Optimal. Leaf size=353 \[ -\frac {\left (\frac {49}{16}+\frac {45 i}{16}\right ) d^{9/2} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} a^2 f}+\frac {\left (\frac {49}{16}+\frac {45 i}{16}\right ) d^{9/2} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}+1\right )}{\sqrt {2} a^2 f}+\frac {\left (\frac {49}{32}-\frac {45 i}{32}\right ) d^{9/2} \log \left (\sqrt {d} \tan (e+f x)-\sqrt {2} \sqrt {d \tan (e+f x)}+\sqrt {d}\right )}{\sqrt {2} a^2 f}-\frac {\left (\frac {49}{32}-\frac {45 i}{32}\right ) d^{9/2} \log \left (\sqrt {d} \tan (e+f x)+\sqrt {2} \sqrt {d \tan (e+f x)}+\sqrt {d}\right )}{\sqrt {2} a^2 f}-\frac {45 i d^4 \sqrt {d \tan (e+f x)}}{8 a^2 f}-\frac {49 d^3 (d \tan (e+f x))^{3/2}}{24 a^2 f}+\frac {9 i d^2 (d \tan (e+f x))^{5/2}}{8 a^2 f (1+i \tan (e+f x))}-\frac {d (d \tan (e+f x))^{7/2}}{4 f (a+i a \tan (e+f x))^2} \]
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Rubi [A] time = 0.54, antiderivative size = 353, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 10, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {3558, 3595, 3528, 3534, 1168, 1162, 617, 204, 1165, 628} \[ -\frac {\left (\frac {49}{16}+\frac {45 i}{16}\right ) d^{9/2} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} a^2 f}+\frac {\left (\frac {49}{16}+\frac {45 i}{16}\right ) d^{9/2} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}+1\right )}{\sqrt {2} a^2 f}-\frac {45 i d^4 \sqrt {d \tan (e+f x)}}{8 a^2 f}-\frac {49 d^3 (d \tan (e+f x))^{3/2}}{24 a^2 f}+\frac {9 i d^2 (d \tan (e+f x))^{5/2}}{8 a^2 f (1+i \tan (e+f x))}+\frac {\left (\frac {49}{32}-\frac {45 i}{32}\right ) d^{9/2} \log \left (\sqrt {d} \tan (e+f x)-\sqrt {2} \sqrt {d \tan (e+f x)}+\sqrt {d}\right )}{\sqrt {2} a^2 f}-\frac {\left (\frac {49}{32}-\frac {45 i}{32}\right ) d^{9/2} \log \left (\sqrt {d} \tan (e+f x)+\sqrt {2} \sqrt {d \tan (e+f x)}+\sqrt {d}\right )}{\sqrt {2} a^2 f}-\frac {d (d \tan (e+f x))^{7/2}}{4 f (a+i a \tan (e+f x))^2} \]
Antiderivative was successfully verified.
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Rule 204
Rule 617
Rule 628
Rule 1162
Rule 1165
Rule 1168
Rule 3528
Rule 3534
Rule 3558
Rule 3595
Rubi steps
\begin {align*} \int \frac {(d \tan (e+f x))^{9/2}}{(a+i a \tan (e+f x))^2} \, dx &=-\frac {d (d \tan (e+f x))^{7/2}}{4 f (a+i a \tan (e+f x))^2}-\frac {\int \frac {(d \tan (e+f x))^{5/2} \left (-\frac {7 a d^2}{2}+\frac {11}{2} i a d^2 \tan (e+f x)\right )}{a+i a \tan (e+f x)} \, dx}{4 a^2}\\ &=\frac {9 i d^2 (d \tan (e+f x))^{5/2}}{8 a^2 f (1+i \tan (e+f x))}-\frac {d (d \tan (e+f x))^{7/2}}{4 f (a+i a \tan (e+f x))^2}+\frac {\int (d \tan (e+f x))^{3/2} \left (-\frac {45}{2} i a^2 d^3-\frac {49}{2} a^2 d^3 \tan (e+f x)\right ) \, dx}{8 a^4}\\ &=-\frac {49 d^3 (d \tan (e+f x))^{3/2}}{24 a^2 f}+\frac {9 i d^2 (d \tan (e+f x))^{5/2}}{8 a^2 f (1+i \tan (e+f x))}-\frac {d (d \tan (e+f x))^{7/2}}{4 f (a+i a \tan (e+f x))^2}+\frac {\int \sqrt {d \tan (e+f x)} \left (\frac {49 a^2 d^4}{2}-\frac {45}{2} i a^2 d^4 \tan (e+f x)\right ) \, dx}{8 a^4}\\ &=-\frac {45 i d^4 \sqrt {d \tan (e+f x)}}{8 a^2 f}-\frac {49 d^3 (d \tan (e+f x))^{3/2}}{24 a^2 f}+\frac {9 i d^2 (d \tan (e+f x))^{5/2}}{8 a^2 f (1+i \tan (e+f x))}-\frac {d (d \tan (e+f x))^{7/2}}{4 f (a+i a \tan (e+f x))^2}+\frac {\int \frac {\frac {45}{2} i a^2 d^5+\frac {49}{2} a^2 d^5 \tan (e+f x)}{\sqrt {d \tan (e+f x)}} \, dx}{8 a^4}\\ &=-\frac {45 i d^4 \sqrt {d \tan (e+f x)}}{8 a^2 f}-\frac {49 d^3 (d \tan (e+f x))^{3/2}}{24 a^2 f}+\frac {9 i d^2 (d \tan (e+f x))^{5/2}}{8 a^2 f (1+i \tan (e+f x))}-\frac {d (d \tan (e+f x))^{7/2}}{4 f (a+i a \tan (e+f x))^2}+\frac {\operatorname {Subst}\left (\int \frac {\frac {45}{2} i a^2 d^6+\frac {49}{2} a^2 d^5 x^2}{d^2+x^4} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{4 a^4 f}\\ &=-\frac {45 i d^4 \sqrt {d \tan (e+f x)}}{8 a^2 f}-\frac {49 d^3 (d \tan (e+f x))^{3/2}}{24 a^2 f}+\frac {9 i d^2 (d \tan (e+f x))^{5/2}}{8 a^2 f (1+i \tan (e+f x))}-\frac {d (d \tan (e+f x))^{7/2}}{4 f (a+i a \tan (e+f x))^2}+-\frac {\left (\left (\frac {49}{16}-\frac {45 i}{16}\right ) d^5\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 {49}{16}+\frac {45 i}{16}\right ) d^5\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 {45 i d^4 \sqrt {d \tan (e+f x)}}{8 a^2 f}-\frac {49 d^3 (d \tan (e+f x))^{3/2}}{24 a^2 f}+\frac {9 i d^2 (d \tan (e+f x))^{5/2}}{8 a^2 f (1+i \tan (e+f x))}-\frac {d (d \tan (e+f x))^{7/2}}{4 f (a+i a \tan (e+f x))^2}+\frac {\left (\left (\frac {49}{32}-\frac {45 i}{32}\right ) d^{9/2}\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 {49}{32}-\frac {45 i}{32}\right ) d^{9/2}\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 {49}{32}+\frac {45 i}{32}\right ) d^5\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 {49}{32}+\frac {45 i}{32}\right ) d^5\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 {49}{32}-\frac {45 i}{32}\right ) d^{9/2} \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 {49}{32}-\frac {45 i}{32}\right ) d^{9/2} \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 {45 i d^4 \sqrt {d \tan (e+f x)}}{8 a^2 f}-\frac {49 d^3 (d \tan (e+f x))^{3/2}}{24 a^2 f}+\frac {9 i d^2 (d \tan (e+f x))^{5/2}}{8 a^2 f (1+i \tan (e+f x))}-\frac {d (d \tan (e+f x))^{7/2}}{4 f (a+i a \tan (e+f x))^2}+-\frac {\left (\left (\frac {49}{16}+\frac {45 i}{16}\right ) d^{9/2}\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 {49}{16}+\frac {45 i}{16}\right ) d^{9/2}\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 {49}{16}+\frac {45 i}{16}\right ) d^{9/2} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} a^2 f}+\frac {\left (\frac {49}{16}+\frac {45 i}{16}\right ) d^{9/2} \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} a^2 f}+\frac {\left (\frac {49}{32}-\frac {45 i}{32}\right ) d^{9/2} \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 {49}{32}-\frac {45 i}{32}\right ) d^{9/2} \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 {45 i d^4 \sqrt {d \tan (e+f x)}}{8 a^2 f}-\frac {49 d^3 (d \tan (e+f x))^{3/2}}{24 a^2 f}+\frac {9 i d^2 (d \tan (e+f x))^{5/2}}{8 a^2 f (1+i \tan (e+f x))}-\frac {d (d \tan (e+f x))^{7/2}}{4 f (a+i a \tan (e+f x))^2}\\ \end {align*}
