Optimal. Leaf size=166 \[ -\frac {\left (3 b c d-a \left (2 c^2+d^2\right )\right ) \tanh ^{-1}\left (\frac {\sqrt {c-d} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c+d}}\right )}{f (c-d)^{5/2} (c+d)^{5/2}}-\frac {\left (3 a c d-b \left (c^2+2 d^2\right )\right ) \tan (e+f x)}{2 f \left (c^2-d^2\right )^2 (c+d \sec (e+f x))}+\frac {(b c-a d) \tan (e+f x)}{2 f \left (c^2-d^2\right ) (c+d \sec (e+f x))^2} \]
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Rubi [A] time = 0.30, antiderivative size = 166, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {4003, 12, 3831, 2659, 208} \[ -\frac {\left (3 b c d-a \left (2 c^2+d^2\right )\right ) \tanh ^{-1}\left (\frac {\sqrt {c-d} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c+d}}\right )}{f (c-d)^{5/2} (c+d)^{5/2}}-\frac {\left (3 a c d-b \left (c^2+2 d^2\right )\right ) \tan (e+f x)}{2 f \left (c^2-d^2\right )^2 (c+d \sec (e+f x))}+\frac {(b c-a d) \tan (e+f x)}{2 f \left (c^2-d^2\right ) (c+d \sec (e+f x))^2} \]
Antiderivative was successfully verified.
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Rule 12
Rule 208
Rule 2659
Rule 3831
Rule 4003
Rubi steps
\begin {align*} \int \frac {\sec (e+f x) (a+b \sec (e+f x))}{(c+d \sec (e+f x))^3} \, dx &=\frac {(b c-a d) \tan (e+f x)}{2 \left (c^2-d^2\right ) f (c+d \sec (e+f x))^2}-\frac {\int \frac {\sec (e+f x) (-2 (a c-b d)-(b c-a d) \sec (e+f x))}{(c+d \sec (e+f x))^2} \, dx}{2 \left (c^2-d^2\right )}\\ &=\frac {(b c-a d) \tan (e+f x)}{2 \left (c^2-d^2\right ) f (c+d \sec (e+f x))^2}-\frac {\left (3 a c d-b \left (c^2+2 d^2\right )\right ) \tan (e+f x)}{2 \left (c^2-d^2\right )^2 f (c+d \sec (e+f x))}+\frac {\int \frac {\left (-3 b c d+a \left (2 c^2+d^2\right )\right ) \sec (e+f x)}{c+d \sec (e+f x)} \, dx}{2 \left (c^2-d^2\right )^2}\\ &=\frac {(b c-a d) \tan (e+f x)}{2 \left (c^2-d^2\right ) f (c+d \sec (e+f x))^2}-\frac {\left (3 a c d-b \left (c^2+2 d^2\right )\right ) \tan (e+f x)}{2 \left (c^2-d^2\right )^2 f (c+d \sec (e+f x))}-\frac {\left (3 b c d-a \left (2 c^2+d^2\right )\right ) \int \frac {\sec (e+f x)}{c+d \sec (e+f x)} \, dx}{2 \left (c^2-d^2\right )^2}\\ &=\frac {(b c-a d) \tan (e+f x)}{2 \left (c^2-d^2\right ) f (c+d \sec (e+f x))^2}-\frac {\left (3 a c d-b \left (c^2+2 d^2\right )\right ) \tan (e+f x)}{2 \left (c^2-d^2\right )^2 f (c+d \sec (e+f x))}-\frac {\left (3 b c d-a \left (2 c^2+d^2\right )\right ) \int \frac {1}{1+\frac {c \cos (e+f x)}{d}} \, dx}{2 d \left (c^2-d^2\right )^2}\\ &=\frac {(b c-a d) \tan (e+f x)}{2 \left (c^2-d^2\right ) f (c+d \sec (e+f x))^2}-\frac {\left (3 a c d-b \left (c^2+2 d^2\right )\right ) \tan (e+f x)}{2 \left (c^2-d^2\right )^2 f (c+d \sec (e+f x))}-\frac {\left (3 b c d-a \left (2 c^2+d^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{1+\frac {c}{d}+\left (1-\frac {c}{d}\right ) x^2} \, dx,x,\tan \left (\frac {1}{2} (e+f x)\right )\right )}{d \left (c^2-d^2\right )^2 f}\\ &=\frac {\left (2 a c^2-3 b c d+a d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c-d} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c+d}}\right )}{(c-d)^{5/2} (c+d)^{5/2} f}+\frac {(b c-a d) \tan (e+f x)}{2 \left (c^2-d^2\right ) f (c+d \sec (e+f x))^2}-\frac {\left (3 a c d-b \left (c^2+2 d^2\right )\right ) \tan (e+f x)}{2 \left (c^2-d^2\right )^2 f (c+d \sec (e+f x))}\\ \end {align*}
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Mathematica [A] time = 0.98, size = 172, normalized size = 1.04 \[ \frac {\frac {\left (a d \left (d^2-4 c^2\right )+b c \left (2 c^2+d^2\right )\right ) \sin (e+f x)}{c (c-d)^2 (c+d)^2 (c \cos (e+f x)+d)}-\frac {2 \left (a \left (2 c^2+d^2\right )-3 b c d\right ) \tanh ^{-1}\left (\frac {(d-c) \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c^2-d^2}}\right )}{\left (c^2-d^2\right )^{5/2}}+\frac {d (a d-b c) \sin (e+f x)}{c (c-d) (c+d) (c \cos (e+f x)+d)^2}}{2 f} \]
Antiderivative was successfully verified.
