Optimal. Leaf size=105 \[ \frac {2 \sqrt [4]{-1} a (A-i B) \tan ^{-1}\left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )}{d}+\frac {2 a (B+i A) \tan ^{\frac {3}{2}}(c+d x)}{3 d}+\frac {2 a (A-i B) \sqrt {\tan (c+d x)}}{d}+\frac {2 i a B \tan ^{\frac {5}{2}}(c+d x)}{5 d} \]
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Rubi [A] time = 0.17, antiderivative size = 105, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 34, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {3592, 3528, 3533, 205} \[ \frac {2 a (B+i A) \tan ^{\frac {3}{2}}(c+d x)}{3 d}+\frac {2 \sqrt [4]{-1} a (A-i B) \tan ^{-1}\left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )}{d}+\frac {2 a (A-i B) \sqrt {\tan (c+d x)}}{d}+\frac {2 i a B \tan ^{\frac {5}{2}}(c+d x)}{5 d} \]
Antiderivative was successfully verified.
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Rule 205
Rule 3528
Rule 3533
Rule 3592
Rubi steps
\begin {align*} \int \tan ^{\frac {3}{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 {5}{2}}(c+d x)}{5 d}+\int \tan ^{\frac {3}{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 {3}{2}}(c+d x)}{3 d}+\frac {2 i a B \tan ^{\frac {5}{2}}(c+d x)}{5 d}+\int \sqrt {\tan (c+d x)} (-a (i A+B)+a (A-i B) \tan (c+d x)) \, dx\\ &=\frac {2 a (A-i B) \sqrt {\tan (c+d x)}}{d}+\frac {2 a (i A+B) \tan ^{\frac {3}{2}}(c+d x)}{3 d}+\frac {2 i a B \tan ^{\frac {5}{2}}(c+d x)}{5 d}+\int \frac {-a (A-i B)-a (i A+B) \tan (c+d x)}{\sqrt {\tan (c+d x)}} \, dx\\ &=\frac {2 a (A-i B) \sqrt {\tan (c+d x)}}{d}+\frac {2 a (i A+B) \tan ^{\frac {3}{2}}(c+d x)}{3 d}+\frac {2 i a B \tan ^{\frac {5}{2}}(c+d x)}{5 d}+\frac {\left (2 a^2 (A-i B)^2\right ) \operatorname {Subst}\left (\int \frac {1}{-a (A-i B)+a (i A+B) x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{d}\\ &=\frac {2 \sqrt [4]{-1} a (A-i B) \tan ^{-1}\left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )}{d}+\frac {2 a (A-i B) \sqrt {\tan (c+d x)}}{d}+\frac {2 a (i A+B) \tan ^{\frac {3}{2}}(c+d x)}{3 d}+\frac {2 i a B \tan ^{\frac {5}{2}}(c+d x)}{5 d}\\ \end {align*}
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Mathematica [B] time = 3.10, size = 266, normalized size = 2.53 \[ \frac {\cos ^2(c+d x) (\cos (d x)-i \sin (d x)) (a+i a \tan (c+d x)) (A+B \tan (c+d x)) \left (\frac {2 e^{-i c} (B+i A) \sqrt {\frac {-1+e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}} \tanh ^{-1}\left (\sqrt {\frac {-1+e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}}\right )}{\sqrt {-\frac {i \left (-1+e^{2 i (c+d x)}\right )}{1+e^{2 i (c+d x)}}}}+\frac {1}{15} (\cos (c)-i \sin (c)) \sqrt {\tan (c+d x)} \sec ^2(c+d x) (5 (B+i A) \sin (2 (c+d x))+3 (5 A-6 i B) \cos (2 (c+d x))+3 (5 A-4 i B))\right )}{d (A \cos (c+d x)+B \sin (c+d x))} \]
Antiderivative was successfully verified.
