Optimal. Leaf size=269 \[ \frac{24 i (a+i a \tan (c+d x))^{n+3} (e \sec (c+d x))^{-n-4}}{a^3 d (4-n) n \left (4-n^2\right )}-\frac{24 i (a+i a \tan (c+d x))^{n+4} (e \sec (c+d x))^{-n-4}}{a^4 d n \left (n^4-20 n^2+64\right )}-\frac{12 i (a+i a \tan (c+d x))^{n+2} (e \sec (c+d x))^{-n-4}}{a^2 d (2-n) (4-n) n}+\frac{4 i (a+i a \tan (c+d x))^{n+1} (e \sec (c+d x))^{-n-4}}{a d \left (n^2-6 n+8\right )}+\frac{i (a+i a \tan (c+d x))^n (e \sec (c+d x))^{-n-4}}{d (4-n)} \]
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Rubi [A] time = 0.406169, antiderivative size = 269, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 2, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.067, Rules used = {3504, 3488} \[ \frac{24 i (a+i a \tan (c+d x))^{n+3} (e \sec (c+d x))^{-n-4}}{a^3 d (4-n) n \left (4-n^2\right )}-\frac{24 i (a+i a \tan (c+d x))^{n+4} (e \sec (c+d x))^{-n-4}}{a^4 d n \left (n^4-20 n^2+64\right )}-\frac{12 i (a+i a \tan (c+d x))^{n+2} (e \sec (c+d x))^{-n-4}}{a^2 d (2-n) (4-n) n}+\frac{4 i (a+i a \tan (c+d x))^{n+1} (e \sec (c+d x))^{-n-4}}{a d \left (n^2-6 n+8\right )}+\frac{i (a+i a \tan (c+d x))^n (e \sec (c+d x))^{-n-4}}{d (4-n)} \]
Antiderivative was successfully verified.
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Rule 3504
Rule 3488
Rubi steps
\begin{align*} \int (e \sec (c+d x))^{-4-n} (a+i a \tan (c+d x))^n \, dx &=\frac{i (e \sec (c+d x))^{-4-n} (a+i a \tan (c+d x))^n}{d (4-n)}+\frac{4 \int (e \sec (c+d x))^{-4-n} (a+i a \tan (c+d x))^{1+n} \, dx}{a (4-n)}\\ &=\frac{i (e \sec (c+d x))^{-4-n} (a+i a \tan (c+d x))^n}{d (4-n)}+\frac{4 i (e \sec (c+d x))^{-4-n} (a+i a \tan (c+d x))^{1+n}}{a d \left (8-6 n+n^2\right )}+\frac{12 \int (e \sec (c+d x))^{-4-n} (a+i a \tan (c+d x))^{2+n} \, dx}{a^2 (2-n) (4-n)}\\ &=\frac{i (e \sec (c+d x))^{-4-n} (a+i a \tan (c+d x))^n}{d (4-n)}+\frac{4 i (e \sec (c+d x))^{-4-n} (a+i a \tan (c+d x))^{1+n}}{a d \left (8-6 n+n^2\right )}-\frac{12 i (e \sec (c+d x))^{-4-n} (a+i a \tan (c+d x))^{2+n}}{a^2 d (2-n) (4-n) n}-\frac{24 \int (e \sec (c+d x))^{-4-n} (a+i a \tan (c+d x))^{3+n} \, dx}{a^3 (2-n) (4-n) n}\\ &=\frac{i (e \sec (c+d x))^{-4-n} (a+i a \tan (c+d x))^n}{d (4-n)}+\frac{4 i (e \sec (c+d x))^{-4-n} (a+i a \tan (c+d x))^{1+n}}{a d \left (8-6 n+n^2\right )}-\frac{12 i (e \sec (c+d x))^{-4-n} (a+i a \tan (c+d x))^{2+n}}{a^2 d (2-n) (4-n) n}+\frac{24 i (e \sec (c+d x))^{-4-n} (a+i a \tan (c+d x))^{3+n}}{a^3 d (2-n) (4-n) n (2+n)}+\frac{24 \int (e \sec (c+d x))^{-4-n} (a+i a \tan (c+d x))^{4+n} \, dx}{a^4 (2-n) (4-n) n (2+n)}\\ &=\frac{i (e \sec (c+d x))^{-4-n} (a+i a \tan (c+d x))^n}{d (4-n)}+\frac{4 i (e \sec (c+d x))^{-4-n} (a+i a \tan (c+d x))^{1+n}}{a d \left (8-6 n+n^2\right )}-\frac{12 i (e \sec (c+d x))^{-4-n} (a+i a \tan (c+d x))^{2+n}}{a^2 d (2-n) (4-n) n}+\frac{24 i (e \sec (c+d x))^{-4-n} (a+i a \tan (c+d x))^{3+n}}{a^3 d (2-n) (4-n) n (2+n)}-\frac{24 i (e \sec (c+d x))^{-4-n} (a+i a \tan (c+d x))^{4+n}}{a^4 d n \left (64-20 n^2+n^4\right )}\\ \end{align*}
Mathematica [A] time = 0.622421, size = 165, normalized size = 0.61 \[ -\frac{i (a+i a \tan (c+d x))^n (e \sec (c+d x))^{-n} \left (-8 i n^3 \sin (2 (c+d x))-4 i n^3 \sin (4 (c+d x))+4 \left (n^2-16\right ) n^2 \cos (2 (c+d x))+\left (n^2-4\right ) n^2 \cos (4 (c+d x))+128 i n \sin (2 (c+d x))+16 i n \sin (4 (c+d x))+3 n^4-60 n^2+192\right )}{8 d e^4 (n-4) (n-2) n (n+2) (n+4)} \]
Antiderivative was successfully verified.
