Optimal. Leaf size=343 \[ \frac {\left (\frac {5}{16}-\frac {7 i}{16}\right ) d^{7/2} \text {ArcTan}\left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} a^3 f}-\frac {\left (\frac {5}{16}-\frac {7 i}{16}\right ) d^{7/2} \text {ArcTan}\left (1+\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} a^3 f}+\frac {\left (\frac {5}{32}+\frac {7 i}{32}\right ) d^{7/2} \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)-\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{\sqrt {2} a^3 f}-\frac {\left (\frac {5}{32}+\frac {7 i}{32}\right ) d^{7/2} \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)+\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{\sqrt {2} a^3 f}-\frac {d (d \tan (e+f x))^{5/2}}{6 f (a+i a \tan (e+f x))^3}+\frac {i d^2 (d \tan (e+f x))^{3/2}}{3 a f (a+i a \tan (e+f x))^2}+\frac {5 d^3 \sqrt {d \tan (e+f x)}}{8 f \left (a^3+i a^3 \tan (e+f x)\right )} \]
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Rubi [A]
time = 0.39, antiderivative size = 343, normalized size of antiderivative = 1.00, number of steps
used = 13, number of rules used = 9, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.321, Rules used = {3639, 3676,
3615, 1182, 1176, 631, 210, 1179, 642} \begin {gather*} \frac {\left (\frac {5}{16}-\frac {7 i}{16}\right ) d^{7/2} \text {ArcTan}\left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} a^3 f}-\frac {\left (\frac {5}{16}-\frac {7 i}{16}\right ) d^{7/2} \text {ArcTan}\left (\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}+1\right )}{\sqrt {2} a^3 f}+\frac {\left (\frac {5}{32}+\frac {7 i}{32}\right ) d^{7/2} \log \left (\sqrt {d} \tan (e+f x)-\sqrt {2} \sqrt {d \tan (e+f x)}+\sqrt {d}\right )}{\sqrt {2} a^3 f}-\frac {\left (\frac {5}{32}+\frac {7 i}{32}\right ) d^{7/2} \log \left (\sqrt {d} \tan (e+f x)+\sqrt {2} \sqrt {d \tan (e+f x)}+\sqrt {d}\right )}{\sqrt {2} a^3 f}+\frac {5 d^3 \sqrt {d \tan (e+f x)}}{8 f \left (a^3+i a^3 \tan (e+f x)\right )}+\frac {i d^2 (d \tan (e+f x))^{3/2}}{3 a f (a+i a \tan (e+f x))^2}-\frac {d (d \tan (e+f x))^{5/2}}{6 f (a+i a \tan (e+f x))^3} \end {gather*}
Antiderivative was successfully verified.
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Rule 210
Rule 631
Rule 642
Rule 1176
Rule 1179
Rule 1182
Rule 3615
Rule 3639
Rule 3676
Rubi steps
\begin {align*} \int \frac {(d \tan (e+f x))^{7/2}}{(a+i a \tan (e+f x))^3} \, dx &=-\frac {d (d \tan (e+f x))^{5/2}}{6 f (a+i a \tan (e+f x))^3}-\frac {\int \frac {(d \tan (e+f x))^{3/2} \left (-\frac {5 a d^2}{2}+\frac {11}{2} i a d^2 \tan (e+f x)\right )}{(a+i a \tan (e+f x))^2} \, dx}{6 a^2}\\ &=-\frac {d (d \tan (e+f x))^{5/2}}{6 f (a+i a \tan (e+f x))^3}+\frac {i d^2 (d \tan (e+f x))^{3/2}}{3 a f (a+i a \tan (e+f x))^2}+\frac {\int \frac {\sqrt {d \tan (e+f x)} \left (-12 i a^2 d^3-18 a^2 d^3 \tan (e+f x)\right )}{a+i a \tan (e+f x)} \, dx}{24 a^4}\\ &=-\frac {d (d \tan (e+f x))^{5/2}}{6 f (a+i a \tan (e+f x))^3}+\frac {i d^2 (d \tan (e+f x))^{3/2}}{3 a f (a+i a \tan (e+f x))^2}+\frac {5 d^3 \sqrt {d \tan (e+f x)}}{8 f \left (a^3+i a^3 \tan (e+f x)\right )}-\frac {\int \frac {15 a^3 d^4-21 i a^3 d^4 \tan (e+f x)}{\sqrt {d \tan (e+f x)}} \, dx}{48 a^6}\\ &=-\frac {d (d \tan (e+f x))^{5/2}}{6 f (a+i a \tan (e+f