {2}\, \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )}{d}-\frac {a^{2} 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 )}{2 d}-\frac {a^{2} A \sqrt {2}\, \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )}{d}+\frac {i a^{2} B \sqrt {2}\, \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )}{d}-\frac {i a^{2} A \sqrt {2}\, \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )}{d}-\frac {i a^{2} 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 )}{2 d}-\frac {a^{2} 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 )}{2 d}-\frac {a^{2} B \sqrt {2}\, \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )}{d}-\frac {a^{2} B \sqrt {2}\, \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )}{d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.91, size = 212, normalized size = 1.36 \[ -\frac {60 \, B a^{2} \tan \left (d x + c\right )^{\frac {7}{2}} + 84 \, {\left (A - 2 i \, B\right )} a^{2} \tan \left (d x + c\right )^{\frac {5}{2}} - 4 \, {\left (70 i \, A + 70 \, B\right )} a^{2} \tan \left (d x + c\right )^{\frac {3}{2}} - 840 \, {\left (A - i \, B\right )} a^{2} \sqrt {\tan \left (d x + c\right )} - 105 \, {\left (2 \, \sqrt {2} {\left (-\left (i + 1\right ) \, A + \left (i - 1\right ) \, B\right )} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) + 2 \, \sqrt {2} {\left (-\left (i + 1\right ) \, A + \left (i - 1\right ) \, B\right )} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) + \sqrt {2} {\left (\left (i - 1\right ) \, A + \left (i + 1\right ) \, B\right )} \log \left (\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right ) - \sqrt {2} {\left (\left (i - 1\right ) \, A + \left (i + 1\right ) \, B\right )} \log \left (-\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right )\right )} a^{2}}{210 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 9.28, size = 291, normalized size = 1.87 \[ \frac {4\,A\,a^2\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}}{d}+\frac {A\,a^2\,{\mathrm {tan}\left (c+d\,x\right )}^{3/2}\,4{}\mathrm {i}}{3\,d}-\frac {2\,A\,a^2\,{\mathrm {tan}\left (c+d\,x\right )}^{5/2}}{5\,d}-\frac {B\,a^2\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,4{}\mathrm {i}}{d}+\frac {4\,B\,a^2\,{\mathrm {tan}\left (c+d\,x\right )}^{3/2}}{3\,d}+\frac {B\,a^2\,{\mathrm {tan}\left (c+d\,x\right )}^{5/2}\,4{}\mathrm {i}}{5\,d}-\frac {2\,B\,a^2\,{\mathrm {tan}\left (c+d\,x\right )}^{7/2}}{7\,d}+\frac {\sqrt {2}\,A\,a^2\,\ln \left (-A\,a^2\,d\,4{}\mathrm {i}+\sqrt {2}\,A\,a^2\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\left (-2+2{}\mathrm {i}\right )\right )\,\left (1-\mathrm {i}\right )}{d}-\frac {\sqrt {-4{}\mathrm {i}}\,A\,a^2\,\ln \left (-A\,a^2\,d\,4{}\mathrm {i}+2\,\sqrt {-4{}\mathrm {i}}\,A\,a^2\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\right )}{d}+\frac {\sqrt {2}\,B\,a^2\,\ln \left (-4\,B\,a^2\,d+\sqrt {2}\,B\,a^2\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\left (-2-2{}\mathrm {i}\right )\right )\,\left (1+1{}\mathrm {i}\right )}{d}-\frac {\sqrt {4{}\mathrm {i}}\,B\,a^2\,\ln \left (-4\,B\,a^2\,d+2\,\sqrt {4{}\mathrm {i}}\,B\,a^2\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\right )}{d} \]
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 \[ - a^{2} \left (\int \left (- A \tan ^{\frac {3}{2}}{\left (c + d x \right )}\right )\, dx + \int A \tan ^{\frac {7}{2}}{\left (c + d x \right )}\, dx + \int \left (- B \tan ^{\frac {5}{2}}{\left (c + d x \right )}\right )\, dx + \int B \tan ^{\frac {9}{2}}{\left (c + d x \right )}\, dx + \int \left (- 2 i A \tan ^{\frac {5}{2}}{\left (c + d x \right )}\right )\, dx + \int \left (- 2 i B \tan ^{\frac {7}{2}}{\left (c + d x \right )}\right )\, dx\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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