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fricas [B] time = 0.54, size = 752, normalized size = 4.53 \[ \left [\frac {{\left (2 \, a c^{2} d^{2} - 3 \, b c d^{3} + a d^{4} + {\left (2 \, a c^{4} - 3 \, b c^{3} d + a c^{2} d^{2}\right )} \cos \left (f x + e\right )^{2} + 2 \, {\left (2 \, a c^{3} d - 3 \, b c^{2} d^{2} + a c d^{3}\right )} \cos \left (f x + e\right )\right )} \sqrt {c^{2} - d^{2}} \log \left (\frac {2 \, c d \cos \left (f x + e\right ) - {\left (c^{2} - 2 \, d^{2}\right )} \cos \left (f x + e\right )^{2} + 2 \, \sqrt {c^{2} - d^{2}} {\left (d \cos \left (f x + e\right ) + c\right )} \sin \left (f x + e\right ) + 2 \, c^{2} - d^{2}}{c^{2} \cos \left (f x + e\right )^{2} + 2 \, c d \cos \left (f x + e\right ) + d^{2}}\right ) + 2 \, {\left (b c^{4} d - 3 \, a c^{3} d^{2} + b c^{2} d^{3} + 3 \, a c d^{4} - 2 \, b d^{5} + {\left (2 \, b c^{5} - 4 \, a c^{4} d - b c^{3} d^{2} + 5 \, a c^{2} d^{3} - b c d^{4} - a d^{5}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{4 \, {\left ({\left (c^{8} - 3 \, c^{6} d^{2} + 3 \, c^{4} d^{4} - c^{2} d^{6}\right )} f \cos \left (f x + e\right )^{2} + 2 \, {\left (c^{7} d - 3 \, c^{5} d^{3} + 3 \, c^{3} d^{5} - c d^{7}\right )} f \cos \left (f x + e\right ) + {\left (c^{6} d^{2} - 3 \, c^{4} d^{4} + 3 \, c^{2} d^{6} - d^{8}\right )} f\right )}}, \frac {{\left (2 \, a c^{2} d^{2} - 3 \, b c d^{3} + a d^{4} + {\left (2 \, a c^{4} - 3 \, b c^{3} d + a c^{2} d^{2}\right )} \cos \left (f x + e\right )^{2} + 2 \, {\left (2 \, a c^{3} d - 3 \, b c^{2} d^{2} + a c d^{3}\right )} \cos \left (f x + e\right )\right )} \sqrt {-c^{2} + d^{2}} \arctan \left (-\frac {\sqrt {-c^{2} + d^{2}} {\left (d \cos \left (f x + e\right ) + c\right )}}{{\left (c^{2} - d^{2}\right )} \sin \left (f x + e\right )}\right ) + {\left (b c^{4} d - 3 \, a c^{3} d^{2} + b c^{2} d^{3} + 3 \, a c d^{4} - 2 \, b d^{5} + {\left (2 \, b c^{5} - 4 \, a c^{4} d - b c^{3} d^{2} + 5 \, a c^{2} d^{3} - b c d^{4} - a d^{5}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{2 \, {\left ({\left (c^{8} - 3 \, c^{6} d^{2} + 3 \, c^{4} d^{4} - c^{2} d^{6}\right )} f \cos \left (f x + e\right )^{2} + 2 \, {\left (c^{7} d - 3 \, c^{5} d^{3} + 3 \, c^{3} d^{5} - c d^{7}\right )} f \cos \left (f x + e\right ) + {\left (c^{6} d^{2} - 3 \, c^{4} d^{4} + 3 \, c^{2} d^{6} - d^{8}\right )} f\right )}}\right ] \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 1.95, size = 418, normalized size = 2.52 \[ \frac {\frac {{\left (2 \, a c^{2} - 3 \, b c d + a d^{2}\right )} {\left (\pi \left \lfloor \frac {f x + e}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, c + 2 \, d\right ) + \arctan \left (-\frac {c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )}{\sqrt {-c^{2} + d^{2}}}\right )\right )}}{{\left (c^{4} - 2 \, c^{2} d^{2} + d^{4}\right )} \sqrt {-c^{2} + d^{2}}} - \frac {2 \, b c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 4 \, a c^{2} d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - b c^{2} d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 3 \, a c d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + