Optimal. Leaf size=170 \[ \frac {\left (4 a^2 (2 A+C)+b^2 (4 A+3 C)\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {a \left (12 A b^2-a^2 C+8 b^2 C\right ) \tan (c+d x)}{6 b d}-\frac {\left (2 a^2 C-3 b^2 (4 A+3 C)\right ) \sec (c+d x) \tan (c+d x)}{24 d}-\frac {a C (a+b \sec (c+d x))^2 \tan (c+d x)}{12 b d}+\frac {C (a+b \sec (c+d x))^3 \tan (c+d x)}{4 b d} \]
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Rubi [A]
time = 0.22, antiderivative size = 170, normalized size of antiderivative = 1.00, number of steps
used = 7, 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 (a^2 (-C)+12 A b^2+8 b^2 C\right ) \tan (c+d x)}{6 b d}+\frac {\left (4 a^2 (2 A+C)+b^2 (4 A+3 C)\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}-\frac {\left (2 a^2 C-3 b^2 (4 A+3 C)\right ) \tan (c+d x) \sec (c+d x)}{24 d}+\frac {C \tan (c+d x) (a+b \sec (c+d x))^3}{4 b d}-\frac {a C \tan (c+d x) (a+b \sec (c+d x))^2}{12 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))^2 \left (A+C \sec ^2(c+d x)\right ) \, dx &=\frac {C (a+b \sec (c+d x))^3 \tan (c+d x)}{4 b d}+\frac {\int \sec (c+d x) (a+b \sec (c+d x))^2 (b (4 A+3 C)-a C \sec (c+d x)) \, dx}{4 b}\\ &=-\frac {a C (a+b \sec (c+d x))^2 \tan (c+d x)}{12 b d}+\frac {C (a+b \sec (c+d x))^3 \tan (c+d x)}{4 b d}+\frac {\int \sec (c+d x) (a+b \sec (c+d x)) \left (a b (12 A+7 C)-\left (2 a^2 C-3 b^2 (4 A+3 C)\right ) \sec (c+d x)\right ) \, dx}{12 b}\\ &=-\frac {\left (2 a^2 C-3 b^2 (4 A+3 C)\right ) \sec (c+d x) \tan (c+d x)}{24 d}-\frac {a C (a+b \sec (c+d x))^2 \tan (c+d x)}{12 b d}+\frac {C (a+b \sec (c+d x))^3 \tan (c+d x)}{4 b d}+\frac {\int \sec (c+d x) \left (3 b \left (4 a^2 (2 A+C)+b^2 (4 A+3 C)\right )+4 a \left (12 A b^2-\left (a^2-8 b^2\right ) C\right ) \sec (c+d x)\right ) \, dx}{24 b}\\ &=-\frac {\left (2 a^2 C-3 b^2 (4 A+3 C)\right ) \sec (c+d x) \tan (c+d x)}{24 d}-\frac {a C (a+b \sec (c+d x))^2 \tan (c+d x)}{12 b d}+\frac {C (a+b \sec (c+d x))^3 \tan (c+d x)}{4 b d}+\frac {\left (a \left (12 A b^2-a^2 C+8 b^2 C\right )\right ) \int \sec ^2(c+d x) \, dx}{6 b}+\frac {1}{8} \left (4 a^2 (2 A+C)+b^2 (4 A+3 C)\right ) \int \sec (c+d x) \, dx\\ &=\frac {\left (4 a^2 (2 A+C)+b^2 (4 A+3 C)\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}-\frac {\left (2 a^2 C-3 b^2 (4 A+3 C)\right ) \sec (c+d x) \tan (c+d x)}{24 d}-\frac {a C (a+b \sec (c+d x))^2 \tan (c+d x)}{12 b d}+\frac {C (a+b \sec (c+d x))^3 \tan (c+d x)}{4 b d}-\frac {\left (a \left (12 A b^2-a^2 C+8 b^2 C\right )\right ) \text {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{6 b d}\\ &=\frac {\left (4 a^2 (2 A+C)+b^2 (4 A+3 C)\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {a \left (12 A b^2-a^2 C+8 b^2 C\right ) \tan (c+d x)}{6 b d}-\frac {\left (2 a^2 C-3 b^2 (4 A+3 C)\right ) \sec (c+d x) \tan (c+d x)}{24 d}-\frac {a C (a+b \sec (c+d x))^2 \tan (c+d x)}{12 b d}+\frac {C (a+b \sec (c+d x))^3 \tan (c+d x)}{4 b d}\\ \end {align*}
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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(1123\) vs. \(2(170)=340\).
