Integrand size = 21, antiderivative size = 87 \[ \int (a+b \sec (c+d x))^2 \tan ^3(c+d x) \, dx=\frac {a^2 \log (\cos (c+d x))}{d}-\frac {2 a b \sec (c+d x)}{d}+\frac {\left (a^2-b^2\right ) \sec ^2(c+d x)}{2 d}+\frac {2 a b \sec ^3(c+d x)}{3 d}+\frac {b^2 \sec ^4(c+d x)}{4 d} \]
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Time = 0.08 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {3970, 908} \[ \int (a+b \sec (c+d x))^2 \tan ^3(c+d x) \, dx=\frac {\left (a^2-b^2\right ) \sec ^2(c+d x)}{2 d}+\frac {a^2 \log (\cos (c+d x))}{d}+\frac {2 a b \sec ^3(c+d x)}{3 d}-\frac {2 a b \sec (c+d x)}{d}+\frac {b^2 \sec ^4(c+d x)}{4 d} \]
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Rule 908
Rule 3970
Rubi steps \begin{align*} \text {integral}& = -\frac {\text {Subst}\left (\int \frac {(a+x)^2 \left (b^2-x^2\right )}{x} \, dx,x,b \sec (c+d x)\right )}{b^2 d} \\ & = -\frac {\text {Subst}\left (\int \left (2 a b^2+\frac {a^2 b^2}{x}-\left (a^2-b^2\right ) x-2 a x^2-x^3\right ) \, dx,x,b \sec (c+d x)\right )}{b^2 d} \\ & = \frac {a^2 \log (\cos (c+d x))}{d}-\frac {2 a b \sec (c+d x)}{d}+\frac {\left (a^2-b^2\right ) \sec ^2(c+d x)}{2 d}+\frac {2 a b \sec ^3(c+d x)}{3 d}+\frac {b^2 \sec ^4(c+d x)}{4 d} \\ \end{align*}
Time = 0.38 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.85 \[ \int (a+b \sec (c+d x))^2 \tan ^3(c+d x) \, dx=\frac {12 a^2 \log (\cos (c+d x))-24 a b \sec (c+d x)+6 \left (a^2-b^2\right ) \sec ^2(c+d x)+8 a b \sec ^3(c+d x)+3 b^2 \sec ^4(c+d x)}{12 d} \]
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Time = 1.51 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.86
method | result | size |
parts | \(\frac {a^{2} \left (\frac {\tan \left (d x +c \right )^{2}}{2}-\frac {\ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2}\right )}{d}+\frac {b^{2} \tan \left (d x +c \right )^{4}}{4 d}+\frac {2 a b \left (\frac {\sec \left (d x +c \right )^{3}}{3}-\sec \left (d x +c \right )\right )}{d}\) | \(75\) |
derivativedivides | \(\frac {\frac {b^{2} \sec \left (d x +c \right )^{4}}{4}+\frac {2 a b \sec \left (d x +c \right )^{3}}{3}+\frac {a^{2} \sec \left (d x +c \right )^{2}}{2}-\frac {\sec \left (d x +c \right )^{2} b^{2}}{2}-2 a b \sec \left (d x +c \right )-a^{2} \ln \left (\sec \left (d x +c \right )\right )}{d}\) | \(79\) |
default | \(\frac {\frac {b^{2} \sec \left (d x +c \right )^{4}}{4}+\frac {2 a b \sec \left (d x +c \right )^{3}}{3}+\frac {a^{2} \sec \left (d x +c \right )^{2}}{2}-\frac {\sec \left (d x +c \right )^{2} b^{2}}{2}-2 a b \sec \left (d x +c \right )-a^{2} \ln \left (\sec \left (d x +c \right )\right )}{d}\) | \(79\) |
risch | \(-i a^{2} x -\frac {2 i a^{2} c}{d}-\frac {2 \left (6 a b \,{\mathrm e}^{7 i \left (d x +c \right )}-3 a^{2} {\mathrm e}^{6 i \left (d x +c \right )}+3 b^{2} {\mathrm e}^{6 i \left (d x +c \right )}+10 a b \,{\mathrm e}^{5 i \left (d x +c \right )}-6 a^{2} {\mathrm e}^{4 i \left (d x +c \right )}+10 a b \,{\mathrm e}^{3 i \left (d x +c \right )}-3 a^{2} {\mathrm e}^{2 i \left (d x +c \right )}+3 b^{2} {\mathrm e}^{2 i \left (d x +c \right )}+6 a b \,{\mathrm e}^{i \left (d x +c \right )}\right )}{3 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}+\frac {a^{2} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{d}\) | \(179\) |