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fricas [B] time = 0.63, size = 428, normalized size = 4.08 \[ -\frac {15 \, {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {\frac {{\left (-4 i \, A^{2} - 8 \, A B + 4 i \, B^{2}\right )} a^{2}}{d^{2}}} \log \left (\frac {{\left (2 \, {\left (A - i \, B\right )} a e^{\left (2 i \, d x + 2 i \, c\right )} + {\left (i \, d e^{\left (2 i \, d x + 2 i \, c\right )} + i \, d\right )} \sqrt {\frac {{\left (-4 i \, A^{2} - 8 \, A B + 4 i \, B^{2}\right )} a^{2}}{d^{2}}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{{\left (i \, A + B\right )} a}\right ) - 15 \, {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {\frac {{\left (-4 i \, A^{2} - 8 \, A B + 4 i \, B^{2}\right )} a^{2}}{d^{2}}} \log \left (\frac {{\left (2 \, {\left (A - i \, B\right )} a e^{\left (2 i \, d x + 2 i \, c\right )} + {\left (-i \, d e^{\left (2 i \, d x + 2 i \, c\right )} - i \, d\right )} \sqrt {\frac {{\left (-4 i \, A^{2} - 8 \, A B + 4 i \, B^{2}\right )} a^{2}}{d^{2}}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{{\left (i \, A + B\right )} a}\right ) - 8 \, {\left ({\left (20 \, A - 23 i \, B\right )} a e^{\left (4 i \, d x + 4 i \, c\right )} + 6 \, {\left (5 \, A - 4 i \, B\right )} a e^{\left (2 i \, d x + 2 i \, c\right )} + {\left (10 \, A - 13 i \, B\right )} a\right )} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}}{60 \, {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]
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 \[ \int {\left (B \tan \left (d x + c\right ) + A\right )} {\left (i \, a \tan \left (d x + c\right ) + a\right )} \tan \left (d x + c\right )^{\frac {3}{2}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.10, size = 506, normalized size = 4.82 \[ -\frac {i a A \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}\, \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 B \left (\tan ^{\frac {3}{2}}\left (d x +c \right )\right )}{3 d}+\frac {2 i a A \left (\tan ^{\frac {3}{2}}\left (d x +c \right )\right )}{3 d}+\frac {2 a A \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 {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 {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}-\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 {2 i a B \left (\sqrt {\tan }\left (d x +c \right )\right )}{d}+\frac {2 i a B \left (\tan ^{\frac {5}{2}}\left (d x +c \right )\right )}{5 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 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} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.81, size = 186, normalized size = 1.77 \[ -\frac {-24 i \, B a \tan \left (d x + c\right )^{\frac {5}{2}} - 8 \, {\left (5 i \, A + 5 \, B\right )} a \tan \left (d x + c\right )^{\frac {3}{2}} - 120 \, {\left (A - i \, B\right )} a \sqrt {\tan \left (d x + c\right )} - 15 \, {\left (2 \, \sqrt {2} {\left (-\left (i + 1\right ) \, A + \left (i - 1\right ) \, B\right )} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) + 2 \, \sqrt {2} {\left (-\left (i + 1\right ) \, A + \left (i - 1\right ) \, B\right )} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) + \sqrt {2} {\left (\left (i - 1\right ) \, A + \left (i + 1\right ) \, B\right )} \log \left (\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right ) - \sqrt {2} {\left (\left (i - 1\right ) \, A + \left (i + 1\right ) \, B\right )} \log \left (-\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right )\right )} a}{60 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 8.40, size = 130, normalized size = 1.24 \[ \frac {2\,A\,a\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}}{d}+\frac {A\,a\,{\mathrm {tan}\left (c+d\,x\right )}^{3/2}\,2{}\mathrm {i}}{3\,d}-\frac {B\,a\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,2{}\mathrm {i}}{d}+\frac {2\,B\,a\,{\mathrm {tan}\left (c+d\,x\right )}^{3/2}}{3\,d}+\frac {B\,a\,{\mathrm {tan}\left (c+d\,x\right )}^{5/2}\,2{}\mathrm {i}}{5\,d}+\frac {\sqrt {2}\,A\,a\,\mathrm {atan}\left (\sqrt {2}\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\left (\frac {1}{2}-\frac {1}{2}{}\mathrm {i}\right )\right )\,\left (-1-\mathrm {i}\right )}{d}-\frac {{\left (-1\right )}^{1/4}\,B\,a\,\mathrm {atan}\left ({\left (-1\right )}^{1/4}\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,1{}\mathrm {i}\right )\,2{}\mathrm {i}}{d} \]
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 \[ i a \left (\int A \tan ^{\frac {5}{2}}{\left (c + d x \right )}\, dx + \int B \tan ^{\frac {7}{2}}{\left (c + d x \right )}\, dx + \int \left (- i A \tan ^{\frac {3}{2}}{\left (c + d x \right )}\right )\, dx + \int \left (- i B \tan ^{\frac {5}{2}}{\left (c + d x \right )}\right )\, dx\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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