time = 6.35, size = 1123, normalized size = 6.61 \begin {gather*} \frac {\left (-8 a^2 A-4 A b^2-4 a^2 C-3 b^2 C\right ) \cos ^4(c+d x) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right ) (a+b \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right )}{4 d (b+a \cos (c+d x))^2 (A+2 C+A \cos (2 c+2 d x))}+\frac {\left (8 a^2 A+4 A b^2+4 a^2 C+3 b^2 C\right ) \cos ^4(c+d x) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right ) (a+b \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right )}{4 d (b+a \cos (c+d x))^2 (A+2 C+A \cos (2 c+2 d x))}+\frac {b^2 C \cos ^4(c+d x) (a+b \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right )}{8 d (b+a \cos (c+d x))^2 (A+2 C+A \cos (2 c+2 d x)) \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^4}+\frac {\left (12 A b^2+12 a^2 C+8 a b C+9 b^2 C\right ) \cos ^4(c+d x) (a+b \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right )}{24 d (b+a \cos (c+d x))^2 (A+2 C+A \cos (2 c+2 d x)) \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {2 a b C \cos ^4(c+d x) (a+b \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right ) \sin \left (\frac {1}{2} (c+d x)\right )}{3 d (b+a \cos (c+d x))^2 (A+2 C+A \cos (2 c+2 d x)) \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^3}-\frac {b^2 C \cos ^4(c+d x) (a+b \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right )}{8 d (b+a \cos (c+d x))^2 (A+2 C+A \cos (2 c+2 d x)) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^4}+\frac {2 a b C \cos ^4(c+d x) (a+b \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right ) \sin \left (\frac {1}{2} (c+d x)\right )}{3 d (b+a \cos (c+d x))^2 (A+2 C+A \cos (2 c+2 d x)) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^3}+\frac {\left (-12 A b^2-12 a^2 C-8 a b C-9 b^2 C\right ) \cos ^4(c+d x) (a+b \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right )}{24 d (b+a \cos (c+d x))^2 (A+2 C+A \cos (2 c+2 d x)) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {4 \cos ^4(c+d x) (a+b \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right ) \left (3 a A b \sin \left (\frac {1}{2} (c+d x)\right )+2 a b C \sin \left (\frac {1}{2} (c+d x)\right )\right )}{3 d (b+a \cos (c+d x))^2 (A+2 C+A \cos (2 c+2 d x)) \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}+\frac {4 \cos ^4(c+d x) (a+b \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right ) \left (3 a A b \sin \left (\frac {1}{2} (c+d x)\right )+2 a b C \sin \left (\frac {1}{2} (c+d x)\right )\right )}{3 d (b+a \cos (c+d x))^2 (A+2 C+A \cos (2 c+2 d x)) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.10, size = 180, normalized size = 1.06
method | result | size |
derivativedivides | \(\frac {A \,b^{2} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+b^{2} C \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+2 a A b \tan \left (d x +c \right )-2 a b C \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+A \,a^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+a^{2} C \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\) | \(180\) |
default | \(\frac {A \,b^{2} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+b^{2} C \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+2 a A b \tan \left (d x +c \right )-2 a b C \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+A \,a^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+a^{2} C \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\) | \(180\) |
norman | \(\frac {-\frac {\left (16 a A b -4 A \,b^{2}-4 a^{2} C +16 a b C -5 b^{2} C \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}+\frac {\left (16 a A b +4 A \,b^{2}+4 a^{2} C +16 a b C +5 b^{2} C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}+\frac {\left (144 a A b -12 A \,b^{2}-12 a^{2} C +80 a b C +9 b^{2} C \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{12 d}-\frac {\left (144 a A b +12 A \,b^{2}+12 a^{2} C +80 a b C -9 b^{2} C \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{12 d}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{4}}-\frac {\left (8 A \,a^{2}+4 A \,b^{2}+4 a^{2} C +3 b^{2} C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 d}+\frac {\left (8 A \,a^{2}+4 A \,b^{2}+4 a^{2} C +3 b^{2} C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 d}\) | \(281\) |
