Optimal. Leaf size=242 \[ \frac {105 i \tanh ^{-1}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{256 \sqrt {2} a^3 c^{3/2} f}-\frac {35 i}{128 a^3 f (c-i c \tan (e+f x))^{3/2}}+\frac {i}{6 a^3 f (1+i \tan (e+f x))^3 (c-i c \tan (e+f x))^{3/2}}+\frac {3 i}{16 a^3 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{3/2}}+\frac {21 i}{64 a^3 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}}-\frac {105 i}{256 a^3 c f \sqrt {c-i c \tan (e+f x)}} \]
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Rubi [A]
time = 0.17, antiderivative size = 242, normalized size of antiderivative = 1.00, number of steps
used = 9, number of rules used = 6, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {3603, 3568, 44,
53, 65, 212} \begin {gather*} \frac {105 i \tanh ^{-1}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{256 \sqrt {2} a^3 c^{3/2} f}-\frac {105 i}{256 a^3 c f \sqrt {c-i c \tan (e+f x)}}-\frac {35 i}{128 a^3 f (c-i c \tan (e+f x))^{3/2}}+\frac {21 i}{64 a^3 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}}+\frac {3 i}{16 a^3 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{3/2}}+\frac {i}{6 a^3 f (1+i \tan (e+f x))^3 (c-i c \tan (e+f x))^{3/2}} \end {gather*}
Antiderivative was successfully verified.
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Rule 44
Rule 53
Rule 65
Rule 212
Rule 3568
Rule 3603
Rubi steps
\begin {align*} \int \frac {1}{(a+i a \tan (e+f x))^3 (c-i c \tan (e+f x))^{3/2}} \, dx &=\frac {\int \cos ^6(e+f x) (c-i c \tan (e+f x))^{3/2} \, dx}{a^3 c^3}\\ &=\frac {\left (i c^4\right ) \text {Subst}\left (\int \frac {1}{(c-x)^4 (c+x)^{5/2}} \, dx,x,-i c \tan (e+f x)\right )}{a^3 f}\\ &=\frac {i}{6 a^3 f (1+i \tan (e+f x))^3 (c-i c \tan (e+f x))^{3/2}}+\frac {\left (3 i c^3\right ) \text {Subst}\left (\int \frac {1}{(c-x)^3 (c+x)^{5/2}} \, dx,x,-i c \tan (e+f x)\right )}{4 a^3 f}\\ &=\frac {i}{6 a^3 f (1+i \tan (e+f x))^3 (c-i c \tan (e+f x))^{3/2}}+\frac {3 i}{16 a^3 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{3/2}}+\frac {\left (21 i c^2\right ) \text {Subst}\left (\int \frac {1}{(c-x)^2 (c+x)^{5/2}} \, dx,x,-i c \tan (e+f x)\right )}{32 a^3 f}\\ &=\frac {i}{6 a^3 f (1+i \tan (e+f x))^3 (c-i c \tan (e+f x))^{3/2}}+\frac {3 i}{16 a^3 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{3/2}}+\frac {21 i}{64 a^3 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}}+\frac {(105 i c) \text {Subst}\left (\int \frac {1}{(c-x) (c+x)^{5/2}} \, dx,x,-i c \tan (e+f x)\right )}{128 a^3 f}\\ &=-\frac {35 i}{128 a^3 f (c-i c \tan (e+f x))^{3/2}}+\frac {i}{6 a^3 f (1+i \tan (e+f x))^3 (c-i c \tan (e+f x))^{3/2}}+\frac {3 i}{16 a^3 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{3/2}}+\frac {21 i}{64 a^3 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}}+\frac {(105 i) \text {Subst}\left (\int \frac {1}{(c-x) (c+x)^{3/2}} \, dx,x,-i c \tan (e+f x)\right )}{256 a^3 f}\\ &=-\frac {35 i}{128 a^3 f (c-i c \tan (e+f x))^{3/2}}+\frac {i}{6 a^3 f (1+i \tan (e+f x))^3 (c-i c \tan (e+f x))^{3/2}}+\frac {3 i}{16 a^3 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{3/2}}+\frac {21 i}{64 a^3 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}}-\frac {105 i}{256 a^3 c f \sqrt {c-i c \tan (e+f x)}}+\frac {(105 i) \text {Subst}\left (\int \frac {1}{(c-x) \sqrt {c+x}} \, dx,x,-i c \tan (e+f x)\right )}{512 a^3 c f}\\ &=-\frac {35 i}{128 a^3 f (c-i c \tan (e+f x))^{3/2}}+\frac {i}{6 a^3 f (1+i \tan (e+f x))^3 (c-i c \tan (e+f x))^{3/2}}+\frac {3 i}{16 a^3 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{3/2}}+\frac {21 i}{64 a^3 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}}-\frac {105 i}{256 a^3 c f \sqrt {c-i c \tan (e+f x)}}+\frac {(105 i) \text {Subst}\left (\int \frac {1}{2 c-x^2} \, dx,x,\sqrt {c-i c \tan (e+f x)}\right )}{256 a^3 c f}\\ &=\frac {105 i \tanh ^{-1}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{256 \sqrt {2} a^3 c^{3/2} f}-\frac {35 i}{128 a^3 f (c-i c \tan (e+f x))^{3/2}}+\frac {i}{6 a^3 f (1+i \tan (e+f x))^3 (c-i c \tan (e+f x))^{3/2}}+\frac {3 i}{16 a^3 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{3/2}}+\frac {21 i}{64 a^3 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}}-\frac {105 i}{256 a^3 c f \sqrt {c-i c \tan (e+f x)}}\\ \end {align*}
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Mathematica [A]
time = 2.67, size = 160, normalized size = 0.66 \begin {gather*} \frac {(\cos (e+f x)-i \sin (e+f x)) \left (315 i e^{i (e+f x)} \sqrt {1+e^{2 i (e+f x)}} \tanh ^{-1}\left (\sqrt {1+e^{2 i (e+f x)}}\right )+172 i \cos (e+f x)-166 i \cos (3 (e+f x))-8 i \cos (5 (e+f x))+258 \sin (e+f x)+282 \sin (3 (e+f x))+24 \sin (5 (e+f x))\right ) \sqrt {c-i c \tan (e+f x)}}{1536 a^3 c^2 f} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.38, size = 158, normalized size = 0.65
method | result | size |
derivativedivides | \(\frac {2 i c^{4} \left (-\frac {1}{8 c^{5} \sqrt {c -i c \tan \left (f x +e \right )}}-\frac {1}{48 c^{4} \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}+\frac {\frac {\frac {41 \left (c -i c \tan \left (f x +e \right )\right )^{\frac {5}{2}}}{32}-\frac {35 c \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{6}+\frac {55 c^{2} \sqrt {c -i c \tan \left (f x +e \right )}}{8}}{\left (c +i c \tan \left (f x +e \right )\right )^{3}}+\frac {105 \sqrt {2}\, \arctanh \left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{64 \sqrt {c}}}{16 c^{5}}\right )}{f \,a^{3}}\) | \(158\) |
default | \(\frac {2 i c^{4} \left (-\frac {1}{8 c^{5} \sqrt {c -i c \tan \left (f x +e \right )}}-\frac {1}{48 c^{4} \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}+\frac {\frac {\frac {41 \left (c -i c \tan \left (f x +e \right )\right )^{\frac {5}{2}}}{32}-\frac {35 c \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{6}+\frac {55 c^{2} \sqrt {c -i c \tan \left (f x +e \right )}}{8}}{\left (c +i c \tan \left (f x +e \right )\right )^{3}}+\frac {105 \sqrt {2}\, \arctanh \left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{64 \sqrt {c}}}{16 c^{5}}\right )}{f \,a^{3}}\) | \(158\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.54, size = 233, normalized size = 0.96 \begin {gather*} -\frac {i \, {\left (\frac {4 \, {\left (315 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{4} - 1680 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{3} c + 2772 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{2} c^{2} - 1152 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )} c^{3} - 256 \, c^{4}\right )}}{{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {9}{2}} a^{3} - 6 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {7}{2}} a^{3} c + 12 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {5}{2}} a^{3} c^{2} - 