Optimal. Leaf size=99 \[ -\frac{\cos ^6(e+f x) (B-A \tan (e+f x))}{6 a^3 c^3 f}+\frac{5 A \sin (e+f x) \cos ^3(e+f x)}{24 a^3 c^3 f}+\frac{5 A \sin (e+f x) \cos (e+f x)}{16 a^3 c^3 f}+\frac{5 A x}{16 a^3 c^3} \]
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Rubi [A] time = 0.14586, antiderivative size = 99, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 41, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.122, Rules used = {3588, 73, 639, 199, 205} \[ -\frac{\cos ^6(e+f x) (B-A \tan (e+f x))}{6 a^3 c^3 f}+\frac{5 A \sin (e+f x) \cos ^3(e+f x)}{24 a^3 c^3 f}+\frac{5 A \sin (e+f x) \cos (e+f x)}{16 a^3 c^3 f}+\frac{5 A x}{16 a^3 c^3} \]
Antiderivative was successfully verified.
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Rule 3588
Rule 73
Rule 639
Rule 199
Rule 205
Rubi steps
\begin{align*} \int \frac{A+B \tan (e+f x)}{(a+i a \tan (e+f x))^3 (c-i c \tan (e+f x))^3} \, dx &=\frac{(a c) \operatorname{Subst}\left (\int \frac{A+B x}{(a+i a x)^4 (c-i c x)^4} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{(a c) \operatorname{Subst}\left (\int \frac{A+B x}{\left (a c+a c x^2\right )^4} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac{\cos ^6(e+f x) (B-A \tan (e+f x))}{6 a^3 c^3 f}+\frac{(5 A) \operatorname{Subst}\left (\int \frac{1}{\left (a c+a c x^2\right )^3} \, dx,x,\tan (e+f x)\right )}{6 f}\\ &=\frac{5 A \cos ^3(e+f x) \sin (e+f x)}{24 a^3 c^3 f}-\frac{\cos ^6(e+f x) (B-A \tan (e+f x))}{6 a^3 c^3 f}+\frac{(5 A) \operatorname{Subst}\left (\int \frac{1}{\left (a c+a c x^2\right )^2} \, dx,x,\tan (e+f x)\right )}{8 a c f}\\ &=\frac{5 A \cos (e+f x) \sin (e+f x)}{16 a^3 c^3 f}+\frac{5 A \cos ^3(e+f x) \sin (e+f x)}{24 a^3 c^3 f}-\frac{\cos ^6(e+f x) (B-A \tan (e+f x))}{6 a^3 c^3 f}+\frac{(5 A) \operatorname{Subst}\left (\int \frac{1}{a c+a c x^2} \, dx,x,\tan (e+f x)\right )}{16 a^2 c^2 f}\\ &=\frac{5 A x}{16 a^3 c^3}+\frac{5 A \cos (e+f x) \sin (e+f x)}{16 a^3 c^3 f}+\frac{5 A \cos ^3(e+f x) \sin (e+f x)}{24 a^3 c^3 f}-\frac{\cos ^6(e+f x) (B-A \tan (e+f x))}{6 a^3 c^3 f}\\ \end{align*}
Mathematica [A] time = 0.144557, size = 63, normalized size = 0.64 \[ \frac{A (45 \sin (2 (e+f x))+9 \sin (4 (e+f x))+\sin (6 (e+f x))+60 e+60 f x)-32 B \cos ^6(e+f x)}{192 a^3 c^3 f} \]
Antiderivative was successfully verified.