x))^3}+\frac {i d^2 (d \tan (e+f x))^{3/2}}{3 a f (a+i a \tan (e+f x))^2}+\frac {5 d^3 \sqrt {d \tan (e+f x)}}{8 f \left (a^3+i a^3 \tan (e+f x)\right )}-\frac {\text {Subst}\left (\int \frac {15 a^3 d^5-21 i a^3 d^4 x^2}{d^2+x^4} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{24 a^6 f}\\ &=-\frac {d (d \tan (e+f x))^{5/2}}{6 f (a+i a \tan (e+f x))^3}+\frac {i d^2 (d \tan (e+f x))^{3/2}}{3 a f (a+i a \tan (e+f x))^2}+\frac {5 d^3 \sqrt {d \tan (e+f x)}}{8 f \left (a^3+i a^3 \tan (e+f x)\right )}-\frac {\left (\left (\frac {5}{16}-\frac {7 i}{16}\right ) d^4\right ) \text {Subst}\left (\int \frac {d+x^2}{d^2+x^4} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{a^3 f}-\frac {\left (\left (\frac {5}{16}+\frac {7 i}{16}\right ) d^4\right ) \text {Subst}\left (\int \frac {d-x^2}{d^2+x^4} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{a^3 f}\\ &=-\frac {d (d \tan (e+f x))^{5/2}}{6 f (a+i a \tan (e+f x))^3}+\frac {i d^2 (d \tan (e+f x))^{3/2}}{3 a f (a+i a \tan (e+f x))^2}+\frac {5 d^3 \sqrt {d \tan (e+f x)}}{8 f \left (a^3+i a^3 \tan (e+f x)\right )}--\frac {\left (\left (\frac {5}{32}+\frac {7 i}{32}\right ) d^{7/2}\right ) \text {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^3 f}--\frac {\left (\left (\frac {5}{32}+\frac {7 i}{32}\right ) d^{7/2}\right ) \text {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^3 f}-\frac {\left (\left (\frac {5}{32}-\frac {7 i}{32}\right ) d^4\right ) \text {Subst}\left (\int \frac {1}{d-\sqrt {2} \sqrt {d} x+x^2} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{a^3 f}-\frac {\left (\left (\frac {5}{32}-\frac {7 i}{32}\right ) d^4\right ) \text {Subst}\left (\int \frac {1}{d+\sqrt {2} \sqrt {d} x+x^2} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{a^3 f}\\ &=\frac {\left (\frac {5}{32}+\frac {7 i}{32}\right ) d^{7/2} \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)-\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{\sqrt {2} a^3 f}-\frac {\left (\frac {5}{32}+\frac {7 i}{32}\right ) d^{7/2} \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)+\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{\sqrt {2} a^3 f}-\frac {d (d \tan (e+f x))^{5/2}}{6 f (a+i a \tan (e+f x))^3}+\frac {i d^2 (d \tan (e+f x))^{3/2}}{3 a f (a+i a \tan (e+f x))^2}+\frac {5 d^3 \sqrt {d \tan (e+f x)}}{8 f \left (a^3+i a^3 \tan (e+f x)\right )}--\frac {\left (\left (\frac {5}{16}-\frac {7 i}{16}\right ) d^{7/2}\right ) \text {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^3 f}-\frac {\left (\left (\frac {5}{16}-\frac {7 i}{16}\right ) d^{7/2}\right ) \text {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^3 f}\\ &=\frac {\left (\frac {5}{16}-\frac {7 i}{16}\right ) d^{7/2} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} a^3 f}-\frac {\left (\frac {5}{16}-\frac {7 i}{16}\right ) d^{7/2} \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} a^3 f}+\frac {\left (\frac {5}{32}+\frac {7 i}{32}\right ) d^{7/2} \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)-\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{\sqrt {2} a^3 f}-\frac {\left (\frac {5}{32}+\frac {7 i}{32}\right ) d^{7/2} \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)+\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{\sqrt {2} a^3 f}-\frac {d (d \tan (e+f x))^{5/2}}{6 f (a+i a \tan (e+f x))^3}+\frac {i d^2 (d \tan (e+f x))^{3/2}}{3 a f (a+i a \tan (e+f x))^2}+\frac {5 d^3 \sqrt {d \tan (e+f x)}}{8 f \left (a^3+i a^3 \tan (e+f x)\right )}\\ \end {align*}
