Optimal. Leaf size=105 \[ \frac {i 2^{\frac {m}{2}+n} \, _2F_1\left (\frac {m}{2},1-\frac {m}{2}-n;\frac {2+m}{2};\frac {1}{2} (1-i \tan (c+d x))\right ) (e \sec (c+d x))^m (1+i \tan (c+d x))^{-\frac {m}{2}-n} (a+i a \tan (c+d x))^n}{d m} \]
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Rubi [A]
time = 0.11, antiderivative size = 105, normalized size of antiderivative = 1.00, number of steps
used = 4, number of rules used = 4, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {3586, 3604, 72,
71} \begin {gather*} \frac {i 2^{\frac {m}{2}+n} (a+i a \tan (c+d x))^n (e \sec (c+d x))^m (1+i \tan (c+d x))^{-\frac {m}{2}-n} \, _2F_1\left (\frac {m}{2},-\frac {m}{2}-n+1;\frac {m+2}{2};\frac {1}{2} (1-i \tan (c+d x))\right )}{d m} \end {gather*}
Antiderivative was successfully verified.
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Rule 71
Rule 72
Rule 3586
Rule 3604
Rubi steps
\begin {align*} \int (e \sec (c+d x))^m (a+i a \tan (c+d x))^n \, dx &=\left ((e \sec (c+d x))^m (a-i a \tan (c+d x))^{-m/2} (a+i a \tan (c+d x))^{-m/2}\right ) \int (a-i a \tan (c+d x))^{m/2} (a+i a \tan (c+d x))^{\frac {m}{2}+n} \, dx\\ &=\frac {\left (a^2 (e \sec (c+d x))^m (a-i a \tan (c+d x))^{-m/2} (a+i a \tan (c+d x))^{-m/2}\right ) \text {Subst}\left (\int (a-i a x)^{-1+\frac {m}{2}} (a+i a x)^{-1+\frac {m}{2}+n} \, dx,x,\tan (c+d x)\right )}{d}\\ &=\frac {\left (2^{-1+\frac {m}{2}+n} a (e \sec (c+d x))^m (a-i a \tan (c+d x))^{-m/2} (a+i a \tan (c+d x))^n \left (\frac {a+i a \tan (c+d x)}{a}\right )^{-\frac {m}{2}-n}\right ) \text {Subst}\left (\int \left (\frac {1}{2}+\frac {i x}{2}\right )^{-1+\frac {m}{2}+n} (a-i a x)^{-1+\frac {m}{2}} \, dx,x,\tan (c+d x)\right )}{d}\\ &=\frac {i 2^{\frac {m}{2}+n} \, _2F_1\left (\frac {m}{2},1-\frac {m}{2}-n;\frac {2+m}{2};\frac {1}{2} (1-i \tan (c+d x))\right ) (e \sec (c+d x))^m (1+i \tan (c+d x))^{-\frac {m}{2}-n} (a+i a \tan (c+d x))^n}{d m}\\ \end {align*}
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Mathematica [A]
time = 9.42, size = 165, normalized size = 1.57 \begin {gather*} -\frac {i 2^{m+n} \left (e^{i d x}\right )^n \left (\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}\right )^{m+n} \left (1+e^{2 i (c+d x)}\right )^{m+n} \, _2F_1\left (\frac {m}{2}+n,m+n;1+\frac {m}{2}+n;-e^{2 i (c+d x)}\right ) \sec ^{-m-n}(c+d x) (e \sec (c+d x))^m (\cos (d x)+i \sin (d x))^{-n} (a+i a \tan (c+d x))^n}{d (m+2 n)} \end {gather*}
Antiderivative was successfully verified.
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Maple [F]
time = 0.69, size = 0, normalized size = 0.00 \[\int \left (e \sec \left (d x +c \right )\right )^{m} \left (a +i a \tan \left (d x +c \right )\right )^{n}\, dx\]
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}
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 \begin {gather*} \int \left (e \sec {\left (c + d x \right )}\right )^{m} \left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{n}\, dx \end {gather*}
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 \begin {gather*} \text {could not integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\left (\frac {e}{\cos \left (c+d\,x\right )}\right )}^m\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^n \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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