Optimal. Leaf size=198 \[ \frac {\left (4 a^2 B+3 b^2 B+6 a b C\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {\left (4 b^2 C+5 a (2 b B+a C)\right ) \tan (c+d x)}{5 d}+\frac {\left (4 a^2 B+3 b^2 B+6 a b C\right ) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {b (5 b B+6 a C) \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac {b C \sec ^3(c+d x) (a+b \sec (c+d x)) \tan (c+d x)}{5 d}+\frac {\left (4 b^2 C+5 a (2 b B+a C)\right ) \tan ^3(c+d x)}{15 d} \]
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Rubi [A]
time = 0.26, antiderivative size = 198, normalized size of antiderivative = 1.00, number of steps
used = 8, number of rules used = 7, integrand size = 40, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.175, Rules used = {4157, 4111,
4132, 3852, 4131, 3853, 3855} \begin {gather*} \frac {\left (4 a^2 B+6 a b C+3 b^2 B\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {\left (4 a^2 B+6 a b C+3 b^2 B\right ) \tan (c+d x) \sec (c+d x)}{8 d}+\frac {\left (5 a (a C+2 b B)+4 b^2 C\right ) \tan ^3(c+d x)}{15 d}+\frac {\left (5 a (a C+2 b B)+4 b^2 C\right ) \tan (c+d x)}{5 d}+\frac {b (6 a C+5 b B) \tan (c+d x) \sec ^3(c+d x)}{20 d}+\frac {b C \tan (c+d x) \sec ^3(c+d x) (a+b \sec (c+d x))}{5 d} \end {gather*}
Antiderivative was successfully verified.
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Rule 3852
Rule 3853
Rule 3855
Rule 4111
Rule 4131
Rule 4132
Rule 4157
Rubi steps
\begin {align*} \int \sec ^2(c+d x) (a+b \sec (c+d x))^2 \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx &=\int \sec ^3(c+d x) (a+b \sec (c+d x))^2 (B+C \sec (c+d x)) \, dx\\ &=\frac {b C \sec ^3(c+d x) (a+b \sec (c+d x)) \tan (c+d x)}{5 d}+\frac {1}{5} \int \sec ^3(c+d x) \left (a (5 a B+3 b C)+\left (4 b^2 C+5 a (2 b B+a C)\right ) \sec (c+d x)+b (5 b B+6 a C) \sec ^2(c+d x)\right ) \, dx\\ &=\frac {b C \sec ^3(c+d x) (a+b \sec (c+d x)) \tan (c+d x)}{5 d}+\frac {1}{5} \int \sec ^3(c+d x) \left (a (5 a B+3 b C)+b (5 b B+6 a C) \sec ^2(c+d x)\right ) \, dx+\frac {1}{5} \left (4 b^2 C+5 a (2 b B+a C)\right ) \int \sec ^4(c+d x) \, dx\\ &=\frac {b (5 b B+6 a C) \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac {b C \sec ^3(c+d x) (a+b \sec (c+d x)) \tan (c+d x)}{5 d}+\frac {1}{4} \left (4 a^2 B+3 b^2 B+6 a b C\right ) \int \sec ^3(c+d x) \, dx-\frac {\left (4 b^2 C+5 a (2 b B+a C)\right ) \text {Subst}\left (\int \left (1+x^2\right ) \, dx,x,-\tan (c+d x)\right )}{5 d}\\ &=\frac {\left (4 b^2 C+5 a (2 b B+a C)\right ) \tan (c+d x)}{5 d}+\frac {\left (4 a^2 B+3 b^2 B+6 a b C\right ) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {b (5 b B+6 a C) \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac {b C \sec ^3(c+d x) (a+b \sec (c+d x)) \tan (c+d x)}{5 d}+\frac {\left (4 b^2 C+5 a (2 b B+a C)\right ) \tan ^3(c+d x)}{15 d}+\frac {1}{8} \left (4 a^2 B+3 b^2 B+6 a b C\right ) \int \sec (c+d x) \, dx\\ &=\frac {\left (4 a^2 B+3 b^2 B+6 a b C\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {\left (4 b^2 C+5 a (2 b B+a C)\right ) \tan (c+d x)}{5 d}+\frac {\left (4 a^2 B+3 b^2 B+6 a b C\right ) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {b (5 b B+6 a C) \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac {b C \sec ^3(c+d x) (a+b \sec (c+d x)) \tan (c+d x)}{5 d}+\frac {\left (4 b^2 C+5 a (2 b B+a C)\right ) \tan ^3(c+d x)}{15 d}\\ \end {align*}
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Mathematica [A]
time = 1.55, size = 150, normalized size = 0.76 \begin {gather*} \frac {15 \left (4 a^2 B+3 b^2 B+6 a b C\right ) \tanh ^{-1}(\sin (c+d x))+\tan (c+d x) \left (15 \left (4 a^2 B+3 b^2 B+6 a b C\right ) \sec (c+d x)+30 b (b B+2 a C) \sec ^3(c+d x)+8 \left (15 \left (2 a b B+a^2 C+b^2 C\right )+5 \left (2 a b B+a^2 C+2 b^2 C\right ) \tan ^2(c+d x)+3 b^2 C \tan ^4(c+d x)\right )\right )}{120 d} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.11, size = 221, normalized size = 1.12
method | result | size |