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Mathematica [A]
time = 1.56, size = 234, normalized size = 0.68 \begin {gather*} \frac {d^4 \sec ^4(e+f x) \left (-19+19 \cos (4 (e+f x))+(21-15 i) \cos (3 (e+f x)) \log \left (\cos (e+f x)+\sin (e+f x)+\sqrt {\sin (2 (e+f x))}\right ) \sqrt {\sin (2 (e+f x))}-12 i \sin (2 (e+f x))+(21+15 i) \text {ArcSin}(\cos (e+f x)-\sin (e+f x)) \sqrt {\sin (2 (e+f x))} (\cos (3 (e+f x))+i \sin (3 (e+f x)))+(15+21 i) \log \left (\cos (e+f x)+\sin (e+f x)+\sqrt {\sin (2 (e+f x))}\right ) \sqrt {\sin (2 (e+f x))} \sin (3 (e+f x))+21 i \sin (4 (e+f x))\right )}{96 a^3 f \sqrt {d \tan (e+f x)} (-i+\tan (e+f x))^3} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.18, size = 125, normalized size = 0.36
method | result | size |
derivativedivides | \(\frac {2 d^{4} \left (-\frac {9 \left (d \tan \left (f x +e \right )\right )^{\frac {5}{2}}-\frac {38 i d \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}-5 d^{2} \sqrt {d \tan \left (f x +e \right )}}{16 \left (i d \tan \left (f x +e \right )+d \right )^{3}}+\frac {3 i \arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {-i d}}\right )}{8 \sqrt {-i d}}+\frac {i \arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {i d}}\right )}{16 \sqrt {i d}}\right )}{f \,a^{3}}\) | \(125\) |
default | \(\frac {2 d^{4} \left (-\frac {9 \left (d \tan \left (f x +e \right )\right )^{\frac {5}{2}}-\frac {38 i d \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}-5 d^{2} \sqrt {d \tan \left (f x +e \right )}}{16 \left (i d \tan \left (f x +e \right )+d \right )^{3}}+\frac {3 i \arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {-i d}}\right )}{8 \sqrt {-i d}}+\frac {i \arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {i d}}\right )}{16 \sqrt {i d}}\right )}{f \,a^{3}}\) | \(125\) |
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. 614 vs. \(2 (267) = 534\).
time = 0.40, size = 614, normalized size = 1.79 \begin {gather*} -\frac {{\left (12 \, a^{3} f \sqrt {-\frac {i \, d^{7}}{64 \, a^{6} f^{2}}} e^{\left (6 i \, f x + 6 i \, e\right )} \log \left (-\frac {2 \, {\left (i \, d^{4} e^{\left (2 i \, f x + 2 i \, e\right )} + 8 \, {\left (a^{3} f e^{\left (2 i \, f x + 2 i \, e\right )} + a^{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 {i \, d^{7}}{64 \, a^{6} f^{2}}}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{d^{3}}\right ) - 12 \, a^{3} f \sqrt {-\frac {i \, d^{7}}{64 \, a^{6} f^{2}}} e^{\left (6 i \, f x + 6 i \, e\right )} \log \left (-\frac {2 \, {\left (i \, d^{4} e^{\left (2 i \, f x + 2 i \, e\right )} - 8 \, {\left (a^{3} f e^{\left (2 i \, f x + 2 i \, e\right )} + a^{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 {i \, d^{7}}{64 \, a^{6} f^{2}}}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{d^{3}}\right ) - 12 \, a^{3} f \sqrt {\frac {9 i \, d^{7}}{16 \, a^{6} f^{2}}} e^{\left (6 i \, f x + 6 i \, e\right )} \log \left (\frac {{\left (-3 i \, d^{4} + 4 \, {\left (a^{3} f e^{\left (2 i \, f x + 2 i \, e\right )} + a^{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 {9 i \, d^{7}}{16 \, a^{6} f^{2}}}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{4 \, a^{3} f}\right ) + 12 \, a^{3} f \sqrt {\frac {9 i \, d^{7}}{16 \, a^{6} f^{2}}} e^{\left (6 i \, f x + 6 i \, e\right )} \log \left (\frac {{\left (-3 i \, d^{4} - 4 \, {\left (a^{3} f e^{\left (2 i \, f x + 2 i \, e\right )} + a^{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 {9 i \, d^{7}}{16 \, a^{6} f^{2}}}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{4 \, a^{3} f}\right ) - {\left (20 \, d^{3} e^{\left (6 i \, f x + 6 i \, e\right )} + 14 \, d^{3} e^{\left (4 i \, f x + 4 i \, e\right )} - 5 \, d^{3} e^{\left (2 i \, f x + 2 i \, e\right )} + d^{3}\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 (-6 i \, f x - 6 i \, e\right )}}{48 \, a^{3} 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 {\left (d \tan {\left (e + f x \right )}\right )^{\frac {7}{2}}}{\tan ^{3}{\left (e + f x \right )} - 3 i \tan ^{2}{\left (e + f x \right )} - 3 \tan {\left (e + f x \right )} + i}\, dx}{a^{3}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A]
time = 0.77, size = 231, normalized size = 0.67 \begin {gather*} -\frac {1}{24} \, d^{3} {\left (-\frac {3 i \, \sqrt {2} \sqrt {d} \arctan \left (\frac {8 \, \sqrt {d^{2}} \sqrt {d \tan \left (f x + e\right )}}{4 i \, \sqrt {2} d^{\frac {3}{2}} + 4 \, \sqrt {2} \sqrt {d^{2}} \sqrt {d}}\right )}{a^{3} f {\left (\frac {i \, d}{\sqrt {d^{2}}} + 1\right )}} - \frac {18 i \, \sqrt {2} \sqrt {d} \arctan \left (\frac {8 \, \sqrt {d^{2}} \sqrt {d \tan \left (f x + e\right )}}{-4 i \, \sqrt {2} d^{\frac {3}{2}} + 4 \, \sqrt {2} \sqrt {d^{2}} \sqrt {d}}\right )}{a^{3} f {\left (-\frac {i \, d}{\sqrt {d^{2}}} + 1\right )}} + \frac {27 i \, \sqrt {d \tan \left (f x + e\right )} d^{3} \tan \left (f x + e\right )^{2} + 38 \, \sqrt {d \tan \left (f x + e\right )} d^{3} \tan \left (f x + e\right ) - 15 i \, \sqrt {d \tan \left (f x + e\right )} d^{3}}{{\left (d \tan \left (f x + e\right ) - i \, d\right )}^{3} a^{3} f}\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 5.78, size = 217, normalized size = 0.63 \begin {gather*} \mathrm {atan}\left (\frac {8\,a^3\,f\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,\sqrt {\frac {d^7\,9{}\mathrm {i}}{64\,a^6\,f^2}}}{3\,d^4}\right )\,\sqrt {\frac {d^7\,9{}\mathrm {i}}{64\,a^6\,f^2}}\,2{}\mathrm {i}+\mathrm {atan}\left (\frac {16\,a^3\,f\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,\sqrt {-\frac {d^7\,1{}\mathrm {i}}{256\,a^6\,f^2}}}{d^4}\right )\,\sqrt {-\frac {d^7\,1{}\mathrm {i}}{256\,a^6\,f^2}}\,2{}\mathrm {i}+\frac {\frac {19\,d^5\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}}{12\,a^3\,f}-\frac {d^6\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,5{}\mathrm {i}}{8\,a^3\,f}+\frac {d^4\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{5/2}\,9{}\mathrm {i}}{8\,a^3\,f}}{-d^3\,{\mathrm {tan}\left (e+f\,x\right )}^3+d^3\,{\mathrm {tan}\left (e+f\,x\right )}^2\,3{}\mathrm {i}+3\,d^3\,\mathrm {tan}\left (e+f\,x\right )-d^3\,1{}\mathrm {i}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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