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Mathematica [A] time = 2.25, size = 346, normalized size = 0.98 \[ \frac {d^5 \sec ^4(e+f x) \left (142 i \sin (2 (e+f x))+199 i \sin (4 (e+f x))+64 \cos (2 (e+f x))+205 \cos (4 (e+f x))+(294+270 i) \sqrt {\sin (2 (e+f x))} \cos (e+f x) \sin ^{-1}(\cos (e+f x)-\sin (e+f x)) (\cos (2 (e+f x))+i \sin (2 (e+f x)))+(135+147 i) \sqrt {\sin (2 (e+f x))} \sin (3 (e+f x)) \log \left (\sin (e+f x)+\sqrt {\sin (2 (e+f x))}+\cos (e+f x)\right )+(147-135 i) \sqrt {\sin (2 (e+f x))} \cos (e+f x) \log \left (\sin (e+f x)+\sqrt {\sin (2 (e+f x))}+\cos (e+f x)\right )+(147-135 i) \sqrt {\sin (2 (e+f x))} \cos (3 (e+f x)) \log \left (\sin (e+f x)+\sqrt {\sin (2 (e+f x))}+\cos (e+f x)\right )+(135+147 i) \sin (e+f x) \sqrt {\sin (2 (e+f x))} \log \left (\sin (e+f x)+\sqrt {\sin (2 (e+f x))}+\cos (e+f x)\right )-269\right )}{192 a^2 f (\tan (e+f x)-i)^2 \sqrt {d \tan (e+f x)}} \]
Antiderivative was successfully verified.
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fricas [B] time = 0.54, size = 668, normalized size = 1.89 \[ -\frac {12 \, {\left (a^{2} f e^{\left (6 i \, f x + 6 i \, e\right )} + a^{2} f e^{\left (4 i \, f x + 4 i \, e\right )}\right )} \sqrt {-\frac {2209 i \, d^{9}}{64 \, a^{4} f^{2}}} \log \left (-\frac {{\left (47 \, d^{5} + 8 \, {\left (a^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + a^{2} f\right )} \sqrt {-\frac {2209 i \, d^{9}}{64 \, a^{4} f^{2}}} \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 (-2 i \, f x - 2 i \, e\right )}}{8 \, a^{2} f}\right ) - 12 \, {\left (a^{2} f e^{\left (6 i \, f x + 6 i \, e\right )} + a^{2} f e^{\left (4 i \, f x + 4 i \, e\right )}\right )} \sqrt {-\frac {2209 i \, d^{9}}{64 \, a^{4} f^{2}}} \log \left (-\frac {{\left (47 \, d^{5} - 8 \, {\left (a^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + a^{2} f\right )} \sqrt {-\frac {2209 i \, d^{9}}{64 \, a^{4} f^{2}}} \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 (-2 i \, f x - 2 i \, e\right )}}{8 \, a^{2} f}\right ) - 12 \, {\left (a^{2} f e^{\left (6 i \, f x + 6 i \, e\right )} + a^{2} f e^{\left (4 i \, f x + 4 i \, e\right )}\right )} \sqrt {\frac {i \, d^{9}}{16 \, a^{4} f^{2}}} \log \left (\frac {{\left (-2 i \, d^{5} e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (8 i \, a^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + 8 i \, a^{2} f\right )} \sqrt {\frac {i \, d^{9}}{16 \, a^{4} f^{2}}} \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 (-2 i \, f x - 2 i \, e\right )}}{d^{4}}\right ) + 12 \, {\left (a^{2} f e^{\left (6 i \, f x + 6 i \, e\right )} + a^{2} f e^{\left (4 i \, f x + 4 i \, e\right )}\right )} \sqrt {\frac {i \, d^{9}}{16 \, a^{4} f^{2}}} \log \left (\frac {{\left (-2 i \, d^{5} e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (-8 i \, a^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} - 8 i \, a^{2} f\right )} \sqrt {\frac {i \, d^{9}}{16 \, a^{4} f^{2}}} \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 (-2 i \, f x - 2 i \, e\right )}}{d^{4}}\right ) - {\left (-202 i \, d^{4} e^{\left (6 i \, f x + 6 i \, e\right )} - 305 i \, d^{4} e^{\left (4 i \, f x + 4 i \, e\right )} - 36 i \, d^{4} e^{\left (2 i \, f x + 2 i \, e\right )} + 3 i \, d^{4}\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}}}{48 \, {\left (a^{2} f e^{\left (6 i \, f x + 6 i \, e\right )} + a^{2} f e^{\left (4 i \, f x + 4 i \, e\right )}\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 1.52, size = 269, normalized size = 0.76 \[ \frac {1}{24} \, d^{4} {\left (-\frac {141 i \, \sqrt {2} \sqrt {d} \arctan \left (-\frac {16 i \, \sqrt {d^{2}} \sqrt {d \tan \left (f x + e\right )}}{8 i \, \sqrt {2} d^{\frac {3}{2}} + 8 \, \sqrt {2} \sqrt {d^{2}} \sqrt {d}}\right )}{a^{2} f {\left (\frac {i \, d}{\sqrt {d^{2}}} + 1\right )}} - \frac {6 i \, \sqrt {2} \sqrt {d} \arctan \left (-\frac {16 i \, \sqrt {d^{2}} \sqrt {d \tan \left (f x + e\right )}}{-8 i \, \sqrt {2} d^{\frac {3}{2}} + 8 \, \sqrt {2} \sqrt {d^{2}} \sqrt {d}}\right )}{a^{2} f {\left (-\frac {i \, d}{\sqrt {d^{2}}} + 1\right )}} - \frac {3 \, {\left (15 \, \sqrt {d \tan \left (f x + e\right )} d^{2} \tan \left (f x + e\right ) - 13 i \, \sqrt {d \tan \left (f x + e\right )} d^{2}\right )}}{{\left (d \tan \left (f x + e\right ) - i \, d\right )}^{2} a^{2} f} - \frac {16 \, {\left (\sqrt {d \tan \left (f x + e\right )} a^{4} d^{3} f^{2} \tan \left (f x + e\right ) + 6 i \, \sqrt {d \tan \left (f x + e\right )} a^{4} d^{3} f^{2}\right )}}{a^{6} d^{3} f^{3}}\right )} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.28, size = 188, normalized size = 0.53 \[ -\frac {2 d^{3} \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3 a^{2} f}-\frac {4 i d^{4} \sqrt {d \tan \left (f x +e \right )}}{f \,a^{2}}-\frac {15 d^{5} \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{8 f \,a^{2} \left (d \tan \left (f x +e \right )-i d \right )^{2}}+\frac {13 i d^{6} \sqrt {d \tan \left (f x +e \right )}}{8 f \,a^{2} \left (d \tan \left (f x +e \right )-i d \right )^{2}}+\frac {47 d^{5} \arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {-i d}}\right )}{8 f \,a^{2} \sqrt {-i d}}+\frac {d^{5} \arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {i d}}\right )}{4 f \,a^{2} \sqrt {i d}} \]
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 = 5.64, size = 225, normalized size = 0.64 \[ -\frac {-\frac {15\,d^5\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}}{8\,a^2\,f}+\frac {d^6\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,13{}\mathrm {i}}{8\,a^2\,f}}{-d^2\,{\mathrm {tan}\left (e+f\,x\right )}^2+d^2\,\mathrm {tan}\left (e+f\,x\right )\,2{}\mathrm {i}+d^2}-\frac {2\,d^3\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}}{3\,a^2\,f}+\mathrm {atan}\left (\frac {a^2\,f\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,\sqrt {\frac {d^9\,1{}\mathrm {i}}{64\,a^4\,f^2}}\,8{}\mathrm {i}}{d^5}\right )\,\sqrt {\frac {d^9\,1{}\mathrm {i}}{64\,a^4\,f^2}}\,2{}\mathrm {i}+\mathrm {atan}\left (\frac {a^2\,f\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,\sqrt {-\frac {d^9\,2209{}\mathrm {i}}{256\,a^4\,f^2}}\,16{}\mathrm {i}}{47\,d^5}\right )\,\sqrt {-\frac {d^9\,2209{}\mathrm {i}}{256\,a^4\,f^2}}\,2{}\mathrm {i}-\frac {d^4\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,4{}\mathrm {i}}{a^2\,f} \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
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