Optimal. Leaf size=178 \[ -\frac {b \left (28 a^2+b^2\right ) \sin (c+d x)}{6 a d}-\frac {\left (12 a^2+b^2\right ) \sin (c+d x) (a \cos (c+d x)+b)^2}{12 a b d}-\frac {\left (39 a^2+2 b^2\right ) \sin (c+d x) \cos (c+d x)}{24 d}+\frac {3}{8} x \left (a^2-4 b^2\right )+\frac {2 a b \tanh ^{-1}(\sin (c+d x))}{d}+\frac {\sin (c+d x) (a \cos (c+d x)+b)^3}{4 a d}+\frac {\tan (c+d x) (a \cos (c+d x)+b)^3}{b d} \]
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Rubi [A] time = 0.56, antiderivative size = 178, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3872, 2894, 3049, 3033, 3023, 2735, 3770} \[ -\frac {b \left (28 a^2+b^2\right ) \sin (c+d x)}{6 a d}-\frac {\left (12 a^2+b^2\right ) \sin (c+d x) (a \cos (c+d x)+b)^2}{12 a b d}-\frac {\left (39 a^2+2 b^2\right ) \sin (c+d x) \cos (c+d x)}{24 d}+\frac {3}{8} x \left (a^2-4 b^2\right )+\frac {2 a b \tanh ^{-1}(\sin (c+d x))}{d}+\frac {\sin (c+d x) (a \cos (c+d x)+b)^3}{4 a d}+\frac {\tan (c+d x) (a \cos (c+d x)+b)^3}{b d} \]
Antiderivative was successfully verified.
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Rule 2735
Rule 2894
Rule 3023
Rule 3033
Rule 3049
Rule 3770
Rule 3872
Rubi steps
\begin {align*} \int (a+b \sec (c+d x))^2 \sin ^4(c+d x) \, dx &=\int (-b-a \cos (c+d x))^2 \sin ^2(c+d x) \tan ^2(c+d x) \, dx\\ &=\frac {(b+a \cos (c+d x))^3 \sin (c+d x)}{4 a d}+\frac {(b+a \cos (c+d x))^3 \tan (c+d x)}{b d}-\frac {\int (-b-a \cos (c+d x))^2 \left (-8 a^2+5 a b \cos (c+d x)+\left (12 a^2+b^2\right ) \cos ^2(c+d x)\right ) \sec (c+d x) \, dx}{4 a b}\\ &=-\frac {\left (12 a^2+b^2\right ) (b+a \cos (c+d x))^2 \sin (c+d x)}{12 a b d}+\frac {(b+a \cos (c+d x))^3 \sin (c+d x)}{4 a d}+\frac {(b+a \cos (c+d x))^3 \tan (c+d x)}{b d}-\frac {\int (-b-a \cos (c+d x)) \left (24 a^2 b-17 a b^2 \cos (c+d x)-b \left (39 a^2+2 b^2\right ) \cos ^2(c+d x)\right ) \sec (c+d x) \, dx}{12 a b}\\ &=-\frac {\left (39 a^2+2 b^2\right ) \cos (c+d x) \sin (c+d x)}{24 d}-\frac {\left (12 a^2+b^2\right ) (b+a \cos (c+d x))^2 \sin (c+d x)}{12 a b d}+\frac {(b+a \cos (c+d x))^3 \sin (c+d x)}{4 a d}+\frac {(b+a \cos (c+d x))^3 \tan (c+d x)}{b d}-\frac {\int \left (-48 a^2 b^2-9 a b \left (a^2-4 b^2\right ) \cos (c+d x)+4 b^2 \left (28 a^2+b^2\right ) \cos ^2(c+d x)\right ) \sec (c+d x) \, dx}{24 a b}\\ &=-\frac {b \left (28 a^2+b^2\right ) \sin (c+d x)}{6 a d}-\frac {\left (39 a^2+2 b^2\right ) \cos (c+d x) \sin (c+d x)}{24 d}-\frac {\left (12 a^2+b^2\right ) (b+a \cos (c+d x))^2 \sin (c+d x)}{12 a b d}+\frac {(b+a \cos (c+d x))^3 \sin (c+d x)}{4 a d}+\frac {(b+a \cos (c+d x))^3 \tan (c+d x)}{b d}-\frac {\int \left (-48 a^2 b^2-9 a b \left (a^2-4 b^2\right ) \cos (c+d x)\right ) \sec (c+d x) \, dx}{24 a b}\\ &=\frac {3}{8} \left (a^2-4 b^2\right ) x-\frac {b \left (28 a^2+b^2\right ) \sin (c+d x)}{6 a d}-\frac {\left (39 a^2+2 b^2\right ) \cos (c+d x) \sin (c+d x)}{24 d}-\frac {\left (12 a^2+b^2\right ) (b+a \cos (c+d x))^2 \sin (c+d x)}{12 a b d}+\frac {(b+a \cos (c+d x))^3 \sin (c+d x)}{4 a d}+\frac {(b+a \cos (c+d x))^3 \tan (c+d x)}{b d}+(2 a b) \int \sec (c+d x) \, dx\\ &=\frac {3}{8} \left (a^2-4 b^2\right ) x+\frac {2 a b \tanh ^{-1}(\sin (c+d x))}{d}-\frac {b \left (28 a^2+b^2\right ) \sin (c+d x)}{6 a d}-\frac {\left (39 a^2+2 b^2\right ) \cos (c+d x) \sin (c+d x)}{24 d}-\frac {\left (12 a^2+b^2\right ) (b+a \cos (c+d x))^2 \sin (c+d x)}{12 a b d}+\frac {(b+a \cos (c+d x))^3 \sin (c+d x)}{4 a d}+\frac {(b+a \cos (c+d x))^3 \tan (c+d x)}{b d}\\ \end {align*}
