Optimal. Leaf size=103 \[ \frac {2 \sqrt [4]{-1} a (B+i A) \tan ^{-1}\left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )}{d}-\frac {2 a (B+i A)}{3 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {2 a (A-i B)}{d \sqrt {\tan (c+d x)}}-\frac {2 a A}{5 d \tan ^{\frac {5}{2}}(c+d x)} \]
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Rubi [A] time = 0.15, antiderivative size = 103, 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 = {3591, 3529, 3533, 205} \[ \frac {2 \sqrt [4]{-1} a (B+i A) \tan ^{-1}\left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )}{d}-\frac {2 a (B+i A)}{3 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {2 a (A-i B)}{d \sqrt {\tan (c+d x)}}-\frac {2 a A}{5 d \tan ^{\frac {5}{2}}(c+d x)} \]
Antiderivative was successfully verified.
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Rule 205
Rule 3529
Rule 3533
Rule 3591
Rubi steps
\begin {align*} \int \frac {(a+i a \tan (c+d x)) (A+B \tan (c+d x))}{\tan ^{\frac {7}{2}}(c+d x)} \, dx &=-\frac {2 a A}{5 d \tan ^{\frac {5}{2}}(c+d x)}+\int \frac {a (i A+B)-a (A-i B) \tan (c+d x)}{\tan ^{\frac {5}{2}}(c+d x)} \, dx\\ &=-\frac {2 a A}{5 d \tan ^{\frac {5}{2}}(c+d x)}-\frac {2 a (i A+B)}{3 d \tan ^{\frac {3}{2}}(c+d x)}+\int \frac {-a (A-i B)-a (i A+B) \tan (c+d x)}{\tan ^{\frac {3}{2}}(c+d x)} \, dx\\ &=-\frac {2 a A}{5 d \tan ^{\frac {5}{2}}(c+d x)}-\frac {2 a (i A+B)}{3 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {2 a (A-i B)}{d \sqrt {\tan (c+d x)}}+\int \frac {-a (i A+B)+a (A-i B) \tan (c+d x)}{\sqrt {\tan (c+d x)}} \, dx\\ &=-\frac {2 a A}{5 d \tan ^{\frac {5}{2}}(c+d x)}-\frac {2 a (i A+B)}{3 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {2 a (A-i B)}{d \sqrt {\tan (c+d x)}}+\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 A}{5 d \tan ^{\frac {5}{2}}(c+d x)}-\frac {2 a (i A+B)}{3 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {2 a (A-i B)}{d \sqrt {\tan (c+d x)}}\\ \end {align*}
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Mathematica [B] time = 4.26, size = 265, normalized size = 2.57 \[ \frac {\cos ^2(c+d x) (\cos (d x)-i \sin (d x)) (a+i a \tan (c+d x)) (A+B \tan (c+d x)) \left (-\frac {2 i e^{-i c} (A-i B) \sqrt {-\frac {i \left (-1+e^{2 i (c+d x)}\right )}{1+e^{2 i (c+d x)}}} \tanh ^{-1}\left (\sqrt {\frac {-1+e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}}\right )}{\sqrt {\frac {-1+e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}}}-\frac {(\cos (c)-i \sin (c)) \csc ^2(c+d x) (5 (B+i A) \sin (2 (c+d x))+3 (6 A-5 i B) \cos (2 (c+d x))-12 A+15 i B)}{15 \sqrt {\tan (c+d x)}}\right )}{d (A \cos (c+d x)+B \sin (c+d x))} \]
Antiderivative was successfully verified.
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fricas [B] time = 0.79, size = 482, normalized size = 4.68 \[ -\frac {15 \, {\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 ) - 15 \, {\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 (184 i \, A + 160 \, B\right )} a e^{\left (6 i \, d x + 6 i \, c\right )} + {\left (-8 i \, A - 80 \, B\right )} a e^{\left (4 i \, d x + 4 i \, c\right )} + {\left (-88 i \, A - 160 \, B\right )} a e^{\left (2 i \, d x + 2 i \, c\right )} + {\left (104 i \, A + 80 \, 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 (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 \[ \int \frac {{\left (B \tan \left (d x + c\right ) + A\right )} {\left (i \, a \tan \left (d x + c\right ) + a\right )}}{\tan \left (d x + c\right )^{\frac {7}{2}}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.11, size = 505, normalized size = 4.90 \[ -\frac {2 a A}{5 d \tan \left (d x +c \right )^{\frac {5}{2}}}-\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 a A}{d \sqrt {\tan \left (d x +c \right )}}-\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 {2 a B}{3 d \tan \left (d x +c \right )^{\frac {3}{2}}}-\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 {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 {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 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 {2 i a B}{d \sqrt {\tan \left (d x +c \right )}}-\frac {2 i a A}{3 d \tan \left (d x +c \right )^{\frac {3}{2}}}-\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}\, \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} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.87, size = 187, normalized size = 1.82 \[ -\frac {15 \, {\left (2 \, \sqrt {2} {\left (\left (i - 1\right ) \, A + \left (i + 1\right ) \, B\right )} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) + 2 \, \sqrt {2} {\left (\left (i - 1\right ) \, A + \left (i + 1\right ) \, B\right )} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) - \sqrt {2} {\left (-\left (i + 1\right ) \, A + \left (i - 1\right ) \, B\right )} \log \left (\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right ) + \sqrt {2} {\left (-\left (i + 1\right ) \, A + \left (i - 1\right ) \, B\right )} \log \left (-\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right )\right )} a - \frac {8 \, {\left (15 \, {\left (A - i \, B\right )} a \tan \left (d x + c\right )^{2} - {\left (5 i \, A + 5 \, B\right )} a \tan \left (d x + c\right ) - 3 \, A a\right )}}{\tan \left (d x + c\right )^{\frac {5}{2}}}}{60 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 8.74, size = 123, normalized size = 1.19 \[ -\frac {\frac {2\,B\,a}{3\,d}+\frac {B\,a\,\mathrm {tan}\left (c+d\,x\right )\,2{}\mathrm {i}}{d}}{{\mathrm {tan}\left (c+d\,x\right )}^{3/2}}+\frac {\sqrt {2}\,B\,a\,\mathrm {atan}\left (\sqrt {2}\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\left (\frac {1}{2}-\frac {1}{2}{}\mathrm {i}\right )\right )\,\left (-1-\mathrm {i}\right )}{d}-\frac {2\,A\,a\,\left (15\,{\left (-1\right )}^{1/4}\,{\mathrm {tan}\left (c+d\,x\right )}^{5/2}\,\mathrm {atanh}\left ({\left (-1\right )}^{1/4}\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\right )-15\,{\mathrm {tan}\left (c+d\,x\right )}^2+3+\mathrm {tan}\left (c+d\,x\right )\,5{}\mathrm {i}\right )}{15\,d\,{\mathrm {tan}\left (c+d\,x\right )}^{5/2}} \]
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 \[ i a \left (\int \frac {A}{\tan ^{\frac {5}{2}}{\left (c + d x \right )}}\, dx + \int \frac {B}{\tan ^{\frac {3}{2}}{\left (c + d x \right )}}\, dx + \int \left (- \frac {i A}{\tan ^{\frac {7}{2}}{\left (c + d x \right )}}\right )\, dx + \int \left (- \frac {i B}{\tan ^{\frac {5}{2}}{\left (c + d x \right )}}\right )\, dx\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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