Optimal. Leaf size=343 \[ -\frac {\left (\frac {7}{16}-\frac {5 i}{16}\right ) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} a^3 \sqrt {d} f}+\frac {\left (\frac {7}{16}-\frac {5 i}{16}\right ) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}+1\right )}{\sqrt {2} a^3 \sqrt {d} f}+\frac {5 \sqrt {d \tan (e+f x)}}{8 d f \left (a^3+i a^3 \tan (e+f x)\right )}-\frac {\left (\frac {7}{32}+\frac {5 i}{32}\right ) \log \left (\sqrt {d} \tan (e+f x)-\sqrt {2} \sqrt {d \tan (e+f x)}+\sqrt {d}\right )}{\sqrt {2} a^3 \sqrt {d} f}+\frac {\left (\frac {7}{32}+\frac {5 i}{32}\right ) \log \left (\sqrt {d} \tan (e+f x)+\sqrt {2} \sqrt {d \tan (e+f x)}+\sqrt {d}\right )}{\sqrt {2} a^3 \sqrt {d} f}+\frac {\sqrt {d \tan (e+f x)}}{3 a d f (a+i a \tan (e+f x))^2}+\frac {\sqrt {d \tan (e+f x)}}{6 d f (a+i a \tan (e+f x))^3} \]
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Rubi [A] time = 0.59, 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 = {3559, 3596, 3534, 1168, 1162, 617, 204, 1165, 628} \[ -\frac {\left (\frac {7}{16}-\frac {5 i}{16}\right ) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} a^3 \sqrt {d} f}+\frac {\left (\frac {7}{16}-\frac {5 i}{16}\right ) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}+1\right )}{\sqrt {2} a^3 \sqrt {d} f}+\frac {5 \sqrt {d \tan (e+f x)}}{8 d f \left (a^3+i a^3 \tan (e+f x)\right )}-\frac {\left (\frac {7}{32}+\frac {5 i}{32}\right ) \log \left (\sqrt {d} \tan (e+f x)-\sqrt {2} \sqrt {d \tan (e+f x)}+\sqrt {d}\right )}{\sqrt {2} a^3 \sqrt {d} f}+\frac {\left (\frac {7}{32}+\frac {5 i}{32}\right ) \log \left (\sqrt {d} \tan (e+f x)+\sqrt {2} \sqrt {d \tan (e+f x)}+\sqrt {d}\right )}{\sqrt {2} a^3 \sqrt {d} f}+\frac {\sqrt {d \tan (e+f x)}}{3 a d f (a+i a \tan (e+f x))^2}+\frac {\sqrt {d \tan (e+f x)}}{6 d f (a+i a \tan (e+f x))^3} \]
Antiderivative was successfully verified.
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Rule 204
Rule 617
Rule 628
Rule 1162
Rule 1165
Rule 1168
Rule 3534
Rule 3559
Rule 3596
Rubi steps
\begin {align*} \int \frac {1}{\sqrt {d \tan (e+f x)} (a+i a \tan (e+f x))^3} \, dx &=\frac {\sqrt {d \tan (e+f x)}}{6 d f (a+i a \tan (e+f x))^3}+\frac {\int \frac {\frac {11 a d}{2}-\frac {5}{2} i a d \tan (e+f x)}{\sqrt {d \tan (e+f x)} (a+i a \tan (e+f x))^2} \, dx}{6 a^2 d}\\ &=\frac {\sqrt {d \tan (e+f x)}}{6 d f (a+i a \tan (e+f x))^3}+\frac {\sqrt {d \tan (e+f x)}}{3 a d f (a+i a \tan (e+f x))^2}+\frac {\int \frac {18 a^2 d^2-12 i a^2 d^2 \tan (e+f x)}{\sqrt {d \tan (e+f x)} (a+i a \tan (e+f x))} \, dx}{24 a^4 d^2}\\ &=\frac {\sqrt {d \tan (e+f x)}}{6 d f (a+i a \tan (e+f x))^3}+\frac {\sqrt {d \tan (e+f x)}}{3 a d f (a+i a \tan (e+f x))^2}+\frac {5 \sqrt {d \tan (e+f x)}}{8 d f \left (a^3+i a^3 \tan (e+f x)\right )}+\frac {\int \frac {21 a^3 d^3-15 i a^3 d^3 \tan (e+f x)}{\sqrt {d \tan (e+f x)}} \, dx}{48 a^6 d^3}\\ &=\frac {\sqrt {d \tan (e+f x)}}{6 d f (a+i a \tan (e+f x))^3}+\frac {\sqrt {d \tan (e+f x)}}{3 a d f (a+i a \tan (e+f x))^2}+\frac {5 \sqrt {d \tan (e+f x)}}{8 d f \left (a^3+i a^3 \tan (e+f x)\right )}+\frac {\operatorname {Subst}\left (\int \frac {21 a^3 d^4-15 i a^3 d^3 x^2}{d^2+x^4} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{24 a^6 d^3 f}\\ &=\frac {\sqrt {d \tan (e+f x)}}{6 d f (a+i a \tan (e+f x))^3}+\frac {\sqrt {d \tan (e+f x)}}{3 a d f (a+i a \tan (e+f x))^2}+\frac {5 \sqrt {d \tan (e+f x)}}{8 d f \left (a^3+i a^3 \tan (e+f x)\right )}+\frac {\left (\frac {7}{16}-\frac {5 i}{16}\right ) \operatorname {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 (\frac {7}{16}+\frac {5 i}{16}\right ) \operatorname {Subst}\left (\int \frac {d-x^2}{d^2+x^4} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{a^3 f}\\ &=\frac {\sqrt {d \tan (e+f x)}}{6 d f (a+i a \tan (e+f x))^3}+\frac {\sqrt {d \tan (e+f x)}}{3 a d f (a+i a \tan (e+f x))^2}+\frac {5 \sqrt {d \tan (e+f x)}}{8 d f \left (a^3+i a^3 \tan (e+f x)\right )}+\frac {\left (\frac {7}{32}-\frac {5 i}{32}\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^3 f}+\frac {\left (\frac {7}{32}-\frac {5 i}{32}\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^3 f}+-\frac {\left (\frac {7}{32}+\frac {5 i}{32}\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^3 \sqrt {d} f}+-\frac {\left (\frac {7}{32}+\frac {5 i}{32}\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^3 \sqrt {d} f}\\ &=-\frac {\left (\frac {7}{32}+\frac {5 i}{32}\right ) \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)-\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{\sqrt {2} a^3 \sqrt {d} f}+\frac {\left (\frac {7}{32}+\frac {5 i}{32}\right ) \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)+\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{\sqrt {2} a^3 \sqrt {d} f}+\frac {\sqrt {d \tan (e+f x)}}{6 d f (a+i a \tan (e+f x))^3}+\frac {\sqrt {d \tan (e+f x)}}{3 a d f (a+i a \tan (e+f x))^2}+\frac {5 \sqrt {d \tan (e+f x)}}{8 d f \left (a^3+i a^3 \tan (e+f x)\right )}+-\frac {\left (\frac {7}{16}-\frac {5 i}{16}\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^3 \sqrt {d} f}+\frac {\left (\frac {7}{16}-\frac {5 i}{16}\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^3 \sqrt {d} f}\\ &=-\frac {\left (\frac {7}{16}-\frac {5 i}{16}\right ) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} a^3 \sqrt {d} f}+\frac {\left (\frac {7}{16}-\frac {5 i}{16}\right ) \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} a^3 \sqrt {d} f}-\frac {\left (\frac {7}{32}+\frac {5 i}{32}\right ) \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)-\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{\sqrt {2} a^3 \sqrt {d} f}+\frac {\left (\frac {7}{32}+\frac {5 i}{32}\right ) \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)+\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{\sqrt {2} a^3 \sqrt {d} f}+\frac {\sqrt {d \tan (e+f x)}}{6 d f (a+i a \tan (e+f x))^3}+\frac {\sqrt {d \tan (e+f x)}}{3 a d f (a+i a \tan (e+f x))^2}+\frac {5 \sqrt {d \tan (e+f x)}}{8 d f \left (a^3+i a^3 \tan (e+f x)\right )}\\ \end {align*}
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Mathematica [A] time = 1.23, size = 234, normalized size = 0.68 \[ \frac {\sec ^4(e+f x) \left (19 \cos (4 (e+f x))-(15+21 i) \sqrt {\sin (2 (e+f x))} \sin ^{-1}(\cos (e+f x)-\sin (e+f x)) (\cos (3 (e+f x))+i \sin (3 (e+f x)))+i \left (12 \sin (2 (e+f x))+21 \sin (4 (e+f x))+(-15+21 i) \sqrt {\sin (2 (e+f x))} \sin (3 (e+f x)) \log \left (\sin (e+f x)+\sqrt {\sin (2 (e+f x))}+\cos (e+f x)\right )+(21+15 i) \sqrt {\sin (2 (e+f x))} \cos (3 (e+f x)) \log \left (\sin (e+f x)+\sqrt {\sin (2 (e+f x))}+\cos (e+f x)\right )+19 i\right )\right )}{96 a^3 f (\tan (e+f x)-i)^3 \sqrt {d \tan (e+f x)}} \]
Antiderivative was successfully verified.
