Optimal. Leaf size=263 \[ \frac {a (A b-a B) \tan (c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^3}-\frac {\left (a^3 (-B)+4 a^2 A b-4 a b^2 B+A b^3\right ) \tanh ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{d (a-b)^{7/2} (a+b)^{7/2}}+\frac {\left (a^3 B+2 a^2 A b-6 a b^2 B+3 A b^3\right ) \tan (c+d x)}{6 b d \left (a^2-b^2\right )^2 (a+b \sec (c+d x))^2}+\frac {\left (a^4 B+2 a^3 A b-10 a^2 b^2 B+13 a A b^3-6 b^4 B\right ) \tan (c+d x)}{6 b d \left (a^2-b^2\right )^3 (a+b \sec (c+d x))} \]
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Rubi [A] time = 0.62, antiderivative size = 263, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.194, Rules used = {4009, 4003, 12, 3831, 2659, 208} \[ -\frac {\left (4 a^2 A b+a^3 (-B)-4 a b^2 B+A b^3\right ) \tanh ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{d (a-b)^{7/2} (a+b)^{7/2}}+\frac {\left (2 a^3 A b-10 a^2 b^2 B+a^4 B+13 a A b^3-6 b^4 B\right ) \tan (c+d x)}{6 b d \left (a^2-b^2\right )^3 (a+b \sec (c+d x))}+\frac {\left (2 a^2 A b+a^3 B-6 a b^2 B+3 A b^3\right ) \tan (c+d x)}{6 b d \left (a^2-b^2\right )^2 (a+b \sec (c+d x))^2}+\frac {a (A b-a B) \tan (c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^3} \]
Antiderivative was successfully verified.
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Rule 12
Rule 208
Rule 2659
Rule 3831
Rule 4003
Rule 4009
Rubi steps
\begin {align*} \int \frac {\sec ^2(c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^4} \, dx &=\frac {a (A b-a B) \tan (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^3}+\frac {\int \frac {\sec (c+d x) \left (-3 b (A b-a B)+\left (2 a A b+a^2 B-3 b^2 B\right ) \sec (c+d x)\right )}{(a+b \sec (c+d x))^3} \, dx}{3 b \left (a^2-b^2\right )}\\ &=\frac {a (A b-a B) \tan (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^3}+\frac {\left (2 a^2 A b+3 A b^3+a^3 B-6 a b^2 B\right ) \tan (c+d x)}{6 b \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))^2}-\frac {\int \frac {\sec (c+d x) \left (2 b \left (5 a A b-2 a^2 B-3 b^2 B\right )-\left (2 a^2 A b+3 A b^3+a^3 B-6 a b^2 B\right ) \sec (c+d x)\right )}{(a+b \sec (c+d x))^2} \, dx}{6 b \left (a^2-b^2\right )^2}\\ &=\frac {a (A b-a B) \tan (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^3}+\frac {\left (2 a^2 A b+3 A b^3+a^3 B-6 a b^2 B\right ) \tan (c+d x)}{6 b \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))^2}+\frac {\left (2 a^3 A b+13 a A b^3+a^4 B-10 a^2 b^2 B-6 b^4 B\right ) \tan (c+d x)}{6 b \left (a^2-b^2\right )^3 d (a+b \sec (c+d x))}+\frac {\int -\frac {3 b \left (4 a^2 A b+A b^3-a^3 B-4 a b^2 B\right ) \sec (c+d x)}{a+b \sec (c+d x)} \, dx}{6 b \left (a^2-b^2\right )^3}\\ &=\frac {a (A b-a B) \tan (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^3}+\frac {\left (2 a^2 A b+3 A b^3+a^3 B-6 a b^2 B\right ) \tan (c+d x)}{6 b \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))^2}+\frac {\left (2 a^3 A b+13 a A b^3+a^4 B-10 a^2 b^2 B-6 b^4 B\right ) \tan (c+d x)}{6 b \left (a^2-b^2\right )^3 d (a+b \sec (c+d x))}-\frac {\left (4 a^2 A b+A b^3-a^3 B-4 a b^2 B\right ) \int \frac {\sec (c+d x)}{a+b \sec (c+d x)} \, dx}{2 \left (a^2-b^2\right )^3}\\ &=\frac {a (A b-a B) \tan (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^3}+\frac {\left (2 a^2 A b+3 A b^3+a^3 B-6 a b^2 B\right ) \tan (c+d x)}{6 b \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))^2}+\frac {\left (2 a^3 A b+13 a A b^3+a^4 B-10 a^2 b^2 B-6 b^4 B\right ) \tan (c+d x)}{6 b \left (a^2-b^2\right )^3 d (a+b \sec (c+d x))}-\frac {\left (4 a^2 A b+A b^3-a^3 B-4 a b^2 B\right ) \int \frac {1}{1+\frac {a \cos (c+d x)}{b}} \, dx}{2 b \left (a^2-b^2\right )^3}\\ &=\frac {a (A b-a B) \tan (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^3}+\frac {\left (2 a^2 A b+3 A b^3+a^3 B-6 a b^2 B\right ) \tan (c+d x)}{6 b \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))^2}+\frac {\left (2 a^3 A b+13 a A b^3+a^4 B-10 a^2 b^2 B-6 b^4 B\right ) \tan (c+d x)}{6 b \left (a^2-b^2\right )^3 d (a+b \sec (c+d x))}-\frac {\left (4 a^2 A b+A b^3-a^3 B-4 a b^2 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 )^3 d}\\ &=-\frac {\left (4 a^2 A b+A b^3-a^3 B-4 a b^2 B\right ) \tanh ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{(a-b)^{7/2} (a+b)^{7/2} d}+\frac {a (A b-a B) \tan (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^3}+\frac {\left (2 a^2 A b+3 A b^3+a^3 B-6 a b^2 B\right ) \tan (c+d x)}{6 b \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))^2}+\frac {\left (2 a^3 A b+13 a A b^3+a^4 B-10 a^2 b^2 B-6 b^4 B\right ) \tan (c+d x)}{6 b \left (a^2-b^2\right )^3 d (a+b \sec (c+d x))}\\ \end {align*}
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Mathematica [A] time = 1.27, size = 252, normalized size = 0.96 \[ \frac {\frac {24 \left (a^3 B-4 a^2 A b+4 a b^2 B-A b^3\right ) \tanh ^{-1}\left (\frac {(b-a) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}}+\frac {2 \sin (c+d x) \left (-6 a^5 A+11 a^4 b B-14 a^3 A b^2+22 a^2 b^3 B+a \left (-6 a^4 A+13 a^3 b B-10 a^2 A b^2+2 a b^3 B+A b^4\right ) \cos (2 (c+d x))-6 \left (a^5 B+2 a^4 A b-9 a^3 b^2 B+9 a^2 A b^3-2 a b^4 B-A b^5\right ) \cos (c+d x)-25 a A b^4+12 b^5 B\right )}{(a \cos (c+d x)+b)^3}}{24 d \left (b^2-a^2\right )^3} \]
Antiderivative was successfully verified.
