Optimal. Leaf size=164 \[ \frac {\left (2 a^2 A-3 a b B+A b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{d (a-b)^{5/2} (a+b)^{5/2}}-\frac {\left (a^2 (-B)+3 a A b-2 b^2 B\right ) \tan (c+d x)}{2 d \left (a^2-b^2\right )^2 (a+b \sec (c+d x))}-\frac {(A b-a B) \tan (c+d x)}{2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2} \]
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Rubi [A] time = 0.26, antiderivative size = 164, 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} \[ \frac {\left (2 a^2 A-3 a b B+A b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{d (a-b)^{5/2} (a+b)^{5/2}}-\frac {\left (a^2 (-B)+3 a A b-2 b^2 B\right ) \tan (c+d x)}{2 d \left (a^2-b^2\right )^2 (a+b \sec (c+d x))}-\frac {(A b-a B) \tan (c+d x)}{2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2} \]
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 (c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^3} \, dx &=-\frac {(A b-a B) \tan (c+d x)}{2 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}-\frac {\int \frac {\sec (c+d x) (-2 (a A-b B)+(A b-a B) \sec (c+d x))}{(a+b \sec (c+d x))^2} \, dx}{2 \left (a^2-b^2\right )}\\ &=-\frac {(A b-a B) \tan (c+d x)}{2 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}-\frac {\left (3 a A b-a^2 B-2 b^2 B\right ) \tan (c+d x)}{2 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))}+\frac {\int \frac {\left (2 a^2 A+A b^2-3 a b B\right ) \sec (c+d x)}{a+b \sec (c+d x)} \, dx}{2 \left (a^2-b^2\right )^2}\\ &=-\frac {(A b-a B) \tan (c+d x)}{2 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}-\frac {\left (3 a A b-a^2 B-2 b^2 B\right ) \tan (c+d x)}{2 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))}+\frac {\left (2 a^2 A+A b^2-3 a b B\right ) \int \frac {\sec (c+d x)}{a+b \sec (c+d x)} \, dx}{2 \left (a^2-b^2\right )^2}\\ &=-\frac {(A b-a B) \tan (c+d x)}{2 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}-\frac {\left (3 a A b-a^2 B-2 b^2 B\right ) \tan (c+d x)}{2 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))}+\frac {\left (2 a^2 A+A b^2-3 a b B\right ) \int \frac {1}{1+\frac {a \cos (c+d x)}{b}} \, dx}{2 b \left (a^2-b^2\right )^2}\\ &=-\frac {(A b-a B) \tan (c+d x)}{2 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}-\frac {\left (3 a A b-a^2 B-2 b^2 B\right ) \tan (c+d x)}{2 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))}+\frac {\left (2 a^2 A+A b^2-3 a b B\right ) \operatorname {Subst}\left (\int \frac {1}{1+\frac {a}{b}+\left (1-\frac {a}{b}\right ) x^2} \, dx,x,\tan \left (\frac {1}{2} (c+d x)\right )\right )}{b \left (a^2-b^2\right )^2 d}\\ &=\frac {\left (2 a^2 A+A b^2-3 a b B\right ) \tanh ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{(a-b)^{5/2} (a+b)^{5/2} d}-\frac {(A b-a B) \tan (c+d x)}{2 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}-\frac {\left (3 a A b-a^2 B-2 b^2 B\right ) \tan (c+d x)}{2 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))}\\ \end {align*}
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Mathematica [A] time = 0.92, size = 172, normalized size = 1.05 \[ \frac {-\frac {2 \left (2 a^2 A-3 a b B+A b^2\right ) \tanh ^{-1}\left (\frac {(b-a) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{5/2}}+\frac {\left (2 a^3 B-4 a^2 A b+a b^2 B+A b^3\right ) \sin (c+d x)}{a (a-b)^2 (a+b)^2 (a \cos (c+d x)+b)}+\frac {b (A b-a B) \sin (c+d x)}{a (a-b) (a+b) (a \cos (c+d x)+b)^2}}{2 d} \]
Antiderivative was successfully verified.
