Optimal. Leaf size=205 \[ \frac {\tan (e+f x) \left (2 c^4+8 c^3 d-d^2 \left (2 c^2+10 c d-27 d^2\right ) \sec (e+f x)+21 c^2 d^2-88 c d^3+72 d^4\right )}{15 f \left (a^3 \sec (e+f x)+a^3\right )}+\frac {d^3 (4 c-3 d) \tanh ^{-1}(\sin (e+f x))}{a^3 f}+\frac {(c-d) \tan (e+f x) (c+d \sec (e+f x))^3}{5 f (a \sec (e+f x)+a)^3}+\frac {(c-d) (2 c+9 d) \tan (e+f x) (c+d \sec (e+f x))^2}{15 a f (a \sec (e+f x)+a)^2} \]
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Rubi [A] time = 0.28, antiderivative size = 265, normalized size of antiderivative = 1.29, number of steps used = 7, number of rules used = 7, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.226, Rules used = {3987, 98, 150, 143, 63, 217, 203} \[ \frac {\tan (e+f x) \left (-d^2 \left (2 c^2+10 c d-27 d^2\right ) \sec (e+f x)+21 c^2 d^2+8 c^3 d+2 c^4-88 c d^3+72 d^4\right )}{15 f \left (a^3 \sec (e+f x)+a^3\right )}+\frac {2 d^3 (4 c-3 d) \tan (e+f x) \tan ^{-1}\left (\frac {\sqrt {a-a \sec (e+f x)}}{\sqrt {a (\sec (e+f x)+1)}}\right )}{a^2 f \sqrt {a-a \sec (e+f x)} \sqrt {a \sec (e+f x)+a}}+\frac {(c-d) \tan (e+f x) (c+d \sec (e+f x))^3}{5 f (a \sec (e+f x)+a)^3}+\frac {(c-d) (2 c+9 d) \tan (e+f x) (c+d \sec (e+f x))^2}{15 a f (a \sec (e+f x)+a)^2} \]
Antiderivative was successfully verified.
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Rule 63
Rule 98
Rule 143
Rule 150
Rule 203
Rule 217
Rule 3987
Rubi steps
\begin {align*} \int \frac {\sec (e+f x) (c+d \sec (e+f x))^4}{(a+a \sec (e+f x))^3} \, dx &=-\frac {\left (a^2 \tan (e+f x)\right ) \operatorname {Subst}\left (\int \frac {(c+d x)^4}{\sqrt {a-a x} (a+a x)^{7/2}} \, dx,x,\sec (e+f x)\right )}{f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}}\\ &=\frac {(c-d) (c+d \sec (e+f x))^3 \tan (e+f x)}{5 f (a+a \sec (e+f x))^3}+\frac {\tan (e+f x) \operatorname {Subst}\left (\int \frac {(c+d x)^2 \left (-a^2 \left (2 c^2+6 c d-3 d^2\right )+a^2 (c-6 d) d x\right )}{\sqrt {a-a x} (a+a x)^{5/2}} \, dx,x,\sec (e+f x)\right )}{5 a f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}}\\ &=\frac {(c-d) (2 c+9 d) (c+d \sec (e+f x))^2 \tan (e+f x)}{15 a f (a+a \sec (e+f x))^2}+\frac {(c-d) (c+d \sec (e+f x))^3 \tan (e+f x)}{5 f (a+a \sec (e+f x))^3}+\frac {\tan (e+f x) \operatorname {Subst}\left (\int \frac {(c+d x) \left (-a^4 \left (2 c^3+8 c^2 d+23 c d^2-18 d^3\right )+a^4 d \left (2 c^2+10 c d-27 d^2\right ) x\right )}{\sqrt {a-a x} (a+a x)^{3/2}} \, dx,x,\sec (e+f x)\right )}{15 a^4 f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}}\\ &=\frac {(c-d) (2 c+9 d) (c+d \sec (e+f x))^2 \tan (e+f x)}{15 a f (a+a \sec (e+f x))^2}+\frac {(c-d) (c+d \sec (e+f x))^3 \tan (e+f x)}{5 f (a+a \sec (e+f x))^3}+\frac {\left (2 c^4+8 c^3 d+21 c^2 d^2-88 c d^3+72 d^4-d^2 \left (2 c^2+10 c d-27 d^2\right ) \sec (e+f x)\right ) \tan (e+f x)}{15 f \left (a^3+a^3 \sec (e+f x)\right )}-\frac {\left ((4 c-3 d) d^3 \tan (e+f x)\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a-a x} \sqrt {a+a x}} \, dx,x,\sec (e+f x)\right )}{a f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}}\\ &=\frac {(c-d) (2 c+9 d) (c+d \sec (e+f x))^2 \tan (e+f x)}{15 a f (a+a \sec (e+f x))^2}+\frac {(c-d) (c+d \sec (e+f x))^3 \tan (e+f x)}{5 f (a+a \sec (e+f x))^3}+\frac {\left (2 c^4+8 c^3 d+21 c^2 d^2-88 c d^3+72 d^4-d^2 \left (2 c^2+10 c d-27 d^2\right ) \sec (e+f x)\right ) \tan (e+f x)}{15 f \left (a^3+a^3 \sec (e+f x)\right )}+\frac {\left (2 (4 c-3 d) d^3 \tan (e+f x)\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {2 a-x^2}} \, dx,x,\sqrt {a-a \sec (e+f x)}\right )}{a^2 f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}}\\ &=\frac {(c-d) (2 c+9 d) (c+d \sec (e+f x))^2 \tan (e+f x)}{15 a f (a+a \sec (e+f x))^2}+\frac {(c-d) (c+d \sec (e+f x))^3 \tan (e+f x)}{5 f (a+a \sec (e+f x))^3}+\frac {\left (2 c^4+8 c^3 d+21 c^2 d^2-88 c d^3+72 d^4-d^2 \left (2 c^2+10 c d-27 d^2\right ) \sec (e+f x)\right ) \tan (e+f x)}{15 f \left (a^3+a^3 \sec (e+f x)\right )}+\frac {\left (2 (4 c-3 d) d^3 \tan (e+f x)\right ) \operatorname {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\frac {\sqrt {a-a \sec (e+f x)}}{\sqrt {a+a \sec (e+f x)}}\right )}{a^2 f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}}\\ &=\frac {2 (4 c-3 d) d^3 \tan ^{-1}\left (\frac {\sqrt {a-a \sec (e+f x)}}{\sqrt {a+a \sec (e+f x)}}\right ) \tan (e+f x)}{a^2 f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}}+\frac {(c-d) (2 c+9 d) (c+d \sec (e+f x))^2 \tan (e+f x)}{15 a f (a+a \sec (e+f x))^2}+\frac {(c-d) (c+d \sec (e+f x))^3 \tan (e+f x)}{5 f (a+a \sec (e+f x))^3}+\frac {\left (2 c^4+8 c^3 d+21 c^2 d^2-88 c d^3+72 d^4-d^2 \left (2 c^2+10 c d-27 d^2\right ) \sec (e+f x)\right ) \tan (e+f x)}{15 f \left (a^3+a^3 \sec (e+f x)\right )}\\ \end {align*}
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Mathematica [A] time = 2.39, size = 292, normalized size = 1.42 \[ \frac {2 \cos \left (\frac {1}{2} (e+f x)\right ) \left (4 (c-d)^2 \left (7 c^2+26 c d+57 d^2\right ) \sec \left (\frac {e}{2}\right ) \sin \left (\frac {f x}{2}\right ) \cos ^4\left (\frac {1}{2} (e+f x)\right )-60 d^3 \cos ^5\left (\frac {1}{2} (e+f x)\right ) \left ((4 c-3 d) \left (\log \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )-\log \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )\right )-d \sec (e) \sin (f x) \sec (e+f x)\right )-8 (c-d)^3 (2 c+3 d) \tan \left (\frac {e}{2}\right ) \cos ^3\left (\frac {1}{2} (e+f x)\right )+3 (c-d)^4 \tan \left (\frac {e}{2}\right ) \cos \left (\frac {1}{2} (e+f x)\right )+3 (c-d)^4 \sec \left (\frac {e}{2}\right ) \sin \left (\frac {f x}{2}\right )-8 (c-d)^3 (2 c+3 d) \sec \left (\frac {e}{2}\right ) \sin \left (\frac {f x}{2}\right ) \cos ^2\left (\frac {1}{2} (e+f x)\right )\right )}{15 a^3 f (\cos (e+f x)+1)^3} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.46, size = 385, normalized size = 1.88 \[ \frac {15 \, {\left ({\left (4 \, c d^{3} - 3 \, d^{4}\right )} \cos \left (f x + e\right )^{4} + 3 \, {\left (4 \, c d^{3} - 3 \, d^{4}\right )} \cos \left (f x + e\right )^{3} + 3 \, {\left (4 \, c d^{3} - 3 \, d^{4}\right )} \cos \left (f x + e\right )^{2} + {\left (4 \, c d^{3} - 3 \, d^{4}\right )} \cos \left (f x + e\right )\right )} \log \left (\sin \left (f x + e\right ) + 1\right ) - 15 \, {\left ({\left (4 \, c d^{3} - 3 \, d^{4}\right )} \cos \left (f x + e\right )^{4} + 3 \, {\left (4 \, c d^{3} - 3 \, d^{4}\right )} \cos \left (f x + e\right )^{3} + 3 \, {\left (4 \, c d^{3} - 3 \, d^{4}\right )} \cos \left (f x + e\right )^{2} + {\left (4 \, c d^{3} - 3 \, d^{4}\right )} \cos \left (f x + e\right )\right )} \log \left (-\sin \left (f x + e\right ) + 1\right ) + 2 \, {\left (15 \, d^{4} + {\left (7 \, c^{4} + 12 \, c^{3} d + 12 \, c^{2} d^{2} - 88 \, c d^{3} + 72 \, d^{4}\right )} \cos \left (f x + e\right )^{3} + 3 \, {\left (2 \, c^{4} + 12 \, c^{3} d + 12 \, c^{2} d^{2} - 68 \, c d^{3} + 57 \, d^{4}\right )} \cos \left (f x + e\right )^{2} + {\left (2 \, c^{4} + 12 \, c^{3} d + 42 \, c^{2} d^{2} - 128 \, c d^{3} + 117 \, d^{4}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{30 \, {\left (a^{3} f \cos \left (f x + e\right )^{4} + 3 \, a^{3} f \cos \left (f x + e\right )^{3} + 3 \, a^{3} f \cos \left (f x + e\right )^{2} + a^{3} f \cos \left (f x + e\right )\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: NotImplementedError} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.71, size = 454, normalized size = 2.21 \[ \frac {\left (\tan ^{5}\left (\frac {e}{2}+\frac {f x}{2}\right )\right ) c^{4}}{20 f \,a^{3}}-\frac {\left (\tan ^{5}\left (\frac {e}{2}+\frac {f x}{2}\right )\right ) c^{3} d}{5 f \,a^{3}}+\frac {3 \left (\tan ^{5}\left (\frac {e}{2}+\frac {f x}{2}\right )\right ) c^{2} d^{2}}{10 f \,a^{3}}-\frac {\left (\tan ^{5}\left (\frac {e}{2}+\frac {f x}{2}\right )\right ) c \,d^{3}}{5 f \,a^{3}}+\frac {\left (\tan ^{5}\left (\frac {e}{2}+\frac {f x}{2}\right )\right ) d^{4}}{20 f \,a^{3}}-\frac {\left (\tan ^{3}\left (\frac {e}{2}+\frac {f x}{2}\right )\right ) c^{4}}{6 f \,a^{3}}+\frac {\left (\tan ^{3}\left (\frac {e}{2}+\frac {f x}{2}\right )\right ) c^{2} d^{2}}{f \,a^{3}}-\frac {4 \left (\tan ^{3}\left (\frac {e}{2}+\frac {f x}{2}\right )\right ) c \,d^{3}}{3 f \,a^{3}}+\frac {\left (\tan ^{3}\left (\frac {e}{2}+\frac {f x}{2}\right )\right ) d^{4}}{2 f \,a^{3}}+\frac {\tan \left (\frac {e}{2}+\frac {f x}{2}\right ) c^{4}}{4 f \,a^{3}}+\frac {\tan \left (\frac {e}{2}+\frac {f x}{2}\right ) c^{3} d}{f \,a^{3}}+\frac {3 \tan \left (\frac {e}{2}+\frac {f x}{2}\right ) c^{2} d^{2}}{2 f \,a^{3}}-\frac {7 \tan \left (\frac {e}{2}+\frac {f x}{2}\right ) c \,d^{3}}{f \,a^{3}}+\frac {17 \tan \left (\frac {e}{2}+\frac {f x}{2}\right ) d^{4}}{4 f \,a^{3}}-\frac {d^{4}}{f \,a^{3} \left (\tan \left (\frac {e}{2}+\frac {f x}{2}\right )-1\right )}-\frac {4 d^{3} \ln \left (\tan \left (\frac {e}{2}+\frac {f x}{2}\right )-1\right ) c}{f \,a^{3}}+\frac {3 d^{4} \ln \left (\tan \left (\frac {e}{2}+\frac {f x}{2}\right )-1\right )}{f \,a^{3}}-\frac {d^{4}}{f \,a^{3} \left (\tan \left (\frac {e}{2}+\frac {f x}{2}\right )+1\right )}+\frac {4 d^{3} \ln \left (\tan \left (\frac {e}{2}+\frac {f x}{2}\right )+1\right ) c}{f \,a^{3}}-\frac {3 d^{4} \ln \left (\tan \left (\frac {e}{2}+\frac {f x}{2}\right )+1\right )}{f \,a^{3}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.36, size = 475, normalized size = 2.32 \[ \frac {3 \, d^{4} {\left (\frac {40 \, \sin \left (f x + e\right )}{{\left (a^{3} - \frac {a^{3} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (f x + e\right ) + 1\right )}} + \frac {\frac {85 