Optimal. Leaf size=133 \[ -\frac {i \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{4 \sqrt {2} a^{5/2} d}+\frac {i}{4 a^2 d \sqrt {a+i a \tan (c+d x)}}+\frac {i}{6 a d (a+i a \tan (c+d x))^{3/2}}+\frac {i}{5 d (a+i a \tan (c+d x))^{5/2}} \]
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Rubi [A] time = 0.08, antiderivative size = 133, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {3479, 3480, 206} \[ \frac {i}{4 a^2 d \sqrt {a+i a \tan (c+d x)}}-\frac {i \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{4 \sqrt {2} a^{5/2} d}+\frac {i}{6 a d (a+i a \tan (c+d x))^{3/2}}+\frac {i}{5 d (a+i a \tan (c+d x))^{5/2}} \]
Antiderivative was successfully verified.
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Rule 206
Rule 3479
Rule 3480
Rubi steps
\begin {align*} \int \frac {1}{(a+i a \tan (c+d x))^{5/2}} \, dx &=\frac {i}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {\int \frac {1}{(a+i a \tan (c+d x))^{3/2}} \, dx}{2 a}\\ &=\frac {i}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {i}{6 a d (a+i a \tan (c+d x))^{3/2}}+\frac {\int \frac {1}{\sqrt {a+i a \tan (c+d x)}} \, dx}{4 a^2}\\ &=\frac {i}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {i}{6 a d (a+i a \tan (c+d x))^{3/2}}+\frac {i}{4 a^2 d \sqrt {a+i a \tan (c+d x)}}+\frac {\int \sqrt {a+i a \tan (c+d x)} \, dx}{8 a^3}\\ &=\frac {i}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {i}{6 a d (a+i a \tan (c+d x))^{3/2}}+\frac {i}{4 a^2 d \sqrt {a+i a \tan (c+d x)}}-\frac {i \operatorname {Subst}\left (\int \frac {1}{2 a-x^2} \, dx,x,\sqrt {a+i a \tan (c+d x)}\right )}{4 a^2 d}\\ &=-\frac {i \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{4 \sqrt {2} a^{5/2} d}+\frac {i}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {i}{6 a d (a+i a \tan (c+d x))^{3/2}}+\frac {i}{4 a^2 d \sqrt {a+i a \tan (c+d x)}}\\ \end {align*}
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Mathematica [A] time = 0.76, size = 137, normalized size = 1.03 \[ \frac {i e^{-6 i (c+d x)} \left (1+e^{2 i (c+d x)}\right )^{3/2} \sec ^2(c+d x) \left (\sqrt {1+e^{2 i (c+d x)}} \left (11 e^{2 i (c+d x)}+23 e^{4 i (c+d x)}+3\right )-15 e^{5 i (c+d x)} \sinh ^{-1}\left (e^{i (c+d x)}\right )\right )}{240 a^2 d \sqrt {a+i a \tan (c+d x)}} \]
Antiderivative was successfully verified.
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fricas [B] time = 0.45, size = 283, normalized size = 2.13 \[ \frac {{\left (-15 i \, \sqrt {\frac {1}{2}} a^{3} d \sqrt {\frac {1}{a^{5} d^{2}}} e^{\left (5 i \, d x + 5 i \, c\right )} \log \left (4 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{3} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a^{5} d^{2}}} + a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) + 15 i \, \sqrt {\frac {1}{2}} a^{3} d \sqrt {\frac {1}{a^{5} d^{2}}} e^{\left (5 i \, d x + 5 i \, c\right )} \log \left (-4 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{3} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a^{5} d^{2}}} - a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) + \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (23 i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 34 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 14 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 3 i\right )}\right )} e^{\left (-5 i \, d x - 5 i \, c\right )}}{120 \, a^{3} d} \]
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 \[ \int \frac {1}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.12, size = 97, normalized size = 0.73 \[ \frac {2 i a \left (\frac {1}{8 a^{3} \sqrt {a +i a \tan \left (d x +c \right )}}+\frac {1}{12 a^{2} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}+\frac {1}{10 a \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}}}-\frac {\sqrt {2}\, \arctanh \left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{16 a^{\frac {7}{2}}}\right )}{d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.60, size = 119, normalized size = 0.89 \[ \frac {i \, {\left (\frac {15 \, \sqrt {2} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {i \, a \tan \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {i \, a \tan \left (d x + c\right ) + a}}\right )}{a^{\frac {3}{2}}} + \frac {4 \, {\left (15 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} + 10 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )} a + 12 \, a^{2}\right )}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}} a}\right )}}{240 \, a d} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 3.81, size = 106, normalized size = 0.80 \[ \frac {\frac {1{}\mathrm {i}}{5\,d}+\frac {{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^2\,1{}\mathrm {i}}{4\,a^2\,d}+\frac {\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{6\,a\,d}}{{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2}}+\frac {\sqrt {2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {-a}}\right )\,1{}\mathrm {i}}{8\,{\left (-a\right )}^{5/2}\,d} \]
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 \[ \int \frac {1}{\left (i a \tan {\left (c + d x \right )} + a\right )^{\frac {5}{2}}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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