b c d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + a d^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 2 \, b d^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 2 \, b c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 4 \, a c^{2} d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - b c^{2} d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 3 \, a c d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - b c d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - a d^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 2 \, b d^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )}{{\left (c^{4} - 2 \, c^{2} d^{2} + d^{4}\right )} {\left (c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - c - d\right )}^{2}}}{f} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.59, size = 236, normalized size = 1.42 \[ \frac {-\frac {2 \left (-\frac {\left (4 a c d +a \,d^{2}-2 c^{2} b -b c d -2 d^{2} b \right ) \left (\tan ^{3}\left (\frac {e}{2}+\frac {f x}{2}\right )\right )}{2 \left (c -d \right ) \left (c^{2}+2 c d +d^{2}\right )}+\frac {\left (4 a c d -a \,d^{2}-2 c^{2} b +b c d -2 d^{2} b \right ) \tan \left (\frac {e}{2}+\frac {f x}{2}\right )}{2 \left (c +d \right ) \left (c^{2}-2 c d +d^{2}\right )}\right )}{\left (\left (\tan ^{2}\left (\frac {e}{2}+\frac {f x}{2}\right )\right ) c -\left (\tan ^{2}\left (\frac {e}{2}+\frac {f x}{2}\right )\right ) d -c -d \right )^{2}}+\frac {\left (2 a \,c^{2}+a \,d^{2}-3 b c d \right ) \arctanh \left (\frac {\tan \left (\frac {e}{2}+\frac {f x}{2}\right ) \left (c -d \right )}{\sqrt {\left (c +d \right ) \left (c -d \right )}}\right )}{\left (c^{4}-2 c^{2} d^{2}+d^{4}\right ) \sqrt {\left (c +d \right ) \left (c -d \right )}}}{f} \]
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 \[ \text {Exception raised: ValueError} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 4.99, size = 250, normalized size = 1.51 \[ \frac {\frac {\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (a\,d^2+2\,b\,c^2+2\,b\,d^2-4\,a\,c\,d-b\,c\,d\right )}{\left (c+d\right )\,\left (c^2-2\,c\,d+d^2\right )}-\frac {{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3\,\left (2\,b\,c^2-a\,d^2+2\,b\,d^2-4\,a\,c\,d+b\,c\,d\right )}{{\left (c+d\right )}^2\,\left (c-d\right )}}{f\,\left (2\,c\,d-{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,\left (2\,c^2-2\,d^2\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4\,\left (c^2-2\,c\,d+d^2\right )+c^2+d^2\right )}+\frac {\mathrm {atanh}\left (\frac {\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (2\,c-2\,d\right )\,\left (c^2-2\,c\,d+d^2\right )}{2\,\sqrt {c+d}\,{\left (c-d\right )}^{5/2}}\right )\,\left (2\,a\,c^2-3\,b\,c\,d+a\,d^2\right )}{f\,{\left (c+d\right )}^{5/2}\,{\left (c-d\right )}^{5/2}} \]
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 \[ \int \frac {\left (a + b \sec {\left (e + f x \right )}\right ) \sec {\left (e + f x \right )}}{\left (c + d \sec {\left (e + f x \right )}\right )^{3}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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