Optimal. Leaf size=55 \[ \frac {i (a+i a \tan (c+d x))^5}{5 a^3 d}-\frac {i (a+i a \tan (c+d x))^4}{2 a^2 d} \]
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Rubi [A] time = 0.04, antiderivative size = 55, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {3487, 43} \[ \frac {i (a+i a \tan (c+d x))^5}{5 a^3 d}-\frac {i (a+i a \tan (c+d x))^4}{2 a^2 d} \]
Antiderivative was successfully verified.
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Rule 43
Rule 3487
Rubi steps
\begin {align*} \int \sec ^4(c+d x) (a+i a \tan (c+d x))^2 \, dx &=-\frac {i \operatorname {Subst}\left (\int (a-x) (a+x)^3 \, dx,x,i a \tan (c+d x)\right )}{a^3 d}\\ &=-\frac {i \operatorname {Subst}\left (\int \left (2 a (a+x)^3-(a+x)^4\right ) \, dx,x,i a \tan (c+d x)\right )}{a^3 d}\\ &=-\frac {i (a+i a \tan (c+d x))^4}{2 a^2 d}+\frac {i (a+i a \tan (c+d x))^5}{5 a^3 d}\\ \end {align*}
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Mathematica [A] time = 0.50, size = 77, normalized size = 1.40 \[ \frac {a^2 \sec (c) \sec ^5(c+d x) (-5 \sin (2 c+d x)+5 \sin (2 c+3 d x)+\sin (4 c+5 d x)+5 i \cos (2 c+d x)+5 \sin (d x)+5 i \cos (d x))}{20 d} \]
Antiderivative was successfully verified.
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fricas [B] time = 0.51, size = 113, normalized size = 2.05 \[ \frac {80 i \, a^{2} e^{\left (6 i \, d x + 6 i \, c\right )} + 80 i \, a^{2} e^{\left (4 i \, d x + 4 i \, c\right )} + 40 i \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + 8 i \, a^{2}}{5 \, {\left (d e^{\left (10 i \, d x + 10 i \, c\right )} + 5 \, d e^{\left (8 i \, d x + 8 i \, c\right )} + 10 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 10 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 5 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.95, size = 56, normalized size = 1.02 \[ -\frac {2 \, a^{2} \tan \left (d x + c\right )^{5} - 5 i \, a^{2} \tan \left (d x + c\right )^{4} - 10 i \, a^{2} \tan \left (d x + c\right )^{2} - 10 \, a^{2} \tan \left (d x + c\right )}{10 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.42, size = 85, normalized size = 1.55 \[ \frac {-a^{2} \left (\frac {\sin ^{3}\left (d x +c \right )}{5 \cos \left (d x +c \right )^{5}}+\frac {2 \left (\sin ^{3}\left (d x +c \right )\right )}{15 \cos \left (d x +c \right )^{3}}\right )+\frac {i a^{2}}{2 \cos \left (d x +c \right )^{4}}-a^{2} \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )}{d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.41, size = 56, normalized size = 1.02 \[ -\frac {6 \, a^{2} \tan \left (d x + c\right )^{5} - 15 i \, a^{2} \tan \left (d x + c\right )^{4} - 30 i \, a^{2} \tan \left (d x + c\right )^{2} - 30 \, a^{2} \tan \left (d x + c\right )}{30 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 3.22, size = 56, normalized size = 1.02 \[ \frac {-\frac {a^2\,{\mathrm {tan}\left (c+d\,x\right )}^5}{5}+\frac {a^2\,{\mathrm {tan}\left (c+d\,x\right )}^4\,1{}\mathrm {i}}{2}+a^2\,{\mathrm {tan}\left (c+d\,x\right )}^2\,1{}\mathrm {i}+a^2\,\mathrm {tan}\left (c+d\,x\right )}{d} \]
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 \[ - a^{2} \left (\int \tan ^{2}{\left (c + d x \right )} \sec ^{4}{\left (c + d x \right )}\, dx + \int \left (- 2 i \tan {\left (c + d x \right )} \sec ^{4}{\left (c + d x \right )}\right )\, dx + \int \left (- \sec ^{4}{\left (c + d x \right )}\right )\, dx\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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