Optimal. Leaf size=257 \[ \frac {5 (-1)^{3/4} \text {ArcTan}\left (\frac {(-1)^{3/4} \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{a^{5/2} d}-\frac {\left (\frac {1}{8}+\frac {i}{8}\right ) \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{a^{5/2} d}-\frac {\tan ^{\frac {7}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {19 i \tan ^{\frac {5}{2}}(c+d x)}{30 a d (a+i a \tan (c+d x))^{3/2}}+\frac {41 \tan ^{\frac {3}{2}}(c+d x)}{12 a^2 d \sqrt {a+i a \tan (c+d x)}}+\frac {21 i \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{4 a^3 d} \]
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Rubi [A]
time = 0.56, antiderivative size = 257, normalized size of antiderivative = 1.00, number of steps
used = 11, number of rules used = 10, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {3639, 3676,
3678, 3682, 3625, 211, 3680, 65, 223, 209} \begin {gather*} \frac {5 (-1)^{3/4} \text {ArcTan}\left (\frac {(-1)^{3/4} \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{a^{5/2} d}-\frac {\left (\frac {1}{8}+\frac {i}{8}\right ) \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{a^{5/2} d}+\frac {21 i \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{4 a^3 d}+\frac {41 \tan ^{\frac {3}{2}}(c+d x)}{12 a^2 d \sqrt {a+i a \tan (c+d x)}}-\frac {\tan ^{\frac {7}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {19 i \tan ^{\frac {5}{2}}(c+d x)}{30 a d (a+i a \tan (c+d x))^{3/2}} \end {gather*}
Antiderivative was successfully verified.
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Rule 65
Rule 209
Rule 211
Rule 223
Rule 3625
Rule 3639
Rule 3676
Rule 3678
Rule 3680
Rule 3682
Rubi steps
\begin {align*} \int \frac {\tan ^{\frac {9}{2}}(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx &=-\frac {\tan ^{\frac {7}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}-\frac {\int \frac {\tan ^{\frac {5}{2}}(c+d x) \left (-\frac {7 a}{2}+6 i a \tan (c+d x)\right )}{(a+i a \tan (c+d x))^{3/2}} \, dx}{5 a^2}\\ &=-\frac {\tan ^{\frac {7}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {19 i \tan ^{\frac {5}{2}}(c+d x)}{30 a d (a+i a \tan (c+d x))^{3/2}}+\frac {\int \frac {\tan ^{\frac {3}{2}}(c+d x) \left (-\frac {95 i a^2}{4}-\frac {55}{2} a^2 \tan (c+d x)\right )}{\sqrt {a+i a \tan (c+d x)}} \, dx}{15 a^4}\\ &=-\frac {\tan ^{\frac {7}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {19 i \tan ^{\frac {5}{2}}(c+d x)}{30 a d (a+i a \tan (c+d x))^{3/2}}+\frac {41 \tan ^{\frac {3}{2}}(c+d x)}{12 a^2 d \sqrt {a+i a \tan (c+d x)}}-\frac {\int \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)} \left (\frac {615 a^3}{8}-\frac {315}{4} i a^3 \tan (c+d x)\right ) \, dx}{15 a^6}\\ &=-\frac {\tan ^{\frac {7}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {19 i \tan ^{\frac {5}{2}}(c+d x)}{30 a d (a+i a \tan (c+d x))^{3/2}}+\frac {41 \tan ^{\frac {3}{2}}(c+d x)}{12 a^2 d \sqrt {a+i a \tan (c+d x)}}+\frac {21 i \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{4 a^3 d}-\frac {\int \frac {\sqrt {a+i a \tan (c+d x)} \left (\frac {315 i a^4}{8}+\frac {75}{2} a^4 \tan (c+d x)\right )}{\sqrt {\tan (c+d x)}} \, dx}{15 a^7}\\ &=-\frac {\tan ^{\frac {7}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {19 i \tan ^{\frac {5}{2}}(c+d x)}{30 a d (a+i a \tan (c+d x))^{3/2}}+\frac {41 \tan ^{\frac {3}{2}}(c+d x)}{12 a^2 d \sqrt {a+i a \tan (c+d x)}}+\frac {21 i \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{4 a^3 d}-\frac {(5 i) \int \frac {(a-i a \tan (c+d x)) \sqrt {a+i a \tan (c+d x)}}{\sqrt {\tan (c+d x)}} \, dx}{2 a^4}-\frac {i \int \frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {\tan (c+d x)}} \, dx}{8 a^3}\\ &=-\frac {\tan ^{\frac {7}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {19 i \tan ^{\frac {5}{2}}(c+d x)}{30 a d (a+i a \tan (c+d x))^{3/2}}+\frac {41 \tan ^{\frac {3}{2}}(c+d x)}{12 a^2 d \sqrt {a+i a \tan (c+d x)}}+\frac {21 i \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{4 a^3 d}-\frac {(5 i) \text {Subst}\left (\int \frac {1}{\sqrt {x} \sqrt {a+i a x}} \, dx,x,\tan (c+d x)\right )}{2 a^2 d}-\frac {\text {Subst}\left (\int \frac {1}{-i a-2 a^2 x^2} \, dx,x,\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{4 a d}\\ &=-\frac {\left (\frac {1}{8}+\frac {i}{8}\right ) \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{a^{5/2} d}-\frac {\tan ^{\frac {7}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {19 i \tan ^{\frac {5}{2}}(c+d x)}{30 a d (a+i a \tan (c+d x))^{3/2}}+\frac {41 \tan ^{\frac {3}{2}}(c+d x)}{12 a^2 d \sqrt {a+i a \tan (c+d x)}}+\frac {21 i \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{4 a^3 d}-\frac {(5 