risch | \(-\frac {i \left (12 A \,b^{2} {\mathrm e}^{7 i \left (d x +c \right )}+12 C \,a^{2} {\mathrm e}^{7 i \left (d x +c \right )}+9 C \,b^{2} {\mathrm e}^{7 i \left (d x +c \right )}-48 A a b \,{\mathrm e}^{6 i \left (d x +c \right )}+12 A \,b^{2} {\mathrm e}^{5 i \left (d x +c \right )}+12 C \,a^{2} {\mathrm e}^{5 i \left (d x +c \right )}+33 C \,b^{2} {\mathrm e}^{5 i \left (d x +c \right )}-144 A a b \,{\mathrm e}^{4 i \left (d x +c \right )}-96 C a b \,{\mathrm e}^{4 i \left (d x +c \right )}-12 A \,b^{2} {\mathrm e}^{3 i \left (d x +c \right )}-12 C \,a^{2} {\mathrm e}^{3 i \left (d x +c \right )}-33 C \,b^{2} {\mathrm e}^{3 i \left (d x +c \right )}-144 a A b \,{\mathrm e}^{2 i \left (d x +c \right )}-128 C a b \,{\mathrm e}^{2 i \left (d x +c \right )}-12 A \,b^{2} {\mathrm e}^{i \left (d x +c \right )}-12 C \,a^{2} {\mathrm e}^{i \left (d x +c \right )}-9 C \,b^{2} {\mathrm e}^{i \left (d x +c \right )}-48 a A b -32 a b C \right )}{12 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A \,a^{2}}{d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A \,b^{2}}{2 d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) a^{2} C}{2 d}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) b^{2} C}{8 d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A \,a^{2}}{d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A \,b^{2}}{2 d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) a^{2} C}{2 d}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) b^{2} C}{8 d}\) | \(457\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.28, size = 225, normalized size = 1.32 \begin {gather*} \frac {32 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C a b - 3 \, C 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 )} - 12 \, C 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 )} - 12 \, 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 )} + 48 \, A a^{2} \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) + 96 \, A a b \tan \left (d x + c\right )}{48 \, d} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 2.92, size = 171, normalized size = 1.01 \begin {gather*} \frac {3 \, {\left (4 \, {\left (2 \, A + C\right )} a^{2} + {\left (4 \, A + 3 \, C\right )} b^{2}\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (4 \, {\left (2 \, A + C\right )} a^{2} + {\left (4 \, A + 3 \, C\right )} b^{2}\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (16 \, {\left (3 \, A + 2 \, C\right )} a b \cos \left (d x + c\right )^{3} + 16 \, C a b \cos \left (d x + c\right ) + 6 \, C b^{2} + 3 \, {\left (4 \, C a^{2} + {\left (4 \, A + 3 \, C\right )} b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{48 \, d \cos \left (d x + c\right )^{4}} \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 )^{2} \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. 426 vs.
\(2 (159) = 318\).
time = 0.49, size = 426, normalized size = 2.51 \begin {gather*} \frac {3 \, {\left (8 \, A a^{2} + 4 \, C a^{2} + 4 \, A b^{2} + 3 \, C b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, {\left (8 \, A a^{2} + 4 \, C a^{2} + 4 \, A b^{2} + 3 \, C b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + \frac {2 \, {\left (12 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 48 \, A a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 48 \, C a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 12 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 15 \, C b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 12 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 144 \, A a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 80 \, C a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 12 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 9 \, C b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 12 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 144 \, A a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 80 \, C a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 12 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 9 \, C b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 12 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 48 \, A a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 48 \, C a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 12 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 15 \, 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 )}^{4}}}{24 \, d} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 7.37, size = 307, normalized size = 1.81 \begin {gather*} \frac {\left (A\,b^2+C\,a^2+\frac {5\,C\,b^2}{4}-4\,A\,a\,b-4\,C\,a\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {3\,C\,b^2}{4}-C\,a^2-A\,b^2+12\,A\,a\,b+\frac {20\,C\,a\,b}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (\frac {3\,C\,b^2}{4}-C\,a^2-A\,b^2-12\,A\,a\,b-\frac {20\,C\,a\,b}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (A\,b^2+C\,a^2+\frac {5\,C\,b^2}{4}+4\,A\,a\,b+4\,C\,a\,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 )}^8-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}+\frac {\mathrm {atanh}\left (\frac {4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (A\,a^2+\frac {A\,b^2}{2}+\frac {C\,a^2}{2}+\frac {3\,C\,b^2}{8}\right )}{4\,A\,a^2+2\,A\,b^2+2\,C\,a^2+\frac {3\,C\,b^2}{2}}\right )\,\left (2\,A\,a^2+A\,b^2+C\,a^2+\frac {3\,C\,b^2}{4}\right )}{d} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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