Optimal. Leaf size=179 \[ \frac {\left (8 a A b+4 a^2 B+3 b^2 B\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {\left (4 a^2 A b+4 A b^3-a^3 B+8 a b^2 B\right ) \tan (c+d x)}{6 b d}+\frac {\left (8 a A b-2 a^2 B+9 b^2 B\right ) \sec (c+d x) \tan (c+d x)}{24 d}+\frac {(4 A b-a B) (a+b \sec (c+d x))^2 \tan (c+d x)}{12 b d}+\frac {B (a+b \sec (c+d x))^3 \tan (c+d x)}{4 b d} \]
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Rubi [A]
time = 0.22, antiderivative size = 179, 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 = {4095, 4087,
4082, 3872, 3855, 3852, 8} \begin {gather*} \frac {\left (4 a^2 B+8 a A b+3 b^2 B\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {\left (-2 a^2 B+8 a A b+9 b^2 B\right ) \tan (c+d x) \sec (c+d x)}{24 d}+\frac {\left (a^3 (-B)+4 a^2 A b+8 a b^2 B+4 A b^3\right ) \tan (c+d x)}{6 b d}+\frac {(4 A b-a B) \tan (c+d x) (a+b \sec (c+d x))^2}{12 b d}+\frac {B \tan (c+d x) (a+b \sec (c+d x))^3}{4 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 4095
Rubi steps
\begin {align*} \int \sec ^2(c+d x) (a+b \sec (c+d x))^2 (A+B \sec (c+d x)) \, dx &=\frac {B (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 (3 b B+(4 A b-a B) \sec (c+d x)) \, dx}{4 b}\\ &=\frac {(4 A b-a B) (a+b \sec (c+d x))^2 \tan (c+d x)}{12 b d}+\frac {B (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 (b (8 A b+7 a B)+\left (8 a A b-2 a^2 B+9 b^2 B\right ) \sec (c+d x)\right ) \, dx}{12 b}\\ &=\frac {\left (8 a A b-2 a^2 B+9 b^2 B\right ) \sec (c+d x) \tan (c+d x)}{24 d}+\frac {(4 A b-a B) (a+b \sec (c+d x))^2 \tan (c+d x)}{12 b d}+\frac {B (a+b \sec (c+d x))^3 \tan (c+d x)}{4 b d}+\frac {\int \sec (c+d x) \left (3 b \left (8 a A b+4 a^2 B+3 b^2 B\right )+4 \left (4 a^2 A b+4 A b^3-a^3 B+8 a b^2 B\right ) \sec (c+d x)\right ) \, dx}{24 b}\\ &=\frac {\left (8 a A b-2 a^2 B+9 b^2 B\right ) \sec (c+d x) \tan (c+d x)}{24 d}+\frac {(4 A b-a B) (a+b \sec (c+d x))^2 \tan (c+d x)}{12 b d}+\frac {B (a+b \sec (c+d x))^3 \tan (c+d x)}{4 b d}+\frac {1}{8} \left (8 a A b+4 a^2 B+3 b^2 B\right ) \int \sec (c+d x) \, dx+\frac {\left (4 a^2 A b+4 A b^3-a^3 B+8 a b^2 B\right ) \int \sec ^2(c+d x) \, dx}{6 b}\\ &=\frac {\left (8 a A b+4 a^2 B+3 b^2 B\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {\left (8 a A b-2 a^2 B+9 b^2 B\right ) \sec (c+d x) \tan (c+d x)}{24 d}+\frac {(4 A b-a B) (a+b \sec (c+d x))^2 \tan (c+d x)}{12 b d}+\frac {B (a+b \sec (c+d x))^3 \tan (c+d x)}{4 b d}-\frac {\left (4 a^2 A b+4 A b^3-a^3 B+8 a b^2 B\right ) \text {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{6 b d}\\ &=\frac {\left (8 a A b+4 a^2 B+3 b^2 B\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {\left (4 a^2 A b+4 A b^3-a^3 B+8 a b^2 B\right ) \tan (c+d x)}{6 b d}+\frac {\left (8 a A b-2 a^2 B+9 b^2 B\right ) \sec (c+d x) \tan (c+d x)}{24 d}+\frac {(4 A b-a B) (a+b \sec (c+d x))^2 \tan (c+d x)}{12 b d}+\frac {B (a+b \sec (c+d x))^3 \tan (c+d x)}{4 b d}\\ \end {align*}
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Mathematica [A]
time = 0.78, size = 120, normalized size = 0.67 \begin {gather*} \frac {3 \left (8 a A b+4 a^2 B+3 b^2 B\right ) \tanh ^{-1}(\sin (c+d x))+\tan (c+d x) \left (24 \left (a^2 A+A b^2+2 a b B\right )+3 \left (8 a A b+4 a^2 B+3 b^2 B\right ) \sec (c+d x)+6 b^2 B \sec ^3(c+d x)+8 b (A b+2 a B) \tan ^2(c+d x)\right )}{24 d} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.45, size = 185, normalized size = 1.03
method | result | size |
derivativedivides | \(\frac {a^{2} A \tan \left (d x +c \right )+a^{2} B \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+2 A b a \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-2 B a b \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )-A \,b^{2} \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+b^{2} B \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\) | \(185\) |
default | \(\frac {a^{2} A \tan \left (d x +c \right )+a^{2} B \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+2 A b a \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-2 B a b \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )-A \,b^{2} \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+b^{2} B \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\) | \(185\) |
norman | \(\frac {-\frac {\left (8 a^{2} A -8 A b a +8 A \,b^{2}-4 a^{2} B +16 B a b -5 b^{2} B \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}+\frac {\left (8 a^{2} A +8 A b a +8 A \,b^{2}+4 a^{2} B +16 B a b +5 b^{2} B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}+\frac {\left (72 a^{2} A -24 A b a +40 A \,b^{2}-12 a^{2} B +80 B a b +9 b^{2} B \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{12 d}-\frac {\left (72 a^{2} A +24 A b a +40 A \,b^{2}+12 a^{2} B +80 B a b -9 b^{2} B \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{12 d}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{4}}-\frac {\left (8 A b a +4 a^{2} B +3 b^{2} B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 d}+\frac {\left (8 A b a +4 a^{2} B +3 b^{2} B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 d}\) | \(291\) |
