Optimal. Leaf size=118 \[ \frac {i 2^{\frac {1+n}{2}} \, _2F_1\left (\frac {1-n}{2},\frac {1-n}{2};\frac {3-n}{2};\frac {1}{2} (1-i \tan (c+d x))\right ) (e \sec (c+d x))^{1-n} (1+i \tan (c+d x))^{\frac {1}{2} (-1-n)} (a+i a \tan (c+d x))^n}{d (1-n)} \]
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Rubi [A]
time = 0.16, antiderivative size = 118, normalized size of antiderivative = 1.00, number of steps
used = 4, number of rules used = 4, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {3586, 3604, 72,
71} \begin {gather*} \frac {i 2^{\frac {n+1}{2}} (1+i \tan (c+d x))^{\frac {1}{2} (-n-1)} (a+i a \tan (c+d x))^n (e \sec (c+d x))^{1-n} \, _2F_1\left (\frac {1-n}{2},\frac {1-n}{2};\frac {3-n}{2};\frac {1}{2} (1-i \tan (c+d x))\right )}{d (1-n)} \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))^{1-n} (a+i a \tan (c+d x))^n \, dx &=\left ((e \sec (c+d x))^{1-n} (a-i a \tan (c+d x))^{\frac {1}{2} (-1+n)} (a+i a \tan (c+d x))^{\frac {1}{2} (-1+n)}\right ) \int (a-i a \tan (c+d x))^{\frac {1-n}{2}} (a+i a \tan (c+d x))^{\frac {1-n}{2}+n} \, dx\\ &=\frac {\left (a^2 (e \sec (c+d x))^{1-n} (a-i a \tan (c+d x))^{\frac {1}{2} (-1+n)} (a+i a \tan (c+d x))^{\frac {1}{2} (-1+n)}\right ) \text {Subst}\left (\int (a-i a x)^{-1+\frac {1-n}{2}} (a+i a x)^{-1+\frac {1-n}{2}+n} \, dx,x,\tan (c+d x)\right )}{d}\\ &=\frac {\left (2^{-\frac {1}{2}+\frac {n}{2}} a (e \sec (c+d x))^{1-n} (a-i a \tan (c+d x))^{\frac {1}{2} (-1+n)} (a+i a \tan (c+d x))^{\frac {1}{2}+\frac {1}{2} (-1+n)+\frac {n}{2}} \left (\frac {a+i a \tan (c+d x)}{a}\right )^{-\frac {1}{2}-\frac {n}{2}}\right ) \text {Subst}\left (\int \left (\frac {1}{2}+\frac {i x}{2}\right )^{-1+\frac {1-n}{2}+n} (a-i a x)^{-1+\frac {1-n}{2}} \, dx,x,\tan (c+d x)\right )}{d}\\ &=\frac {i 2^{\frac {1+n}{2}} \, _2F_1\left (\frac {1-n}{2},\frac {1-n}{2};\frac {3-n}{2};\frac {1}{2} (1-i \tan (c+d x))\right ) (e \sec (c+d x))^{1-n} (1+i \tan (c+d x))^{\frac {1}{2} (-1-n)} (a+i a \tan (c+d x))^n}{d (1-n)}\\ \end {align*}
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Mathematica [A]
time = 5.14, size = 102, normalized size = 0.86 \begin {gather*} -\frac {2 e \cos (d x) \, _2F_1\left (1,\frac {1-n}{2};\frac {3-n}{2};-\cos (2 (c+d x))+i \sin (2 (c+d x))\right ) (e \sec (c+d x))^{-n} (\cos (c)-i \sin (c)) (i+\tan (d x)) (a+i a \tan (c+d x))^n}{d (-1+n)} \end {gather*}
Antiderivative was successfully verified.
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Maple [F]
time = 0.84, size = 0, normalized size = 0.00 \[\int \left (e \sec \left (d x +c \right )\right )^{1-n} \left (a +i a \tan \left (d x +c \right )\right )^{n}\, dx\]
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 )^{1 - n} \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 )}^{1-n}\,{\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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