Optimal. Leaf size=202 \[ -\frac {\left (3 a b B-a^2 (2 A+C)-b^2 (A+2 C)\right ) \tanh ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{(a-b)^{5/2} (a+b)^{5/2} d}-\frac {\left (A b^2-a (b B-a C)\right ) \tan (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}+\frac {\left (a^2 b B+2 b^3 B+a^3 C-a b^2 (3 A+4 C)\right ) \tan (c+d x)}{2 b \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))} \]
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Rubi [A]
time = 0.33, antiderivative size = 202, normalized size of antiderivative = 1.00, number of steps
used = 6, number of rules used = 6, integrand size = 39, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {4165, 4088, 12,
3916, 2738, 214} \begin {gather*} -\frac {\left (-\left (a^2 (2 A+C)\right )+3 a b B-b^2 (A+2 C)\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 {\tan (c+d x) \left (A b^2-a (b B-a C)\right )}{2 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}+\frac {\tan (c+d x) \left (a^3 C+a^2 b B-a b^2 (3 A+4 C)+2 b^3 B\right )}{2 b d \left (a^2-b^2\right )^2 (a+b \sec (c+d x))} \end {gather*}
Antiderivative was successfully verified.
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Rule 12
Rule 214
Rule 2738
Rule 3916
Rule 4088
Rule 4165
Rubi steps
\begin {align*} \int \frac {\sec (c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^3} \, dx &=-\frac {\left (A b^2-a (b B-a C)\right ) \tan (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}-\frac {\int \frac {\sec (c+d x) \left (2 b (b B-a (A+C))+\left (A b^2-a b B-a^2 C+2 b^2 C\right ) \sec (c+d x)\right )}{(a+b \sec (c+d x))^2} \, dx}{2 b \left (a^2-b^2\right )}\\ &=-\frac {\left (A b^2-a (b B-a C)\right ) \tan (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}+\frac {\left (a^2 b B+2 b^3 B+a^3 C-a b^2 (3 A+4 C)\right ) \tan (c+d x)}{2 b \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))}-\frac {\int \frac {b \left (3 a b B-a^2 (2 A+C)-b^2 (A+2 C)\right ) \sec (c+d x)}{a+b \sec (c+d x)} \, dx}{2 b \left (a^2-b^2\right )^2}\\ &=-\frac {\left (A b^2-a (b B-a C)\right ) \tan (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}+\frac {\left (a^2 b B+2 b^3 B+a^3 C-a b^2 (3 A+4 C)\right ) \tan (c+d x)}{2 b \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))}-\frac {\left (3 a b B-a^2 (2 A+C)-b^2 (A+2 C)\right ) \int \frac {\sec (c+d x)}{a+b \sec (c+d x)} \, dx}{2 \left (a^2-b^2\right )^2}\\ &=-\frac {\left (A b^2-a (b B-a C)\right ) \tan (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}+\frac {\left (a^2 b B+2 b^3 B+a^3 C-a b^2 (3 A+4 C)\right ) \tan (c+d x)}{2 b \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))}-\frac {\left (3 a b B-a^2 (2 A+C)-b^2 (A+2 C)\right ) \int \frac {1}{1+\frac {a \cos (c+d x)}{b}} \, dx}{2 b \left (a^2-b^2\right )^2}\\ &=-\frac {\left (A b^2-a (b B-a C)\right ) \tan (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}+\frac {\left (a^2 b B+2 b^3 B+a^3 C-a b^2 (3 A+4 C)\right ) \tan (c+d x)}{2 b \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))}-\frac {\left (3 a b B-a^2 (2 A+C)-b^2 (A+2 C)\right ) \text {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+a^2 C+2 b^2 C\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 {\left (A b^2-a (b B-a C)\right ) \tan (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}+\frac {\left (a^2 b B+2 b^3 B+a^3 C-a b^2 (3 A+4 C)\right ) \tan (c+d x)}{2 b \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))}\\ \end {align*}
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Mathematica [C] Result contains complex when optimal does not.
