Optimal. Leaf size=111 \[ \frac {6 i a^3}{5 d e^2 \sqrt {e \sec (c+d x)}}-\frac {6 a^3 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d e^2 \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}-\frac {4 i a (a+i a \tan (c+d x))^2}{5 d (e \sec (c+d x))^{5/2}} \]
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Rubi [A] time = 0.10, antiderivative size = 111, 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 = {3496, 3486, 3771, 2639} \[ \frac {6 i a^3}{5 d e^2 \sqrt {e \sec (c+d x)}}-\frac {6 a^3 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d e^2 \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}-\frac {4 i a (a+i a \tan (c+d x))^2}{5 d (e \sec (c+d x))^{5/2}} \]
Antiderivative was successfully verified.
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Rule 2639
Rule 3486
Rule 3496
Rule 3771
Rubi steps
\begin {align*} \int \frac {(a+i a \tan (c+d x))^3}{(e \sec (c+d x))^{5/2}} \, dx &=-\frac {4 i a (a+i a \tan (c+d x))^2}{5 d (e \sec (c+d x))^{5/2}}-\frac {\left (3 a^2\right ) \int \frac {a+i a \tan (c+d x)}{\sqrt {e \sec (c+d x)}} \, dx}{5 e^2}\\ &=\frac {6 i a^3}{5 d e^2 \sqrt {e \sec (c+d x)}}-\frac {4 i a (a+i a \tan (c+d x))^2}{5 d (e \sec (c+d x))^{5/2}}-\frac {\left (3 a^3\right ) \int \frac {1}{\sqrt {e \sec (c+d x)}} \, dx}{5 e^2}\\ &=\frac {6 i a^3}{5 d e^2 \sqrt {e \sec (c+d x)}}-\frac {4 i a (a+i a \tan (c+d x))^2}{5 d (e \sec (c+d x))^{5/2}}-\frac {\left (3 a^3\right ) \int \sqrt {\cos (c+d x)} \, dx}{5 e^2 \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}\\ &=\frac {6 i a^3}{5 d e^2 \sqrt {e \sec (c+d x)}}-\frac {6 a^3 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d e^2 \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}-\frac {4 i a (a+i a \tan (c+d x))^2}{5 d (e \sec (c+d x))^{5/2}}\\ \end {align*}
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Mathematica [C] time = 1.45, size = 108, normalized size = 0.97 \[ -\frac {4 i a^3 e^{2 i (c+d x)} \left (-\sqrt {1+e^{2 i (c+d x)}} \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};-e^{2 i (c+d x)}\right )+e^{2 i (c+d x)}+1\right )}{5 d e^2 \left (1+e^{2 i (c+d x)}\right ) \sqrt {e \sec (c+d x)}} \]
Antiderivative was successfully verified.
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fricas [F] time = 0.67, size = 0, normalized size = 0.00 \[ \frac {\sqrt {2} {\left (-2 i \, a^{3} e^{\left (4 i \, d x + 4 i \, c\right )} + 2 i \, a^{3} e^{\left (3 i \, d x + 3 i \, c\right )} + 4 i \, a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + 2 i \, a^{3} e^{\left (i \, d x + i \, c\right )} + 6 i \, a^{3}\right )} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )} + 5 \, {\left (d e^{3} e^{\left (i \, d x + i \, c\right )} - d e^{3}\right )} {\rm integral}\left (\frac {\sqrt {2} {\left (3 i \, a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + 6 i \, a^{3} e^{\left (i \, d x + i \, c\right )} + 3 i \, a^{3}\right )} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )}}{5 \, {\left (d e^{3} e^{\left (3 i \, d x + 3 i \, c\right )} - 2 \, d e^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + d e^{3} e^{\left (i \, d x + i \, c\right )}\right )}}, x\right )}{5 \, {\left (d e^{3} e^{\left (i \, d x + i \, c\right )} - d e^{3}\right )}} \]
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 \[ \int \frac {{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3}}{\left (e \sec \left (d x + c\right )\right )^{\frac {5}{2}}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.96, size = 1086, normalized size = 9.78 \[ \text {result too large to display} \]
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 \[ \int \frac {{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3}}{\left (e \sec \left (d x + c\right )\right )^{\frac {5}{2}}}\,{d x} \]
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 \[ \int \frac {{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^3}{{\left (\frac {e}{\cos \left (c+d\,x\right )}\right )}^{5/2}} \,d x \]
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 \[ - i a^{3} \left (\int \frac {i}{\left (e \sec {\left (c + d x \right )}\right )^{\frac {5}{2}}}\, dx + \int \left (- \frac {3 \tan {\left (c + d x \right )}}{\left (e \sec {\left (c + d x \right )}\right )^{\frac {5}{2}}}\right )\, dx + \int \frac {\tan ^{3}{\left (c + d x \right )}}{\left (e \sec {\left (c + d x \right )}\right )^{\frac {5}{2}}}\, dx + \int \left (- \frac {3 i \tan ^{2}{\left (c + d x \right )}}{\left (e \sec {\left (c + d x \right )}\right )^{\frac {5}{2}}}\right )\, dx\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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