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Maple [C] time = 1.361, size = 5866, normalized size = 21.8 \begin{align*} \text{output too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 2.0563, size = 583, normalized size = 2.17 \begin{align*} \frac{{\left (-i \, a^{n} n^{4} + 4 i \, a^{n} n^{3} + 4 i \, a^{n} n^{2} - 16 i \, a^{n} n\right )} \cos \left ({\left (d x + c\right )}{\left (n + 4\right )}\right ) +{\left (-4 i \, a^{n} n^{4} + 8 i \, a^{n} n^{3} + 64 i \, a^{n} n^{2} - 128 i \, a^{n} n\right )} \cos \left ({\left (d x + c\right )}{\left (n + 2\right )}\right ) +{\left (-4 i \, a^{n} n^{4} - 8 i \, a^{n} n^{3} + 64 i \, a^{n} n^{2} + 128 i \, a^{n} n\right )} \cos \left ({\left (d x + c\right )}{\left (n - 2\right )}\right ) +{\left (-i \, a^{n} n^{4} - 4 i \, a^{n} n^{3} + 4 i \, a^{n} n^{2} + 16 i \, a^{n} n\right )} \cos \left ({\left (d x + c\right )}{\left (n - 4\right )}\right ) +{\left (-6 i \, a^{n} n^{4} + 120 i \, a^{n} n^{2} - 384 i \, a^{n}\right )} \cos \left ({\left (d x + c\right )} n\right ) +{\left (a^{n} n^{4} - 4 \, a^{n} n^{3} - 4 \, a^{n} n^{2} + 16 \, a^{n} n\right )} \sin \left ({\left (d x + c\right )}{\left (n + 4\right )}\right ) + 4 \,{\left (a^{n} n^{4} - 2 \, a^{n} n^{3} - 16 \, a^{n} n^{2} + 32 \, a^{n} n\right )} \sin \left ({\left (d x + c\right )}{\left (n + 2\right )}\right ) + 4 \,{\left (a^{n} n^{4} + 2 \, a^{n} n^{3} - 16 \, a^{n} n^{2} - 32 \, a^{n} n\right )} \sin \left ({\left (d x + c\right )}{\left (n - 2\right )}\right ) +{\left (a^{n} n^{4} + 4 \, a^{n} n^{3} - 4 \, a^{n} n^{2} - 16 \, a^{n} n\right )} \sin \left ({\left (d x + c\right )}{\left (n - 4\right )}\right ) + 6 \,{\left (a^{n} n^{4} - 20 \, a^{n} n^{2} + 64 \, a^{n}\right )} \sin \left ({\left (d x + c\right )} n\right )}{16 \,{\left (e^{n + 4} n^{5} - 20 \, e^{n + 4} n^{3} + 64 \, e^{n + 4} n\right )} d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 2.20184, size = 844, normalized size = 3.14 \begin{align*} \frac{{\left (-i \, n^{4} - 4 i \, n^{3} + 4 i \, n^{2} +{\left (-i \, n^{4} + 4 i \, n^{3} + 4 i \, n^{2} - 16 i \, n\right )} e^{\left (8 i \, d x + 8 i \, c\right )} +{\left (-4 i \, n^{4} + 8 i \, n^{3} + 64 i \, n^{2} - 128 i \, n\right )} e^{\left (6 i \, d x + 6 i \, c\right )} +{\left (-6 i \, n^{4} + 120 i \, n^{2} - 384 i\right )} e^{\left (4 i \, d x + 4 i \, c\right )} +{\left (-4 i \, n^{4} - 8 i \, n^{3} + 64 i \, n^{2} + 128 i \, n\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + 16 i \, n\right )} \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} \left (\frac{2 \, e e^{\left (i \, d x + i \, c\right )}}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{-n - 4}}{d n^{5} - 20 \, d n^{3} + 64 \, d n +{\left (d n^{5} - 20 \, d n^{3} + 64 \, d n\right )} e^{\left (8 i \, d x + 8 i \, c\right )} + 4 \,{\left (d n^{5} - 20 \, d n^{3} + 64 \, d n\right )} e^{\left (6 i \, d x + 6 i \, c\right )} + 6 \,{\left (d n^{5} - 20 \, d n^{3} + 64 \, d n\right )} e^{\left (4 i \, d x + 4 i \, c\right )} + 4 \,{\left (d n^{5} - 20 \, d n^{3} + 64 \, d n\right )} e^{\left (2 i \, d x + 2 i \, c\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 (e \sec \left (d x + c\right )\right )^{-n - 4}{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{n}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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