Optimal. Leaf size=90 \[ \frac {\left (2 a^2-b^2\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}-\frac {3 a^2 \sin (c+d x)}{2 d}+\frac {a b \tan (c+d x)}{d}+\frac {\tan (c+d x) \sec (c+d x) (a \cos (c+d x)+b)^2}{2 d}-2 a b x \]
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Rubi [A] time = 0.43, antiderivative size = 90, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.269, Rules used = {4397, 2889, 3048, 3031, 3023, 2735, 3770} \[ \frac {\left (2 a^2-b^2\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}-\frac {3 a^2 \sin (c+d x)}{2 d}+\frac {a b \tan (c+d x)}{d}+\frac {\tan (c+d x) \sec (c+d x) (a \cos (c+d x)+b)^2}{2 d}-2 a b x \]
Antiderivative was successfully verified.
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Rule 2735
Rule 2889
Rule 3023
Rule 3031
Rule 3048
Rule 3770
Rule 4397
Rubi steps
\begin {align*} \int \sec (c+d x) (a \sin (c+d x)+b \tan (c+d x))^2 \, dx &=\int (b+a \cos (c+d x))^2 \sec (c+d x) \tan ^2(c+d x) \, dx\\ &=\int (b+a \cos (c+d x))^2 \left (1-\cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx\\ &=\frac {(b+a \cos (c+d x))^2 \sec (c+d x) \tan (c+d x)}{2 d}+\frac {1}{2} \int (b+a \cos (c+d x)) \left (2 a-b \cos (c+d x)-3 a \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx\\ &=\frac {a b \tan (c+d x)}{d}+\frac {(b+a \cos (c+d x))^2 \sec (c+d x) \tan (c+d x)}{2 d}-\frac {1}{2} \int \left (-2 a^2+b^2+4 a b \cos (c+d x)+3 a^2 \cos ^2(c+d x)\right ) \sec (c+d x) \, dx\\ &=-\frac {3 a^2 \sin (c+d x)}{2 d}+\frac {a b \tan (c+d x)}{d}+\frac {(b+a \cos (c+d x))^2 \sec (c+d x) \tan (c+d x)}{2 d}-\frac {1}{2} \int \left (-2 a^2+b^2+4 a b \cos (c+d x)\right ) \sec (c+d x) \, dx\\ &=-2 a b x-\frac {3 a^2 \sin (c+d x)}{2 d}+\frac {a b \tan (c+d x)}{d}+\frac {(b+a \cos (c+d x))^2 \sec (c+d x) \tan (c+d x)}{2 d}-\frac {1}{2} \left (-2 a^2+b^2\right ) \int \sec (c+d x) \, dx\\ &=-2 a b x+\frac {\left (2 a^2-b^2\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}-\frac {3 a^2 \sin (c+d x)}{2 d}+\frac {a b \tan (c+d x)}{d}+\frac {(b+a \cos (c+d x))^2 \sec (c+d x) \tan (c+d x)}{2 d}\\ \end {align*}
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Mathematica [A] time = 0.19, size = 75, normalized size = 0.83 \[ \frac {\left (2 a^2-b^2\right ) \tanh ^{-1}(\sin (c+d x))-2 a^2 \sin (c+d x)-4 a b \tan ^{-1}(\tan (c+d x))+4 a b \tan (c+d x)+b^2 \tan (c+d x) \sec (c+d x)}{2 d} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.71, size = 126, normalized size = 1.40 \[ -\frac {8 \, a b d x \cos \left (d x + c\right )^{2} - {\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) + {\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (2 \, a^{2} \cos \left (d x + c\right )^{2} - 4 \, a b \cos \left (d x + c\right ) - b^{2}\right )} \sin \left (d x + c\right )}{4 \, d \cos \left (d x + c\right )^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 1.58, size = 171, normalized size = 1.90 \[ -\frac {4 \, {\left (d x + c\right )} a b - {\left (2 \, a^{2} - b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) + {\left (2 \, a^{2} - b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + \frac {4 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1} + \frac {2 \, {\left (4 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 4 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2}}}{2 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.09, size = 123, normalized size = 1.37 \[ -\frac {a^{2} \sin \left (d x +c \right )}{d}+\frac {a^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}-2 a b x +\frac {2 a b \tan \left (d x +c \right )}{d}-\frac {2 a b c}{d}+\frac {b^{2} \left (\sin ^{3}\left (d x +c \right )\right )}{2 d \cos \left (d x +c \right )^{2}}+\frac {b^{2} \sin \left (d x +c \right )}{2 d}-\frac {b^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2 d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.46, size = 102, normalized size = 1.13 \[ -\frac {8 \, {\left (d x + c - \tan \left (d x + c\right )\right )} a b + b^{2} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} + \log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 2 \, a^{2} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right ) - 2 \, \sin \left (d x + c\right )\right )}}{4 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 0.79, size = 147, normalized size = 1.63 \[ \frac {2\,a^2\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}-\frac {a^2\,\sin \left (c+d\,x\right )}{d}-\frac {b^2\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {b^2\,\sin \left (c+d\,x\right )}{2\,d\,{\cos \left (c+d\,x\right )}^2}-\frac {4\,a\,b\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {2\,a\,b\,\sin \left (c+d\,x\right )}{d\,\cos \left (c+d\,x\right )} \]
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 \[ \int \left (a \sin {\left (c + d x \right )} + b \tan {\left (c + d x \right )}\right )^{2} \sec {\left (c + d x \right )}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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