derivativedivides | \(\frac {b^{2} B \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )-b^{2} C \left (-\frac {8}{15}-\frac {\left (\sec ^{4}\left (d x +c \right )\right )}{5}-\frac {4 \left (\sec ^{2}\left (d x +c \right )\right )}{15}\right ) \tan \left (d x +c \right )-2 a b B \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+2 a b C \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+a^{2} B \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-a^{2} C \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )}{d}\) | \(221\) |
default | \(\frac {b^{2} B \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )-b^{2} C \left (-\frac {8}{15}-\frac {\left (\sec ^{4}\left (d x +c \right )\right )}{5}-\frac {4 \left (\sec ^{2}\left (d x +c \right )\right )}{15}\right ) \tan \left (d x +c \right )-2 a b B \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+2 a b C \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+a^{2} B \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-a^{2} C \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )}{d}\) | \(221\) |
norman | \(\frac {-\frac {4 \left (50 a b B +25 a^{2} C +29 b^{2} C \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{15 d}+\frac {\left (4 a^{2} B -16 a b B +5 b^{2} B -8 a^{2} C +10 a b C -8 b^{2} C \right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}-\frac {\left (4 a^{2} B +16 a b B +5 b^{2} B +8 a^{2} C +10 a b C +8 b^{2} C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}-\frac {\left (12 a^{2} B -64 a b B +3 b^{2} B -32 a^{2} C +6 a b C -16 b^{2} C \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 d}+\frac {\left (12 a^{2} B +64 a b B +3 b^{2} B +32 a^{2} C +6 a b C +16 b^{2} C \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 d}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{5}}-\frac {\left (4 a^{2} B +3 b^{2} B +6 a b C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 d}+\frac {\left (4 a^{2} B +3 b^{2} B +6 a b C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 d}\) | \(325\) |
risch | \(-\frac {i \left (60 a^{2} B \,{\mathrm e}^{9 i \left (d x +c \right )}+45 B \,b^{2} {\mathrm e}^{9 i \left (d x +c \right )}+90 C a b \,{\mathrm e}^{9 i \left (d x +c \right )}+120 B \,a^{2} {\mathrm e}^{7 i \left (d x +c \right )}+210 B \,b^{2} {\mathrm e}^{7 i \left (d x +c \right )}+420 C a b \,{\mathrm e}^{7 i \left (d x +c \right )}-480 B a b \,{\mathrm e}^{6 i \left (d x +c \right )}-240 C \,a^{2} {\mathrm e}^{6 i \left (d x +c \right )}-1120 B a b \,{\mathrm e}^{4 i \left (d x +c \right )}-560 C \,a^{2} {\mathrm e}^{4 i \left (d x +c \right )}-640 C \,b^{2} {\mathrm e}^{4 i \left (d x +c \right )}-120 a^{2} B \,{\mathrm e}^{3 i \left (d x +c \right )}-210 B \,b^{2} {\mathrm e}^{3 i \left (d x +c \right )}-420 C a b \,{\mathrm e}^{3 i \left (d x +c \right )}-800 B a b \,{\mathrm e}^{2 i \left (d x +c \right )}-400 C \,a^{2} {\mathrm e}^{2 i \left (d x +c \right )}-320 C \,b^{2} {\mathrm e}^{2 i \left (d x +c \right )}-60 a^{2} B \,{\mathrm e}^{i \left (d x +c \right )}-45 B \,b^{2} {\mathrm e}^{i \left (d x +c \right )}-90 C b a \,{\mathrm e}^{i \left (d x +c \right )}-160 a b B -80 a^{2} C -64 b^{2} C \right )}{60 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{5}}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) a^{2} B}{2 d}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) b^{2} B}{8 d}-\frac {3 b a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{4 d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) a^{2} B}{2 d}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) b^{2} B}{8 d}+\frac {3 b a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{4 d}\) | \(462\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.29, size = 276, normalized size = 1.39 \begin {gather*} \frac {80 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C a^{2} + 160 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B a b + 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^{2} - 30 \, C a b {\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 )} - 15 \, B 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 \, B a^{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 \, d} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 