Optimal. Leaf size=109 \[ \frac{i 2^{\frac{m-5}{2}} (1+i \tan (c+d x))^{\frac{1-m}{2}} (e \sec (c+d x))^m \text{Hypergeometric2F1}\left (\frac{7-m}{2},\frac{m}{2},\frac{m+2}{2},\frac{1}{2} (1-i \tan (c+d x))\right )}{a^2 d m \sqrt{a+i a \tan (c+d x)}} \]
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Rubi [A] time = 0.210445, 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} \[ \frac{i 2^{\frac{m-5}{2}} (1+i \tan (c+d x))^{\frac{1-m}{2}} (e \sec (c+d x))^m \text{Hypergeometric2F1}\left (\frac{7-m}{2},\frac{m}{2},\frac{m+2}{2},\frac{1}{2} (1-i \tan (c+d x))\right )}{a^2 d m \sqrt{a+i a \tan (c+d x)}} \]
Antiderivative was successfully verified.
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Rule 3505
Rule 3523
Rule 70
Rule 69
Rubi steps
\begin{align*} \int \frac{(e \sec (c+d x))^m}{(a+i a \tan (c+d x))^{5/2}} \, 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{5}{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{7}{2}+\frac{m}{2}} \, dx,x,\tan (c+d x)\right )}{d}\\ &=\frac{\left (2^{-\frac{7}{2}+\frac{m}{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{7}{2}+\frac{m}{2}} (a-i a x)^{-1+\frac{m}{2}} \, dx,x,\tan (c+d x)\right )}{a d \sqrt{a+i a \tan (c+d x)}}\\ &=\frac{i 2^{\frac{1}{2} (-5+m)} \, _2F_1\left (\frac{7-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}}}{a^2 d m \sqrt{a+i a \tan (c+d x)}}\\ \end{align*}
Mathematica [A] time = 3.14941, size = 178, normalized size = 1.63 \[ -\frac{i 2^{m-\frac{5}{2}} \sqrt{e^{i d x}} e^{-3 i (c+2 d x)} \left (1+e^{2 i (c+d x)}\right )^4 \left (\frac{e^{i (c+d x)}}{1+e^{2 i (c+d x)}}\right )^{m+\frac{1}{2}} (\cos (d x)+i \sin (d x))^{5/2} \sec ^{\frac{5}{2}-m}(c+d x) \text{Hypergeometric2F1}\left (1,1-\frac{m}{2},\frac{m-3}{2},-e^{2 i (c+d x)}\right ) (e \sec (c+d x))^m}{d (m-5) (a+i a \tan (c+d x))^{5/2}} \]
Warning: Unable to verify antiderivative.
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Maple [F] time = 0.398, 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{5}{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 \frac{\left (e \sec \left (d x + c\right )\right )^{m}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac{5}{2}}}\,{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{\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}}{\left (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\right )} e^{\left (-5 i \, d x - 5 i \, c\right )}}{8 \, a^{3}}, 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 \frac{\left (e \sec \left (d x + c\right )\right )^{m}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac{5}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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