time = 4.62, size = 410, normalized size = 2.03 \begin {gather*} \frac {(b+a \cos (c+d x)) \sec (c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (-\frac {4 i \left (-3 a b B+a^2 (2 A+C)+b^2 (A+2 C)\right ) \text {ArcTan}\left (\frac {(i \cos (c)+\sin (c)) \left (a \sin (c)+(-b+a \cos (c)) \tan \left (\frac {d x}{2}\right )\right )}{\sqrt {a^2-b^2} \sqrt {(\cos (c)-i \sin (c))^2}}\right ) (b+a \cos (c+d x))^2 (\cos (c)-i \sin (c))}{\left (a^2-b^2\right )^{5/2} \sqrt {(\cos (c)-i \sin (c))^2}}+\frac {a \sec (c) \left (\left (2 A b^4+5 a^3 b B+4 a b^3 B+a^4 C-a^2 b^2 (11 A+10 C)\right ) \sin (d x)+\left (-2 A b^4-3 a^3 b B+a^4 C+a^2 b^2 (5 A+2 C)\right ) \sin (2 c+d x)+a \left (A b^3+2 a^3 B+a b^2 B-a^2 b (4 A+3 C)\right ) \sin (c+2 d x)\right )-\left (a^2+2 b^2\right ) \left (A b^3+2 a^3 B+a b^2 B-a^2 b (4 A+3 C)\right ) \tan (c)}{\left (a^3-a b^2\right )^2}\right )}{2 d (A+2 C+2 B \cos (c+d x)+A \cos (2 (c+d x))) (a+b \sec (c+d x))^3} \end {gather*}
Warning: Unable to verify antiderivative.
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Maple [A]
time = 0.57, size = 268, normalized size = 1.33 Too large to display
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 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 395 vs.
\(2 (185) = 370\).
time = 2.85, size = 848, normalized size = 4.20 \begin {gather*} \left [\frac {{\left ({\left (2 \, A + C\right )} a^{2} b^{2} - 3 \, B a b^{3} + {\left (A + 2 \, C\right )} b^{4} + {\left ({\left (2 \, A + C\right )} a^{4} - 3 \, B a^{3} b + {\left (A + 2 \, C\right )} a^{2} b^{2}\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left ({\left (2 \, A + C\right )} a^{3} b - 3 \, B a^{2} b^{2} + {\left (A + 2 \, C\right )} 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 (C a^{5} + B a^{4} b - {\left (3 \, A + 5 \, C\right )} a^{3} b^{2} + B a^{2} b^{3} + {\left (3 \, A + 4 \, C\right )} a b^{4} - 2 \, B b^{5} + {\left (2 \, B a^{5} - {\left (4 \, A + 3 \, C\right )} a^{4} b - B a^{3} b^{2} + {\left (5 \, A + 3 \, C\right )} 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 ({\left (2 \, A + C\right )} a^{2} b^{2} - 3 \, B a b^{3} + {\left (A + 2 \, C\right )} b^{4} + {\left ({\left (2 \, A + C\right )} a^{4} - 3 \, B a^{3} b + {\left (A + 2 \, C\right )} a^{2} b^{2}\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left ({\left (2 \, A + C\right )} a^{3} b - 3 \, B a^{2} b^{2} + {\left (A + 2 \, C\right )} 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 (C a^{5} + B a^{4} b - {\left (3 \, A + 5 \, C\right )} a^{3} b^{2} + B a^{2} b^{3} + {\left (3 \, A + 4 \, C\right )} a b^{4} - 2 \, B b^{5} + {\left (2 \, B a^{5} - {\left (4 \, A + 3 \, C\right )} a^{4} b - B a^{3} b^{2} + {\left (5 \, A + 3 \, C\right )} 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 ] \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*} \int \frac {\left (A + B \sec {\left (c + d x \right )} + C \sec ^{2}{\left (c + d x \right )}\right ) \sec {\left (c + d x \right )}}{\left (a + b \sec {\left (c + d x \right )}\right )^{3}}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 510 vs.
\(2 (185) = 370\).
time = 0.56, size = 510, normalized size = 2.52 \begin {gather*} \frac {\frac {{\left (2 \, A a^{2} + C a^{2} - 3 \, B a b + A b^{2} + 2 \, C 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} - C 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 \, C 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} + 4 \, C 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 ) - C 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 \, C 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 ) + 4 \, C 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} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 7.84, size = 280, normalized size = 1.39 \begin {gather*} \frac {\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (A\,b^2-2\,B\,a^2-2\,B\,b^2+C\,a^2+4\,A\,a\,b-B\,a\,b+4\,C\,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+C\,a^2-4\,A\,a\,b-B\,a\,b-4\,C\,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 )}+\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+A\,b^2+C\,a^2+2\,C\,b^2-3\,B\,a\,b\right )}{d\,{\left (a+b\right )}^{5/2}\,{\left (a-b\right )}^{5/2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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