Optimal. Leaf size=86 \[ \frac {5 a^3 c \tanh ^{-1}(\sin (e+f x))}{8 f}-\frac {3 a^3 c \sec (e+f x) \tan (e+f x)}{8 f}-\frac {a^3 c \sec ^3(e+f x) \tan (e+f x)}{4 f}-\frac {2 a^3 c \tan ^3(e+f x)}{3 f} \]
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Rubi [A]
time = 0.11, antiderivative size = 86, normalized size of antiderivative = 1.00, number of steps
used = 9, number of rules used = 6, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {4043, 2691,
3855, 2687, 30, 3853} \begin {gather*} -\frac {2 a^3 c \tan ^3(e+f x)}{3 f}+\frac {5 a^3 c \tanh ^{-1}(\sin (e+f x))}{8 f}-\frac {a^3 c \tan (e+f x) \sec ^3(e+f x)}{4 f}-\frac {3 a^3 c \tan (e+f x) \sec (e+f x)}{8 f} \end {gather*}
Antiderivative was successfully verified.
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Rule 30
Rule 2687
Rule 2691
Rule 3853
Rule 3855
Rule 4043
Rubi steps
\begin {align*} \int \sec (e+f x) (a+a \sec (e+f x))^3 (c-c \sec (e+f x)) \, dx &=-\left ((a c) \int \left (a^2 \sec (e+f x) \tan ^2(e+f x)+2 a^2 \sec ^2(e+f x) \tan ^2(e+f x)+a^2 \sec ^3(e+f x) \tan ^2(e+f x)\right ) \, dx\right )\\ &=-\left (\left (a^3 c\right ) \int \sec (e+f x) \tan ^2(e+f x) \, dx\right )-\left (a^3 c\right ) \int \sec ^3(e+f x) \tan ^2(e+f x) \, dx-\left (2 a^3 c\right ) \int \sec ^2(e+f x) \tan ^2(e+f x) \, dx\\ &=-\frac {a^3 c \sec (e+f x) \tan (e+f x)}{2 f}-\frac {a^3 c \sec ^3(e+f x) \tan (e+f x)}{4 f}+\frac {1}{4} \left (a^3 c\right ) \int \sec ^3(e+f x) \, dx+\frac {1}{2} \left (a^3 c\right ) \int \sec (e+f x) \, dx-\frac {\left (2 a^3 c\right ) \text {Subst}\left (\int x^2 \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {a^3 c \tanh ^{-1}(\sin (e+f x))}{2 f}-\frac {3 a^3 c \sec (e+f x) \tan (e+f x)}{8 f}-\frac {a^3 c \sec ^3(e+f x) \tan (e+f x)}{4 f}-\frac {2 a^3 c \tan ^3(e+f x)}{3 f}+\frac {1}{8} \left (a^3 c\right ) \int \sec (e+f x) \, dx\\ &=\frac {5 a^3 c \tanh ^{-1}(\sin (e+f x))}{8 f}-\frac {3 a^3 c \sec (e+f x) \tan (e+f x)}{8 f}-\frac {a^3 c \sec ^3(e+f x) \tan (e+f x)}{4 f}-\frac {2 a^3 c \tan ^3(e+f x)}{3 f}\\ \end {align*}
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Mathematica [A]
time = 0.61, size = 70, normalized size = 0.81 \begin {gather*} \frac {a^3 c \left (60 \tanh ^{-1}(\sin (e+f x))-\sec ^4(e+f x) (33 \sin (e+f x)+16 \sin (2 (e+f x))+9 \sin (3 (e+f x))-8 \sin (4 (e+f x)))\right )}{96 f} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.21, size = 111, normalized size = 1.29
method | result | size |
derivativedivides | \(\frac {-a^{3} c \left (-\left (-\frac {\left (\sec ^{3}\left (f x +e \right )\right )}{4}-\frac {3 \sec \left (f x +e \right )}{8}\right ) \tan \left (f x +e \right )+\frac {3 \ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )}{8}\right )+2 a^{3} c \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (f x +e \right )\right )}{3}\right ) \tan \left (f x +e \right )+2 a^{3} c \tan \left (f x +e \right )+a^{3} c \ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )}{f}\) | \(111\) |
default | \(\frac {-a^{3} c \left (-\left (-\frac {\left (\sec ^{3}\left (f x +e \right )\right )}{4}-\frac {3 \sec \left (f x +e \right )}{8}\right ) \tan \left (f x +e \right )+\frac {3 \ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )}{8}\right )+2 a^{3} c \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (f x +e \right )\right )}{3}\right ) \tan \left (f x +e \right )+2 a^{3} c \tan \left (f x +e \right )+a^{3} c \ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )}{f}\) | \(111\) |
norman | \(\frac {-\frac {5 a^{3} c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{4 f}-\frac {73 a^{3} c \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{12 f}+\frac {55 a^{3} c \left (\tan ^{5}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{12 f}-\frac {5 a^{3} c \left (\tan ^{7}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{4 f}}{\left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{4}}-\frac {5 a^{3} c \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}{8 f}+\frac {5 a^{3} c \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}{8 f}\) | \(139\) |
