Optimal. Leaf size=105 \[ -\frac {i a 2^{\frac {1}{2}-\frac {m}{2}} (1+i \tan (c+d x))^{\frac {m+1}{2}} (e \cos (c+d x))^m \, _2F_1\left (-\frac {m}{2},\frac {m+1}{2};1-\frac {m}{2};\frac {1}{2} (1-i \tan (c+d x))\right )}{d m \sqrt {a+i a \tan (c+d x)}} \]
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Rubi [A] time = 0.29, antiderivative size = 105, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {3515, 3505, 3523, 70, 69} \[ -\frac {i a 2^{\frac {1}{2}-\frac {m}{2}} (1+i \tan (c+d x))^{\frac {m+1}{2}} (e \cos (c+d x))^m \, _2F_1\left (-\frac {m}{2},\frac {m+1}{2};1-\frac {m}{2};\frac {1}{2} (1-i \tan (c+d x))\right )}{d m \sqrt {a+i a \tan (c+d x)}} \]
Antiderivative was successfully verified.
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Rule 69
Rule 70
Rule 3505
Rule 3515
Rule 3523
Rubi steps
\begin {align*} \int (e \cos (c+d x))^m \sqrt {a+i a \tan (c+d x)} \, dx &=\left ((e \cos (c+d x))^m (e \sec (c+d x))^m\right ) \int (e \sec (c+d x))^{-m} \sqrt {a+i a \tan (c+d x)} \, 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))^{\frac {1}{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 ) \operatorname {Subst}\left (\int (a-i a x)^{-1-\frac {m}{2}} (a+i a x)^{-\frac {1}{2}-\frac {m}{2}} \, dx,x,\tan (c+d x)\right )}{d}\\ &=\frac {\left (2^{-\frac {1}{2}-\frac {m}{2}} a^2 (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 )^{\frac {1}{2}+\frac {m}{2}}\right ) \operatorname {Subst}\left (\int \left (\frac {1}{2}+\frac {i x}{2}\right )^{-\frac {1}{2}-\frac {m}{2}} (a-i a x)^{-1-\frac {m}{2}} \, dx,x,\tan (c+d x)\right )}{d \sqrt {a+i a \tan (c+d x)}}\\ &=-\frac {i 2^{\frac {1}{2}-\frac {m}{2}} a (e \cos (c+d x))^m \, _2F_1\left (-\frac {m}{2},\frac {1+m}{2};1-\frac {m}{2};\frac {1}{2} (1-i \tan (c+d x))\right ) (1+i \tan (c+d x))^{\frac {1+m}{2}}}{d m \sqrt {a+i a \tan (c+d x)}}\\ \end {align*}
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Mathematica [A] time = 0.88, size = 106, normalized size = 1.01 \[ \frac {i 2^{-m} \left (1+e^{2 i (c+d x)}\right ) \sqrt {a+i a \tan (c+d x)} \left (e e^{-i (c+d x)} \left (1+e^{2 i (c+d x)}\right )\right )^m \, _2F_1\left (1,\frac {m+2}{2};\frac {3-m}{2};-e^{2 i (c+d x)}\right )}{d (m-1)} \]
Warning: Unable to verify antiderivative.
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fricas [F] time = 0.56, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\sqrt {2} \left (\frac {1}{2} \, {\left (e e^{\left (2 i \, d x + 2 i \, c\right )} + e\right )} e^{\left (-i \, d x - i \, c\right )}\right )^{m} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (i \, d x + i \, c\right )}, x\right ) \]
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 \[ \int \sqrt {i \, a \tan \left (d x + c\right ) + a} \left (e \cos \left (d x + c\right )\right )^{m}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 1.47, size = 0, normalized size = 0.00 \[ \int \left (e \cos \left (d x +c \right )\right )^{m} \sqrt {a +i a \tan \left (d x +c \right )}\, dx \]
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 \[ \int \sqrt {i \, a \tan \left (d x + c\right ) + a} \left (e \cos \left (d x + c\right )\right )^{m}\,{d x} \]
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 \[ \int {\left (e\,\cos \left (c+d\,x\right )\right )}^m\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}} \,d x \]
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 \[ \int \left (e \cos {\left (c + d x \right )}\right )^{m} \sqrt {i a \left (\tan {\left (c + d x \right )} - i\right )}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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