Optimal. Leaf size=107 \[ \frac{i a 2^{\frac{m+1}{2}} (1+i \tan (c+d x))^{\frac{1-m}{2}} (e \sec (c+d x))^m \text{Hypergeometric2F1}\left (\frac{1-m}{2},\frac{m}{2},\frac{m+2}{2},\frac{1}{2} (1-i \tan (c+d x))\right )}{d m \sqrt{a+i a \tan (c+d x)}} \]
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Rubi [A] time = 0.178756, antiderivative size = 107, normalized size of antiderivative = 1., 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 3505
Rule 3523
Rule 70
Rule 69
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*}
Mathematica [A] time = 0.657975, size = 162, normalized size = 1.51 \[ -\frac{i 2^{m+\frac{1}{2}} \sqrt{e^{i d x}} e^{i (c+d x)} \left (\frac{e^{i (c+d x)}}{1+e^{2 i (c+d x)}}\right )^{m-\frac{1}{2}} \sqrt{a+i a \tan (c+d x)} \sec ^{-m-\frac{1}{2}}(c+d x) \text{Hypergeometric2F1}\left (1,1-\frac{m}{2},\frac{m+3}{2},-e^{2 i (c+d x)}\right ) (e \sec (c+d x))^m}{d (m+1) \sqrt{\cos (d x)+i \sin (d x)}} \]
Warning: Unable to verify antiderivative.
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Maple [F] time = 0.403, size = 0, normalized size = 0. \begin{align*} \int \left ( e\sec \left ( dx+c \right ) \right ) ^{m}\sqrt{a+ia\tan \left ( dx+c \right ) }\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{i \, a \tan \left (d x + c\right ) + a} \left (e \sec \left (d x + c\right )\right )^{m}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\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 ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{a \left (i \tan{\left (c + d x \right )} + 1\right )} \left (e \sec{\left (c + d x \right )}\right )^{m}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{i \, a \tan \left (d x + c\right ) + a} \left (e \sec \left (d x + c\right )\right )^{m}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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