Optimal. Leaf size=168 \[ -\frac {\left (\frac {1}{8}+\frac {i}{8}\right ) \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{a^{5/2} d}-\frac {i \sqrt {\tan (c+d x)}}{20 a^2 d \sqrt {a+i a \tan (c+d x)}}+\frac {i \sqrt {\tan (c+d x)}}{10 a d (a+i a \tan (c+d x))^{3/2}}+\frac {i \sqrt {\tan (c+d x)}}{5 d (a+i a \tan (c+d x))^{5/2}} \]
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Rubi [A] time = 0.42, antiderivative size = 168, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {3557, 3596, 12, 3544, 205} \[ -\frac {i \sqrt {\tan (c+d x)}}{20 a^2 d \sqrt {a+i a \tan (c+d x)}}-\frac {\left (\frac {1}{8}+\frac {i}{8}\right ) \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{a^{5/2} d}+\frac {i \sqrt {\tan (c+d x)}}{10 a d (a+i a \tan (c+d x))^{3/2}}+\frac {i \sqrt {\tan (c+d x)}}{5 d (a+i a \tan (c+d x))^{5/2}} \]
Antiderivative was successfully verified.
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Rule 12
Rule 205
Rule 3544
Rule 3557
Rule 3596
Rubi steps
\begin {align*} \int \frac {\sqrt {\tan (c+d x)}}{(a+i a \tan (c+d x))^{5/2}} \, dx &=\frac {i \sqrt {\tan (c+d x)}}{5 d (a+i a \tan (c+d x))^{5/2}}-\frac {\int \frac {i a-4 a \tan (c+d x)}{\sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^{3/2}} \, dx}{10 a^2}\\ &=\frac {i \sqrt {\tan (c+d x)}}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {i \sqrt {\tan (c+d x)}}{10 a d (a+i a \tan (c+d x))^{3/2}}-\frac {\int \frac {\frac {9 i a^2}{2}-3 a^2 \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}} \, dx}{30 a^4}\\ &=\frac {i \sqrt {\tan (c+d x)}}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {i \sqrt {\tan (c+d x)}}{10 a d (a+i a \tan (c+d x))^{3/2}}-\frac {i \sqrt {\tan (c+d x)}}{20 a^2 d \sqrt {a+i a \tan (c+d x)}}-\frac {\int \frac {15 i a^3 \sqrt {a+i a \tan (c+d x)}}{4 \sqrt {\tan (c+d x)}} \, dx}{30 a^6}\\ &=\frac {i \sqrt {\tan (c+d x)}}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {i \sqrt {\tan (c+d x)}}{10 a d (a+i a \tan (c+d x))^{3/2}}-\frac {i \sqrt {\tan (c+d x)}}{20 a^2 d \sqrt {a+i a \tan (c+d x)}}-\frac {i \int \frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {\tan (c+d x)}} \, dx}{8 a^3}\\ &=\frac {i \sqrt {\tan (c+d x)}}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {i \sqrt {\tan (c+d x)}}{10 a d (a+i a \tan (c+d x))^{3/2}}-\frac {i \sqrt {\tan (c+d x)}}{20 a^2 d \sqrt {a+i a \tan (c+d x)}}-\frac {\operatorname {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 {i \sqrt {\tan (c+d x)}}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {i \sqrt {\tan (c+d x)}}{10 a d (a+i a \tan (c+d x))^{3/2}}-\frac {i \sqrt {\tan (c+d x)}}{20 a^2 d \sqrt {a+i a \tan (c+d x)}}\\ \end {align*}
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Mathematica [A] time = 1.78, size = 175, normalized size = 1.04 \[ \frac {i e^{-4 i (c+d x)} \sqrt {\tan (c+d x)} \left (\sqrt {-1+e^{2 i (c+d x)}} \left (3 e^{2 i (c+d x)}+e^{4 i (c+d x)}+1\right )-5 e^{5 i (c+d x)} \tanh ^{-1}\left (\frac {e^{i (c+d x)}}{\sqrt {-1+e^{2 i (c+d x)}}}\right )\right )}{20 \sqrt {2} a^2 d \sqrt {-1+e^{2 i (c+d x)}} \sqrt {\frac {a e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}}} \]
Antiderivative was successfully verified.
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fricas [B] time = 0.58, size = 331, normalized size = 1.97 \[ \frac {{\left (10 \, a^{3} d \sqrt {\frac {i}{8 \, a^{5} d^{2}}} e^{\left (5 i \, d x + 5 i \, c\right )} \log \left (i \, a^{3} d \sqrt {\frac {i}{8 \, a^{5} d^{2}}} e^{\left (i \, d x + i \, c\right )} + \frac {1}{4} \, \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )}\right ) - 10 \, a^{3} d \sqrt {\frac {i}{8 \, a^{5} d^{2}}} e^{\left (5 i \, d x + 5 i \, c\right )} \log \left (-i \, a^{3} d \sqrt {\frac {i}{8 \, a^{5} d^{2}}} e^{\left (i \, d x + i \, c\right )} + \frac {1}{4} \, \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )}\right ) + \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 4 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 4 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i\right )}\right )} e^{\left (-5 i \, d x - 5 i \, c\right )}}{40 \, a^{3} d} \]
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 \[ \int \frac {\sqrt {\tan \left (d x + c\right )}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.21, size = 575, normalized size = 3.42 \[ \frac {\left (\sqrt {\tan }\left (d x +c \right )\right ) \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, \left (20 i \sqrt {2}\, \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) \left (\tan ^{3}\left (d x +c \right )\right ) a -5 \sqrt {2}\, \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) \left (\tan ^{4}\left (d x +c \right )\right ) a -4 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {-i a}\, \left (\tan ^{3}\left (d x +c \right )\right )-20 i \sqrt {2}\, \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) \tan \left (d x +c \right ) a +30 \sqrt {2}\, \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) \left (\tan ^{2}\left (d x +c \right )\right ) a +4 i \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \left (\tan ^{2}\left (d x +c \right )\right )-5 \sqrt {2}\, \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a +20 i \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {-i a}-20 \tan \left (d x +c \right ) \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {-i a}\right )}{80 d \,a^{3} \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {-i a}\, \left (-\tan \left (d x +c \right )+i\right )^{4}} \]
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 \[ \text {Exception raised: RuntimeError} \]
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 \[ \int \frac {\sqrt {\mathrm {tan}\left (c+d\,x\right )}}{{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2}} \,d x \]
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 \[ \int \frac {\sqrt {\tan {\left (c + d x \right )}}}{\left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{\frac {5}{2}}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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