Optimal. Leaf size=93 \[ \frac {4 \sqrt [4]{-1} a^2 \tan ^{-1}\left (\frac {(-1)^{3/4} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{d^{5/2} f}-\frac {4 i a^2}{d^2 f \sqrt {d \tan (e+f x)}}-\frac {2 a^2}{3 d f (d \tan (e+f x))^{3/2}} \]
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Rubi [A] time = 0.15, antiderivative size = 93, 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 = {3542, 3529, 3533, 205} \[ \frac {4 \sqrt [4]{-1} a^2 \tan ^{-1}\left (\frac {(-1)^{3/4} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{d^{5/2} f}-\frac {4 i a^2}{d^2 f \sqrt {d \tan (e+f x)}}-\frac {2 a^2}{3 d f (d \tan (e+f x))^{3/2}} \]
Antiderivative was successfully verified.
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Rule 205
Rule 3529
Rule 3533
Rule 3542
Rubi steps
\begin {align*} \int \frac {(a+i a \tan (e+f x))^2}{(d \tan (e+f x))^{5/2}} \, dx &=-\frac {2 a^2}{3 d f (d \tan (e+f x))^{3/2}}+\frac {\int \frac {2 i a^2 d-2 a^2 d \tan (e+f x)}{(d \tan (e+f x))^{3/2}} \, dx}{d^2}\\ &=-\frac {2 a^2}{3 d f (d \tan (e+f x))^{3/2}}-\frac {4 i a^2}{d^2 f \sqrt {d \tan (e+f x)}}+\frac {\int \frac {-2 a^2 d^2-2 i a^2 d^2 \tan (e+f x)}{\sqrt {d \tan (e+f x)}} \, dx}{d^4}\\ &=-\frac {2 a^2}{3 d f (d \tan (e+f x))^{3/2}}-\frac {4 i a^2}{d^2 f \sqrt {d \tan (e+f x)}}+\frac {\left (8 a^4\right ) \operatorname {Subst}\left (\int \frac {1}{-2 a^2 d^3+2 i a^2 d^2 x^2} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{f}\\ &=\frac {4 \sqrt [4]{-1} a^2 \tan ^{-1}\left (\frac {(-1)^{3/4} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{d^{5/2} f}-\frac {2 a^2}{3 d f (d \tan (e+f x))^{3/2}}-\frac {4 i a^2}{d^2 f \sqrt {d \tan (e+f x)}}\\ \end {align*}
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Mathematica [A] time = 2.05, size = 87, normalized size = 0.94 \[ -\frac {2 a^2 \left (\cot (e+f x)-6 i \sqrt {i \tan (e+f x)} \tanh ^{-1}\left (\sqrt {\frac {-1+e^{2 i (e+f x)}}{1+e^{2 i (e+f x)}}}\right )+6 i\right )}{3 d^2 f \sqrt {d \tan (e+f x)}} \]
Antiderivative was successfully verified.
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fricas [B] time = 0.58, size = 403, normalized size = 4.33 \[ -\frac {3 \, {\left (d^{3} f e^{\left (4 i \, f x + 4 i \, e\right )} - 2 \, d^{3} f e^{\left (2 i \, f x + 2 i \, e\right )} + d^{3} f\right )} \sqrt {-\frac {16 i \, a^{4}}{d^{5} f^{2}}} \log \left (\frac {{\left (-4 i \, a^{2} d e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (d^{3} f e^{\left (2 i \, f x + 2 i \, e\right )} + d^{3} f\right )} \sqrt {\frac {-i \, d e^{\left (2 i \, f x + 2 i \, e\right )} + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {-\frac {16 i \, a^{4}}{d^{5} f^{2}}}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{2 \, a^{2}}\right ) - 3 \, {\left (d^{3} f e^{\left (4 i \, f x + 4 i \, e\right )} - 2 \, d^{3} f e^{\left (2 i \, f x + 2 i \, e\right )} + d^{3} f\right )} \sqrt {-\frac {16 i \, a^{4}}{d^{5} f^{2}}} \log \left (\frac {{\left (-4 i \, a^{2} d e^{\left (2 i \, f x + 2 i \, e\right )} - {\left (d^{3} f e^{\left (2 i \, f x + 2 i \, e\right )} + d^{3} f\right )} \sqrt {\frac {-i \, d e^{\left (2 i \, f x + 2 i \, e\right )} + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {-\frac {16 i \, a^{4}}{d^{5} f^{2}}}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{2 \, a^{2}}\right ) - 8 \, {\left (7 \, a^{2} e^{\left (4 i \, f x + 4 i \, e\right )} + 2 \, a^{2} e^{\left (2 i \, f x + 2 i \, e\right )} - 5 \, a^{2}\right )} \sqrt {\frac {-i \, d e^{\left (2 i \, f x + 2 i \, e\right )} + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{12 \, {\left (d^{3} f e^{\left (4 i \, f x + 4 i \, e\right )} - 2 \, d^{3} f e^{\left (2 i \, f x + 2 i \, e\right )} + d^{3} f\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 2.18, size = 117, normalized size = 1.26 \[ -\frac {4 i \, \sqrt {2} a^{2} \arctan \left (\frac {16 \, \sqrt {d^{2}} \sqrt {d \tan \left (f x + e\right )}}{8 i \, \sqrt {2} d^{\frac {3}{2}} + 8 \, \sqrt {2} \sqrt {d^{2}} \sqrt {d}}\right )}{d^{\frac {5}{2}} f {\left (\frac {i \, d}{\sqrt {d^{2}}} + 1\right )}} - \frac {2 \, {\left (6 i \, a^{2} d \tan \left (f x + e\right ) + a^{2} d\right )}}{3 \, \sqrt {d \tan \left (f x + e\right )} d^{3} f \tan \left (f x + e\right )} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.17, size = 393, normalized size = 4.23 \[ -\frac {a^{2} \left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )}{2 f \,d^{3}}-\frac {a^{2} \left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )}{f \,d^{3}}+\frac {a^{2} \left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )}{f \,d^{3}}-\frac {i a^{2} \sqrt {2}\, \ln \left (\frac {d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )}{2 f \,d^{2} \left (d^{2}\right )^{\frac {1}{4}}}-\frac {i a^{2} \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )}{f \,d^{2} \left (d^{2}\right )^{\frac {1}{4}}}+\frac {i a^{2} \sqrt {2}\, \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )}{f \,d^{2} \left (d^{2}\right )^{\frac {1}{4}}}-\frac {2 a^{2}}{3 d f \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}-\frac {4 i a^{2}}{d^{2} f \sqrt {d \tan \left (f x +e \right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.83, size = 198, normalized size = 2.13 \[ \frac {\frac {3 \, a^{2} {\left (-\frac {\left (2 i + 2\right ) \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} + 2 \, \sqrt {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {d}}\right )}{\sqrt {d}} - \frac {\left (2 i + 2\right ) \, \sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} - 2 \, \sqrt {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {d}}\right )}{\sqrt {d}} + \frac {\left (i - 1\right ) \, \sqrt {2} \log \left (d \tan \left (f x + e\right ) + \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {d} + d\right )}{\sqrt {d}} - \frac {\left (i - 1\right ) \, \sqrt {2} \log \left (d \tan \left (f x + e\right ) - \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {d} + d\right )}{\sqrt {d}}\right )}}{d} - \frac {4 \, {\left (6 i \, a^{2} d \tan \left (f x + e\right ) + a^{2} d\right )}}{\left (d \tan \left (f x + e\right )\right )^{\frac {3}{2}} d}}{6 \, d f} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 4.38, size = 80, normalized size = 0.86 \[ -\frac {\frac {2\,a^2}{3\,d\,f}+\frac {a^2\,\mathrm {tan}\left (e+f\,x\right )\,4{}\mathrm {i}}{d\,f}}{{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}}-\frac {\sqrt {4{}\mathrm {i}}\,a^2\,\mathrm {atan}\left (\frac {\sqrt {4{}\mathrm {i}}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{2\,\sqrt {-d}}\right )\,2{}\mathrm {i}}{{\left (-d\right )}^{5/2}\,f} \]
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 \[ - a^{2} \left (\int \left (- \frac {1}{\left (d \tan {\left (e + f x \right )}\right )^{\frac {5}{2}}}\right )\, dx + \int \frac {\tan ^{2}{\left (e + f x \right )}}{\left (d \tan {\left (e + f x \right )}\right )^{\frac {5}{2}}}\, dx + \int \left (- \frac {2 i \tan {\left (e + f x \right )}}{\left (d \tan {\left (e + f x \right )}\right )^{\frac {5}{2}}}\right )\, dx\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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