Optimal. Leaf size=107 \[ \frac {i a 2^{\frac {m+1}{2}} (1+i \tan (c+d x))^{\frac {1-m}{2}} (e \sec (c+d x))^m \, _2F_1\left (\frac {1-m}{2},\frac {m}{2};\frac {m+2}{2};\frac {1}{2} (1-i \tan (c+d x))\right )}{d m \sqrt {a+i a \tan (c+d x)}} \]
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Rubi [A] time = 0.18, antiderivative size = 107, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3505, 3523, 70, 69} \[ \frac {i a 2^{\frac {m+1}{2}} (1+i \tan (c+d x))^{\frac {1-m}{2}} (e \sec (c+d x))^m \text {Hypergeometric2F1}\left (\frac {1-m}{2},\frac {m}{2},\frac {m+2}{2},\frac {1}{2} (1-i \tan (c+d x))\right )}{d m \sqrt {a+i a \tan (c+d x)}} \]
Antiderivative was successfully verified.
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Rule 69
Rule 70
Rule 3505
Rule 3523
Rubi steps
\begin {align*} \int (e \sec (c+d x))^m \sqrt {a+i a \tan (c+d x)} \, 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 {1}{2}+\frac {m}{2}} \, 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 ) \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 \sec (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+m}{2}} a \, _2F_1\left (\frac {1-m}{2},\frac {m}{2};\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 {1-m}{2}}}{d m \sqrt {a+i a \tan (c+d x)}}\\ \end {align*}
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Mathematica [A] time = 0.73, size = 162, normalized size = 1.51 \[ -\frac {i 2^{m+\frac {1}{2}} \sqrt {e^{i d x}} e^{i (c+d x)} \left (\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}\right )^{m-\frac {1}{2}} \sqrt {a+i a \tan (c+d x)} \, _2F_1\left (1,1-\frac {m}{2};\frac {m+3}{2};-e^{2 i (c+d x)}\right ) \sec ^{-m-\frac {1}{2}}(c+d x) (e \sec (c+d x))^m}{d (m+1) \sqrt {\cos (d x)+i \sin (d x)}} \]
Warning: Unable to verify antiderivative.
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fricas [F] time = 0.57, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\sqrt {2} \left (\frac {2 \, e e^{\left (i \, d x + i \, c\right )}}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{m} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (i \, d x + i \, c\right )}, x\right ) \]
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 \[ \int \sqrt {i \, a \tan \left (d x + c\right ) + a} \left (e \sec \left (d x + c\right )\right )^{m}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 1.51, size = 0, normalized size = 0.00 \[ \int \left (e \sec \left (d x +c \right )\right )^{m} \sqrt {a +i a \tan \left (d x +c \right )}\, dx \]
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 \[ \int \sqrt {i \, a \tan \left (d x + c\right ) + a} \left (e \sec \left (d x + c\right )\right )^{m}\,{d x} \]
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 \[ \int {\left (\frac {e}{\cos \left (c+d\,x\right )}\right )}^m\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}} \,d x \]
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 \[ \int \left (e \sec {\left (c + d x \right )}\right )^{m} \sqrt {i a \left (\tan {\left (c + d x \right )} - i\right )}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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