Optimal. Leaf size=77 \[ -\frac {a^2 \sin (c+d x) \cos (c+d x)}{2 d}+\frac {a^2 x}{2}-\frac {2 a b \sin (c+d x)}{d}+\frac {2 a b \tanh ^{-1}(\sin (c+d x))}{d}+\frac {b^2 \tan (c+d x)}{d}-b^2 x \]
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Rubi [A] time = 0.13, antiderivative size = 77, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 8, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.381, Rules used = {3872, 2722, 2635, 8, 2592, 321, 206, 3473} \[ -\frac {a^2 \sin (c+d x) \cos (c+d x)}{2 d}+\frac {a^2 x}{2}-\frac {2 a b \sin (c+d x)}{d}+\frac {2 a b \tanh ^{-1}(\sin (c+d x))}{d}+\frac {b^2 \tan (c+d x)}{d}-b^2 x \]
Antiderivative was successfully verified.
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Rule 8
Rule 206
Rule 321
Rule 2592
Rule 2635
Rule 2722
Rule 3473
Rule 3872
Rubi steps
\begin {align*} \int (a+b \sec (c+d x))^2 \sin ^2(c+d x) \, dx &=\int (-b-a \cos (c+d x))^2 \tan ^2(c+d x) \, dx\\ &=\int \left (a^2 \sin ^2(c+d x)+2 a b \sin (c+d x) \tan (c+d x)+b^2 \tan ^2(c+d x)\right ) \, dx\\ &=a^2 \int \sin ^2(c+d x) \, dx+(2 a b) \int \sin (c+d x) \tan (c+d x) \, dx+b^2 \int \tan ^2(c+d x) \, dx\\ &=-\frac {a^2 \cos (c+d x) \sin (c+d x)}{2 d}+\frac {b^2 \tan (c+d x)}{d}+\frac {1}{2} a^2 \int 1 \, dx-b^2 \int 1 \, dx+\frac {(2 a b) \operatorname {Subst}\left (\int \frac {x^2}{1-x^2} \, dx,x,\sin (c+d x)\right )}{d}\\ &=\frac {a^2 x}{2}-b^2 x-\frac {2 a b \sin (c+d x)}{d}-\frac {a^2 \cos (c+d x) \sin (c+d x)}{2 d}+\frac {b^2 \tan (c+d x)}{d}+\frac {(2 a b) \operatorname {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\sin (c+d x)\right )}{d}\\ &=\frac {a^2 x}{2}-b^2 x+\frac {2 a b \tanh ^{-1}(\sin (c+d x))}{d}-\frac {2 a b \sin (c+d x)}{d}-\frac {a^2 \cos (c+d x) \sin (c+d x)}{2 d}+\frac {b^2 \tan (c+d x)}{d}\\ \end {align*}
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Mathematica [A] time = 0.60, size = 121, normalized size = 1.57 \[ -\frac {a^2 \sin (2 (c+d x))-2 a^2 c-2 a^2 d x+8 a b \sin (c+d x)+8 a b \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-8 a b \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )-4 b^2 \tan (c+d x)+4 b^2 c+4 b^2 d x}{4 d} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.50, size = 108, normalized size = 1.40 \[ \frac {{\left (a^{2} - 2 \, b^{2}\right )} d x \cos \left (d x + c\right ) + 2 \, a b \cos \left (d x + c\right ) \log \left (\sin \left (d x + c\right ) + 1\right ) - 2 \, a b \cos \left (d x + c\right ) \log \left (-\sin \left (d x + c\right ) + 1\right ) - {\left (a^{2} \cos \left (d x + c\right )^{2} + 4 \, a b \cos \left (d x + c\right ) - 2 \, b^{2}\right )} \sin \left (d x + c\right )}{2 \, d \cos \left (d x + c\right )} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.28, size = 159, normalized size = 2.06 \[ \frac {4 \, a b \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 4 \, a b \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + {\left (a^{2} - 2 \, b^{2}\right )} {\left (d x + c\right )} - \frac {4 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1} + \frac {2 \, {\left (a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 4 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 4 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2}}}{2 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.39, size = 99, normalized size = 1.29 \[ -\frac {a^{2} \cos \left (d x +c \right ) \sin \left (d x +c \right )}{2 d}+\frac {a^{2} x}{2}+\frac {a^{2} c}{2 d}-\frac {2 a b \sin \left (d x +c \right )}{d}+\frac {2 a b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}-b^{2} x +\frac {b^{2} \tan \left (d x +c \right )}{d}-\frac {c \,b^{2}}{d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.91, size = 80, normalized size = 1.04 \[ \frac {{\left (2 \, d x + 2 \, c - \sin \left (2 \, d x + 2 \, c\right )\right )} a^{2} - 4 \, {\left (d x + c - \tan \left (d x + c\right )\right )} b^{2} + 4 \, a b {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right ) - 2 \, \sin \left (d x + c\right )\right )}}{4 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 1.17, size = 143, normalized size = 1.86 \[ \frac {a^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}-\frac {2\,b^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {b^2\,\sin \left (c+d\,x\right )}{d\,\cos \left (c+d\,x\right )}-\frac {2\,a\,b\,\sin \left (c+d\,x\right )}{d}+\frac {4\,a\,b\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}-\frac {a^2\,\cos \left (c+d\,x\right )\,\sin \left (c+d\,x\right )}{2\,d} \]
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 \[ \int \left (a + b \sec {\left (c + d x \right )}\right )^{2} \sin ^{2}{\left (c + d x \right )}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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