i) \text {Subst}\left (\int \frac {1}{\sqrt {a+i a x^2}} \, dx,x,\sqrt {\tan (c+d x)}\right )}{a^2 d}\\ &=-\frac {\left (\frac {1}{8}+\frac {i}{8}\right ) \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{a^{5/2} d}-\frac {\tan ^{\frac {7}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {19 i \tan ^{\frac {5}{2}}(c+d x)}{30 a d (a+i a \tan (c+d x))^{3/2}}+\frac {41 \tan ^{\frac {3}{2}}(c+d x)}{12 a^2 d \sqrt {a+i a \tan (c+d x)}}+\frac {21 i \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{4 a^3 d}-\frac {(5 i) \text {Subst}\left (\int \frac {1}{1-i a x^2} \, dx,x,\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{a^2 d}\\ &=\frac {5 (-1)^{3/4} \tan ^{-1}\left (\frac {(-1)^{3/4} \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{a^{5/2} d}-\frac {\left (\frac {1}{8}+\frac {i}{8}\right ) \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{a^{5/2} d}-\frac {\tan ^{\frac {7}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {19 i \tan ^{\frac {5}{2}}(c+d x)}{30 a d (a+i a \tan (c+d x))^{3/2}}+\frac {41 \tan ^{\frac {3}{2}}(c+d x)}{12 a^2 d \sqrt {a+i a \tan (c+d x)}}+\frac {21 i \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{4 a^3 d}\\ \end {align*}
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Mathematica [A]
time = 3.97, size = 270, normalized size = 1.05 \begin {gather*} -\frac {i e^{-6 i (c+d x)} \sqrt {\frac {a e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}} \left (-\sqrt {-1+e^{2 i (c+d x)}} \left (3-28 e^{2 i (c+d x)}+252 e^{4 i (c+d x)}+403 e^{6 i (c+d x)}\right )+15 e^{5 i (c+d x)} \left (1+e^{2 i (c+d x)}\right ) \tanh ^{-1}\left (\frac {e^{i (c+d x)}}{\sqrt {-1+e^{2 i (c+d x)}}}\right )+300 \sqrt {2} e^{5 i (c+d x)} \left (1+e^{2 i (c+d x)}\right ) \tanh ^{-1}\left (\frac {\sqrt {2} e^{i (c+d x)}}{\sqrt {-1+e^{2 i (c+d x)}}}\right )\right ) \sqrt {\tan (c+d x)}}{60 \sqrt {2} a^3 d \sqrt {-1+e^{2 i (c+d x)}}} \end {gather*}
Antiderivative was successfully verified.
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Maple [B] Both result and optimal contain complex but leaf count of result is larger than twice
the leaf count of optimal. 1006 vs. \(2 (201 ) = 402\).
time = 0.19, size = 1007, normalized size = 3.92 Too large to display
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: RuntimeError} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [B] Both result and optimal contain complex but leaf count of result is larger than twice
the leaf count of optimal. 623 vs. \(2 (189) = 378\).
time = 0.52, size = 623, normalized size = 2.42 \begin {gather*} \frac {{\left (30 \, a^{3} d \sqrt {\frac {i}{8 \, a^{5} d^{2}}} e^{\left (5 i \, d x + 5 i \, c\right )} \log \left (i \, a^{3} d \sqrt {\frac {i}{8 \, a^{5} d^{2}}} e^{\left (i \, d x + i \, c\right )} + \frac {1}{4} \, \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )}\right ) - 30 \, a^{3} d \sqrt {\frac {i}{8 \, a^{5} d^{2}}} e^{\left (5 i \, d x + 5 i \, c\right )} \log \left (-i \, a^{3} d \sqrt {\frac {i}{8 \, a^{5} d^{2}}} e^{\left (i \, d x + i \, c\right )} + \frac {1}{4} \, \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )}\right ) - 30 \, a^{3} d \sqrt {\frac {25 i}{a^{5} d^{2}}} e^{\left (5 i \, d x + 5 i \, c\right )} \log \left (\frac {104 \, {\left (10 \, \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (e^{\left (3 i \, d x + 3 i \, c\right )} + e^{\left (i \, d x + i \, c\right )}\right )} - {\left (3 i \, a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} - i \, a^{3} d\right )} \sqrt {\frac {25 i}{a^{5} d^{2}}}\right )}}{3025 \, {\left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )}}\right ) + 30 \, a^{3} d \sqrt {\frac {25 i}{a^{5} d^{2}}} e^{\left (5 i \, d x + 5 i \, c\right )} \log \left (\frac {104 \, {\left (10 \, \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (e^{\left (3 i \, d x + 3 i \, c\right )} + e^{\left (i \, d x + i \, c\right )}\right )} - {\left (-3 i \, a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} + i \, a^{3} d\right )} \sqrt {\frac {25 i}{a^{5} d^{2}}}\right )}}{3025 \, {\left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )}}\right ) + \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (403 i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 252 i \, e^{\left (4 i \, d x + 4 i \, c\right )} - 28 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} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\mathrm {tan}\left (c+d\,x\right )}^{9/2}}{{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2}} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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