risch | \(\frac {i a^{3} c \left (9 \,{\mathrm e}^{7 i \left (f x +e \right )}+48 \,{\mathrm e}^{6 i \left (f x +e \right )}+33 \,{\mathrm e}^{5 i \left (f x +e \right )}+48 \,{\mathrm e}^{4 i \left (f x +e \right )}-33 \,{\mathrm e}^{3 i \left (f x +e \right )}+16 \,{\mathrm e}^{2 i \left (f x +e \right )}-9 \,{\mathrm e}^{i \left (f x +e \right )}+16\right )}{12 f \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{4}}-\frac {5 a^{3} c \ln \left ({\mathrm e}^{i \left (f x +e \right )}-i\right )}{8 f}+\frac {5 a^{3} c \ln \left ({\mathrm e}^{i \left (f x +e \right )}+i\right )}{8 f}\) | \(148\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.30, size = 144, normalized size = 1.67 \begin {gather*} -\frac {32 \, {\left (\tan \left (f x + e\right )^{3} + 3 \, \tan \left (f x + e\right )\right )} a^{3} c - 3 \, a^{3} c {\left (\frac {2 \, {\left (3 \, \sin \left (f x + e\right )^{3} - 5 \, \sin \left (f x + e\right )\right )}}{\sin \left (f x + e\right )^{4} - 2 \, \sin \left (f x + e\right )^{2} + 1} - 3 \, \log \left (\sin \left (f x + e\right ) + 1\right ) + 3 \, \log \left (\sin \left (f x + e\right ) - 1\right )\right )} - 48 \, a^{3} c \log \left (\sec \left (f x + e\right ) + \tan \left (f x + e\right )\right ) - 96 \, a^{3} c \tan \left (f x + e\right )}{48 \, f} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 3.70, size = 126, normalized size = 1.47 \begin {gather*} \frac {15 \, a^{3} c \cos \left (f x + e\right )^{4} \log \left (\sin \left (f x + e\right ) + 1\right ) - 15 \, a^{3} c \cos \left (f x + e\right )^{4} \log \left (-\sin \left (f x + e\right ) + 1\right ) + 2 \, {\left (16 \, a^{3} c \cos \left (f x + e\right )^{3} - 9 \, a^{3} c \cos \left (f x + e\right )^{2} - 16 \, a^{3} c \cos \left (f x + e\right ) - 6 \, a^{3} c\right )} \sin \left (f x + e\right )}{48 \, f \cos \left (f x + e\right )^{4}} \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*} - a^{3} c \left (\int \left (- \sec {\left (e + f x \right )}\right )\, dx + \int \left (- 2 \sec ^{2}{\left (e + f x \right )}\right )\, dx + \int 2 \sec ^{4}{\left (e + f x \right )}\, dx + \int \sec ^{5}{\left (e + f x \right )}\, dx\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A]
time = 0.49, size = 128, normalized size = 1.49 \begin {gather*} \frac {15 \, a^{3} c \log \left ({\left | \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1 \right |}\right ) - 15 \, a^{3} c \log \left ({\left | \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1 \right |}\right ) - \frac {2 \, {\left (15 \, a^{3} c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{7} - 55 \, a^{3} c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} + 73 \, a^{3} c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 15 \, a^{3} c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 1\right )}^{4}}}{24 \, f} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 4.79, size = 146, normalized size = 1.70 \begin {gather*} \frac {5\,a^3\,c\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\right )}{4\,f}-\frac {\frac {5\,c\,a^3\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^7}{4}-\frac {55\,c\,a^3\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^5}{12}+\frac {73\,c\,a^3\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3}{12}+\frac {5\,c\,a^3\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}{4}}{f\,\left ({\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^8-4\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^6+6\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4-4\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+1\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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