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fricas [B] time = 0.58, size = 1242, normalized size = 4.72 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 1.99, size = 726, normalized size = 2.76 \[ \frac {\frac {3 \, {\left (B a^{3} - 4 \, A a^{2} b + 4 \, B a b^{2} - A b^{3}\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^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} \sqrt {-a^{2} + b^{2}}} - \frac {6 \, A a^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, B a^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 6 \, A a^{4} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 12 \, B a^{4} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 12 \, A a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 27 \, B a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 27 \, A a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 12 \, B a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 12 \, A a b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6 \, B a b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, A b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 6 \, B b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 12 \, A a^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 28 \, B a^{4} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 16 \, A a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 16 \, B a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 28 \, A a b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 12 \, B b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 6 \, A a^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, B a^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, A a^{4} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 12 \, B a^{4} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 12 \, A a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 27 \, B a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 27 \, A a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 12 \, B a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 12 \, A a b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 6 \, B a b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 3 \, A b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 6 \, B b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\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 )}^{3}}}{3 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.79, size = 388, normalized size = 1.48 \[ \frac {\frac {-\frac {\left (2 A \,a^{3}+2 A \,a^{2} b +6 A a \,b^{2}+A \,b^{3}-a^{3} B -6 a^{2} b B -2 B a \,b^{2}-2 b^{3} B \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{\left (a -b \right ) \left (a^{3}+3 a^{2} b +3 b^{2} a +b^{3}\right )}+\frac {4 \left (3 A \,a^{3}+7 A a \,b^{2}-7 a^{2} b B -3 b^{3} B \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 \left (a^{2}+2 a b +b^{2}\right ) \left (a^{2}-2 a b +b^{2}\right )}-\frac {\left (2 A \,a^{3}-2 A \,a^{2} b +6 A a \,b^{2}-A \,b^{3}+a^{3} B -6 a^{2} b B +2 B a \,b^{2}-2 b^{3} B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (a +b \right ) \left (a^{3}-3 a^{2} b +3 b^{2} a -b^{3}\right )}}{\left (a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b -a -b \right )^{3}}-\frac {\left (4 A \,a^{2} b +A \,b^{3}-a^{3} B -4 B a \,b^{2}\right ) \arctanh \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (a -b \right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{\left (a^{6}-3 a^{4} b^{2}+3 b^{4} a^{2}-b^{6}\right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}}{d} \]
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 = 6.65, size = 451, normalized size = 1.71 \[ \frac {\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (2\,A\,a^3-A\,b^3+B\,a^3-2\,B\,b^3+6\,A\,a\,b^2-2\,A\,a^2\,b+2\,B\,a\,b^2-6\,B\,a^2\,b\right )}{\left (a+b\right )\,\left (a^3-3\,a^2\,b+3\,a\,b^2-b^3\right )}-\frac {4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (3\,A\,a^3-7\,B\,a^2\,b+7\,A\,a\,b^2-3\,B\,b^3\right )}{3\,{\left (a+b\right )}^2\,\left (a^2-2\,a\,b+b^2\right )}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (2\,A\,a^3+A\,b^3-B\,a^3-2\,B\,b^3+6\,A\,a\,b^2+2\,A\,a^2\,b-2\,B\,a\,b^2-6\,B\,a^2\,b\right )}{{\left (a+b\right )}^3\,\left (a-b\right )}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (-3\,a^3-3\,a^2\,b+3\,a\,b^2+3\,b^3\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (-3\,a^3+3\,a^2\,b+3\,a\,b^2-3\,b^3\right )+3\,a\,b^2+3\,a^2\,b+a^3+b^3-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (a^3-3\,a^2\,b+3\,a\,b^2-b^3\right )\right )}-\frac {\mathrm {atanh}\left (\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (2\,a-2\,b\right )\,\left (a^3-3\,a^2\,b+3\,a\,b^2-b^3\right )}{2\,\sqrt {a+b}\,{\left (a-b\right )}^{7/2}}\right )\,\left (-B\,a^3+4\,A\,a^2\,b-4\,B\,a\,b^2+A\,b^3\right )}{d\,{\left (a+b\right )}^{7/2}\,{\left (a-b\right )}^{7/2}} \]
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 \[ \int \frac {\left (A + B \sec {\left (c + d x \right )}\right ) \sec ^{2}{\left (c + d x \right )}}{\left (a + b \sec {\left (c + d x \right )}\right )^{4}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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