Optimal. Leaf size=297 \[ \frac {\left (\frac {1}{16}+\frac {3 i}{16}\right ) d^{3/2} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} a^2 f}-\frac {\left (\frac {1}{16}+\frac {3 i}{16}\right ) d^{3/2} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}+1\right )}{\sqrt {2} a^2 f}+\frac {\left (\frac {1}{32}-\frac {3 i}{32}\right ) d^{3/2} \log \left (\sqrt {d} \tan (e+f x)-\sqrt {2} \sqrt {d \tan (e+f x)}+\sqrt {d}\right )}{\sqrt {2} a^2 f}-\frac {\left (\frac {1}{32}-\frac {3 i}{32}\right ) d^{3/2} \log \left (\sqrt {d} \tan (e+f x)+\sqrt {2} \sqrt {d \tan (e+f x)}+\sqrt {d}\right )}{\sqrt {2} a^2 f}+\frac {3 d \sqrt {d \tan (e+f x)}}{8 a^2 f (1+i \tan (e+f x))}-\frac {d \sqrt {d \tan (e+f x)}}{4 f (a+i a \tan (e+f x))^2} \]
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Rubi [A] time = 0.43, antiderivative size = 297, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 9, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.321, Rules used = {3558, 3596, 3534, 1168, 1162, 617, 204, 1165, 628} \[ \frac {\left (\frac {1}{16}+\frac {3 i}{16}\right ) d^{3/2} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} a^2 f}-\frac {\left (\frac {1}{16}+\frac {3 i}{16}\right ) d^{3/2} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}+1\right )}{\sqrt {2} a^2 f}+\frac {\left (\frac {1}{32}-\frac {3 i}{32}\right ) d^{3/2} \log \left (\sqrt {d} \tan (e+f x)-\sqrt {2} \sqrt {d \tan (e+f x)}+\sqrt {d}\right )}{\sqrt {2} a^2 f}-\frac {\left (\frac {1}{32}-\frac {3 i}{32}\right ) d^{3/2} \log \left (\sqrt {d} \tan (e+f x)+\sqrt {2} \sqrt {d \tan (e+f x)}+\sqrt {d}\right )}{\sqrt {2} a^2 f}+\frac {3 d \sqrt {d \tan (e+f x)}}{8 a^2 f (1+i \tan (e+f x))}-\frac {d \sqrt {d \tan (e+f x)}}{4 f (a+i a \tan (e+f x))^2} \]
Antiderivative was successfully verified.
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Rule 204
Rule 617
Rule 628
Rule 1162
Rule 1165
Rule 1168
Rule 3534
Rule 3558
Rule 3596
Rubi steps
\begin {align*} \int \frac {(d \tan (e+f x))^{3/2}}{(a+i a \tan (e+f x))^2} \, dx &=-\frac {d \sqrt {d \tan (e+f x)}}{4 f (a+i a \tan (e+f x))^2}-\frac {\int \frac {-\frac {a d^2}{2}+\frac {5}{2} i a d^2 \tan (e+f x)}{\sqrt {d \tan (e+f x)} (a+i a \tan (e+f x))} \, dx}{4 a^2}\\ &=\frac {3 d \sqrt {d \tan (e+f x)}}{8 a^2 f (1+i \tan (e+f x))}-\frac {d \sqrt {d \tan (e+f x)}}{4 f (a+i a \tan (e+f x))^2}-\frac {\int \frac {\frac {a^2 d^3}{2}+\frac {3}{2} i a^2 d^3 \tan (e+f x)}{\sqrt {d \tan (e+f x)}} \, dx}{8 a^4 d}\\ &=\frac {3 d \sqrt {d \tan (e+f x)}}{8 a^2 f (1+i \tan (e+f x))}-\frac {d \sqrt {d \tan (e+f x)}}{4 f (a+i a \tan (e+f x))^2}-\frac {\operatorname {Subst}\left (\int \frac {\frac {a^2 d^4}{2}+\frac {3}{2} i a^2 d^3 x^2}{d^2+x^4} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{4 a^4 d f}\\ &=\frac {3 d \sqrt {d \tan (e+f x)}}{8 a^2 f (1+i \tan (e+f x))}-\frac {d \sqrt {d \tan (e+f x)}}{4 f (a+i a \tan (e+f x))^2}-\frac {\left (\left (\frac {1}{16}-\frac {3 i}{16}\right ) d^2\right ) \operatorname {Subst}\left (\int \frac {d-x^2}{d^2+x^4} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{a^2 f}-\frac {\left (\left (\frac {1}{16}+\frac {3 i}{16}\right ) d^2\right ) \operatorname {Subst}\left (\int \frac {d+x^2}{d^2+x^4} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{a^2 