Optimal. Leaf size=109 \[ \frac{i a^3 2^{\frac{m+7}{2}} \sqrt{a+i a \tan (c+d x)} (1+i \tan (c+d x))^{\frac{1}{2} (-m-1)} (e \sec (c+d x))^m \text{Hypergeometric2F1}\left (\frac{1}{2} (-m-5),\frac{m}{2},\frac{m+2}{2},\frac{1}{2} (1-i \tan (c+d x))\right )}{d m} \]
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Rubi [A] time = 0.206088, antiderivative size = 109, 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} \[ \frac{i a^3 2^{\frac{m+7}{2}} \sqrt{a+i a \tan (c+d x)} (1+i \tan (c+d x))^{\frac{1}{2} (-m-1)} (e \sec (c+d x))^m \text{Hypergeometric2F1}\left (\frac{1}{2} (-m-5),\frac{m}{2},\frac{m+2}{2},\frac{1}{2} (1-i \tan (c+d x))\right )}{d m} \]
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 (a+i a \tan (c+d x))^{7/2} \, 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{7}{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{5}{2}+\frac{m}{2}} \, dx,x,\tan (c+d x)\right )}{d}\\ &=\frac{\left (2^{\frac{5}{2}+\frac{m}{2}} a^4 (e \sec (c+d x))^m (a-i a \tan (c+d x))^{-m/2} \sqrt{a+i a \tan (c+d x)} \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{5}{2}+\frac{m}{2}} (a-i a x)^{-1+\frac{m}{2}} \, dx,x,\tan (c+d x)\right )}{d}\\ &=\frac{i 2^{\frac{7+m}{2}} a^3 \, _2F_1\left (\frac{1}{2} (-5-m),\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}{2} (-1-m)} \sqrt{a+i a \tan (c+d x)}}{d m}\\ \end{align*}
Mathematica [A] time = 2.40871, size = 178, normalized size = 1.63 \[ -\frac{i 2^{m+\frac{7}{2}} \sqrt{e^{i d x}} e^{3 i (c+2 d x)} \left (\frac{e^{i (c+d x)}}{1+e^{2 i (c+d x)}}\right )^{m+\frac{1}{2}} (a+i a \tan (c+d x))^{7/2} \sec ^{-m-\frac{7}{2}}(c+d x) \text{Hypergeometric2F1}\left (1,1-\frac{m}{2},\frac{m+9}{2},-e^{2 i (c+d x)}\right ) (e \sec (c+d x))^m}{d (m+7) \left (1+e^{2 i (c+d x)}\right )^2 (\cos (d x)+i \sin (d x))^{7/2}} \]
Warning: Unable to verify antiderivative.
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Maple [F] time = 0.335, size = 0, normalized size = 0. \begin{align*} \int \left ( e\sec \left ( dx+c \right ) \right ) ^{m} \left ( a+ia\tan \left ( dx+c \right ) \right ) ^{{\frac{7}{2}}}\, 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{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac{7}{2}} \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 (\frac{8 \, \sqrt{2} a^{3} \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 (7 i \, d x + 7 i \, c\right )}}{e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, e^{\left (2 i \, d x + 2 i \, c\right )} + 1}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \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{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac{7}{2}} \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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