risch | \(-\frac {i \left (24 A a b \,{\mathrm e}^{7 i \left (d x +c \right )}+12 B \,a^{2} {\mathrm e}^{7 i \left (d x +c \right )}+9 B \,b^{2} {\mathrm e}^{7 i \left (d x +c \right )}-24 A \,a^{2} {\mathrm e}^{6 i \left (d x +c \right )}+24 A a b \,{\mathrm e}^{5 i \left (d x +c \right )}+12 B \,a^{2} {\mathrm e}^{5 i \left (d x +c \right )}+33 B \,b^{2} {\mathrm e}^{5 i \left (d x +c \right )}-72 A \,a^{2} {\mathrm e}^{4 i \left (d x +c \right )}-48 A \,b^{2} {\mathrm e}^{4 i \left (d x +c \right )}-96 B a b \,{\mathrm e}^{4 i \left (d x +c \right )}-24 A a b \,{\mathrm e}^{3 i \left (d x +c \right )}-12 B \,a^{2} {\mathrm e}^{3 i \left (d x +c \right )}-33 B \,b^{2} {\mathrm e}^{3 i \left (d x +c \right )}-72 a^{2} A \,{\mathrm e}^{2 i \left (d x +c \right )}-64 A \,b^{2} {\mathrm e}^{2 i \left (d x +c \right )}-128 B a b \,{\mathrm e}^{2 i \left (d x +c \right )}-24 A a b \,{\mathrm e}^{i \left (d x +c \right )}-12 B \,a^{2} {\mathrm e}^{i \left (d x +c \right )}-9 B \,b^{2} {\mathrm e}^{i \left (d x +c \right )}-24 a^{2} A -16 A \,b^{2}-32 B a b \right )}{12 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A b a}{d}-\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{2 d}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) b^{2} B}{8 d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A b a}{d}+\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{2 d}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) b^{2} B}{8 d}\) | \(447\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.28, size = 228, normalized size = 1.27 \begin {gather*} \frac {32 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B a b + 16 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A b^{2} - 3 \, 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 )} - 12 \, 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 )} - 24 \, A a b {\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} \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 = 4.80, size = 180, normalized size = 1.01 \begin {gather*} \frac {3 \, {\left (4 \, B a^{2} + 8 \, A a b + 3 \, B b^{2}\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (4 \, B a^{2} + 8 \, A a b + 3 \, B b^{2}\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (8 \, {\left (3 \, A a^{2} + 4 \, B a b + 2 \, A b^{2}\right )} \cos \left (d x + c\right )^{3} + 6 \, B b^{2} + 3 \, {\left (4 \, B a^{2} + 8 \, A a b + 3 \, B b^{2}\right )} \cos \left (d x + c\right )^{2} + 8 \, {\left (2 \, B a b + A b^{2}\right )} \cos \left (d x + c\right )\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 + B \sec {\left (c + d x \right )}\right ) \left (a + b \sec {\left (c + d x \right )}\right )^{2} \sec ^{2}{\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. 478 vs.
\(2 (167) = 334\).
time = 0.52, size = 478, normalized size = 2.67 \begin {gather*} \frac {3 \, {\left (4 \, B a^{2} + 8 \, A 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 ) - 3 \, {\left (4 \, B a^{2} + 8 \, A 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 (24 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 12 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 24 \, A a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 48 \, B a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 24 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 15 \, B b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 72 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 12 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 24 \, A a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 80 \, B a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 40 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 9 \, B b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 72 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 12 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 24 \, A a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 80 \, B a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 40 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 9 \, B b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 24 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 12 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 24 \, A a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 48 \, B a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 24 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 15 \, B 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 = 5.69, size = 317, normalized size = 1.77 \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}+A\,a\,b+\frac {3\,B\,b^2}{8}\right )}{2\,B\,a^2+4\,A\,a\,b+\frac {3\,B\,b^2}{2}}\right )\,\left (B\,a^2+2\,A\,a\,b+\frac {3\,B\,b^2}{4}\right )}{d}-\frac {\left (2\,A\,a^2+2\,A\,b^2-B\,a^2-\frac {5\,B\,b^2}{4}-2\,A\,a\,b+4\,B\,a\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (B\,a^2-\frac {10\,A\,b^2}{3}-6\,A\,a^2-\frac {3\,B\,b^2}{4}+2\,A\,a\,b-\frac {20\,B\,a\,b}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (6\,A\,a^2+\frac {10\,A\,b^2}{3}+B\,a^2-\frac {3\,B\,b^2}{4}+2\,A\,a\,b+\frac {20\,B\,a\,b}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (-2\,A\,a^2-2\,A\,b^2-B\,a^2-\frac {5\,B\,b^2}{4}-2\,A\,a\,b-4\,B\,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 )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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