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fricas [B] time = 0.59, size = 568, normalized size = 1.66 \[ -\frac {{\left (12 \, a^{3} d f \sqrt {-\frac {i}{64 \, a^{6} d f^{2}}} e^{\left (6 i \, f x + 6 i \, e\right )} \log \left (-2 \, {\left (8 \, {\left (a^{3} d f e^{\left (2 i \, f x + 2 i \, e\right )} + a^{3} d 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}{64 \, a^{6} d f^{2}}} + i \, d e^{\left (2 i \, f x + 2 i \, e\right )}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}\right ) - 12 \, a^{3} d f \sqrt {-\frac {i}{64 \, a^{6} d f^{2}}} e^{\left (6 i \, f x + 6 i \, e\right )} \log \left (2 \, {\left (8 \, {\left (a^{3} d f e^{\left (2 i \, f x + 2 i \, e\right )} + a^{3} d 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}{64 \, a^{6} d f^{2}}} - i \, d e^{\left (2 i \, f x + 2 i \, e\right )}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}\right ) - 12 \, a^{3} d f \sqrt {\frac {9 i}{16 \, a^{6} d f^{2}}} e^{\left (6 i \, f x + 6 i \, e\right )} \log \left (\frac {{\left (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}{16 \, a^{6} d f^{2}}} + 3 i\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{4 \, a^{3} f}\right ) + 12 \, a^{3} d f \sqrt {\frac {9 i}{16 \, a^{6} d f^{2}}} e^{\left (6 i \, f x + 6 i \, e\right )} \log \left (-\frac {{\left (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}{16 \, a^{6} d f^{2}}} - 3 i\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{4 \, 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}} {\left (20 \, e^{\left (6 i \, f x + 6 i \, e\right )} + 26 \, e^{\left (4 i \, f x + 4 i \, e\right )} + 7 \, e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right )}\right )} e^{\left (-6 i \, f x - 6 i \, e\right )}}{48 \, a^{3} d f} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 5.05, size = 227, normalized size = 0.66 \[ \frac {i \, \sqrt {2} \arctan \left (\frac {16 \, \sqrt {d^{2}} \sqrt {d \tan \left (f x + e\right )}}{8 i \, \sqrt {2} d^{\frac {3}{2}} + 8 \, \sqrt {2} \sqrt {d^{2}} \sqrt {d}}\right )}{8 \, a^{3} \sqrt {d} f {\left (\frac {i \, d}{\sqrt {d^{2}}} + 1\right )}} - \frac {3 i \, \sqrt {2} \arctan \left (\frac {16 \, \sqrt {d^{2}} \sqrt {d \tan \left (f x + e\right )}}{-8 i \, \sqrt {2} d^{\frac {3}{2}} + 8 \, \sqrt {2} \sqrt {d^{2}} \sqrt {d}}\right )}{4 \, a^{3} \sqrt {d} f {\left (-\frac {i \, d}{\sqrt {d^{2}}} + 1\right )}} - \frac {15 i \, \sqrt {d \tan \left (f x + e\right )} d^{2} \tan \left (f x + e\right )^{2} + 38 \, \sqrt {d \tan \left (f x + e\right )} d^{2} \tan \left (f x + e\right ) - 27 i \, \sqrt {d \tan \left (f x + e\right )} d^{2}}{24 \, {\left (d \tan \left (f x + e\right ) - i \, d\right )}^{3} a^{3} f} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.45, size = 173, normalized size = 0.50 \[ -\frac {5 i \left (d \tan \left (f x +e \right )\right )^{\frac {5}{2}}}{8 f \,a^{3} \left (d \tan \left (f x +e \right )-i d \right )^{3}}-\frac {19 d \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{12 f \,a^{3} \left (d \tan \left (f x +e \right )-i d \right )^{3}}+\frac {9 i d^{2} \sqrt {d \tan \left (f x +e \right )}}{8 f \,a^{3} \left (d \tan \left (f x +e \right )-i d \right )^{3}}-\frac {3 i \arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {-i d}}\right )}{4 f \,a^{3} \sqrt {-i d}}+\frac {i \arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {i d}}\right )}{8 f \,a^{3} \sqrt {i d}} \]
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 = 5.68, size = 206, normalized size = 0.60 \[ \frac {\frac {19\,d\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}}{12\,a^3\,f}+\frac {{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{5/2}\,5{}\mathrm {i}}{8\,a^3\,f}-\frac {d^2\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,9{}\mathrm {i}}{8\,a^3\,f}}{-d^3\,{\mathrm {tan}\left (e+f\,x\right )}^3+d^3\,{\mathrm {tan}\left (e+f\,x\right )}^2\,3{}\mathrm {i}+3\,d^3\,\mathrm {tan}\left (e+f\,x\right )-d^3\,1{}\mathrm {i}}-\mathrm {atan}\left (\frac {8\,a^3\,f\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,\sqrt {\frac {9{}\mathrm {i}}{64\,a^6\,d\,f^2}}}{3}\right )\,\sqrt {\frac {9{}\mathrm {i}}{64\,a^6\,d\,f^2}}\,2{}\mathrm {i}+\mathrm {atan}\left (16\,a^3\,f\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,\sqrt {-\frac {1{}\mathrm {i}}{256\,a^6\,d\,f^2}}\right )\,\sqrt {-\frac {1{}\mathrm {i}}{256\,a^6\,d\,f^2}}\,2{}\mathrm {i} \]
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 \[ \frac {i \int \frac {1}{\sqrt {d \tan {\left (e + f x \right )}} \tan ^{3}{\left (e + f x \right )} - 3 i \sqrt {d \tan {\left (e + f x \right )}} \tan ^{2}{\left (e + f x \right )} - 3 \sqrt {d \tan {\left (e + f x \right )}} \tan {\left (e + f x \right )} + i \sqrt {d \tan {\left (e + f x \right )}}}\, dx}{a^{3}} \]
Verification of antiderivative is not currently implemented for this CAS.
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