Optimal. Leaf size=86 \[ -\frac {i 2^{2-\frac {m}{2}} a^2 (e \cos (c+d x))^m \, _2F_1\left (\frac {1}{2} (-2+m),-\frac {m}{2};1-\frac {m}{2};\frac {1}{2} (1-i \tan (c+d x))\right ) (1+i \tan (c+d x))^{m/2}}{d m} \]
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Rubi [A]
time = 0.16, antiderivative size = 86, normalized size of antiderivative = 1.00, number of steps
used = 5, number of rules used = 5, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {3596, 3586,
3604, 72, 71} \begin {gather*} -\frac {i a^2 2^{2-\frac {m}{2}} (1+i \tan (c+d x))^{m/2} (e \cos (c+d x))^m \, _2F_1\left (\frac {m-2}{2},-\frac {m}{2};1-\frac {m}{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 3596
Rule 3604
Rubi steps
\begin {align*} \int (e \cos (c+d x))^m (a+i a \tan (c+d x))^2 \, dx &=\left ((e \cos (c+d x))^m (e \sec (c+d x))^m\right ) \int (e \sec (c+d x))^{-m} (a+i a \tan (c+d x))^2 \, dx\\ &=\left ((e \cos (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))^{2-\frac {m}{2}} \, dx\\ &=\frac {\left (a^2 (e \cos (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}} \, dx,x,\tan (c+d x)\right )}{d}\\ &=\frac {\left (2^{1-\frac {m}{2}} a^3 (e \cos (c+d x))^m (a-i a \tan (c+d x))^{m/2} \left (\frac {a+i a \tan (c+d x)}{a}\right )^{m/2}\right ) \text {Subst}\left (\int \left (\frac {1}{2}+\frac {i x}{2}\right )^{1-\frac {m}{2}} (a-i a x)^{-1-\frac {m}{2}} \, dx,x,\tan (c+d x)\right )}{d}\\ &=-\frac {i 2^{2-\frac {m}{2}} a^2 (e \cos (c+d x))^m \, _2F_1\left (\frac {1}{2} (-2+m),-\frac {m}{2};1-\frac {m}{2};\frac {1}{2} (1-i \tan (c+d x))\right ) (1+i \tan (c+d x))^{m/2}}{d m}\\ \end {align*}
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Mathematica [A]
time = 1.56, size = 144, normalized size = 1.67 \begin {gather*} -\frac {i 2^{2-m} a^2 e^{2 i (c+d x)} \left (1+e^{2 i (c+d x)}\right )^{-m} \left (e^{-i (c+d x)} \left (1+e^{2 i (c+d x)}\right )\right )^m \cos ^{2-m}(c+d x) (e \cos (c+d x))^m \, _2F_1\left (2-m,2-\frac {m}{2};3-\frac {m}{2};-e^{2 i (c+d x)}\right ) (-i+\tan (c+d x))^2}{d (-4+m)} \end {gather*}
Antiderivative was successfully verified.
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Maple [F]
time = 0.40, size = 0, normalized size = 0.00 \[\int \left (e \cos \left (d x +c \right )\right )^{m} \left (a +i a \tan \left (d x +c \right )\right )^{2}\, dx\]
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*} - a^{2} \left (\int \left (- \left (e \cos {\left (c + d x \right )}\right )^{m}\right )\, dx + \int \left (e \cos {\left (c + d x \right )}\right )^{m} \tan ^{2}{\left (c + d x \right )}\, dx + \int \left (- 2 i \left (e \cos {\left (c + d x \right )}\right )^{m} \tan {\left (c + d x \right )}\right )\, dx\right ) \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 (e\,\cos \left (c+d\,x\right )\right )}^m\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^2 \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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