1.67, size = 208, normalized size = 1.05 \begin {gather*} \frac {15 \, {\left (4 \, B a^{2} + 6 \, C a b + 3 \, B b^{2}\right )} \cos \left (d x + c\right )^{5} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, {\left (4 \, B a^{2} + 6 \, C a b + 3 \, B b^{2}\right )} \cos \left (d x + c\right )^{5} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (16 \, {\left (5 \, C a^{2} + 10 \, B a b + 4 \, C b^{2}\right )} \cos \left (d x + c\right )^{4} + 15 \, {\left (4 \, B a^{2} + 6 \, C a b + 3 \, B b^{2}\right )} \cos \left (d x + c\right )^{3} + 24 \, C b^{2} + 8 \, {\left (5 \, C a^{2} + 10 \, B a b + 4 \, C b^{2}\right )} \cos \left (d x + c\right )^{2} + 30 \, {\left (2 \, C a b + B b^{2}\right )} \cos \left (d x + c\right )\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 (B + C \sec {\left (c + d x \right )}\right ) \left (a + b \sec {\left (c + d x \right )}\right )^{2} \sec ^{3}{\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. 528 vs.
\(2 (186) = 372\).
time = 0.52, size = 528, normalized size = 2.67 \begin {gather*} \frac {15 \, {\left (4 \, B a^{2} + 6 \, C a b + 3 \, B b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 15 \, {\left (4 \, B a^{2} + 6 \, C a b + 3 \, B 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 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 120 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 240 \, B a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 150 \, C a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 75 \, B b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 120 \, C b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 120 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 320 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 640 \, B a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 60 \, C a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 30 \, B b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 160 \, C b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 400 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 800 \, B a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 464 \, C b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 120 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 320 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 640 \, B a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 60 \, C a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 30 \, B b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 160 \, C b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 60 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 120 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 240 \, B a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 150 \, C a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 75 \, B b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 120 \, C b^{2} \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 = 7.68, size = 359, normalized size = 1.81 \begin {gather*} \frac {\mathrm {atanh}\left (\frac {4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {B\,a^2}{2}+\frac {3\,C\,a\,b}{4}+\frac {3\,B\,b^2}{8}\right )}{2\,B\,a^2+3\,C\,a\,b+\frac {3\,B\,b^2}{2}}\right )\,\left (B\,a^2+\frac {3\,C\,a\,b}{2}+\frac {3\,B\,b^2}{4}\right )}{d}-\frac {\left (2\,C\,a^2-\frac {5\,B\,b^2}{4}-B\,a^2+2\,C\,b^2+4\,B\,a\,b-\frac {5\,C\,a\,b}{2}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (2\,B\,a^2+\frac {B\,b^2}{2}-\frac {16\,C\,a^2}{3}-\frac {8\,C\,b^2}{3}-\frac {32\,B\,a\,b}{3}+C\,a\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {20\,C\,a^2}{3}+\frac {40\,B\,a\,b}{3}+\frac {116\,C\,b^2}{15}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (-2\,B\,a^2-\frac {B\,b^2}{2}-\frac {16\,C\,a^2}{3}-\frac {8\,C\,b^2}{3}-\frac {32\,B\,a\,b}{3}-C\,a\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (B\,a^2+\frac {5\,B\,b^2}{4}+2\,C\,a^2+2\,C\,b^2+4\,B\,a\,b+\frac {5\,C\,a\,b}{2}\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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