8 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}} a^{3} c^{3}} + \frac {315 \, \sqrt {2} \log \left (-\frac {\sqrt {2} \sqrt {c} - \sqrt {-i \, c \tan \left (f x + e\right ) + c}}{\sqrt {2} \sqrt {c} + \sqrt {-i \, c \tan \left (f x + e\right ) + c}}\right )}{a^{3} \sqrt {c}}\right )}}{3072 \, c f} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 1.52, size = 346, normalized size = 1.43 \begin {gather*} \frac {{\left (-315 i \, \sqrt {\frac {1}{2}} a^{3} c^{2} f \sqrt {\frac {1}{a^{6} c^{3} f^{2}}} e^{\left (6 i \, f x + 6 i \, e\right )} \log \left (-\frac {105 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (i \, a^{3} c f e^{\left (2 i \, f x + 2 i \, e\right )} + i \, a^{3} c f\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {1}{a^{6} c^{3} f^{2}}} - i\right )} e^{\left (-i \, f x - i \, e\right )}}{128 \, a^{3} c f}\right ) + 315 i \, \sqrt {\frac {1}{2}} a^{3} c^{2} f \sqrt {\frac {1}{a^{6} c^{3} f^{2}}} e^{\left (6 i \, f x + 6 i \, e\right )} \log \left (-\frac {105 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (-i \, a^{3} c f e^{\left (2 i \, f x + 2 i \, e\right )} - i \, a^{3} c f\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {1}{a^{6} c^{3} f^{2}}} - i\right )} e^{\left (-i \, f x - i \, e\right )}}{128 \, a^{3} c f}\right ) + \sqrt {2} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} {\left (-16 i \, e^{\left (10 i \, f x + 10 i \, e\right )} - 224 i \, e^{\left (8 i \, f x + 8 i \, e\right )} - 43 i \, e^{\left (6 i \, f x + 6 i \, e\right )} + 215 i \, e^{\left (4 i \, f x + 4 i \, e\right )} + 58 i \, e^{\left (2 i \, f x + 2 i \, e\right )} + 8 i\right )}\right )} e^{\left (-6 i \, f x - 6 i \, e\right )}}{1536 \, a^{3} c^{2} f} \end {gather*}
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 \begin {gather*} \frac {i \int \frac {1}{- i c \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{4}{\left (e + f x \right )} - 2 c \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{3}{\left (e + f x \right )} - 2 c \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan {\left (e + f x \right )} + i c \sqrt {- i c \tan {\left (e + f x \right )} + c}}\, dx}{a^{3}} \end {gather*}
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 \begin {gather*} \text {could not integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 5.07, size = 229, normalized size = 0.95 \begin {gather*} -\frac {\frac {{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^3\,35{}\mathrm {i}}{16\,a^3\,f}+\frac {c^3\,1{}\mathrm {i}}{3\,a^3\,f}-\frac {{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^4\,105{}\mathrm {i}}{256\,a^3\,c\,f}-\frac {c\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^2\,231{}\mathrm {i}}{64\,a^3\,f}+\frac {c^2\,\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )\,3{}\mathrm {i}}{2\,a^3\,f}}{6\,c\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{7/2}-{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{9/2}+8\,c^3\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{3/2}-12\,c^2\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{5/2}}+\frac {\sqrt {2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {-c}}\right )\,105{}\mathrm {i}}{512\,a^3\,{\left (-c\right )}^{3/2}\,f} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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