Optimal. Leaf size=234 \[ \frac {a \left (4 a^2 (2 A+C)+3 b^2 (4 A+3 C)\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}-\frac {\left (3 a^4 C-4 b^4 (5 A+4 C)-4 a^2 b^2 (20 A+13 C)\right ) \tan (c+d x)}{30 b d}+\frac {a \left (100 A b^2-6 a^2 C+71 b^2 C\right ) \sec (c+d x) \tan (c+d x)}{120 d}-\frac {\left (3 a^2 C-4 b^2 (5 A+4 C)\right ) (a+b \sec (c+d x))^2 \tan (c+d x)}{60 b d}-\frac {a C (a+b \sec (c+d x))^3 \tan (c+d x)}{20 b d}+\frac {C (a+b \sec (c+d x))^4 \tan (c+d x)}{5 b d} \]
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Rubi [A]
time = 0.34, antiderivative size = 234, normalized size of antiderivative = 1.00, number of steps
used = 8, number of rules used = 7, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.226, Rules used = {4168, 4087,
4082, 3872, 3855, 3852, 8} \begin {gather*} \frac {a \left (4 a^2 (2 A+C)+3 b^2 (4 A+3 C)\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}-\frac {\left (3 a^2 C-4 b^2 (5 A+4 C)\right ) \tan (c+d x) (a+b \sec (c+d x))^2}{60 b d}+\frac {a \left (-6 a^2 C+100 A b^2+71 b^2 C\right ) \tan (c+d x) \sec (c+d x)}{120 d}-\frac {\left (3 a^4 C-4 a^2 b^2 (20 A+13 C)-4 b^4 (5 A+4 C)\right ) \tan (c+d x)}{30 b d}+\frac {C \tan (c+d x) (a+b \sec (c+d x))^4}{5 b d}-\frac {a C \tan (c+d x) (a+b \sec (c+d x))^3}{20 b d} \end {gather*}
Antiderivative was successfully verified.
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Rule 8
Rule 3852
Rule 3855
Rule 3872
Rule 4082
Rule 4087
Rule 4168
Rubi steps
\begin {align*} \int \sec (c+d x) (a+b \sec (c+d x))^3 \left (A+C \sec ^2(c+d x)\right ) \, dx &=\frac {C (a+b \sec (c+d x))^4 \tan (c+d x)}{5 b d}+\frac {\int \sec (c+d x) (a+b \sec (c+d x))^3 (b (5 A+4 C)-a C \sec (c+d x)) \, dx}{5 b}\\ &=-\frac {a C (a+b \sec (c+d x))^3 \tan (c+d x)}{20 b d}+\frac {C (a+b \sec (c+d x))^4 \tan (c+d x)}{5 b d}+\frac {\int \sec (c+d x) (a+b \sec (c+d x))^2 \left (a b (20 A+13 C)-\left (3 a^2 C-4 b^2 (5 A+4 C)\right ) \sec (c+d x)\right ) \, dx}{20 b}\\ &=-\frac {\left (3 a^2 C-4 b^2 (5 A+4 C)\right ) (a+b \sec (c+d x))^2 \tan (c+d x)}{60 b d}-\frac {a C (a+b \sec (c+d x))^3 \tan (c+d x)}{20 b d}+\frac {C (a+b \sec (c+d x))^4 \tan (c+d x)}{5 b d}+\frac {\int \sec (c+d x) (a+b \sec (c+d x)) \left (b \left (8 b^2 (5 A+4 C)+a^2 (60 A+33 C)\right )+a \left (100 A b^2-6 a^2 C+71 b^2 C\right ) \sec (c+d x)\right ) \, dx}{60 b}\\ &=\frac {a \left (100 A b^2-6 a^2 C+71 b^2 C\right ) \sec (c+d x) \tan (c+d x)}{120 d}-\frac {\left (3 a^2 C-4 b^2 (5 A+4 C)\right ) (a+b \sec (c+d x))^2 \tan (c+d x)}{60 b d}-\frac {a C (a+b \sec (c+d x))^3 \tan (c+d x)}{20 b d}+\frac {C (a+b \sec (c+d x))^4 \tan (c+d x)}{5 b d}+\frac {\int \sec (c+d x) \left (15 a b \left (4 a^2 (2 A+C)+3 b^2 (4 A+3 C)\right )-4 \left (3 a^4 C-4 b^4 (5 A+4 C)-4 a^2 b^2 (20 A+13 C)\right ) \sec (c+d x)\right ) \, dx}{120 b}\\ &=\frac {a \left (100 A b^2-6 a^2 C+71 b^2 C\right ) \sec (c+d x) \tan (c+d x)}{120 d}-\frac {\left (3 a^2 C-4 b^2 (5 A+4 C)\right ) (a+b \sec (c+d x))^2 \tan (c+d x)}{60 b d}-\frac {a C (a+b \sec (c+d x))^3 \tan (c+d x)}{20 b d}+\frac {C (a+b \sec (c+d x))^4 \tan (c+d x)}{5 b d}+\frac {1}{8} \left (a \left (4 a^2 (2 A+C)+3 b^2 (4 A+3 C)\right )\right ) \int \sec (c+d x) \, dx-\frac {\left (3 a^4 C-4 b^4 (5 A+4 C)-4 a^2 b^2 (20 A+13 C)\right ) \int \sec ^2(c+d x) \, dx}{30 b}\\ &=\frac {a \left (4 a^2 (2 A+C)+3 b^2 (4 A+3 C)\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {a \left (100 A b^2-6 a^2 C+71 b^2 C\right ) \sec (c+d x) \tan (c+d x)}{120 d}-\frac {\left (3 a^2 C-4 b^2 (5 A+4 C)\right ) (a+b \sec (c+d x))^2 \tan (c+d x)}{60 b d}-\frac {a C (a+b \sec (c+d x))^3 \tan (c+d x)}{20 b d}+\frac {C (a+b \sec (c+d x))^4 \tan (c+d x)}{5 b d}+\frac {\left (3 a^4 C-4 b^4 (5 A+4 C)-4 a^2 b^2 (20 A+13 C)\right ) \text {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{30 b d}\\ &=\frac {a \left (4 a^2 (2 A+C)+3 b^2 (4 A+3 C)\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}-\frac {\left (3 a^4 C-4 b^4 (5 A+4 C)-4 a^2 b^2 (20 A+13 C)\right ) \tan (c+d x)}{30 b d}+\frac {a \left (100 A b^2-6 a^2 C+71 b^2 C\right ) \sec (c+d x) \tan (c+d x)}{120 d}-\frac {\left (3 a^2 C-4 b^2 (5 A+4 C)\right ) (a+b \sec (c+d x))^2 \tan (c+d x)}{60 b d}-\frac {a C (a+b \sec (c+d x))^3 \tan (c+d x)}{20 b d}+\frac {C (a+b \sec (c+d x))^4 \tan (c+d x)}{5 b d}\\ \end {align*}
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Mathematica [A]
time = 1.89, size = 324, normalized size = 1.38 \begin {gather*} -\frac {\left (C+A \cos ^2(c+d x)\right ) \sec ^5(c+d x) \left (120 a \left (4 a^2 (2 A+C)+3 b^2 (4 A+3 C)\right ) \cos ^5(c+d x) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )-2 \left (540 a^2 A b+200 A b^3+600 a^2 b C+256 b^3 C+45 a \left (12 A b^2+4 a^2 C+17 b^2 C\right ) \cos (c+d x)+48 b \left (15 a^2 (A+C)+b^2 (5 A+4 C)\right ) \cos (2 (c+d x))+180 a A b^2 \cos (3 (c+d x))+60 a^3 C \cos (3 (c+d x))+135 a b^2 C \cos (3 (c+d x))+180 a^2 A b \cos (4 (c+d x))+40 A b^3 \cos (4 (c+d x))+120 a^2 b C \cos (4 (c+d x))+32 b^3 C \cos (4 (c+d x))\right ) \sin (c+d x)\right )}{480 d (A+2 C+A \cos (2 (c+d x)))} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.12, size = 246, normalized size = 1.05 Too large to display
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.29, size = 289, normalized size = 1.24 \begin {gather*} \frac {240 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C a^{2} b + 80 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A b^{3} + 16 \, {\left (3 \, \tan \left (d x + c\right )^{5} + 10 \, \tan \left (d x + c\right )^{3} + 15 \, \tan \left (d x + c\right )\right )} C b^{3} - 45 \, C a b^{2} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 60 \, C a^{3} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 180 \, A a b^{2} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 240 \, A a^{3} \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) + 720 \, A a^{2} b \tan \left (d x + c\right )}{240 \, d} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 4.50, size = 225, normalized size = 0.96 \begin {gather*} \frac {15 \, {\left (4 \, {\left (2 \, A + C\right )} a^{3} + 3 \, {\left (4 \, A + 3 \, C\right )} a b^{2}\right )} \cos \left (d x + c\right )^{5} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, {\left (4 \, {\left (2 \, A + C\right )} a^{3} + 3 \, {\left (4 \, A + 3 \, C\right )} a b^{2}\right )} \cos \left (d x + c\right )^{5} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (90 \, C a b^{2} \cos \left (d x + c\right ) + 8 \, {\left (15 \, {\left (3 \, A + 2 \, C\right )} a^{2} b + 2 \, {\left (5 \, A + 4 \, C\right )} b^{3}\right )} \cos \left (d x + c\right )^{4} + 24 \, C b^{3} + 15 \, {\left (4 \, C a^{3} + 3 \, {\left (4 \, A + 3 \, C\right )} a b^{2}\right )} \cos \left (d x + c\right )^{3} + 8 \, {\left (15 \, C a^{2} b + {\left (5 \, A + 4 \, C\right )} b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{240 \, d \cos \left (d x + c\right )^{5}} \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 \left (A + C \sec ^{2}{\left (c + d x \right )}\right ) \left (a + b \sec {\left (c + d x \right )}\right )^{3} \sec {\left (c + d x \right )}\, 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. 656 vs.
\(2 (222) = 444\).