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fricas [B] time = 0.51, size = 752, normalized size = 4.59 \[ \left [\frac {{\left (2 \, A a^{2} b^{2} - 3 \, B a b^{3} + A b^{4} + {\left (2 \, A a^{4} - 3 \, B a^{3} b + A a^{2} b^{2}\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (2 \, A a^{3} b - 3 \, B a^{2} b^{2} + A a b^{3}\right )} \cos \left (d x + c\right )\right )} \sqrt {a^{2} - b^{2}} \log \left (\frac {2 \, a b \cos \left (d x + c\right ) - {\left (a^{2} - 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {a^{2} - b^{2}} {\left (b \cos \left (d x + c\right ) + a\right )} \sin \left (d x + c\right ) + 2 \, a^{2} - b^{2}}{a^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + b^{2}}\right ) + 2 \, {\left (B a^{4} b - 3 \, A a^{3} b^{2} + B a^{2} b^{3} + 3 \, A a b^{4} - 2 \, B b^{5} + {\left (2 \, B a^{5} - 4 \, A a^{4} b - B a^{3} b^{2} + 5 \, A a^{2} b^{3} - B a b^{4} - A b^{5}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{4 \, {\left ({\left (a^{8} - 3 \, a^{6} b^{2} + 3 \, a^{4} b^{4} - a^{2} b^{6}\right )} d \cos \left (d x + c\right )^{2} + 2 \, {\left (a^{7} b - 3 \, a^{5} b^{3} + 3 \, a^{3} b^{5} - a b^{7}\right )} d \cos \left (d x + c\right ) + {\left (a^{6} b^{2} - 3 \, a^{4} b^{4} + 3 \, a^{2} b^{6} - b^{8}\right )} d\right )}}, \frac {{\left (2 \, A a^{2} b^{2} - 3 \, B a b^{3} + A b^{4} + {\left (2 \, A a^{4} - 3 \, B a^{3} b + A a^{2} b^{2}\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (2 \, A a^{3} b - 3 \, B a^{2} b^{2} + A a b^{3}\right )} \cos \left (d x + c\right )\right )} \sqrt {-a^{2} + b^{2}} \arctan \left (-\frac {\sqrt {-a^{2} + b^{2}} {\left (b \cos \left (d x + c\right ) + a\right )}}{{\left (a^{2} - b^{2}\right )} \sin \left (d x + c\right )}\right ) + {\left (B a^{4} b - 3 \, A a^{3} b^{2} + B a^{2} b^{3} + 3 \, A a b^{4} - 2 \, B b^{5} + {\left (2 \, B a^{5} - 4 \, A a^{4} b - B a^{3} b^{2} + 5 \, A a^{2} b^{3} - B a b^{4} - A b^{5}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{2 \, {\left ({\left (a^{8} - 3 \, a^{6} b^{2} + 3 \, a^{4} b^{4} - a^{2} b^{6}\right )} d \cos \left (d x + c\right )^{2} + 2 \, {\left (a^{7} b - 3 \, a^{5} b^{3} + 3 \, a^{3} b^{5} - a b^{7}\right )} d \cos \left (d x + c\right ) + {\left (a^{6} b^{2} - 3 \, a^{4} b^{4} + 3 \, a^{2} b^{6} - b^{8}\right )} d\right )}}\right ] \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.38, size = 399, normalized size = 2.43 \[ \frac {\frac {{\left (2 \, A a^{2} - 3 \, B a b + A b^{2}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {-a^{2} + b^{2}}}\right )\right )}}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sqrt {-a^{2} + b^{2}}} - \frac {2 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 4 \, A a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - B a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 3 \, A a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + B a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + A b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, B b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 4 \, A a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - B a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, A a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - B a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - A b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, B b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a - b\right )}^{2}}}{d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.72, size = 236, normalized size = 1.44 \[ \frac {-\frac {2 \left (-\frac {\left (4 A a b +A \,b^{2}-2 a^{2} B -B a b -2 b^{2} B \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 \left (a -b \right ) \left (a^{2}+2 a b +b^{2}\right )}+\frac {\left (4 A a b -A \,b^{2}-2 a^{2} B +B a b -2 b^{2} B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 \left (a +b \right ) \left (a^{2}-2 a b +b^{2}\right )}\right )}{\left (a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b -a -b \right )^{2}}+\frac {\left (2 a^{2} A +A \,b^{2}-3 B a b \right ) \arctanh \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (a -b \right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{\left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}}{d} \]
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 \[ \text {Exception raised: ValueError} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 5.35, size = 251, normalized size = 1.53 \[ \frac {\mathrm {atanh}\left (\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (2\,a-2\,b\right )\,\left (a^2-2\,a\,b+b^2\right )}{2\,\sqrt {a+b}\,{\left (a-b\right )}^{5/2}}\right )\,\left (2\,A\,a^2-3\,B\,a\,b+A\,b^2\right )}{d\,{\left (a+b\right )}^{5/2}\,{\left (a-b\right )}^{5/2}}-\frac {\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (2\,B\,a^2-A\,b^2+2\,B\,b^2-4\,A\,a\,b+B\,a\,b\right )}{{\left (a+b\right )}^2\,\left (a-b\right )}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (A\,b^2+2\,B\,a^2+2\,B\,b^2-4\,A\,a\,b-B\,a\,b\right )}{\left (a+b\right )\,\left (a^2-2\,a\,b+b^2\right )}}{d\,\left (2\,a\,b-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (2\,a^2-2\,b^2\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (a^2-2\,a\,b+b^2\right )+a^2+b^2\right )} \]
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 \[ \int \frac {\left (A + B \sec {\left (c + d x \right )}\right ) \sec {\left (c + d x \right )}}{\left (a + b \sec {\left (c + d x \right )}\right )^{3}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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