f}\\ &=\frac {3 d \sqrt {d \tan (e+f x)}}{8 a^2 f (1+i \tan (e+f x))}-\frac {d \sqrt {d \tan (e+f x)}}{4 f (a+i a \tan (e+f x))^2}--\frac {\left (\left (\frac {1}{32}-\frac {3 i}{32}\right ) d^{3/2}\right ) \operatorname {Subst}\left (\int \frac {\sqrt {2} \sqrt {d}+2 x}{-d-\sqrt {2} \sqrt {d} x-x^2} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{\sqrt {2} a^2 f}--\frac {\left (\left (\frac {1}{32}-\frac {3 i}{32}\right ) d^{3/2}\right ) \operatorname {Subst}\left (\int \frac {\sqrt {2} \sqrt {d}-2 x}{-d+\sqrt {2} \sqrt {d} x-x^2} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{\sqrt {2} a^2 f}-\frac {\left (\left (\frac {1}{32}+\frac {3 i}{32}\right ) d^2\right ) \operatorname {Subst}\left (\int \frac {1}{d-\sqrt {2} \sqrt {d} x+x^2} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{a^2 f}-\frac {\left (\left (\frac {1}{32}+\frac {3 i}{32}\right ) d^2\right ) \operatorname {Subst}\left (\int \frac {1}{d+\sqrt {2} \sqrt {d} x+x^2} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{a^2 f}\\ &=\frac {\left (\frac {1}{32}-\frac {3 i}{32}\right ) d^{3/2} \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)-\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{\sqrt {2} a^2 f}-\frac {\left (\frac {1}{32}-\frac {3 i}{32}\right ) d^{3/2} \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)+\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{\sqrt {2} a^2 f}+\frac {3 d \sqrt {d \tan (e+f x)}}{8 a^2 f (1+i \tan (e+f x))}-\frac {d \sqrt {d \tan (e+f x)}}{4 f (a+i a \tan (e+f x))^2}--\frac {\left (\left (\frac {1}{16}+\frac {3 i}{16}\right ) d^{3/2}\right ) \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} a^2 f}-\frac {\left (\left (\frac {1}{16}+\frac {3 i}{16}\right ) d^{3/2}\right ) \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} a^2 f}\\ &=\frac {\left (\frac {1}{16}+\frac {3 i}{16}\right ) d^{3/2} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} a^2 f}-\frac {\left (\frac {1}{16}+\frac {3 i}{16}\right ) d^{3/2} \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} a^2 f}+\frac {\left (\frac {1}{32}-\frac {3 i}{32}\right ) d^{3/2} \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)-\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{\sqrt {2} a^2 f}-\frac {\left (\frac {1}{32}-\frac {3 i}{32}\right ) d^{3/2} \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)+\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{\sqrt {2} a^2 f}+\frac {3 d \sqrt {d \tan (e+f x)}}{8 a^2 f (1+i \tan (e+f x))}-\frac {d \sqrt {d \tan (e+f x)}}{4 f (a+i a \tan (e+f x))^2}\\ \end {align*}
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Mathematica [A] time = 1.07, size = 231, normalized size = 0.78 \[ \frac {d^2 \sec ^3(e+f x) \left (-\sin (e+f x)-\sin (3 (e+f x))-3 i \cos (e+f x)+3 i \cos (3 (e+f x))+(3-i) \sqrt {\sin (2 (e+f x))} \sin ^{-1}(\cos (e+f x)-\sin (e+f x)) (\sin (2 (e+f x))-i \cos (2 (e+f x)))+(3+i) \sin ^{\frac {3}{2}}(2 (e+f x)) \log \left (\sin (e+f x)+\sqrt {\sin (2 (e+f x))}+\cos (e+f x)\right )+(1-3 i) \sqrt {\sin (2 (e+f x))} \cos (2 (e+f x)) \log \left (\sin (e+f x)+\sqrt {\sin (2 (e+f x))}+\cos (e+f x)\right )\right )}{32 a^2 f (\tan (e+f x)-i)^2 \sqrt {d \tan (e+f x)}} \]
Antiderivative was successfully verified.