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Maple [C] time = 0.07, size = 330, normalized size = 3.3 \begin{align*}{\frac{-{\frac{5\,i}{32}}A\ln \left ( \tan \left ( fx+e \right ) -i \right ) }{f{a}^{3}{c}^{3}}}+{\frac{5\,A}{32\,f{a}^{3}{c}^{3} \left ( \tan \left ( fx+e \right ) -i \right ) }}+{\frac{{\frac{i}{32}}B}{f{a}^{3}{c}^{3} \left ( \tan \left ( fx+e \right ) -i \right ) }}-{\frac{{\frac{i}{48}}B}{f{a}^{3}{c}^{3} \left ( \tan \left ( fx+e \right ) -i \right ) ^{3}}}-{\frac{A}{48\,f{a}^{3}{c}^{3} \left ( \tan \left ( fx+e \right ) -i \right ) ^{3}}}+{\frac{B}{32\,f{a}^{3}{c}^{3} \left ( \tan \left ( fx+e \right ) -i \right ) ^{2}}}-{\frac{{\frac{i}{16}}A}{f{a}^{3}{c}^{3} \left ( \tan \left ( fx+e \right ) -i \right ) ^{2}}}+{\frac{{\frac{5\,i}{32}}A\ln \left ( \tan \left ( fx+e \right ) +i \right ) }{f{a}^{3}{c}^{3}}}+{\frac{5\,A}{32\,f{a}^{3}{c}^{3} \left ( \tan \left ( fx+e \right ) +i \right ) }}-{\frac{{\frac{i}{32}}B}{f{a}^{3}{c}^{3} \left ( \tan \left ( fx+e \right ) +i \right ) }}-{\frac{A}{48\,f{a}^{3}{c}^{3} \left ( \tan \left ( fx+e \right ) +i \right ) ^{3}}}+{\frac{{\frac{i}{48}}B}{f{a}^{3}{c}^{3} \left ( \tan \left ( fx+e \right ) +i \right ) ^{3}}}+{\frac{B}{32\,f{a}^{3}{c}^{3} \left ( \tan \left ( fx+e \right ) +i \right ) ^{2}}}+{\frac{{\frac{i}{16}}A}{f{a}^{3}{c}^{3} \left ( \tan \left ( fx+e \right ) +i \right ) ^{2}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: RuntimeError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [C] time = 1.10016, size = 363, normalized size = 3.67 \begin{align*} \frac{{\left (120 \, A f x e^{\left (6 i \, f x + 6 i \, e\right )} +{\left (-i \, A - B\right )} e^{\left (12 i \, f x + 12 i \, e\right )} +{\left (-9 i \, A - 6 \, B\right )} e^{\left (10 i \, f x + 10 i \, e\right )} +{\left (-45 i \, A - 15 \, B\right )} e^{\left (8 i \, f x + 8 i \, e\right )} +{\left (45 i \, A - 15 \, B\right )} e^{\left (4 i \, f x + 4 i \, e\right )} +{\left (9 i \, A - 6 \, B\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + i \, A - B\right )} e^{\left (-6 i \, f x - 6 i \, e\right )}}{384 \, a^{3} c^{3} f} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [A] time = 5.92886, size = 510, normalized size = 5.15 \begin{align*} \frac{5 A x}{16 a^{3} c^{3}} + \begin{cases} \frac{\left (\left (103079215104 i A a^{15} c^{15} f^{5} e^{6 i e} - 103079215104 B a^{15} c^{15} f^{5} e^{6 i e}\right ) e^{- 6 i f x} + \left (927712935936 i A a^{15} c^{15} f^{5} e^{8 i e} - 618475290624 B a^{15} c^{15} f^{5} e^{8 i e}\right ) e^{- 4 i f x} + \left (4638564679680 i A a^{15} c^{15} f^{5} e^{10 i e} - 1546188226560 B a^{15} c^{15} f^{5} e^{10 i e}\right ) e^{- 2 i f x} + \left (- 4638564679680 i A a^{15} c^{15} f^{5} e^{14 i e} - 1546188226560 B a^{15} c^{15} f^{5} e^{14 i e}\right ) e^{2 i f x} + \left (- 927712935936 i A a^{15} c^{15} f^{5} e^{16 i e} - 618475290624 B a^{15} c^{15} f^{5} e^{16 i e}\right ) e^{4 i f x} + \left (- 103079215104 i A a^{15} c^{15} f^{5} e^{18 i e} - 103079215104 B a^{15} c^{15} f^{5} e^{18 i e}\right ) e^{6 i f x}\right ) e^{- 12 i e}}{39582418599936 a^{18} c^{18} f^{6}} & \text{for}\: 39582418599936 a^{18} c^{18} f^{6} e^{12 i e} \neq 0 \\x \left (- \frac{5 A}{16 a^{3} c^{3}} + \frac{\left (A e^{12 i e} + 6 A e^{10 i e} + 15 A e^{8 i e} + 20 A e^{6 i e} + 15 A e^{4 i e} + 6 A e^{2 i e} + A - i B e^{12 i e} - 4 i B e^{10 i e} - 5 i B e^{8 i e} + 5 i B e^{4 i e} + 4 i B e^{2 i e} + i B\right ) e^{- 6 i e}}{64 a^{3} c^{3}}\right ) & \text{otherwise} \end{cases} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A] time = 1.33634, size = 107, normalized size = 1.08 \begin{align*} \frac{\frac{15 \,{\left (f x + e\right )} A}{a^{3} c^{3}} + \frac{15 \, A \tan \left (f x + e\right )^{5} + 40 \, A \tan \left (f x + e\right )^{3} + 33 \, A \tan \left (f x + e\right ) - 8 \, B}{{\left (\tan \left (f x + e\right )^{2} + 1\right )}^{3} a^{3} c^{3}}}{48 \, f} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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