Optimal. Leaf size=94 \[ \frac {i 2^{\frac {5}{2}+n} a^2 \, _2F_1\left (\frac {5}{2},-\frac {3}{2}-n;\frac {7}{2};\frac {1}{2} (1-i \tan (c+d x))\right ) \sec ^5(c+d x) (1+i \tan (c+d x))^{-\frac {1}{2}-n} (a+i a \tan (c+d x))^{-2+n}}{5 d} \]
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Rubi [A]
time = 0.14, antiderivative size = 94, normalized size of antiderivative = 1.00, number of steps
used = 4, number of rules used = 4, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {3586, 3604, 72,
71} \begin {gather*} \frac {i a^2 2^{n+\frac {5}{2}} \sec ^5(c+d x) (1+i \tan (c+d x))^{-n-\frac {1}{2}} (a+i a \tan (c+d x))^{n-2} \, _2F_1\left (\frac {5}{2},-n-\frac {3}{2};\frac {7}{2};\frac {1}{2} (1-i \tan (c+d x))\right )}{5 d} \end {gather*}
Antiderivative was successfully verified.
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Rule 71
Rule 72
Rule 3586
Rule 3604
Rubi steps
\begin {align*} \int \sec ^5(c+d x) (a+i a \tan (c+d x))^n \, dx &=\frac {\sec ^5(c+d x) \int (a-i a \tan (c+d x))^{5/2} (a+i a \tan (c+d x))^{\frac {5}{2}+n} \, dx}{(a-i a \tan (c+d x))^{5/2} (a+i a \tan (c+d x))^{5/2}}\\ &=\frac {\left (a^2 \sec ^5(c+d x)\right ) \text {Subst}\left (\int (a-i a x)^{3/2} (a+i a x)^{\frac {3}{2}+n} \, dx,x,\tan (c+d x)\right )}{d (a-i a \tan (c+d x))^{5/2} (a+i a \tan (c+d x))^{5/2}}\\ &=\frac {\left (2^{\frac {3}{2}+n} a^3 \sec ^5(c+d x) (a+i a \tan (c+d x))^{-2+n} \left (\frac {a+i a \tan (c+d x)}{a}\right )^{-\frac {1}{2}-n}\right ) \text {Subst}\left (\int \left (\frac {1}{2}+\frac {i x}{2}\right )^{\frac {3}{2}+n} (a-i a x)^{3/2} \, dx,x,\tan (c+d x)\right )}{d (a-i a \tan (c+d x))^{5/2}}\\ &=\frac {i 2^{\frac {5}{2}+n} a^2 \, _2F_1\left (\frac {5}{2},-\frac {3}{2}-n;\frac {7}{2};\frac {1}{2} (1-i \tan (c+d x))\right ) \sec ^5(c+d x) (1+i \tan (c+d x))^{-\frac {1}{2}-n} (a+i a \tan (c+d x))^{-2+n}}{5 d}\\ \end {align*}
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Mathematica [A]
time = 14.34, size = 153, normalized size = 1.63 \begin {gather*} -\frac {i 2^{5+n} e^{5 i (c+d x)} \left (e^{i d x}\right )^n \left (\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}\right )^n \left (1+e^{2 i (c+d x)}\right )^n \, _2F_1\left (\frac {5}{2}+n,5+n;\frac {7}{2}+n;-e^{2 i (c+d x)}\right ) \sec ^{-n}(c+d x) (\cos (d x)+i \sin (d x))^{-n} (a+i a \tan (c+d x))^n}{d (5+2 n)} \end {gather*}
Antiderivative was successfully verified.
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Maple [F]
time = 0.34, size = 0, normalized size = 0.00 \[\int \left (\sec ^{5}\left (d x +c \right )\right ) \left (a +i a \tan \left (d x +c \right )\right )^{n}\, dx\]
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 \left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{n} \sec ^{5}{\left (c + d x \right )}\, 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 (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^n}{{\cos \left (c+d\,x\right )}^5} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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