Optimal. Leaf size=94 \[ \frac {i a^2 2^{n+\frac {5}{2}} \sec ^5(c+d x) (1+i \tan (c+d x))^{-n-\frac {1}{2}} (a+i a \tan (c+d x))^{n-2} \, _2F_1\left (\frac {5}{2},-n-\frac {3}{2};\frac {7}{2};\frac {1}{2} (1-i \tan (c+d x))\right )}{5 d} \]
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Rubi [A] time = 0.19, 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 = {3505, 3523, 70, 69} \[ \frac {i a^2 2^{n+\frac {5}{2}} \sec ^5(c+d x) (1+i \tan (c+d x))^{-n-\frac {1}{2}} (a+i a \tan (c+d x))^{n-2} \text {Hypergeometric2F1}\left (\frac {5}{2},-n-\frac {3}{2},\frac {7}{2},\frac {1}{2} (1-i \tan (c+d x))\right )}{5 d} \]
Antiderivative was successfully verified.
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Rule 69
Rule 70
Rule 3505
Rule 3523
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 ) \operatorname {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 ) \operatorname {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 = 14.01, size = 149, normalized size = 1.59 \[ -\frac {i 2^{n+5} e^{5 i (c+d x)} \left (e^{i d x}\right )^n \left (\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}\right )^n \, _2F_1\left (-\frac {3}{2},1;n+\frac {7}{2};-e^{2 i (c+d x)}\right ) \sec ^{-n}(c+d x) (\cos (d x)+i \sin (d x))^{-n} (a+i a \tan (c+d x))^n}{d (2 n+5) \left (1+e^{2 i (c+d x)}\right )^4} \]
Warning: Unable to verify antiderivative.
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fricas [F] time = 0.47, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {32 \, \left (\frac {2 \, a e^{\left (2 i \, d x + 2 i \, c\right )}}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{n} e^{\left (5 i \, d x + 5 i \, c\right )}}{e^{\left (10 i \, d x + 10 i \, c\right )} + 5 \, e^{\left (8 i \, d x + 8 i \, c\right )} + 10 \, e^{\left (6 i \, d x + 6 i \, c\right )} + 10 \, e^{\left (4 i \, d x + 4 i \, c\right )} + 5 \, e^{\left (2 i \, d x + 2 i \, c\right )} + 1}, x\right ) \]
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 \[ \int {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{n} \sec \left (d x + c\right )^{5}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 1.05, 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 \[ \int {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{n} \sec \left (d x + c\right )^{5}\,{d x} \]
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 \[ \int \frac {{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^n}{{\cos \left (c+d\,x\right )}^5} \,d x \]
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 \[ \int \left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{n} \sec ^{5}{\left (c + d x \right )}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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