Optimal. Leaf size=329 \[ \frac {d^{5/2} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{8 \sqrt {2} a^3 f}-\frac {d^{5/2} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}+1\right )}{8 \sqrt {2} a^3 f}-\frac {d^{5/2} \log \left (\sqrt {d} \tan (e+f x)-\sqrt {2} \sqrt {d \tan (e+f x)}+\sqrt {d}\right )}{16 \sqrt {2} a^3 f}+\frac {d^{5/2} \log \left (\sqrt {d} \tan (e+f x)+\sqrt {2} \sqrt {d \tan (e+f x)}+\sqrt {d}\right )}{16 \sqrt {2} a^3 f}-\frac {i d^2 \sqrt {d \tan (e+f x)}}{4 f \left (a^3+i a^3 \tan (e+f x)\right )}+\frac {i d^2 \sqrt {d \tan (e+f x)}}{4 a f (a+i a \tan (e+f x))^2}-\frac {d (d \tan (e+f x))^{3/2}}{6 f (a+i a \tan (e+f x))^3} \]
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Rubi [A] time = 0.60, antiderivative size = 329, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 13, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.464, Rules used = {3558, 3595, 3596, 12, 16, 3476, 329, 297, 1162, 617, 204, 1165, 628} \[ \frac {d^{5/2} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{8 \sqrt {2} a^3 f}-\frac {d^{5/2} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}+1\right )}{8 \sqrt {2} a^3 f}-\frac {i d^2 \sqrt {d \tan (e+f x)}}{4 f \left (a^3+i a^3 \tan (e+f x)\right )}-\frac {d^{5/2} \log \left (\sqrt {d} \tan (e+f x)-\sqrt {2} \sqrt {d \tan (e+f x)}+\sqrt {d}\right )}{16 \sqrt {2} a^3 f}+\frac {d^{5/2} \log \left (\sqrt {d} \tan (e+f x)+\sqrt {2} \sqrt {d \tan (e+f x)}+\sqrt {d}\right )}{16 \sqrt {2} a^3 f}+\frac {i d^2 \sqrt {d \tan (e+f x)}}{4 a f (a+i a \tan (e+f x))^2}-\frac {d (d \tan (e+f x))^{3/2}}{6 f (a+i a \tan (e+f x))^3} \]
Antiderivative was successfully verified.
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Rule 12
Rule 16
Rule 204
Rule 297
Rule 329
Rule 617
Rule 628
Rule 1162
Rule 1165
Rule 3476
Rule 3558
Rule 3595
Rule 3596
Rubi steps
\begin {align*} \int \frac {(d \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^3} \, dx &=-\frac {d (d \tan (e+f x))^{3/2}}{6 f (a+i a \tan (e+f x))^3}-\frac {\int \frac {\sqrt {d \tan (e+f x)} \left (-\frac {3 a d^2}{2}+\frac {9}{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))^{3/2}}{6 f (a+i a \tan (e+f x))^3}+\frac {i d^2 \sqrt {d \tan (e+f x)}}{4 a f (a+i a \tan (e+f x))^2}+\frac {\int \frac {-3 i a^2 d^3-9 a^2 d^3 \tan (e+f x)}{\sqrt {d \tan (e+f x)} (a+i a \tan (e+f x))} \, dx}{24 a^4}\\ &=-\frac {d (d \tan (e+f x))^{3/2}}{6 f (a+i a \tan (e+f x))^3}+\frac {i d^2 \sqrt {d \tan (e+f x)}}{4 a f (a+i a \tan (e+f x))^2}-\frac {i d^2 \sqrt {d \tan (e+f x)}}{4 f \left (a^3+i a^3 \tan (e+f x)\right )}+\frac {\int -\frac {6 a^3 d^4 \tan (e+f x)}{\sqrt {d \tan (e+f x)}} \, dx}{48 a^6 d}\\ &=-\frac {d (d \tan (e+f x))^{3/2}}{6 f (a+i a \tan (e+f x))^3}+\frac {i d^2 \sqrt {d \tan (e+f x)}}{4 a f (a+i a \tan (e+f x))^2}-\frac {i d^2 \sqrt {d \tan (e+f x)}}{4 f \left (a^3+i a^3 \tan (e+f x)\right )}-\frac {d^3 \int \frac {\tan (e+f