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Mathematica [A] time = 1.05, size = 157, normalized size = 0.88 \[ \frac {\tan (c+d x) \left (-6 \left (3 a^2-4 b^2\right ) \cos (2 (c+d x))+3 \left (a^2 \cos (4 (c+d x))-7 a^2+40 b^2\right )+16 a b \cos (3 (c+d x))\right )+12 \left (3 \left (a^2-4 b^2\right ) (c+d x)-16 a b \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+16 a b \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )\right )-208 a b \sin (c+d x)}{96 d} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.62, size = 142, normalized size = 0.80 \[ \frac {9 \, {\left (a^{2} - 4 \, b^{2}\right )} d x \cos \left (d x + c\right ) + 24 \, a b \cos \left (d x + c\right ) \log \left (\sin \left (d x + c\right ) + 1\right ) - 24 \, a b \cos \left (d x + c\right ) \log \left (-\sin \left (d x + c\right ) + 1\right ) + {\left (6 \, a^{2} \cos \left (d x + c\right )^{4} + 16 \, a b \cos \left (d x + c\right )^{3} - 64 \, a b \cos \left (d x + c\right ) - 3 \, {\left (5 \, a^{2} - 4 \, b^{2}\right )} \cos \left (d x + c\right )^{2} + 24 \, b^{2}\right )} \sin \left (d x + c\right )}{24 \, d \cos \left (d x + c\right )} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.30, size = 285, normalized size = 1.60 \[ \frac {48 \, a b \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 48 \, a b \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + 9 \, {\left (a^{2} - 4 \, b^{2}\right )} {\left (d x + c\right )} - \frac {48 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1} + \frac {2 \, {\left (9 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 48 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 12 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 33 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 208 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 12 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 33 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 208 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 12 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 9 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 48 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 12 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{4}}}{24 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.64, size = 187, normalized size = 1.05 \[ -\frac {a^{2} \cos \left (d x +c \right ) \left (\sin ^{3}\left (d x +c \right )\right )}{4 d}-\frac {3 a^{2} \cos \left (d x +c \right ) \sin \left (d x +c \right )}{8 d}+\frac {3 a^{2} x}{8}+\frac {3 a^{2} c}{8 d}-\frac {2 a b \left (\sin ^{3}\left (d x +c \right )\right )}{3 d}-\frac {2 a b \sin \left (d x +c \right )}{d}+\frac {2 a b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {b^{2} \left (\sin ^{5}\left (d x +c \right )\right )}{d \cos \left (d x +c \right )}+\frac {b^{2} \cos \left (d x +c \right ) \left (\sin ^{3}\left (d x +c \right )\right )}{d}+\frac {3 b^{2} \cos \left (d x +c \right ) \sin \left (d x +c \right )}{2 d}-\frac {3 b^{2} x}{2}-\frac {3 c \,b^{2}}{2 d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.88, size = 125, normalized size = 0.70 \[ \frac {3 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) - 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} a^{2} - 32 \, {\left (2 \, \sin \left (d x + c\right )^{3} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right ) + 6 \, \sin \left (d x + c\right )\right )} a b - 48 \, {\left (3 \, d x + 3 \, c - \frac {\tan \left (d x + c\right )}{\tan \left (d x + c\right )^{2} + 1} - 2 \, \tan \left (d x + c\right )\right )} b^{2}}{96 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 1.23, size = 207, normalized size = 1.16 \[ \frac {3\,a^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{4\,d}-\frac {3\,b^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {a^2\,{\cos \left (c+d\,x\right )}^3\,\sin \left (c+d\,x\right )}{4\,d}+\frac {b^2\,\sin \left (c+d\,x\right )}{d\,\cos \left (c+d\,x\right )}-\frac {8\,a\,b\,\sin \left (c+d\,x\right )}{3\,d}+\frac {4\,a\,b\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}-\frac {5\,a^2\,\cos \left (c+d\,x\right )\,\sin \left (c+d\,x\right )}{8\,d}+\frac {b^2\,\cos \left (c+d\,x\right )\,\sin \left (c+d\,x\right )}{2\,d}+\frac {2\,a\,b\,{\cos \left (c+d\,x\right )}^2\,\sin \left (c+d\,x\right )}{3\,d} \]
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 \[ \int \left (a + b \sec {\left (c + d x \right )}\right )^{2} \sin ^{4}{\left (c + d x \right )}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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