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {10 \, \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {\sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}}{a^{3}} - \frac {60 \, \log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right )}{a^{3}} + \frac {60 \, \log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - 1\right )}{a^{3}}\right )} - 4 \, c d^{3} {\left (\frac {\frac {105 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {20 \, \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}}{a^{3}} - \frac {60 \, \log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right )}{a^{3}} + \frac {60 \, \log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - 1\right )}{a^{3}}\right )} + \frac {6 \, c^{2} d^{2} {\left (\frac {15 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {10 \, \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}\right )}}{a^{3}} + \frac {c^{4} {\left (\frac {15 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac {10 \, \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}\right )}}{a^{3}} + \frac {12 \, c^{3} d {\left (\frac {5 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac {\sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}\right )}}{a^{3}}}{60 \, f} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 1.82, size = 195, normalized size = 0.95 \[ \frac {\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (\frac {3\,{\left (c-d\right )}^4}{4\,a^3}+\frac {3\,{\left (c^2-d^2\right )}^2}{2\,a^3}-\frac {2\,\left (c+d\right )\,{\left (c-d\right )}^3}{a^3}\right )}{f}+\frac {{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3\,\left (\frac {{\left (c-d\right )}^4}{6\,a^3}-\frac {\left (c+d\right )\,{\left (c-d\right )}^3}{3\,a^3}\right )}{f}-\frac {2\,d^4\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}{f\,\left (a^3\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2-a^3\right )}+\frac {{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^5\,{\left (c-d\right )}^4}{20\,a^3\,f}+\frac {2\,d^3\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\right )\,\left (4\,c-3\,d\right )}{a^3\,f} \]
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 \[ \frac {\int \frac {c^{4} \sec {\left (e + f x \right )}}{\sec ^{3}{\left (e + f x \right )} + 3 \sec ^{2}{\left (e + f x \right )} + 3 \sec {\left (e + f x \right )} + 1}\, dx + \int \frac {d^{4} \sec ^{5}{\left (e + f x \right )}}{\sec ^{3}{\left (e + f x \right )} + 3 \sec ^{2}{\left (e + f x \right )} + 3 \sec {\left (e + f x \right )} + 1}\, dx + \int \frac {4 c d^{3} \sec ^{4}{\left (e + f x \right )}}{\sec ^{3}{\left (e + f x \right )} + 3 \sec ^{2}{\left (e + f x \right )} + 3 \sec {\left (e + f x \right )} + 1}\, dx + \int \frac {6 c^{2} d^{2} \sec ^{3}{\left (e + f x \right )}}{\sec ^{3}{\left (e + f x \right )} + 3 \sec ^{2}{\left (e + f x \right )} + 3 \sec {\left (e + f x \right )} + 1}\, dx + \int \frac {4 c^{3} d \sec ^{2}{\left (e + f x \right )}}{\sec ^{3}{\left (e + f x \right )} + 3 \sec ^{2}{\left (e + f x \right )} + 3 \sec {\left (e + f x \right )} + 1}\, dx}{a^{3}} \]
Verification of antiderivative is not currently implemented for this CAS.
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