Optimal. Leaf size=109 \[ \frac {i 2^{\frac {1}{2} (-5+m)} \, _2F_1\left (\frac {7-m}{2},\frac {m}{2};\frac {2+m}{2};\frac {1}{2} (1-i \tan (c+d x))\right ) (e \sec (c+d x))^m (1+i \tan (c+d x))^{\frac {1-m}{2}}}{a^2 d m \sqrt {a+i a \tan (c+d x)}} \]
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Rubi [A]
time = 0.16, antiderivative size = 109, 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 = {3586, 3604, 72,
71} \begin {gather*} \frac {i 2^{\frac {m-5}{2}} (1+i \tan (c+d x))^{\frac {1-m}{2}} (e \sec (c+d x))^m \, _2F_1\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)}} \end {gather*}
Antiderivative was successfully verified.
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Rule 71
Rule 72
Rule 3586
Rule 3604
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 ) \text {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 ) \text {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*}
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Mathematica [A]
time = 3.64, size = 178, normalized size = 1.63 \begin {gather*} -\frac {i 2^{-\frac {5}{2}+m} e^{-3 i (c+2 d x)} \sqrt {e^{i d x}} \left (\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}\right )^{\frac {1}{2}+m} \left (1+e^{2 i (c+d x)}\right )^4 \, _2F_1\left (1,1-\frac {m}{2};\frac {1}{2} (-3+m);-e^{2 i (c+d x)}\right ) \sec ^{\frac {5}{2}-m}(c+d x) (e \sec (c+d x))^m (\cos (d x)+i \sin (d x))^{5/2}}{d (-5+m) (a+i a \tan (c+d x))^{5/2}} \end {gather*}
Warning: Unable to verify antiderivative.
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Maple [F]
time = 0.88, size = 0, normalized size = 0.00 \[\int \frac {\left (e \sec \left (d x +c \right )\right )^{m}}{\left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}}}\, dx\]
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 \begin {gather*} \text {Failed to integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}
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 \begin {gather*} \int \frac {\left (e \sec {\left (c + d x \right )}\right )^{m}}{\left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{\frac {5}{2}}}\, dx \end {gather*}
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 \begin {gather*} \text {could not integrate} \end {gather*}
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 \begin {gather*} \int \frac {{\left (\frac {e}{\cos \left (c+d\,x\right )}\right )}^m}{{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2}} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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