time = 0.55, size = 656, normalized size = 2.80 \begin {gather*} \frac {15 \, {\left (8 \, A a^{3} + 4 \, C a^{3} + 12 \, A a b^{2} + 9 \, C a b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 15 \, {\left (8 \, A a^{3} + 4 \, C a^{3} + 12 \, A a b^{2} + 9 \, C a b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + \frac {2 \, {\left (60 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 360 \, A a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 360 \, C a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 180 \, A a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 225 \, C a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 120 \, A b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 120 \, C b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 120 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 1440 \, A a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 960 \, C a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 360 \, A a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 90 \, C a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 320 \, A b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 160 \, C b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 2160 \, A a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 1200 \, C a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 400 \, A b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 464 \, C b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 120 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 1440 \, A a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 960 \, C a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 360 \, A a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 90 \, C a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 320 \, A b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 160 \, C b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 60 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 360 \, A a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 360 \, C a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 180 \, A a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 225 \, C a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 120 \, A b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 120 \, C b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{5}}}{120 \, d} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 7.52, size = 445, normalized size = 1.90 \begin {gather*} \frac {a\,\mathrm {atanh}\left (\frac {a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (8\,A\,a^2+12\,A\,b^2+4\,C\,a^2+9\,C\,b^2\right )}{2\,\left (4\,A\,a^3+2\,C\,a^3+6\,A\,a\,b^2+\frac {9\,C\,a\,b^2}{2}\right )}\right )\,\left (8\,A\,a^2+12\,A\,b^2+4\,C\,a^2+9\,C\,b^2\right )}{4\,d}-\frac {\left (2\,A\,b^3-C\,a^3+2\,C\,b^3-3\,A\,a\,b^2+6\,A\,a^2\,b-\frac {15\,C\,a\,b^2}{4}+6\,C\,a^2\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (2\,C\,a^3-\frac {16\,A\,b^3}{3}-\frac {8\,C\,b^3}{3}+6\,A\,a\,b^2-24\,A\,a^2\,b+\frac {3\,C\,a\,b^2}{2}-16\,C\,a^2\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {20\,A\,b^3}{3}+\frac {116\,C\,b^3}{15}+36\,A\,a^2\,b+20\,C\,a^2\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (-\frac {16\,A\,b^3}{3}-2\,C\,a^3-\frac {8\,C\,b^3}{3}-6\,A\,a\,b^2-24\,A\,a^2\,b-\frac {3\,C\,a\,b^2}{2}-16\,C\,a^2\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (2\,A\,b^3+C\,a^3+2\,C\,b^3+3\,A\,a\,b^2+6\,A\,a^2\,b+\frac {15\,C\,a\,b^2}{4}+6\,C\,a^2\,b\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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