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fricas [B] time = 0.59, size = 565, normalized size = 1.90 \[ -\frac {{\left (4 \, a^{2} f \sqrt {-\frac {i \, d^{3}}{16 \, a^{4} f^{2}}} e^{\left (4 i \, f x + 4 i \, e\right )} \log \left (\frac {{\left (-2 i \, d^{2} e^{\left (2 i \, f x + 2 i \, e\right )} + 8 \, {\left (a^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + a^{2} 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 {i \, d^{3}}{16 \, a^{4} f^{2}}}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{d}\right ) - 4 \, a^{2} f \sqrt {-\frac {i \, d^{3}}{16 \, a^{4} f^{2}}} e^{\left (4 i \, f x + 4 i \, e\right )} \log \left (\frac {{\left (-2 i \, d^{2} e^{\left (2 i \, f x + 2 i \, e\right )} - 8 \, {\left (a^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + a^{2} 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 {i \, d^{3}}{16 \, a^{4} f^{2}}}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{d}\right ) - 4 \, a^{2} f \sqrt {\frac {i \, d^{3}}{64 \, a^{4} f^{2}}} e^{\left (4 i \, f x + 4 i \, e\right )} \log \left (\frac {{\left (i \, d^{2} + 8 \, {\left (a^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + a^{2} 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 {i \, d^{3}}{64 \, a^{4} f^{2}}}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{8 \, a^{2} f}\right ) + 4 \, a^{2} f \sqrt {\frac {i \, d^{3}}{64 \, a^{4} f^{2}}} e^{\left (4 i \, f x + 4 i \, e\right )} \log \left (\frac {{\left (i \, d^{2} - 8 \, {\left (a^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + a^{2} 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 {i \, d^{3}}{64 \, a^{4} f^{2}}}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{8 \, a^{2} f}\right ) - {\left (2 \, d e^{\left (4 i \, f x + 4 i \, e\right )} + d e^{\left (2 i \, f x + 2 i \, e\right )} - d\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}}\right )} e^{\left (-4 i \, f x - 4 i \, e\right )}}{16 \, a^{2} f} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 1.49, size = 203, normalized size = 0.68 \[ \frac {1}{8} \, d {\left (\frac {\sqrt {2} \sqrt {d} \arctan \left (\frac {16 i \, \sqrt {d^{2}} \sqrt {d \tan \left (f x + e\right )}}{8 i \, \sqrt {2} d^{\frac {3}{2}} + 8 \, \sqrt {2} \sqrt {d^{2}} \sqrt {d}}\right )}{a^{2} f {\left (\frac {i \, d}{\sqrt {d^{2}}} + 1\right )}} + \frac {2 \, \sqrt {2} \sqrt {d} \arctan \left (\frac {16 i \, \sqrt {d^{2}} \sqrt {d \tan \left (f x + e\right )}}{-8 i \, \sqrt {2} d^{\frac {3}{2}} + 8 \, \sqrt {2} \sqrt {d^{2}} \sqrt {d}}\right )}{a^{2} f {\left (-\frac {i \, d}{\sqrt {d^{2}}} + 1\right )}} + \frac {-3 i \, \sqrt {d \tan \left (f x + e\right )} d^{2} \tan \left (f x + e\right ) - \sqrt {d \tan \left (f x + e\right )} d^{2}}{{\left (d \tan \left (f x + e\right ) - i \, d\right )}^{2} a^{2} f}\right )} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.27, size = 147, normalized size = 0.49 \[ -\frac {3 i d^{2} \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{8 f \,a^{2} \left (d \tan \left (f x +e \right )-i d \right )^{2}}-\frac {d^{3} \sqrt {d \tan \left (f x +e \right )}}{8 f \,a^{2} \left (d \tan \left (f x +e \right )-i d \right )^{2}}-\frac {i d^{2} \arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {-i d}}\right )}{8 f \,a^{2} \sqrt {-i d}}-\frac {i d^{2} \arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {i d}}\right )}{4 f \,a^{2} \sqrt {i d}} \]
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 [B] time = 5.55, size = 179, normalized size = 0.60 \[ \frac {\frac {d^3\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{8\,a^2\,f}+\frac {d^2\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}\,3{}\mathrm {i}}{8\,a^2\,f}}{-d^2\,{\mathrm {tan}\left (e+f\,x\right )}^2+d^2\,\mathrm {tan}\left (e+f\,x\right )\,2{}\mathrm {i}+d^2}-\mathrm {atan}\left (\frac {8\,a^2\,f\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,\sqrt {-\frac {d^3\,1{}\mathrm {i}}{64\,a^4\,f^2}}}{d^2}\right )\,\sqrt {-\frac {d^3\,1{}\mathrm {i}}{64\,a^4\,f^2}}\,2{}\mathrm {i}-\mathrm {atan}\left (\frac {16\,a^2\,f\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,\sqrt {\frac {d^3\,1{}\mathrm {i}}{256\,a^4\,f^2}}}{d^2}\right )\,\sqrt {\frac {d^3\,1{}\mathrm {i}}{256\,a^4\,f^2}}\,2{}\mathrm {i} \]
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 \[ - \frac {\int \frac {\left (d \tan {\left (e + f x \right )}\right )^{\frac {3}{2}}}{\tan ^{2}{\left (e + f x \right )} - 2 i \tan {\left (e + f x \right )} - 1}\, dx}{a^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
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