x)}{\sqrt {d \tan (e+f x)}} \, dx}{8 a^3}\\ &=-\frac {d (d \tan (e+f x))^{3/2}}{6 f (a+i a \tan (e+f x))^3}+\frac {i d^2 \sqrt {d \tan (e+f x)}}{4 a f (a+i a \tan (e+f x))^2}-\frac {i d^2 \sqrt {d \tan (e+f x)}}{4 f \left (a^3+i a^3 \tan (e+f x)\right )}-\frac {d^2 \int \sqrt {d \tan (e+f x)} \, dx}{8 a^3}\\ &=-\frac {d (d \tan (e+f x))^{3/2}}{6 f (a+i a \tan (e+f x))^3}+\frac {i d^2 \sqrt {d \tan (e+f x)}}{4 a f (a+i a \tan (e+f x))^2}-\frac {i d^2 \sqrt {d \tan (e+f x)}}{4 f \left (a^3+i a^3 \tan (e+f x)\right )}-\frac {d^3 \operatorname {Subst}\left (\int \frac {\sqrt {x}}{d^2+x^2} \, dx,x,d \tan (e+f x)\right )}{8 a^3 f}\\ &=-\frac {d (d \tan (e+f x))^{3/2}}{6 f (a+i a \tan (e+f x))^3}+\frac {i d^2 \sqrt {d \tan (e+f x)}}{4 a f (a+i a \tan (e+f x))^2}-\frac {i d^2 \sqrt {d \tan (e+f x)}}{4 f \left (a^3+i a^3 \tan (e+f x)\right )}-\frac {d^3 \operatorname {Subst}\left (\int \frac {x^2}{d^2+x^4} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{4 a^3 f}\\ &=-\frac {d (d \tan (e+f x))^{3/2}}{6 f (a+i a \tan (e+f x))^3}+\frac {i d^2 \sqrt {d \tan (e+f x)}}{4 a f (a+i a \tan (e+f x))^2}-\frac {i d^2 \sqrt {d \tan (e+f x)}}{4 f \left (a^3+i a^3 \tan (e+f x)\right )}+\frac {d^3 \operatorname {Subst}\left (\int \frac {d-x^2}{d^2+x^4} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{8 a^3 f}-\frac {d^3 \operatorname {Subst}\left (\int \frac {d+x^2}{d^2+x^4} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{8 a^3 f}\\ &=-\frac {d (d \tan (e+f x))^{3/2}}{6 f (a+i a \tan (e+f x))^3}+\frac {i d^2 \sqrt {d \tan (e+f x)}}{4 a f (a+i a \tan (e+f x))^2}-\frac {i d^2 \sqrt {d \tan (e+f x)}}{4 f \left (a^3+i a^3 \tan (e+f x)\right )}-\frac {d^{5/2} \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 )}{16 \sqrt {2} a^3 f}-\frac {d^{5/2} \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 )}{16 \sqrt {2} a^3 f}-\frac {d^3 \operatorname {Subst}\left (\int \frac {1}{d-\sqrt {2} \sqrt {d} x+x^2} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{16 a^3 f}-\frac {d^3 \operatorname {Subst}\left (\int \frac {1}{d+\sqrt {2} \sqrt {d} x+x^2} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{16 a^3 f}\\ &=-\frac {d^{5/2} \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)-\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{16 \sqrt {2} a^3 f}+\frac {d^{5/2} \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)+\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{16 \sqrt {2} a^3 f}-\frac {d (d \tan (e+f x))^{3/2}}{6 f (a+i a \tan (e+f x))^3}+\frac {i d^2 \sqrt {d \tan (e+f x)}}{4 a f (a+i a \tan (e+f x))^2}-\frac {i d^2 \sqrt {d \tan (e+f x)}}{4 f \left (a^3+i a^3 \tan (e+f x)\right )}-\frac {d^{5/2} \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{8 \sqrt {2} a^3 f}+\frac {d^{5/2} \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{8 \sqrt {2} a^3 f}\\ &=\frac {d^{5/2} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{8 \sqrt {2} a^3 f}-\frac {d^{5/2} \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{8 \sqrt {2} a^3 f}-\frac {d^{5/2} \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)-\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{16 \sqrt {2} a^3 f}+\frac {d^{5/2} \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)+\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{16 \sqrt {2} a^3 f}-\frac {d (d \tan (e+f x))^{3/2}}{6 f (a+i a \tan (e+f x))^3}+\frac {i d^2 \sqrt {d \tan (e+f x)}}{4 a f (a+i a \tan (e+f x))^2}-\frac {i d^2 \sqrt {d \tan (e+f x)}}{4 f \left (a^3+i a^3 \tan (e+f x)\right )}\\ \end {align*}
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Mathematica [A] time = 1.09, size = 232, normalized size = 0.71 \[ \frac {d^3 \sec ^4(e+f x) \left (-6 \sin (2 (e+f x))+3 \sin (4 (e+f x))-i \cos (4 (e+f x))+6 i \sqrt {\sin (2 (e+f x))} \sin ^{-1}(\cos (e+f x)-\sin (e+f x)) (\cos (3 (e+f x))+i \sin (3 (e+f x)))-6 \sqrt {\sin (2 (e+f x))} \sin (3 (e+f x)) \log \left (\sin (e+f x)+\sqrt {\sin (2 (e+f x))}+\cos (e+f x)\right )+6 i \sqrt {\sin (2 (e+f x))} \cos (3 (e+f x)) \log \left (\sin (e+f x)+\sqrt {\sin (2 (e+f x))}+\cos (e+f x)\right )+i\right )}{96 a^3 f (\tan (e+f x)-i)^3 \sqrt {d \tan (e+f x)}} \]
Antiderivative was successfully verified.
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fricas [B] time = 0.53, size = 584, normalized size = 1.78 \[ -\frac {{\left (12 \, a^{3} f \sqrt {\frac {i \, d^{5}}{64 \, a^{6} f^{2}}} e^{\left (6 i \, f x + 6 i \, e\right )} \log \left (\frac {{\left (-2 i \, d^{3} e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (16 i \, a^{3} f e^{\left (2 i \, f x + 2 i \, e\right )} + 16 i \, 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^{5}}{64 \, a^{6} f^{2}}}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{d^{2}}\right ) - 12 \, a^{3} f \sqrt {\frac {i \, d^{5}}{64 \, a^{6} f^{2}}} e^{\left (6 i \, f x + 6 i \, e\right )} \log \left (\frac {{\left (-2 i \, d^{3} e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (-16 i \, a^{3} f e^{\left (2 i \, f x + 2 i \, e\right )} - 16 i \, 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^{5}}{64 \, a^{6} f^{2}}}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{d^{2}}\right ) - 12 \, a^{3} f \sqrt {-\frac {i \, d^{5}}{64 \, a^{6} f^{2}}} e^{\left (6 i \, f x + 6 i \, e\right )} \log \left (\frac {{\left (d^{3} + 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^{5}}{64 \, a^{6} f^{2}}}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{8 \, a^{3} f}\right ) + 12 \, a^{3} f \sqrt {-\frac {i \, d^{5}}{64 \, a^{6} f^{2}}} e^{\left (6 i \, f x + 6 i \, e\right )} \log \left (\frac {{\left (d^{3} - 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^{5}}{64 \, a^{6} f^{2}}}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{8 \, a^{3} f}\right ) - {\left (-2 i \, d^{2} e^{\left (6 i \, f x + 6 i \, e\right )} + i \, d^{2} e^{\left (4 i \, f x + 4 i \, e\right )} + 2 i \, d^{2} e^{\left (2 i \, f x + 2 i \, e\right )} - i \, d^{2}\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} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 3.23, size = 216, normalized size = 0.66 \[ -\frac {1}{24} \, d^{2} {\left (\frac {3 \, \sqrt {2} \sqrt {d} \arctan \left (\frac {16 \, \sqrt {d^{2}} \sqrt {d \tan \left (f x + e\right )}}{8 i \, \sqrt {2} d^{\frac {3}{2}} + 8 \, \sqrt {2} \sqrt {d^{2}} \sqrt {d}}\right )}{a^{3} f {\left (\frac {i \, d}{\sqrt {d^{2}}} + 1\right )}} + \frac {3 \, \sqrt {2} \sqrt {d} \arctan \left (\frac {16 \, \sqrt {d^{2}} \sqrt {d \tan \left (f x + e\right )}}{-8 i \, \sqrt {2} d^{\frac {3}{2}} + 8 \, \sqrt {2} \sqrt {d^{2}} \sqrt {d}}\right )}{a^{3} f {\left (-\frac {i \, d}{\sqrt {d^{2}}} + 1\right )}} + \frac {2 \, {\left (3 \, \sqrt {d \tan \left (f x + e\right )} d^{3} \tan \left (f x + e\right )^{2} - i \, \sqrt {d \tan \left (f x + e\right )} d^{3} \tan \left (f x + e\right )\right )}}{{\left (d \tan \left (f x + e\right ) - i \, d\right )}^{3} a^{3} f}\right )} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.35, size = 145, normalized size = 0.44 \[ -\frac {d^{3} \left (d \tan \left (f x +e \right )\right )^{\frac {5}{2}}}{4 f \,a^{3} \left (d \tan \left (f x +e \right )-i d \right )^{3}}+\frac {i d^{4} \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{12 f \,a^{3} \left (d \tan \left (f x +e \right )-i d \right )^{3}}-\frac {d^{3} \arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {-i d}}\right )}{8 f \,a^{3} \sqrt {-i d}}-\frac {d^{3} \arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {i d}}\right )}{8 f \,a^{3} \sqrt {i d}} \]
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 = 4.16, size = 158, normalized size = 0.48 \[ \frac {{\left (-1\right )}^{1/4}\,d^{5/2}\,\mathrm {atanh}\left (\frac {{\left (-1\right )}^{1/4}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{\sqrt {d}}\right )}{8\,a^3\,f}-\frac {{\left (-1\right )}^{1/4}\,d^{5/2}\,\mathrm {atan}\left (\frac {{\left (-1\right )}^{1/4}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{\sqrt {d}}\right )}{8\,a^3\,f}-\frac {-\frac {d^3\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{5/2}}{4\,a^3\,f}+\frac {d^4\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}\,1{}\mathrm {i}}{12\,a^3\,f}}{-d^3\,{\mathrm {tan}\left (e+f\,x\right )}^3+d^3\,{\mathrm {tan}\left (e+f\,x\right )}^2\,3{}\mathrm {i}+3\,d^3\,\mathrm {tan}\left (e+f\,x\right )-d^3\,1{}\mathrm {i}} \]
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 \[ \frac {i \int \frac {\left (d \tan {\left (e + f x \right )}\right )^{\frac {5}{2}}}{\tan ^{3}{\left (e + f x \right )} - 3 i \tan ^{2}{\left (e + f x \right )} - 3 \tan {\left (e + f x \right )} + i}\, dx}{a^{3}} \]
Verification of antiderivative is not currently implemented for this CAS.
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