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Time = 0.30 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.94 \[ \int (a+b \sec (c+d x))^2 \tan ^3(c+d x) \, dx=\frac {12 \, a^{2} \cos \left (d x + c\right )^{4} \log \left (-\cos \left (d x + c\right )\right ) - 24 \, a b \cos \left (d x + c\right )^{3} + 8 \, a b \cos \left (d x + c\right ) + 6 \, {\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + 3 \, b^{2}}{12 \, d \cos \left (d x + c\right )^{4}} \]
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Time = 0.27 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.45 \[ \int (a+b \sec (c+d x))^2 \tan ^3(c+d x) \, dx=\begin {cases} - \frac {a^{2} \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} + \frac {a^{2} \tan ^{2}{\left (c + d x \right )}}{2 d} + \frac {2 a b \tan ^{2}{\left (c + d x \right )} \sec {\left (c + d x \right )}}{3 d} - \frac {4 a b \sec {\left (c + d x \right )}}{3 d} + \frac {b^{2} \tan ^{2}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}}{4 d} - \frac {b^{2} \sec ^{2}{\left (c + d x \right )}}{4 d} & \text {for}\: d \neq 0 \\x \left (a + b \sec {\left (c \right )}\right )^{2} \tan ^{3}{\left (c \right )} & \text {otherwise} \end {cases} \]
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Time = 0.20 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.86 \[ \int (a+b \sec (c+d x))^2 \tan ^3(c+d x) \, dx=\frac {12 \, a^{2} \log \left (\cos \left (d x + c\right )\right ) - \frac {24 \, a b \cos \left (d x + c\right )^{3} - 8 \, a b \cos \left (d x + c\right ) - 6 \, {\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 3 \, b^{2}}{\cos \left (d x + c\right )^{4}}}{12 \, d} \]
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Leaf count of result is larger than twice the leaf count of optimal. 267 vs. \(2 (81) = 162\).
Time = 0.74 (sec) , antiderivative size = 267, normalized size of antiderivative = 3.07 \[ \int (a+b \sec (c+d x))^2 \tan ^3(c+d x) \, dx=-\frac {12 \, a^{2} \log \left ({\left | -\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 1 \right |}\right ) - 12 \, a^{2} \log \left ({\left | -\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} - 1 \right |}\right ) + \frac {25 \, a^{2} + 32 \, a b + \frac {124 \, a^{2} {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac {128 \, a b {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac {198 \, a^{2} {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {96 \, a b {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {48 \, b^{2} {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {124 \, a^{2} {\left (\cos \left (d x + c\right ) - 1\right )}^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {25 \, a^{2} {\left (\cos \left (d x + c\right ) - 1\right )}^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}}{{\left (\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 1\right )}^{4}}}{12 \, d} \]
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Time = 16.45 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.74 \[ \int (a+b \sec (c+d x))^2 \tan ^3(c+d x) \, dx=-\frac {\frac {8\,a\,b}{3}-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (2\,a^2+\frac {32\,b\,a}{3}\right )-2\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (4\,a^2+8\,a\,b-4\,b^2\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}-\frac {2\,a^2\,\mathrm {atanh}\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\right )}{d} \]
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